University Physics for JEE Mains and Advance [2] 9353433673, 9789353433673

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About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide. Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world. We believe learning opens up opportunities, creates fulfilling careers and hence better lives. We hence collaborate with the best of minds to deliver you class-leading products, spread across the Higher Education and K12 spectrum. Superior learning experience and improved outcomes are at the heart of everything we do. This product is the result of one such effort. Your feedback plays a critical role in the evolution of our products and you can contact us – [email protected]. We look forward to it.

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Sears and Zemansky’s

UNIVERSITY PHYSICS FOR THE JEE

VOLUME II

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Copyright © 2019 Pearson India Education Services Pvt. Ltd Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128. No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 978-93-534-3367-3 eISBN 978-93-539-4048-5

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Brief Contents Electromagnetism

Optics



1 Electric Charge and Electric Field



2 Gauss’s Law

39

14 Geometric Optics

444



3 Electric Potential

70

15 Interference

497



4 Capacitance and Dielectrics

106

16 Diffraction

524



5 Current, Resistance, and Electromotive Force

138

Modern Physics



6 Direct-Current Circuits

170



7 Magnetic Field and Magnetic Forces

207

17 Photons: Light Waves Behaving as Particles

545



8 Sources of Magnetic Field

249

18 Particles Behaving as Waves

570



9 Electromagnetic Induction

285

19 Nuclear Physics

599

1

10 Inductance

323

11 Alternating Current

353

12 Electromagnetic Waves

383

13 The Nature and Propagation of Light 412

Photo Credits C-1

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About the Authors Hugh D. Young is Emeritus Professor of Physics at Carnegie Mellon University. He earned both his undergraduate and graduate degrees from that university. He earned his Ph.D. in fundamental particle theory under the direction of the late Richard Cutkosky. He joined the faculty of Carnegie Mellon in 1956 and retired in 2004. He also had two visiting professorships at the University of California, Berkeley. Dr. Young’s career has centered entirely on undergraduate education. He has written several undergraduate-level textbooks, and in 1973 he became a coauthor with Francis Sears and Mark Zemansky for their well-known introductory texts. In addition to his role on Sears and Zemansky’s University Physics, he is also author of Sears and Zemansky’s College Physics. Dr. Young earned a bachelor’s degree in organ performance from Carnegie Mellon in 1972 and spent several years as Associate Organist at St. Paul’s Cathedral in Pittsburgh. He has played numerous organ recitals in the Pittsburgh area. Dr. Young and his wife, Alice, usually travel extensively in the summer, especially overseas and in the desert canyon country of southern Utah. Roger A. Freedman is a Lecturer in Physics at the University of California, Santa Barbara. Dr. Freedman was an undergraduate at the University of California campuses in San Diego and Los Angeles, and did his doctoral research in nuclear theory at Stanford University under the direction of Professor J. Dirk Walecka. He came to UCSB in 1981 after three years teaching and doing research at the University of Washington. At UCSB, Dr. Freedman has taught in both the Department of Physics and the College of Creative Studies, a branch of the university intended for highly gifted and motivated undergraduates. He has published research in nuclear physics, elementary particle physics, and laser physics. In recent years, he has worked to make physics lectures a more interactive experience through the use of classroom response systems. In the 1970s Dr. Freedman worked as a comic book letterer and helped organize the San Diego Comic-Con (now the world’s largest popular culture convention) during its first few years. Today, when not in the classroom or slaving over a computer, Dr. Freedman can be found either flying (he holds a commercial pilot’s license) or driving with his wife, Caroline, in their 1960 Nash Metropolitan convertible. A. Lewis Ford is Professor of Physics at Texas A&M University. He received a B.A. from Rice University in 1968 and a Ph.D. in chemical physics from the University of Texas at Austin in 1972. After a one-year postdoc at Harvard University, he joined the Texas A&M physics faculty in 1973 and has been there ever since. Professor Ford’s research area is theoretical atomic physics, with a specialization in atomic collisions. At Texas A&M he has taught a variety of undergraduate and graduate courses, but primarily introductory physics.

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ABOUT THE ADAPTERS Nitin Jain Nitin Jain, B.E. (Hons.) from BITS Pilani, has more than two decades of experience in imparting proficiency in the subject to aspirants of IIT-JEE and Physics Olympiads. His association with apex institutions at Kota enabled him to cultivate a distinct expertise to identify the peculiar problems of students, which he puts to good use to reveal to them the nuances of cracking the IIT-JEE and the Physics Olympiads. Jain is now Director of Vibrant Academy (India) Private Limited, Kota. An avid reader, his interest extends to topics as diverse as technology and psychology. This book is a result of his immense experience and diligence to ensure that the learners of physics steer clear of every conceivable obstacle. Email: [email protected]. Neel Kamal Sethia Neel Kamal Sethia, B.Tech. (Hons.) from IIT Kharagpur, has been teaching physics for the past sixteen years. His passion for the intricacies of the subject took him to the premier institution of this country as faculty. His stint at Delhi and Kota helped innumerable students get through the IIT-JEE and the Physics Olympiads. With in-depth knowledge of the subject, he has honed his skills to foresee the obstacles of students in their quest to prepare for these examinations and successfully weeds them out. This book is the result of his passion and toil for years to see the students rise above the ordinary and achieve the pinnacles of glory. Sethia is currently Director of Vibrant Academy (India) Private Limited, Kota. He regularly plays badminton and evinces keen interest in other sports as well. Email: [email protected]

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PREFACE

This book is the product of more than six decades of leadership and innovation in physics education. When the first edition of University Physics by Francis W. Sears and Mark W. Zemansky was published in 1949, it was revolutionary among calculus-based physics textbooks in its emphasis on the fundamental principles of physics and how to apply them. The success of University Physics with generations of several million students and educators around the world is a testament to the merits of this approach, and to the many innovations it has introduced subsequently. In preparing this new Thirteenth Edition, we have further enhanced and developed University Physics to assimilate the best ideas from education research with enhanced problem-solving instruction, pioneering visual and conceptual pedagogy, the first systematically enhanced problems, and the most pedagogically proven and widely used online homework and tutorial system in the world.

Key Features of University Physics

• Deep and extensive problem sets cover a wide range of difficulty and exercise both physical understanding and problem-solving expertise. Many problems are based on complex real-life situations. • This text offers a larger number of Examples and Conceptual Examples than any other leading calculus-based text, allowing it to explore problem-solving challenges not addressed in other texts. • A research-based problem-solving approach (Identify, Set Up, Execute, Evaluate) is used not just in every Example but also in the Problem-Solving Strategies and throughout the Student and Instructor Solutions Manuals and the Study Guide. This consistent approach teaches students to tackle problems thoughtfully rather than cutting straight to the math. • Problem-Solving Strategies coach students in how to approach specific types of problems. • The Figures use a simplified graphical style to focus on the physics of a situation, and they incorporate explanatory annotation. Both techniques have been demonstrated to have a strong positive effect on learning. • Figures that illustrate Example solutions often take the form of black-andwhite pencil sketches, which directly represent what a student should draw in solving such a problem. • The popular Caution paragraphs focus on typical misconceptions and student problem areas. • End-of-section Test Your Understanding questions let students check their grasp of the material and use a multiple-choice or ranking-task format to probe for common misconceptions. • Visual Summaries at the end of each chapter present the key ideas in words, equations, and thumbnail pictures, helping students to review more effectively.

New to This Edition

University Physics for the JEE caters to the needs of aspirants seeking admission in various engineering and medical courses. This book, Volume II, is the second of the two volumes designed for this purpose and comprises 19 chapters. These chapters thoroughly cover the syllabus for class XII. The book begins with study of electrostatics which is covered in first three chapters. Capacitors and dielectric effect is discussed in Chapter 4. The current electricity and analysis of electric circuits is described in Chapters 5 and 6. Chapters 7 and 8 cover magnetic effect of current and magnetic forces on charges and currents. Electromagnetic induction has been analysed in Chapters 9 and 10. Alternating current is described in Chapter 11. Electromagnetism ends with vii

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viii    PREFACE study of electromagnetic waves in Chapter 12. Chapter 13 introduces the basic nature of light. Geometrical optics has been taken up in Chapter 14 using new Cartesian sign convention. Chapters 15 and 16 are devoted to wave properties of light, that is, Interference and Diffraction respectively. Concepts of modern physics have been described in Chapters 17, 18, and 19 of the book. Chapter 17 examines the particle behaviour of waves whereas Chapter 18 is on Wave like behaviour of particles. In Chapter 19 of the book atomic nucleus has been elaborated upon. Any entrance examination requires the practical application of concepts involving calculations to be performed within a short duration of time. This calls for a fair amount of practice of multiple-choice questions (MCQs). Hence, we have included an ample number of problems at the end of each chapter. The MCQs have been carefully selected, keeping in view the difficulty level as well as the requirements of various examinations. Comments and suggestions from readers will be highly appreciated. Nitin Jain Neel Kamal Sethia

Acknowledgments We would like to thank the hundreds of reviewers and colleagues who have offered valuable comments and suggestions over the life of this textbook. The continuing success of University Physics is due in large measure to their contributions. Edward Adelson (Ohio State University), Ralph Alexander (University of Missouri at Rolla), J. G. Anderson, R. S. Anderson, Wayne Anderson (Sacramento City College), Alex Azima (Lansing Community College), Dilip Balamore (Nassau Community College), Harold Bale (University of North Dakota), Arun Bansil (Northeastern University), John Barach (Vanderbilt University), J. D. Barnett, H. H. Barschall, Albert Bartlett (University of Colorado), Marshall Bartlett (Hollins University), Paul Baum (CUNY, Queens College), Frederick Becchetti (University of Michigan), B. Bederson, David Bennum (University of Nevada, Reno), Lev I. Berger (San Diego State University), Robert Boeke (William Rainey Harper College), S. Borowitz, A. C. Braden, James Brooks (Boston University), Nicholas E. Brown (California Polytechnic State University, San Luis Obispo), Tony Buffa (California Polytechnic State University, San Luis Obispo), A. Capecelatro, Michael Cardamone (Pennsylvania State University), Duane Carmony (Purdue University), Troy Carter (UCLA), P. Catranides, John Cerne (SUNY at Buffalo), Tim Chupp (University of Michigan), Shinil Cho (La Roche College), Roger Clapp (University of South Florida), William M. Cloud (Eastern Illinois University), Leonard Cohen (Drexel University), W. R. Coker (University of Texas, Austin), Malcolm D. Cole (University of Missouri at Rolla), H. Conrad, David Cook (Lawrence University), Gayl Cook (University of Colorado), Hans Courant (University of Minnesota), Bruce A. Craver (University of Dayton), Larry Curtis (University of Toledo), Jai Dahiya (Southeast Missouri State University), Steve Detweiler (University of Florida), George Dixon (Oklahoma State University), Donald S. Duncan, Boyd Edwards (West Virginia University), Robert Eisenstein (Carnegie Mellon University), Amy Emerson Missourn (Virginia Institute of Technology), William Faissler (Northeastern University), William Fasnacht (U.S. Naval Academy), Paul Feldker (St. Louis Community College), Carlos Figueroa (Cabrillo College), L. H. Fisher, Neil Fletcher (Florida State University), Robert Folk, Peter Fong (Emory University), A. Lewis Ford (Texas A&M University), D. Frantszog, James R. Gaines (Ohio State University), Solomon Gartenhaus (Purdue University), Ron Gautreau (New Jersey Institute of Technology), J. David Gavenda (University of Texas, Austin), Dennis Gay (University of North Florida), James Gerhart (University of Washington), N. S. Gingrich, J. L. Glathart, S. Goodwin, Rich Gottfried (Frederick Community College), Walter S. Gray (University of Michigan), Paul Gresser (University of Maryland), Benjamin Grinstein (UC San Diego), Howard Grotch (Pennsylvania State University), John Gruber (San Jose State University), Graham D. Gutsche (U.S. Naval Academy), Michael J. Harrison (Michigan State University), Harold Hart (Western Illinois University), Howard Hayden (University of Connecticut), Carl Helrich (Goshen College), Laurent Hodges (Iowa State University), C. D. Hodgman, Michael Hones (Villanova University), Keith Honey (West Virginia Institute of Technology), Gregory Hood (Tidewater Community College), John Hubisz (North Carolina State University), M. Iona, John Jaszczak (Michigan Technical University), Alvin Jenkins (North Carolina State University), Robert P. Johnson (UC Santa Cruz), Lorella Jones (University of Illinois), John Karchek (GMI Engineering & Management Institute), Thomas Keil (Worcester Polytechnic Institute), Robert Kraemer (Carnegie Mellon University), Jean P. Krisch (University of Michigan), Robert A. Kromhout, Andrew Kunz (Marquette University), Charles Lane (Berry College), Thomas N. Lawrence (Texas State University), Robert J. Lee, Alfred Leitner (Rensselaer Polytechnic University), Gerald P. Lietz (De Paul University), Gordon Lind (Utah State University), S. Livingston, Elihu Lubkin (University of Wisconsin, Milwaukee), Robert Luke (Boise State University), David Lynch (Iowa State University), Michael

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PREFACE    ix Lysak (San Bernardino Valley College), Jeffrey Mallow (Loyola University), Robert Mania (Kentucky State University), Robert Marchina (University of Memphis), David Markowitz (University of Connecticut), R. J. Maurer, Oren Maxwell (Florida International University), Joseph L. McCauley (University of Houston), T. K. McCubbin, Jr. (Pennsylvania State University), Charles McFarland (University of Missouri at Rolla), James Mcguire (Tulane University), Lawrence McIntyre (University of Arizona), Fredric Messing (Carnegie-Mellon University), Thomas Meyer (Texas A&M University), Andre Mirabelli (St. Peter’s College, New Jersey), Herbert Muether (S.U.N.Y., Stony Brook), Jack Munsee (California State University, Long Beach), Lorenzo Narducci (Drexel University), Van E. Neie (Purdue University), David A. Nordling (U. S. Naval Academy), Benedict Oh (Pennsylvania State University), L. O. Olsen, Jim Pannell (DeVry Institute of Technology), W. F. Parks (University of Missouri), Robert Paulson (California State University, Chico), Jerry Peacher (University of Missouri at Rolla), Arnold Perlmutter (University of Miami), Lennart Peterson (University of Florida), R. J. Peterson (University of Colorado, Boulder), R. Pinkston, Ronald Poling (University of Minnesota), J. G. Potter, C. W. Price (Millersville University), Francis Prosser (University of Kansas), Shelden H. Radin, Roberto Ramos (Drexel University), Michael Rapport (Anne Arundel Community College), R. Resnick, James A. Richards, Jr., John S. Risley (North Carolina State University), Francesc Roig (University of California, Santa Barbara), T. L. Rokoske, Richard Roth (Eastern Michigan University), Carl Rotter (University of West Virginia), S. Clark Rowland (Andrews University), Rajarshi Roy (Georgia Institute of Technology), Russell A. Roy (Santa Fe Community College), Dhiraj Sardar (University of Texas, San Antonio), Bruce Schumm (UC Santa Cruz), Melvin Schwartz (St. John’s University), F. A. Scott, L. W. Seagondollar, Paul Shand (University of Northern Iowa), Stan Shepherd (Pennsylvania State University), Douglas Sherman (San Jose State), Bruce Sherwood (Carnegie Mellon University), Hugh Siefkin (Greenville College), Tomasz Skwarnicki (Syracuse University), C. P. Slichter, Charles W. Smith (University of Maine, Orono), Malcolm Smith (University of Lowell), Ross Spencer (Brigham Young University), Julien Sprott (University of Wisconsin), Victor ­Stanionis (Iona College), James Stith (American Institute of Physics), Chuck Stone (North Carolina A&T State University), Edward Strother (Florida Institute of Technology), Conley Stutz (Bradley University), Albert Stwertka (U.S. Merchant Marine Academy), Kenneth Szpara-DeNisco (Harrisburg Area Community College), Martin Tiersten (CUNY, City College), David Toot (Alfred University), Somdev Tyagi (Drexel University), F. Verbrugge, Helmut Vogel (Carnegie Mellon University), Robert Webb (Texas A & M), Thomas Weber (Iowa State University), M. Russell Wehr, (Pennsylvania State University), Robert Weidman (Michigan Technical University), Dan Whalen (UC San Diego), Lester V. Whitney, Thomas Wiggins (Pennsylvania State University), David Willey (University of Pittsburgh, Johnstown), George Williams (University of Utah), John Williams (Auburn University), Stanley Williams (Iowa State University), Jack Willis, Suzanne Willis (Northern Illinois University), Robert Wilson (San Bernardino Valley College), L. Wolfenstein, James Wood (Palm Beach Junior College), Lowell Wood (University of Houston), R. E. Worley, D. H. Ziebell (Manatee Community College), George O. Zimmerman (Boston University)

In addition, we both have individual acknowledgments we would like to make. I want to extend my heartfelt thanks to my colleagues at Carnegie Mellon, especially Professors Robert Kraemer, Bruce Sherwood, Ruth Chabay, Helmut Vogel, and Brian Quinn, for many stimulating discussions about physics pedagogy and for their support and encouragement during the writing of several successive editions of this book. I am equally indebted to the many generations of Carnegie Mellon students who have helped me learn what good teaching and good writing are, by showing me what works and what doesn’t. It is always a joy and a privilege to express my gratitude to my wife Alice and our children Gretchen and Rebecca for their love, support, and emotional sustenance during the writing of several successive editions of this book. May all men and women be blessed with love such as theirs. — H. D. Y. I would like to thank my past and present colleagues at UCSB, including Rob Geller, Carl Gwinn, Al Nash, Elisabeth Nicol, and Francesc Roig, for their wholehearted support and for many helpful discussions. I owe a special debt of gratitude to my early teachers Willa Ramsay, Peter Zimmerman, William Little, Alan Schwettman, and Dirk Walecka for showing me what clear and engaging physics teaching is all about, and to Stuart Johnson for inviting me to become a co-author of University Physics beginning with the 9th edition. I want to express special thanks to the editorial staff at Addison-Wesley and their partners: to Nancy Whilton for her editorial vision; to Margot Otway for her superb graphic sense and careful development of this edition; to Peter Murphy for his contributions to the worked examples; to Jason J. B. Harlow for his careful reading of the page proofs; and to Chandrika Madhavan, Steven Le, and Cindy Johnson for keeping the editorial and production pipeline flowing. Most of all, I want to express my gratitude and love to my wife Caroline, to whom I dedicate my contribution to this book. Hey, Caroline, the new edition’s done at last — let’s go flying! — R. A. F.

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x    PREFACE

Please Tell Us What You Think! We welcome communications from students and professors, especially concerning errors or deficiencies that you find in this edition. We have devoted a lot of time and effort to writing the best book we know how to write, and we hope it will help you to teach and learn physics. In turn, you can help us by letting us know what still needs to be improved! Please feel free to contact us either electronically or by ordinary mail. Your comments will be greatly appreciated. December 2010

Hugh D. Young

Roger A. Freedman

Department of Physics Department of Physics Carnegie Mellon University University of California, Santa Barbara Pittsburgh, PA 15213 Santa Barbara, CA 93106-9530 [email protected] [email protected] http://www.physics.ucsb.edu/airboy/

We take this opportunity to express our gratitude to our colleagues M. S. Chouhan, Narendra Awasthi, Pankaj Joshi, V. K. Jaiswal, and Vikas Gupta for their valuable suggestions. We are indebted to Vivek Prakash Gaur as well as Satyanarayan Sharma and his team for their useful contributions. Finally, we thank our families for their constant support and encouragement. Nitin Jain Neel Kamal Sethia

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Detailed Contents ELECTROMAGNETISM

1

Electric Charge and Electric Field

1

1.1 Electric Charge 1.2 Conductors, Insulators, and Induced Charges 1.3 Coulomb’s Law 1.4 Electric Field and Electric Forces 1.5 Electric-Field Calculations 1.6 Electric Field Lines 1.7 Electric Dipoles Summary Questions/Exercises/Problems



2

Gauss’s Law

2.1 Charge and Electric Flux 2.2 Calculating Electric Flux 2.3 Gauss’s Law 2.4 Applications of Gauss’s Law 2.5 Charges on Conductors Summary Questions/Exercises/Problems



3

Electric Potential

3.1 Electric Potential Energy 3.2 Electric Potential 3.3 Calculating Electric Potential 3.4 Equipotential Surfaces 3.5 Potential Gradient Summary Questions/Exercises/Problems

2 5 7 12 17 22 23 28 29

39 39 42 46 50 55 60 61

70 70 77 83 87 90 93 94

Capacitance and Dielectrics 106

4.1 4.2 4.3

107 111

Capacitors and Capacitance Capacitors in Series and Parallel Energy Storage in Capacitors and Electric-Field Energy 4.4 Dielectrics 4.5 Molecular Model of Induced Charge 4.6 Gauss’s Law in Dielectrics Summary Questions/Exercises/Problems



5



Current, Resistance, and Electromotive Force

5.1 Current 5.2 Resistivity

114 118 123 125 127 128

6



7

Magnetic Field and Magnetic Forces

145 148 154 158 161 162

170 170 175 180 184 188 193 194

207

7.1 7.2 7.3 7.4

Magnetism 207 Magnetic Field 209 Magnetic Field Lines and Magnetic Flux 213 Motion of Charged Particles in a Magnetic Field 216 7.5 Applications of Motion of Charged Particles 220 7.6 Magnetic Force on a Current-Carrying Conductor 222 7.7 Force and Torque on a Current Loop 225 7.8 The Direct-Current Motor 231 7.9 The Hall Effect 233 Summary 235 Questions/Exercises/Problems 236

8

Sources of Magnetic Field 249

8.1 8.2 8.3

249 252

Magnetic Field of a Moving Charge Magnetic Field of a Current Element Magnetic Field of a Straight Current-Carrying Conductor 8.4 Force Between Parallel Conductors 8.5 Magnetic Field of a Circular Current Loop 8.6 Ampere’s Law 8.7 Applications of Ampere’s Law 8.8 Magnetic Materials Summary Questions/Exercises/Problems



254 257 258 261 264 267 273 274

9 Electromagnetic

138 139 142

Direct-Current Circuits

6.1 Resistors in Series and Parallel 6.2 Kirchhoff’s Rules 6.3 Electrical Measuring Instruments 6.4 R-C Circuits 6.5 Power Distribution Systems Summary Questions/Exercises/Problems



4



5.3 Resistance 5.4 Electromotive Force and Circuits 5.5 Energy and Power in Electric Circuits 5.6 Theory of Metallic Conduction Summary Questions/Exercises/Problems

9.1 9.2

Induction 285

Induction Experiments Faraday’s Law

286 287 xi

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xii  DETAILED CONTENTS

9.3 9.4 9.5 9.6 9.7

Lenz’s Law 295 Motional Electromotive Force 297 Induced Electric Fields 299 Eddy Currents 302 Displacement Current and Maxwell’s Equations 303 9.8 Superconductivity 307 Summary 309 Questions/Exercises/Problems 310

10 Inductance

323

10.1 Mutual Inductance 10.2 Self-Inductance and Inductors 10.3 Magnetic-Field Energy 10.4 The R-L Circuit 10.5 The L-C Circuit 10.6 The L-R-C Series Circuit Summary Questions/Exercises/Problems

323 326 330 333 337 341 344 345

11 Alternating Current

353

11.1 Phasors and Alternating Currents 11.2 Resistance and Reactance 11.3 The L-R-C Series Circuit 11.4 Power in Alternating-Current Circuits 11.5 Resonance in Alternating-Current Circuits 11.6 Transformers Summary Questions/Exercises/Problems

12 Electromagnetic Waves

353 356 362 366 369 372 375 376

383

12.1

Maxwell’s Equations and Electromagnetic Waves 384 12.2 Plane Electromagnetic Waves and the Speed of Light 387 12.3 Sinusoidal Electromagnetic Waves 392 12.4 Energy and Momentum in Electromagnetic Waves 396 12.5 Standing Electromagnetic Waves 401 Summary 405 Questions/Exercises/Problems 406

14 Geometric Optics

13 The Nature and Propagation of Light The Nature of Light Reflection and Refraction Total Internal Reflection Dispersion

412 412 414 420 423

425 432 434 437 438

444

14.1 Reflection and Refraction at a Plane Surface 14.2 Reflection at a Spherical Surface 14.3 Refraction at a Spherical Surface 14.4 Thin Lenses 14.5 Cameras 14.6 The Eye 14.7 The Magnifier 14.8 Microscopes and Telescopes Summary Questions/Exercises/Problems

444 447 457 461 469 472 476 478 482 483

15 Interference 497 15.1 Interference and Coherent Sources 15.2 Two-Source Interference of Light 15.3 Intensity in Interference Patterns 15.4 Interference in Thin Films 15.5 The Michelson Interferometer Summary Questions/Exercises/Problems

16 Diffraction 16.1 Fresnel and Fraunhofer Diffraction 16.2 Diffraction from a Single Slit 16.3 Intensity in the Single-Slit Pattern 16.4 Circular Apertures and Resolving Power 16.5 Holography Summary Questions/Exercises/Problems

498 500 504 507 513 516 517

524 525 526 529 533 536 539 540

Modern Physics 17 Photons: Light Waves Behaving as Particles 17.1

optics

13.1 13.2 13.3 13.4

13.5 Polarization 13.6 Scattering of Light 13.7 Huygens’s Principle Summary Questions/Exercises/Problems

545

Light Absorbed as Photons: The Photoelectric Effect 545 17.2 Light Emitted as Photons: X-Ray Production 550 17.3 Light Scattered as Photons: Compton Scattering and Pair Production 553 17.4 Wave–Particle Duality, Probability, and Uncertainty 557 Summary 564 Questions/Exercises/Problems 565

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DETAILED CONTENTS  xiii

18 Particles Behaving as Waves 570 18.1 18.2 18.3

Electron Waves The Nuclear Atom and Atomic Spectra Energy Levels and the Bohr Model of the Atom Summary Questions/Exercises/Problems

570 575 580 591 592

19.3 Nuclear Stability and Radioactivity 19.4 Activities and Half-Lives 19.5 Biological Effects of Radiation 19.6 Nuclear Reactions 19.7 Nuclear Fission 19.8 Nuclear Fusion Summary Questions/Exercises/Problems

607 613 617 619 622 626 629 630

Photo CreditsC-1

19 Nuclear Physics 19.1 19.2

Properties of Nuclei Nuclear Binding and Nuclear Structure

599 599 602

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1

ELECTRIC CHARGE AND ELECTRIC FIELD

LEARNING GOALS By studying this chapter, you will learn: • The nature of electric charge, and how we know that electric charge is conserved. • How objects become electrically charged.

?

Water makes life possible: The cells of your body could not function without water in which to dissolve essential biological molecules. What electrical properties of water make it such a good solvent?

n Chapter 5 we mentioned the four kinds of fundamental forces. To this point the only one of these forces that we have examined in any detail is gravity. Now we are ready to examine the force of electromagnetism, which encompasses both electricity and magnetism. Electromagnetic phenomena will occupy our attention for most of the remainder of this book. Electromagnetic interactions involve particles that have a property called electric charge, an attribute that is as fundamental as mass. Just as objects with mass are accelerated by gravitational forces, so electrically charged objects are accelerated by electric forces. The shock you feel when you scuff your shoes across a carpet and then reach for a metal doorknob is due to charged particles leaping between your finger and the doorknob. Electric currents are simply streams of charged particles flowing within wires in response to electric forces. Even the forces that hold atoms together to form solid matter, and that keep the atoms of solid objects from passing through each other, are fundamentally due to electric interactions between the charged particles within atoms. We begin our study of electromagnetism in this chapter by examining the nature of electric charge. We’ll find that charge is quantized and obeys a conservation principle. When charges are at rest in our frame of reference, they exert electrostatic forces on each other. These forces are of tremendous importance in chemistry and biology and have many technological applications. Electrostatic forces are governed by a simple relationship known as Coulomb’s law and are most conveniently described by using the concept of electric field. In later chapters we’ll expand our discussion to include electric charges in motion. This will lead us to an understanding of magnetism and, remarkably, of the nature of light. While the key ideas of electromagnetism are conceptually simple, applying them to practical problems will make use of many of your mathematical skills, especially your knowledge of geometry and integral calculus. For this reason you may find this chapter and those that follow to be more mathematically demanding

I

• How to use Coulomb’s law to calculate the electric force between charges. • The distinction between electric force and electric field. • How to calculate the electric field due to a collection of charges. • How to use the idea of electric field lines to visualize and interpret electric fields. • How to calculate the properties of electric dipoles.

1

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2

CHAPTER 1 Electric Charge and Electric Field

than earlier chapters. The reward for your extra effort will be a deeper understanding of principles that are at the heart of modern physics and technology.

1.1

Electric Charge

The ancient Greeks discovered as early as 600 B.C. that after they rubbed amber with wool, the amber could attract other objects. Today we say that the amber has acquired a net electric charge, or has become charged. The word “electric” is derived from the Greek word elektron, meaning amber. When you scuff your shoes across a nylon carpet, you become electrically charged, and you can charge a comb by passing it through dry hair. Plastic rods and fur (real or fake) are particularly good for demonstrating electrostatics, the interactions between electric charges that are at rest (or nearly so). After we charge both plastic rods in Fig. 1.1a by rubbing them with the piece of fur, we find that the rods repel each other. When we rub glass rods with silk, the glass rods also become charged and repel each other (Fig. 1.1b). But a charged plastic rod attracts a charged glass rod; furthermore, the plastic rod and the fur attract each other, and the glass rod and the silk attract each other (Fig. 1.1c). These experiments and many others like them have shown that there are exactly two kinds of electric charge: the kind on the plastic rod rubbed with fur and the kind on the glass rod rubbed with silk. Benjamin Franklin (1706–1790) suggested calling these two kinds of charge negative and positive, respectively, and these names are still used. The plastic rod and the silk have negative charge; the glass rod and the fur have positive charge. Two positive charges or two negative charges repel each other. A positive charge and a negative charge attract each other. CAUTION Electric attraction and repulsion The attraction and repulsion of two charged objects are sometimes summarized as “Like charges repel, and opposite charges attract.” But keep in mind that the phrase “like charges” does not mean that the two charges are exactly identical, only that both charges have the same algebraic sign (both positive or both negative). “Opposite charges” means that both objects have an electric charge, and those charges have different signs (one positive and the other negative). y 1.1 Experiments in electrostatics. (a) Negatively charged objects repel each other. (b) Positively charged objects repel each other. (c) Positvely charged objects and negatively charged objects attract each other. (a) Interaction between plastic rods rubbed on fur

(b) Interaction between glass rods rubbed on silk

Plain plastic rods neither attract nor repel each other ...

Plain glass rods neither attract nor repel each other ...

(c) Interaction between objects with opposite charges The fur-rubbed plastic rod and the silkrubbed glass rod attract each other ...

– – – – – Fur

Plastic

Silk ... but after being rubbed with fur, the rods repel each other.

– – – – – –

+ + + ++

Glass ... but after being rubbed with silk, the rods repel each other.

... and the fur and silk each attracts the rod it rubbed.

+ + + ++

+ + + ++ –

+ –

–

–

+

+

+

+

+ ++++ ++ ++

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1.1 Electric Charge

3

1.2 Schematic diagram of the operation of a laser printer. 2 Laser beam “writes” on the drum, leaving negatively

charged areas where the image will be. 1 Wire sprays ions onto drum, giving the drum

+ – – +

a positive charge. 6 Lamp discharges the drum, readying

it to start the process over. 5 Fuser rollers heat paper so toner

remains permanently attached.

+ + – + +

+

+

+ +

–

Rotating imaging drum

–

+ + +

+

Toner (positively charged)

+ + +

– + + 3 Roller applies positively charged toner to drum. – Toner adheres only to negatively charged areas + – + of the drum “written” by the laser. – + – +

+ – + – – – + – ++ – – – – – – – – –

Paper (feeding to left) 4 Wires spray a stronger negative charge

on paper so toner will adhere to it.

One application of forces between charged bodies is in a laser printer (Fig. 1.2). The printer’s light-sensitive imaging drum is given a positive charge. As the drum rotates, a laser beam shines on selected areas of the drum, leaving those areas with a negative charge. Positively charged particles of toner adhere only to the areas of the drum “written” by the laser. When a piece of paper is placed in contact with the drum, the toner particles stick to the paper and form an image.

Electric Charge and the Structure of Matter When you charge a rod by rubbing it with fur or silk as in Fig. 1.1, there is no visible change in the appearance of the rod. What, then, actually happens to the rod when you charge it? To answer this question, we must look more closely at the structure of atoms, the building blocks of ordinary matter. The structure of atoms can be described in terms of three particles: the negatively charged electron, the positively charged proton, and the uncharged neutron (Fig. 1.3). The proton and neutron are combinations of other entities called quarks, which have charges of 6 13 and 6 23 times the electron charge. Isolated quarks have not been observed, and there are theoretical reasons to believe that it is impossible in principle to observe a quark in isolation. The protons and neutrons in an atom make up a small, very dense core called the nucleus, with dimensions of the order of 10-15 m. Surrounding the nucleus are the electrons, extending out to distances of the order of 10-10 m from the nucleus. If an atom were a few kilometers across, its nucleus would be the size of a tennis ball. The negatively charged electrons are held within the atom by the attractive electric forces exerted on them by the positively charged nucleus. (The protons and neutrons are held within stable atomic nuclei by an attractive interaction, called the strong nuclear force, that overcomes the electric repulsion of the protons. The strong nuclear force has a short range, and its effects do not extend far beyond the nucleus.) The masses of the individual particles, to the precision that they are presently known, are Mass of electron = me = 9.109382151452 * 10-31 kg Mass of proton = mp = 1.6726216371832 * 10-27 kg Mass of neutron = mn = 1.6749272111842 * 10-27 kg The numbers in parentheses are the uncertainties in the last two digits. Note that the masses of the proton and neutron are nearly equal and are roughly 2000 times

1.3 The structure of an atom. The particular atom depicted here is lithium (see Fig. 1.4a). Atom

~10210 m

+ Nucleus

+

+

Most of the atom’s volume is occupied sparsely by electrons.

Tiny compared with the rest of the atom, the nucleus contains over 99.9% of the atom’s mass.

~10215 m

+

Proton:

Positive charge Mass 5 1.673 3 10227 kg

Neutron: No charge Mass 5 1.675 3 10227 kg Electron: Negative charge Mass 5 9.109 3 10231 kg The charges of the electron and proton are equal in magnitude.

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4

CHAPTER 1 Electric Charge and Electric Field

1.4 (a) A neutral atom has as many electrons as it does protons. (b) A positive ion has a deficit of electrons. (c) A negative ion has an excess of electrons. (The electron “shells” are a schematic representation of the actual electron distribution, a diffuse cloud many times larger than the nucleus.)

Protons (1) Neutrons Electrons (2)

(a) Neutral lithium atom (Li): (b) Positive lithium ion (Li 1): (c) Negative lithium ion (Li 2): 3 protons (31) 3 protons (31) 3 protons (31) 4 neutrons 4 neutrons 4 neutrons 2 electrons (22) 4 electrons (42) 3 electrons (32) More electrons than protons: Electrons equal protons: Fewer electrons than protons: Negative net charge Zero net charge Positive net charge

the mass of the electron. Over 99.9% of the mass of any atom is concentrated in its nucleus. The negative charge of the electron has (within experimental error) exactly the same magnitude as the positive charge of the proton. In a neutral atom the number of electrons equals the number of protons in the nucleus, and the net electric charge (the algebraic sum of all the charges) is exactly zero (Fig. 1.4a). The number of protons or electrons in a neutral atom of an element is called the atomic number of the element. If one or more electrons are removed from an atom, what remains is called a positive ion (Fig. 1.4b). A negative ion is an atom that has gained one or more electrons (Fig. 1.4c). This gain or loss of electrons is called ionization. When the total number of protons in a macroscopic body equals the total number of electrons, the total charge is zero and the body as a whole is electrically neutral. To give a body an excess negative charge, we may either add negative charges to a neutral body or remove positive charges from that body. Similarly, we can create an excess positive charge by either adding positive charge or removing negative charge. In most cases, negatively charged (and highly mobile) electrons are added or removed, and a “positively charged body” is one that has lost some of its normal complement of electrons. When we speak of the charge of a body, we always mean its net charge. The net charge is always a very small fraction (typically no more than 10-12 ) of the total positive charge or negative charge in the body.

Electric Charge Is Conserved Implicit in the foregoing discussion are two very important principles. First is the principle of conservation of charge: The algebraic sum of all the electric charges in any closed system is constant.

If we rub together a plastic rod and a piece of fur, both initially uncharged, the rod acquires a negative charge (since it takes electrons from the fur) and the fur acquires a positive charge of the same magnitude (since it has lost as many electrons as the rod has gained). Hence the total electric charge on the two bodies together does not change. In any charging process, charge is not created or destroyed; it is merely transferred from one body to another. Conservation of charge is thought to be a universal conservation law. No experimental evidence for any violation of this principle has ever been observed. Even in high-energy interactions in which particles are created and destroyed, such as the creation of electron–positron pairs, the total charge of any closed system is exactly constant.

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1.2 Conductors, Insulators, and Induced Charges

The second important principle is: The magnitude of charge of the electron or proton is a natural unit of charge.

Every observable amount of electric charge is always an integer multiple of this basic unit. We say that charge is quantized. A familiar example of quantization is money. When you pay cash for an item in a store, you have to do it in one-cent increments. Cash can’t be divided into amounts smaller than one cent, and electric charge can’t be divided into amounts smaller than the charge of one electron or proton. (The quark charges, 6 13 and 6 23 of the electron charge, are probably not observable as isolated charges.) Thus the charge on any macroscopic body is always either zero or an integer multiple (negative or positive) of the electron charge. Understanding the electric nature of matter gives us insight into many aspects of the physical world (Fig. 1.5). The chemical bonds that hold atoms together to form molecules are due to electric interactions between the atoms. They include the strong ionic bonds that hold sodium and chlorine atoms together to make table salt and the relatively weak bonds between the strands of DNA that record your body’s genetic code. The normal force exerted on you by the chair in which you’re sitting arises from electric forces between charged particles in the atoms of your seat and in the atoms of your chair. The tension force in a stretched string and the adhesive force of glue are likewise due to the electric interactions of atoms.

1.5 Most of the forces on this water skier are electric. Electric interactions between adjacent molecules give rise to the force of the water on the ski, the tension in the tow rope, and the resistance of the air on the skier’s body. Electric interactions also hold the atoms of the skier’s body together. Only one wholly nonelectric force acts on the skier: the force of gravity.

Test Your Understanding of Section 1.1 (a) Strictly speaking, does the plastic rod in Fig. 1.1 weigh more, less, or the same after rubbing it with fur? (b) What about the glass rod after rubbing it with silk? What about (c) the fur and (d) the silk? y

1.2

Conductors, Insulators, and Induced Charges

5

PhET: Balloons and Static Electricity PhET: John Travoltage

Some materials permit electric charge to move easily from one region of the material to another, while others do not. For example, Fig. 1.6a shows a copper wire supported by a nylon thread. Suppose you touch one end of the wire to a charged plastic rod and attach the other end to a metal ball that is initially uncharged; you then remove the charged rod and the wire. When you bring another charged body up close to the ball (Figs. 1.6b and 1.6c), the ball is attracted or repelled, showing that the ball has become electrically charged. Electric charge has been transferred through the copper wire between the ball and the surface of the plastic rod. The copper wire is called a conductor of electricity. If you repeat the experiment using a rubber band or nylon thread in place of the wire, you find that no charge is transferred to the ball. These materials are called insulators. Conductors permit the easy movement of charge through them, while insulators do not. (The supporting nylon threads shown in Fig. 1.6 are insulators, which prevents charge from leaving the metal ball and copper wire.) As an example, carpet fibers on a dry day are good insulators. As you walk across a carpet, the rubbing of your shoes against the fibers causes charge to build up on you, and this charge remains on you because it can’t flow through the insulating fibers. If you then touch a conducting object such as a doorknob, a rapid charge transfer takes place between your finger and the doorknob, and you feel a shock. One way to prevent this is to wind some of the carpet fibers around conducting cores so that any charge that builds up on you can be transferred harmlessly to the carpet. Another solution is to coat the carpet fibers with an antistatic layer that does not easily transfer electrons to or from your shoes; this prevents any charge from building up on you in the first place.

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6

CHAPTER 1 Electric Charge and Electric Field

Most metals are good conductors, while most nonmetals are insulators. Within a solid metal such as copper, one or more outer electrons in each atom become detached and can move freely throughout the material, just as the molecules of a gas can move through the spaces between the grains in a bucket of sand. The other electrons remain bound to the positively charged nuclei, which themselves are bound in nearly fixed positions within the material. In an insulator there are no, or very few, free electrons, and electric charge cannot move freely through the material. Some materials called semiconductors are intermediate in their properties between good conductors and good insulators.

1.6 Copper is a good conductor of electricity; nylon is a good insulator. (a) The copper wire conducts charge between the metal ball and the charged plastic rod to charge the ball negatively. Afterward, the metal ball is (b) repelled by a negatively charged plastic rod and (c) attracted to a positively charged glass rod. (a) Insulating nylon threads – – Metal ball

Copper wire

Charging by Induction

Charged plastic rod – – –

The wire conducts charge from the negatively charged plastic rod to the metal ball. (b)

A negatively charged plastic rod now repels the ball ... – –

– – – – – Charged plastic rod

(c) ... and a positively charged glass rod attracts the ball. – –

+ + + + + Charged glass rod

We can charge a metal ball using a copper wire and an electrically charged plastic rod, as in Fig. 1.6a. In this process, some of the excess electrons on the rod are transferred from it to the ball, leaving the rod with a smaller negative charge. But there is a different technique in which the plastic rod can give another body a charge of opposite sign without losing any of its own charge. This process is called charging by induction. Figure 1.7 shows an example of charging by induction. An uncharged metal ball is supported on an insulating stand (Fig. 1.7a). When you bring a negatively charged rod near it, without actually touching it (Fig. 1.7b), the free electrons in the metal ball are repelled by the excess electrons on the rod, and they shift toward the right, away from the rod. They cannot escape from the ball because the supporting stand and the surrounding air are insulators. So we get excess negative charge at the right surface of the ball and a deficiency of negative charge (that is, a net positive charge) at the left surface. These excess charges are called induced charges. Not all of the free electrons move to the right surface of the ball. As soon as any induced charge develops, it exerts forces toward the left on the other free electrons. These electrons are repelled by the negative induced charge on the right and attracted toward the positive induced charge on the left. The system reaches an equilibrium state in which the force toward the right on an electron, due to the charged rod, is just balanced by the force toward the left due to the induced charge. If we remove the charged rod, the free electrons shift back to the left, and the original neutral condition is restored. What happens if, while the plastic rod is nearby, you touch one end of a conducting wire to the right surface of the ball and the other end to the earth (Fig. 1.7c)? The earth is a conductor, and it is so large that it can act as a practically infinite source of extra electrons or sink of unwanted electrons. Some of the negative charge flows through the wire to the earth. Now suppose you disconnect the wire (Fig. 1.7d) and then remove the rod (Fig. 1.7e); a net positive charge is left on the ball. The charge on the negatively charged rod has not changed during this process. The earth acquires a negative charge that is equal in magnitude to the induced positive charge remaining on the ball.

1.7 Charging a metal ball by induction. Electron Electron buildup deficiency ++ – Negatively charged – + – +– rod – – ––

Metal ball

– –

Insulating stand (a) Uncharged metal ball

(b) Negative charge on rod repels electrons, creating zones of negative and positive induced charge.

++ + – – +

Wire

– – – –

Ground (c) Wire lets electron buildup (induced negative charge) flow into ground.

–

– –– –

++ + + –

++ ++

Negative charge in ground

–

–

(d) Wire removed; ball now has only an electrondeficient region of positive charge.

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–

–

–

–

(e) Rod removed; electrons rearrange themselves, ball has overall electron deficiency (net positive charge).

1.3 Coulomb’s Law

7

1.8 The charges within the molecules of an insulating material can shift slightly. As a result, a comb with either sign of charge attracts a neutral insulator. By Newton’s third law the neutral insulator exerts an equal-magnitude attractive force on the comb. (b) How a negatively charged comb attracts an insulator

(c) How a positively charged comb attracts an insulator

Electrons in each molecule of the neutral insulator shift away from the comb.

This time, electrons in the molecules shift toward the comb ...

– – – – –– –– – – – – – –– – Negatively – charged comb S F

+– + – +– + + – – +– + + – – + + + – – +– + + –

S

2F

As a result, the (1) charges in each molecule are closer to the comb than are the (2) charges and so feel a stronger force from the comb. Therefore the net force is attractive.

+ + + + + + ++ + ++++ ++ + Positively + charged comb S F S ... so that the 2F (2) charges in each molecule are closer to the comb, and feel a stronger force from it, than the (+) charges. Again, the net force is attractive.

– +– +– + – – +– + + +– + + – – – – +– +– + –

(a) A charged comb picking up uncharged pieces of plastic

Electric Forces on Uncharged Objects Finally, we note that a charged body can exert forces even on objects that are not charged themselves. If you rub a balloon on the rug and then hold the balloon against the ceiling, it sticks, even though the ceiling has no net electric charge. After you electrify a comb by running it through your hair, you can pick up uncharged bits of paper or plastic with the comb (Fig. 1.8a). How is this possible? This interaction is an induced-charge effect. Even in an insulator, electric charges can shift back and forth a little when there is charge nearby. This is shown in Fig. 1.8b; the negatively charged plastic comb causes a slight shifting of charge within the molecules of the neutral insulator, an effect called polarization. The positive and negative charges in the material are present in equal amounts, but the positive charges are closer to the plastic comb and so feel an attraction that is stronger than the repulsion felt by the negative charges, giving a net attractive force. (In Section 1.3 we will study how electric forces depend on distance.) Note that a neutral insulator is also attracted to a positively charged comb (Fig. 1.8c). Now the charges in the insulator shift in the opposite direction; the negative charges in the insulator are closer to the comb and feel an attractive force that is stronger than the repulsion felt by the positive charges in the insulator. Hence a charged object of either sign exerts an attractive force on an uncharged insulator. Figure 1.9 shows an industrial application of this effect. Test Your Understanding of Section 1.2 You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative charge, while the other sphere has no net charge. (a) If the spheres are close together but do not touch, will they (i) attract each other, (ii) repel each other, or (iii) exert no force on each other? (b) You now allow the two spheres to touch. Once they have touched, will the two spheres (i) attract each other, (ii) repel each other, or (iii) exert no force on each other? y

1.3

Coulomb’s Law

Charles Augustin de Coulomb (1736–1806) studied the interaction forces of charged particles in detail in 1784. He used a torsion balance (Fig. 1.10a) similar to the one used 13 years later by Cavendish to study the much weaker gravitational interaction, as we discussed in Section 13.1. For point charges, charged

1.9 The electrostatic painting process (compare Figs. 1.7b and 1.7c). A metal object to be painted is connected to the earth (“ground”), and the paint droplets are given an electric charge as they exit the sprayer nozzle. Induced charges of the opposite sign appear in the object as the droplets approach, just as in Fig. 1.7b, and they attract the droplets to the surface. This process minimizes overspray from clouds of stray paint particles and gives a particularly smooth finish.

Metal object to be painted

Spray of negatively charged paint droplets

+

– –

–

–

+ +

–

Positive charge

– + is induced on – surface of metal. – – – + +

Paint sprayer

Ground

ActivPhysics 11.1: Electric Force: Coulomb's Law ActivPhysics 11.2: Electric Force: Superposition Principle ActivPhysics 11.3: Electric Force: Superposition (Quantitative)

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8

CHAPTER 1 Electric Charge and Electric Field

Electric Forces, Sweat, and Cystic Fibrosis

Application

One way to test for the genetic disease cystic fibrosis (CF) is by measuring the salt content of a person’s sweat. Sweat is a mixture of water and ions, including the sodium (Na + ) and chloride (Cl - ) ions that make up ordinary salt (NaCl). When sweat is secreted by epithelial cells, some of the Cl - ions flow from the sweat back into these cells (a process called reabsorption). The electric attraction between negative and positive charges pulls Na + ions along with the Cl - . Water molecules cannot flow back into the epithelial cells, so sweat on the skin has a low salt content. However, in persons with CF the reabsorption of Cl - ions is blocked. Hence the sweat of persons with CF is unusually salty, with up to four times the normal concentration of Cl - and Na + .

bodies that are very small in comparison with the distance r between them, Coulomb found that the electric force is proportional to 1>r2. That is, when the distance r doubles, the force decreases to one-quarter of its initial value; when the distance is halved, the force increases to four times its initial value. The electric force between two point charges also depends on the quantity of charge on each body, which we will denote by q or Q. To explore this dependence, Coulomb divided a charge into two equal parts by placing a small charged spherical conductor into contact with an identical but uncharged sphere; by symmetry, the charge is shared equally between the two spheres. (Note the essential role of the principle of conservation of charge in this procedure.) Thus he could obtain one-half, one-quarter, and so on, of any initial charge. He found that the forces that two point charges q1 and q2 exert on each other are proportional to each charge and therefore are proportional to the product q1 q2 of the two charges. Thus Coulomb established what we now call Coulomb’s law: The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

In mathematical terms, the magnitude F of the force that each of two point charges q1 and q2 a distance r apart exerts on the other can be expressed as F = k

ƒ q1 q2 ƒ

(1.1)

r2

where k is a proportionality constant whose numerical value depends on the system of units used. The absolute value bars are used in Eq. (1.1) because the charges q1 and q2 can be either positive or negative, while the force magnitude F is always positive. The directions of the forces the two charges exert on each other are always along the line joining them. When the charges q1 and q2 have the same sign, either both positive or both negative, the forces are repulsive; when the charges have opposite signs, the forces are attractive (Fig. 1.10b). The two forces obey Newton’s third law; they are always equal in magnitude and opposite in direction, even when the charges are not equal in magnitude.

1.10 (a) Measuring the electric force between point charges. (b) The electric forces between point charges obey NewS S ton’s third law: F1 on 2 5 - F2 on 1.

(a) A torsion balance of the type used by Coulomb to measure the electric force

(b) Interactions between point charges S

F2 on 1

The negatively charged ball attracts the positively charged one; the positive ball moves until the elastic forces in the torsion fiber balance the electrostatic attraction.

Torsion fiber

Charges of the r same sign repel.

q1 S

F1 on 2 S

S

F1 on 2 5 2F2 on 1 F1 on 2 5 F2 on 1 5 k

q1

0q1q2 0

q2

r2 Charges of opposite r sign attract.

S

F2 on 1 Charged pith balls

S

F1 on 2

+

+

Scale

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q2

1.3 Coulomb’s Law

The proportionality of the electric force to 1>r2 has been verified with great precision. There is no reason to suspect that the exponent is different from precisely 2. Thus the form of Eq. (1.1) is the same as that of the law of gravitation. But electric and gravitational interactions are two distinct classes of phenomena. Electric interactions depend on electric charges and can be either attractive or repulsive, while gravitational interactions depend on mass and are always attractive (because there is no such thing as negative mass).

Fundamental Electric Constants The value of the proportionality constant k in Coulomb’s law depends on the system of units used. In our study of electricity and magnetism we will use SI units exclusively. The SI electric units include most of the familiar units such as the volt, the ampere, the ohm, and the watt. (There is no British system of electric units.) The SI unit of electric charge is called one coulomb (1 C). In SI units the constant k in Eq. (1.1) is k = 8.987551787 * 109 N # m2>C2 > 8.988 * 109 N # m2>C2

The value of k is known to such a large number of significant figures because this value is closely related to the speed of light in vacuum. (We will show this in Chapter 12 when we study electromagnetic radiation.) As we discussed in Section 1.3, this speed is defined to be exactly c = 2.99792458 * 108 m>s. The numerical value of k is defined in terms of c to be precisely k = 110-7 N # s2>C22c2

You should check this expression to confirm that k has the right units. In principle we can measure the electric force F between two equal charges q at a measured distance r and use Coulomb’s law to determine the charge. Thus we could regard the value of k as an operational definition of the coulomb. For reasons of experimental precision it is better to define the coulomb instead in terms of a unit of electric current (charge per unit time), the ampere, equal to 1 coulomb per second. We will return to this definition in Chapter 8. In SI units we usually write the constant k in Eq. (1.1) as 1>4pP0 , where P0 (“epsilon-nought” or “epsilon-zero”) is another constant. This appears to complicate matters, but it actually simplifies many formulas that we will encounter in later chapters. From now on, we will usually write Coulomb’s law as F =

1 ƒ q1 q2 ƒ 4pP0 r2

(Coulomb’s law: force between two point charges)

(1.2)

The constants in Eq. (1.2) are approximately P0 = 8.854 * 10-12 C2>N # m2

and

1 = k = 8.988 * 109 N # m2>C2 4pP0

In examples and problems we will often use the approximate value 1 = 9.0 * 109 N # m2>C2 4pP0 which is within about 0.1% of the correct value. As we mentioned in Section 1.1, the most fundamental unit of charge is the magnitude of the charge of an electron or a proton, which is denoted by e. The most precise value available as of the writing of this book is e = 1.6021764871402 * 10-19 C One coulomb represents the negative of the total charge of about 6 * 1018 electrons. For comparison, a copper cube 1 cm on a side contains about 2.4 * 1024

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9

10

CHAPTER 1 Electric Charge and Electric Field

electrons. About 1019 electrons pass through the glowing filament of a flashlight bulb every second. In electrostatics problems (that is, problems that involve charges at rest), it’s very unusual to encounter charges as large as 1 coulomb. Two 1-C charges separated by 1 m would exert forces on each other of magnitude 9 * 109 N (about 1 million tons)! The total charge of all the electrons in a copper one-cent coin is even greater, about 1.4 * 105 C, which shows that we can’t disturb electric neutrality very much without using enormous forces. More typical values of charge range from about 10-9 to about 10-6 C. The microcoulomb 11 mC = 10-6 C2 and the nanocoulomb 11 nC = 10-9 C2 are often used as practical units of charge.

Example 1.1

Electric force versus gravitational force

An a particle (the nucleus of a helium atom) has mass m = 6.64 * 10-27 kg and charge q = + 2e = 3.2 * 10-19 C. Compare the magnitude of the electric repulsion between two a (“alpha”) particles with that of the gravitational attraction between them. SOLUTION IDENTIFY and SET UP: This problem involves Newton’s law for the gravitational force Fg between particles (see Section 13.1) and Coulomb’s law for the electric force Fe between point charges. To compare these forces, we make our target variable the ratio Fe>Fg. We use Eq. (1.2) for Fe and Eq. (13.1) for Fg.

1.11 Our sketch for this problem.

EXECUTE: Figure 1.11 shows our sketch. From Eqs. (1.2) and (13.1), Fe =

q2 1 4pP0 r2

Fg = G

m2 r2

These are both inverse-square forces, so the r2 factors cancel when we take the ratio: q2 Fe 1 = Fg 4pP0G m2 =

9.0 * 109 N # m2>C2

13.2 * 10-19 C22

6.67 * 10-11 N # m2>kg2 16.64 * 10-27 kg22

= 3.1 * 1035

EVALUATE: This astonishingly large number shows that the gravitational force in this situation is completely negligible in comparison to the electric force. This is always true for interactions of atomic and subnuclear particles. But within objects the size of a person or a planet, the positive and negative charges are nearly equal in magnitude, and the net electric force is usually much smaller than the gravitational force.

Superposition of Forces Coulomb’s law as we have stated it describes only the interaction of two point charges. Experiments show that when two charges exert forces simultaneously on a third charge, the total force acting on that charge is the vector sum of the forces that the two charges would exert individually. This important property, called the principle of superposition of forces, holds for any number of charges. By using this principle, we can apply Coulomb’s law to any collection of charges. Two of the examples at the end of this section use the superposition principle. Strictly speaking, Coulomb’s law as we have stated it should be used only for point charges in a vacuum. If matter is present in the space between the charges, the net force acting on each charge is altered because charges are induced in the molecules of the intervening material. We will describe this effect later. As a practical matter, though, we can use Coulomb’s law unaltered for point charges in air. At normal atmospheric pressure, the presence of air changes the electric force from its vacuum value by only about one part in 2000.

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1.3 Coulomb’s Law

Problem-Solving Strategy 1.1

Coulomb’s Law

IDENTIFY the relevant concepts: Coulomb’s law describes the electric force between charged particles. SET UP the problem using the following steps: 1. Sketch the locations of the charged particles and label each particle with its charge. 2. If the charges do not all lie on a single line, set up an xycoordinate system. 3. The problem will ask you to find the electric force on one or more particles. Identify which these are. EXECUTE the solution as follows: 1. For each particle that exerts an electric force on a given particle of interest, use Eq. (1.2) to calculate the magnitude of that force. 2. Using those magnitudes, sketch a free-body diagram showing the electric force vectors acting on each particle of interest. The force exerted by particle 1 on particle 2 points from particle 2 toward particle 1 if the charges have opposite signs, but points from particle 2 directly away from particle 1 if the charges have the same sign. 3. Use the principle of superposition to calculate the total electric force—a vector sum—on each particle of interest. (Review the

Example 1.2

vector algebra in Sections 1.7 through 1.9. The method of components is often helpful.) 4. Use consistent units; SI units are completely consistent. With 1>4pP0 5 9.0 * 109 N # m2>C2, distances must be in meters, charges in coulombs, and forces in newtons. 5. Some examples and problems in this and later chapters involve continuous distributions of charge along a line, over a surface, or throughout a volume. In these cases the vector sum in step 3 becomes a vector integral. We divide the charge distribution into infinitesimal pieces, use Coulomb’s law for each piece, and integrate to find the vector sum. Sometimes this can be done without actual integration. 6. Exploit any symmetries in the charge distribution to simplify your problem solving. For example, two identical charges q exert zero net electric force on a charge Q midway between them, because the forces on Q have equal magnitude and opposite direction. EVALUATE your answer: Check whether your numerical results are reasonable. Confirm that the direction of the net electric force agrees with the principle that charges of the same sign repel and charges of opposite sign attract.

Force between two point charges

Two point charges, q1 = + 25 nC and q2 = - 75 nC, are separated by a distance r = 3.0 cm (Fig. 1.12a). Find the magnitude and direction of the electric force (a) that q1 exerts on q2 and (b) that q2 exerts on q1. S O L U T IO N

1.12 What force does q1 exert on q2 , and what force does q2 exert on q1 ? Gravitational forces are negligible. '

'

(a) The two charges q1

IDENTIFY and SET UP: This problem asks for the electric forces that two charges exert on each other. We use Coulomb’s law, Eq. (1.2), to calculate the magnitudes of the forces. The signs of the charges will determine the directions of the forces. EXECUTE: (a) After converting the units of r to meters and the units of q1 and q2 to coulombs, Eq. (1.2) gives us F1

on 2

=

1 ƒ q1q2 ƒ 4pP0 r2

= 19.0 * 109 N # m2>C22

q2

ƒ 1+ 25 * 10-9 C21- 75 * 10-9 C2 ƒ 10.030 m2

2

(b) Free-body diagram for charge q2 S

F1 on 2

q2

(c) Free-body diagram for charge q1 q1

S

F2 on 1

r

(b) Proceeding as in part (a), we have F1

= 0.019 N The charges have opposite signs, so the force is attractive (to the left in Fig. 1.12b); that is, the force that acts on q2 is directed toward q1 along the line joining the two charges.

Example 1.3

11

on 2

=

1 ƒ q2q1 ƒ = F2 4pP0 r2

on 1

= 0.019 N

The attractive force that acts on q1 is to the right, toward q2 (Fig. 1.12c). EVALUATE: Newton’s third law applies to the electric force. Even though the charges have different magnitudes, the magnitude of the force that q2 exerts on q1 is the same as the magnitude of the force that q1 exerts on q2, and these two forces are in opposite directions.

Vector addition of electric forces on a line

Two point charges are located on the x-axis of a coordinate system: q1 = 1.0 nC is at x = + 2.0 cm, and q2 = - 3.0 nC is at x = + 4.0 cm. What is the total electric force exerted by q1 and q2 on a charge q3 = 5.0 nC at x = 0?

SOLUTION IDENTIFY and SET UP: Figure 1.13a shows the situation. To find the total force on q3, our target variable, we find the vector sum of the two electric forces on it. Continued

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12

CHAPTER 1 Electric Charge and Electric Field

EXECUTE: Figure 1.13b is a free-body diagram for q3, which is repelled by q1 (which has the same sign) and attracted to q2 (which S S has the opposite sign): F1 on 3 is in the –x-direction and F2 on 3 is in the +x-direction. After unit conversions, we have from Eq. (1.2) F1 on 3 =

1 ƒ q1q3 ƒ 4pP0 r213

= 19.0 * 109 N # m2>C22

10.020 m22

In the same way you can show that F2 on 3 = 84 mN. We thus have S S F1 on 3 5 1 - 112 mN2≤n and F2 on 3 5 184 mN2≤n. The net force on q3 is F3 5 F1 on 3 1 F 2 on 3 5 1- 112 mN2≤n 1 184 mN2≤n 5 1 -28 mN2≤n S

Example 1.4

(a) Our diagram of the situation

(b) Free-body diagram for q3

11.0 * 10-9 C215.0 * 10-9 C2

= 1.12 * 10-4 N = 112 mN

S

1.13 Our sketches for this problem.

S

EVALUATE: As a check, note that the magnitude of q2 is three times that of q1, but q2 is twice as far from q3 as q1. Equation (1.2) then says that F2 on 3 must be 3>22 = 3>4 = 0.75 as large as F1 on 3. This agrees with our calculated values: F2 on 3>F1 on 3 = 184 mN2> 1112 mN2 = 0.75. Because F2 on 3 is the weaker force, S the direction of the net force is that of F1 on 3—that is, in the negative x-direction.

Vector addition of electric forces in a plane S

Two equal positive charges q1 = q2 = 2.0 mC are located at x = 0, y = 0.30 m and x = 0, y = - 0.30 m, respectively. What are the magnitude and direction of the total electric force that q1 and q2 exert on a third charge Q = 4.0 mC at x = 0.40 m, y = 0?

F1

or 2 on Q

= 19.0 * 109 N # m2>C22

SOLUTION IDENTIFY and SET UP: As in Example 1.3, we must compute the force that each charge exerts on Q and then find the vector sum of those forces. Figure 1.14 shows the situation. Since the three charges do not all lie on a line, the best way to calculate the forces is to use components. 1.14 Our sketch for this problem.

S

EXECUTE: Figure 1.14 shows the forces F1 on Q and F2 on Q due to the identical charges q1 and q2, which are at equal distances from Q. From Coulomb’s law, both forces have magnitude

*

14.0 * 10-6 C212.0 * 10-6 C2 10.50 m22

= 0.29 N

The x-components of the two forces are equal: 1F1

or 2 on Q2x

= 1F1

or 2 on Q2cos

a = 10.29 N2

0.40 m = 0.23 N 0.50 m

From symmetry we see that the y-components of the two forces are S equal and opposite. Hence their sum is zero and the total force F on Q has only an x-component Fx = 0.23 N + 0.23 N = 0.46 N. The total force on Q is in the + x-direction, with magnitude 0.46 N. EVALUATE: The total force on Q points neither directly away from q1 nor directly away from q2. Rather, this direction is a compromise that points away from the system of charges q1 and q2. Can you see that the total force would not be in the +x-direction if q1 and q2 were not equal or if the geometrical arrangement of the changes were not so symmetric?

Test Your Understanding of Section 1.3 Suppose that charge q2 in Example 1.4 were - 2.0 mC. In this case, the total electric force on Q would be (i) in the positive x-direction; (ii) in the negative x-direction; (iii) in the positive y-direction; (iv) in the negative y-direction; (v) zero; (vi) none of these.

1.4

y

Electric Field and Electric Forces

When two electrically charged particles in empty space interact, how does each one know the other is there? We can begin to answer this question, and at the same time reformulate Coulomb’s law in a very useful way, by using the concept of electric field.

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1.4 Electric Field and Electric Forces

13

Electric Field To introduce this concept, let’s look at the mutual repulsion of two positively S charged bodies A and B (Fig. 1.15a). Suppose B has charge q0 , and let F0 be the electric force of A on B. One way to think about this force is as an “action-at-adistance” force—that is, as a force that acts across empty space without needing any matter (such as a push rod or a rope) to transmit it through the intervening space. (Gravity can also be thought of as an “action-at-a-distance” force.) But a more fruitful way to visualize the repulsion between A and B is as a two-stage process. We first envision that body A, as a result of the charge that it carries, somehow modifies the properties of the space around it. Then body B, as a result of the charge that it carries, senses how space hasSbeen modified at its position. The response of body B is to experience the force F0. To elaborate how this two-stage process occurs, we first consider body A by itself: We remove body B and label its former position as point P (Fig. 1.15b). We say that the charged body A produces or causes an electric field at point P (and at all other points in the neighborhood). This electric field is present at P even if there is no charge at P; it is a consequence of the charge on Sbody A only. If a point charge q0 is then placed at point P, it experiences the force F0. We take the point of view that this force is exerted on q0 by the field at P (Fig. 1.15c). Thus the electric field is the intermediary through which A communicates its presence to q0. Because the point charge q0 would experience a force at any point in the neighborhood of A, the electric field that A produces exists at all points in the region around A. We can likewise say that the point charge q0 produces an electric field in the S space around it and that this electric field exerts the force 2 F0 on body A. For each force (the force of A on q0 and the force of q0 on A), one charge sets up an electric field that exerts a force on the second charge. We emphasize that this is an interaction between two charged bodies. A single charge produces an electric field in the surrounding space, but this electric field cannot exert a net force on the charge that created it; as we discussed in Section 4.3, a body cannot exert a net force on itself. (If this wasn’t true, you would be able to lift yourself to the ceiling by pulling up on your belt!) '

The electric force on a charged body is exerted by the electric field created by other charged bodies.

To find out experimentally whether there is an electric field at a particular point, we place a small charged body, which we call a test charge, at the point (Fig. 1.15c). If the test charge experiences an electric force, then there is an electric field at that point. This field is produced by charges other than q0. Force is a vector quantity, so electric field is also a vector quantity. (Note the use of vector signs as well as boldface letters and plus, Sminus, and equals signs in the following discussion.) We define the electric field E at a point as the electric S force F0 experienced by a test charge q0 at the point, divided by the charge q0. That is, the electric field at a certain point is equal to the electric force per unit charge experienced by a charge at that point: S

S

E5

(definition of electric field as electric force per unit charge)

F0 q0

(1.3)

1.15 A charged body creates an electric field in the space around it. (a) A and B exert electric forces on each other.

B

(b) Remove body B ... ... and label its former position as P. P A S

(c) Body A sets up an electric field E at point P. Test charge q0 S

S

S

'

(force exerted on a point charge q0 S by an electric field E)

F0 S E5 q 0

A S

E is the force per unit charge exerted by A on a test charge at P.

Sharks and the “Sixth Sense”

Application

Sharks have the ability to locate prey (such as flounder and other bottom-dwelling fish) that are completely hidden beneath the sand at the bottom of the ocean. They do this by sensing the weak electric fields produced by muscle contractions in their prey. Sharks derive their sensitivity to electric fields (a “sixth sense”) from jelly-filled canals in their bodies. These canals end in pores on the shark’s skin (shown in this photograph). An electric field as weak as 5 * 10-7 N> C causes charge flow within the canals and triggers a signal in the shark’s nervous system. Because the shark has canals with different orientations, it can measure different components of the electric-field vector and hence determine the direction of the field.

'

F0 5 q0 E

S

F0

A

In SI units, in which the unit of force is 1 N and the unit of charge is 1 C, the unit of electric field Smagnitude is 1 newton per coulomb 11 N>C2. If the field E at a certain point is known, rearS ranging Eq. (1.3) gives the force F0 experienced by a point charge q0 placed at S that point. This force is just equal to the electric field E produced at that point by charges other than q0 , multiplied by the charge q0 : '

q0

S

2F0

(1.4)

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14

CHAPTER 1 Electric Charge and Electric Field S

S

S

1.16 The force F0 5 q0 E exerted on a S point charge q0 placed in an electric field E.

The charge q0 can be either positive or negative. If q0 is positive, the force F0 S S S experienced by the charge is the same direction as E; if q0 is negative, F0 and E are in opposite directions (Fig. 1.16). While the electric field concept may be new to you, the basic idea—that one body sets up a field in the space around it and a second body responds to that field—is one that you’ve actually usedS before. Compare Eq. (1.4) to the familiar expression for the gravitational force Fg that the earth exerts on a mass m0 :

'

Q

S

E (due to charge Q) q0 S

F0 The force on a positive test charge q0 points in the direction of the electric field.

'

S

S

Fg 5 m0 g

Q

(1.5)

'

'

S

S

F0

E (due to charge Q) S

In this expression, g is the acceleration due to gravity. If we divide both sides of Eq. (1.5) by the mass m0 , we obtain

q0

'

The force on a negative test charge q0 points opposite to the electric field.

S

S

g5

Fg m0

S

Thus g can be regarded as the gravitational force per unit mass. By analogy to S Eq. (1.3), we can interpret g as the gravitational field. Thus we treat the gravitational interaction between the earth and the mass m0 as a two-stage process: The S earth sets up a gravitational field g in the space around it, and this gravitational field exerts a force given by Eq. (1.5) on the mass m0 (which we can regard as a S test mass). The gravitational field g , or gravitational force per unit mass, is a useful concept because it does not depend on the mass of the body on which the S gravitational force is exerted; likewise, the electric field E, or electric force per unit charge, is useful because it does not depend on the charge of the body on which the electric force is exerted.

ActivPhysics 11.4: Electric Field: Point Charge ActivPhysics 11.9: Motion of a Charge in an Electric Field: Introduction ActivPhysics 11.10: Motion in an Electric Field: Problems

S

S

CAUTION F0 5 q0 E0 is for point test charges only The electric force experienced by a test charge q0 can vary from point to point, so the electric field can also be different at different points. For this reason, Eq. (1.4) can be used only to find the electric force on a S point charge. If a charged body is large enough in size, the electric field E may be noticeably different in magnitude and direction at different points on the body, and calculating the net electric force on the body can become rather complicated. y '

Electric Field of a Point Charge If the source distribution is a point charge q, it is easy to find the electric field that it produces. We call the location of the charge the source point, and we call the point P where we are determining the field the field point. It is also useful to introduce a unit vector rN that points along the line from source point to field point S (Fig. 1.17a). This unit vector is equal to the displacement vector r from the S source point to the field point, divided by the distance r = ƒ r ƒ between these two S points; that is, rN 5 r >r. If we place a small test charge q0 at the field point P, at a

S

S

1.17 The electric field E produced at point P by an isolated point charge q at S. Note that in both (b) and (c), E is produced by q [see Eq. (1.7)] but acts on the charge q0 at point P [see Eq. (1.4)]. (a)

S

(b) q0

q0 P r^

q S

r Unit vector r^ points from source point S to field point P.

E

P

q

r^ S

At each point P, the electric field set up by an isolated positive point charge q points directly away from the ^ charge in the same direction as r.

(c) S

q0

E q S

P r^ At each point P, the electric field set up by an isolated negative point charge q points directly toward the ^ charge in the opposite direction from r.

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1.4 Electric Field and Electric Forces

15

distance r from the source point, the magnitude F0 of the force is given by Coulomb’s law, Eq. (1.2): F0 =

1 ƒ qq0 ƒ 4pP0 r2

From Eq. (1.3) the magnitude E of the electric field at P is E =

1 ƒqƒ 4pP0 r2

(magnitude of electric field of a point charge)

(1.6)

Using the unit vector rN , we can write a vector equation that gives both the magniS tude and direction of the electric field E: S

E5

1 q rN 4pP0 r2

(electric field of a point charge)

(1.7)

By definition, the electric field of a point charge always points away from a posiq tive charge (that is, in the same direction as rN ; see Fig. 1.17b) but toward a nega- 1.18 ASpoint charge produces an electric field E at all points in space. The field tive charge (that is, in the direction opposite rN ; see Fig. 1.17c). strength decreases with increasing distance. S We have emphasized calculating the electric field E at a certain point. But (a) The field produced by a positive point S since E can vary from point to point, it is not a single vector quantity but rather charge points away from the charge. an infinite set of vector quantities, one associated with each point in space. This is an example of a vector field. Figure 1.18 shows a number of the field vectors produced by a positive or negative point charge. If we use a rectangular 1x, y, z2 S S E coordinate system, each component of E at any point is in general a function of the coordinates 1x, y, z2 of the point. We can represent the functions as q Ex1x, y, z2, Ey1x, y, z2, and Ez1x, y, z2. Vector fields are an important part of the language of physics, not just in electricity and magnetism. One everyday example S of a vector field is the velocity Y of wind currents; the magnitude and direction of S Y, and hence its vector components, vary from point to point in the atmosphere. In some situations the magnitude and direction of the field (and hence its vector components) have the same values everywhere throughout a certain region; we then say that the field is uniform in this region. An important example of this is the electric field inside a conductor. If there is an electric field within a conductor, the field exerts a force on every charge in the conductor, giving the free (b) The field produced by a negative point charges a net motion. By definition an electrostatic situation is one in which the charge points toward the charge. charges have no net motion. We conclude that in electrostatics the electric field at every point within the material of a conductor must be zero. (Note that we are not S E saying that the field is necessarily zero in a hole inside a conductor.) In summary, our description of electric interactions has two parts. First, a given charge distribution acts as a source of electric field. Second, the electric field q exerts a force on any charge that is present in the field. Our analysis often has two corresponding steps: first, calculating the field caused by a source charge distribution; second, looking at the effect of the field in terms of force and motion. The second step often involves Newton’s laws as well as the principles of electric interactions. In the next section we show how to calculate fields caused by various source distributions, but first here are three examples of calculating the S field due to a point charge and of finding the force on a charge due to a given field E. Example 1.5

Electric-field magnitude for a point charge S

What is the magnitude of the electric field E at a field point 2.0 m from a point charge q = 4.0 nC?

SOLUTION IDENTIFY and SET UP: This problem concerns the electric field due to a point charge. We are given the magnitude of the charge Continued

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16

CHAPTER 1 Electric Charge and Electric Field

and the distance from the charge to the field point, so we use Eq. (1.6) to calculate the field magnitude E. EXECUTE: From Eq. (1.6), E =

4.0 * 10-9 C 1 ƒqƒ = 19.0 * 109 N # m2 > C22 2 4pP0 r 12.0 m22 '

EVALUATE: Our result E = 9.0 N>C means that if we placed a 1.0-C charge at a point 2.0 m from q, it would experience a 9.0-N force. The force on a 2.0-C charge at that point would be 12.0 C219.0 N>C2 = 18 N, and so on.

'

= 9.0 N>C

Example 1.6

Electric-field vector for a point charge

A point charge q = - 8.0 nC is located at the origin. Find the electric-field vector at the field point x = 1.2 m, y = -1.6 m. '

gin O) to the field point P, and we must obtain an expression for S the unit vector rN 5 r >r that points from S to P. EXECUTE: The distance from S to P is

SOLUTION S

IDENTIFY and SET UP: We must find the electric-field vector E due to a point charge. Figure 1.19 shows the situation. We use Eq. (1.7); to do this, we must find the distance r from the source point S (the position of the charge q, which in this example is at the ori1.19 Our sketch for this problem.

r = 2x2 + y2 = 211.2 m22 + 1-1.6 m22 = 2.0 m

The unit vector rN is then S x≤n 1 y≥n r rN 5 5 r r 11.2 m2≤n 1 1-1.6 m2≥n 5 5 0.60≤n 2 0.80≥n 2.0 m Then, from Eq. (1.7), S 1 q rN E5 4pP0 r2 5 19.0 * 109 N # m2>C22

1 -8.0 * 10-9 C2 12.0 m22

5 1 -11 N>C2≤n 1 114 N>C2≥n

10.60≤n 2 0.80≥n2

S

EVALUATE: Since q is negative, E points from the field point to the charge (the source point), in the direction opposite to rN (compare Fig. 1.17c). We leave the calculation of the magnitude and direcS tion of E to you (see Exercise 1.36).

Example 1.7

Electron in a uniform field

When the terminals of a battery are connected to two parallel conducting plates with a small gap between them, the resulting charges S on the plates produce a nearly uniform electric field E between the plates. (In the next section we’ll see why this is.) If the plates are 1.0 cm apart and are connected to a 100-volt battery as shown in Fig. 1.20, the field is vertically upward and has magnitude 1.20 A uniform electric field between two parallel conducting plates connected to a 100-volt battery. (The separation of the plates is exaggerated in this figure relative to the dimensions of the plates.) The thin arrows represent the uniform electric field.

–

–

– S

–

–

E

+

+

–

–

S

S

+

+

+

+

x 1.0 cm

F 5 2eE

100 V +

–

O

+

E = 1.00 * 104 N>C. (a) If an electron (charge - e = - 1.60 * 10-9 C, mass m = 9.11 * 10-31 kg) is released from rest at the upper plate, what is its acceleration? (b) What speed and kinetic energy does it acquire while traveling 1.0 cm to the lower plate? (c) How long does it take to travel this distance? SOLUTION IDENTIFY and SET UP: This example involves the relationship between electric field and electric force. It also involves the relationship between force and acceleration, the definition of kinetic energy, and the kinematic relationships among acceleration, distance, velocity, and time. Figure 1.20 shows our coordinate system. We are given the electric field, so we use Eq. (1.4) to find the force on the electron and Newton’s second law to find its acceleration. Because the field is uniform, the force is constant and we can use the constant-acceleration formulas from Chapter 2 to find the electron’s velocity and travel time. We find the kinetic energy using K = 12 mv2. '

y

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1.5 Electric-Field Calculations S

17

S

EXECUTE: (a) Although E is upward (in the +y-direction), F is The velocity is downward, so vy = - 5.9 * 106 m>s. The elecdownward (because the electron’s charge is negative) and so Fy tron’s kinetic energy is is negative. Because Fy is constant, the electron’s acceleration is K = 12 mv2 = 12 19.11 * 10-31 kg215.9 * 106 m>s22 constant: = 1.6 * 10-17 J Fy 1 -1.60 * 10-19 C211.00 * 104 N>C2 - eE ay = = = (c) From Eq. (2.8) for constant acceleration, vy = v0y + ayt, m m 9.11 * 10-31 kg = - 1.76 * 1015 m>s2 t =

vy - v0y

=

1- 5.9 * 106 m>s2 - 10 m>s2

ay (b) The electron starts from rest, so its motion is in the -1.76 * 1015 m>s2 y-direction only (the direction of the acceleration). We can find the = 3.4 * 10-9 s electron’s speed at any position y using the constant-acceleration Eq. (2.13), vy2 = v0 y2 + 2ay1y - y02. We have v0y = 0 and EVALUATE: Our results show that in problems concerning suby0 = 0, so at y = - 1.0 cm = - 1.0 * 10-2 m we have atomic particles such as electrons, many quantities—including 15 2 -2 ƒ vy ƒ = 22ayy = 221 - 1.76 * 10 m>s 21-1.0 * 10 m2 acceleration, speed, kinetic energy, and time—will have very different values from those typical of everyday objects such as base= 5.9 * 106 m>s balls and automobiles. '

'

Test Your Understanding of Section 1.4 (a) A negative point charge moves along a straight-line path directly toward a stationary positive point charge. Which aspect(s) of the electric force on the negative point charge will remain constant as it moves? (i) magnitude; (ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor direction. (b) A negative point charge moves along a circular orbit around a positive point charge. Which aspect(s) of the electric force on the negative point charge will remain constant as it moves? (i) magnitude; (ii) direction; (iii) both magnitude and direction; (iv) neither magnitude nor direction. y

1.5

Electric-Field Calculations

Equation (1.7) gives the electric field caused by a single point charge. But in most realistic situations that involve electric fields and forces, we encounter charge that is distributed over space. The charged plastic and glass rods in Fig. 1.1 have electric charge distributed over their surfaces, as does the imaging drum of a laser printer (Fig. 1.2). In this section we’ll learn to calculate electric fields caused by various distributions of electric charge. Calculations of this kind are of tremendous importance for technological applications of electric forces. To determine the trajectories of atomic nuclei in an accelerator for cancer radiotherapy or of charged particles in a semiconductor electronic device, you have to know the detailed nature of the electric field acting on the charges.

ActivPhysics 11.5: Electric Field Due to a Dipole ActivPhysics 11.6: Electric Field: Problems

The Superposition of Electric Fields To find the field caused by a charge distribution, we imagine the distribution to be made up of many point charges q1 , q2 , q3 , . . . . (This is actually quite a realistic description, since we have seen that charge is carried by electrons and protons that are so small as to be almost pointlike.) At any given point P, each point S S S charge produces its own electric field E1 , E2 , E3 , . . . , so a test charge q0 placed at S S S S P experiences a force F1 5 q0 E1 from charge q1 , a force F2 5 q0 E2 from charge q2 , and so on. From the principle of superposition of forces discussed in S Section 1.3, the total force F0 that the charge distribution exerts on q0 is the vector sum of these individual forces: '

'

'

'

'

'

'

'

'

'

S S S S S S S F0 5 F1 1 F2 1 F3 1 Á 5 q0 E1 1 q0 E2 1 q0 E3 1 Á '

'

'

The combinedS effect of all the charges in the distribution is described by the total electric field E at point P. From the definition of electric field, Eq. (1.3), this is S

S

E5

F0 S S S 5 E1 1 E2 1 E3 1 Á q0 '

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18

CHAPTER 1 Electric Charge and Electric Field

1.21 Illustrating the principle of superposition of electric fields. q1

P

Electric field at P due to q2 S

Electric field at P due to q1 S

E2

E1 S

q2

E

S

The total electric field E at point S S P is the vector sum of E1 and E2.

Problem-Solving Strategy 1.2

The total electric field at P is the vector sum of the fields at P due to each point charge in the charge distribution (Fig. 1.21). This is the principle of superposition of electric fields. When charge is distributed along a line, over a surface, or through a volume, a few additional terms are useful. For a line charge distribution (such as a long, thin, charged plastic rod), we use l (the Greek letter lambda) to represent the linear charge density (charge per unit length, measured in C>m). When charge is distributed over a surface (such as the surface of the imaging drum of a laser printer), we use s (sigma) to represent the surface charge density (charge per unit area, measured in C>m2). And when charge is distributed through a volume, we use r (rho) to represent the volume charge density (charge per unit volume, C>m3). Some of the calculations in the following examples may look fairly intricate. After you’ve worked through the examples one step at a time, the process will seem less formidable. We will use many of the calculational techniques in these examples in Chapter 8 to calculate the magnetic fields caused by charges in motion.

Electric-Field Calculations

IDENTIFY the relevant concepts: Use the principle of superposition to calculate the electric field due to a discrete or continuous charge distribution. SET UP the problem using the following steps: 1. Make a drawing showing the locations of the charges and your choice of coordinate axes. 2. On your drawing, indicate the position of the field point P (the S point at which you want to calculate the electric field E). EXECUTE the solution as follows: 1. Use consistent units. Distances must be in meters and charge must be in coulombs. If you are given centimeters or nanocoulombs, don’t forget to convert. 2. Distinguish between the source point S and the field point P. The field produced by a point charge always points from S to P if the charge is positive, and from P to S if the charge is negative. 3. Use vector addition when applying the principle of superposition; review the treatment of vector addition in Chapter 1 if necessary.

Example 1.8

4. Simplify your calculations by exploiting any symmetries in the charge distribution. 5. If the charge distribution is continuous, define a small element of charge that can be considered as a point, find its electric field at P, and find a way to add the fields of all the charge elements by doing an integral. Usually it is easiest to do this for each S component of E separately, so you may need to evaluate more than one integral. Ensure that the limits on your integrals are correct; especially when the situation has symmetry, don’t count a charge twice. S

EVALUATE your answer: Check that the direction of E is reasonable. If your result for the electric-field magnitude E is a function of position (say, the coordinate x), check your result in any limits for which you know what the magnitude should be. When possible, check your answer by calculating it in a different way.

Field of an electric dipole

Point charges q1 5 + 12 nC and q2 5 -12 nC are 0.100 m apart (Fig. 1.22). (Such pairs of point charges with equal magnitude and opposite sign are called electric dipoles.) Compute the electric field caused by q1, the field caused by q2, and the total field (a) at point a; (b) at point b; and (c) at point c. SOLUTION IDENTIFY and SET UP: We must find the total electric field at various points due toStwo point charges. We use the principle of superS S position: E 5 E1 1 E2. Figure 1.22 shows the coordinate system and the locations of the field points a, b, and c. S

S

S

EXECUTE: At each field point, E depends on E1 and E2 there; we first calculate the magnitudes E1 and E2 at each field point. At a S the magnitude of the field E1a caused by q1 is

1 ƒ q1 ƒ 12 * 10-9 C = 19.0 * 109 N # m2>C22 2 4pP0 r 10.060 m22 '

E1a =

= 3.0 * 104 N>C

We calculate the other field magnitudes in a similar way. The results are E1a = 3.0 * 104 N>C

E1b = 6.8 * 104 N>C

E1c = 6.39 * 103 N>C E2a = 6.8 * 104 N>C

E2b = 0.55 * 104 N>C

E2c = E1c = 6.39 * 103 N>C The directions of the corresponding fields are in all cases away from the positive charge q1 and toward the negative charge q2.

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1.5 Electric-Field Calculations 1.22 Electric field at three points, a, b, and c, set up by charges q1 and q2 , which form an electric dipole. '

S

y

E1 a

5 E1cx = E2cx = E1c cos a = 16.39 * 103 N>C2 A 13 B

Ec

a

S

(c) Figure 1.22 shows the directions of E1 and E2 at c. Both vectors have the same x-component:

= 2.46 * 103 N>C

S

c

S

19

From symmetry, E1y and E2y are equal and opposite, so their sum is zero. Hence

S

E2

Ec 5 212.46 * 103 N>C2≤n 5 14.9 * 103 N>C2≤n S

S

13.0 cm

13.0 cm

a

q1

S

Eb b

–

x

S

(a) At a, E1a and E2a are both directed to the right, so Ea 5 E1an≤ 1 E2an≤ 5 19.8 * 104 N>C2≤n S

S

S

(b) At b, E1b is directed to the left and E2b is directed to the right, so Eb 5 - E1bn≤ 1 E2bn≤ 5 1- 6.2 * 104 N>C2≤n S

Example 1.9

S O L U T IO N

1.23 Calculating the electric field on the axis of a ring of charge. In this figure, the charge is assumed to be positive. y

5 219.0 * 109 N # m2>C22 5 14.9 * 103 N>C2≤n

12 * 10-9 C 10.13 m22

A 135 B n≤

EXECUTE: We divide the ring into infinitesimal segments ds as shown in Fig. 1.23. In terms of the linear charge density l = Q>2pa, the charge in a segment of length ds is dQ = l ds. Consider two identical segments, one as shown in the figure at y = a and another halfway around the ring at y = -a. From S Example 1.4, we see that the net force dFS they exert on a point test charge at P, and thus their net field dE, are directed along the x-axis. The same is true for any such pair ofSsegments around the ring, so the net field at P is along the x-axis: E 5 Exn≤ . To calculate Ex, note that the square of the distance r from a single ring segment to the point P is r2 S= x2 + a2. Hence the magnitude of this segment’s contribution dE to the electric field at P is dE =

dQ 1 2 4pP0 x + a2

The x-component of this field is dEx = dE cos a. We know dQ 5 l ds and Fig. 1.23 shows that cos a = x>r = x>1x2 + a221>2, so

dQ ds r5

x 21

x

dQ 1 x 2 4pP0 x + a2 2x2 + a2 1 lx = ds 4pP0 1x2 + a223>2

dEx = dE cos a = a2

a

Q

S

Field of a ring of charge

IDENTIFY and SET UP: This is a problem in the superposition of electric fields. Each bit of charge around the ring produces an electric field at an arbitrary point on the x-axis; our target variable is the total field at this point due to all such bits of charge.

O

1 q1 1 q2 rN 1 rN 4pP0 r2 1 4pP0 r2 2 q1 1 5 1q1rN 1 1 q2rN 22 5 1rN 1 2 rN 22 4pP0r2 4pP0r2 1 q1 5 12 cosa≤n2 4pP0 r2 S

This is the same as we calculated in part (c).

Charge Q is uniformly distributed around a conducting ring of radius a (Fig. 1.23). Find the electric field at a point P on the ring axis at a distance x from its center.

a

S

Ec 5 E1c 1 E2c 5

4.0 cm

6.0 cm

S

q2

S

Ea

a

+ 4.0 cm

EVALUATE: We can also find Ec using Eq. (1.7) for the field of a S point charge. The displacement vector r 1 from q1 to point c is S n n r 1 5 r cos a ≤ 1 r sin a ≥ . Hence the unit vector that points from S q1 to point c is rN 1 5 r 1>r 5 cos a n≤ + sin a n≥ . By symmetry, the unit vector that points from q2 to point c has the opposite xcomponent but the same y-component: rN 2 5 - cos a n≤ 1 sin a n≥ . S S We can now use Eq. (1.7) to write the fields E1c and E2c at c in vector form, then find their sum. Since q2 = - q1 and the distance r to c is the same for both charges,

dEy

P

dEx

x

a S

dE

Continued

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20

CHAPTER 1 Electric Charge and Electric Field

To find Ex we integrate this expression over the entire ring—that is, for s from 0 to 2pa (the circumference of the ring). The integrand has the same value for all points on the ring, so it can be taken outside the integral. Hence we get 2pa

lx 1 Ex = dEx = ds 2 4pP0 1x + a223>2 3 2

S

EVALUATE: Equation (1.8) shows that E 5 0 at the center of the ring 1x 5 02. This makes sense; charges on opposite sides of the ring push in opposite directions on a test charge at the center, and the vector sum of each such pair of forces is zero. When the field point P is much farther from the ring than the ring’s radius, we have x W a and the denominator in Eq. (1.8) becomes approximately equal to x3. In this limit the electric field at P is

0

1 lx = 12pa2 4pP0 1x2 + a223>2

S

E 5 Exn≤ 5

Example 1.10

Qx 1 n≤ 4pP0 1x2 + a223>2

S

E5

(1.8)

1 Q n≤ 4pP0 x2

That is, when the ring is so far away that its radius is negligible in comparison to the distance x, its field is the same as that of a point charge.

Field of a charged line segment

Positive charge Q is distributed uniformly along the y-axis between y = -a and y = +a. Find the electric field at point P on the x-axis at a distance x from the origin.

1.24 Our sketch for this problem.

SOLUTION IDENTIFY and SET UP: Figure 1.24 shows the situation. As in Example 1.9, we must find the electric field due to a continuous distribution of charge. Our target variable is an expression for the electric field at P as a function of x. The x-axis is a perpendicular bisector of the segment, so we can use a symmetry argument. EXECUTE: We divide the line charge of length 2a into infinitesimal segments of length dy. The linear charge density is l = Q>2a, and the charge in a segment is dQ = l dy = 1Q>2a2dy. The distance r from a segment at height y to the field point P is r = 1x2 + y221>2, so the magnitude of the field at P due to the segment at height y is dE =

dy 1 dQ 1 Q = 4pP0 r2 4pP0 2a 1x2 + y22

S

E points away from the line of charge if l is positive and toward the line of charge if l is negative.

Figure 1.24 shows that the x- and y-components of this field are dEx = dE cos a and dEy = - dE sin a, where cos a 5 x> r and sin a = y> r. Hence dEx = dEy =

x dy 1 Q 4pP0 2a 1x2 + y223>2 y dy 1 Q 2 4pP0 2a 1x + y223>2

To find the total field at P, we must sum the fields from all segments along the line—that is, we must integrate from y = - a to y = +a. You should work out the details of the integration (a table of integrals will help). The results are Ex =

+a x dy Q 1 Q 1 = 4pP0 2a 2-a 1x2 + y223>2 4pP0 x2x2 + a2

E5

Q 1 n≤ 4pP0 x2x2 + a2

1 l n≤ 2 2pP0 x 21x >a22 + 1

(1.10)

In the limit a W x we can neglect x2>a2 in the denominator of Eq. (1.10), so

or, in vector form, S

1 Q n≤ 4pP0 x2 To see what happens if the segment is very long (or the field point is very close to it) so that a W x, we first rewrite Eq. (1.9) slightly: S

E5

S

+a y dy 1 Q Ey = = 0 2 2 4pP0 2a -a 1x + y223>2

E5

EVALUATE: Using a symmetry argument as in Example 1.9, we could have guessed that Ey would be zero; if we place a positive test charge at P, the upper half of the line of charge pushes downward on it, and the lower half pushes up with equal magnitude. Symmetry also tells us that the upper and lower halves of the segment contribute equally to the total field at P. If the segment is very short (or the field point is very far from the segment) so that x W a, we can neglect a in the denominator of Eq. (1.9). Then the field becomes that of a point charge, just as in Example 1.9:

(1.9)

S

E5

l n≤ 2pP0x

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21

1.5 Electric-Field Calculations P This is the field of an infinitely long line of charge. At any point S at a perpendicular distance r from the line in any direction, E has magnitude l (infinite line of charge) E = 2pP0r Note that this field is proportional to 1>r rather than to 1>r2 as for a point charge.

Example 1.11

There’s really no such thing in nature as an infinite line of charge. But when the field point is close enough to the line, there’s very little difference between the result for an infinite line and the real-life finite case. For example, if the distance r of the field point from the center of the line is 1% of the length of the line, the value of E differs from the infinite-length value by less than 0.02%.

Field of a uniformly charged disk

A nonconducting disk of radius R has a uniform positive surface charge density s. Find the electric field at a point along the axis of the disk a distance x from its center. Assume that x is positive.

To find the total field due to all the rings, we integrate dEx over r from r = 0 to r = R (not from –R to R):

S O L U T IO N IDENTIFY and SET UP: Figure 1.25 shows the situation. We represent the charge distribution as a collection of concentric rings of charge dQ. In Example 1.9 we obtained Eq. (1.8) for the field on the axis of a single uniformly charged ring, so all we need do here is integrate the contributions of our rings. EXECUTE: A typical ring has charge dQ, inner radius r, and outer radius r + dr. Its area is approximately equal to its width dr times its circumference 2pr, or dA = 2pr dr. The charge per unit area is s = dQ>dA, so the charge of the ring is dQ = s dA = 2psr dr. We use dQ in place of Q in Eq. (1.8), the expression for the field due to a ring that we found in Example 1.9, and replace the ring radius a with r. Then the field component dEx at point P due to this ring is dEx =

1 2psrx dr 4pP0 1x2 + r223>2

1.25 Our sketch for this problem.

Example 1.12

1 12psr dr2x sx 2r dr = 20 4pP0 1x2 + r223>2 4P0 20 1x2 + r223>2 R

Ex =

R

You can evaluate this integral by making the substitution t = x2 + r2 (which yields dt = 2r dr); you can work out the details. The result is Ex = =

sx 1 1 c + d 2 2 x 2P0 2x + R 1 s c1 d 2 2 2P0 21R >x 2 + 1

(1.11)

EVALUATE: If the disk is very large (or if we are very close to it), so that R W x, the term 1> 21R2>x22 + 1 in Eq. (1.11) is very much less than 1. Then Eq. (1.11) becomes s E = (1.12) 2P0 Our final result does not contain the distance x from the plane. Hence the electric field produced by an infinite plane sheet of charge is independent of the distance from the sheet. The field direction is everywhere perpendicular to the sheet, away from it. There is no such thing as an infinite sheet of charge, but if the dimensions of the sheet are much larger than the distance x of the field point P from the sheet, the field is very nearly given by Eq. (1.12). If P is to the left of the plane 1x 6 02, the result is the same S except that the direction of E is to the left instead of the right. If the surface charge density is negative, the directions of the fields on both sides of the plane are toward it rather than away from it.

Field of two oppositely charged infinite sheets

Two infinite plane sheets with uniform surface charge densities 1.26 Finding the electric field due to two oppositely charged + s and - s are placed parallel to each other with separation d infinite sheets. The sheets are seen edge-on; only a portion of the (Fig. 1.26). Find the electric field between the sheets, above the infinite sheets can be shown! upper sheet, and below the lower sheet. y S O L U T IO N IDENTIFY and SET UP: Equation (1.12) gives the electric field due to a single infinite plane sheet of charge. To find the field due to two such sheets, we combine the fields using the principle of superposition (Fig. 1.26).

S

S

S

E1

E2

E 5 E1 1 E2 5 0

S

S

S

S

S

E1

E2

E 5 E1 1 E2

Sheet 2 2s

x d

S

S

Sheet 1 1s S

S

S

E1

E2

E 5 E1 1 E2 5 0

S

S

Continued

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22

CHAPTER 1 Electric Charge and Electric Field S

S

EXECUTE: From Eq. (1.12), both E1 and E2 have the same magnitude at all points, independent of distance from either sheet: s E1 = E2 = 2P0 S

FromS Example 1.11, E1 is everywhere directed away from sheet 1, and E2 is everywhere directed toward sheet 2. S S Between the sheets, E1 and E2 reinforce each other; above the upper sheet and below the lower sheet, they cancel each other. Thus the total field is 0 s E 5 E1 1 E2 5 d n≥ P0 0 S

S

S

above the upper sheet between the sheets below the lower sheet

EVALUATE: Because we considered the sheets to be infinite, our result does not depend on the separation d. Our result shows that the field between oppositely charged plates is essentially uniform if the plate separation is much smaller than the dimensions of the plates. We actually used this result in Example 1.7 (Section 1.4). CAUTION Electric fields are not “flows” You may have thought S that the field E1 of sheet 1 would be unable to “penetrate” sheet 2, S and that field E2 caused by sheet 2 would be unable to “penetrate” sheet 1. You might conclude this if you think of the electric field as some kind of physical substance that “flows” into or out of charges. But inS fact there is no such substance, and the electric S fields E1 and E2 depend only on the individual charge distribu tions that create them. The total field at every point is just the vecS S tor sum of E1 and E2. y

Test Your Understanding of Section 1.5 Suppose that the line of charge in Fig. 1.25 (Example 1.11) had charge + Q distributed uniformly between y = 0 and y = +a and had charge -Q distributed uniformly between y = 0 and y = -a. In this situation, the electric field at P would be (i) in the positive x-direction; (ii) in the negative x-direction; (iii) in the positive y-direction; (iv) in the negative y-direction; (v) zero; (vi) none of these.

1.6 1.27 The direction of the electric field at any point is tangent to the field line through that point. Field at ES P point P P Electric field line

PhET: Charges and Fields PhET: Electric Field of Dreams PhET: Electric Field Hockey

Field at point R S ER R

y

Electric Field Lines

The concept of an electric field can be a little elusive because you can’t see an electric field directly. Electric field lines can be a big help for visualizing electric fields and making them seem more real. An electric field line is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric-field vector at that point. Figure 1.27 shows the basic idea. (We used a similar concept in our discussion of fluid flow in Section 12.5. A streamline is a line or curve whose tangent at any point is in the direction of the velocity of the fluid at that point. However, the similarity between electric field lines and fluid streamlines is a mathematical one only; there is nothing “flowing” in an electric field.) The English scientist Michael Faraday (1791–1867) first introduced the concept of field lines. He called them “lines of force,” but the term “field lines” is preferable. S Electric field lines show the directionSof E at each point, andS their spacing gives a general idea of the magnitude of E atSeach point. Where E is strong, we draw lines bunched closely together; where E is weaker, they are farther apart. At any particular point, the electric field has a unique direction, so only one field line can pass through each point of the field. In other words, field lines never intersect. Figure 1.28 shows some of the electric field lines in a plane containing (a) a single positive charge; (b) two equal-magnitude charges, one positive and one negative (a dipole); and (c) two equal positive charges. Diagrams such as these are sometimes called field maps; they are cross sections of the actual threedimensional patterns. The direction of the total electric field at every point in each diagram is along the tangent to the electricSfield line passing through the point. Arrowheads indicate the direction of the E-field vector along each field line. The actual field vectors have been drawn at several points in each pattern. Notice that in general, the magnitude of the electric field is different at different points on a given field line; a field line is not a curve of constant electric-field magnitude! Figure 1.28 shows that field lines are Sdirected away from positive charges (since close to a positive point charge, E points away from the charge) and

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1.7 Electric Dipoles

23

S

1.28 Electric field lines for three different charge distributions. In general, the magnitude of E is different at different points along a given field line. (a) A single positive charge

(b) Two equal and opposite charges (a dipole)

(c) Two equal positive charges

S

E

S

E S

E

S

E

–

+

+

+

+

S

E

S

E

S

E Field lines always point away from (1) charges and toward (2) charges.

S

E

At each point in space, the electric field vector is tangent to the field line passing through that point.

Field lines are close together where the field is strong, farther apart where it is weaker.

S

toward negative charges (since close to a negative point charge, E points toward the charge). In regions where the field magnitude is large, such as between the positive and negative charges in Fig. 1.28b, the field lines are drawn close together. In regions where the field magnitude is small, such as between the two positive charges in Fig. 1.28c, the lines are widely separated. In a uniform field, the field lines are straight, parallel, and uniformly spaced, as in Fig. 1.20. Figure 1.29 is a view from above of a demonstration setup for visualizing electric field lines. In the arrangement shown here, the tips of two positively charged wires are inserted in a container of insulating liquid, and some grass seeds are floated on the liquid. The grass seeds are electrically neutral insulators, but the electric field of the two charged wires causes polarization of the grass seeds; there is a slight shifting of the positive and negative charges within the molecules of each seed, like that shown in Fig.S1.8. The positively charged end of each grass seed isSpulled in the direction of E and the negatively charged end is pulled opposite E. Hence the long axis of each grass seed tends to orient parallel to the electric field, in the direction of the field line that passes through the position of the seed (Fig. 1.29b). CAUTION Electric field lines are not the same as trajectories It’s a common misconception that if a charged particle of charge q is in motion where there is an electric field, the S particle must move along an electric field line. Because E at anySpoint S is tangent to the field line that passes through that point, it is indeed true that the force F 5 qE on the particle, and hence the particle’s acceleration, are tangent to the field line. But we learned in Chapter 3 that when a particle moves on a curved path, its acceleration cannot be tangent to the path. So in general, the trajectory of a charged particle is not the same as a field line. y

1.29 (a) Electric field lines produced by two equal point charges. The pattern is formed by grass seeds floating on a liquid above two charged wires. Compare this pattern with Fig. 1.28c. (b) The electric field causes polarization of the grass seeds, which in turn causes the seeds to align with the field. (a)

(b)

Test Your Understanding of Section 1.6 Suppose the electric field lines in a region of space are straight lines. If a charged particle is released from rest in that region, will the trajectory of the particle be along a field line? y

S

E

1 2

1.7

Electric Dipoles

An electric dipole is a pair of point charges with equal magnitude and opposite sign (a positive charge q and a negative charge - q) separated by a distance d. We introduced electric dipoles in Example 1.8 (Section 1.5); the concept is worth exploring further because many physical systems, from molecules to TV antennas, can be described as electric dipoles. We will also use this concept extensively in our discussion of dielectrics in Chapter 24.

Field line

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Grass seed

24

CHAPTER 1 Electric Charge and Electric Field

1.30 (a) A water molecule is an example of an electric dipole. (b) Each test tube contains a solution of a different substance in water. The large electric dipole moment of water makes it an excellent solvent. (a) A water molecule, showing positive charge as red and negative charge as blue

1 H

H

O

S

p

2 S

The electric dipole moment p is directed from the negative end to the positive end of the molecule. (b) Various substances dissolved in water

Figure 1.30 a shows a molecule of water 1H2O2, which in many ways behaves like an electric dipole. The water molecule as a whole is electrically neutral, but the chemical bonds within the molecule cause a displacement of charge; the result is a net negative charge on the oxygen end of the molecule and a net positive charge on the hydrogen end, forming an electric dipole. The effect is equivalent to shifting one electron only about 4 * 10-11 m (about the radius of a hydrogen atom), but the consequences of this shift are profound. Water is an excellent solvent for ionic substances such as table salt (sodium chlo ride, NaCl) precisely because the water molecule is an electric dipole (Fig. 1.30b). When dissolved in water, salt dissociates into a positive sodium ion 1Na+2 and a negative chlorine ion 1Cl-2, which tend to be attracted to the negative and positive ends, respectively, of water molecules; this holds the ions in solution. If water molecules were not electric dipoles, water would be a poor solvent, and almost all of the chemistry that occurs in aqueous solutions would be impossible. This includes all of the biochemical reactions that occur in all of the life on earth. In a very real sense, your existence as a living being depends on electric dipoles! We examine two questions about electric dipoles. First, what forces and torques does an electric dipole experience when placed in an external electric field (that is, a field set up by charges outside the dipole)? Second, what electric field does an electric dipole itself produce?

?

Force and Torque on an Electric Dipole

1.31 The net force on this electric dipole is zero, but there is a torque directed into the page that tends to rotate the dipole clockwise. S

+

S

p S

d

E

– S

S

F25 2qE

2q

1q f

S

F15 qE

d sin f

To start with the first question, let’s place an electric dipole in a Suniform exterS S nal electric field E, as shown in Fig. 1.31. The forces F + and F- on the two charges both have magnitude qE, but their directions are opposite, and they add to zero. The net force on an electric dipole in a uniform external electric field is zero. However, the two forces don’t act along the same line, so their torques don’t add to zero. We calculate torques with respect to the center of the dipole. Let the S angleSbetweenS the electric field E and the dipole axis be f; then the Slever arm for S both F + and F- is 1d>22 sin f. The torque of F + and the torque of F- both have the same magnitude of 1qE21d>22 sin f, and both torques tend to rotate the dipole S clockwise (that is, T is directed into the page in Fig. 1.31). Hence the magnitude of the net torque is twice the magnitude of either individual torque: t = 1qE21d sin f2

(1.13)

where d sin f is the perpendicular distance between the lines of action of the two forces. The product of the charge q and the separation d is the magnitude of a quantity called the electric dipole moment, denoted by p: p = qd

(magnitude of electric dipole moment)

(1.14)

The units of p are charge times distance 1C # m2. For example, the magnitude of the electric dipole moment of a water molecule is p = 6.13 * 10-30 C # m. CAUTION The symbol p has multiple meanings Be careful not to confuse dipole moment with momentum or pressure. There aren’t as many letters in the alphabet as there are physical quantities, so some letters are used several times. The context usually makes it clear what we mean, but be careful. y S

We further define the electric dipole moment to be a vector quantity p . The S magnitude of p is given by Eq. (1.14), and its direction is along the dipole axis from the negative charge to the positive charge as shown in Fig. 1.31. In terms of p, Eq. (1.13) for the magnitude t of the torque exerted by the field becomes

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1.7 Electric Dipoles

t = pE sin f

(magnitude of the torque on an electric dipole)

25

(1.15) S

SinceSthe angle f in Fig. 1.31 is the angle between the directions of the vectors p and E, this is reminiscent of the expression for the magnitude of the vector product discussed in Section 1.10. (You may want to review that discussion.) Hence we can write the torque on the dipole in vector form as S

S

S

T5p : E

(torque on an electric dipole, in vector form)

(1.16)

You can use the right-hand rule for the vector product to verify that in the situation S S S shown in Fig. 1.31, T is directed into the page. The torque is greatest when p and E are perpendicular and is zero when theyS are parallel or antiparallel.SThe torque S p to line it up with E. The position f = 0, with p parallel to always tends to turn S S E, is a position of stable equilibrium, and the position f = p, with S p and E antiparallel, is a position of unstable equilibrium. The polarization of a grass seed in the apparatus of Fig. 1.29b gives it anSelectric dipole moment; the torque exerted S by E then causes the seed to align with E and hence with the field lines.

Potential Energy of an Electric Dipole When a dipole changes direction in an electric field, the electric-field torque does work on it, with a corresponding change in potential energy. The work dW done by a torque t during an infinitesimal displacement df is given by Eq. (10.19): dW = t df. Because the torque is in the direction of decreasing f, we must write the torque as t = -pE sin f, and

PhET: Microwaves

dW = t df = -pE sin f df In a finite displacement from f1 to f2 the total work done on the dipole is f2

W =

2f1

1 -pE sin f2 df

= pE cos f2 - pE cos f1 The work is the negative of the change of potential energy, just as in Chapter 7: W = U1 - U2. So a suitable definition of potential energy U for this system is U1f2 = - pE cos f

(1.17)

In this expression we recognize the scalar product p # E = pE cos f, so we can also write S

S

#

S

U = -p E

S

(potential energy for a dipole in an electric field)

(1.18)

The potential energy has its minimum (most negative) value U =S - pE at the staS ble equilibrium position, where f = 0 and p is parallel to E. The potential S S energy is maximum when f = p and p is antiparallel to E; then U = +pE. At S S f = p>2, where p is perpendicular to E, U is zero. We could define U differently S so that it is zero at some other orientation of p , but our definition is simplest. Equation (1.18) givesS us another way to look at the effect shown in Fig. 1.29. The electric field E gives eachS grass seed an electric dipole moment, and the grass seed then aligns itself with E to minimize the potential energy.

Example 1.13

Force and torque on an electric dipole

Figure 1.32a shows an electric dipole in a uniform electric field of magnitude 5.0 * 105 N>C that is directed parallel to the plane of the figure. The charges are 61.6 * 10-19 C; both lie in the plane

and are separated by 0.125 nm = 0.125 * 10-9 m. Find (a) the net force exerted by the field on the dipole; (b) the magnitude and

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Continued

26

CHAPTER 1 Electric Charge and Electric Field

1.32 (a) An electric dipole. (b) Directions of the electric dipole S moment, electric field, and torque (T points out of the page). (b)

(a)

p = qd = 11.6 * 10-19 C210.125 * 10-9 m2 = 2.0 * 10-29 C # m

2q S

t

S

E

35°

EXECUTE: (a) The field is uniform, so the forces on the two charges are equal and opposite. Hence the total force on the dipole is zero. S (b) The magnitude p of the electric dipole moment p is

S

E

S

p 145°

145°

S

The direction of p is from the negative to the positive charge, 145° clockwise from the electric-field direction (Fig. 1.32b). (c) The magnitude of the torque is t = pE sin f = 12.0 * 10-29 C215.0 * 105 N>C21sin 145°2

1q

= 5.7 * 10-24 N # m

direction of the electric dipole moment; (c) the magnitude and direction of the torque; (d) the potential energy of the system in the position shown. SOLUTION

From the right-hand rule for vector products (see Section 1.10), the S S S direction of the torque T 5 p : E is out of the page. This correS S sponds to a counterclockwise torque that tends to align p with E. (d) The potential energy U = -pE cos f

IDENTIFY and SET UP: This problem uses the ideas of this section about an electric dipole placed in an electric field. We use the relaS S tionship F 5 qE for each point charge to find the force on the dipole as a whole. Equation (1.14) gives the dipole moment, Eq. (1.16) gives the torque on the dipole, and Eq. (1.18) gives the potential energy of the system.

= - 12.0 * 10-29 C # m215.0 * 105 N>C21cos 145°2 = 8.2 * 10-24 J

EVALUATE: The charge magnitude, the distance between the charges, the dipole moment, and the potential energy are all very small, but are all typical of molecules.

S

In this discussion we have assumed that E is uniform, so there is no net force S on the dipole. If E is not uniform, the forces at the ends may not cancel completely, and the net force may not be zero. Thus a body with zero net charge but an electric dipole moment can experience a net force in a nonuniform electric field. As we mentioned in Section 1.1, an uncharged body can be polarized by an electric field, giving rise to a separation of charge and an electric dipole moment. This is how uncharged bodies can experience electrostatic forces (see Fig. 1.8).

Field of an Electric Dipole Now let’s think of an electric dipole as a source of electric field. What does the field look like? The general shape of things is shown by the field map of Fig. S 1.28b. At each point in the pattern the total E field is the vector sum of the fields from the two individual charges, as in Example 1.8 (Section 1.5). Try drawing diagrams showing this vector sum for several points. To get quantitative information about the field of an electric dipole, we have to do some calculating, as illustrated in the next example. Notice the use of the principle of superposition of electric fields to add up the contributions to the field of the individual charges. Also notice that we need to use approximation techniques even for the relatively simple case of a field due to two charges. Field calcula tions often become very complicated, and computer analysis is typically used to determine the field due to an arbitrary charge distribution.

Example 1.14

Field of an electric dipole, revisited S

An electric dipole is centered at the origin, with p in the direction of the + y-axis (Fig. 1.33). Derive an approximate expression for the electric field at a point P on the y-axis for which y is much larger than d. To do this, use the binomial expansion 11 + x2n > 1 + nx + n1n - 12x2>2 + Á (valid for the case ƒ x ƒ 6 1).

SOLUTION IDENTIFY and SET UP: We use the principle of superposition: The total electric field is the vector sum of the field produced by the positive charge and the field produced by the negative charge. At S the field point P shown in Fig. 1.33, the field E + of the positive S charge has a positive (upward) y-component and the field E - of

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1.7 Electric Dipoles 1.33 Finding the electric field of an electric dipole at a point on its axis. y

27

We used this same approach in Example 1.8 (Section 1.5). Now the approximation: When we are far from the dipole compared to its size, so y W d, we have d>2y V 1. With n = - 2 and with d>2y replacing x in the binomial expansion, we keep only the first two terms (the terms we discard are much smaller). We then have

S

E1 P

a1 -

S

E2

d -2 d d d -2 b >1 + b >1 and a1 + y y 2y 2y

Hence Ey is given approximately by Ey >

/

y 2d2

/

y1d2

q 4pP0y

2

'

c1 +

qd p d d = - a1 - b d = 3 y y 2pP0y 2pP0y3

EVALUATE: An alternative route to this result is to put the fractions in the first expression for Ey over a common denominator, add, and then approximate the denominator 1y - d>222 1y + d>222 as y4. We leave the details to you (see Exercise 1.60). For points P off the coordinate axes, the expressions are more complicated, but at all points far away from the dipole (in any direction) the field drops off as 1>r3. We can compare this with the 1>r2 behavior of a point charge, the 1>r behavior of a long line charge, and the independence of r for a large sheet of charge. There are charge distributions for which the field drops off even more quickly. At large distances, the field of an electric quadrupole, which consists of two equal dipoles with opposite orientation, separated by a small distance, drops off as 1>r4. '

1q d

S

p

x

O 2q

the negative charge has a negative (downward) y-component. We add these components to find the total field and then apply the approximation that y is much greater than d. EXECUTE: The total y-component Ey of electric field from the two charges is Ey =

=

q 1 1 c d 4pP0 1y - d>222 1y + d>222 '

q 4pP0y

2

'

c a1 -

d -2 d -2 b - a1 + b d 2y 2y

Test Your Understanding of S Section 1.7 An electric dipole is placed S in a region of uniform electric field E , with the electric dipole moment p , pointing S in the direction opposite to E. Is the dipole (i) in stable equilibrium, (ii) in unstable equilibrium, or (iii) neither? (Hint: You many want to review Section 7.5.)

y

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CHAPTER

1

SUMMARY

Electric charge, conductors, and insulators: The fundamental quantity in electrostatics is electric charge. There are two kinds of charge, positive and negative. Charges of the same sign repel each other; charges of opposite sign attract. Charge is conserved; the total charge in an isolated system is constant. All ordinary matter is made of protons, neutrons, and electrons. The positive protons and electrically neutral neutrons in the nucleus of an atom are bound together by the nuclear force; the negative electrons surround the nucleus at distances much greater than the nuclear size. Electric interactions are chiefly responsible for the structure of atoms, molecules, and solids. Conductors are materials in which charge moves easily; in insulators, charge does not move easily. Most metals are good conductors; most nonmetals are insulators.

Coulomb’s law: For charges q1 and q2 separated by a distance r, the magnitude of the electric force on either charge is proportional to the product q1 q2 and inversely proportional to r2. The force on each charge is along the line joining the two charges—repulsive if q1 and q2 have the same sign, attractive if they have opposite signs. In SI units the unit of electric charge is the coulomb, abbreviated C. (See Examples 1.1 and 1.2.) When two or more charges each exert a force on a charge, the total force on that charge is the vector sum of the forces exerted by the individual charges. (See Examples 1.3 and 1.4.)

1 ƒ q1 q2 ƒ 4pP0 r2

+

+

S

F2 on 1

'

F =

(1.2)

r

q1

'

1 = 8.988 * 109 N # m2>C2 4pP0

S

F1 on 2 q2 q1 S F2 on 1

r

S

F1 on 2 q2

S

S

Electric field: Electric field E, a vector quantity, is the force per unit charge exerted on a test charge at any point. The electric field produced by a point charge is directed radially away from or toward the charge. (See Examples 1.5–1.7.)

+

–

F0 q0

(1.3)

1 q Nr E5 4pP0 r2

(1.7)

S

E5

S

E q

S

S

Superposition of electric fields: The electric field E of any combination of charges is the vector sum of the fields caused by the individual charges. To calculate the electric field caused by a continuous distribution of charge, divide the distribution into small elements, calculate the field caused by each element, and then carry out the vector sum, usually by integrating. Charge distributions are described by linear charge density l, surface charge density s, and volume charge density r. (See Examples 1.8–1.12.)

y dQ ds r

5

x 21

a

a P a x dEy

O

2

dEx x a r

dE

Q

S

Electric field lines: Field lines provide a graphical representation of electric fields. At any point on a S field line, the tangent to the line is in the direction of E at that point. The number of lines per unit S area (perpendicular to their direction) is proportional to the magnitude of E at the point.

S

E

E

+

–

S

E

Electric dipoles: An electric dipole is a pair of electric charges of equal magnitude q but opposite sign, separated S by a distance d. The electric dipole moment p has magniS tude p = qd. The direction of p is from negative toward S positive charge. An electric dipole in an electric field E S S experiences a torque T equal to the vector product of p S and E. The magnitude of the torque depends on the angle f S S between p and E. The potential energy U for an electric dipole in an electric field also depends on the relative oriS S entation of p and E. (See Examples 1.13 and 1.14.)

t = pE sin f

(1.15)

S

S

S

S

T5p : E S

#

S

(1.16)

d

E

1q d sin f

f S

U = -p E

(1.18)

S

S

F25 2qE

28

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S

F15 qE

S

p

2q

Discussion Questions

BRIDGING PROBLEM

29

Calculating Electric Field: Half a Ring of Charge

Positive charge Q is uniformly distributed around a semicircle of radius a as shown in Fig. 1.34. Find the magnitude and direction of the resulting electric field at point P, the center of curvature of the semicircle. SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. The target variables are the components of the electric field at P. 2. Divide the semicircle into infinitesimal segments, each of which is a short circular arc of radius a and angle du. What is the length of such a segment? How much charge is on a segment?

EXECUTE 4. Integrate your expressions for dEx and dEy from u 5 0 to u 5 p. The results will be the x-component and y-component of the electric field at P. 5. Use your results from step 4 to find the magnitude and direction of the field at P. EVALUATE 6. Does your result for the electric-field magnitude have the correct units? 7. Explain how you could have found the x-component of the electric field using a symmetry argument. 8. What would be the electric field at P if the semicircle were extended to a full circle centered at P?

1.34 y Q

a P

3. Consider an infinitesimal segment located at an angular position u on the semicircle, measured from the lower right cor ner of the semicircle at x 5 a, y 5 0. (Thus u 5 p> 2 at x 5 0, y 5 a and u = p at x = -a, y = 0.) What are the x- and y-components of the electric field at P (dEx and dEy) produced by just this segment?

x

Problems

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q1.1 If you peel two strips of transparent tape off the same roll and immediately let them hang near each other, they will repel each other. If you then stick the sticky side of one to the shiny side of the other and rip them apart, they will attract each other. Give a plausible explanation, involving transfer of electrons between the strips of tape, for this sequence of events. Q1.2 Two metal spheres are hanging from nylon threads. When you bring the spheres close to each other, they tend to attract. Based on this information alone, discuss all the possible ways that the spheres could be charged. Is it possible that after the spheres touch, they will cling together? Explain. Q1.3 The electric force between two charged particles becomes weaker with increasing distance. Suppose instead that the electric force were independent of distance. In this case, would a charged comb still cause a neutral insulator to become polarized as in Fig. 1.8? Why or why not? Would the neutral insulator still be attracted to the comb? Again, why or why not? Q1.4 Your clothing tends to cling together after going through the dryer. Why? Would you expect more or less clinging if all your clothing were made of the same material (say, cotton) than if you dried different kinds of clothing together? Again, why? (You may want to experiment with your next load of laundry.) Q1.5 An uncharged metal sphere hangs from a nylon thread. When a positively charged glass rod is brought close to the metal sphere,

the sphere is drawn toward the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain why the sphere is first attracted and then repelled. Q1.6 The free electrons in a metal are gravitationally attracted toward the earth. Why, then, don’t they all settle to the bottom of the conductor, like sediment settling to the bottom of a river? Q1.7 . Figure Q1.7 shows some of the Figure Q1.7 electric field lines due to three point charges arranged along the vertical axis. All three charges have the same magnitude. (a) What are the signs of the three charges? Explain your reasoning. (b) At what point(s) is the magnitude of the electric field the smallest? Explain your reasoning. Explain how the fields produced by each individual point charge combine to give a small net field at this point or points. Q1.8 Good electrical conductors, such as metals, are typically good conductors of heat; electrical insulators, such as wood, are typically poor conductors of heat. Explain why there should be a relationship between electrical conduction and heat conduction in these materials.

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30

CHAPTER 1 Electric Charge and Electric Field

Q1.9 . Suppose the charge shown in Fig. 1.28a is fixed in position. A small, positively charged particle is then placed at some point in the figure and released. Will the trajectory of the particle follow an electric field line? Why or why not? Suppose instead that the particle is placed at some point in Fig. 1.28b and released (the positive and negative charges shown in the figure are fixed in position). Will its trajectory follow an electric field line? Again, why or why not? Explain any differences between your answers for the two different situations. Q1.10 Two identical metal objects are mounted on insulating stands. Describe how you could place charges of opposite sign but exactly equal magnitude on the two objects. Q1.11 You can use plastic food wrap to cover a container by stretching the material across the top and pressing the overhanging material against the sides. What makes it stick? (Hint: The answer involves the electric force.) Does the food wrap stick to itself with equal tenacity? Why or why not? Does it work with metallic containers? Again, why or why not? Q1.12 If you walk across a nylon rug and then touch a large metal object such as a doorknob, you may get a spark and a shock. Why does this tend to happen more on dry days than on humid days? (Hint: See Fig. 1.30.) Why are you less likely to get a shock if you touch a small metal object, such as a paper clip? Q1.13 You have a negatively charged object. How can you use it to place a net negative charge on an insulated metal sphere? To place a net positive charge on the sphere? Q1.14 When two point charges of equal mass and charge are released on a frictionless table, each has an initial acceleration a0. If instead you keep one fixed and release the other one, what will be its initial acceleration: a0, 2a0, or a0 > 2? Explain. Q1.15 A point charge of mass m and charge Q and another point charge of mass m but charge 2Q are released on a frictionless table. If the charge Q has an initial acceleration a0, what will be the acceleration of 2Q: a0, 2a0, 4a0, a0 > 2, or a0 > 4? Explain. Q1.16 A proton is placed in a uniform electric field and then released. Then an electron is placed at this same point and released. Do these two particles experience the same force? The same acceleration? Do they move in the same direction when released? Q1.17 In Example 1.1 (Section 1.3) we saw that the electric force between two a particles is of the order of 1035 times as strong as the gravitational force. So why do we readily feel the gravity of the earth but no electrical force from it? Q1.18 What similarities do electrical forces have with gravitational forces? What are the most significant differences? Q1.19 Two irregular objects A Figure Q1.19 and B carry charges of opposite sign. Figure Q1.19 shows the electric field lines near each of these objects. (a) Which object S is positive, A or B? How do you E B know? (b) Where is the electric field stronger, close to A or close to B? How do you know? A Q1.20 Atomic nuclei are made of protons and neutrons. This shows that there must be another kind of interaction in addition to gravitational and electric forces. Explain. Q1.21 Sufficiently strong electric fields can cause atoms to become positively ionized—that is, to lose one or more electrons. Explain how this can happen. What determines how strong the field must be to make this happen? '

'

'

'

'

'

Q1.22 The electric fields at point P due to Figure Q1.22 the positive charges q1 and q2 are shown in S S E2 E1 Fig. Q1.22. Does the fact that they cross each other violate the statement in Section 1.6 that electric field lines never cross? Explain. P Q1.23 The air temperature and the velocity of the air have different values at different q2 places in the earth’s atmosphere. Is the air q1 velocity a vector field? Why or why not? Is the air temperature a vector field? Again, why or why not?

SINGLE CORRECT Q1.1 Two small charged metal spheres hung on insulating threads attract one another as shown. It is known that a positively charged rod will attract ball A. i. Ball A has a positive charge ii. Ball B has a negative charge A B iii. Ball A and B have opposite charges Which of the above can be correctly concluded about the charge on the balls? (a) I only (b) II only (c) III only (d) all of these Q1.2 An electroscope is given a positive charge, causing its foil leaves to separate. When an object is brought near the top plate of the electroscope, the foils separate even further. We could conclude (a) that the object is positively charged. (b) that the object is electrically netural. (c) that the object is negatively charged. (d) only that the object is charged. Q1.3 Given are four arrangements of three fixed charges. In each arrangement, a point labeled P is also identified — test charge, + q, is placed at point P. All of the charges are of the same magnitude, Q, but they can be either positive or negative as indicated. The charges and point P all lie on straight line. The distances between adjacent items, either between two charges or between a charge and point P, are all the same. • I. { { { P II. { { P• | III. { { | P • IV. { | { P • Correct order of choices in a decreasing order of magnitude of force on P is (a) II > I > III > IV (b) I > II > III> IV (c) II > I > IV > III (d) III > IV > I > II Q1.4 The figure below shows the forces that three charged particles exert on each other. Which of the four situations shown can be correct.

(I)

(II)

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One or More Correct

(III)

(IV)

(a) all of the above (b) none of the above (c) II, III (d) II, III & IV Q1.5 In a milikan-type experiment there are two oil droplets P and Q between the charged horizontal plates, as shown in the diagram. Droplet P is in rest while droplet Q is moving upwards. The polarity of the charges on P and Q is + + + + + + P Q (a) + + Q P (b) neutral − (c) − − − − − − − − (d) + − Q1.6 A small object of mass M and charge Q is connected to an insulating massless string in a vaccum on Earth. A uniform electric field exists throughout the region of the vaccum as indicated. The mass remains q (IV) in static equilibrium at an angle of u with the vertical as shown in the figure. When the string is cut, which of the Electric (I)(II)(III) field illustrated paths best indicates the trajectory of the mass? (a) I (b) II (c) III (d) IV Q1.7 Two charges q1&q2 are kept on x-axis q1 q2 x and electric field at different points an x-axis is plotted against x. Choose correct statement about nature and magnitude of q1&q2. (a) q1 + ve, q2 - ve ; |q1| > |q2| (b) q1 + ve, q2 - ve ; |q1| < |q2| (c) q1 - ve, q2 + ve ; |q1| > |q2| (d) q1 - ve, q2 + ve ; |q1| < |q2| Q1.8 An electric dipole is placed in an electric field generated by a point charge (a) The net force on the dipole must be zero. (b) The net force on the dipole may be zero. (c) The torque on the dipole due to the field must be zero. (d) The torque on the dipole due to the field may be zero. Z Q1.9 A distribution of charges is held fixed by rigid insulators as shown in figure. A +Q charge Q at ( - a, 0, 0) and at (a, 0, 0) and a O charge - 2Q at (0, 0, 0). Which of the folY −2Q lowing electric fields will cause a net torque +Q to be exerted on the system of charges. ^

S

X

31

placed along the x-axis at a large distance r apart oriented as shown below: Q1.10 The dipole on the left (a) will feel a force upwards and a torque trying to make it rotate clockwise. (b) will feel a force upwards and a torque trying to make it rotate counterclockwise. (c) will feel a force upwards and no torque about its centre. (d) will feel a force downwards and a torque trying to make it rotate clockwise. Q1.11 The dipole on the right (a) will feel a force downwards and a torque trying to make it rotate clockwise. (b) will feel a force downwards and a torque trying to make it rotate counterclockwise. (c) will feel a force upwards and no torque about its centre. (d) will feel no force and a torque trying to make it rotate counterclokwise. Q1.12 Questions given below consist of two statements each printed as Assertion (A) and Reason (R); while anwering these questions you are required to choose any one of the following four responses: (a) both (A) and (R) are true and (R) is the correct explanation of (A) (b) (A) is correct and (R) is incorrect (c) (A) is incorrect and (R) is correct (d) (A) and (R) both incorrect Assertion (A): Angular momentum of the two dipole system is not conserved. Reason (R):There is a net torque on the system.

ONE OR MORE CORRECT Q1.13 Three charged particles are in equilibrium under their electrostatic forces only (a) The particles must be collinear. (b) All the charges cannot have the same magnitude. (c) All the charges cannot have the same sign. (d) The equilibrium is unstable. Y Q1.14 Consider two identical charges placed distance 2d apart, along x-axis. The Q Q equilibrium of a positive test charge placed X 2d at the point O midway between them is (a) Neutral (b) Stable for displacements along the x-axis (c) Stable for displacements along the y-axis. (d) Unstable for displacements along the y-axis. Q1.15 An electric dipole is placed in an electric field generated by an infinitely long uniformly charged wire (a) The net electric force on the dipole must be zero (b) The net electric force on the dipole may be zero (c) The torque on the dipole due to the field must be zero (d) The torque on the dipole due to the field may be zero

(a) E = (constant) j S

(b) E S

(c) E

= (E0x + constant) î = (E0x + constant)kN

S

(d) E = E0 ƒ x ƒ î

Paragraph Question (3 questions) We have two electric dipoles. Each dipole − + − d consists of two equal and opposite point d + charges at the end of an insulating rod of length d. The dipoles are

EXERCISES Section 1.3 Coulomb’s Law

1.1 .. Excess electrons are placed on a small lead sphere with mass 20.7 g so that its net charge is - 3.20 * 10-9 C. (a) Find the number of excess electrons on the sphere. (b) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g > mol. 1.2 . Lightning occurs when there is a flow of electric charge (principally electrons) between the ground and a thundercloud. '

'

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32

CHAPTER 1 Electric Charge and Electric Field

The maximum rate of charge flow in a lightning bolt is about 20,000 C/s; this lasts for 100 ms or less. How much charge flows between the ground and the cloud in this time? How many electrons flow during this time? 1.3 . Particles in a Gold Ring. You have a pure (24-karat) gold ring with mass 19.7 g. Gold has an atomic mass of 197 g > mol and an atomic number of 79. (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it? 1.4 . Two small spheres spaced 20.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57 * 10-21 N ? 1.5 .. An average human weighs about 650 N. If two such generic humans each carried 1.0 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650-N weight? 1.6 . In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 2.50 mm away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration–time and velocity–time graphs of the released proton’s motion. 1.7 . A negative charge of - 0.550 mC exerts an upward 0.200-N force on an unknown charge 0.300 m directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the - 0.550-mC charge? 1.8 .. Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = - 3.00 nC and is at x = +4.00 cm. Charge q1 is at x = + 2.00 cm. What is q1 (magnitude and sign) if the net force on q3 is zero? 1.9 .. Two point charges are located on the y-axis as follows: charge q1 = - 1.50 nC at y = - 0.600 m, and charge q2 = +3.20 nC at the origin 1y = 02. What is the total force (magnitude and direction) exerted by these two charges on a third charge q3 = +5.00 nC located at y = - 0.400 m ? 1.10 .. Two point charges are placed on the x-axis as follows: Charge q1 = +4.00 nC is located at x = 0.200 m, and charge is at What are the magnitude and direction of the total force exerted by these two charges on a negative point charge that is placed at the origin? '

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'

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Section 1.4 Electric Field and Electric Forces

1.11 . CP A proton is placed in a uniform electric field of 2.75 × 103 N/C. Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton’s acceleration; (c) the proton’s speedafter 1.00 μs in the field, assuming it starts from rest. 1.12 . A particle has charge -3.00 n C. (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 m directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 N/C? 1.13 . CP proton is traveling horizontally to the right at 4.50 × 106 m/s. (a) Find the magnitude and direction of theweakest electric field that can bring the proton uniformly to rest over a distance of 3.20 cm. (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)? 1.14 . CP An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, travelling 4.50 m in the first 3.00 ms after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

1.15 .. (a) What must the charge (sign and magnitude) of a 1.3 g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N > C ? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? 1.16 .. A point charge is placed at Figure E1.16 each corner of a square with side 1q 1q a length a. The charges all have the same magnitude q. Two of the charges are positive and two are negative, as shown in Fig. E1.16. What is the a direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of q and a? 2q 2q 1.17 . Two point charges are separated by 25.0 cm (Fig. E1.17). Find the net electric field these charges produce at (a) point A and (b) point B. (c) What would be the magnitude and direction of the electric force this combination of charges would produce on a proton at A? '

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Figure E1.17 B

26.25 nC

A

10.0 cm

212.5 nC 10.0 cm

25.0 cm

1.18 .. Electric Field of the Earth. The earth has a net electric charge that causes a field at points near its surface equal to 150 N > C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m ? Is use of the earth’s electric field a feasible means of flight? Why or why not? 1.19 .. CP An electron is projected Figure E1.19 with an initial speed v0 = 1.60 * 2.00 cm 6 10 m > s into the uniform field v0 between the parallel plates in Fig. – S E1.20. Assume that the field between 1.00 cm E the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. (a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. (b) Suppose that in Fig. E1.19 the electron is replaced by a proton with the same initial speed v0. Would the proton hit one of the plates? If the proton would not hit one of the plates, what would be the magnitude and direction of its vertical displacement as it exits the region between the plates? (c) Compare the paths traveled by the electron and the proton and explain the differences. (d) Discuss whether it is reasonable to ignore the effects of gravity for each particle. 1.20 .. If two electrons are each 1.50 * Figure E1.20 10-10 m from a proton, as shown in e Fig. E1.20, find the magnitude and direction of the net electric force they will exert on the proton. 74° 1.21 .. CP A uniform electric field p e exists in the region between two oppo'

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Problems

sitely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 2.25 cm distant from the first, in a time interval of 1.50 * 10-6 s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Section 1.5 Electric-Field Calculations

1.22 . Two point charges Q and + q (where q Figure E1.22 is positive) produce the net electric field shown Q at point P in Fig. E1.22. The field points parallel to the line connecting the two charges. (a) S E What can you conclude about the sign and d magnitude of Q? Explain your reasoning. (b) If the lower charge were negative instead, would P it be possible for the field to have the direction d shown in the figure? Explain your reasoning. 1.23 .. Two positive point charges q are 1q placed on the x-axis, one at x = a and one at x = -a. (a) Find the magnitude and direction of the electric field at x = 0. (b) Derive an expression for the electric field at points on the x-axis. Use your result to graph the xcomponent of the electric field as a function of x, for values of x between - 4a and + 4a. Figure E1.24 1.24 . The two charges q1 and q2 B shown in Fig. E1.24 have equal magnitudes. What is the direction of the net electric field due to these two charges at points A (midway A between the charges), B, and C if (a) q1 q2 both charges are negative, (b) both charges are positive, (c) q1 is positive and q2 is negative. C 1.25 . A + 2.00-nC point charge is at the origin, and a second - 5.00-nC point charge is on the x-axis at x = 0.800 m. (a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: (i) x = 0.200 m; (ii) x = - 0.200 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a). 1.26 . Three negative point charges lie Figure E1.26 along a line as shown in Fig. E1.26. Find the magnitude and direction of the electric 25.00 mC field this combination of charges produces at point P, which lies 6.00 cm from the 8.00 cm 6.00 cm - 2.00-mC charge measured perpendicular P to the line connecting the three charges. 22.00 mC 1.27 .. A point charge q1 = - 4.00 nC is at the point x = 0.600 m, y = 0.800 m, 8.00 cm and a second point charge q2 = + 6.00 nC 25.00 mC is at the point x = 0.600 m, y = 0. CalcuN N late the net electric field vector (in i - j form) at the origin due to these two point charges. 1.28 .. A very long, straight wire has charge per unit length 1.50 * 10-10 C > m. At what distance from the wire is the electricfield magnitude equal to 2.50 N > C ? 1.29 . A ring-shaped conductor with radius a = 2.50 cm has a total positive charge Q = + 0.125 nC uniformly distributed around it, as shown in Fig. 1.23. The center of the ring is at the origin of coordinates O. (a) What is the electric field (magnitude and direction) at point P, which is on the x-axis at x = 40.0 cm ? '

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(b) A point charge q = - 2.50 mC is placed at the point P described in part (a). What are the magnitude and direction of the force exerted by the charge q on the ring?

Section 1.7 Electric Dipoles

1.30 . The ammonia molecule 1NH32 has a dipole moment of 5.0 * 10-30 C # m. Ammonia molecules in the gas phase are S placed in a uniform electric field E with magnitude 1.6 * 106 N > C. What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect S to E from parallel to perpendicular? 1.31 . Point charges q1 = -5 nC and q2 = +5 nC are separated by 3 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of 37° with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.5 * 10-9 N # m ? 1.32 . Torque on a Dipole. An electric dipole with dipole S S moment p is in a uniform electric field E. (a) Find the orientations of the dipole for which the torque on the dipole is zero. (b) Which of the orientations in part (a) is stable, and which is unstable? (Hint: Consider a small displacement away from the equilibrium position and see what happens.) (c) Show that for the stable orientation in part (b), the dipole’s own electric field tends to oppose the external field. '

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PROBLEMS 1.33 ... Four identical charges Q are placed at the corners of a square of side L. (a) In a free-body diagram, show all of the forces that act on one of the charges. (b) Find the magnitude and direction of the total force exerted on one charge by the other three charges. 1.34 .. A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive x-axis at x = 4.00 cm. If a third charge q3 = +6.00 nC is now placed at the point x = 4.00 cm, y = 3.00 cm, find the x- and y-components of the total force exerted on this charge by the other two. 1.35 .. CP Two positive point charges Q are held fixed on the x-axis at x = a and x = - a. A third positive point charge q, with mass m, is placed on the x-axis away from the origin at a coordinate x such that ƒ x ƒ V a. The charge q, which is free to move along the x-axis, is then released. (a) Find the frequency of oscillation of the charge q. (Hint: Review the definition of simple harmonic motion in Section 14.2. Use the binomial expansion 11 + z2n = 1 + nz + n 1n - 12z2 > 2 + Á , valid for the case ƒ z ƒ 6 1.) (b) Suppose instead that the charge q were placed on the y-axis at a coordinate y such that Figure P1.36 ƒ y ƒ V a, and then released. If this charge is free to move anywhere in the xy-plane, what will happen to it? Explain your answer. L L 1.36 .. CP Two identical spheres with u u mass m are hung from silk threads of length L, as shown in Fig. P1.36. Each sphere has the same charge, so q1 = q2 = q. The radius of each sphere is very small compared to the distance mass m mass m between the spheres, so they may be charge q1 charge q2 treated as point charges. Show that if the '

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CHAPTER 1 Electric Charge and Electric Field

angle u is small, the equilibrium separation d between the spheres is d = 1q2L > 2pP0mg21 > 3. (Hint: If u is small, then tan u > sin u.) 1.37 ... CP Two small spheres with mass m = 15.0 g are hung by silk threads of length L = 1.20 m from a common point (Fig. P1.37). When the spheres are given equal quantities of negative charge, so that q1 = q2 = q, each thread hangs at u = 37° from the vertical. (a) Draw a diagram showing the forces on each sphere. Treat the spheres as point charges. (b) Find the magnitude of q. by using trial values for u and adjusting the values of u until a self-consistent answer is obtained.) 1.38 .. CP At t = 0 a very small object with mass 0.400 mg and charge +9.00 mC is traveling at 125 m> s in the -x-direction. The charge is moving in a uniform electric field that is in the +y-direction and that has magnitude E = 895 N>C. The gravitational force on the particle can be neglected. How far is the particle from the origin at t = 7.00 ms? 1.39 .. Two particles having charges q1 = 0.500 nC and q2 = 8.00 nC are separated by a distance of 1.20 m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero? 1.40 ... Two point charges q1 and q2 Figure P1.40 are held in place 4.50 cm apart. Another q1 point charge Q = - 1.75 mC of mass 5.00 g is initially located 3.00 cm from S a each of these charges (Fig. P1.40) and released from rest. You observe that the initial acceleration of Q is 324 m > s2 Q 4.50 cm upward, parallel to the line connecting the two point charges. Find q1 and q2. 1.41 . Three identical point charges q are placed at each of three corners of a q2 square of side L. Find the magnitude and direction of the net force on a point charge - 3q placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the -3q charge by each of the other three charges. 1.42 ... Three point charges are placed on the y-axis: a charge q at y = a, a charge -2q at the origin, and a charge q at y = -a. Such an arrangement is called an electric quadrupole. (a) Find the magnitude and direction of the electric field at points on the positive x-axis. (b) Use the binomial expansion to find an approximate expression for the electric field valid for x W a. Contrast this behavior to that of the electric field of a point charge and that of the electric field of a dipole. 1.43 . Electric Force Within the Nucleus. Typical dimensions of atomic nuclei are of the order of 10-15 m (1 fm2. (a) If two protons in a nucleus are 2.0 fm apart, find the magnitude of the electric force each one exerts on the other. Express the answer in newtons. Would this force be large enough for a person to feel? (b) Since the protons repel each other so strongly, why don’t they shoot out of the nucleus? 1.44 .. CP A small sphere with mass 9.00 mg and charge - 4.30 mC is moving in a circular orbit around a stationary sphere that has charge + 7.50 mC. If the speed of the small sphere is 5.90 * 103 m>s, what is the radius of its orbit? Treat the spheres as point charges and ignore gravity. 1.45 ... CP Operation of an Inkjet Printer. In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of 1.4 * 10-8 g each, leave the nozzle and travel toward the paper '

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at 20 m > s, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 cm long where there is a uniform vertical electric field with magnitude 8.0 * 104 N > C. If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? 1.46 .. CP A proton is projected into a uniform electric field that points vertically upward and has magnitude E. The initial velocity of the proton has a magnitude v0 and is directed at an angle a below the horizontal. (a) Find the maximum distance hmax that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance d does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of hmax and d if E = 500 N > C, v0 = 4.00 * 105 m > s, and a = 30.0°. 1.47 .. CALC Positive charge Q Figure P1.47 is distributed uniformly along y the x-axis from x = 0 to x = a. A positive point charge q is located on the positive x-axis at q x = a + r, a distance r to the Q x right of the end of Q (Fig. P1.47). O a r (a) Calculate the x- and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis where x 7 a. (b) Calculate the force (magnitude and direction) that the charge distribution Q exerts on q. (c) Show that if r W a, the magnitude of the force in part (b) is approximately Qq > 4pP0r2. Explain why this result is obtained. 1.48 .. CALC Positive charge Q is Figure P1.48 distributed uniformly along the posy itive y-axis between y = 0 and a y = a. A negative point charge -q lies on the positive x-axis, a disQ tance x from the origin (Fig. P1.48). (a) Calculate the x- and y-compox nents of the electric field produced O 2q by the charge distribution Q at points on the positive x-axis. (b) Calculate the x- and y-components of the force that the charge distribution Q exerts on q. (c) Show that if x W a, Fx > -Qq > 4pP0x2 and Fy > + Qqa > 8pP0x3. Explain why this result is obtained. 1.49 . CALC Positive charge +Q is distributed uniformly along the + x-axis from x = 0 to x = a. Negative charge - Q is distributed uniformly along the -x-axis from x = 0 to x = -a. (a) A positive point charge q lies on the positive y-axis, a distance y from the origin. Find the force (magnitude and direction) that the positive and negative charge distributions together exert on q. Show that this force is proportional to y-3 for y W a. (b) Suppose instead that the positive point charge q lies on the positive x-axis, a distance x 7 a from the origin. Find the force (magnitude and direction) that the charge distribution exerts on q. Show that this force is proportional to x-3 for x W a. 1.50 .. CP A small sphere with mass m carries a positive charge q and is attached to one end of a silk fiber of length L. The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density s. Show that when the sphere is in equilibrium, the fiber makes an angle equal to tan-1 1qs > 2mgP02 with the vertical sheet. '

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Problems

1.51 .. CALC Negative charge -Q is distributed uniformly around a quarter-circle of radius a that lies in the first quadrant, with the center of curvature at the origin. Find the x- and y-components of the net electric field at the origin. 1.52 .. CALC A semicircle of Figure P1.52 radius a is in the first and second y quadrants, with the center of curvature at the origin. Positive 2Q charge +Q is distributed uni- 1Q formly around the left half of the a semicircle, and negative charge -Q is distributed uniformly x around the right half of the semicircle (Fig. P1.52). What are the magnitude and direction of the net electric field at the origin produced by this distribution of charge? 1.53 .. Two 1.20-m nonconducting wires meet at a right Figure P1.53 angle. One segment carries 1.20 m + + + + + + + + + + 2.50 mC of charge distributed – uniformly along its length, and – the other carries -2.50 mC dis– tributed uniformly along it, as – shown in Fig. P1.53. (a) Find 1.20 m – P – the magnitude and direction of – the electric field these wires – produce at point P, which is – 60.0 cm from each wire. (b) If an electron is released at P, what are the magnitude and direction of the net force that these wires exert on it? 1.54 . Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of - 9.50mC > m2, and sheet B, which is to the right of A, carries a uniform charge density of - 11.6 mC>m2,. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B.of the net force that these wires exert on it? 1.55 . Two very large horizontal sheets are 4.25 cm apart and carry equal but opposite uniform surface charge densities of magnitude s. You want to use these sheets to hold stationary in the region between them an oil droplet of mass 324 mg that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should s be? 1.56 .. An infinite sheet with positive charge per unit area s lies in the xy-plane. A second infinite sheet with negative charge per unit area - s lies in the yz-plane. Find the net electric field at all points that do not lie in either of these planes. Express your answer in terms of the unit vectors n≤ , n≥ , and kN . 1.57 .. CP A thin disk with a circular hole at its center, called Figure P1.57 x an annulus, has inner radius R1 and outer radius R2 (Fig. P1.57). R2 The disk has a uniform positive surface charge density s on its R1 z surface. (a) Determine the total electric charge on the annulus. (b) y s O The annulus lies in the yz-plane, with its center at the origin. For an arbitrary point on the x-axis (the '

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axis of theS annulus), find the magnitude and direction of the electric field E. Consider points both above and below the annulus in Fig. P1.56. (c) Show that at points on the x-axis that are sufficiently close to the origin, the magnitude of the electric field is approximately proportional to the distance between the center of the annulus and the point. How close is “sufficiently close”? (d) A point particle with mass m and negative charge -q is free to move along the x-axis (but cannot move off the axis). The particle is originally placed at rest at x = 0.01R1 and released. Find the frequency of oscillation of the particle. (Hint: Review Section 14.2. The annulus is held stationary.) 24 cm 1.58 The two ends of a rubber string of negligible mass and having unstretched length 24 h cm are fixed at the same height as shown. A small object is attached to the string in its midpoint, thus the depression (h) of the object in equilibrium is 5 cm. Then the small object is charged and vertical electric field (E1) is applied. The equlibrium depression of the object increases to 9 cm, then the electric field is changed to E2 and the depression of object in equilibrium increases to 16 cm. What is the ratio of electric field in the second case to that of in the first case (E2/E1) ? S 1.59 A tiny electric dipole (H2O modecule) of dipole moment p is S placed at a distance r form an infinitely long wire, with its p normal to the wire in the same plane. If the linear charge density of the wire is l, find the electrostatic force acting on the dipole is equal to 1.60 Two point charges are placed at point a and b. The field strength to the right of the charge Qb on the line that passes through the two charges varies according to a law that is represented graphically in the figure. Find the signs of the charges and ratio of magnitudes of charges Qa/Qb and the distance x2 of the point from b where the field is maximum, in terms of l and x1 .

E x1 l

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CHALLENGE PROBLEMS 1.61 ... CALC Two thin rods of length L lie along the x-axis, one

between x = a > 2 and x = a > 2 + L and the other between x = -a > 2 and x = -a > 2 - L. Each rod has positive charge Q distributed uniformly along its length. (a) Calculate the electric field produced by the second rod at points along the positive x-axis. (b) Show that the magnitude of the force that one rod exerts on the other is Q2 (a + L)2 ln c d F = 2 a 1a + 2L) 4pP0L '

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1.62 A thin non-conducting ring of radius R has a linear charge density l = l0 cos u, where l0 is the value of l at u = 0. Find net electric dipole moment for this charge distribution.

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CHAPTER 1 Electric Charge and Electric Field

Answers Chapter Opening Question

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Water molecules have a permanent electric dipole moment: One end of the molecule has a positive charge and the other end has a negative charge. These ends attract negative and positive ions, respectively, holding the ions apart in solution. Water is less effective as a solvent for materials whose molecules do not ionize (called nonionic substances), such as oils.

Test Your Understanding Questions 1.1 (a) the plastic rod weighs more, (b) the glass rod weighs less, (c) the fur weighs less, (d) the silk weighs more The plastic rod gets a negative charge by taking electrons from the fur, so the rod weighs a little more and the fur weighs a little less after the rubbing. By contrast, the glass rod gets a positive charge by giving electrons to the silk. Hence, after they are rubbed together, the glass rod weighs a little less and the silk weighs a little more. The weight change is very small: The number of electrons transferred is a small fraction of a mole, and a mole of electrons has a mass of only 16.02 * 1023 electrons219.11 * 10 -31 kg>electron2 = 5.48 * 10-7 kg = 0.548 milligram! 1.2 (a) (i), (b) (ii) Before the two spheres touch, the negatively charged sphere exerts a repulsive force on the electrons in the other sphere, causing zones of positive and negative induced charge (see Fig. 1.7b). The positive zone is closer to the negatively charged sphere than the negative zone, so there is a net force of attraction that pulls the spheres together, like the comb and insulator in Fig. 1.8b. Once the two metal spheres touch, some of the excess electrons on the negatively charged sphere will flow onto the other sphere (because metals are conductors). Then both spheres will have a net negative charge and will repel each other. 1.3 (iv) The force exerted by q1 on Q is still as in Example 1.4. The magnitude of the force exerted by q2 on Q is still equal to F1 on Q, but the direction of the force is now toward q2 at an angle a below the x-axis. Hence the x-components of the two forces cancel while the (negative) y-components add together, and the total electric force is in the negative y-direction. S 1.4 (a) (ii), (b) (i) The electric field E produced by a positive point charge points directly away from the charge (see Fig. 1.18a) and has a magnitude that depends on the distance rfrom the charge to the field point. Hence a second, negative point charge q 6 0 will S S feel a force F 5 qE that points directly toward the positive charge and has a magnitude that depends on the distance r between the two charges. If the negative charge moves directly toward the positive charge, the direction of the force remains the same but the force magnitude increases as the distance r decreases. If the negative charge moves in a circle around the positive charge, the force magnitude stays the same (because the distance r is constant) but the force direction changes. 1.5 (iv) Think of a pair of segments of length dy, one at coordinate y 7 0 and the other at coordinate -y 6 0. The upper segment has S a positive charge and produces an electric field dE at P that points S away from the segment, so this dE has a positive S x-component and a negative y-component, like the vector dE in Fig. 1.24. The lower segment has the same amount of negative S charge. It produces a dE that has the same magnitude but points toward the lower segment, so it has a negative x-component and a

negative y-component. By symmetry, the two x-components are equal but opposite, so they cancel. Thus the total electric field has only a negative y-component. S 1.6 yes If the field lines are straight, E must point inS the same S direction throughout the region. Hence the force F = qE on a particle of charge q is always in the same direction. A particle released S from rest accelerates in a straight line in the direction of F, and so its trajectory is a straight line along a field line. 1.7 (ii) Equations (1.17) and (1.18) tell us that the potential energy S S for a dipole in an electric field is U = - p E =S- pE cos f,Swhere S f is the angle between the directions of p and E. If S p and E point in opposite directions, so that f = 180°, we have cos f = - 1 and U = + pE. This is the maximum value that U can have. From our discussion of energy diagrams in Section 7.5, it follows that this is a situation of unstable equilibrium.

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Bridging Problem E = 2kQ>pa2 in the –y-direction

Discussion Questions 1.1 When the strips were peeled off the same roll, they had identical charges (sign) and hence repelled each other. But on ripping apart the sticky and shiny sides there was transfer of electrons from one to other (due to friction) there by giving the strips, charges of opposite signs. This causes attraction. 1.2 Both spheres being neutral initially, when brought nearer, attraction will be caused by equal and opposite charges being induced on their facing sides. When the spheres touch each other, charge transfer occur between them occurs till they attain a common potential and get identically charged (sign). So they repel. 1.3 Yes, No. 1.4 Due to static charge. Less clinging in case of same material. 1.5 Same as 1.2. 1.6 Due to mutual electrostatic repulsion. 1.7 (a) Upper most and lower most are positive and the middle one is negative. It is because field lines emanate from a positive and enter into a negative charge. (b) Near the location of negative charge because the fields tend to oppose each other vectorially. 1.8 It is because both thermal conductivity and electrical conductivity is determined largely by free electrons which are abundant in conductors. 1.9 Yes, No. Curved electric field lines indicates direction of force only instead of direction of velocity. 1.10 By the method of electrostatic induction and earthing away the extra charges (not held by induction). 1.11 Static charge makes cling film stick with insulating glass bowl. 1.12 Humid air conducts and we lose charge. No, due to smaller flow of charge. 1.13 Same as in 1.10 1.14 Acceleration remains the same as a0. It is because the free charge is still acted upon by the same electric force as earlier. 1.15 Force will be same for both (in magnitude) F ‹ aQ = = a2Q ( { their masses are same) m = a0

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Answers S

1.16 Both experience the same force |F| = eE but in exactly opposite direction. F 1 a of proton < a of electron ‹ C { a = a D m m 1.17 Atoms constituting earth are electrically neutral. 1.18 Both the forces are central forces and conservative. Both obey Newton's 3rd law and also inverse square law for point charges/masses. The differences are as follows : (i) Gravitational force is medium independent while electrical forces depend on medium. (ii) Gravitational force is always attractive while electric forces are repulsive also. (iii) Electric forces are stronger. 1.19 (a) B is positive as field lines emanate from it. (b) It is stronger near A as the density of field lines through any area is more near A. 1.20 They are attracted by nuclear forces which are due to exchange of p mesons between protons and neutrons. Within nucleus, the forces are stronger than electric/gravitational forces. 1.21 Strong electric fields provide kinetic energy to the electrons of the atom, greater than their binding energies (due to attraction by nucleus). Hence the electrons leave the atom and it gets ionised. 1.22 No, as electric field lines are drawn only for net electric field. 1.23 Yes, No.

Single correct Q1.1 (c) Q1.2 (a) Q1.3 (c) Q1.4 (c) Q1.5 (c) Q1.6 (b) Q1.7 (c) Q1.8 (d) Q1.9 (c) Q1.10 (b) Q1.11 (b) Q1.12 (d)

37

1.13 (a) 3.3 * 106 N/C, to the left (b) 14.2 ns (c) 1.8 * 103 N/C, to the right 1.14 (a) 5.7 N/C (b) acn. of electron is about 1011 times move than acn. due to gravity. 1.15 (a) - 20mC (b) 1.04 * 10 - 7 N/C 4 22 kq 1.16 , in - y direction a2 1.17 (a) 8.8 * 103 N/C rightwards (b) 6.54 * 103 N/C rightwards (c) 1.4 * 10 - 15 N rightwards 1.18 (a) - 4C (b) 1.44 * 107 N 1.19 (a) 364 N/C (b) 2.73 * 10 - 6 m downward (c) yes 1.20 1.64 * 10-8 N, directed toward a point midway between the two electrons. 1.21 (a) 200 N/C (b) 3 * 104 m/s 1.22 (a) Q is negative. Q and q must have same magnitude (b) No 1.23 (a) Electric field is zero at x = 0 (b) For a > ƒ x ƒ ; - 4q ax Ex = 4pe0 (x2 - a2)2 For x > a, Ex =

2q x2 + a2 4pe0 (x2 - a2)2

- 2q x2 + a2 4pe0 (x2 - a2)2 Ey is zero everywhere on x axis.

For - a > x, Ex =

1.24 (a) zero at A1 toward A at points B and C (b) zero at A1 away from A at points B and C (c) horizontal and to the right at points A, B, C 1.25 (a) (i) 575 N/C, in the + x direction (ii) 405 N/C, in the - x direction (b) (i) 9.2 * 10 - 17 N, in the - x direction (ii) 6.48 * 10 - 17 N, in the + x direction

One or More Correct Q1.13 (a), (b), (c), (d) Q1.14 (b), (d) Q1.15 (b), (d)

Exercise

1.26 1.04 * 107 N/C, toward - 2 µC charge 1.27 ( - 128.4 N/C) î + (28.8 N/C) Nj 1.28 1.08 m 1.29 (a) (7 N/C) î (b) (1.75 * 10 - 5 N) î 1.30 8 * 10 - 24 J

2500 1.31 (a) 1.5 * 10 - 11 ( - m, fromSq1 to q2) (b) a 3 bN/C 1.32 (a) The torque is zero when p S is aligned either in the same direction as E or in the opposite direction S S (b) When p is aligned in the same direction as E

1.1 (a) 2 * 1010 (b) 3.3 * 10 - 13 10

1.2 2C, 1.25 * 10 1.3 (a) 4.76 * 1024 protons, total positive charge = 7.6 * 105C (b) 4.76 * 1024 electrons 1.4 890 electrons 1.5 3.7 km 1.6 (a) 2.21 * 104 m/s2 1.7 (a) + 3.64 * 10 - 6C (b) 0.2 N downward 1.8 q1 = + 0.750 nC 1.9 2.6 * 10 - 6N, in - y direction 1.10 2.4 * 10 - 6N, in + x direction 1.11 (a) 4.4 * 10 - 16N (b) 2.63 * 1011m/s2 (c) 2.63 * 105m/s 1.12 (a) 432 N/C, downward (b) 1.5 m

1.33 (b)

Q2 8pe0L2

(1 + 2 22) , directed away from the opposite

corner. 1.34 Fx = 8.64 * 10 - 5 N; Fy = - 5.52 * 10 - 5 N qQ 1 Ä 2p pe0ma3 (b)The charge moves away from the origin on the y-axis and never returns. 1.37 (b) 5.1 * 10 - 6 C 1.38 1m 1.39 Electric Field is zero at a point between the two charges, 0.24 m from the 0.5 nC charge 1.35 (a)

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38

CHAPTER 1 Electric Charge and Electric Field

1.40 q2 = - q1 = - 6.17 * 10 - 8 C 6q2 1.41 (a) , away from the vacant corner 4pe0L2 2

1 a !2 + b , and is directed towards the centre of the 2 4pe0L2 square 2q 1 1.42 (a) E = c1 d , in the - x direction 2 4pe0x a2 3/2 a1 + 2 b x 1.43 (a) 58 N (b) Something must be holding the nucleus together by opposing this enormous repulsion. This is strong nuclear force. 1.44 0.926 m 1.45 1.05 * 10 - 13 C 3q

(b)

1.46 (a) hmax =

mV20 sin2a

2eE mV20 sin(2a) (b) d = eE 1 Q 1 1 1.47 (a) Ex = a - b ; Ey = 0 x 4pe0 a x - a S

(b) F =

1 qQ 1 1 bî a 4pe0 a r a + r

Q 1 1.48 (a) Ex = 4pe0x 2x2 + a2 - Q 1 1 Ey = a b 2 4pe0a x 2x + a2 1.49 (a) F = S

- 2Qq 1 1 1 a b Tî b a 2 2 1/2 a y 4pe0 (y + a ) Qq S 1 1 2 1 (b) F = c a b dî b a - + x - a x 4pe0 a (x + a) S

1.51 Ex = 1.52 Ex =

Q 2

2

2p e0a + Q 2

p e0a2

Q

; Ey =

2

2p e0a2

in the + x direction; Ey = 0

1.53 (a) Enet = 6.25 * 104 N/C, direction is 225° counterclockwise from an axis pointing to the right at point P. (b) 10 - 14 N, opposite to the direction of electric field 1.54(a) 1.19 * 105 N/C, to the right 1.55 (a) downward (b) 35.4 C/m2 s x z 1.56 C - a b Ni + a b kN D 2e0 ƒxƒ ƒzƒ 1.57 (a) Q = p(R2 2 - R1 2) s S

(b) E(x) =

(d) f =

s c 2e0

1

2(R1/x)

+ 1

2(R2/x)

2

+ 1

qs 1 1 1 b a 2p Å2e0m R1 R2

1.58 4.25 lP 1.59 2pe0r2

Q1.60 Qb is negative, Qa is positive x2 =

1

2

l

Qa l + x1 2 = a b , x1 Qb

(Qa)/(Qb)1/3 - 1

Challenging Problems 1.61 (a) Ex =

2KQ 1 1 a b L 2x + a 2L + 2x + a

1.62 pR2l0

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2

GAUSS’S LAW

LEARNING GOALS By studying this chapter, you will learn: • How you can determine the amount of charge within a closed surface by examining the electric field on the surface. • What is meant by electric flux, and how to calculate it.

?

This child acquires an electric charge by touching the charged metal sphere. The charged hairs on the child’s head repel and stand out. If the child stands inside a large, charged metal sphere, will her hair stand on end?

ften, there are both an easy way and a hard way to do a job; the easy way may involve nothing more than using the right tools. In physics, an important tool for simplifying problems is the symmetry properties of systems. Many physical systems have symmetry; for example, a cylindrical body doesn’t look any different after you’ve rotated it around its axis, and a charged metal sphere looks just the same after you’ve turned it about any axis through its center. Gauss’s law is part of the key to using symmetry considerations to simplify electric-field calculations. For example, the field of a straight-line or plane-sheet charge distribution, which we derived in Section 1.5 using some fairly strenuous integrations, can be obtained in a few lines with the help of Gauss’s law. But Gauss’s law is more than just a way to make certain calculations easier. Indeed, it is a fundamental statement about the relationship between electric charges and electric fields. Among other things, Gauss’s law can help us understand how electric charge distributes itself over conducting bodies. Here’s what Gauss’s law is all about. Given any general distribution of charge, we surround it with an imaginary surface that encloses the charge. Then we look at the electric field at various points on this imaginary surface. Gauss’s law is a relationship between the field at all the points on the surface and the total charge enclosed within the surface. This may sound like a rather indirect way of expressing things, but it turns out to be a tremendously useful relationship. Above and beyond its use as a calculational tool, Gauss’s law can help us gain deeper insights into electric fields. We will make use of these insights repeatedly in the next several chapters as we pursue our study of electromagnetism.

O

2.1

Charge and Electric Flux

In Chapter 1 we asked the question, “Given a charge distribution, what is the electric field produced by that distribution at a point P?” We saw that the answer could be found by representing the distribution as an assembly of point charges,

• How Gauss’s law relates the electric flux through a closed surface to the charge enclosed by the surface. • How to use Gauss’s law to calculate the electric field due to a symmetric charge distribution. • Where the charge is located on a charged conductor.

The discussion of Gauss’s law in this section is based on and inspired by the innovative ideas of Ruth W. Chabay and Bruce A. Sherwood in Electric and Magnetic Interactions (John Wiley & Sons, 1994).

39

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40

CHAPTER 2 Gauss’s Law S

ActivPhysics 11.7: Electric Flux

2.1 How can you measure the charge inside a box without opening it? (a) A box containing an unknown amount of charge

?

(b) Using a test charge outside the box to probe the amount of charge inside the box S

E S S

E

1

E

Test charge q0 S

E q S

E S

each of which produces an electric field E given by Eq. (1.7). The total field at P is then the vector sum of the fields due to all the point charges. But there is an alternative relationship between charge distributions and electric fields. To discover this relationship, let’s stand the question of Chapter 1 on its head and ask, “If the electric field pattern is known in a given region, what can we determine about the charge distribution in that region?” Here’s an example. Consider the box shown in Fig. 2.1a, which may or may not contain electric charge. We’ll imagine that the box is made of a material that has no effect on any electric fields; it’s of the same breed as the massless rope and the frictionless incline. Better still, let the box represent an imaginary surface that may or may not enclose some charge. We’ll refer to the box as a closed surface because it completely encloses a volume. How can you determine how much (if any) electric charge lies within the box? Knowing that a charge distribution produces an electric field and that an electric field exerts a force on a test charge, youSmove a test charge q0 around the vicinity of the box. By measuring the force F experienced by the test charge at different positions, you make a three-dimensional map of the electric field S S E 5 F>q0 outside the box. In the case shown in Fig. 2.1b, the map turns out to be the same as that of the electric field produced by a positive point charge (Fig. 1.28a). From the details of the map, you can find the exact value of the point charge inside the box. S To determine the contents of the box, we actually need to measure E only on the surface of the box. In Fig. 2.2a there is a single positive point charge inside the box, and in Fig. 2.2b there are two such charges. The field patterns on the surfaces of the boxes are different in detail, but in each case the electric field points out of the box. Figures 2.2c and 2.2d show cases with one and two negative point S charges, respectively, inside the box. Again, the details of E are different for the two cases, but the electric field points into each box.

E

Electric Flux and Enclosed Charge S

E

S

E S

E

In Section 1.4 we mentioned the analogy between electric-field vectors and the velocity vectors of a fluid in motion. This analogy can be helpful, even though an electric field does not actually “flow.” Using this analogy, in Figs. 2.2a and 2.2b, in which the electric field vectors point out of the surface, we say that there is an outward electric flux. (The wordS“flux” comes from a Latin word meaning “flow.”) In Figs. 2.2c and 2.2d the E vectors point into the surface, and the electric flux is inward. Figure 2.2 suggests a simple relationship: Positive charge inside the box goes with an outward electric flux through the box’s surface, and negative charge inside goes with an inward electric flux. What happens if there is zero charge

2.2 The electric field on the surface of boxes containing (a) a single positive point charge, (b) two positive point charges, (c) a single negative point charge, or (d) two negative point charges. (a) Positive charge inside box, outward flux

(b) Positive charges inside box, outward flux

S

(c) Negative charge inside box, inward flux

(d) Negative charges inside box, inward flux

S

E

E

S

E

S

E 1q

1q

1q

2q

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2q 2q

2.1 Charge and Electric Flux

41

S

inside the box? In Fig. 2.3a the box is empty and E 5 0 everywhere, so there is no electric flux into or out of the box. In Fig. 2.3b, one positive and one negative point charge of equal magnitude are enclosed within the box, so the net charge inside the box is zero. There is an electric field, but it “flows into” the box on half of its surface and “flows out of” the box on the other half. Hence there is no net electric flux into or out of the box. The box is again empty in Fig. 2.3c. However, there is charge present outside the box; the box has been placed with one end parallel to a uniformly charged infinite sheet, which produces a uniform electric field perpendicular to the sheet S (as we learned in Example 1.11 of SSection 1.5). On one end of the box, E points S into the box; on the opposite end, E points out of the box; and on the sides, E is parallel to the surface and so points neither into nor out of the box. As in Fig. 2.3b, the inward electric flux on one part of the box exactly compensates for the outward electric flux on the other part. So in all of the cases shown in Fig. 2.3, there is no net electric flux through the surface of the box, and no net charge is enclosed in the box. Figures 2.2 and 2.3 demonstrate a connection between the sign (positive, negative, or zero) of the net charge enclosed by a closed surface and the sense (outward, inward, or none) of the net electric flux through the surface. There is also a connection between the magnitudeS of the net charge inside the closed surface and the strength of the net “flow” of E over the surface. In both Figs. 2.4a and 2.4b there is a single point charge inside the box, but in Fig. 2.4b the magnitude of the S charge is twice as great, and so E is everywhere twice as great in magnitude as in Fig. 2.4a. If we keep in mind the fluid-flow analogy, this means that the net outward electric flux is also twice as great in Fig. 2.4b as in Fig. 2.4a. This suggests that the net electric flux through the surface of the box is directly proportional to the magnitude of the net charge enclosed by the box. This conclusion is independent of the size of the box. In Fig. 2.4c the point charge +q is enclosed by a box with twice the linear dimensions of the box in Fig. 2.4a. The magnitude of the electric field of a point charge decreases with distance S according to 1>r2, so the average magnitude of E on each face of the large box in Fig. 2.4c is just 14 of the average magnitude on the corresponding face in Fig. 2.4a. But each face of the large box has exactly four times the area of the corresponding face of the small box. Hence the outward electric flux is the same for the two boxes if we define electric flux as follows: For each face of the box, take the prodS uct of the average perpendicular component of E and the area of that face; then add up the results from all faces of the box. With this definition the net electric flux due to a single point charge inside the box is independent of the size of the box and depends only on the net charge inside the box.

2.3 SThree cases in which there is zero net charge inside a box and no net electric flux through the surface of the box. (a) An empty box with E 5 0. (b) A box containing one positive and one equal-magnitude negative point charge. (c) An empty box immersed in a uniform electric field. (a) No charge inside box, zero flux

(b) Zero net charge inside box, inward flux cancels outward flux.

(c) No charge inside box, inward flux cancels outward flux. 1s

S

E S

Uniformly charged sheet

E50 1q 2q

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S

E

42

CHAPTER 2 Gauss’s Law

2.4 (a) A box enclosing a positive point charge +q. (b) Doubling the charge causes S the magnitude of E to double, and it doubles the electric flux through the surface. (c) If the charge stays the same but the dimensions of the box are doubled,S the flux stays the same. The magnitude of E on the surface decreases bySa factor of 14, but the area through which E “flows” increases by a factor of 4. (a) A box containing a charge S

To summarize, for the special cases of a closed surface in the shape of a rectangular box and charge distributions made up of point charges or infinite charged sheets, we have found: 1. Whether there is a net outward or inward electric flux through a closed surface depends on the sign of the enclosed charge. 2. Charges outside the surface do not give a net electric flux through the surface. 3. The net electric flux is directly proportional to the net amount of charge enclosed within the surface but is otherwise independent of the size of the closed surface.

E

These observations are a qualitative statement of Gauss’s law. Do these observations hold true for other kinds of charge distributions and for closed surfaces of arbitrary shape? The answer to these questions will prove to be yes. But to explain why this is so, we need a precise mathematical statement of what we mean by electric flux. We develop this in the next section.

1q

(b) Doubling the enclosed charge doubles the flux. S

E

Test Your Understanding of Section 2.1 If all of the dimensions of the box in Fig. 2.2a are increased by a factor of 3, what effect will this change have on the electric flux through the box? (i) The flux will be 32 = 9 times greater; (ii) the flux will be 3 times greater; (iii) the flux will be unchanged; (iv) the flux will be 13 as great; (v) the flux will be A 13 B 2 = 19 as great; (vi) not enough information is given to decide.

y

12q

2.2

(c) Doubling the box dimensions does not change the flux. S

E

Calculating Electric Flux

In the preceding section we introduced the concept of electric flux. We used this to give a rough qualitative statement of Gauss’s law: The net electric flux through a closed surface is directly proportional to the net charge inside that surface. To be able to make full use of this law, we need to know how to calculate electric S flux. To do this, let’s again make use of the analogy between an electric field E S and the field of velocity vectors v in a flowing fluid. (Again, keep in mind that this is only an analogy; an electric field is not a flow.)

Flux: Fluid-Flow Analogy 1q

Figure 2.5 shows a fluid flowing steadily from left to right. Let’s examine the volume flow rate dV>dt (in, say, cubic meters per second) through the wire rectangle S with area A. When the area is perpendicular to the flow velocity v (Fig. 2.5a) and the flow velocity is the same at all points in the fluid, the volume flow rate dV>dt is the area A multiplied by the flow speed v: dV = vA dt When the rectangle is tilted at an angle f (Fig. 2.5b) so that its face is not perS pendicular to v, the area that counts is the silhouette area that we see when we S look in the direction of v. This area, which is outlined in red and labeled A' in S Fig. 2.5b, is the projection of the area A onto a surface perpendicular to v. Two sides of the projected rectangle have the same length as the original one, but the other two are foreshortened by a factor of cos f, so the projected area A' is equal to A cos f. Then the volume flow rate through A is dV = vA cos f dt If f = 90°, dV>dt = 0; the wire rectangle is edge-on to the flow, and no fluid passes through the rectangle.

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2.2 Calculating Electric Flux S

Also, v cos f is the component of the vector v perpendicular to the plane of the area A. Calling this component v' , we can rewrite the volume flow rate as '

dV = v' A dt '

We can express the volume flow rate more compactly by using the concept of S vector area A, a vector quantity with magnitude A and a direction perpendicular to S A the plane of the area we are describing. The vector area describes both the size S of an area and its orientation in space. In terms of A, we can write the volume flow rate of fluid through the rectangle in Fig. 2.5b as a scalar (dot) product:

43

2.5 The volume flow rate of fluid through the wire rectangle (a) is vA when the area of the rectangle is perpendicular to S v and (b) is vA cos f when the rectangle is tilted at an angle f. (a) A wire rectangle in a fluid

A

S

v

dV S # S = v A dt (b) The wire rectangle tilted by an angle f

Flux of a Uniform Electric Field Using the analogy between electric field and fluid flow, we now define electric flux in the same way as we have just defined the volume flow rate of a fluid; we S S simply replace the fluid velocity v by the electric field E. The symbol that we use for electric flux is £ E (the capital Greek letter phi; the subscript E is a reminder that this is electric flux). Consider first a flat area A perpendicular to a uniform S electric field E (Fig. 2.6a). We define the electric flux through this area to be the product of the field magnitude E and the area A:

A' 5 A cos f

A S

A

S

v

f

£ E = EA Roughly speaking, we can picture £ E in terms of the field lines passing through S area, increasA. Increasing the area means that more lines of E pass through the S ing the flux; a stronger field means more closely spaced lines of E and therefore more lines per unit area, so again the flux increases. S If the area A is flat but not perpendicular to the field E, then fewer field lines pass through it. In this case the area that counts is the silhouette area that we see S when looking in the direction of E. This is the area A' in Fig. 2.6b and is equal to A cos f (compare to Fig. 2.5b). We generalize our definition of electric flux for a uniform electric field to £ E = EA cos f

S

(electric flux for uniform E, flat surface)

(2.1)

S

Since E cos f is the component of E perpendicular to the area, we can rewrite Eq. (2.1) as £ E = E' A '

S

(electric flux for uniform E, flat surface)

(2.2)

S

In terms of the vector area A Sperpendicular to the area, we can write the elecS tric flux as the scalar product of E and A: £E = E # A S

S

S

(electric flux for uniform E, flat surface)

(2.3)

Flux Through a Basking Shark’s Mouth

Application

Unlike aggressive carnivorous sharks such as great whites, a basking shark feeds passively on plankton in the water that passes through the shark’s gills as it swims. To survive on these tiny organisms requires a huge flux of water through a basking shark’s immense mouth, which can be up to a meter across. The water flux—the product of the shark’s speed through the water and the area of its mouth—can be up to 0.5 m3/s (500 liters per second, or almost 5 * 105 gallons per hour). In a similar way, the flux of electric field through a surface depends on the magnitude of the field and the area of the surface (as well as the relative orientation of the field and surface).

Equations (2.1), (2.2), and (2.3) express the electric flux for a flat surface and a uniform electric field in different but equivalent ways. The SI unit Sfor electric flux S is 1 N # m2>C. Note that if the area is edge-on to the field, E and A are perpendicular and the flux is zero (Fig. 2.6c). S We can represent the direction of a vector area A by using a unit vector nN perpendicular to the area; nN stands for “normal.” Then S

A 5 AnN

(2.4) S

A surface has two sides, so there are two possible directions for nN and A. We must always specify which direction we choose. In Section 2.1 we related the charge inside a closed surface to the electric flux through the surface. With a closed surface we will always choose the direction of nN to be outward, and we

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44

CHAPTER 2 Gauss’s Law S

2.6 A flat surface S in a uniform electric field. The electric flux £ E through the surface equals the scalar product of the electric field E and the area vector A. (a) Surface is face-on to electric field: S S S • E and A are parallel (the angle between E S and A is f 5 0).S S • The flux FE 5 E • A 5 EA.

(b) Surface is tilted from a face-on orientation by an angle f: S S • The angle between E and A is f. S S • The flux FE 5 E • A 5 EA cos f.

(c) Surface is edge-on to electric field: S S • E and A are perpendicular (the angle S S between E and ASis Sf 5 90°). • The flux FE 5 E • A 5 EA cos 90° 5 0. S

A

f50

S

f 5 90°

A

S

E

S S

A

f

f

E

S

E A

A'

A

A

will speak of the flux out of a closed surface. Thus what we called “outward electric flux” in Section 2.1 corresponds to a positive value of £ E , and what we called “inward electric flux” corresponds to a negative value of £ E . '

'

Flux of a Nonuniform Electric Field S

What happens if the electric field E isn’t uniform but varies from point to point over the area A? Or what if A is part of a curved surface? Then we divide A into many small elements dA, each of which has a unit vector nN perpendicular to it S and a vector area dA 5 nN dA. We calculate the electric flux through each element and integrate the results to obtain the total flux: £E =

2

E cos f dA =

2

E' dA =

E # dA S

2

S

(general definition of electric flux)

(2.5)

We call this integral theS surface integral of the component E' over the area, or S the surface integral of E # dA. In specific problems, one form of the integral is sometimes more convenient than another. Example 2.3 at the end of this section illustrates the use of Eq. (2.5). In Eq. (2.5) the electric flux 1 E' dA is equal to the average value of the perpendicular component of the electric field, multiplied by the area of the surface. This is the same definition of electric flux that we were led to in Section 2.1, now expressed more mathematically. In the next section we will see the connection between the total electric flux through any closed surface, no matter what its shape, and the amount of charge enclosed within that surface.

Example 2.1

Electric flux through a disk

A disk of radius 0.10 m is oriented with its normal unit vector nN S at 30° to a uniform electric field E of magnitude 2.0 * 103 N > C (Fig. 2.7). (Since this isn’t a closed surface, it has no “inside” or “outside.” That’s why we have to specify the direction of nN in the figure.) (a) What is the electric flux through the disk? (b) What is the flux through the disk if it is turned so that nN is S perpendicular to (c) What is the flux through the disk if nN is E ? S parallel to E? '

'

2.7 The electric flux £ E through a disk depends on the angle S between its normal nN and the electric field E. r 5 0.10 m

n^ 30°

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S

E

2.2 Calculating Electric Flux S

S O L U T IO N IDENTIFY and SET UP: This problem is about a flat surface in a uniform electric field, so we can apply the ideas of this section. We calculate the electric flux using Eq. (2.1).

(c) The normal to the disk is parallel to E, so f = 0 and cos f = 1: £ E = EA cos f = 12.0 * 103 N > C210.0314 m22112 '

EXECUTE: (a) The area is A = p10.10 m22 = 0.0314 m2 and the S S angle between E and A 5 AnN is f = 30°, so from Eq. (2.1), £ E = EA cos f = 12.0 * 10 N > C210.0314 m 21cos 30°2 3

2

'

'

54 N # m2 > C '

'

= 63 N # m2 > C '

=

45

'

EVALUATE: As a check on our results, note that our answer to part (b) is smaller than that to part (a), which is in turn smaller than that to part (c). Is all this as it should be?

'

S

(b) The normal to the disk is now perpendicular to E, so f = 90°, cos f = 0, and £ E = 0.

Example 2.2

Electric flux through a cube

An imaginaryS cubical surface of side L is in a region of uniform electric field E. Find the electric flux through each face of the cube and the total flux through theScube when (a) it is oriented with two of its faces perpendicular to E (Fig. 2.8a) and (b) the cube is turned by an angle u about a vertical axis (Fig. 2.8b).

S

and nN 2 is 0°, and the angle between E and each of the other four unit vectors is 90°. Each face of the cube has area L2, so the fluxes through the faces are S

£ E1 = E

#

nN 1A = EL2 cos 180° = - EL2

£ E2 = E # nN 2A = EL2 cos 0° = + EL2 S

£ E3 = £ E4 = £ E5 = £ E6 = EL2 cos 90° = 0

S O L U T IO N S

IDENTIFY and SET UP: Since E is uniform and each of the six faces of the cube is flat, we find the flux £ Ei through each face using Eqs. (2.3) and (2.4). The total flux through the cube is the sum of the six individual fluxes. EXECUTE: (a) Figure 2.8a shows the unit vectors nN 1 through nN 6 for each face; each unit vector points outward from the cube’s closed S S surface. The angle between E and nN 1 is 180°, the angle between E

S

2.8 Electric flux of a uniform field E through a cubical box of side L in two orientations. (a)

(b) n^ 5

n^ 3

r

E

n^ 5

n^ 3

r

E

90° 2 u n^ 2 n^ 6

Example 2.3

u

n^ 4

n^ 1

£ E = £ E1 + £ E2 + £ E3 + £ E4 + £ E5 + £ E6 = -EL2 + EL2 + 0 + 0 + 0 + 0 = 0 S

(b) The field E is directed into faces 1 and 3, so the fluxes S through them are negative; E is directed out of faces 2 and 4, so the fluxes through them are positive. We find £ E1 = E # nN 1A = EL2 cos 1180° - u2 = - EL2 cos u S

£ E2 = E # nN 2 A = + EL2 cos u S £ E3 = E # nN 3 A = EL2 cos 190° + u2 = - EL2 sin u S

£ E4 = E # nN 4 A = EL2 cos 190° - u2 = +EL2 sin u S

n^ 2

n^ 4 n^ 1

S

The flux is negative on face 1, Swhere E is directed into the cube, and positive on face 2, where E is directed out of the cube. The total flux through the cube is

n^ 6

£ E5 = £ E6 = EL2 cos 90° = 0

The total flux £ E = £ E1 + £ E2 + £ E3 + £ E4 + £ E5 + £ E6 through the surface of the cube is again zero. EVALUATE: We came to the same conclusion in our discussion of Fig. 2.3c: There is zero net flux of a uniform electric field through a closed surface that contains no electric charge.

Electric flux through a sphere

A point charge q = +3.0 mC is surrounded by an imaginary sphere of radius r = 0.20 m centered on the charge (Fig. 2.9). Find the resulting electric flux through the sphere. S O L U T IO N IDENTIFY and SET UP: The surface is not flat and the electric field is not uniform, so to calculate the electric flux (our target variable)

we must use the general definition, Eq. (2.5). We use Eq. (2.5) to calculate the electric flux (our target variable). Because the sphere is centered on the point charge, at any point on the spherical surS face, E is directed out of the sphere perpendicular to the surface. The positive direction for both nN and E' is outward, so E' = E and S S the flux through a surface element dA is E # dA = E dA. This greatly simplifies the integral in Eq. (2.5). Continued

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46

CHAPTER 2 Gauss’s Law

2.9 Electric flux through a sphere centered on a point charge. S

dA

area of the spherical surface: A = 4pr2. Hence the total flux through the sphere is £ E = EA = =

q

r

S

E

EXECUTE: We must evaluate the integral of Eq. (2.5), £ E = 1 E dA. At any point on the sphere of radius r the electric field has the same magnitude E = q>4pP0r2 . Hence E can be taken outside the integral, which becomes £ E = E 1 dA = EA, where A is the

q 4pP0r

2

4pr2 =

3.0 * 10-6 C

q P0

8.85 * 10-12 C2>N # m2

= 3.4 * 105 N # m2>C

EVALUATE: The radius r of the sphere cancels out of the result for £ E. We would have obtained the same flux with a sphere of radius 2.0 m or 200 m. We came to essentially the same conclusion in our discussion of Fig. 2.4 in Section 2.1, where we considered rectangular closed surfaces of two different sizes enclosing a point charge. S There we found that the flux of E was independent of the size of the surface; the same result holds true for a spherical surface. Indeed, the flux through any surface enclosing a single point charge is independent of the shape or size of the surface, as we’ll soon see.

Test Your Understanding of Section 2.2 Rank the following surfaces in order from most positive to most negative electric flux. (i)Sa flat rectangular surS face with vector area A 5 16.0 m22≤N inSa uniform electric field E 5 14.0 N>C2≥N; (ii) a flat circular surface with vector area A 5 13.0 m22≥N in a uniform electric field S S E 5 14.0 N>C2≤N + 12.0 N>C2≥N; (iii) a flat square surface with vector area A 5 S 2 N 2 N 13.0 m 2≤ 1 17.0 m 2≥ in a uniform electric field E 5 14.0 N>C2≤N 2 12.0 N>C2≥N; S (iv) a flat oval surface with vector area A 5 13.0 m22≤N 2 17.0 m22≥N in a uniform S electric field E 5 14.0 N>C2≤N 2 12.0 N>C2≥N.

2.3 2.10 Carl Friedrich Gauss helped develop several branches of mathematics, including differential geometry, real analysis, and number theory. The “bell curve” of statistics is one of his inventions. Gauss also made state-of-the-art investigations of the earth’s magnetism and calculated the orbit of the first asteroid to be discovered.

y

Gauss’s Law

Gauss’s law is an alternative to Coulomb’s law. While completely equivalent to Coulomb’s law, Gauss’s law provides a different way to express the relationship between electric charge and electric field. It was formulated by Carl Friedrich Gauss (1777–1855), one of the greatest mathematicians of all time (Fig. 2.10).

Point Charge Inside a Spherical Surface Gauss’s law states that the total electric flux through any closed surface (a surface enclosing a definite volume) is proportional to the total (net) electric charge inside the surface. In Section 2.1 we observed this relationship qualitatively for certain special cases; now we’ll develop it more rigorously. We’ll start with the field of a single positive point charge q. The field lines radiate out equally in all directions. We place this charge at the center of an imaginary spherical surface with radius R. The magnitude E of the electric field at every point on the surface is given by E =

1 q 4pP0 R2

S

At each point on the surface, E is perpendicular to the surface, and its magnitude is the same at every point, just as in Example 2.3 (Section 2.2). The total electric flux is the product of the field magnitude E and the total area A = 4pR2 of the sphere: £ E = EA =

q 1 q 14pR22 = 2 P0 4pP0 R

(2.6)

The flux is independent of the radius R of the sphere. It depends only on the charge q enclosed by the sphere.

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47

2.3 Gauss’s Law

We can also interpret this result in terms of field lines. Figure 2.11 shows two spheres with radii R and 2R centered on the point charge q. Every field line that passes through the smaller sphere also passes through the larger sphere, so the total flux through each sphere is the same. What is true of the entire sphere is also true of any portion of its surface. In Fig. 2.11 an area dA is outlined on the sphere of radius R and then projected onto the sphere of radius 2R by drawing lines from the center through points on the boundary of dA. The area projected on the larger sphere is clearly 4 dA. But since the electric field due to a point charge is inversely proportional to r2, the field magnitude is 14 as great on the sphere of radius 2R as on the sphere of radius R. Hence the electric flux is the same for both areas and is independent of the radius of the sphere.

2.11 Projection of an element of area dA of a sphere of radius R onto a concentric sphere of radius 2R. The projection multiplies each linear dimension by 2, so the area element on the larger sphere is 4 dA. The same number of field lines and the same flux pass through both of these area elements. S

E

4 dA

Point Charge Inside a Nonspherical Surface

dA

This projection technique shows us how to extend this discussion to nonspherical surfaces. Instead of a second sphere, let us surround the sphere of radius R by a surface of irregular shape, as in Fig. 2.12a. Consider a small element of area dA on the irregular surface; we note that this area is larger than the corresponding element on a spherical surface at the same distance from q. If a normal to dA makes an angle f with a radial line from q, two sides of the area projected onto the spherical surface are foreshortened by a factor cos f (Fig. 2.12b). The other two sides are unchanged. Thus the electric flux through the spherical surface element is equal to the flux E dA cos f through the corresponding irregular surface element. We can divide the entire irregular surface into elements dA, compute the electric flux E dA cos f for each, and sum the results by integrating, as in Eq. (2.5). Each of the area elements projects onto a corresponding spherical surface element. Thus the total electric flux through the irregular surface, given by any of the forms of Eq. (2.5), must be the same as the total flux through a sphere, which Eq. (2.6) shows is equal to q>P0 . Thus, for the irregular surface,

R

q

2R

'

£E =

S

B

#

S

E dA =

q P0

(2.7)

Equation (2.7) holds for a surface of any shape or size, provided only that it is a closed surface enclosing the charge q. The circle on the integral sign reminds us that the integral is always taken over a closed surface. S The area elements dA and the corresponding unit vectors nN always point out of the volume enclosed by the surface. The electric flux is then positive in areas (a) The outward normal to the surface makes an angle f S with the direction of E. E' f

(b)

2.12 Calculating the electric flux through a nonspherical surface.

S S

E

E' f

E dA

dA

f

r

dA cos f

R

q

q

The projection of the area element dA onto the spherical surface is dA cos f.

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48

CHAPTER 2 Gauss’s Law

where the electric field points out of the surface and negative where it points S E inward. Also, E' is positive at points where points out of the surface and negaS tive at points where E points into the surface. S If the point charge in Fig. 2.12 is negative, the E field is directed radially inward; the angle f is then greater than 90°, its cosine is negative, and the integral in Eq. (2.7) is negative. But since q is also negative, Eq. (2.7) still holds. For a closed surface enclosing no charge, £E = 2.13 A point charge outside a closed surface that encloses no charge. If an electric field line from the external charge enters the surface at one point, it must leave at another. S

E

S

B

#

S

E dA = 0

This is the mathematical statement that when a region contains no charge, any field lines caused by charges outside the region that enter on one side must leave again on the other side. (In Section 2.1 we came to the same conclusion by considering the special case of a rectangular box in a uniform field.) Figure 2.13 illustrates this point. Electric field lines can begin or end inside a region of space only when there is charge in that region.

General Form of Gauss’s Law Now comes the final step in obtaining the general form of Gauss’s law. Suppose the surface encloses not just one point charge q but several charges q1 , qS2 , q3 , Á . S The total (resultant) electric field Eat any point is the vector sum of the E fields of the individual charges. Let Qencl be theS total charge enclosed by the surface: Qencl = q1 + q2 + q3 S+ Á . Also let E be the total field at the position of the surface area element dA, and let E' be its component perpendicular to the plane S of that element (that is, parallel to dA). Then we can write an equation like Eq. (2.7) for each charge and its corresponding field and add the results. When we do, we obtain the general statement of Gauss’s law: '

Field line entering surface

Same field line leaving surface

£E =

S

B

#

S

E dA =

Qencl P0

(Gauss’s law)

'

'

(2.8)

The total electric flux through a closed surface is equal to the total (net) electric charge inside the surface, divided by ` 0 . '

CAUTION Gaussian surfaces are imaginary Remember that the closed surface in Gauss’s law is imaginary; there need not be any material object at the position of the surface. We often refer to a closed surface used in Gauss’s law as a Gaussian surface. y

Using the definition of Qencl and the various ways to express electric flux given in Eq. (2.5), we can express Gauss’s law in the following equivalent forms:

£E =

B

E cos f dA =

B

E' dA =

S

B

#

S

E dA =

Qencl (various forms P0 of Gauss’s law)

(2.9)

As in Eq. (2.5), the various forms of the integral all express the same thing, the total electric flux through the Gaussian surface, in different terms. One form is sometimes more convenient than another. As an example, Fig. 2.14a shows a spherical Gaussian surface of radius r around a positive point charge + q. The electric field points out of the Gaussian surface, so S S at every point on the surface E is in the same direction as dA, f = 0, and E' is equal to the field magnitude E = q>4pP0 r2. Since E is the same at all points on the

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2.3 Gauss’s Law (a) Gaussian surface around positive charge: positive (outward) flux

(b) Gaussian surface around negative charge: negative (inward) flux

S

S

dA

1q

49

2.14 Spherical Gaussian surfaces around (a) a positive point charge and (b) a negative point charge.

dA

r

2q

S

r

S

E

E

surface, we can take it outside the integral in Eq. (2.9). Then the remaining integral is 1 dA = A = 4pr2, the area of the sphere. Hence Eq. (2.9) becomes £E =

B

E' dA =

q

a

'

B 4pP0 r

2

b dA =

'

q 2

4pP0 r B

q

dA =

2

4pP0 r

'

4pr2 =

'

q P0

The enclosed charge Qencl is just the charge +q, so this agrees with Gauss’s law. S If the Gaussian surface encloses a negative point charge as in SFig. 2.14b, then E points into the surface at each point in the direction opposite dA. Then f = 180° and E' is equal to the negative of the field magnitude: E' = - E = - ƒ -q ƒ >4pP0 r2 = -q>4pP0 r2. Equation (2.9) then becomes '

'

-q £E =

B

E' dA =

a

2

B 4pP0 r '

b dA =

-q

-q 2

4pP0 r B '

dA =

4pP0 r

2

4pr2 =

'

-q P0

This again agrees with Gauss’s law because the enclosed charge in Fig. 2.14b is Qencl = -q. In Eqs. (2.8) and (2.9), Qencl is always the algebraic sum of all the positive and S negative charges enclosed by the Gaussian surface, and E is the total field at each point on the surface. Also note that in general, this field is caused partly by charges inside the surface and partly by charges outside. But as Fig. 2.13 shows, the outside charges do not contribute to the total (net) flux through the surface. So Eqs. (2.8) and (2.9) are correct even when there are charges outside the surface that contribute to the electric field at the surface. When Qencl = 0, the total flux through the Gaussian surface must be zero, even though some areas may have positive flux and others may have negative flux (see Fig. 2.3b). Gauss’s law is the definitive answer to the question we posed at the beginning of Section 2.1: “If the electric field pattern is known in a given region, what can we determine about the charge distribution in that region?” It provides a relationship between the electric field on a closed surface and the charge distribution within that surface. But in some cases we can use Gauss’s law to answer the reverse question: “If the charge distribution is known, what can we determine about the electric field that the charge distribution produces?” Gauss’s law may seem like an unappealing way to address this question, since it may look as though evaluating the integral in Eq. (2.8) is a hopeless task. Sometimes it is, but other times it is surprisingly easy. Here’s an example in which no integration is involved at all; we’ll work out several more examples in the next section.

Conceptual Example 2.4

Electric flux and enclosed charge

Figure 2.15 shows the field produced by two point charges +q and -q (an electric dipole). Find the electric flux through each of the closed surfaces A, B, C, and D.

SOLUTION Gauss’s law, Eq. (2.8), says that the total electric flux through a closed surface is equal to the total enclosed charge divided by P0. In Continued

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50

CHAPTER 2 Gauss’s Law

Fig. 2.15, surface A (shown in red) encloses the positive charge, so Qencl = +q; surface B (in blue) encloses the negative charge, so Qencl = -q; surface C (in purple) encloses both charges, so Qencl = +q + 1- q2 = 0; and surface D (in yellow) encloses no charges, so Qencl = 0. Hence, without having to do any integration, we have £ EA = + q>P0, £ EB = - q>P0, and £ EC = £ ED = 0. These results depend only on the charges enclosed within each Gaussian surface, not on the precise shapes of the surfaces. We can draw similar conclusions by examining the electric field lines. All the field lines that cross surface A are directed out of the surface, so the flux through A must be positive. Similarly, the flux through B must be negative since all of the field lines that cross that surface point inward. For both surface C and surface D, there are as many field lines pointing into the surface as there are field lines pointing outward, so the flux through each of these surfaces is zero.

2.16 Five Gaussian surfaces and six point charges.

S1

S2

2.15 The net number of field lines leaving a closed surface is proportional to the total charge enclosed by that surface.

C 1q

2q

D A

B

S

E

Test Your Understanding of Section 2.3 Figure 2.16 shows six point charges that all lie in the same plane. Five Gaussian surfaces—S1 , S2 , S3 , S4 , and S5—each enclose part of this plane, and Fig. 2.16 shows the intersection of each surface with the plane. Rank these five surfaces in order of the electric flux through them, from most positive to most negative. '

'

'

'

15.0 mC S4 27.0 mC 18.0 mC

19.0 mC

S3

S5

11.0 mC 210.0 mC

y

2.4

2.17 Under electrostatic conditions (charges not in motion), any excess charge on a solid conductor resides entirely on the conductor’s surface. Gaussian surface A inside conductor (shown in cross section)

Conductor (shown in cross section)

Charge on surface of conductor

Applications of Gauss’s Law

Gauss’s law is valid for any distribution of charges and for any closed surface. Gauss’s law can be used in two ways. If we know the charge distribution, and if it has enough symmetry to let us evaluate the integral in Gauss’s law, we can find the field. Or if we know the field, we can use Gauss’s law to find the charge distribution, such as charges on conducting surfaces. In this section we present examples of both kinds of applications. As you study them, watch for the role played by the symmetry properties of each system. We will use Gauss’s law to calculate the electric fields caused by several simple charge distributions; the results are collected in a table in the chapter summary. In practical problems we often encounter situations in which we want to know the electric field caused by a charge distribution on a conductor. These calculations are aided by the following remarkable fact: When excess charge is placed on a solid conductor and is at rest, it resides entirely on the surface, not in the interior of the material. (By excess we mean charges other than the ions and free electrons that make up the neutral conductor.) Here’s the proof. We know from Section 1.4 that in an electrostatic situation (with all charges at rest) the electric S S field E at every point in the interior of a conducting material is zero. If E were not zero, the excess charges would move. Suppose we construct aS Gaussian surface inside the conductor, such as surface A in Fig. 2.17. Because E 5 0 everywhere on this surface, Gauss’s law requires that the net charge inside the surface is zero. Now imagine shrinking the surface like a collapsing balloon until it encloses a region so small that we may consider it as a point P; then the charge at that point must be zero. We can do this anywhere inside the conductor, so there can be no excess charge at any point within a solid conductor; any excess charge must reside on the conductor’s surface. (This result is for a solid conductor. In the next section we’ll discuss what can happen if the conductor has cavities in its interior.) We will make use of this fact frequently in the examples that follow.

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2.4 Applications of Gauss’s Law

Problem-Solving Strategy 2.1

Gauss’s Law

IDENTIFY the relevant concepts: Gauss’s law is most useful when the charge distribution has spherical, cylindrical, or planar symmeS try. In these cases the symmetry determines the direction of E. Then S Gauss’s law yields the magnitude of E if we are given the charge distribution, and vice versa. In either case, begin the analysis by asking the question: What is the symmetry? SET UP the problem using the following steps: 1. List the known and unknown quantities and identify the target variable. 2. Select the appropriate closed, imaginary Gaussian surface. For spherical symmetry, use a concentric spherical surface. For cylindrical symmetry, use a coaxial cylindrical surface with flat ends perpendicular to the axis of symmetry (like a soup can). For planar symmetry, use a cylindrical surface (like a tuna can) with its flat ends parallel to the plane. EXECUTE the solution as follows: 1. Determine the appropriate size and placement of your Gaussian surface. To evaluate the field magnitude at a particular point, the surface must include that point. It may help to place one end S of a can-shaped surface within a conductor, where E and therefore £ are zero, or to place its ends equidistant from a charged plane. 2. Evaluate the integral A E' dA in Eq. (2.9). In this equation E' is the perpendicular component of the total electric field at each point on the Gaussian surface. A well-chosen Gaussian surface should make integration trivial or unnecessary. If the surface comprises several separate surfaces, such as the sides and ends

Example 2.5

51

3.

4.

5.

6.

of a cylinder, the integral A E' dA over the entire closed surface is the sum of the integrals 1 E' dA over the separate surfaces. Consider points 3–6 as you work. S If E is perpendicular (normal) at every point to a surface with area A, if it points outward from the interior of the surface, and if it has the same magnitude at every point on the surface, then E' = ES= constant, and 1 E' dA over that surface is equal to EA. (If E is inward, then E' = - E and 1 E' dA = - EA.) This S should be the case for part or all of your Gaussian surface. If E is tangent to a surface at every point, then E' = 0 and the integral over that surface is zero. This may be the case for parts of a S cylindrical Gaussian surface. If E 5 0 at every point on a surface, the integral is zero. Even when there is no charge within a Gaussian surface, the field at any given point on the surface is not necessarily zero. In that case, however, the total electric flux through the surface is always zero. The flux integral A E' dA can be approximated as the difference between the numbers of electric lines of force leaving and entering the Gaussian surface. In this sense the flux gives the sign of the enclosed charge, but is only proportional to it; zero flux corresponds to zero enclosed charge. Once you have evaluated A E' dA, use Eq. (2.9) to solve for your target variable.

EVALUATE your answer: If your result is a function that describes how the magnitude of the electric field varies with position, ensure that it makes sense.

Field of a charged conducting sphere

We place a total positive chargeS q on a solid conducting sphere with radius R (Fig. 2.18). Find E at any point inside or outside the sphere. 2.18 Calculating the electric field of a conducting sphere with positive charge q. Outside the sphere, the field is the same as if all of the charge were concentrated at the center of the sphere. + + +

Gaussian surfaces at r 5 2R and r 5 3R

+

+ + +

R

+

+ + + E Outside the sphere, the magnitude of the electric field decreases with the square of the radial distance from the center of the sphere: q 1 E5 4pP0 r 2

1 q E 1R 2 5 4pP0 R2 Inside the sphere, the electric field is zero: E 5 0. E 1R 2 4 E 1R 2 9

/ /

O

R

2R

3R

r

SOLUTION IDENTIFY and SET UP: As we discussed earlier in this section, all of the charge must be on the surface of the sphere. The charge is free to move on the conductor, and there is no preferred position on the surface; the charge is therefore distributed uniformly over the surface, and the system is spherically symmetric. To exploit this symmetry, we take as our Gaussian surface a sphere of radius r centered on the conductor. We can calculate the field inside or outside the conductor by taking r 6 R or r 7 R, respectively. In S either case, the point at which we want to calculate E lies on the Gaussian surface. EXECUTE: The spherical symmetry means that the direction of the electric field must be radial; that’s Sbecause there is no preferred direction parallel to the surface, so E can have no component parallel to the surface. There is also no preferred orientation of the sphere, so the field magnitude E can depend only on the distance r from the center and must have the same value at all points on the Gaussian surface. For r 7 R the entire conductor is within the Gaussian surface, so the enclosed charge is q. The area of the Gaussian surface is S 4pr2, and E is uniform over the surface and perpendicular to it at each point. The flux integral A E' dA is then just E14pr22, and Eq. (2.8) gives Continued

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52

CHAPTER 2 Gauss’s Law E14pr22 =

q and P0 1 q E = 4pP0 r2

encloses no charge, so Qencl = 0. The electric field inside the conductor is therefore zero. (outside a charged conducting sphere)

This expression is the same as that for a point charge; outside the charged sphere, its field is the same as though the entire charge were concentrated at its center. Just outside the surface of the sphere, where r = R, E =

q 1 (at the surface of a charged conducting sphere) 4pP0 R2

CAUTION Flux can be positive or negative Remember that we have chosen the charge q to be positive. If the charge is negative, the electric field is radially inward instead of radially outward, and the electric flux through the Gaussian surface is negative. The electric-field magnitudes outside and at the surface of the sphere are given by the same expressions as above, except that q denotes the magnitude (absolute value) of the charge. y For r 6 R we again have E14pr22 = Qencl>P0. But now our Gaussian surface (which lies entirely within the conductor)

Example 2.6

S

EVALUATE: We already knew that E 5 0 inside a solid conductor (whether spherical or not) when the charges are at rest. Figure 2.18 shows E as a function of the distance r from the center of the sphere. Note that in the limit as R S 0, the sphere becomes a point charge; there is then only an “outside,” and the field is everywhere given by E = q>4pP0r2. Thus we have deduced Coulomb’s law from Gauss’s law. (In Section 2.3 we deduced Gauss’s law from Coulomb’s law; the two laws are equivalent.) We can also use this method for a conducting spherical shell (a spherical conductor with a concentric spherical hole inside) if there is no charge inside the hole. We use a spherical Gaussian surface with radius r less than the radius of the hole. If there were a field inside the hole, it would have to be radial and spherically symmetric as before, so E = Qencl>4pP0r2. But now there is no enclosed charge, so Qencl = 0 and E = 0 inside the hole. Can you use this same technique to find the electric field in the region between a charged sphere and a concentric hollow conducting sphere that surrounds it?

Field of a uniform line charge E # nN = E' = - E everywhere.) The area of the cylindrical surface is 2prl, so the flux through it—and hence the total flux £ E through the Gaussian surface—is EA = 2prlE. The total enclosed charge is Qencl = ll, and so from Gauss’s law, Eq. (2.8), S

Electric charge is distributed uniformly along an infinitely long, thin wire. The charge per unit length is l (assumed positive). Find the electric field using Gauss’s law. S O L UT I ON IDENTIFY andS SET UP: We found in Example 1.10 (Section 1.5) that the field E of a uniformly charged, infinite wire is radially outward if l is positive and radially inward if l is negative, and that the field magnitude E depends only on the radial distance from the wire. This suggests that we use a cylindrical Gaussian surface, of radius r and arbitrary length l, coaxial with the wire and with its ends perpendicular to the wire (Fig. 2.19). EXECUTE: The flux through the flat ends of our Gaussian surface is zeroSbecause the radial electric field is parallel to these ends, and so E # nN = 0. On the cylindrical part of our surface we have S E # nN = E' = E everywhere. (If l were negative, we would have 2.19 A coaxial cylindrical Gaussian surface is used to find the electric field outside an infinitely long, charged wire. E' 5 E S

dA E' 5 0

r l

Gaussian surface

£ E = 2prlE = E =

1 l 2pP0 r

ll and P0 (field of an infinite line of charge)

We found this same result in Example 1.10 with much more effort. S If l is negative, E is directed radially inward, and in the above expression for E we must interpret l as the absolute value of the charge per unit length. EVALUATE: We saw in Example 1.10 that the entire charge on the wire contributes to the field at any point, and yet we consider only that part of the charge Qencl = ll within the Gaussian surface when we apply Gauss’s law. There’s nothing inconsistent here; it takes the entire charge to give the field the properties that allow us to calculate £ E so easily, and Gauss’s law always applies to the enclosed charge only. If the wire is short, the symmetry of the infinite wire is lost, and E is not uniform over a coaxial, cylindrical Gaussian surface. Gauss’s law then cannot be used to find £ E; we must solve the problem the hard way, as in Example 1.10. We can use the Gaussian surface in Fig. 2.19 to show that the field outside a long, uniformly charged cylinder is the same as though all the charge were concentrated on a line along its axis (see Problem 2.42). We can also calculate the electric field in the space between a charged cylinder and a coaxial hollow conducting cylinder surrounding it (see Problem 2.39).

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53

2.4 Applications of Gauss’s Law

Example 2.7

Field of an infinite plane sheet of charge

Use Gauss’s law to find the electric field caused by a thin, flat, infinite sheet with a uniform positive surface charge density s. S O L U T IO N IDENTIFY and SET UP: In Example 1.11 (Section 1.5) we found S that the field E of a uniformly charged infinite sheet is normal to the sheet, and that its magnitude is independent of the distance from the sheet. To take advantage of these symmetry properties, we use a cylindrical Gaussian surface with ends of area A and with its axis perpendicular to the sheet of charge (Fig. 2.20). 2.20 A cylindrical Gaussian surface is used to find the field of an infinite plane sheet of charge.

E' 5 E

Example 2.8

+ + + + + + A

+ + + + + +

+ + + + + +

+ + + + + +

+ + + + + +

E

EXECUTE: The flux through the cylindrical part of our Gaussian S surface is zero because E # nN = 0 everywhere. The flux through S each flat end of the surface is + EA because E # nN = E' = E everywhere, so the total flux through both ends—and hence the total flux £ E through the Gaussian surface—is + 2EA. The total enclosed charge is Qencl = sA, and so from Gauss’s law, sA and P0 s (field of an infinite sheet of charge) E = 2P0

2EA =

In Example 1.11 we found this same result using a much more complex calculation.S If s is negative, E is directed toward the sheet, the flux through the Gaussian surface in Fig. 2.20 is negative, and s in the expression E = s>2P0 denotes the magnitude (absolute value) of the charge density. EVALUATE: Again we see that, given favorable symmetry, we can deduce electric fields using Gauss’s law much more easily than using Coulomb’s law.

Gaussian surface

Field between oppositely charged parallel conducting plates

Two large plane parallel conducting plates are given charges of equal magnitude and opposite sign; the surface charge densities are + s and -s . Find the electric field in the region between the plates. S O L U T IO N IDENTIFY and SET UP: Figure 2.21a shows the field. Because opposite charges attract, most of the charge accumulates at the opposing faces of the plates. A small amount of charge resides on the outer surfaces of the plates, and there is some spreading or “fringing” of the

field at the edges. But if the plates are very large in comparison to the distance between them, the amount of charge on the outer surfaces is negligibly small, and the fringing can be neglected except near the edges. In this case we can assume that the field is uniform in the interior region between the plates, as in Fig. 2.21b, and that the charges are distributed uniformly over the opposing surfaces. To exploit this symmetry, we can use the shaded Gaussian surfaces S1, S2, S3, and S4. These surfaces are cylinders with flat ends of area A; one end of each surface lies within a plate.

2.21 Electric field between oppositely charged parallel plates. (a) Realistic drawing

Between the two plates the electric field is nearly uniform, pointing from the positive plate toward the negative one.

(b) Idealized model

++ + ++ S + E ++ + ++ + ++ + ++ + ++

–– – –– – –– – –– – –– – –– – ––

1

S

S

E1

E2 a

In the idealized case we ignore “fringing” at the plate edges and treat the field between the plates as uniform. Cylindrical Gaussian surfaces (seen from the side)

+ + + + + + S1

S2

+ + + + + + +

2

b

– S E1 – S E2 – – – – S – E – S4

r

r

E2

E1 c

S3

– – – – –

Continued

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54

CHAPTER 2 Gauss’s Law

EXECUTE: The left-hand end of surface S1 is within the positive plate 1. Since the field is zero within the volume of any solid conductor under electrostatic conditions, there is no electric flux through this end. The electric field between the plates is perpendicular to the right-hand end, so on Sthat end, E' is equal to E and the flux is EA; this is positive, since E is directed out of the Gaussian surface. There is no flux throughSthe side walls of the cylinder, since these walls are parallel to E. So the total flux integral in Gauss’s law is EA. The net charge enclosed by the cylinder is sA, so Eq. (2.8) yields EA = sA>P0; we then have E =

s (field between oppositely charged conducting plates) P0

Example 2.9

The field is uniform and perpendicular to the plates, and its magnitude is independent of the distance from either plate. The Gaussian surface S4 yields the same result. Surfaces S2 and S3 yield E = 0 to the left of plate 1 and to the right of plate 2, respectively. We leave these calculations to you (see Exercise 2.29). '

EVALUATE: We obtained the same results in Example 1.11 by using the principle of superposition of electric fields. The Sfields due to the S two sheets of charge (one on each plate) are E1 and E2 ; from Example 2.7, both of these have magnitude s>2P0. The total electric field S S S at any pointSis the vector sum E 5 E1 1 E2. At points a and c in S Fig. 2.21b, E1 andS E2 point in opposite directions, and their sum is S zero. At point b, E1 and E2 are in the same direction; their sum has magnitude E = s>P0, just as we found above using Gauss’s law. '

Field of a uniformly charged sphere

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. Find the magnitude of the electric field at a point P a distance r from the center of the sphere.

2.22 The magnitude of the electric field of a uniformly charged insulating sphere. Compare this with the field for a conducting sphere (see Fig. 2.18). Spherical insulator + + r+ + + + r + ++ + R + + + ++ + Gaussian + ++ +

SOLUTION IDENTIFY and SET UP: As in Example 2.5, the system is spherically symmetric. Hence we can use the conclusions of that examS ple about the direction and magnitude of E. To make use of the spherical symmetry, we choose as our Gaussian surface a sphere with radius r, concentric with the charge distribution. S

EXECUTE: From symmetry, the direction of E is radial at every point on the Gaussian surface, so E' = E and the field magnitude E is the same at every point on the surface. Hence the total electric flux through the Gaussian surface is the product of E and the total area of the surface A = 4pr2—that is, £ E = 4pr2E. The amount of charge enclosed within the Gaussian surface depends on r. To find E inside the sphere, we choose r 6 R. The volume charge density r is the charge Q divided by the volume of the entire charged sphere of radius R : r =

Q

4pR3>3

The volume Vencl enclosed by the Gaussian surface is 43 pr3, so the total charge Qencl enclosed by that surface is Qencl = rVencl = a

Q

4pR >3 3

b A 43 pr3 B = Q

r3 R

3

Then Gauss’s law, Eq. (2.8), becomes Q r3 or P0 R3 1 Qr (field inside a uniformly charged sphere) E = 4pP0 R3 The field magnitude is proportional to the distance r of the field point from the center of the sphere (see the graph of E versus r in Fig. 2.22). To find E outside the sphere, we take r 7 R. This surface encloses the entire charged sphere, so Qencl = Q, and Gauss’s law gives 4pr2E =

Q or P0 1 Q (field outside a uniformly charged sphere) E = 4pP0 r2 4pr2E =

surface

E E(R) 5

1 Q 4pP0 R2 E5

1 Qr E5 4pP0 R3

O

R

1 Q 4pP0 r 2

r

The field outside any spherically symmetric charged body varies as 1>r2, as though the entire charge were concentrated at the center. This is graphed in Fig. 2.22. S If the charge is negative, E is radially inward and in the expressions for E we interpret Q as the absolute value of the charge. EVALUATE: Notice that if we set r = R in either expression for E, we get the same result E = Q>4pP0R2 for the magnitude of the field at the surface of the sphere. This is because the magnitude E is a continuous function of r. By contrast, for the charged conducting sphere of Example 2.5 the electric-field magnitude is discontinuous at r = R (it jumps from E = 0 just inside the sphere to 2 E = Q>4pP 0 R just outside the sphere). In general, the electric S field E is discontinuous in magnitude, direction, or both wherever there is a sheet of charge, such as at the surface of a charged conducting sphere (Example 2.5), at the surface of an infinite charged sheet (Example 2.7), or at the surface of a charged conducting plate (Example 2.8). The approach used here can be applied to any spherically symmetric distribution of charge, even if it is not radially uniform, as it was here. Such charge distributions occur within many atoms and atomic nuclei, so Gauss’s law is useful in atomic and nuclear physics.

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2.5 Charges on Conductors

Example 2.10

Charge on a hollow sphere

A thin-walled, hollow sphere of radius 0.250 m has an unknown charge distributed uniformly over its surface. At a distance of 0.300 m from the center of the sphere, the electric field points radially inward and has magnitude 1.80 * 102 N > C. How much charge is on the sphere? '

'

S O L U T IO N

field here is directed toward the sphere, so that q must be negative. Furthermore, the electric field is directed into the Gaussian surface, so that E' = -E and £ E = A E' dA = -E14pr22. By Gauss’s law, the flux is equal to the charge q on the sphere (all of which is enclosed by the Gaussian surface) divided by P0. Solving for q, we find q = - E14pP0r22 = -11.80 * 102 N > C214p2 '

IDENTIFY and SET UP: The charge distribution is spherically symmetric. As in Examples 2.5 and 2.9, it follows that the electric field is radial everywhere and its magnitude is a function only of the radial distance r from the center of the sphere. We use a spherical Gaussian surface that is concentric with the charge distribution and has radius r = 0.300 m. Our target variable is Qencl = q. EXECUTE: The charge distribution is the same as if the charge were on the surface of a 0.250-m-radius conducting sphere. Hence we can borrow the results of Example 2.5. We note that the electric

* 18.854 * 10

-12

'

C2>N # m2210.300 m22

= - 1.80 * 10-9 C = -1.80 nC

EVALUATE: To determine the charge, we had to know the electric field at all points on the Gaussian surface so that we could calculate the flux integral. This was possible here because the charge distribution is highly symmetric. If the charge distribution is irregular or lacks symmetry, Gauss’s law is not very useful for calculating the charge distribution from the field, or vice versa.

Test Your Understanding of Section 2.4 You place a known amount of charge Q on the irregularly shaped conductor shown in Fig. 2.17. If you know the size and shape of the conductor, can you use Gauss’s law to calculate the electric field at an arbitrary position outside the conductor?

2.5

55

Application Charge Distribution Inside a Nerve Cell

y

Charges on Conductors

We have learned that in an electrostatic situation (in which there is no net motion of charge) the electric field at every point within a conductor is zero and that any excess charge on a solid conductor is located entirely on its surface (Fig. 2.23a). But what if there is a cavity inside the conductor (Fig. 2.23b)? If there is no charge within the cavity, we can use a Gaussian surface such as A (which lies completely within the material of the conductor) to show that the net charge on S the surface of the cavity must be zero, because E 5 0 everywhere on the Gaussian surface. In fact, we can prove in this situation that there can’t be any charge anywhere on the cavity surface. We will postpone detailed proof of this statement until Chapter 3. Suppose we place a small body with a charge q inside a cavity within a conductor (Fig.S 2.23c). The conductor is uncharged and is insulated from the charge q. Again E 5 0 everywhere on surface A, so according to Gauss’s law the total charge inside this surface must be zero. Therefore there must be a charge - q distributed on the surface of the cavity, drawn there by the charge q inside the cavity. The total charge on the conductor must remain zero, so a charge +q must appear

The interior of a human nerve cell contains both positive potassium ions (K + ) and negatively charged protein molecules (Pr- ). Potassium ions can flow out of the cell through the cell membrane, but the much larger protein molecules cannot. The result is that the interior of the cell has a net negative charge. (The fluid outside the cell has a positive charge that balances this.) The fluid within the cell is a good conductor, so the Pr- molecules distribute themselves on the outer surface of the fluid—that is, on the inner surface of the cell membrane, which is an insulator. This is true no matter what the shape of the cell.

2.23 Finding the electric field within a charged conductor. (a) Solid conductor with charge qC qC

(b) The same conductor with an internal cavity qC

Arbitrary Gaussian surface A

S

E 5 0 within conductor

– –– – –– – – q – – – – – – – –

Cavity

S

The charge qC resides entirely on the surface of the conductor. The situation is electrostatic, so S E 5 0 within the conductor.

(c) An isolated charge q placed in the cavity qC 1 q

Because E 5 0 at all points within the conductor, the electric field at all points on the Gaussian surface must be zero.

S

For E to be zero at all points on the Gaussian surface, the surface of the cavity must have a total charge 2q.

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56

CHAPTER 2 Gauss’s Law

either on its outer surface or inside the material. But we showed that in an electrostatic situation there can’t be any excess charge within the material of a conductor. So we conclude that the charge + q must appear on the outer surface. By the same reasoning, if the conductor originally had a charge qC , then the total charge on the outer surface must be qC + q after the charge q is inserted into the cavity. '

Conceptual Example 2.11

A conductor with a cavity

A solid conductor with a cavity carries a total charge of +7 nC. Within the cavity, insulated from the conductor, is a point charge of -5 nC. How much charge is on each surface (inner and outer) of the conductor?

2.24 Our sketch for this problem. There is zero electric field inside the bulk conductor and hence zero flux through the Gaussian surface shown, so the charge on the cavity wall must be the opposite of the point charge.

SOLUTION Figure 2.24 shows the situation. If the charge in the cavity is q = -5 nC, the charge on the inner cavity surface must be -q = - 1-5 nC2 = + 5 nC. The conductor carries a total charge of + 7 nC, none of which is in the interior of the material. If +5 nC is on the inner surface of the cavity, then there must be 1+ 7 nC2 - 1 +5 nC2 = + 2 nC on the outer surface of the conductor.

Testing Gauss’s Law Experimentally We can now consider a historic experiment, shown in Fig. 2.25. We mount a conducting container on an insulating stand. The container is initially uncharged. Then we hang a charged metal ball from an insulating thread (Fig. 2.25a), lower it into the container, and put the lid on (Fig. 2.25b). Charges are induced on the walls of the container, as shown. But now we let the ball touch the inner wall (Fig. 2.25c). The surface of the ball becomes part of the cavity surface. The situation is now the same as Fig. 2.23b; if Gauss’s law is correct, the net charge on the cavity surface must be zero. Thus the ball must lose all its charge. Finally, we pull the ball out; we find that it has indeed lost all its charge. This experiment was performed in the 19th century by the English scientist Michael Faraday, using a metal icepail with a lid, and it is called Faraday’s icepail experiment. The result confirms the validity of Gauss’s law and therefore of 2.25 (a) A charged conducting ball suspended by an insulating thread outside a conducting container on an insulating stand. (b) The ball is lowered into the container, and the lid is put on. (c) The ball is touched to the inner surface of the container. (a) Insulating thread

(b) ++ + + + + ++

(c)

Charged conducting ball

Metal lid + –

Metal container

Insulating stand

+ –

+ –

+ –

+– –+ +– –+ +– – + ++ + + +– –+ + + ++ +– –+ +– –+ +– – – – – – + + + + +

Charged ball induces charges on the interior and exterior of the container.

+++ + + + + + + +

+

++

+

Metal lid +++ + + + + + + + + +

Once the ball touches the container, it is part of the interior surface; all the charge moves to the container’s exterior.

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2.5 Charges on Conductors 2.26 Cutaway view of the essential parts of a Van de Graaff electrostatic generator. The electron sink at the bottom draws electrons from the belt, giving it a positive charge; at the top the belt attracts electrons away from the conducting shell, giving the shell a positive charge. Conducting shell + +

+

+ + +

+ +

+ – – – +

+ + +

+ + –

+ +

+ – + –

Insulating belt

+ – + – Electron sink

Insulating support

+ – + – – – – –

–

Motor for belt

Coulomb’s law. Faraday’s result was significant because Coulomb’s experimental method, using a torsion balance and dividing of charges, was not very precise; it is very difficult to confirm the 1>r2 dependence of the electrostatic force by direct force measurements. By contrast, experiments like Faraday’s test the validity of Gauss’s law, and therefore of Coulomb’s law, with much greater precision. Modern versions of this experiment have shown that the exponent 2 in the 1>r2 of Coulomb’s law does not differ from precisely 2 by more than 10-16. So there is no reason to believe it is anything other than exactly 2. The same principle behind Faraday’s icepail experiment is used in a Van de Graaff electrostatic generator (Fig. 2.26). A charged belt continuously carries charge to the inside of a conducting shell. By Gauss’s law, there can never be any charge on the inner surface of this shell, so the charge is immediately carried away to the outside surface of the shell. As a result, the charge on the shell and the electric field around it can become very large very rapidly. The Van de Graaff generator is used as an accelerator of charged particles and for physics demonstrations. This principle also forms the basis for electrostatic shielding. Suppose we have a very sensitive electronic instrument that we want to protect from stray electric fields that might cause erroneous measurements. We surround the instrument with a conducting box, or we line the walls, floor, and ceiling of the room with a conducting material such as sheet copper. The external electric field redistributes the free electrons in the conductor, leaving a net positive charge on the outer surface in some regions and a net negative charge in others (Fig. 2.27). This charge distribution causes an additional electric field such that the total field at every point inside the box is zero, as Gauss’s law says it must be. The charge distribution on the box also alters the shapes of the field lines near the box, as the figure shows. Such a setup is often called a Faraday cage. The same physics tells

?

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57

58

CHAPTER 2 Gauss’s Law

2.27 (a) A conducting box (a Faraday cage) immersed in a uniform electric field. The field of the induced charges on the box combines with the uniform field to give zero total field inside the box. (b) This person is inside a Faraday cage, and so is protected from the powerful electric discharge. (a)

(b)

Field pushes electrons Net positive charge toward left side. remains on right side.

S

E

– – – – – – – –

S

E50

+ + + + + + + +

S

E

Field perpendicular to conductor surface

you that one of the safest places to be in a lightning storm is inside an automobile; if the car is struck by lightning, the charge tends to remain on the metal skin of the vehicle, and little or no electric field is produced inside the passenger compartment.

Field at the Surface of a Conductor S

2.28 The field just outside a charged conductor is perpendicular to the surface, and its perpendicular component E' is equal to s/P0 . '

Outer surface of charged conductor

A E' 5 E E' 5 0 +++++ + ++

Finally, we note that there is a direct relationship between the E field at a point just outside any conductor and the surface charge density s at that point. In general, s varies from point to point on the surface. We will show in Chapter 3 that at S any such point, the direction of E is always perpendicular to the surface. (You can see this effect in Fig. 2.27a.) To find a relationship between s at any point on the surface and the perpendicular component of the electric field at that point, we construct a Gaussian surface in the form of a small cylinder (Fig. 2.28). One end face, with area A, lies within the conductor and the other lies just outside. The electric field is zero at all points within the conductor. Outside the conductor the component S of E perpendicular to the side walls of the cylinder is zero, and over the end face the perpendicular component is equal to E' . (If s is positive, the electric field points out of the conductor and E' is positive; if s is negative, the field points inward and E' is negative.) Hence the total flux through the surface is E' A. The charge enclosed within the Gaussian surface is sA, so from Gauss’s law, '

Gaussian surface A

E50

E' A = '

sA P0

and

E' =

s P0

(field at the surface of a conductor)

(2.10)

We can check this with the results we have obtained for spherical, cylindrical, and plane surfaces. We showed in Example 2.8 that the field magnitude between two infinite flat oppositely charged conducting plates also equals s>P0 . In this case the field magnitude E is the same at all distances from the plates, but in all other cases E decreases with increasing distance from the surface. '

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2.5 Charges on Conductors

Conceptual Example 2.12

Field at the surface of a conducting sphere

Verify Eq. (2.10) for a conducting sphere with radius R and total charge q.

The surface charge density is uniform and equal to q divided by the surface area of the sphere: s =

S O L U T IO N In Example 2.5 (Section 2.4) we showed that the electric field just outside the surface is E =

Example 2.13

q 4pR2

Comparing these two expressions, we see that E = s>P0, which verifies Eq. (2.10).

q 1 4pP0 R2

Electric field of the earth

The earth (a conductor) has a net electric charge. The resulting electric field near the surface has an average value of about 150 N > C, directed toward the center of the earth. (a) What is the corresponding surface charge density? (b) What is the total surface charge of the earth? '

59

Q = 4p16.38 * 106 m221- 1.33 * 10-9 C>m22 = - 6.8 * 105 C = - 680 kC

'

S O L U T IO N IDENTIFY and SET UP: We are given the electric-field magnitude at the surface of the conducting earth. We can calculate the surface charge density s using Eq. (2.10). The total charge Q on the earth’s surface is then the product of s and the earth’s surface area. EXECUTE: (a) The Sdirection of the field means that s is negative (corresponding to E being directed into the surface, so E' is negative). From Eq. (2.10), s = P0E' = 18.85 * 10-12 C2>N # m221-150 N > C2 '

= - 1.33 * 10

-9

2

'

2

C>m = - 1.33 nC>m

EVALUATE: You can check our result in part (b) using the result of Example 2.5. Solving for Q, we find Q = 4pP0R2E' 1 = 16.38 * 106 m22 1 -150 N>C2 9.0 * 109 N # m2>C2 = - 6.8 * 105 C

One electron has a charge of - 1.60 * 10-19 C. Hence this much excess negative electric charge corresponds to there being 1 - 6.8 * 105 C2>1- 1.60 * 10-19 C2 = 4.2 * 1024 excess electrons on the earth, or about 7 moles of excess electrons. This is compensated by an equal deficiency of electrons in the earth’s upper atmosphere, so the combination of the earth and its atmosphere is electrically neutral.

(b) The earth’s surface area is 4pRE2, where RE = 6.38 * 10 m is the radius of the earth (see Appendix F). The total charge Q is the product 4pRE2s, or 6

Test Your Understanding of Section 2.5 A hollow conducting sphere has no net charge. There is a positive point charge q at the center of the spherical cavity within the sphere. You connect a conducting wire from the outside of the sphere to ground. Will you measure an electric field outside the sphere? y

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CHAPTER

2

SUMMARY

Electric flux: Electric flux is a measure of the “flow” of electric field through a surface. It is equal to the product of an area element and the perpendicular component of S E, integrated over a surface. (See Examples 2.1–2.3.)

£E =

2

= 2

E cos f dA E' dA =

S

#

S

E dA

2

(2.5)

S

A S

f

f

E

A'

Gauss’s law: Gauss’s law states that the total electric flux through a closed surface, which can Sbe written as the surface integral of the component of E normal to the surface, equals a constant times the total charge Qencl enclosed by the surface. Gauss’s law is logically equivalent to Coulomb’s law, but its use greatly simplifies problems with a high degree of symmetry. (See Examples 2.4–2.10.) When excess charge is placed on a conductor and is S at rest, it resides entirely on the surface, and E 5 0 everywhere in the material of the conductor. (See Examples 2.11–2.13.)

£E =

B

= B

E cos f dA E' dA =

S

B

#

A

Outward normal to surface r E' f E

S

E dA

Qencl = P0

dA

r

(2.8) , (2.9) R

q

Electric field of various symmetric charge distributions: The following table lists electric fields caused by several symmetric charge distributions. In the table, q, Q, l, and s refer to the magnitudes of the quantities.

Charge Distribution

Point in Electric Field

Electric Field Magnitude

Single point charge q

Distance r from q

E =

1 q 4pP0 r2

Charge q on surface of conducting sphere with radius R

Outside sphere, r 7 R

E =

1 q 4pP0 r2

Inside sphere, r 6 R

E = 0

Infinite wire, charge per unit length l

Distance r from wire

E =

1 l 2pP0 r

Infinite conducting cylinder with radius R, charge per unit length l

Outside cylinder, r 7 R

E =

1 l 2pP0 r

Inside cylinder, r 6 R

E = 0

Outside sphere, r 7 R

E =

1 Q 4pP0 r2

Inside sphere, r 6 R

E =

1 Qr 4pP0 R3

Infinite sheet of charge with uniform charge per unit area s

Any point

E =

s 2P0

Two oppositely charged conducting plates with surface charge densities + s and -s

Any point between plates

E =

s P0

Charged conductor

Just outside the conductor

E =

s P0

Solid insulating sphere with radius R, charge Q distributed uniformly throughout volume

60

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Discussion Questions

61

Electric Field Inside a Hydrogen Atom

BRIDGING PROBLEM

A hydrogen atom is made up of a proton of charge + Q = 1.60 * 10-19 C and an electron of charge - Q = - 1.60 * 10-19 C. The proton may be regarded as a point charge at r = 0, the center of the atom. The motion of the electron causes its charge to be “smeared out” into a spherical distribution around the proton, so that the electron is equivalent to a charge per unit volume of r1r2 = - 1Q>pa032e-2r > a0, where a0 = 5.29 * 10-11 m is called the Bohr radius. (a) Find the total amount of the hydrogen atom’s charge that is enclosed within a sphere with radius r centered on the proton. (b) Find the electric field (magnitude and direction) caused by the charge of the hydrogen atom as a function of r. (c) Make a graph as a function of r of the ratio of the electric-field magnitude E to the magnitude of the field due to the proton alone. '

'

SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. The charge distribution in this problem is spherically symmetric, just as in Example 2.9, so you can solve it using Gauss’s law. 2. The charge within a sphere of radius r includes the proton charge 1Q plus the portion of the electron charge distribution that lies within the sphere. The difference from Example 2.9 is that the electron charge distribution is not uniform, so the charge enclosed within a sphere of radius r is not simply the charge density multiplied by the volume 4pr3>3 of the sphere. Instead, you’ll have to do an integral.

Problems

3. Consider a thin spherical shell centered on the proton, with radius r¿ and infinitesimal thickness dr¿ . Since the shell is so thin, every point within the shell is at essentially the same radius from the proton. Hence the amount of electron charge within this shell is equal to the electron charge density r1r¿2 at this radius multiplied by the volume dV of the shell. What is dV in terms of r¿ ? 4. The total electron charge within a radius r equals the integral of r1r¿2dV from r¿ = 0 to r¿ = r. Set up this integral (but don’t solve it yet), and use it to write an expression for the total charge (including the proton) within a sphere of radius r. EXECUTE 5. Integrate your expression from step 4 to find the charge within radius r. Hint: Integrate by substitution: Change the integration variable from r¿ to x = 2r¿>a0. You can calculate the integral 2 -x 1 x e dx using integration by parts, or you can look it up in a table of integrals or on the World Wide Web. 6. Use Gauss’s law and your results from step 5 to find the electric field at a distance r from the proton. 7. Find the ratio referred to in part (c) and graph it versus r. (You’ll actually find it simplest to graph this function versus the quantity r>a0.) EVALUATE 8. How do your results for the enclosed charge and the electricfield magnitude behave in the limit r S 0? In the limit r S q ? Explain your results.

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q2.1 A rubber balloon has a single point charge in its interior. Does the electric flux through the balloon depend on whether or not it is fully inflated? Explain your reasoning. Q2.2 Suppose that in Fig. 2.15 both charges were positive. What would be the fluxes through each of the four surfaces in the example? Q2.3 In Fig. 2.15, suppose a third point charge were placed outside the purple Gaussian surface C. Would this affect the electric flux through any of the surfaces A, B, C, or D in the figure? Why or why not? Q2.4 A certain region of space bounded by an imaginary closed surface contains no charge. Is the electric field always zero everywhere on the surface? If not, under what circumstances is it zero on the surface? Q2.5 A spherical Gaussian surface encloses a point charge q. If the point charge is moved from the center of the sphere to a point away from the center, does the electric field at a point on the surface change? Does the total flux through the Gaussian surface change? Explain.

Q2.6 You find a sealed box on your doorstep. You suspect that the box contains several charged metal spheres packed in insulating material. How can you determine the total net charge inside the box without opening the box? Or isn’t this possible? Q2.7 A solid copper sphere has a net positive charge. The charge is distributed uniformly over the surface of the sphere, and the electric field inside the sphere is zero. Then a negative point charge outside the sphere is brought close to the surface of the sphere. Is all the net charge on the sphere still on its surface? If so, is this charge still distributed uniformly over the surface? If it is not uniform, how is it distributed? Is the electric field inside the sphere still zero? In each case justify your answers. Q2.8 If the electric field of a point charge were proportional to 1 > r3 instead of 1 > r2, would Gauss’s law still be valid? Explain your reasoning. (Hint: Consider a spherical Gaussian surface centered on a single point charge.) Q2.9 In a conductor, one or more electrons from each atom are free to roam throughout the volume of the conductor. Does this contradict the statement that any excess charge on a solid conductor must reside on its surface? Why or why not? '

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CHAPTER 2 Gauss’s Law

Q2.10 You charge up the van de Graaff generator shown in Fig. 2.26, and then bring an identical but uncharged hollow conducting sphere near it, without letting the two spheres touch. Sketch the distribution of charges on the second sphere. What is the net flux through the second sphere? What is the electric field inside the second sphere? Q2.11 A lightning rod is a rounded copper rod mounted on top of a building and welded to a heavy copper cable running down into the ground. Lightning rods are used to protect houses and barns from lightning; the lightning current runs through the copper rather than through the building. Why? Why should the end of the rod be rounded? Q2.12 A solid conductor has a cavity in its interior. Would the presence of a point charge inside the cavity affect the electric field outside the conductor? Why or why not? Would the presence of a point charge outside the conductor affect the electric field inside the cavity? Again, why or why not? Q2.13 Explain this statement: “In a static situation, the electric field at the surface of a conductor can have no component parallel to the surface because this would violate the condition that the charges on the surface are at rest.” Would this same statement be valid for the electric field at the surface of an insulator? Explain your answer and the reason for any differences between the cases of a conductor and an insulator. S Q2.14 In a certain region of space, the electric field E is uniform. (a) Use Gauss’s law to prove that this region of space must be electrically neutral; that is, the volume charge density r must be zero. (b) Is the converse true? That is, in a region of space where there is S no charge, must E be uniform? Explain. Q2.15 (a) In a certain region of space, the volume charge density r S has a uniform positive value. Can E be uniform in this region? Explain. (b) Suppose that in this region Sof uniform positive r there is a “bubble” within which r = 0. Can E be uniform within this bubble? Explain.

SINGLE CORRECT Q2.1 A point charge lies outside a charge free spherical region. The charge is then shifted to another point still outside the spherical region : (a) The electric flux through the spherical region changes (b) The electric field intensity at each point on the surface of the spherical region changes (c) The electric flux through the spherical region remains nonzero and unchanged (d) Neither the electric field strength at each point on the surface changes, nor does the electric flux through it. Q2.2 A rod containing charge +Q is S5 brought near an initially uncharged + ++++ isolated conducting rod as shown. S3 Regions with total surface charge +Q and –Q are induced in the conductor S2 S1 ++ Conductor –– –– – ++ as shown in the figure. The only S4 regions where the net charge in this configuration is non‑zero are indicated by the “+” and “–” signs. Let us denote the total flux of electric field outward through closed surface S1 as Φ1, through S2 as Φ2, etc. Which of the following is necessarily false. (a) Φ1 > 0 (b) Φ2 = Φ1 (c) Φ3 = Φ1 (d) Φ4 = 0 Q2.3 Which of the following is false for a closed Gaussian surface?

(a) If the electric field is zero everywhere on the surface, then there can be no net charge enclosed by the surface. (b) If the total electric flux through the surface is zero, then the total charge enclosed by the surface is zero. (c) If the electric field is zero everywhere on the surface, then the total electric flux through the surface is zero. (d) If the total electric flux through the surface is zero, then the electric field must be zero everywhere on the surface. Q2.4 Which of the following is sufficient condition for finding the electric flux ΦE through Sa closed surface? (a) If the magnitude of E is known everywhere on the surface (b) If the total charge inside the surface is specified (c) If the total charge outside the surface is specified (d) Only if the location of each point charge inside the surface is specified Q2.5 Statement-1: The flux crossing through a closed surface is independent of the location of enclosed charge. Statement-2:S Upon the displacement of charges within a closed surface, the E at any point on surface does not change. (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement‑1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true. Q2.6 Statement-1: Electric field of a dipole can’t be found using only Gauss law. (i.e. without using superposition principle) Statement-2: Gauss law is valid only for symmetrical charge distribution. (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement‑1. (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement‑1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true. B C D Q2.7 The figure below shows an edge-on A view of three extremely large, parallel insulating sheets positively charged with uniform and identical surface charge densities. In which of the regions shown is the electric field maximum? (a) B (b) C (c) A and D (d) The electric field is the same in all four regions of space. Q2.8 A charge Q0 is placed at centre of spherical cavity of a conductor and another charge Q, is placed outside the conductor. Q0 Q The correct statement is: (a) force on Q0 and Q are equal and opposite (b) force on Q0 is zero and force on Q is non‑zero (c) force on Q and Q0 is zero (d) force on Q0 is non‑zero and force on Q is zero

ONE OR MORE CORRECT Q2.9 Consider a gaussian spherical surface, covering a dipole of charge q and –q, then (a) qin = 0 (Net charge enclosed by the spherical surface)

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Exercises

(b) φnet = 0 (Net flux coming out the spherical surface) (c) E = 0 at all points on the spherical surface S S (d) 1 E # s = 0 (Surface integral of E over the spherical surface) Q2.10 If the flux of the electric field through a closed surface is zero, (a) the electric field must be zero everywhere on the surface (b) the net electric field may be zero everywhere on the surface (c) the net charge inside the surface must be zero (d) the charge in the vicinity of the surface must be zero Q2.11 In which of following cases, electric field is uniform? (a) Inside a uniformly charged spherical shell (b) In any cavity inside a uniformly charged sphere (c) In front of an infinite sheet of uniform surface charge density (d) At a distance x from a point charge q. Q2.12 A positive point charge Q is kept (as shown E in the figure) inside a neutral conducting shell C whose centre is at C. An external uniform electric field E is applied. Then Q (a) force on Q due to E is zero (b) net force on Q is zero (c) net force acting on Q and conducting shell considered as a system is zero. (d) net force acting on the shell due to E is zero. Q2.13 Uniformly charged rod is kept on y-axis y with centre at origin, as shown. Which of the following action will increase the electric field strength at the position of the dot? (a) make the rod longer without changing x x' r the charge Charged (b) make the rod shorter without changing rod the charge (c) make the rod shorter without changing the y' linear charge density (d) rotate the rod about yy' Q2.14 An electron is placed just in the middle +l +l between two long fixed line charges of charge y density l each. The wires are in the xy plane. (Do e– x not consider gravity). (a) The equilibrium of the electron will be unstable along x-direction (b) The equilibrium of the electron will be neutral along y-direction (c) The equilibrium of the electron will be stable along z-direction (d) The equilibrium of the electron will be stable along y-direction Q2.15 We have two electric dipoles. Each + – – + d d dipole consists of two equal and opposite point r charges at the ends of an insulating rod of length d. The dipoles sit along the x-axis a distance r apart, oriented as shown below. Their separation r >> d. The dipole ON THE LEFT : (a) will feel a force to the left. (b) will feel a force to the right. (c) will feel a torque trying to make it rotate counterclockwise. (d) will feel no torque

EXERCISES Section 2.2 Calculating Electric Flux

2.1 .. A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform

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electric field of magnitude 75.0 N > C that is directed at 30° from the plane of the sheet (Fig. E2.1). Find the magnitude of the electric flux through the sheet. '

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Figure E2.1 S

E 30° 0.400 m

0.600 m

2.2 .. A hemispherical surface with radius r in a region of uniS form electric field E has its axis aligned parallel to the direction of the field. Calculate the flux through the surface. 2.3 . The cube in Fig. E2.3 has sides of length L = 10.0 cm. The electric field is uniform, has magnitude E = 4.00 * 103 N > C, and is parallel to the xy-plane at an angle of 53.1° measured from the +x-axis toward the + y-axis. (a) What is the electric flux through each of the six cube faces S1 , S2 , S3 ,S4 , S5 , and S6? (b) What is the total electric flux through all faces of the cube? '

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Figure E2.3 z S2 (top) S1 (left side)

S6 (back)

S3 (right side)

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y L L x

S4 (bottom)

S5 (front)

Section 2.3 Gauss’s Law

2.4 . The three small spheres shown in Fig. E2.4 carry charges q1 = 4.00 nC, q2 = - 7.80 nC, and q3 = 2.40 nC. Find the net electric flux through each of the following closed surfaces shown in cross section in the figure: (a) S1 ; (b) S2 ; (c) S3 ; (d) S4 ; (e) S5 . (f) Do your answers to parts (a)–(e) depend on how the charge is distributed over each small sphere? Why or why not? Figure E2.4 S3 q2

q1 S1

S2 q3

S4

S5

Surface S1 S2 S3 S4 S5

What it encloses q1 q2 q1 and q2 q1 and q3 q1 and q2 and q3

2.5 .. A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 12.0 cm, giving it a charge of - 36 mC. Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) 6 cm outside the surface of the paint layer. 2.6 . A point charge of + 4.9 mC is located on the x-axis at x = 4.00 m, next to a spherical surface of radius 3.00 m centered at the origin. (a) Calculate the magnitude of the electric field at x = 3.00 m. (b) Calculate the magnitude of the electric field at x = -3.00 m. (c) According to Gauss’s law, the net flux through

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CHAPTER 2 Gauss’s Law

the sphere is zero because it contains no charge. Yet the field due to the external charge is much stronger on the near side of the sphere (i.e., at x = 3.00 m) than on the far side (at x = -3.00 m). How, then, can the flux into the sphere (on the near side) equal the flux out of it (on the far side)? Explain. A sketch will help.

PROBLEMS 2.13 .. A cube has sides of length L

= 0.300 m. It is placed with one corner at the origin as shown in Fig. E2.6. The electric field is S not uniform but is given by E 5 1-5.00 N > C # m2x n≤ 1 13.00 N > C # m2zkN . (a) Find the electric flux through each of the six cube faces S1 , S2 , S3 , S4 , S5 , and S6 . (b) Find the total electric charge inside the cube. S 2.14 .. The electric field E1 at one Figure P2.14 face of a parallelepiped is uniform S E2 over the entire face and is directed out of the face. AtS the opposite face, the electric field E2 is also uniform 6.00 cm over the entire face and is directed S 5.00 E1 30° into that face (Fig. P2.14). The two cm faces in question are inclined at 30.0° S S from the horizontal, while E1 and E2 S are both horizontal; E1 has a magni- S tude of 2.50 * 104 N > C, and E2 has a magnitude of 7.00 * 104 N > C. (a) Assuming that no other electric field lines cross the surfaces of the parallelepiped, determine the net charge contained within. (b) Is the electric field produced only by the charges within the parallelepiped, or is the field also due to charges outside the parallelepiped? How can you tell? 2.15 . The Coaxial Cable. A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length l. Calculate the electric field (a) at any point between the cylinders a distance r from the axis and (b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r = 0 to r = 2c. (d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder. 2.16 . A very long conducting tube (hollow cylinder) has inner radius a and outer radius b. It carries charge per unit length + a, where a is a positive constant with units of C > m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length + a. (a) Calculate the electric field in terms of a and the distance r from the axis of the tube for (i) r 6 a; (ii) a 6 r 6 b; (iii) r 7 b. Show your results in a graph of E as a function of r. (b) What is the charge per unit length on (i) the inner surface of the tube and (ii) the outer surface of the tube? 2.17 . A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume r. (a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density r. (b) What is the electric field at a point outside the volume in terms of the charge per unit length l in the cylinder? (c) Compare the answers to parts (a) and (b) for r = R. (d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R. 2.18 . A Sphere in a Sphere. A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b and outer radius c. The hollow sphere has no net charge. (a) Derive expressions for the electric-field magnitude in terms of the distance r from the center for the regions r 6 a, a 6 r 6 b, b 6 r 6 c, and r 7 c. (b) Graph the magnitude of the electric field as a function of r from r = 0 to r = 2c. (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. '

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Section 2.4 Applications of Gauss’s Law and Section 2.5 Charges on Conductors

2.7 .. How many excess electrons must be added to an isolated spherical conductor 32.0 cm in diameter to produce an electric field of 1350 N > C just outside the surface? 2.8 . (a) At a distance of 0.200 cm from the center of a charged conducting sphere with radius 0.100 cm, the electric field is 480 N > C. What is the electric field 0.600 cm from the center of the sphere? (b) At a distance of 0.200 cm from the axis of a very long charged conducting cylinder with radius 0.100 cm, the electric field is 480 N > C. What is the electric field 0.600 cm from the axis of the cylinder? (c) At a distance of 0.200 cm from a large uniform sheet of charge, the electric field is 480 N > C. What is the electric field 1.20 cm from the sheet? 2.9 ... CP CALC An insulating sphere of radius R = 0.160 mhas uniform charge density r = + 7.20 * 10-9 C>m3. A small object that can be treated as a point charge is released from rest just outside the surface of the sphere. The small object has positive charge q = 3.40 * 10-6 C. How much work does the electric field of the sphere do on the object as the object moves to a point very far from the sphere? 2.10 . An infinitely long cylindrical conductor has radius R and uniform surface charge density s. (a) In terms of s and R, what is the charge per unit length l for the cylinder? (b) In terms of s, what is the magnitude of the electric field produced by the charged cylinder at a distance r 7 R from its axis? (c) Express the result of part (b) in terms of l and show that the electric field outside the cylinder is the same as if all the charge were on the axis. 2.11 . Two very large, nonconducting Figure E2.11 plastic sheets, each 10.0 cm thick, carry s1 , uniform charge densities s2 s3 s4 s1 s2 , s3 , and s4 on their surfaces, as shown in Fig. E2.11. These surface charge densities have the values s1 = A B C - 6.00 mC > m2, s2 = + 5.00 mC > m2, s3 = + 2.00 mC > m2, and s4 = +4.00 mC > m2. Use Gauss’s law to find the magnitude and direction of the elec10 cm 12 cm 10 cm tric field at the following points, far from the edges of these sheets: (a) point A, 5.00 cm from the left face of the left-hand sheet; (b) point B, 1.25 cm from the inner surface of the right-hand sheet; (c) point C, in the middle of the righthand sheet. 2.12 . A negative charge - Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not? (c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge - Q? Is it reasonable to say that the grounded conductor has shielded the region from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not? '

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Problems

Sketch the field lines of the system within a spherical volume of radius 2 c. 2.19 . A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius 2R that also carries charge Q. The charge Q is distributed uniformly over the insulating shell. (a) Find the electric field (magnitude and direction) in each of the regions 0 6 r 6 R, R 6 r 6 2R, and r 7 2R. (b) Graph the electric-field magnitude as a function of r. 2.20 . A conducting spherical shell with inner radius a and outer radius b has a positive point Figure P2.20 charge Q located at its center. The total charge on the shell is - 3Q, and it is insulated from its a b surroundings (Fig. P2.20). (a) Derive expressions for the electric-field magnitude in terms of Q 23Q the distance r from the center for the regions r 6 a, a 6 r 6 b, and r 7 b. (b) What is the surface charge density on the inner surface of the conducting shell? (c) What is the surface charge density on the outer surface of the conducting shell? (d) Sketch the electric field lines and the location of all charges. (e) Graph the electric-field magnitude as a function of r. 2.21 . Concentric Spherical Shells. A small conducting spherical shell with inner Figure P2.21 radius a and outer radius b is concentric with a larger conducting spherical shell with inner a radius c and outer radius d (Fig. P2.21). The b inner shell has total charge +2q, and the outer c shell has charge + 4q. (a) Calculate the electric field (magnitude and direction) in terms d of q and the distance r from the common center of the two shells for (i) r 6 a; (ii) a 6 r 6 b; (iii) b 6 r 6 c; (iv)c 6 r 6 d; (v) r 7 d. Show your results in a graph of the radial S component of E as a function of r. (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell? 2.22 . A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with inner radius R and outer radius 2R. The insulating shell has a uniform charge density r. (a) Find the value of r so that the net charge of the entire system is zero. (b) If r has the value found in part (a), find the electric field (magnitude and direction) in each of the regions 0 6 r 6 R, R 6 r 6 2R, and rS 7 2R. Show your results in a graph of the radial component of E as a function of r. (c) As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in part (b) agree with this rule. 2.23 . Negative charge - Q is distributed uniformly over the surface of a thin spherical insulating shell with radius R. Calculate the force (magnitude and direction) that the shell exerts on a positive point charge q located (a) a distance r 7 R from the center of the shell (outside the shell) and (b) a distance r 6 R from the center of the shell (inside the shell). 2.24 ... CALC An insulating hollow sphere has inner radius a and outer radius b. Within the insulating material the volume charge density is given by r 1r2 = ar , where a is a positive constant. (a) In terms of a and a, what is the magnitude of the electric field at a distance r from the center of the shell, where a 6 r 6 b? (b) A point charge q is placed at the center of the hollow space, at r = 0. In terms of a and a, what value must q have (sign and magnitude) in order for the electric field to be constant in the region a 6 r 6 b, and what then is the value of the constant field in this region? 2.25 . A Uniformly Charged Slab. A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x = d and x = -d. The y-

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and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density r. (a) Explain why the electric field due to the slab is zero at the center of the slab 1x = 02. (b) Using Gauss’s law, find the electric field due to the slab (magnitude and direction) at all points in space. 2.26 . CALC A Nonuniformly Charged Slab. Repeat Problem 2.56, but now let the charge density of the slab be given by r1x2 = r0 1x > d22, where r0 is a positive constant. 2.27 . CALC A nonuniform, but spherically symmetric, distribution of charge has a charge density r1r2 given as follows: '

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r1r2 = r011 - 4r > 3R2 '

r1r2 = 0

for r … R

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for r Ú R

where r0 is a positive constant. (a) Find the total charge contained in the charge distribution. (b) Obtain an expression for the electric field in the region r Ú R. (c) Obtain an expression for the electric field in the region r … R. (d) Graph the electric-field magnitude E as a function of r. (e) Find the value of r at which the electric field is maximum, and find the value of that maximum field. 2.28 . CP CALC Gauss’s Law for Gravitation. The gravitational force between two point masses separated by a distance r is proportional to 1 > r2, just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss’s law for gravitation. (a) Let S g be the acceleration due to gravity caused by a point mass m at S the origin, so that g 5 - 1Gm > r22rn. Consider a spherical Gaussian surface with radius r centered on this point mass, and show that the S flux of g through this surface is given by '

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g # dA = -4pGm S

S

B (b) By following the same logical steps used in Section 2.3 to S obtain Gauss’s law for the electric field, show that the flux of g through any closed surface is given by g # dA = - 4pGMencl

S

S

B where Mencl is the total mass enclosed within the closed surface. 2.29 . (a) An insulating sphere with radius Figure P2.29 a has a uniform charge density r. The sphere S S is not centered at the origin but at r 5 b . Show that the electric field inside the sphere R S S S is given by E 5 r1r - b 2 > 3P0. (b) An a b insulating sphere of radius R has a spherical Charge density r hole of radius a located within its volume and centered a distance b from the center of the sphere, where a 6 b 6 R (a cross section of the sphere is shown in Fig. P2.29). The solid part of the sphere has a uniform volume charge density r. Find the magnitude and direction of the S S electric field E inside the hole, and show that E is uniform over the entire hole. [Hint: Use the principle of superposition and the result of part (a).] 2.30 . A very long, solid insulating Figure P2.30 cylinder with radius R has a cylindrical hole with radius a bored along its entire length. The axis of the hole is a distance b from the axis of the cylinder, where a 6 b 6 R (Fig. P2.30). The solid material of R the cylinder has a uniform volume b a Charge charge density r. Find the magnitude density r S and direction of the electric field E '

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CHAPTER 2 Gauss’s Law S

inside the hole, and show that E is uniform over the entire hole. (Hint: See Problem 2.61.) 2.31 . Positive charge Q is distributed uniformly over each of two spherical volumes with radius R. One sphere of charge is centered Figure P2.31 at the origin and the other at y x = 2R (Fig. P2.31). Find the magnitude and direction of the R R R net electric field due to these two x O distributions of charge at the following points on the x-axis: Q Q (a) x = 0; (b) x = R > 2; (c) x = R; (d) x = 3R. 2.32 .. CALC A nonuniform, but spherically symmetric, distribution of charge has a charge density r1r2 given as follows: '

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r1r2 = r011 - r > R2 '

for r … R

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r1r2 = 0

for r Ú R

where r0 = 3Q > pR is a positive constant. (a) Show that the total charge contained in the charge distribution is Q. (b) Show that the electric field in the region r Ú R is identical to that produced by a point charge Q at r = 0. (c) Obtain an expression for the electric field in the region r … R. (d) Graph the electric-field magnitude E as a function of r. (e) Find the value of r at which the electric field is maximum, and find the value of that maximum field. Q2.33 A sphere of radius R carries charge such that its volume charge density is proportional to the square of the distance from the centre. What is the ratio of the magnitude of the electric field at a distance 2R from the centre to the magnitude of the electric field at a distance of R/2 from the centre? Q2.34 A fixed circle of radius a and centre O is drawn and a charge +q is placed at a distance 3a/4 from O on a line through O perpendicular to the plane of circle. If a second charge q' is similarly placed at a distance 5a/12 on the opposite side of circle, the net electric flux through circle becomes zero. Find the ratio of q' to q. Q2.35 A charge is spherically symmetrically distributed around origin such that (charge / volume) at distance r from origin is - 2r Q r = e a0 . Find 2 pa0r 3

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(a) the total charge in space. (b) electric field at distance r from the origin.

CHALLENGE PROBLEMS 2.36 ... CP CALC A region in

space contains a total positive charge Q that is distributed spherically such that the volume charge density r1r2 is given by for r … R > 2

r1r2 = a

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r1r2 = 2a11 - r > R2 '

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for R > 2 … r … R

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r1r2 = 0

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for r Ú R

Here a is a positive constant having units of C > m3. (a) Determine a in terms of Q and R.S (b) Using Gauss’s law, derive an expression for the magnitude of E as a function of r. Do this separately for all three regions. Express your answers in terms of the total charge Q. Be sure to check that your results agree on the boundaries of the regions. (c) What fraction of the total charge is contained within the region r … R > 2? (d) If an electron with charge q¿ = - e is oscillating back and forth about r = 0 (the center of the distribution) with an amplitude less than R > 2, show that the motion is simple harmonic. (If, and only if, the net force on the electron is proportional to its displacement from equilibrium, then the motion is simple harmonic.) (e) What is the period of the motion in part (d)? (f) If the amplitude of the motion described in part (e) is greater than R > 2, is the motion still simple harmonic? Why or why not? 2.37 ... CP CALC A region in space contains a total positive charge Q that is distributed spherically such that the volume charge density r1r2 is given by '

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'

'

r1r2 = a31 - 1r > R2 4 '

'

'

for R > 2 … r … R

2

'

'

r1r2 = 0

'

for r Ú R

Here a is a positive constant having units of C > m3. (a) Determine a in terms of Q and R. (b) Using Gauss’s law, derive an expression for the magnitude of the electric field as a function of r. Do this separately for all three regions. Express your answers in terms of the total charge Q. (c) What fraction of the total charge is contained within the S region R > 2 … r … R? (d) What is the magnitude of E at r = R > 2? (e) If an electron with charge q¿ = -e is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why? '

'

'

'

'

'

Answers Chapter Opening Question

?

No. The electric field inside a cavity within a conductor is zero, so there is no electric effect on the child. (See Section 2.5.)

Test Your Understanding Questions 2.1 (iii) Each part of the surface of the box will be three times farther from the charge + q, so the electric field will be 11322 = 19 as strong. But the area of the box will increase by a factor of 32 = 9. Hence the electric flux will be multiplied by a factor of 1192192 = 1. In other words, the flux will be unchanged. 2.2 (iv), (ii), (i), (iii) In each case the electric field is uniform, so the S S flux is £ E = E A. We use the relationships for the scalar products of

#

#

#

#

unit vectors: ≤N ≤N = ≥N ≥N = 1, ≤N ≥N = 0. In case (i) we have £ E = 14.0 N>C216.0 m22≤N ≥N = 0 (the electric field and vector area are perpendicular, so there is zero flux). In case (ii) we have £ E 314.0 N>C2≤N 1 12.0 N>C2≥N4 13.0 m22≥N = 12.0 N>C2 13.0 m22 = 6.0 N # m2>C. Similarly, in case (iii) we have £ E = 314.0 N>C2≤N 2 12.0 N>C2≥N4 313.0 m22≤N 1 17.0 m22≥N4 = 14.0 N>C213.0 m22 - 12.0 N>C217.0 m22 = -2 N # m2>C, and in case (iv) we have £ E = 314.0 N>C2≤N 2 12.0 N>C2≥N4 313.0 m22≤N 2 17.0 m 22 ≥N4 = 14.0 N>C2 13.0 m 22 + 12.0 N>C2 17.0 m 22 = 26 N # m2>C.

#

#

#

#

#

'

'

'

'

'

'

'

'

'

'

#

'

'

2.3 S2 , S5 , S4 , S1 and S3 (tie) Gauss’s law tells us that the flux through a closed surface is proportional to the amount of charge enclosed within that surface. So an ordering of these surfaces by '

'

'

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Answers

their fluxes is the same as an ordering by the amount of enclosed charge. Surface S1 encloses no charge, surface S2 encloses 9.0 mC + 5.0 mC + 1 -7.0 mC2 = 7.0 mC, surface S3 encloses 9.0 mC + 1.0 mC + 1 -10.0 mC2 = 0, surface S4 encloses 8.0 mC + 1- 7.0 mC2 = 1.0 mC, and surface S5 encloses 8.0 mC + 1 - 7.0 mC2 + 1- 10.0 mC2 + 11.0 mC2 + 19.0 mC2 + 15.0 mC2 = 6.0 mC. 2.4 no You might be tempted to draw a Gaussian surface that is an enlarged version of the conductor, with the same shape and placed so that it completely encloses the conductor. While you know the flux through this Gaussian surface (by Gauss’s law, it’s £ E = Q>P0), the direction of the electric field need not be perpendicular to the surface and the magnitude of the field need not be the same at all points on the surface. It’s not possible to do the flux integral A E' dA, and we can’t calculate the electric field. Gauss’s law is useful for calculating the electric field only when the charge distribution is highly symmetric. 2.4 no You might be tempted to draw a Gaussian surface that is an enlarged version of the conductor, with the same shape and placed so that it completely encloses the conductor. While you know the flux through this Gaussian surface (by Gauss’s law, it’s £ E = Q>P0), the direction of the electric field need not be perpendicular to the surface and the magnitude of the field need not be the same at all points on the surface. It’s not possible to do the flux integral A E' dA, and we can’t calculate the electric field. Gauss’s law is useful for calculating the electric field only when the charge distribution is highly symmetric. 2.5 no Before you connect the wire to the sphere, the presence of the point charge will induce a charge - q on the inner surface of the hollow sphere and a charge q on the outer surface (the net charge on the sphere is zero). There will be an electric field outside the sphere due to the charge on the outer surface. Once you touch the conducting wire to the sphere, however, electrons will flow from ground to the outer surface of the sphere to neutralize the charge there (see Fig. 21.7c). As a result the sphere will have no charge on its outer surface and no electric field outside.

Bridging Problem (a) Q1r2 = Qe (b) E =

-2r>a0

kQe-2r>a0 2

r

2.2 fA =

67

q q 2q , fB = , fC = and fD = 0. P0 P0 P0

2.3 No, flux through a closed surface depends only on enclosed charges. S 2.4 No. Because E on surface will be due to both enclosed and external charges. S E = 0 if (i) qext = 0 or (ii) There is symmetrical distibution of qext outside the surface S such that Enet =0. 2.5 Till the charge is enclosed within the surface, f will not qin change. (Qf = is independent of location of qin) P0 2.6 By calculating the electric field intensity on each face of the box and using. qin S S ftotal = a Ek˙Ak = , where Ak = surface area of kth face. P0 2.7 Yes, all the charge will be still distributed on the outer surface, but non uniformly. Change density will be greater near the negative charge and less on the portion away from the charge. 4pa 2.8 Q f = E(4fr2) = is variable r q qin But by Gauss law, f = = constant for a given ‘q’ = P0 P0 ∴ The law will become invalid. r' q E=

a (a isa constant) r3

Gaussian surface

2.9 No the statement is not contradicted. Actually the entire conductor volume can be imagined as a combination of infinite number of conducting shells on whose facing surfaces, equal and opposite charges get induced. So by this neutrality, all charge appears on outer most surface. 2.10 Net flux through the sphere is zero. Electric field inside the second sphere is zero.

321r>a02 + 21r>a02 + 14 2

+ + + + +

321r>a02 + 21r>a02 + 14 2

+

+ +

+ + + + +

– – – –

+ + + +

–

+

(c) E/Eproton 1.00 0.80 0.60 0.40 0.20 0.00 0.00

r/a0 1.00

2.00

3.00

4.00

Discussion Problem 2.1 No. Because

f =

enclosing the charge.

qin is independent of shape of surface P0

2.11 Lightning current runs through the copper rod because it provides a low resistance path for the lightning circuit between air and ground. The end of the rod is rounded because blunt metal rods are better lightning strike receptors than sharper rods. 2.12 Due to induction, all of the charge inside cavity will appear on outer surface of conductor and will hence affect the external field. But external charges will not affect the inner field, due to electrostatic shielding (Net charge inside conductor will be zero always). S 2.13 Since the surface of a conductor is equipotential, henceSE should not have any component along it. Hence no work is done by E to move the charges on the surface, due to which they are at rest. Insulating surfaces may not necessarily be equipotential. S 2.14 (a) Since E is uniform in space, hence no. of field lines entering any imaginary elementary volume in the region will

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68

CHAPTER 2 Gauss’s Law

exactly be equal to the field lines emanating out. Hence net flux dφ = 0 rdV dqin ∴ from Gauss law, df = = = 0 1r = 0 e0 e0 (b) The converse is not true always. If q = 0 ⇒ f = 0 which may still be S possible if E is non uniform but is perpendicular to the area vector to the Gaussian surfaces, everywhere. 2.15 (a) No, as it will violate Gauss’s law. (b) Yes it is possible. For example considering the bubble within a uniformly charged solid sphere.

(d) −λ (inner), +λ (outer), a a 2.16 (a) (i), (ii) 0, (iii) pP0r 2pP0r Graph E

a

Single correct

r

b

(b) (i) −a, (ii) + 2a

Q2.1 (b) Q2.2 (b) Q2.3 (d) Q2.4 (b) Q2.5 (c) Q2.6 (c) Q2.7 (c) Q2.8 (b)

2.17 (a)

dr l , (b) (c) same 2P0 2pP0r

(d) E

One or More Correct

R

Q2.9 (a), (b), (d) Q2.10 (b), (c) Q2.11 (a), (c) Q2.12 (d) Q2.13 (b) Q2.14 (a), (b), (c) Q2.15 (a), (d)

r

2R

2.18 (a) 0,

q 4pP0r2

, 0,

q 4pP0r2

(b) E

Exercise 2.1 9N − m2/C 2.2 πr2E 2.3 −32N − m2/C, 0, 32N − m2/C, 0, 24N − m2/C, −32N − m2/C (b) 0 2.4 (a) 452N − m2/C, (b) −881N − m2/C, (c) −429N − m2/C, (d) 723N − m2/C, (e) −158N − m2/C, (f) No. 2.5 (a) 0, (b) 9 × 107 N/C, (c) 2.25 × 107 N/C 2.6 (a) 4.41 × 104 N/C, (b) 9 × 102 N/C 2.7 (a) 2.4 × 1010 2.8 (a) 53 N/C, (b) 160 N/C, (c) 480 N/C 2.9 (a) 7.92 mJ sR l 2.10 (a) 2pRs, (b) , (c) P0r 2pP0r 2.11 (a) 2.82 × 105 N/C left, (b) 3.95 × 105 N/C left, (c) 1.69 × 105 N/C left 2.12 (a) + Q, (b) No, (c) Yes, (d) No, (e) No, Yes 2.13 (a) 0, 0.081N − m2/C, 0, 0, −0.135N − m2/C, 0 (b) −4.78 × 10−13C 2.14 −67.5 × 10−13N − m2/C l l 2.15 (a) (b) 2pP0r 2pP0r (c)

a

r

c

b

(c) −q (d) q 2.19 (a) 0,

2Q Q , 4pP0r2 4pP0r2

(b) E

R

2.20 (a)

r

2R

Q 2

4pP0r

, 0,

2Q 4pP0r

a

b

λ

c

- Q 2

4pa

(c)

- 2Q 4pb2

E

a

r

b

2.21 (a) (i) 0, (ii) 0, (iii) 2π ∈0 r

(b)

(e)

E λ

2

2q 4pP0r

2

(iv) 0, (v)

2π ∈0 r

r

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6q 4pP0r2

69

Answers

(c) 0

E 6q 4π ∈0 d 2

5Q

(d)

i 18pe0R2 Qr 3r a4 b 2.32 (a) R 4pP0R3 (d)

2q 4π ∈0 b2

b

c

r

d

E

(b) (i) 0, (ii) +2q, (iii) −2q (iv) +6q 2.22 (a)

- 3Q 28pR3

(b) 0,

2Q 7pP0R2

-

Qr 28pP0R3

,0 r 2R 3

Graph: E 7Q 28π ∈0 R 2

(e)

R

2R

Q 2R , 3 3pP0R2

2.33 2:1 2.34 q'/q = 13/20 2.35 (a) Q∞ = 2Q,

r

qQ

2.23 (a)

(b) 0 4pP0r2 a a a2 2.24 (a) a 1 - 2 b (b) 2paa2, 2P0 2P0 r S dx 2.25 (b) E = iN for | x| … d P0 dd = iN for | x| Ú d P0 2.26 E = a

d0d x S ba b i for 3P0 | x| 3 d0x S = i for 3P0d2 r dor 31 - 4 2.27 (a) 0, (b) 0, (c) 3P0 R (d) S

(b) Er = 2.36 (a)

2pP0r

8Q 5pR3 8Qr

| x| < d

4pe0r2 (c)

,

r … R/2

,

R/2 … r … R

,

r Ú R

4 15

(e) 2p

15pP0R3mc Ä 8eQ

(f) No. r R

R doR (e) , 2 12P0

2.30 E =

2

2.37 Answer: (a)

480Q 233pR3

(b) 180Qr2

2.29 (b) E =

(b)

C 1 - e a0 S - 2r

2

15pe0R3 KQ S E = f (64(r/R)3 - 48(r/4)4 - 1) 15r2 Q

| x| > d

f0R 12 ∈0

Q

(b)

E

2.31 (a)

2R

db 3P0

db 2P0 - Q

233pP0R4 480Q 1 r 3 S 1 r 5 23 c a b a b d E = f 2 3 R 5 R 1920 233pP0r Q 4pP0r2

16pP0R2 Q 72pP0R2

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R 2

,

r …

,

R … r … R 2

,

r Ú R

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3

ELECTRIC POTENTIAL

LEARNING GOALS By studying this chapter, you will learn: • How to calculate the electric potential energy of a collection of charges. • The meaning and significance of electric potential. • How to calculate the electric potential that a collection of charges produces at a point in space. • How to use equipotential surfaces to visualize how the electric potential varies in space.

?

In one type of welding, electric charge flows between the welding tool and the metal pieces that are to be joined together. This produces a glowing arc whose high temperature fuses the pieces together. Why must the tool be held close to the pieces being welded?

• How to use electric potential to calculate the electric field.

his chapter is about energy associated with electrical interactions. Every time you turn on a light, listen to an MP3 player, or talk on a mobile phone, you are using electrical energy, an indispensable ingredient of our technological society. In Chapters 6 and 7 we introduced the concepts of work and energy in the context of mechanics; now we’ll combine these concepts with what we’ve learned about electric charge, electric forces, and electric fields. Just as we found for many problems in mechanics, using energy ideas makes it easier to solve a variety of problems in electricity. When a charged particle moves in an electric field, the field exerts a force that can do work on the particle. This work can always be expressed in terms of electric potential energy. Just as gravitational potential energy depends on the height of a mass above the earth’s surface, electric potential energy depends on the position of the charged particle in the electric field. We’ll describe electric potential energy using a new concept called electric potential, or simply potential. In circuits, a difference in potential from one point to another is often called voltage. The concepts of potential and voltage are crucial to understanding how electric circuits work and have equally important applications to electron beams used in cancer radiotherapy, high-energy particle accelerators, and many other devices.

T

3.1

Electric Potential Energy

The concepts of work, potential energy, and conservation of energy proved to be extremely useful in our study of mechanics. In this section we’ll show that these concepts are just as useful for understanding and analyzing electrical interactions. 70

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71

3.1 Electric Potential Energy

Let’s beginSby reviewing three essential points from Chapters 6 and 7. First, when a force F acts on a particle that moves from point a to point b, the work WaSb done by the force is given by a line integral: b

WaSb =

S

2a

#

b S

F dl =

2a

F cos f dl

(work done by a force)

(3.1)

S

where d l is anSinfinitesimal displacement along the particle’s path and f is the S angle between F and d lSat each point along the path. Second, if the force F is conservative, as we defined the term in Section 7.3, S the work done by F can always be expressed in terms of a potential energy U. When the particle moves from a point where the potential energy is Ua to a point where it is Ub , the change in potential energy is ¢U = Ub - Ua and the work WaSb done by the force is '

WaSb = Ua - Ub = -1Ub - Ua2 = - ¢U (work done by a conservative force)

(3.2)

When WaSb is positive, Ua is greater than Ub , ¢U is negative, and the potential energy decreases. That’s what happens when a baseball falls from a high point (a) to a lower point (b) under the influence of the earth’s gravity; the force of gravity does positive work, and the gravitational potential energy decreases (Fig. 3.1). When a tossed ball is moving upward, the gravitational force does negative work during the ascent, and the potential energy increases. Third, the work–energy theorem says that the change in kinetic energy ¢K = Kb - Ka during a displacement equals the total work done on the particle. If only conservative forces do work, then Eq. (3.2) gives the total work, and Kb - Ka = -1Ub - Ua2. We usually write this as '

Ka + Ua = Kb + Ub

3.1 The work done on a baseball moving in a uniform gravitational field. Object moving in a uniform gravitational field

a S

The work done by the gravitational force is the same for any path from a to b: Wa S b 5 2DU 5 mgh

S

w 5 mg

h

(3.3) b

That is, the total mechanical energy (kinetic plus potential) is conserved under these circumstances.

Electric Potential Energy in a Uniform Field Let’s look at an electrical example of these basic concepts. In Fig. 3.2 a pair of charged parallel metal plates sets up a uniform, downward electric field with magnitude E. The field exerts a downward force with magnitude F = q0 E on a positive test charge q0 . As the charge moves downward a distance d from point a to point b, the force on the test charge is constant and independent of its location. So the work done by the electric field is the product of the force magnitude and the component of displacement in the (downward) direction of the force: '

'

WaSb = Fd = q0 Ed

3.2 The work done on a point charge moving in a uniform electric field. Compare with Fig. 3.1. Point charge moving in y a uniform electric field +

+

+

+

+

S

E

q0

(3.4)

'

a

d S

S

F 5 q0E

This work is positive, since the force is in the same direction as the net displacement of the test charge. The y-component of the electric force, Fy = - q0 E, is constant, and there is no x- or z-component. This is exactly analogous to the gravitational force on a mass m near the earth’s surface; for this force, there is a constant y-component Fy = - mg and the x- and z-components are zero. Because of this analogy, we can conclude that the force exerted on q0 by the uniform electric field in Fig. 3.2 is conservative, just as is the gravitational force. This means that the work WaSb done by the field is independent of the path the particle takes from a to b. We can represent this work with a potential-energy function U, just as we did for gravitational potential energy '

y

b –

–

– O

–

–

The work done by the electric force is the same for any path from a to b: Wa Sb 5 2DU 5 q0Ed

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72

CHAPTER 3 Electric Potential S

3.3 A positive charge moving (a) in the direction of the electric field E and (b) in the S direction opposite E. S

S

(a) Positive charge moves in the direction of E: • Field does positive work on charge. y • U decreases. +

+

+

+

(b) Positive charge moves opposite E: • Field does negative work on charge. y • U increases.

+

+

S

+

+

+

+

S

E

E a S

b

S

F 5 q0 E

ya

yb a

b ya

yb

S

O

–

–

O

–

–

–

–

–

S

F 5 q0 E

–

–

–

in Section 7.1. The potential energy for the gravitational force Fy = -mg was U = mgy; hence the potential energy for the electric force Fy = - q0 E is '

U = q0 Ey

(3.5)

'

When the test charge moves from height ya to height yb , the work done on the charge by the field is given by '

WaSb = - ¢U = - 1Ub - Ua2 = - 1q0 Eyb - q0 Eya2 = q0 E1ya - yb2 '

'

'

(3.6)

When ya is greater than yb (Fig. 3.3a), the positive test charge q0 moves downS ward, in the same direction as E; the displacement is in the same direction as the S S force F 5 q0 E, so the field does positive work and U decreases. [In particular, if ya - yb = d as in Fig. 3.2, Eq. (3.6) gives WaSb = q0 Ed, in agreement with Eq. (3.4).] When ya is less than yb (Fig. 3.3b), the positive test charge q0 moves S upward, in the opposite direction to E; the displacement is opposite the force, the field does negative work, and U increases. If the test charge q0 is negative, the potential energy increases when it moves with the field and decreases when it moves against the field (Fig. 3.4). Whether the test charge is positive or negative, the following general rules apply: U increases if the test charge q0 moves in the direction opposite the electric S S force F 5 q0 E (Figs. 3.3b and 3.4a); U decreases if q0 moves in the same direc'

'

'

S

3.4 A negative charge moving (a) in the direction of the electric field E and (b) in the S direction opposite E. Compare with Fig. 3.3. S

(a) Negative charge moves in the direction of E: • Field does negative work on charge. y • U increases. +

+

+

+ S

S

E

+

S

(b) Negative charge moves opposite E: • Field does positive work on charge. y • U decreases. +

S

S

F 5 q0 E

E

+

a

S

yb

+

S

F 5 q0 E a

b ya

yb O

–

+

b

ya

–

+

–

O

–

–

–

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–

–

–

–

3.1 Electric Potential Energy S

73

S

tion as F 5 q0 E (Figs. 3.3a and 3.4b). This is the same behavior as for gravitational potential energy, which increases if a mass m moves upward (opposite the direction of the gravitational force) and decreases if m moves downward (in the same direction as the gravitational force). '

CAUTION Electric potential energy The relationship between electric potential energy change and motion in an electric field is an important one that we’ll use often, but that takes some effort to truly understand. Take the time to carefully study the preceding paragraph as well as Figs. 3.3 and 3.4. Doing so now will help you tremendously later!

Electric Potential Energy of Two Point Charges The idea of electric potential energy isn’t restricted to the special case of a uniform electric field. Indeed, we can apply this concept to a point charge in any electric field caused by a static charge distribution. Recall from Chapter 1 that we can represent any charge distribution as a collection of point charges. Therefore it’s useful to calculate the work done on a test charge q0 moving in the electric field caused by a single, stationary point charge q. We’ll consider first a displacement along the radial line in Fig. 3.5. The force on q0 is given by Coulomb’s law, and its radial component is Fr =

1 qq0 4pP0 r2

3.5 Test charge q0 moves along a straight line extending radially from charge q. As it moves from a to b, the distance varies from ra to rb . '

S

Test charge q0 moves from a to b along a radial line from q.

(3.7)

q0

If q and q0 have the same sign 1+ or -2 the force is repulsive and Fr is positive; if the two charges have opposite signs, the force is attractive and Fr is negative. The force is not constant during the displacement, and we have to integrate to calculate the work WaSb done on q0 by this force as q0 moves from a to b: rb

WaSb =

2ra

qq0 1 1 qq0 1 dr = a - b 2ra 4pP0 r2 rb 4pP0 ra

WaSb

a r ra q

(3.8)

The work done by the electric force for this particular path depends only on the endpoints. Now let’s consider a more general displacement (Fig. 3.6) in which a and b do not lie on the same radial line. From Eq. (3.1) the work done on q0 during this displacement is given by rb

rb

S

E

rb

Fr dr =

3.6 The work done on charge q0 by the electric field of charge q does not depend on the path taken, but only on the distances ra and rb . '

rb

1 qq0 = F cos f dl = cos f dl 2ra 2ra 4pP0 r2

Test charge q0 moves from a to b along an arbitrary path.

But Fig. 3.6 shows that cos f dl = dr. That is, the work done during a small S displacement d l depends only on the change dr in the distance r between the charges, which is the radial component of the displacement. Thus Eq. (3.8) is valid even for this more general displacement; the work done on q0 by the electric S field E produced by q depends only on ra and rb , not on the details of the path. Also, if q0 returns to its starting point a by a different path, the total work done in the round-trip displacement is zero (the integral in Eq. (3.8) is from ra back to ra 2. These are the needed characteristics for a conservative force, as we defined it in Section 7.3. Thus the force on q0 is a conservative force. We see that Eqs. (3.2) and (3.8) are consistent if we define the potential energy to be Ua = qq0>4pP0ra when q0 is a distance ra from q, and to be Ub = qq0>4pP0rb when q0 is a distance rb from q. Thus the potential energy U when the test charge q0 is at any distance r from charge q is (electric potential energy of two point charges q and q0)

E b

S

dr dl q0 f

r

'

1 qq0 4pP0 r

S

S

F

S

'

U =

E

b

a

ra

(3.9)

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q

rb

dr

74

CHAPTER 3 Electric Potential

3.7 Graphs of the potential energy U of two point charges q and q0 versus their separation r.

Equation (3.9) is valid no matter what the signs of the charges q and q0. The potential energy is positive if the charges q and q0 have the same sign (Fig. 3.7a) and negative if they have opposite signs (Fig. 3.7b).

(a) q and q0 have the same sign.

CAUTION Electric potential energy vs. electric force Don’t confuse Eq. (3.9) for the potential energy of two point charges with the similar expression in Eq. (3.7) for the radial component of the electric force that one charge exerts on the other. Potential energy U is proportional to 1>r, while the force component Fr is proportional to 1>r2 . y

U q

q0

q

q0

or r

r

'

•U.0 • As r S 0, U S 1`. • As r S `, U S 0. O

Potential energy is always defined relative to some reference point where U = 0. In Eq. (3.9), U is zero when q and q0 are infinitely far apart and r = q . Therefore U represents the work that would be done on the test charge q0 by the field of q if q0 moved from an initial distance r to infinity. If q and q0 have the same sign, the interaction is repulsive, this work is positive, and U is positive at any finite separation (Fig. 3.7a). If the charges have opposite signs, the interaction is attractive, the work done is negative, and U is negative (Fig. 3.7b). We emphasize that the potential energy U given by Eq. (3.9) is a shared property of the two charges. If the distance between q and q0 is changed from ra to rb , the change in potential energy is the same whether q is held fixed and q0 is moved or q0 is held fixed and q is moved. For this reason, we never use the phrase “the electric potential energy of a point charge.” (Likewise, if a mass m is at a height h above the earth’s surface, the gravitational potential energy is a shared property of the mass m and the earth. We emphasized this in Sections 7.1 and 13.3.) Equation (3.9) also holds if the charge q0 is outside a spherically symmetric charge distribution with total charge q; the distance r is from q0 to the center of the distribution. That’s because Gauss’s law tells us that the electric field outside such a distribution is the same as if all of its charge q were concentrated at its center (see Example 2.9 in Section 2.4).

r

(b) q and q0 have opposite signs. U

'

r

O q

q0

q

q0

or r

r

•U,0 • As r S 0, U S 2`. • As r S `, U S 0.

Example 3.1

Conservation of energy with electric forces

A positron (the electron’s antiparticle) has mass 9.11 * 10-31 kg and charge q0 = + e = + 1.60 * 10-19 C. Suppose a positron moves in the vicinity of an a (alpha) particle, which has charge q = +2e = 3.20 * 10-19 C and mass 6.64 * 10-27 kg. The a particle’s mass is more than 7000 times that of the positron, so we assume that the a particle remains at rest. When the positron is 1.00 * 10-10 m from the a particle, it is moving directly away from the a particle at 3.00 * 106 m > s. (a) What is the positron’s speed when the particles are 2.00 * 10-10 m apart? (b) What is the positron’s speed when it is very far from the a particle? (c) Suppose the initial conditions are the same but the moving particle is an electron (with the same mass as the positron but charge q0 = -e). Describe the subsequent motion. '

'

EXECUTE: (a) Both particles have positive charge, so the positron speeds up as it moves away from the a particle. From the energyconservation equation, Eq. (3.3), the final kinetic energy is Kb = 12 mvb2 = Ka + Ua - Ub In this expression, Ka = 12 mva2 = 12 19.11 * 10-31 kg213.00 * 106 m > s22 '

= 4.10 * 10-18 J 1 qq0 = 19.0 * 109 N # m2 > C22 Ua = 4pP0 ra '

*

SOLUTION IDENTIFY and SET UP: The electric force between a positron (or an electron) and an a particle is conservative, so mechanical energy (kinetic plus potential) is conserved. Equation (3.9) gives the potential energy U at any separation r: The potential-energy function for parts (a) and (b) looks like that of Fig. 3.7a, and the function for part (c) looks like that of Fig. 3.7b. We are given the positron speed va = 3.00 * 106 m>s when the separation between the particles is ra = 1.00 * 10-10 m . In parts (a) and (b) we use Eqs. (3.3) and (3.9) to find the speed for r = rb = 2.00 * 10-10 m and r = rc S q , respectively. In part (c) we replace the positron with an electron and reconsider the problem.

'

'

13.20 * 10-19 C211.60 * 10-19 C2 1.00 * 10-10 m

-18

= 4.61 * 10 J 1 qq0 = 2.30 * 10-18 J Ub = 4pP0 rb Hence the positron kinetic energy and speed at r = rb = 2.00 * 10-10 m are Kb = 12 mvb2 = 4.10 * 10-18 J + 4.61 * 10-18 J - 2.30 * 10-18 J

= 6.41 * 10-18 J vb =

2Kb 216.41 * 10-18 J2 6 = Ä m Ä 9.11 * 10-31 kg = 3.8 * 10 m > s

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'

'

75

3.1 Electric Potential Energy (b) When the positron and a particle are very far apart so that r = rc S q , the final potential energy Uc approaches zero. Again from energy conservation, the final kinetic energy and speed of the positron in this case are Kc = Ka + Ua - Uc = 4.10 * 10-18 J + 4.61 * 10-18 J - 0 = 8.71 * 10-18 J 2Kc 218.71 * 10-18 J2 6 vc = = Ä m Ä 9.11 * 10-31 kg = 4.4 * 10 m > s '

Ud = Ka + Ua - Kd = 1-0.51 * 10-18 J2 - 0 1 qq0 = - 0.51 * 10-18 J 4pP0 rd 1 qq0 rd = Ud 4pP0

Ud =

19.0 * 109 N # m2 > C22 '

=

-0.51 * 10

'

(c) The electron and a particle have opposite charges, so the force is attractive and the electron slows down as it moves away. Changing the moving particle’s sign from +e to - e means that the initial potential energy is now Ua = - 4.61 * 10-18 J, which makes the total mechanical energy negative: Ka + Ua = 14.10 * 10-18 J2 - 14.61 * 10-18 J2 = - 0.51 * 10-18 J

'

J

13.20 * 10-19 C21 - 1.60 * 10-19C2

= 9.0 * 10-10 m For rb = 2.00 * 10-10 m we have Ub = -2.30 * 10-18 J, so the electron kinetic energy and speed at this point are Kb =

vb =

The total mechanical energy would have to be positive for the electron to move infinitely far away from the a particle. Like a rock thrown upward at low speed from the earth’s surface, it will reach a maximum separation r = rd from the a particle before reversing direction. At this point its speed and its kinetic energy Kd are zero, so at separation rd we have

-18

1 2

mvb2 = 4.10 * 10-18 J + 1- 4.61 * 10-18 J2

- 1-2.30 * 10-18 J2 = 1.79 * 10-18 J

2Kb 211.79 * 10-18 J2 = 2.0 * 106 m > s = Ä m Ä 9.11 * 10-31 kg '

'

EVALUATE: Both particles behave as expected as they move away from the a particle: The positron speeds up, and the electron slows down and eventually turns around. How fast would an electron have to be moving at ra = 1.00 * 10-10 m to travel infinitely far from the a particle? (Hint: See Example 13.4 in Section 13.3.)

Electric Potential Energy with Several Point Charges S

Suppose the electric field E in which charge q0 moves is caused by several point charges q1 , q2 , q3 , Á at distances r1 , r2 , r3 , Á from q0 , as in Fig. 3.8. For example, q0 could be a positive ion moving in the presence of other ions (Fig. 3.9). The total electric field at each point is the vector sum of the fields due to the individ ual charges, and the total work done on q0 during any displacement is the sum of the contributions from the individual charges. From Eq. (3.9) we conclude that the potential energy associated with the test charge q0 at point a in Fig. 3.8 is the algebraic sum (not a vector sum): '

'

'

'

'

'

'

3.8 The potential energy associated with a charge q0 at point a depends on the other charges q1 , q2 , and q3 and on their distances r1 , r2 , and r3 from point a. '

'

'

'

q1 q2 r1

(point charge q0 and collection of charges qi)

q0 q3 q0 q1 q2 qi U = a + + + Áb = r2 r3 4pP0 r1 4pP0 a i ri '

r2 (3.10)

When q0 is at a different point b, the potential energy is given by the same expression, but r1 , r2 , Á are the distances from q1 , q2 , Á to point b. The work done on charge q0 when it moves from a to b along any path is equal to the difference Ua - Ub between the potential energies when q0 is at a and at b. We can represent any charge distribution as a collection of point charges, so Eq. (3.10) shows that we can always find a potential-energy function for any static electric field. It follows that for every electric field due to a static charge distribution, the force exerted by that field is conservative. Equations (3.9) and (3.10) define U to be zero when all the distances r1 , r2 , Á are infinite—that is, when the test charge q0 is very far away from all the charges that produce the field. As with any potential-energy function, the point where U = 0 is arbitrary; we can always add a constant to make U equal zero at any point we choose. In electrostatics problems it’s usually simplest to choose this point to be at infinity. When we analyze electric circuits in Chapters 5 and 6, other choices will be more convenient. '

'

'

'

r3

a

'

'

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q0

q3

76

CHAPTER 3 Electric Potential

3.9 This ion engine for spacecraft uses electric forces to eject a stream of positive xenon ions 1Xe+2 at speeds in excess of 30 km/s. The thrust produced is very low (about 0.09 newton) but can be maintained continuously for days, in contrast to chemical rockets, which produce a large thrust for a short time (see Fig. 8.33). Such ion engines have been used for maneuvering interplanetary spacecraft.

Equation (3.10) gives the potential energy associated with the presence of the S test charge q0 in the E field produced by q1 , q2 , q3 , . . . . But there is also potential energy involved in assembling these charges. If we start with charges q1 , q2 , q3 , Á all separated from each other by infinite distances and then bring them together so that the distance between qi and qj is rij , the total potential energy U is the sum of the potential energies of interaction for each pair of charges. We can write this as qi qj 1 U = (3.11) a r 4pP '

'

'

'

'

'

'

'

'

0 i6j

ij

This sum extends over all pairs of charges; we don’t let i = j (because that would be an interaction of a charge with itself ), and we include only terms with i 6 j to make sure that we count each pair only once. Thus, to account for the interaction between q3 and q4 , we include a term with i = 3 and j = 4 but not a term with i = 4 and j = 3. '

Interpreting Electric Potential Energy As a final comment, here are two viewpoints on electric potential energy. We have defined it in terms of the work done by the electric field on a charged particle moving in the field, just as in Chapter 7 we defined potential energy in terms of the work done by gravity or by a spring. When a particle moves from point a to point b, the work done on it by the electric field is WaSb = Ua - Ub . Thus the potential-energy difference Ua - Ub equals the work that is done by the electric force when the particle moves from a to b. When Ua is greater than Ub , the field does positive work on the particle as it “falls” from a point of higher potential energy (a) to a point of lower potential energy (b). An alternative but equivalent viewpoint is to consider how much work we would have to do to “raise” a particle from a point b where the potential energy is Ub to a point a where it has a greater value Ua (pushing two positive charges closer together, for example). To move the particle slowly (so asSnot to give it any kinetic energy), we need to exert an additional external force Fext that is equal and opposite to the electric-field force and does positive work. The potentialenergy difference Ua - Ub is then defined as the work that must be done by an external force to move the particle slowly from b to a against the electric force. S Because Fext is the negative of the electric-field force and the displacement is in the opposite direction, this definition of the potential difference Ua - Ub is equivalent to that given above. This alternative viewpoint also works if Ua is less than Ub , corresponding to “lowering” the particle; an example is moving two positive charges away from each other. In this case, Ua - Ub is again equal to the work done by the external force, but now this work is negative. We will use both of these viewpoints in the next section to interpret what is meant by electric potential, or potential energy per unit charge. '

'

'

Example 3.2

A system of point charges

Two point charges are located on the x-axis, q1 = -e at x = 0 and q2 = +e at x = a. (a) Find the work that must be done by an external force to bring a third point charge q3 = + e from infinity to x = 2a. (b) Find the total potential energy of the system of three charges.

to x = 2a. We do this by using Eq. (3.10) to find the potential energy associated with q3 in the presence of q1 and q2. In part (b) we use Eq. (3.11), the expression for the potential energy of a collection of point charges, to find the total potential energy of the system. 3.10 Our sketch of the situation after the third charge has been brought in from infinity.

SOLUTION IDENTIFY and SET UP: Figure 3.10 shows the final arrangement of the three charges. In part (a) we need to find the work W that must S be done on q3 by an external force Fext to bring q3 in from infinity

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3.2 Electric Potential EXECUTE: (a) The work W equals the difference between (i) the potential energy U associated with q3 when it is at x = 2a and (ii) the potential energy when it is infinitely far away. The second of these is zero, so the work required is equal to U. The distances between the charges are r13 = 2a and r23 = a, so from Eq. (3.10), W = U =

q3 q1 q2 +e +e - e + e2 a + b = a + b = r23 a 4pP0 r13 4pP0 2a 8pP0a

This is positive, just as we should expect. If we bring q3 in from infinity along the + x-axis, it is attracted by q1 but is repelled more strongly by q2. Hence we must do positive work to push q3 to the position at x = 2a. (b) From Eq. (3.11), the total potential energy of the threecharge system is

U =

=

qiqj q1q3 q2q3 q1q2 1 1 = a + + b a r13 r23 4pP0 i 6 j rij 4pP0 r12 1e21e2 1-e21e2 1 1- e21e2 - e2 c + + d = a a 4pP0 2a 8pP0a

EVALUATE: Our negative result in part (b) means that the system has lower potential energy than it would if the three charges were infinitely far apart. An external force would have to do negative work to bring the three charges from infinity to assemble this entire arrangement and would have to do positive work to move the three charges back to infinity.

Test Your Understanding of Section 3.1 Consider the system of three point charges in Example 1.4 (Section 1.3) and shown in Fig. 1.14. (a) What is the sign of the total potential energy of this system? (i) positive; (ii) negative; (iii) zero. (b) What is the sign of the total amount of work you would have to do to move these charges infinitely far from each other? (i) positive; (ii) negative; (iii) zero. y

3.2

Electric Potential

In Section 3.1 we looked at the potential energy U associated with a test charge q0 in an electric field. Now we want to describe this potential energy on a “per unit charge” basis, just as electric field describes the force per unit charge on a charged particle in the field. This leads us to the concept of electric potential, often called simply potential. This concept is very useful in calculations involving energies of charged particles. It also facilitates many electric-field calculaS tions because electric potential is closely related to the electric field E. When we need to determine an electric field, it is often easier to determine the potential first and then find the field from it. Potential is potential energy per unit charge. We define the potential V at any point in an electric field as the potential energy U per unit charge associated with a test charge q0 at that point: V =

77

U or U = q0V q0

PhET: Charges and Fields ActivPhysics 11.13: Electrical Potential Energy and Potential

(3.12)

Potential energy and charge are both scalars, so potential is a scalar. From Eq. (3.12) its units are the units of energy divided by those of charge. The SI unit of potential, called one volt (1 V) in honor of the Italian electrical experimenter Alessandro Volta (1745–1827), equals 1 joule per coulomb: 1 V = 1 volt = 1 J>C = 1 joule>coulomb Let’s put Eq. (3.2), which equates the work done by the electric force during a displacement from a to b to the quantity - ¢U = - 1Ub - Ua2, on a “work per unit charge” basis. We divide this equation by q0 , obtaining '

WaSb Ub Ua ¢U = = -a b = - 1Vb - Va2 = Va - Vb q0 q0 q0 q0

(3.13)

where Va = Ua>q0 is the potential energy per unit charge at point a and similarly for Vb . We call Va and Vb the potential at point a and potential at point b, respectively. Thus the work done per unit charge by the electric force when a charged body moves from a to b is equal to the potential at a minus the potential at b. '

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78

CHAPTER 3 Electric Potential

3.11 The voltage of this battery equals the difference in potential Vab = Va - Vb between its positive terminal (point a) and its negative terminal (point b). Point a (positive terminal)

The difference Va - Vb is called the potential of a with respect to b; we sometimes abbreviate this difference as Vab = Va - Vb (note the order of the subscripts). This is often called the potential difference between a and b, but that’s ambiguous unless we specify which is the reference point. In electric circuits, which we will analyze in later chapters, the potential difference between two points is often called voltage (Fig. 3.11). Equation (3.13) then states: Vab , the potential of a with respect to b, equals the work done by the electric force when a UNIT charge moves from a to b. Another way to interpret the potential difference Vab in Eq. (3.13) is to use the alternative viewpoint mentioned at the end of Section 3.1. In that viewpoint, Ua - Ub is the amount of work that must be done by an external force to move a particle of charge q0 slowly from b to a against the electric force. The work that must be done per unit charge by the external force is then 1Ua - Ub2>q0 = Va - Vb = Vab . In other words: Vab , the potential of a with respect to b, equals the work that must be done to move a UNIT charge slowly from b to a against the electric force. An instrument that measures the difference of potential between two points is called a voltmeter. (In Chapter 6 we’ll discuss how these devices work.) Voltmeters that can measure a potential difference of 1 mV are common, and sensitivities down to 10-12 V can be attained. '

Vab 5 1.5 volts

'

Point b (negative terminal)

'

Calculating Electric Potential To find the potential V due to a single point charge q, we divide Eq. (3.9) by q0: V =

U 1 q = q0 4pP0 r

(potential due to a point charge)

(3.14)

where r is the distance from the point charge q to the point at which the potential is evaluated. If q is positive, the potential that it produces is positive at all points; if q is negative, it produces a potential that is negative everywhere. In either case, V is equal to zero at r = q , an infinite distance from the point charge. Note that potential, like electric field, is independent of the test charge q0 that we use to define it. Similarly, we divide Eq. (3.10) by q0 to find the potential due to a collection of point charges: Application Electrocardiography The electrodes used in an electrocardiogram— EKG or ECG for short—measure the potential differences (typically no greater than 1 mV = 10-3 V) between different parts of the patient’s skin. These are indicative of the potential differences between regions of the heart, and so provide a sensitive way to detect any abnormalities in the electrical activity that drives cardiac function.

qi U 1 = a q0 4pP0 i ri '

V =

(potential due to a collection of point charge)

(3.15)

In this expression, ri is the distance from the ith charge, qi , to the point at which V is evaluated. Just as the electric field due to a collection of point charges is the vector sum of the fields produced by each charge, the electric potential due to a collection of point charges is the scalar sum of the potentials due to each charge. When we have a continuous distribution of charge along a line, over a surface, or through a volume, we divide the charge into elements dq, and the sum in Eq. (3.15) becomes an integral: '

V =

dq 1 4pP0 2 r

(potential due to a continuous distribution of charge)

(3.16)

where r is the distance from the charge element dq to the field point where we are finding V. We’ll work out several examples of such cases. The potential defined by Eqs. (3.15) and (3.16) is zero at points that are infinitely far away from all the charges. Later we’ll encounter cases in which the charge distribution itself

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79

3.2 Electric Potential

extends to infinity. We’ll find that in such cases we cannot set V = 0 at infinity, and we’ll need to exercise care in using and interpreting Eqs. (3.15) and (3.16). CAUTION What is electric potential? Before getting too involved in the details of how to calculate electric potential, you should stop and remind yourself what potential is. The electric potential at a certain point is the potential energy that would be associated with a unit charge placed at that point. That’s why potential is measured in joules per coulomb, or volts. Keep in mind, too, that there doesn’t have to be a charge at a given point for a potential V to exist at that point. (In the same way, an electric field can exist at a given point even if there’s no charge there to respond to it.) y

Finding Electric Potential from Electric Field When we are given a collection of point charges, Eq. (3.15) is usually the easi est way to calculate the potential V. But in some problems in which the electric S field isS known or can be found easily, it is easier to determine V from E. The S S force F on a test charge q0 can be written as F 5 q0 E, so from Eq. (3.1) the work done by the electric force as the test charge moves from a to b is given by '

b

WaSb =

S

2a

#

b S

S

F dl =

#

S

q0 E d l

2a

'

If we divide this by q0 and compare the result with Eq. (3.13), we find b

Va - Vb =

E # dl = 2 S

a

b

S

2a

(potential difference S as an integral of E)

E cos f dl

(3.17)

The value of Va - Vb is independent of the path taken from a to b, just as the value of WaSb is independent of the path. To interpret Eq. (3.17), remember that S E is the electric force per unit charge on a test charge. If the line integral S b S 1a E d l is positive, the electric field does positive work on a positive test charge as it moves from a to b. In this case the electric potential energy decreases as the test charge moves, so the potential energy per unit charge decreases as well; hence Vb is less than Va and Va - Vb is positive. As an illustration, consider a positive point charge (Fig. 3.12a). The electric field is directed away from the charge, and V = q/4pP0 r is positive at any finite distance from the charge. If you move away from the charge, in the direction of S E, you move toward lower values of V; if you move toward the charge, in the S direction opposite E, you move toward greater values of V. For the negative point S charge in Fig. 3.12b, E is directed toward the charge and V = q/4pP0 r is negative at any finite distance from the charge. In this case, if you move toward the S charge, you are moving in the direction of E and in the direction of decreasing S (more negative) V. Moving away from the charge, in the direction opposite E, moves you toward increasing (less negative) values of V.SThe general rule, valid for any electric field, is: Moving with the direction of E means moving in the S direction of decreasing V, and moving against the direction of E means moving in the direction of increasing V. Also, a positive test charge q0 experiences an electric force in the direction of S S E, toward lower values of V; a negative test charge experiences a force opposite E, toward higher values of V. Thus a positive charge tends to “fall” from a high-potential region to a lower-potential region. The opposite is true for a negative charge. Notice that Eq. (3.17) can be rewritten as

#

S

3.12 If you move in the direction of E, electric potential V decreases; if you move S in the direction opposite E, V increases.

'

'

a

Va - Vb = -

S

2b

#

S

E dl

(3.18)

(a) A positive point charge V increases as you move inward.

V decreases as you move outward.

S

E

(b) A negative point charge V decreases as you move inward.

S

E

This has a negative sign compared to the integral in Eq. (3.17), and the limits are reversed; hence Eqs. (3.17) and (3.18) are equivalent. But Eq. (3.18) has a slightly different interpretation. To move a unit charge slowly against the electric

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V increases as you move outward.

80

CHAPTER 3 Electric Potential S

force, we must apply an external force per unit charge equal to 2 E, equal and S opposite to the electric force per unit charge E. Equation (3.18) says that Va - Vb = Vab , the potential of a with respect to b, equals the work done per unit charge by this external force to move a unit charge from b to a. This is the same alternative interpretation we discussed under Eq. (3.13). Equations (3.17) and (3.18) show that the unit of potential difference 11 V2 is equal to the unit of electric field 11 N>C2 multiplied by the unit of distance 11 m2. Hence the unit of electric field can be expressed as 1 volt per meter 11 V>m2, as well as 1 N>C: '

1 V>m = 1 volt>meter = 1 N>C = 1 newton>coulomb

In practice, the volt per meter is the usual unit of electric-field magnitude.

Electron Volts The magnitude e of the electron charge can be used to define a unit of energy that is useful in many calculations with atomic and nuclear systems. When a particle with charge q moves from a point where the potential is Vb to a point where it is Va , the change in the potential energy U is '

Ua - Ub = q1Va - Vb2 = qVab

If the charge q equals the magnitude e of the electron charge, 1.602 * 10-19 C, and the potential difference is Vab = 1 V, the change in energy is Ua - Ub = 11.602 * 10-19 C211 V2 = 1.602 * 10-19 J

This quantity of energy is defined to be 1 electron volt 11 eV2: 1 eV = 1.602 * 10-19 J

The multiples meV, keV, MeV, GeV, and TeV are often used. CAUTION Electron volts vs. volts Remember that the electron volt is a unit of energy, not a unit of potential or potential difference! y

When a particle with charge e moves through a potential difference of 1 volt, the change in potential energy is 1 eV. If the charge is some multiple of e—say Ne—the change in potential energy in electron volts is N times the potential difference in volts. For example, when an alpha particle, which has charge 2e, moves between two points with a potential difference of 1000 V, the change in potential energy is 211000 eV2 = 2000 eV. To confirm this, we write Ua - Ub = qVab = 12e211000 V2 = 12211.602 * 10-19 C211000 V2 = 3.204 * 10-16 J = 2000 eV

Although we have defined the electron volt in terms of potential energy, we can use it for any form of energy, such as the kinetic energy of a moving particle. When we speak of a “one-million-electron-volt proton,” we mean a proton with a kinetic energy of one million electron volts 11 MeV2, equal to 1106211.602 * 10-19 J2 = 1.602 * 10-13 J. The Large Hadron Collider near Geneva, Switzerland, is designed to accelerate protons to a kinetic energy of 7 TeV 17 * 1012 eV2.

Example 3.3

Electric force and electric potential

A proton (charge + e = 1.602 * 10-19 C) moves a distance d = 0.50 m in a straight line between points a and b in a linear accelerator. The electric field is uniform along this line, with mag-

nitude E = 1.5 * 107 V>m = 1.5 * 107 N>C in the direction from a to b. Determine (a) the force on the proton; (b) the work done on it by the field; (c) the potential difference Va - Vb.

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3.2 Electric Potential S O L U T IO N IDENTIFY and SET UP: This problem uses the relationship between electric field and electric force. It also uses the relationship among force, work, and potential-energy difference. We are given the electric field, so it is straightforward to find the electric force on S the proton. Calculating the work is also straightforward because E is uniform, so the force on the proton is constant. Once the work is known, we find Va - Vb using Eq. (3.13). EXECUTE: (a) The force on the proton is in the same direction as the electric field, and its magnitude is F = qE = 11.602 * 10-19 C211.5 * 107 N>C2 = 2.4 * 10-12 N

(b) The force is constant and in the same direction as the displacement, so the work done on the proton is WaSb = Fd = 12.4 * 10 = 11.2 * 10-12 J2

-12

N210.50 m2 = 1.2 * 10

-12

J

1 eV 1.602 * 10-19 J

= 7.5 * 106 eV = 7.5 MeV

Example 3.4

81

(c) From Eq. (3.13) the potential difference is the work per unit charge, which is Va - Vb =

WaSb 1.2 * 10-12 J = q 1.602 * 10-19 C = 7.5 * 106 J>C = 7.5 * 106 V = 7.5 MV

We can get this same result even more easily by remembering that 1 electron volt equals 1 volt multiplied by the charge e. The work done is 7.5 * 106 eV and the charge is e, so the potential difference is 17.5 * 106 eV2>e = 7.5 * 106 V.

EVALUATE: We can check our result in part (c) by using Eq. (3.17) S or Eq. (3.18). The angle f between the constant field E and the displacement is zero, so Eq. (3.17) becomes b

Va - Vb =

2a

b

E cos f dl =

2a

b

E dl = E

2a

dl

The integral of dl from a to b is just the distance d, so we again find Va - Vb = Ed = 11.5 * 107 V>m210.50 m2 = 7.5 * 106 V

Potential due to two point charges

An electric dipole consists of point charges q1 = + 12 nC and q2 = -12 nC placed 10.0 cm apart (Fig. 3.13). Compute the electric potentials at points a, b, and c.

3.13 What are the potentials at points a, b, and c due to this electric dipole? c

S O L U T IO N IDENTIFY and SET UP: This is the same arrangement as in Example 1.8, in which we calculated the electric field at each point by doing a vector sum. Here our target variable is the electric potential V at three points, which we find by doing the algebraic sum in Eq. (3.15). EXECUTE: At point a we have r1 = 0.060 m and r2 = 0.040 m, so Eq. (3.15) becomes Va =

qi 1 1 q1 1 q2 = + r r 4pP0 a 4pP 4pP 0 1 0 r2 i i

= 19.0 * 109 N # m2>C22

12 * 10-9 C 0.060 m

1 - 12 * 10-9 C2 + 19.0 * 109 N # m2>C22 0.040 m

= 1800 N # m>C + 1- 2700 N # m>C2

= 1800 V + 1 - 2700 V2 = - 900 V

In a similar way you can show that the potential at point b (where r1 = 0.040 m and r2 = 0.140 m) is Vb = 1930 V and that the potential at point c (where r1 = r2 = 0.130 m) is Vc = 0.

13.0 cm

13.0 cm

b

a q1 4.0 cm

q2 6.0 cm

4.0 cm

is closer to the 112-nC charge than the 212-nC charge. Finally, point c is equidistant from the 112-nC charge and the 212-nC charge, so the potential there is zero. (The potential is also equal to zero at a point infinitely far from both charges.) Comparing this example with Example 21.8 shows that it’s much easier to calculate electric potential (a scalar) than electric field (a vector). We’ll take advantage of this simplification whenever possible.

EVALUATE: Let’s confirm that these results make sense. Point a is closer to the 212-nC charge than to the 112-nC charge, so the potential at a is negative. The potential is positive at point b, which

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82

CHAPTER 3 Electric Potential

Example 3.5

Potential and potential energy

Compute the potential energy associated with a + 4.0-nC point charge if it is placed at points a, b, and c in Fig. 3.13. SOLUTION IDENTIFY and SET UP: The potential energy U associated with a point charge q at a location where the electric potential is V is U = qV. We use the values of V from Example 3.4.

EVALUATE: Note that zero net work is done on the 4.0-nC charge if it moves from point c to infinity by any path. In particular, let the path be along the perpendicular bisector of the line joining the other two charges q1 and q2 in Fig. 3.13. As shown in Example 1.8 (SecS tion 1.5), at points on the bisector, the direction of E is perpendicular to the bisector. Hence the force on the 4.0-nC charge is perpendicular to the path, and no work is done in any displacement along it.

EXECUTE: At the three points we find Ua = qVa = 14.0 * 10 -9 C21 - 900 J>C2 = - 3.6 * 10 -6 J Ub = qVb = 14.0 * 10 -9 C211930 J>C2 = 7.7 * 10 -6 J Uc = qVc = 0

All of these values correspond to U and V being zero at infinity.

Example 3.6

Finding potential by integration

By integrating the electric field as in Eq. (3.17), find the potential at a distance r from a point charge q.

S

3.14 Calculating the potential by integrating E for a single point charge.

SOLUTION IDENTIFY and SET UP: We let point a in Eq. (3.17) be at distance r and let point b be at infinity (Fig. 3.14). As usual, we choose the potential to be zero at an infinite distance from the charge q. EXECUTE: To carry out the integral, we can choose any path we like between points a and b. The most convenient path is a radial S line as shown in Fig. 3.14, so that d l is in the radial direction and S has magnitude dr. Writing d l 5 rN dr , we have from Eq. (3.17) '

q

V - 0 = V = q

=

E # dl S

2r

S

q

q 1 q # Nr rN dr = dr 2r 4pP0 r2 2r 4pP0r2

q q q = b ` = 0 - a4pP0 r r 4pP0r q V = 4pP0r

Example 3.7

EVALUATE: Our result agrees with Eq. (3.14) and is correct for positive or negative q.

Moving through a potential difference

In Fig. 3.15 a dust particle with mass m = 5.0 * 10-9 kg = 5.0 mg and charge q0 = 2.0 nC starts from rest and moves in a straight line from point a to point b. What is its speed v at point b?

3.15 The particle moves from point a to point b; its acceleration is not constant. Particle 3.0 nC a 1.0 cm

SOLUTION IDENTIFY and SET UP: Only the conservative electric force acts on the particle, so mechanical energy is conserved: Ka + Ua = Kb + Ub. We get the potential energies U from the

b 1.0 cm

23.0 nC 1.0 cm

corresponding potentials V using Eq. (3.12): Ua = q0Va and Ub = q0Vb.

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3.3 Calculating Electric Potential EXECUTE: We have Ka = 0 and Kb = 12 mv2. We substitute these and our expressions for Ua and Ub into the energy-conservation equation, then solve for v. We find

Vb = 19.0 * 109 N # m2>C22a

2q01Va - Vb2 m We calculate the potentials using Eq. (3.15), V = q>4pP0r:

1-3.0 * 10-9 C2 b 0.010 m = - 1350 V Va - Vb = 11350 V2 - 1- 1350 V2 = 2700 V

Ä

Finally,

3.0 * 10-9 C Va = 19.0 * 109 N # m2>C22a 0.010 m +

3.0 * 10-9 C 0.020 m

+

0 + q0Va = 12mv2 + q0Vb v =

83

1- 3.0 * 10-9 C2 b 0.020 m

v =

Ä

212.0 * 10-9 C212700 V2 = 46 m>s 5.0 * 10-9 kg

EVALUATE: Our result makes sense: The positive test charge speeds up as it moves away from the positive charge and toward the negative charge. To check unit consistency in the final line of the calculation, note that 1 V = 1 J>C, so the numerator under the radical has units of J or kg # m2>s2.

= 1350 V

Test Your Understanding of Section 3.2 If the electric potential at a certain point is zero, does the electric field at that point have to be zero? (Hint: Consider point c in Example 3.4 and Example 1.8.) y

3.3

Calculating Electric Potential

When calculating the potential due to a charge distribution, we usually follow one of two routes. If we know the charge distribution, we can use Eq. (3.15) or (3.16). Or if we know how the electric field depends on position, we can use Eq. (3.17), defining the potential to be zero at some convenient place. Some problems require a combination of these approaches. As you read through these examples, compare them with the related examples of calculating electric field in Section 1.5. You’ll see how much easier it is to calculate scalar electric potentials than vector electric fields. The moral is clear: Whenever possible, solve problems using an energy approach (using electric potential and electric potential energy) rather than a dynamics approach (using electric fields and electric forces). Problem-Solving Strategy 3.1

Calculating Electric Potential

IDENTIFY the relevant concepts: Remember that electric potential is potential energy per unit charge. SET UP the problem using the following steps: 1. Make a drawing showing the locations and values of the charges (which may be point charges or a continuous distribution of charge) and your choice of coordinate axes. 2. Indicate on your drawing the position of the point at which you want to calculate the electric potential V. Sometimes this position will be an arbitrary one (say, a point a distance r from the center of a charged sphere). EXECUTE the solution as follows: 1. To find the potential due to a collection of point charges, use Eq. (3.15). If you are given a continuous charge distribution, devise a way to divide it into infinitesimal elements and use Eq. (3.16). Carry out the integration, using appropriate limits to include the entire charge distribution. 2. If you are given the electric field, or if you can find it using any of the methods presented in Chapter 21 or 22, it may be

easier to find the potential difference between points a and b using Eq. (3.17) or (3.18). When appropriate, make use of your freedom to define V to be zero at some convenient place, and choose this place to be point b. (For point charges, this will usually be at infinity. For other distributions of charge—especially those that themselves extend to infinity— it may be necessary to define Vb to be zero at some finite distance from the charge distribution.) Then the potential at any other point, say a, can by found from Eq. (3.17) or (3.18) with Vb = 0. 3. Although potential V is a scalar quantity, you may have to use S S components of the vectors E and d l when you use Eq. (3.17) or (3.18) to calculate V. EVALUATE your answer: Check whether your answer agrees with your intuition. If your result gives V as a function of position, graph the function to see whether it makes sense. If you know the electric field, you can make a rough check of your result Sfor V by verifying that V decreases if you move in the direction of E.

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CHAPTER 3 Electric Potential

Example 3.8

A charged conducting sphere

A solid conducting sphere of radius R has a total charge q. Find the electric potential everywhere, both outside and inside the sphere. SOLUTION IDENTIFY and SET UP: In Example 2.5 (Section 2.4) we used Gauss’s law to find the electric field at all points for this charge distribution. We can use that result to determine the corresponding potential. EXECUTE: From Example 2.5, the field outside the sphere is the same as if the sphere were removed and replaced by a point charge q. We take V = 0 at infinity, as we did for a point charge. Then the potential at a point outside the sphere at a distance r from its center is the same as that due to a point charge q at the center:

from the sphere. As you move away from the sphere, in the direcS tion of E, V decreases (as it should). 3.16 Electric-field magnitude E and potential V at points inside and outside a positively charged spherical conductor. ++ + + + ++ + +

84

++ + + + +R + + ++ E

E5

1 q 4pP0 R2 E5

1 q V = 4pP0 r

O

The potential at the surface of the sphere is Vsurface = q>4pP0R. S Inside the sphere, E is zero everywhere. Hence no work is done on a test charge that moves from any point to any other point inside the sphere. This means that the potential is the same at every point inside the sphere and is equal to its value q>4pP0R at the surface.

1 q 4pP0 r 2

E50

V

r

V5

1 q 4pP0 R V5

1 q 4pP0 r r

O

EVALUATE: Figure 3.16 shows the field and potential for a posi tive charge q. In this case the electric field points radially away

Ionization and Corona Discharge The results of Example 3.8 have numerous practical consequences. One consequence relates to the maximum potential to which a conductor in air can be raised. This potential is limited because air molecules become ionized, and air becomes a conductor, at an electric-field magnitude of about 3 * 106 V>m. Assume for the moment that q is positive. When we compare the expressions in Example 3.8 for the potential Vsurface and field magnitude Esurface at the surface of a charged conducting sphere, we note that Vsurface = Esurface R. Thus, if Em represents the electric-field magnitude at which air becomes conductive (known as the dielectric strength of air), then the maximum potential Vm to which a spherical conductor can be raised is '

Vm = REm For a conducting sphere 1 cm in radius in air, Vm = 110-2 m213 * 106 V>m2 = 30,000 V. No amount of “charging” could raise the potential of a conducting sphere of this size in air higher than about 30,000 V; attempting to raise the potential further by adding extra charge would cause the surrounding air to become ionized and conductive, and the extra added charge would leak into the air. To attain even higher potentials, high-voltage machines such as Van de Graaff generators use spherical terminals with very large radii (see Fig. 22.26 and the photograph that opens Chapter 2). For example, a terminal of radius R = 2 m has a maximum potential Vm = 12 m213 * 106 V>m2 = 6 * 106 V = 6 MV. Our result in Example 3.8 also explains what happens with a charged conductor with a very small radius of curvature, such as a sharp point or thin wire. Because the maximum potential is proportional to the radius, even relatively

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3.3 Calculating Electric Potential

small potentials applied to sharp points in air produce sufficiently high fields just outside the point to ionize the surrounding air, making it become a conductor. The resulting current and its associated glow (visible in a dark room) are called corona. Laser printers and photocopying machines use corona from fine wires to spray charge on the imaging drum (see Fig. 1.2). A large-radius conductor is used in situations where it’s important to prevent corona. An example is the metal ball at the end of a car radio antenna, which prevents the static that would be caused by corona. Another example is the blunt end of a metal lightning rod (Fig. 3.17). If there is an excess charge in the atmosphere, as happens during thunderstorms, a substantial charge of the opposite sign can build up on this blunt end. As a result, when the atmospheric charge is discharged through a lightning bolt, it tends to be attracted to the charged lightning rod rather than to other nearby structures that could be damaged. (A conducting wire connecting the lightning rod to the ground then allows the acquired charge to dissipate harmlessly.) A lightning rod with a sharp end would allow less charge buildup and hence would be less effective.

Example 3.9

85

3.17 The metal mast at the top of the Empire State Building acts as a lightning rod. It is struck by lightning as many as 500 times each year.

Oppositely charged parallel plates

Find the potential at any height y between the two oppositely charged parallel plates discussed in Section 3.1 (Fig. 3.18).

3.18 The charged parallel plates from Fig. 3.2. y

S O L U T IO N IDENTIFY and SET UP: We discussed this situation in Section 3.1. From Eq. (3.5), we know the electric potential energy U for a test charge q0 is U = q0Ey. (We set y 5 0 and U 5 0 at the bottom plate.) We use Eq. (3.12), U = q0V, to find the electric potential V as a function of y.

S

E

y b O

EXECUTE: The potential V1y2 at coordinate y is the potential energy per unit charge: V1y2 =

q0Ey U1y2 = = Ey q0 q0

a q0

S

Va - Vb Vab = d d

where Vab is the potential of the positive plate with respect to the negative plate. That is, the electric field equals the potential difference between the plates divided by the distance between them. For a given potential difference Vab, the smaller the distance d between the two plates, the greater the magnitude E of the electric field. (This relationship between E and Vab holds only for the planar geometry

Example 3.10

x

we have described. It does not work for situations such as concentric cylinders or spheres in which the electric field is not uniform.)

The potential decreases as we move in the direction of E from the upper to the lower plate. At point a, where y = d and V1y2 = Va, Va - Vb = Ed and E =

d

?

EVALUATE: Our result shows that V = 0 at the bottom plate (at y = 0). This is consistent with our choice that U = q0V = 0 for a test charge placed at the bottom plate. CAUTION “Zero potential” is arbitrary You might think that if a conducting body has zero potential, it must necessarily also have zero net charge. But that just isn’t so! As an example, the plate at y = 0 in Fig. 3.18 has zero potential 1V = 02 but has a nonzero charge per unit area -s. There’s nothing particularly special about the place where potential is zero; we can define this place to be wherever we want it to be. y

An infinite line charge or charged conducting cylinder

Find the potential at a distance r from a very long line of charge with linear charge density (charge per unit length) l. S O L U T IO N IDENTIFY and SET UP: In both Example 1.10 (Section 1.5) and Example 2.6 (Section 2.4) we found that the electric field at a

radial distance r from a long straight-line charge (Fig. 3.19a) has only a radial component given by Er = l>2pP0r. We use S this expression to find the potential by integrating E as in Eq. (3.17). EXECUTE: Since the field has only a radial component, we have S S E # d l = Er dr. Hence from Eq. (3.17) the potential of any point a Continued

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86

CHAPTER 3 Electric Potential

with respect to any other point b, at radial distances ra and rb from the line of charge, is b

Va - Vb =

E # dl = S

2a

b

S

2a

Er dr =

rb rb l dr l = ln 2pP0 2ra r 2pP0 ra

If we take point b at infinity and set Vb = 0, we find that Va is infinite for any finite distance ra from the line charge: Va = 1l>2pP02 ln 1 q >ra2 = q . This is not a useful way to define V for this problem! The difficulty is that the charge distribution itself extends to infinity. Instead, as recommended in Problem-Solving Strategy 3.1, we set Vb = 0 at point b at an arbitrary but finite radial distance r0. Then the potential V = Va at point a at a radial distance r is given by V - 0 = 1l>2pP02 ln 1r0>r2, or V =

r0 l ln 2pP0 r

EVALUATE: According to our result, if l is positive, then V decreases as r increases. SThis is as it should be: V decreases as we move in the direction of E. From Example 2.6, the expression for Er with which we started also applies outside a long, charged conducting cylinder with charge per unit length l (Fig. 3.19b). Hence our result also gives the potential for such a cylinder, but only for values

Example 3.11

3.19 Electric field outside (a) a long, positively charged wire and (b) a long, positively charged cylinder. (b)

(a) Er

Er

r

+

+ R

r

+ +

+ + + + + + + + + + + +

+++++ ++++++++

+

of r (the distance from the cylinder axis) equal to or greater than the radius R of the cylinder. If we choose r0 to be the cylinder radius R, so that V = 0 when r = R, then at any point for which r 7 R, V =

l R ln 2pP0 r

S

Inside the cylinder, E 5 0, and V has the same value (zero) as on the cylinder’s surface.

A ring of charge

Electric charge Q is distributed uniformly around a thin ring of radius a (Fig. 3.20). Find the potential at a point P on the ring axis at a distance x from the center of the ring.

3.20 All the charge in a ring of charge Q is the same distance r from a point P on the ring axis. r5

SOLUTION

!x 2

a

IDENTIFY and SET UP: We divide the ring into infinitesimal segments and use Eq. (3.16) to find V. All parts of the ring (and therefore all elements of the charge distribution) are at the same distance from P. EXECUTE: Figure 3.20 shows that the distance from each charge element dq to P is r = 1x2 + a2. Hence we can take the factor 1>r outside the integral in Eq. (3.16), and dq Q 1 1 1 1 V = dq = = 4pP0 2 r 4pP0 2x2 + a2 2 4pP0 2x2 + a2 EVALUATE: When x is much larger than a, our expression for V becomes approximately V = Q>4pP0x, which is the potential at a distance x from a point charge Q. Very far away from a charged

Example 3.12

+

O

1

x

a2

P

Q

ring, its electric potential looks like that of a point charge. We drew a similar conclusion about the electric field of a ring in Example 1.9 (Section 1.5). We know the electric field at all points along the x-axis from Example 1.9S (Section 1.5), so we can also find V along this axis by S integrating E # d l as in Eq. (3.17).

Potential of a line of charge

Positive electric charge Q is distributed uniformly along a line of length 2a lying along the y-axis between y = - a and y = +a (Fig. 3.21). Find the electric potential at a point P on the x-axis at a distance x from the origin.

SOLUTION IDENTIFY and SET UP: This is the same situation as in Example 1.10 (Section 1.5), where we found an expression for the electric

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3.4 Equipotential Surfaces S

field E at an arbitrary point on the x-axis. We can find V at point P by integrating over the charge distribution using Eq. (3.16). Unlike the situation in Example 3.11, each charge element dQ is a different distance from point P, so the integration will take a little more effort.

87

3.21 Our sketch for this problem.

EXECUTE: As in Example 21.10, the element of charge dQ corresponding to an element of length dy on the rod is dQ = 1Q>2a2dy. The distance from dQ to P is 1x2 + y2, so the contribution dV that the charge element makes to the potential at P is dV =

dy 1 Q 4pP0 2a 2x2 + y2

To find the potential at P due to the entire rod, we integrate dV over the length of the rod from y = -a to y = a: V =

a 1 Q 4pP0 2a 2-a

dy

2x

2

+ y2

You can look up the integral in a table. The final result is V =

2a2 + x2 + a 1 Q ln a b 4pP0 2a 2a2 + x2 - a

EVALUATE: We can check our result by letting x approach infinity. In this limit the point P is infinitely far from all of the charge, so we expect V to approach zero; you can verify that it does. We know the electric field at all points along the x-axis from Example 21.10. We invite you to use this information to find V S along this axis by integrating E as in Eq. (3.17).

Test Your Understanding of Section 3.3 If the electric field at a certain point is zero, does the electric potential at that point have to be zero? (Hint: Consider the center of the ring in Example 3.11 and Example 1.9.) y

3.4

Equipotential Surfaces

Field lines (see Section 1.6) help us visualize electric fields. In a similar way, the potential at various points in an electric field can be represented graphically by equipotential surfaces. These use the same fundamental idea as topographic maps like those used by hikers and mountain climbers (Fig. 3.22). On a topographic map, contour lines are drawn through points that are all at the same elevation. Any number of these could be drawn, but typically only a few contour lines are shown at equal spacings of elevation. If a mass m is moved over the terrain along such a contour line, the gravitational potential energy mgy does not change because the elevation y is constant. Thus contour lines on a topographic map are really curves of constant gravitational potential energy. Contour lines are close together where the terrain is steep and there are large changes in elevation over a small horizontal distance; the contour lines are farther apart where the terrain is gently sloping. A ball allowed to roll downhill will experience the greatest downhill gravitational force where contour lines are closest together. By analogy to contour lines on a topographic map, an equipotential surface is a three-dimensional surface on which the electric potential V is the same at every point. If a test charge q0 is moved from point to point on such a surface, the electric potential energy q0 V remains constant. In a region where an electric field is present, we can construct an equipotential surface through any point. In diagrams we usu ally show only a few representative equipotentials, often with equal potential dif ferences between adjacent surfaces. No point can be at two different potentials, so equipotential surfaces for different potentials can never touch or intersect.

3.22 Contour lines on a topographic map are curves of constant elevation and hence of constant gravitational potential energy.

'

Equipotential Surfaces and Field Lines Because potential energy does not change as a test charge moves over an equipoS tential surface, the electric field can do no work on such a charge. It follows that E S must be perpendicular to the surface at every point so that the electric force q0 E is always perpendicular to the displacement of a charge moving on the surface. '

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88

CHAPTER 3 Electric Potential

3.23 Cross sections of equipotential surfaces (blue lines) and electric field lines (red lines) for assemblies of point charges. There are equal potential differences between adjacent surfaces. Compare these diagrams to those in Fig. 1.28, which showed only the electric field lines. (a) A single positive charge

(b) An electric dipole

+

–

(c) Two equal positive charges

+

+

+

V 5 230 V V 5 130 V V 5 150 V V 5 170 V

V 5 130 V V50V V 5 250 V V 5 150 V V 5 270 V V 5 170 V Electric field lines

V 5 130 V

V 5 150 V

V 5 170 V

Cross sections of equipotential surfaces

Field lines and equipotential surfaces are always mutually perpendicular. In general, field lines are curves, and equipotentials are curved surfaces. For the special case of a uniform field, in which the field lines are straight, parallel, and equally spaced, the equipotentials are parallel planes perpendicular to the field lines. Figure 3.23 shows three arrangements of charges. The field lines in the plane of the charges are represented by red lines, and the intersections of the equipotential surfaces with this plane (that is, cross sections of these surfaces) are shown as blue lines. The actual equipotential surfaces are three-dimensional. At each crossing of an equipotential and a field line, the two are perpendicular. In Fig. 3.23 we have drawn equipotentials so that there are equal potential S differences between adjacent surfaces. In regions where the magnitude of E is large, the equipotential surfaces are close together because the field does a relatively large amount of work on a test charge in a relatively small displacement. This is the case near the point charge in Fig. 3.23a or between the two point charges in Fig. 3.23b; note that in these regions the field lines are also closer together. This is directly analogous to the downhill force of gravity being greatest in regions on a topographic map where contour lines are close together. Conversely, in regions where the field is weaker, the equipotential surfaces are farther apart; this happens at larger radii in Fig. 3.23a, to the left of the negative charge or the right of the positive charge in Fig. 3.23b, and at greater distances from both charges in Fig. 3.23c. (It may appear that two equipotential surfaces intersect at the center of Fig. 3.23c, in violation of the rule that this can never happen. In fact this is a single figure-8–shaped equipotential surface.) CAUTION E need not be constant over an equipotential surface On a given equipotential surface, the potential V has the same value at every point. In general, however, the electricfield magnitude E is not the same at all points on an equipotential surface. For instance, on the equipotential surface labeled “V = - 30 V” in Fig. 3.23b, the magnitude E is less to the left of the negative charge than it is between the two charges. On the figure-8–shaped equipotential surface in Fig. 3.23c, E = 0 at the middle point halfway between the two charges; at any other point on this surface, E is nonzero. y

Equipotentials and Conductors Here’s an important statement about equipotential surfaces: When all charges are at rest, the surface of a conductor is always an equipotential surface.

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3.4 Equipotential Surfaces S

Since the electric field E is always perpendicular to an equipotential surface, we can prove this statement by proving that when all charges are at rest, the elec tric field just outside a conductor mustSbe perpendicular to the surface at every point (Fig. 3.24). We know that E 5 0 everywhere inside the conduc tor; otherwise, charges would move. In particular, at any point just inside the S surface the component ofS E tangent to the surface is zero. It follows that the tangential component of E is also zero just outside the surface. If it were not, a charge could move around a rectangular path partly inside and partly outside (Fig. 3.25) and return to its starting point with a net amount of work having been done on it. This would violateSthe conservative nature of electrostatic fields, so the tangential component of SE just outside the surface must be zero at every point on the surface. Thus E is perpendicular to the surface at each point, proving our statement. It also follows that when all charges are at rest, the entire solid volume of a conductor is at the same potential. Equation (3.17) states that the potential difference between two points a and b within the conductor’s solid volume, S bS Va - Vb, is equal to the line integral 1a E d l of the electric field from a to b. S Since E 5 0 everywhere inside the conductor, the integral is guaranteed to be zero for any two such points a and b. Hence the potential is the same for any two points within the solid volume of the conductor. We describe this by saying that the solid volume of the conductor is an equipotential volume. Finally, we can now prove a theorem that we quoted without proof in Section 2.5. The theorem is as follows: In an electrostatic situation, if a conductor contains a cavity and if no charge is present inside the cavity, then there can be no net charge anywhere on the surface of the cavity. This means that if you’re inside a charged conducting box, you can safely touch any point on the inside walls of the box without being shocked. To prove this theorem, we first prove that every point in the cavity is at the same potential. In Fig. 3.26 the conducting surface A of the cavity is an equipotential surface, as we have just proved. Suppose point P in the cavity is at a different potential; then we can construct a different equipotential surface B including point P. Now consider a Gaussian surface, shown in Fig. 3.26, between the two S equipotential surfaces. Because of the relationship between E and the equipotentials, we know that the field at every point between the equipotentials is from A toward B, or else at every point it is from B toward A, depending on which equipotential surface is at higher potential. In either case the flux through this Gaussian surface is certainly not zero. But then Gauss’s law says that the charge enclosed by the Gaussian surface cannot be zero. This contradicts our initial assumption that there is no charge in the cavity. So the potential at P cannot be different from that at the cavity wall. The entire region of the cavity must therefore be at the same potential. But for this to be true, the electric field inside the cavity must be zero everywhere. Finally, Gauss’s law shows that the electric field at any point on the surface of a conductor is proportional to the surface charge density s at that point. We conclude that the surface charge density on the wall of the cavity is zero at every point. This chain of reasoning may seem tortuous, but it is worth careful study.

89

3.24 When charges are at rest, a conducting surface is always an equipotential surface. Field lines are perpendicular to a conducting surface. –

– – + + ++ + + + + ++

– – – – –

#

–

S

E

Cross sections of equipotential surfaces Electric field lines

3.25 At all points on the surface of a conductor, the electric field must be perS pendicular to the surface. If E had a tangential component, a net amount of work would be done on a test charge by moving it around a loop as shown here—which is impossible because the electric force is conservative. An impossible electric field If the electric field just outside a conductor had a tangential component Ei, a charge could move in a loop with net work done. S

E Vacuum Ei

E'

S

Conductor

E50

3.26 A cavity in a conductor. If the cavity contains no charge, every point in the cavity is at the same potential, the electric field is zero everywhere in the cavity, and there is no charge anywhere on the surface of the cavity. Cross section of equipotential surface through P

CAUTION Equipotential surfaces vs. Gaussian surfaces Don’t confuse equipotential surfaces with the Gaussian surfaces we encountered in Chapter 2. Gaussian surfaces have relevance only when we are using Gauss’s law, and we can choose any Gaussian surface that’s convenient. We are not free to choose the shape of equipotential surfaces; the shape is determined by the charge distribution. y

Gaussian surface (in cross section) B Surface of cavity

P A

Test Your Understanding of Section 3.4 Would the shapes of the equipotential surfaces in Fig. 3.23 change if the sign of each charge were reversed? y

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Conductor

90

CHAPTER 3 Electric Potential

3.5 ActivPhysics 11.12.3: Electrical Potential, Field, and Force

Potential Gradient

Electric field and potential are closely related. Equation (3.17), restated here, expresses one aspect of that relationship: b S

V a - Vb =

#

S

E dl

2a

S

If we know E at various points, we can use this equation to calculate potential differences. In this section we show how to turn thisS around; if we know the potential V at various points, we can use it to determine E. Regarding V as a function of S the coordinates 1x, y, z2 of a point in space, we will show that the components of E are related to the partial derivatives of V with respect to x, y, and z. In Eq. (3.17), Va - Vb is the potential of a with respect to b—that is, the change of potential encountered on a trip from b to a. We can write this as a

Va - Vb =

2b

b

dV = -

2a

dV

where dV Sis the infinitesimal change of potential accompanying an infinitesimal element d l of the path from b to a. Comparing to Eq. (3.17), we have b

-

2a

b S

dV =

#

S

E dl

2a

These two integrals must be equal for any pair of limits a and b, and for this to be S true the integrands must be equal. Thus, for any infinitesimal displacement d l , S

#

S

-dV = E d l S

S

To interpret this expression, we write E and d l in terms of their components: S S E 5 ≤N Ex 1 ≥N Ey 1 kN Ez and d l 5 ≤N dx 1 ≥N dy 1 kN dz. Then we have -dV = Ex dx + Ey dy + Ez dz Suppose the displacement is parallel to the x-axis, so dy = dz = 0. Then -dV = Ex dx or Ex = - 1dV>dx2y, z constant , where the subscript reminds us that only x varies in the derivative; recall that V is in general a function of x, y, and z. But this is just what is meant by the partial derivative 0V>0x. The y- and z-components of S E are related to the corresponding derivatives of V in the same way, so we have '

Potential Gradient Across a Cell Membrane

Application

The interior of a human cell is at a lower electric potential V than the exterior. (The potential difference when the cell is inactive is about –70 mV in neurons and about –95 mV in skele tal muscle cells.) Hence there is a potential S gradient §V that points from the interior to the exterior of the cell membrane, and an electric S S field E 5 - §V that points from the exterior to the interior. This field affects how ions flow into or out of the cell through special channels in the membrane.

Ex = -

0V 0x

Ey = -

0V 0y

Ez = -

S

0V 0z

(components of E in terms of V)

(3.19)

This is consistent S with the units of electric field being V>m. In terms of unit vectors we can write E as E 5 2 a≤N S

0V 0V 0V 1 ≥N 1 kN b 0x 0y 0z

1E in terms of V2 S

(3.20)

In vector notation the following operation is called the gradient of the function ƒ: §f 5 a≤N S

0 0 0 1 ≥N 1 kN bƒ 0x 0y 0z

(3.21)

S

The operator denoted by the symbol § is called “grad” or “del.” Thus in vector notation, S

S

E 5 2 §V S

(3.22) S

This is read “E isS the negative of the gradient of V” or “E equals negative grad V.” The quantity §V is called the potential gradient.

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91

3.5 Potential Gradient

At each point, the potential gradient points in the direction in which V increases most rapidly with a change in position. Hence at each point the direcS tion of E is the direction in which V decreases most rapidly and is always perpendicular to the equipotential surface through the point. This agrees with our observation in Section 3.2 that moving in the direction of the electric field means moving in the direction of decreasing potential. Equation (3.22) doesn’t depend on the particular choice of the zero point for V. If we were to change the zero point, the effect would be to change V at every point Sby the same amount; the derivatives of V would be the same. If E is radial with respect to a point or an axis and r is the distance from the point or the axis, the relationship corresponding to Eqs. (3.19) is Er = -

0V 0r

1radial electric field2

(3.23)

Often we can compute the electric fieldScaused by a charge distribution in either of two ways: directly, by adding the E fields of point charges, or by first calculating the potential and then taking its gradient to find the field. The second method is often easier because potential is a scalar quantity, requiring at worst the integration of a scalar function. Electric field is a vector quantity, requiring computation of components for each element of charge and a separate integration for each component. Thus, quite apart from its fundamental significance, potential offers a very useful computational technique in field calculations. Below, we present two examples in which a knowledge of V is used to find the electric field. S We stress once more that if we know E as a function of position, we can calculate V using SEq. (3.17) or (3.18), and if we know V as a function of position, we S can calculate E using Eq.S (3.19), (3.20), or (3.23). Deriving V from E requires integration, and deriving E from V requires differentiation.

Example 3.13

Potential and field of a point charge

From Eq. (3.14) the potential at a radial distance r from a point charge q is V = q>4pP0r. Find the vector electric field from this expression for V. S O L U T IO N IDENTIFY and SET UP: This problem uses the general relationship between the electric potential as a function of position and the electric-field vector. By symmetry, the electric field here has only a radial component Er. We use Eq. (3.23) to find this component. 0V 0 1 q 1 q = - a Er = b = 0r 0r 4pP0 r 4pP0 r2 so the vector electric field is S

and similarly qy 0V = 0y 4pP0 r3

qz 0V = 0z 4pP0r3

Then from Eq. (3.20),

EXECUTE: From Eq. (3.23),

E 5 rN Er 5

q qx 0V 0 1 1 = a b =2 2 2 2 0x 0x 4pP0 2x + y + z 4pP0 1x + y2 + z223>2 qx = 4pP0r3

E 5 2 c≤N a S

5 1 q rN 4pP0 r2

EVALUATE: Our result agrees with Eq. (1.7), as it must. An alternative approach is to ignore the radial symmetry, write the radial distance as r = 2x2 + y2 + z2, and take the derivatives of V with respect to x, y, and z as in Eq. (3.20). We find

qx 4pP0r

3

b 1 ≥N a -

qy 4pP0r

3

b 1 kN a -

qz 4pP0r3

bd

1 q 1 q x≤N 1 y≥N 1 zkN Nr b 5 a r 4pP0 r2 4pP0 r2

This approach gives us the same answer, but with more effort. Clearly it’s best to exploit the symmetry of the charge distribution whenever possible.

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92

CHAPTER 3 Electric Potential

Example 3.14

Potential and field of a ring of charge

In Example 3.11 (Section 3.3) we found that for a ring of charge with radius a and total charge Q, the potential at a point P on the ring’s symmetry axis a distance x from the center is V =

Q 1 4pP0 2x2 + a2

EXECUTE: The x-component of the electric field is Ex = -

Qx 0V 1 = 2 0x 4pP0 1x + a223>2

EVALUATE: This agrees with our result in Example 21.9.

Find the electric field at P. SOLUTION IDENTIFY and SET UP: Figure 3.20 shows the situation. We are given V as a function of x along the x-axis, and we wish to find the electric field at a point on this axis. From the symmetry of the charge distribution, the electric field along the symmetry (x-) axis of the ring can have only an x-component. We find it using the first of Eqs. (3.19).

CAUTION Don’t use expressions where they don’t apply In this example, V is not a function of y or z on the ring axis, so that 0V>0y = 0V>0z = 0 and Ey = Ez = 0. But that does not mean that it’s true everywhere; our expressions for V and Ex are valid only on the ring axis. If we had an expression for V valid at all S points in space, we could use it to find the components of E at any point using Eqs. (3.19). y

Test Your Understanding of Section 3.5 In a certain region of space the potential is given by V = A + Bx + Cy3 + Dxy, where A, B, C, and D are S positive constants. Which of these statements about the electric field E in this region of space is correct? (There may beS more than one correct answer.) (i) Increasing the value of A will increase the value of ESat all points; (ii) increasing the value of A will S decrease the value of E at all points; (iii) E has no z-component; (iv) the electric field is zero at the origin 1x = 0, y = 0, z = 02. y

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CHAPTER

3

SUMMARY

Electric potential energy: The electric force caused by any collection of charges at rest is a conservative force. The work W done by the electric force on a charged particle moving in an electric field can be represented by the change in a potential-energy function U. The electric potential energy for two point charges q and q0 depends on their separation r. The electric potential energy for a charge q0 in the presence of a collection of charges q1 , q2 , q3 depends on the distance from q0 to each of these other charges. (See Examples 3.1 and 3.2.)

WaSb = Ua - Ub 1 qq0 U = 4pP0 r (two point charges)

Electric potential: Potential, denoted by V, is potential energy per unit charge. The potential difference between two points equals the amount of work that would be required to move a unit positive test charge between those points. The potential V due to a quantity of charge can be calculated by summing (if the charge is a collection of point charges) or by integrating (if the charge is a distribution). (See Examples 3.3, 3.4, 3.5, 3.7, 3.11, and 3.12.) The potential difference between two points a and b, also called the potential of a with respect to b, is given S by the line integral of E. The Spotential at a given point can be found by first finding E and then carrying out this integral. (See Examples 3.6, 3.8, 3.9, and 3.10.)

U 1 q = q0 4pP0 r (due to a point charge)

'

'

(3.2)

U5

(3.9)

1

q0 q 1 q 2 q3 1 1 4 pP0 r1 r2 r3

(3.14)

qi 1 U = (3.15) q0 4pP0 a i ri (due to a collection of point charges)

q2

r2

'

V =

2

r1

q0 q1 q3 q2 a + + + Áb r2 r3 4pP0 r1 q0 qi = (3.10) 4pP0 a i ri (q0 in presence of other point charges) U =

q3

r3 q0

q1

V5

1

q1 q2 q3 1 1 1 4 pP0 r1 r2 r3

2

q2

'

V =

b

Va - Vb =

S

2a

#

r1 r2

q3

r3

dq 1 4pP0 2 r (due to a charge distribution) V =

(3.16)

P

b S

E dl =

2a

E cos f dl (3.17)

Equipotential surfaces: An equipotential surface is a surface on which the potential has the same value at every point. At a point where a field line crosses an equipotential surface, the two are perpendicular. When all charges are at rest, the surface of a conductor is always an equipotential surface and all points in the interior of a conductor are at the same potential. When a cavity within a conductor contains no charge, the entire cavity is an equipotential region and there is no surface charge anywhere on the surface of the cavity.

Finding electric field from electric potential: If the potential V is known as a function of theScoordinates x, y, and z, the components of electric field E at any point are given by partial derivatives of V. (See Examples 3.13 and 3.14.)

q1

Ex = -

Electric field line

–

Cross section of equipotential surface

+

0V 0V 0V Ey = Ez = 0x 0y 0z (3.19)

E 5 2 a≤N

0V 0V 0V 1 ≥N 1 kN b (3.20) 0x 0y 0z (vector form) S

93

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94

CHAPTER 3 Electric Potential

BRIDGING PROBLEM

A Point Charge and a Line of Charge

Positive electric charge Q is distributed uniformly along a thin rod of length 2a. The rod lies along the x-axis between x 5 2a and x 5 1a. Calculate how much work you must do to bring a positive point charge q from infinity to the point x 5 1L on the x-axis, where L . a. SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. In this problem you must first calculate the potential V at x 5 1L due to the charged rod. You can then find the change in potential energy involved in bringing the point charge q from infinity (where V = 0) to x 5 1L. 2. To find V, divide the rod into infinitesimal segments of length dx¿ . How much charge is on such a segment? Consider one such segment located at x = x¿ , where - a … x¿ … a. What is the potential dV at x 5 1L due to this segment?

Problems

3. The total potential at x 5 1L is the integral of dV, including contributions from all of the segments for x¿ from 2a to 1a. Set up this integral. EXECUTE 4. Integrate your expression from step 3 to find the potential V at x 5 1L. A simple, standard substitution will do the trick; use a table of integrals only as a last resort. 5. Use your result from step 4 to find the potential energy for a point charge q placed at x 5 1L. 6. Use your result from step 5 to find the work you must do to bring the point charge from infinity to x 5 1L. EVALUATE 7. What does your result from step 5 become in the limit a S 0? Does this make sense? 8. Suppose the point charge q were negative rather than positive. How would this affect your result in step 4? In step 5?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems. DISCUSSION QUESTIONS Q3.1 A student asked, “Since electrical potential is always proportional to potential energy, why bother with the concept of potential at all?” How would you respond? Q3.2 The potential (relative to a point at infinity) midway between two charges of equal magnitude and opposite sign is zero. Is it possible to bring a test charge from infinity to this midpoint in such a way that no work is done in any part of the displacement? If so, describe how it can be done. If it is not possible, explain why. Q3.3 Is it possible to have an arrangement of two point charges separated by a finite distance such that the electric potential energy of the arrangement is the same as if the two charges were infinitely far apart? Why or why not? What if there are three charges? Explain your reasoning. Q3.4 Since potential can have any value you want depending on the choice of the reference level of zero potential, how does a voltmeter know what to read when you connect it between two points? S Q3.5 If E is zero everywhere along a certain path that leads from point A to point B, what is the potential difference between those S two points? Does this mean that E is zero everywhere along any path from A to B? Explain. S Q3.6 If E is zero throughout a cer- Figure Q23.7 tain region of space, is the potential S necessarily also zero in this region? dl Why or why not? If not, what can be said about the potential? S Q3.7 If you carry out the integral of E S S the electric field 1 E # d l for a closed path like that shown in Fig. Q3.7, the integral will always be equal to zero, independent of the shape of the path

and independent of where charges may be located relative to the path. Explain why. Q3.8 The potential difference between the two terminals of an AA battery (used in flashlights and portable stereos) is 1.5 V. If two AA batteries are placed end to end with the positive terminal of one battery touching the negative terminal of the other, what is the potential difference between the terminals at the exposed ends of the combination? What if the two positive terminals are touching each other? Explain your reasoning. Q3.9 It is easy to produce a potential difference of several thousand volts between your body and the floor by scuffing your shoes across a nylon carpet. When you touch a metal doorknob, you get a mild shock. Yet contact with a power line of comparable voltage would probably be fatal. Why is there a difference? S Q3.10 If the electric potential at a single point is known, can E at that point be determined? If so, how? If not, why not? Q3.11 Because electric field lines and equipotential surfaces are always perpendicular, two equipotential surfaces can never cross; S if they did, the direction of E would be ambiguous at the crossing points. Yet two equipotential surfaces appear to cross at the center of Fig. S3.23c. Explain why there is no ambiguity about the direction of E in this particular case. Q3.12 A uniform electric field is directed due east. Point B is 2.00 m west of point A, point C is 2.00 m east of point A, and point D is 2.00 m south of A. For each point, B, C, and D, is the potential at that point larger, smaller, or the same as at point A ? Give the reasoning behind your answers. Q3.13 We often say that if point A is at a higher potential than point B, A is at positive potential and B is at negative potential. Does it necessarily follow that a point at positive potential is positively charged, or that a point at negative potential is negatively charged? Illustrate your answers with clear, simple examples.

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'

Single Correct

Q3.14 A conducting sphere is to be charged by bringing in positive charge a little at a time until the total charge is Q. The total work required for this process is alleged to be proportional to Q2. Is this correct? Why or why not? Q3.15 Three pairs of parallel Figure Q3.15 metal plates (A, B, and C) are A connected as shown in Fig. Q23.15, and a battery maintains a B a b potential of 1.5 V across ab. What can you say about the potential difference across each C pair of plates? Why? Q3.16 A conducting sphere is placed between two charged parallel plates such as those shown in Fig. 23.2. Does the electric field inside the sphere depend on precisely where between the plates the sphere is placed? What about the electric potential inside the sphere? Do the answers to these questions depend on whether or not there is a net charge on the sphere? Explain your reasoning. Q3.17 A conductor that carries a net charge Q has a hollow, empty cavity in its interior. Does the potential vary from point to point within the material of the conductor? What about within the cavity? How does the potential inside the cavity compare to the potential within the material of the conductor? Q3.18 A high-voltage dc power line falls on a car, so the entire metal body of the car is at a potential of 10,000 V with respect to the ground. What happens to the occupants (a) when they are sitting in the car and (b) when they step out of the car? Explain your reasoning. Q3.19 In electronics it is customary to define the potential of ground (thinking of the earth as a large conductor) as zero. Is this consistent with the fact that the earth has a net electric charge that is not zero?

95

Q3.4 The electric field intensity at all points in space is given S E = 23 i - j volts/metre. The nature of equipotential lines in x - y plane is given by (a) High potential (b) Low potential y

y

30˚

30˚

x

x

High potential

Low potential

(c) High potential

(d) Low potential

y

30˚

y

60˚

x

Low potential

x

High potential

Q3.5 In an electrical field shown in figure three equipotential surfaces are shown. If function of electric field is E = 2x2V/m, and given that V1 - V2 = V2 - V3, then we have

y

E x

(a) x1 = x2 x1 x2 V3 V1 V2 (b) x1 > x2 z (c) x2 > x1 (d) data insufficient Q3.6 Suppose a region of space has a uniform electric field, directed towards the right, as shown below. Which statement is true? E A

SINGLE CORRECT

C

Q3.1 Two concentric spherical shells are given conducting positive charges : (a) outer shell will be at higher potential (b) inner shell will be at higher potential (c) outer will always be at higher potential irrespective of the sign of charges given to two spheres. (d) no prediction can be made using given data. Q3.2 Consider the two large oppositely A charged plates as shown in the diagram. At which of the marked points shown in the B diagram would a negatively charged particle C have the greatest electrical potential energy? +++++++++++ (a) A (b) B (c) C (d) D D

Q3.3 Two charges - q and + q are on points A and B as shown. Which graph shows the correct distribution of potential V(x) due to two point charges on AB? (a)

–q A

B

+q

(b) V(x)

X

(c) V(x)

X

V(x)

distance from the charge on left

distance from the charge on left

(d) X distance from the charge on left

V(x)

X distance from the charge on left

B

(a) The potential at all three location is the same (b) The potential at points A and B are equal, the potential at point C is higher than the potential at point A (c) The potential at points A and B are equal, and the potential at point C is lower than the potential at point A (d) The potential at point A is the highest, the potential at point B is the second highest, and the potential at point C is the lowest. Q3.7 The electrostatic potential is measured to be v(x,y,z) = 4 |x| + v0, where v0 is constant. The charge distribution responsible for this potential is (a) a uniformly charged thread in the xy plane. (b) a point charge at origin. (c) a uniformly charged sheet in the y−z plane. (d) a uniformly charged sphere of radius 1/p at the origin. Q3.8 In a certain region of space, the electric field is zero. From this we can conclude that the electric potential in this region is: (a) constant (b) zero (c) positive (d) negative Q3.9 Statement-1 : If electric potential while moving in a certain path is constant, then the electric field must be zero. ∂V Statement-2 : Component of electric field Er = . ∂r (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for ostatement-1.

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96

CHAPTER 3 Electric Potential

(b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true. Q3.10 Statement-1 : The electric potential +q −q and the electric field intensity at the centre of a square having four fixed point charges at their vertices as shown in figure are zero. +q Statement-2 : If electric potential at a point −q is zero then the magnitude of electric field at that point must be zero. (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (b) Statement-2 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true. Q3.11 Consider the arrangement of two Q concentric conducting shells with radii C P S a a & b respectively as shown in the diagram. b A The inner shell carries a charge Q. The outer shell is grounded (a) potential at C equals zero (b) potential at S equals zero (c) potential at P equals zero (d) none Q3.12 On a conductor of nonuniform curvature, the charge (a) has the greatest concentration on the part of greatest radius of curvature. (b) has the greatest concentration on the part of least radius of curvature. (c) is distributed uniformly on the whole surface. (d) is distributed uniformly over its volume. Q3.13 When two charged concentric spherical conductors have electric potential V1 and V2 respectively. Statement-1 : The potential at centre is V1 + V2. Statement-2 : Potential is scalar quantity. (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true.

Paragraph Questions Two very large parallel disks of charge have their centres on the x-axis and their planes perpendicular to the x-axis. The disk that intersects x = - R has uniform positive surface charge density +s; the disk that intersects x = + R has uniform negative surface charge density −s. (i)

(ii) O

O

(iii)

(iv) O

O

Q3.14 Which graph best represents the plot of the x-component of the electric field vector on the x-axis? (a) (i) (b) (ii) (c) (iii) (d) (iv)

Q3.15 Which graph best represent the plot of the electric potential (V) as a function of x (treating V = 0 at x = 0) ? (a) (i) (b) (ii) (c) (iii) (d) (iv)

ONE OR MORE CORRECT Q3.16 Particle A having positive charge is moving directly (headon) towards initially stationary positively charged particle B. At the instant when A and B are closest together. (a) the momenta of A and B must be equal (b) the velocities of A and B must be equal (c) B would have gained less kinetic energy than A would have lost. (d) B would have gained the same momentum as A would have lost a1 Q3.17 A wire having a uniform linear charge density l , is bent in the form of radius R. Point r1 A as shown in the figure, is in the plane of the A ring but not at the centre. Two elements of the r2 ring of lengths a1 and a2 subtend very small and equal angles at the point A. They are at distances a2 r1 and r2 from the point A respectively. (a) The ratio of charge on elements a1 and a2 is r1/r2. (b) The element a1 produces greater magnitude of electric field at A than element a2. (c) The elements a1 and a2 produce same potential at A (d) The element a1 produces greater potential at A than element a2 Q3.18 A positive charge Q is located at the centre O of a thin metallic neutral spherical shell. Select the correct statement(s) from the following : (a) The electric field at any point outside the shell is zero. (b) The charge on the outer surface of the shell is - Q. (c) The outer surface of the spherical shell is an equipotential surface. (d) The electric field at any point inside the shell, other than O, is zero. Q3.19 Which of the following statements about a charged conductor in electrostatic equilibrium is/are true? (a) The electric field just outside the surface is zero (b) The electric field just inside the surface is perpendicular to the surface (c) The electric charge is distributed uniformly throughout the volume of the conductor (d) The electric field just outside the surface is perpendicular to the surface Q3.20 Which of the following statements are true? (a) The rate of change of potential with distance along any direction is constant in a uniform electric field. (b) The electric field is zero where the potential is zero. (c) The potential arising from a single point charge may vary from positive to negative in different regions. (d) The force on a charged particle located on an equipotential surface is zero. Q3.21 A conductor A is given a charge of amount + Q and then placed inside an uncharged deep metal can B, without touching it. (a) The potential of A does not change when it is placed inside B. (b) if B is earthed, + Q amount of charge flows from it into the earth. (c) if B is earthed, the potential of A is reduced. (d) if B is earthed, the potential of A and B both becomes zero.

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97

Match the Columns

Q3.22 Two thin conducting shells of radii R and 2R are shown in figure. The outer shell carries a charge + Q and inner shell is neutral. Choose S the correct option(s) (a) With switch open potential of inner sphere 3R is same as that of point at distance from 2 center (b) When switch S is closed, potential of inner sphere becomes zero Q (c) With swith S closed charge on inner sphere is a- b 2 (d) With swith S closed charge on inner sphere is ( - Q) S Q3.23 In an electric field E = ayiN + (ax + bz)jN + bykN , where a and b are constants. (a) The field can be conservation only if a = b (b) The field is conservative for every value of a and b (c) In the xy plane, electric field on the x axis is in the y direction only (d) In the xz plane, electric field on the z axis is 0. Q3.24 An ideal dipole of dipole S R moment p is placed in front of an A B uncharged conducting sphere of radius P O r R as shown. KP (a) The potential at point A is (r - R)2 KP (b) The potential at point A is 2 r KP (c) The potential due to dipole at point B is (r + R)2 KP (d) The potential due to dipole at point B is 2 r Q3.25 A conducting sphere of radius R, carrying charge Q, lies inside an uncharged conducting shell of radius 2R. If they are joined by a metal wire, (a) Q/3 amount of charge will flow from the sphere to the shell (b) 2Q/3 amount of charge will flow from the sphere to the shell (c) Q amount of charge will flow from the sphere to the shell Q2 amount of heat will be produced 4R Q3.26 Three equal point changes (Q) are kept at the three corners of an equilateral triangle ABC of side a. P is a point having equal distance a from A, B and C. If E is the magnitude of electric field and V is the potential at point P, then 3Q 26Q (a) E = (b) E = 2 4pe0a 4pe0a2 (d) k

(c) V =

3Q 4pe0a

(d) E =

3 26Q

Column I

(b) The coordinates of point on circle where potential is zero

(Q)

a -

R - !3 R , b 2 2

(c) The coordinates of point on circle where electric field is directed towards origin.

(R)

a -

R R , b !2 !2

(S)

a

Column I (Cause)

(p) distribution of charge on inner surface of cavity changes

(b) If inside charge is shifted to other position within cavity

(q) distribution of charge on outer surface of conductor changes

(c) If magnitude of charge inside cavity is increased

(r) electric potential at centre of conductor due to charges present on outer surface of conductor changes

(d) If conductor is earthed

(s) force on the charge placed inside cavity changes

Q3.29 Fig shows three concentric thin spherical shells A, B and C of radii R, 2R and 3R. The shell B is earthed and A and C are given charges q and 2q, respectively. If the charge appearing on surfaces 1, 2, 3 and 4 are q1, q2, q3, and q4 respectively, then match the following columns.

(c) q3

Column-II 2 q (p) 3 4 q (q) 3 - q (r)

(d) q4

(s) None of these

(b) q2

3p$j R

x pi$

Column II (Effect)

(a) If outside charge is shifted to other position

(a) q1 y

!3R R , - b 2 2

Q3.28 A spherical metallic conductor has a spherical cavity. A positive charge is placed inside the cavity at its centre. Another positive charge is placed outside it. The conductor is initially electrically neutral.

MATCH THE COLUMNS Q2.27 Column-I gives a sitution in which two dipoles of dipole moment p i and 23 p j are placed at origin. A circle of radius R with centre at origin is drawn as shown in figure. Column-II gives coordinates of certain positions on the circle.

R !3 R a , b 2 2

(a) The coordinates of point on circle where potential is maximum

Column-I

4pe0a2

Column II

(P)

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4 3 1 2 A B C

98

CHAPTER 3 Electric Potential

EXERCISES Section 3.1 Electric Potential Energy

3.1 . A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy of the pair of charges is + 5.4 * 10-8 J. When the second charge is moved to point b, the electric force on the charge does -1.9 * 10-8 J of work. What is the electric potential energy of the pair of charges when the second charge is at point b? 3.2 .. (a) How much work would it take to push two protons very slowly from a separation of 2.00 * 10-10 m (a typical atomic distance) to 3.00 * 10-15 m (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation? 3.3 .. Three equal 1.20-mC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) 3.4 .. Two protons are released from rest when they are 0.256 nm apart. What is the maximum speed they will reach? When does this speed occur? 3.5 .. Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square? 3.6 .. Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides d. Two of the point charges are identical and have charge q. If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?

Section 3.2 Electric Potential

3.7 . A small particle has charge - 5.00 mC and mass 2.00 * 10-4 kg. It moves from point A, where the electric potential is VA = +200 V, to point B, where the electric potential is VB = +800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m> s at point A. What is its speed at point B? Is it moving faster or slower at B than at A? Explain. 3.8 .SA particle with a charge of + 4.20 nC is in a uniform electric field E directed to the left. It is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is found to be +1.50 * 10-6 J. (a) What work was done by the electric force? (b) What is the potential of the startingS point with respect to the end point? (c) What is the magnitude of E? 3.9 . A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.00 * 104 V> m. What work is done by the electric force when the charge moves (a) 0.450 m to the right; (b) 2 m upward; (c) 1.414 m at an angle of 45.0° downward from the horizontal? 3.10 . Two charges of equal magnitude Q are held a distance d apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two charges having opposite signs. 3.11 . A positive charge + q is located at the point x = 0, y = -a, and a negative charge -q is located at the point x = 0, y = +a. (a) Derive an expression for the potential V at points on '

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the y-axis as a function of the coordinate y. Take V to be zero at an infinite distance from the charges. (b) Graph V at points on the y-axis as a function of y over the range from y = - 4a to y = +4a. (c) Show that for y 7 a, the potential at a point on the positive y-axis is given by V = -11> 4pP022qa> y2. (d) What are the answers to parts (a) and (c) if the two charges are interchanged so that + q is at y = +a and - q is at y = -a? 3.12 .. A positive charge q is fixed at the point x = 0, y = 0, and a negative charge -2q is fixed at the point x = a, y = 0. (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential V at points on the x-axis as a function of the coordinate x. Take V to be zero at an infinite distance from the charges. (c) At which positions on the x-axis is V = 0? (d) Graph V at points on the x-axis as a function of x in the range from x = - 2a to x = + 2a. (e) What does the answer to part (b) become when x >> a? Explain why this result is obtained. 3.13 .. Consider the arrangement of point charges described in the previous Question. (a) Derive an expression for the potential V at points on the y-axis as a function of the coordinate y. Take V to be zero at an infinite distance from the charges. (b) At which positions on the y-axis is V = 0? (c) Graph V at points on the y-axis as a function of y in the range from y = -2a to y = +2a. (d) What does the answer to part (a) become when y 7 a? Explain why this result is obtained. 3.14 .. (a) An electron is to be accelerated from 3.00 * 106 m> s to 8.00 * 106 m> s. Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from 8.00 * 106 m> s to a halt? 3.15 . For each of the following arrangements of two point charges, find all the points along the line passing through both charges for which the electric potential V is zero (take V = 0 infinitely far from the charges) and for which the electric field E is zero: (a) charges + Q and +2Q separated by a distance d, and (b) charges -Q and + 2Q separated by a distance d. (c) Are both V and E zero at the same places? Explain. '

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Section 3.3 Calculating Electric Potential 3.16 .. A thin spherical shell with radius R1 = 3.00 cm is concentric with a larger thin spherical shell with radius R2 = 5.00 cm. Both shells are made of insulating material. The smaller shell has charge q1 = +6.00 nC distributed uniformly over its surface, and the larger shell has charge q2 = -9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) r = 0; (ii) r = 4.00 cm; (iii) r = 6.00 cm? (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell? 3.17 . A total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius of 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm; (b) 24.0 cm; (c) 12.0 cm. 3.18 .. A uniformly charged, thin ring has radius 40 cm and total charge +24.0 nC. An electron is placed on the ring’s axis a distance 30.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a)

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99

Exercises

Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring. 3.19 . A very long wire carries a uniform linear charge density l. Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 cm from the wire and the other probe is 1.00 cm farther from the wire, the meter reads 575 V. (a) What is l? (b) If you now place one probe at 3.50 cm from the wire and the other probe 1.00 cm farther away, will the voltmeter read 575 V? If not, will it read more or less than 575 V? Why? (c) If you place both probes 3.50 cm from the wire but 17.0 cm from each other, what will the voltmeter read? '

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7 Given: ln a b = 0.337 5

ter of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Section 3.4 Equipotential Surfaces and Section 3.5 Potential Gradient

3.27 . A very large plastic sheet carries a uniform charge density of - 6.00 nC> m2 on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by 1.00 V. What type of surfaces are these? 3.28 . CALC In a certain region of space, the electric potential isV (x, y, z) = Axy - Bx2 + Cy, where A, B, and C are positive constants. (a) Calculate the x-, y-, and z-components of the electric field. (b) At which points is the electric field equal to zero? 3.29 .. CALC A metal sphere with radius ra is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb. There is charge +q on the inner sphere and charge - q on the outer spherical shell. (a) Calculate the potential V (r) for (i) r 6 ra; (ii) ra 6 r 6 rb; (iii) r 7 rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take V to be zero when r is infinite. (b) Show that the potential of the inner sphere with respect to the outer is q 1 1 Vab = a - b rb 4pP0 ra '

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3.20 . A ring of diameter 8.00 cm is fixed in place and carries a charge of + 5.00 mC uniformly spread over its circumference. (a) How much work does it take to move a tiny +3 mC charged ball of mass 1.50 g from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach? 3.21 .. A very small sphere with positive charge q = + 8.00 mC is released from rest at a point 1.50 cm from a very long line of uniform linear charge density l = +3.00 mC/m. What is the kinetic energy of the sphere when it is 4.50 cm from the line of charge if the only force on it is the force exerted by the line of charge? 3.22 . Charge Q = 5.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R = 12.0 cm. A small sphere with charge q = +3.00 mC and mass 6.00 * 10-5 kg is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 cm of the surface of the large sphere? 3.23 . CP Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. (a) If the surface charge density for each plate has magnitude 47.0 nC > m2, S what is the magnitude of E in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference? 3.24 . Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potential difference between them is 360 V. (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge + 2.40 nC? (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential. 3.25 .. (a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 1.50 kV ? (b) What is the potential of the sphere’s surface relative to infinity? 3.26 . The electric field at the surface of a charged, solid, copper sphere with radius 0.200 m is 3600 N > C, directed toward the cen'

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(c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude E 1r2 = '

Vab 1 11 > ra - 1 > rb2 r2 '

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(d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance r from the center, where r 7 rb. (e) Suppose the charge on the outer sphere is not - q but a negative charge of different magnitude, say - Q. Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different. 3.30 . A very long cylinder of radius 2.00 cm carries a uniform charge density of 1.50 nC > m. (a) Describe the shape of the equipotential surfaces for this cylinder. (b) Taking the reference level for the zero of potential to be the surface of the cylinder, having potentials of 10 V, 20 V, and 30 V. are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as potential increases? '

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PROBLEMS 3.31 . CP A point charge q1

= + 5.00 mC is held fixed in space. From a horizontal distance of 6.00 cm, a small sphere with mass 4.00 * 10-3 kg and charge q2 = +2.00 mC is fired toward the fixed charge with an initial speed of 40.0 m/s. Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is 25.0 m/s? 3.32 ... A point charge q1 = 4.00 nC is placed at the origin, and a second point charge q1 = -3.00 nC is placed on the x-axis at x = +20.0 cm. A third point charge q3 = 2.00 nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.) (a) What is the

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100

CHAPTER 3 Electric Potential

potential energy of the system of the three charges if q3 is placed at x = +10.0 cm? (b) Where should q3 be placed to make the potential energy of the system equal to zero? 3.33 ... A small sphere with mass 5.00 * 10 -7 kg and charge + 3.00 mC is released from rest a distance of 0.400 m above a large horizontal insulating sheet of charge that has uniform surface charge density s = + 8.00 pC/m2. Using energy methods, calculate the speed of the sphere when it is 0.100 m above the sheet of charge? 3.34 .. CALC A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is not a linear function of the position, even with planar geometry, but is given by V 1x2 = Cx4 > 3 '

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where x is the distance from the cathode and C is a constant, characteristic of a particular diode and operating conditions. Obtain a formula for the electric field between the electrodes as a function of x. 3.35 .. Two oppositely charged, Figure P3.35 identical insulating spheres, each 50.0 cm in diameter and carrying V a b a uniform charge of magnitude 250 mC, are placed 1.00 m apart center to center (Fig. P3.35). (a) If a voltmeter is connected between the nearest points (a and b) on their surfaces, what will it read? (b) Which point, a or b, is at the higher potential? How can you know this without any calculations? 3.36 .. An Ionic Crystal. Figure P3.36 Figure P3.59 shows eight point 2q 1q charges arranged at the corners of a cube with sides of length 2q d. The values of the charges are 1q +q and -q, as shown. This is a model of one cell of a cubic d ionic crystal. In sodium chloride (NaCl), for instance, the 2q positive ions are Na+ and the 1q d negative ions are Cl . (a) Cald 2q 1q culate the potential energy U of this arrangement. (Take as zero the potential energy of the eight charges when they are infinitely far apart.) (b) In part (a), you should have found that U 6 0. Explain the relationship between this result and the observation that such ionic crystals exist in nature. 3.37 . (a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00 mC and the other a charge of - 3.50 mC, with their centers separated by a distance of 0.250 m. Assume zero potential energy when the charges are infinitely separated. (b) Suppose that one of the spheres is held in place and the other sphere, which has a mass of 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need in order to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it was infinitely distant from the fixed sphere.) 3.38 .. CP A small sphere with mass 1.50 g hangs by a thread between two parallel vertical plates 5.00 cm apart (Fig. P3.38).

The plates are insulating and Figure P3.38 have uniform surface charge densities +s and - s. The charge on the sphere is What q = 8.70 * 10-6 C. potential difference between 30.0° the plates will cause the thread to assume an angle of 30.0° with the vertical? q 3.39 . CALC Coaxial 5.00 cm Cylinders. A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b. The positive charge per unit length on the inner cylinder is l, and there is an equal negative charge per unit length on the outer cylinder. (a) Calculate the potential V 1r2 for (i) r 6 a; (ii) a 6 r 6 b; (iii) r 7 b. (Hint: The net potential is the sum of the potentials due to the individual conductors.) Take V = 0 at r = b. (b) Show that the potential of the inner cylinder with respect to the outer is l b Vab = ln 2pP0 a '

(c) The result from part (a) to show that the electric field at any point between the cylinders has magnitude Vab 1 E 1r2 = ln 1b > a2 r '

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(d) What is the potential difference between the two cylinders if the outer cylinder has no net charge? 3.40 .. CALC A disk with radius R has uniform surface charge density s. (a) By regarding the disk as a series of thin concentric rings, calculate the electric potential V at a point on the disk’s axis a distance x from the center of the disk. Assume that the potential dV is zero at infinity. Calculate En = - dx . 3.41 . CALC A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along the rod. Calculate the potential at the center of curvature of the arc if the potential is assumed to be zero at infinity. 3.42 ... CALC Self-Energy of a Sphere of Charge. A solid sphere of radius R contains a total charge Q distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the “self-energy” of the charge distribution. (Hint: After you have assembled a charge q in a sphere of radius r, how much energy would it take to add a spherical shell of thickness dr having charge dq? Then integrate to get the total energy.) 3.43 .. Charge Q = +4.00 mC is distributed uniformly over the volume of an insulating sphere that has radius R = 5.00 cm. What is the potential difference between the center of the sphere and the surface of the sphere? 3.44 .. CP Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 cm in diameter, has mass 50.0 g, and contains -10.0 mC of charge. The other sphere is 40.0 cm in diameter, has mass 150.0 g, and contains - 30.0 mC of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on

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Exercises

them. (Hint: The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.) 3.45 . CALC Electric charge is Figure P3.45 distributed uniformly along a R thin rod of length a, with total charge Q. Take the potential to be y Q P zero at infinity. Find the potential x at the following points (Fig. a P3.45): (a) point P, a distance x to the right of the rod, and (b) point R, a distance y above the righthand end of the rod. (c) In parts (a) and (b), what does your result reduce to as x or y becomes much larger than a? 3.46 .. Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere A has a radius three times that of sphere B. Let QA and QB be the charges on the two spheres, and let EA and EB be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio QB > QA and (b) the ratio EB > EA ? 3.47 . A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius R2 that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere. 3.48 . CALC The electric potential V in a region of space is given by '

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V 1x, y, z2 = A 1x2 - 3y2 + z22 '

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where A is a constant. (a) Derive an expression for the electric field S E at any point in this region. (b) The work done by the field when a 1.50-mC test charge moves from the point (x, y, z) = (0, 0, 0.250 m) to the origin is measured to be 6.00 * 10-5 J. Determine A. (c) Determine the electric field at the point (0, 0, 0.250 m). (d) Show that in every plane parallel to the xz-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to V = 1280 V and y = 2.00 m ? 3.49 Charges + q, - q, + q, - q are placed at corners ABCD of a square of a side a. A charge + q is placed at its centre. Find the interaction energy of the system 3.50 A nonconducting ring of radius R has two charges q and q/2 distributed uniformly on two half of it. A negative charge q/2 (mass = m) is placed at a distance R from the centre of the ring on its axis at a point P and released. If this charge is restricted to move freely only the axis of the ring, find the speed of this charge when it crosses the centre of the ring 3.51 Between two infinitely long wires having λ −λ linear charge densities l and - l there are A B two points A and B as shown in the figure. Find the amount of work done by the electric field in a a a moving a point charge q0 from A to B '

101

3.52 There are two non-conducting spheres +ρ −ρ having uniform volume charge densities r R R and - r. Both spheres have equal radius R. O The spheres are now laid down such that they overlaps as shown in the figure. d S (a) Find the electric field E in the overlap region (b) Find the potential difference dV between the centers of the two spheres for d = R. 3.53 (a) Five large identical metal plates of area A are arranged horizontally, parallel to each other, with gap d in between adjacent plates. Counted from the top, they are given charge + q, + 2q, - 3q, - q, + 3q respectively. Find the potential difference between the topmost and bottommost plates. (b) What charge will flow through a conducting wire if the topmost and bottommost plates are shorted? Q 3.54 In the figure shown, both the 3a shells are conducting. The outer shell is thick & has a total charge of Q, inner shell is thin & is earthed. a 2a (a) Find charge on each surface. (b) Draw graph of potential (V) v/s distance from centre (r). 3.55 Two concentric conducting spherical shells of radii a1 and a2 (a2>a1) are charged to potentials f1 and f2 respectively. Find the charge on the inner shell. 3.56 The point charge q is within an electrically neutral conducting shell whose outer surface has r P O spherical shape. Find the potential V at a point P q lying outside the shell at a distance r from the centre O of the outer sphere. 3.57 A conducting sphere of radius R has two spherical cavities of radius a and b. The cavities a qa have charges qa and qb respectively at their cenR tres. Find : b qb (a) The electric field and electric potential at a distance r (i) r (distance from O, the centre of sphere > R) (ii) r (distance from B, the centre of cavity b) < b (b) Surface charge densities on the surface of radius R, radius a and radius b. (c) What is the force on qa and qb? 3.58 A little, positively charged ball of mass m = 5gm is suspended on an isolating thread l h of negligible mass. Another positively charged small ball is moved very slowly from a very large distance and placed at the original position of the first ball which makes the first ball rise by h = 8 cm. How much work has been done? Take l>>h. 3.59 A non-conducting slab infinite in the Y x - y direction and thickness L in the z direction + + + + carries uniform positive volume charge density + + + + + + + + r(z) = r0. X (a) Find and sketch the electric field as a function ++ ++ ++ ++ Z of ‘z’ both outside and inside the slab. + + + + (b) If potential at z = 0 is taken to be zero, then + + + + find and sketch the electrostatic potential as a L 2 L 2 function of z both outside and inside the slab.

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102

CHAPTER 3 Electric Potential

CHALLENGE PROBLEMS 3.60 ... CP CALC In a certain region, a charge distribution exists that is spherically symmetric but nonuniform. That is, the volume charge density r 1r2 depends on the distance r from the center of the distribution but not on the spherical polar angles u and f. The electric potential V (r) due to this charge distribution is '

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r0a2 r 2 r 3 s 1 - 3 a b + 2 a b t for r … a a a V(r) = µ 18P0 0 for r Ú a '

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where r0 is a constant having units of C > m3 and a is a constant S having units of meters. (a) Derive expressions for E for the regions S r … a and r Ú a. [Hint: Use Eq. (23.23).] Explain why E has only a radial component. (b) Derive an expression for r 1r2 in each of '

'

'

the two regions r … a and r Ú a. [Hint: Use Gauss’s law for two spherical shells, one of radius r and the other of radius r + dr. The charge contained in the infinitesimal spherical shell of radius dr is dq = 4pr2r 1r2 dr.] (c) Show that the net charge contained in the volume of a sphere of radius greater than or equal to a is zero. [Hint: Integrate the expressions derived in part (b) for r(r) over a spherical volume of radius greater than or equal to a.] Is this result consistent with the electric field for r 7 a that you calculated in part (a)? Y 3.61 ... CP CALC A uniform surface charge of density s is given to quarter infinite non conducting plane in first quardrant of x - y plane. Find the z-component of the electric field at the point (0, 0, z). Hence or otherwise X O find the potential difference between the points (0, 0, d) & (0, 0, 2d). '

Answers Chapter Opening Question

?

A large, constant potential difference Vab is maintained between the welding tool (a) and the metal pieces to be welded (b). From Example 23.9 (Section 23.3) the electric field between two conductors separated by a distance d has magnitude E = Vab /d. Hence d must be small in order for the field magnitude E to be large enough to ionize the gas between the conductors a and b (see Section 23.3) and produce an arc through this gas.

Test Your Understanding Questions 3.1 (a) (i), (b) (ii) The three charges q1, q2, and q3 are all positive, so all three of the terms in the sum in Eq. (23.11)— q1q2/r12, q1q3/r13, and q2q3/r23—are positive. Hence the total electric potential energy U is positive. This means that it would take positive work to bring the three charges from infinity to the positions shown in Fig. 21.14, and hence negative work to move the three charges from these positions back to infinity. S 3.2 no If V = 0 at a certain point, E does not have to be zero at that point. An example is point c in Figs. 21.23 and 23.13, for which there is an electric field in the + x-direction (see Example 21.9 in Section 21.5) even though V = S0 (see Example 23.4). This isn’t a surprising result because V and E are quite different quantities: V is the net amount of work required to bring a unit charge S from infinity to the point in question, whereas E is the electric force that acts on a unit charge when it arrives at that point. S 3.3 no If E 5 0 at a certain point, V does not have to be zero at that point. An example is point O at the center of the charged ring in Figs. 21.23 and 23.21. From Example 21.9 (Section 21.5), the electric field is zero at O because the electric-field contributions from different parts of the ring completely cancel. From Example 23.11, however, the potential at O is not zero: This point corresponds to x = 0, so V = (1/4pP0)(Q/a). This value of V corresponds to the work that would have to be done to move a unit positive test charge along a path from infinity to point O; it is nonzero because the charged ring repels the test charge, so positive work must be done to move the test charge toward the ring. '

'

'

'

'

'

3.4 no If the positive charges in Fig. 23.23 were replaced by negative charges, and vice versa, the equipotential surfaces would be the same but the sign of the potential would be reversed. For example, the surfaces in Fig. 23.23b with potential V = + 30 V and V = - 50 V would have potential V = - 30 V and V = + 50 V, respectively. 3.5 (iii) From Eqs. (23.19), the components of the electric field are Ex = -0V/0x = B + Dy, Ey = -0V/0y = 3Cy2 + Dx, and Ez = - 0V/0z = 0. The value of A has no effect, which means that we can addS a constant to the electric potential at all points without changing E or the potential difference between two points. The S potential does not depend on z, so the z-component of E is zero. Note that at the origin the electric field is not zero because it has a nonzero x-component: Ex = B, Ey = 0, Ez = 0.

'

Bridging Problem qQ L + a ln a b 8pP0a L - a

Discussion Questions 3.1 Since isolation of a single charge is possible and potential energy requires its interaction with some external electric field, hence for the isolated charge, concept of potential is required. 3.2 Yes it is possible to bring the test charge along the perpendicular bisector of the line joining the two charges. The line is equipotential (as V = 0 everywhere) 3.3 It is not possible for two charges but it is possible for 3 charges. 3.4 A voltmeter reads the potential difference between the end points and not the potential of a point. Potential difference is independent of a reference. 3.5 The p.d. between the points will be zero. No it is not necessary S for E = 0 for all paths from A to B. AS S

S

S

{ VB - VA = 3 E. dl = 0 , it is possible also if E ' dl i.e., B

perpendicular to the path at every point.

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Answers S

3.6 E = 0 everywhere implies a constant potential and not necessarily zero potential. B

3.7 Since, VB - VA =

S

-

S

3 E. dl

for two points A and B,

A

Hence for a closed path, moving from point A to A (say) AS S

S S

VA - VA = 0 = - 3 E. dl 1 D E dl = 0 A

+ + + ++ +

++

++

++

+ + + + ++

+

3.8 In the first case the p.d will be 3V , as their electric fields get added. In second case, p.d. will be zero as the fields cancel each other. 3.9 Energy released is small as the amount of charge is small in the first case. 3.10 No. Because potential difference between the point and infinity 1V = 02 will depend also on the path followed in between, which can be varried. 3.11 Electric field at that point is zero. 3.12 VB 7 VA = VD 7 VC It is so because feild lines are always directed from higher to lower potential. 3.13 No. For example (i) both A and B are negatively charged and B is more negative (ii) both A and B are positively charged and A is more positive. 3.14 The total work done will be stored as the + +++ + ++ self energy of the conductor, which can be found q as shown : R q V. of sphere = 4pP0R If further charge dq is put on qdq sphere, dU = 1dq2V = 4pP0R + ++ +++

Q qdq Q2 r Q2 = { Uself = 3 8pP0R 0 4pP0R

3.15 They will have the same p.d. because the surface of a conductor is equipotential always. Hence potential of plates connected to a will be Va and of the plates connected to b will be Vb . Also Va - Vb = 1.5 volts . 3.16 Electric field does not depend on positions as it is zero. Potential will decrease as the sphere is taken down along the external electric field. The answers do not depend on the charge of sphere. 3.17 { Field inside the conductor = 0 Hence potential will be same everywhere. Also, as all charge resides on outersurface, there is no charge inside cavity and no field also. Hence Vcavity = Vconductor . 3.18 (a) Due to electrostatic shielding, the occupants will not receive any field. Hence no shock, as their potential will be same as that of the power line. (b) When they step out, their potential will become equal to that of earth. Hence due to a large potential difference created, they will receive shock. 3.19 During thunderstorms, the high voltage developed between clouds and ground, ionises the air molecules and the air gases begin to glow. This is called St. Elmo's fire. Since an electric field is intensified in regions of high curvatures, hence the discharge is more intensified at the ends of pointed metal rods. Even if the tip of mass is not conductor, yet presence of sea water can cause it to conduct.

Single Correct Q3.1 (b) Q3.2 (a) Q3.3 (d) Q3.4 (c) Q3.5 (b) Q3.6 (c) Q3.7 (c) Q3.8 (a) Q3.9 (d) Q3.10 (c) Q3.11 (b) Q3.12 (b) Q3.13 (d) Q3.14 (c) Q3.15 (b)

One or More Correct Q3.16 (b), (c), (d) Q3.17 (a), (b), (c) Q3.18 (c) Q3.19 (d) Q3.20 (a) Q3.21 (b), (c) Q3.22 (a), (b), (c) Q3.23 (b), (c), (d) Q3.24 (b), (c) Q3.25 (c), (d) Q3.26 (b), (c)

Match the Column Q3.27 (a) P (b) S (c) Q Q3.28 (a) Q (b) P, S (c) P, Q, R (d) Q, R Q3.29 (a) R (b) S (c) Q (d) P

EXERCISES Section 3.1 Electric Potential Energy 3.1 7.3 * 10 - 8 J 3.2 (a) 7.68 * 10 - 14 J (b) 6.78 * 10 - 6 m/s 3.3 0.078 J 3.4 2.32 * 10 - 4 m/s, when the protons are far apart 3.5 6.08 * 10 - 21 J - q 3.6 2

Section 3.2 Electric Potential 3.7 7.42 m/s, faster 3.8 (a) 1.5 * 10 - 6J (b) 357 V (c) 5.95 * 103 v/m 3.9 (a) zero (b) 2.24 * 10 - 3 J (c) - 1.12 * 10 - 3 J

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103

104

CHAPTER 3 Electric Potential

3.10 (a) (i) no such point (ii) midway between the charger potential is nowhere zero (b) (i) potential is zero midway between the charger but electric field is not zero anywhere (ii) Electric field is not zero anywhere

3.25 (a) 20.8 nc (b) 1.50 kv 3.26 1.6 * 10 - 8 C

3.11 (a) V = R c

Section 3.4 Equipotential Surfaces and Section 3.5 Potential Gradient

- q q + d |y - a| |y + a|

(d) If the charger are interchanged, then the potential is of opposite sign q 2q 3.12 (b) V = R c d |x| |x - a| a (c) at x = - a and 3 - kq (e) v = x 3.13 (a) V = kq s (b) y = ± (d) V = -

a

- 1 |y|

2

2a

2

+ y2

t

23

3.28 (a) - Ay + 2Bx, - Ax - c, zero - 2BC - C (b) at x = ,y = , any value of z A A2 1 1 3.29 (a) (i) V = kq a - b ra rb 1 1 (ii) V = kq a - b r rb (iii) V = zero (d) Electric field = zero

PROBLEMS

kq y

3.14 (a) 156 V (b) - 182 V 3.15 (a) V is zero nowhere except for infinitely far from the charge; d E is zero between the charge at a distance from charge Q. 1 + 22 d (b) V is zero at a point between two charge at a distance x = 3 from charger ( - Q) and at another poit at a distance d from charge ( - Q) in the direction opposite to that of ( + 2Q, E is zero d at a point at a distance of from charge ( - Q) in the 22 - 1 direction opposite to that of ( + 2Q)

Section 3.3 Calculating Electric Potential 3.16 (a) 180 V, - 270 V, - 450 V (b) 720 V, inner shell 3.17 (a) 65.6 V (b) 131 V (c) 131 V 3.18 (a) Oscillatory motion along the axis of the ring with amplitude 40 cm. (b) 6.17 * 10 6 m/s 3.19 (a) 9.48 * 10 - 8 c/m (b) less than 575 V (c) zero 3.20 (a) 3.38 J (b) No (c) 67.1 m/s 3.21 [(0.432) ln3] Joules 3.22 150 m/s 3.23 (a) 5310 N/C (b) 117 V (c) electric field same, potential difference doubles 3.24 (a) 8000 V/m (b) 1.92 * 10 - 5 N (c) 8.64 * 10 - 7 J

3.27 (a) Potential increases, no matter what the reference point (b) 2.95 mm

3.30 Equipotential surfaces are (a) cylinder with the given cylinders (b) gets further apart 3.31 3.33 * 104 m/s2 3.32 (a) - 3.6 * 10 - 7 J (b) 7.4 cm 3.33 2.09 m/s 3.34 Ex = - (1.05 * 105 V/m4/3) x1/3 3.35 (a) 12 MV (b) point a 3.36 (a) U = - 1.46 q2/pP.d 3.37 (a) - 0.252 J (b) 18.3 m/s 3.38 50 V l l 3.39 (a) (i) V = a b ln a b for a > r a 2lP0 b l (ii) V = a b ln a b for b > r > a r 2lP0 (iii) V = 0 for r > e l b (d) a b ln a b a 2pP0 s 3.40 (a) V = a x2 + R2 - xb 2P0 2 sx 1 1 (b) a b 2 2P0 x x + R2

2

3 1 Q2 3.42 a b 5 4pP0 R 5 3.43 3.6 * 10 V 3.44 216 m/s2, 72 m/s2, 12.7 m/s, 4.24 m/s Q x + a 3.45 (a) ln a b x 4pP0a a + 2a2 + y2 Q bt sln a y 4pP0a (c) charge on rod behaves as a point charge. (b)

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Answers

3.46 (a)

(b)

1 3

V

(b) 3 1 Q1 1 Q1 ,V = 4pP0 R21 4pP0 R1 R1 R2 d Q1 , c dQ (b) c R1 + R 2 R1 + R2 1 Q1 Q1 (c) , 4pP0(R1 + R2) 4pP0(R1 + R2) Q1 Q1 (d) , 4pP0R1(R1 + R2) 4pP0R2(R1 + R2) S 3.48 (a) E = - 2 AxiN + 6 AyjN - 2 AzkN 3.47 (a) E =

(b) 640 V/m2 (c) - (320 V/m)kN (e) 3.74 m Kq2 3.49 2 - 4) ( a 2 1/2 3Kq2 ( 22 - 1) t 3.50 s 2 22Rm lq0 ln2 3.51 pP0 r S d 3.52 (a) 3P0 r 2 d (b) 3P0 qd 3.53 (a) AP0 q (b) 4 2Q 2Q 3Q 2QK 1 1 , + , b c - d 3.54 (a) 0, a a r 5 5 5 5

KQ 5a 2a 3a

a r

3.55 q1 = 4pP0 a

f1 - f2 ba a a2 - a1 1 2

1 q . 4pP0 r k(qa + qb) k(qa + qb) ; ,E = 3.57 (a) (i) V = r r2 k(qa + qb) kqb kqb kqb + (ii) ; r R b r2 qa + qb qa qb b , sa = ; sb = (b) sR = a ; 4pR2 4pa2 4pb2 (c) f = 0 3.58 1.2 * 10 - 2 J r0Z r0L , 3.59 (a) P0 2P0 rZ2 rL L , a - Zb (b) 2P0 4P0 4 3.56 V =

CHALLENGE PROBLEM r0a r r2 s - 2 ti; for r Ú a, E = 0 3P0 a a 4r r, r(r) = r0 c1 d for r Ú a, r(r) = 0 3a sd 8P0

3.60 (a) for a Ú r, E = (b) for a Ú Q3.61

s , 8P0

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105

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4

CAPACITANCE AND DIELECTRICS

LEARNING GOALS By studying this chapter, you will learn: • The nature of capacitors, and how to calculate a quantity that measures their ability to store charge. • How to analyze capacitors connected in a network. • How to calculate the amount of energy stored in a capacitor. • What dielectrics are, and how they make capacitors more effective.

?

The energy used in a camera’s flash unit is stored in a capacitor, which consists of two closely spaced conductors that carry opposite charges. If the amount of charge on the conductors is doubled, by what factor does the stored energy increase?

hen you set an old-fashioned spring mousetrap or pull back the string of an archer’s bow, you are storing mechanical energy as elastic potential energy. A capacitor is a device that stores electric potential energy and electric charge. To make a capacitor, just insulate two conductors from each other. To store energy in this device, transfer charge from one conductor to the other so that one has a negative charge and the other has an equal amount of positive charge. Work must be done to move the charges through the resulting potential difference between the conductors, and the work done is stored as electric potential energy. Capacitors have a tremendous number of practical applications in devices such as electronic flash units for photography, pulsed lasers, air bag sensors for cars, and radio and television receivers. We’ll encounter many of these applications in later chapters (particularly Chapter 11, in which we’ll see the crucial role played by capacitors in the alternating-current circuits that pervade our technological society). In this chapter, however, our emphasis is on the fundamental properties of capacitors. For a particular capacitor, the ratio of the charge on each conductor to the potential difference between the conductors is a constant, called the capacitance. The capacitance depends on the sizes and shapes of the conductors and on the insulating material (if any) between them. Compared to the case in which there is only vacuum between the conductors, the capacitance increases when an insulating material (a dielectric) is present. This happens because a redistribution of charge, called polarization, takes place within the insulating material. Studying polarization will give us added insight into the electrical properties of matter. Capacitors also give us a new way to think about electric potential energy. The energy stored in a charged capacitor is related to the electric field in the space between the conductors. We will see that electric potential energy can be regarded as being stored in the field itself. The idea that the electric field is itself a storehouse of energy is at the heart of the theory of electromagnetic waves and our modern understanding of the nature of light, to be discussed in Chapter 12.

W

106

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4.1 Capacitors and Capacitance

4.1

107

Capacitors and Capacitance

Any two conductors separated by an insulator (or a vacuum) form a capacitor (Fig. 4.1). In most practical applications, each conductor initially has zero net charge and electrons are transferred from one conductor to the other; this is called charging the capacitor. Then the two conductors have charges with equal magnitude and opposite sign, and the net charge on the capacitor as a whole remains zero. We will assume throughout this chapter that this is the case. When we say that a capacitor has charge Q, or that a charge Q is stored on the capacitor, we mean that the conductor at higher potential has charge +Q and the conductor at lower potential has charge -Q (assuming that Q is positive). Keep this in mind in the following discussion and examples. In circuit diagrams a capacitor is represented by either of these symbols:

4.1 Any two conductors a and b insulated from each other form a capacitor.

Conductor a 1Q S

E 2Q Conductor b

In either symbol the vertical lines (straight or curved) represent the conductors and the horizontal lines represent wires connected to either conductor. One common way to charge a capacitor is to connect these two wires to opposite terminals of a battery. Once the charges Q and - Q are established on the conductors, the battery is disconnected. This gives a fixed potential difference Vab between the conductors (that is, the potential of the positively charged conductor a with respect to the negatively charged conductor b) that is just equal to the voltage of the battery. The electric field at any point in the region between the conductors is proportional to the magnitude Q of charge on each conductor. It follows that the potential difference Vab between the conductors is also proportional to Q. If we double the magnitude of charge on each conductor, the charge density at each point doubles, the electric field at each point doubles, and the potential difference between conductors doubles; however, the ratio of charge to potential difference does not change. This ratio is called the capacitance C of the capacitor: C =

Q Vab

(definition of capacitance)

ActivPhysics 11.11.6: Electric Potential: Qualitative Introduction ActivPhysics 11.12.1 and 11.12.3: Electric Potential, Field, and Force

(4.1)

The SI unit of capacitance is called one farad (1 F), in honor of the 19th-century English physicist Michael Faraday. From Eq. (4.1), one farad is equal to one coulomb per volt (1 C>V): 1 F = 1 farad = 1 C>V = 1 coulomb>volt CAUTION Capacitance vs. coulombs Don’t confuse the symbol C for capacitance (which is always in italics) with the abbreviation C for coulombs (which is never italicized). y

The greater the capacitance C of a capacitor, the greater the magnitude Q of charge on either conductor for a given potential difference Vab and hence the greater the amount of stored energy. (Remember that potential is potential energy per unit charge.) Thus capacitance is a measure of the ability of a capacitor to store energy. We will see that the value of the capacitance depends only on the shapes and sizes of the conductors and on the nature of the insulating material between them. (The above remarks about capacitance being independent of Q and Vab do not apply to certain special types of insulating materials. We won’t discuss these materials in this book, however.)

Calculating Capacitance: Capacitors in Vacuum We can calculate the capacitance C of a given capacitor by finding the potential difference Vab between the conductors for a given magnitude of charge Q and

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108

CHAPTER 4 Capacitance and Dielectrics

4.2 A charged parallel-plate capacitor. (a) Arrangement of the capacitor plates Wire

Plate a, area A

+Q –Q Potential difference 5 Vab

d Wire

Plate b, area A

then using Eq. (4.1). For now we’ll consider only capacitors in vacuum; that is, we’ll assume that the conductors that make up the capacitor are separated by empty space. The simplest form of capacitor consists of two parallel conducting plates, each with area A, separated by a distance d that is small in comparison with their dimensions (Fig. 4.2a). When the plates are charged, the electric field is almost completely localized in the region between the plates (Fig. 4.2b). As we discussed in Example 2.8 (Section 2.4), the field between such plates is essentially uniform, and the charges on the plates are uniformly distributed over their opposing surfaces. We call this arrangement a parallel-plate capacitor. We worked out the electric-field magnitude E for this arrangement in Example 1.12 (Section 1.5) using the principle of superposition of electric fields and again in Example 2.8 (Section 2.4) using Gauss’s law. It would be a good idea to review those examples. We found that E = s>P0 , where s is the magnitude (absolute value) of the surface charge density on each plate. This is equal to the magnitude of the total charge Q on each plate divided by the area A of the plate, or s = Q>A, so the field magnitude E can be expressed as '

S

(b) Side view of the electric field E

S

E

E =

Q s = P0 P0 A

The field is uniform and the distance between the plates is d, so the potential difference (voltage) between the two plates is When the separation of the plates is small compared to their size, the fringing of the field is slight.

Vab = Ed =

1 Qd P0 A

From this we see that the capacitance C of a parallel-plate capacitor in vacuum is C =

4.3 Inside a condenser microphone is a capacitor with one rigid plate and one flexible plate. The two plates are kept at a constant potential difference Vab. Sound waves cause the flexible plate to move back and forth, varying the capacitance C and causing charge to flow to and from the capacitor in accordance with the relationship C = Q/Vab. Thus a sound wave is converted to a charge flow that can be amplified and recorded digitally.

Q A = P0 Vab d

(capacitance of a parallel-plate capacitor in vacuum)

(4.2)

The capacitance depends only on the geometry of the capacitor; it is directly proportional to the area A of each plate and inversely proportional to their separation d. The quantities A and d are constants for a given capacitor, and P0 is a universal constant. Thus in vacuum the capacitance C is a constant independent of the charge on the capacitor or the potential difference between the plates. If one of the capacitor plates is flexible, the capacitance C changes as the plate separation d changes. This is the operating principle of a condenser microphone (Fig. 4.3). When matter is present between the plates, its properties affect the capacitance. We will return to this topic in Section 4.4. Meanwhile, we remark that if the space contains air at atmospheric pressure instead of vacuum, the capacitance differs from the prediction of Eq. (4.2) by less than 0.06%. In Eq. (4.2), if A is in square meters and d in meters, C is in farads. The units of P0 are C2>N # m2, so we see that 1 F = 1 C2>N # m = 1 C2>J

Because 1 V = 1 J>C (energy per unit charge), this is consistent with our definition 1 F = 1 C>V. Finally, the units of P0 can be expressed as 1 C2>N # m2 = 1 F>m, so P0 = 8.85 * 10-12 F>m This relationship is useful in capacitance calculations, and it also helps us to verify that Eq. (4.2) is dimensionally consistent. One farad is a very large capacitance, as the following example shows. In many applications the most convenient units of capacitance are the microfarad

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4.1 Capacitors and Capacitance

11 mF = 10-6 F2 and the picofarad 11 pF = 10-12 F2. For example, the flash unit in a point-and-shoot camera uses a capacitor of a few hundred microfarads (Fig. 4.4), while capacitances in a radio tuning circuit are typically from 10 to 100 picofarads. For any capacitor in vacuum, the capacitance C depends only on the shapes, dimensions, and separation of the conductors that make up the capacitor. If the conductor shapes are more complex than those of the parallel-plate capacitor, the expression for capacitance is more complicated than in Eq. (4.2). In the following examples we show how to calculate C for two other conductor geometries.

Example 4.1

109

4.4 A commercial capacitor is labeled with the value of its capacitance. For these capacitors, C = 2200 mF, 1000 mF, and 470 mF.

Size of a 1-F capacitor

The parallel plates of a 1.0-F capacitor are 1.0 mm apart. What is their area? S O L U T IO N IDENTIFY and SET UP: This problem uses the relationship among the capacitance C, plate separation d, and plate area A (our target variable) for a parallel-plate capacitor. We solve Eq. (4.2) for A.

EVALUATE: This corresponds to a square about 10 km (about 6 miles) on a side! The volume of such a capacitor would be at least Ad = 1.1 * 105 m3, equivalent to that of a cube about 50 m on a side. In fact, it’s possible to make 1-F capacitors a few centimeters on a side. The trick is to have an appropriate substance between the plates rather than a vacuum, so that (among other things) the plate separation d can greatly reduced. We’ll explore this further in Section 4.4.

EXECUTE: From Eq. (4.2), A =

11.0 F)11.0 * 10-3 m2 Cd = = 1.1 * 108 m2 P0 8.85 * 10-12 F>m

Example 4.2

Properties of a parallel-plate capacitor

The plates of a parallel-plate capacitor in vacuum are 5.00 mm apart and 2.00 m2 in area. A 10.0-kV potential difference is applied across the capacitor. Compute (a) the capacitance; (b) the charge on each plate; and (c) the magnitude of the electric field between the plates. S O L U T IO N IDENTIFY and SET UP: We are given the plate area A, the plate spacing d, and the potential difference Vab = 1.00 * 104 V for this parallel-plate capacitor. Our target variables are the capacitance C, the charge Q on each plate, and the electric-field magnitude E. We use Eq. (4.2) to calculate C and then use Eq. (4.1) and Vab to find Q. We use E = Q>P0 A to find E. '

EXECUTE: (a) From Eq. (4.2), 12.00 m22 A = 18.85 * 10-12 F>m2 d 5.00 * 10-3 m -9 = 3.54 * 10 F = 0.00354 mF

C = P0

(b) The charge on the capacitor is Q = CVab = 13.54 * 10-9 C>V211.00 * 104 V2 = 3.54 * 10-5 C = 35.4 mC

The plate at higher potential has charge +35.4 mC, and the other plate has charge - 35.4 mC. (c) The electric-field magnitude is E =

Q s 3.54 * 10-5 C = = P0 P0 A 18.85 * 10-12 C2>N # m2212.00 m22

= 2.00 * 106 N>C

EVALUATE: We can also find E by recalling that the electric field is equal in magnitude to the potential gradient [Eq. (23.22)]. The field between the plates is uniform, so E =

Vab 1.00 * 104 V = = 2.00 * 106 V>m d 5.00 * 10-3 m

(Remember that 1 N>C = 1 V>m.)

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110

CHAPTER 4 Capacitance and Dielectrics

Example 4.3

A spherical capacitor

Two concentric spherical conducting shells are separated by vacuum (Fig. 4.5). The inner shell has total charge +Q and outer radius ra, and the outer shell has charge - Q and inner radius rb. Find the capacitance of this spherical capacitor. SOLUTION IDENTIFY and SET UP: By definition, the capacitance C is the magnitude Q of the charge on either sphere divided by the potential difference Vab between the spheres. We first find Vab, and then use Eq. (4.1) to find the capacitance C = Q>Vab. EXECUTE: Using a Gaussian surface such as that shown in Fig. 4.5, we found in Example 2.5 (Section 2.4) that the charge on a conducting sphere produces zero field inside the sphere, so the outer sphere makes no contribution to the field between the spheres. Therefore the electric field and the electric potential between the

shells are the same as those outside a charged conducting sphere with charge +Q. We considered that problem in Example 3.8 (Section 3.3), so the same result applies here: The potential at any point between the spheres is V = Q>4pP0r. Hence the potential of the inner (positive) conductor at r = ra with respect to that of the outer (negative) conductor at r = rb is Q Q 4pP0ra 4pP0rb Q Q rb - r a 1 1 = a - b = rb 4pP0 ra 4pP0 rarb

Vab = Va - Vb =

The capacitance is then C =

rarb Q = 4pP0 rb - ra Vab

As an example, if ra = 9.5 cm and rb = 10.5 cm, C = 4p18.85 * 10-12 F>m2

4.5 A spherical capacitor. Inner shell, charge 1Q

10.095 m)10.105 m2 0.010 m

= 1.1 * 10-10 F = 110 pF

Gaussian surface Outer shell, charge 2Q ra

Example 4.4

r

rb

EVALUATE: We can relate our expression for C to that for a parallelplate capacitor. The quantity 4prarb is intermediate between the areas 4pra2 and 4prb2 of the two spheres; in fact, it’s the geometric mean of these two areas, which we can denote by Agm. The distance between spheres is d = rb - ra, so we can write C = 4pP0rarb>1rb - ra2 = P0 Agm>d. This has the same form as for parallel plates: C = P0 A>d. If the distance between spheres is very small in comparison to their radii, their capacitance is the same as that of parallel plates with the same area and spacing.

A cylindrical capacitor

Two long, coaxial cylindrical conductors are separated by vacuum (Fig. 4.6). The inner cylinder has radius ra and linear charge density + l. The outer cylinder has inner radius rb and linear charge density -l. Find the capacitance per unit length for this capacitor. SOLUTION IDENTIFY and SET UP: As in Example 4.3, we use the definition of capacitance, C = Q>Vab. We use the result of Example 3.10 4.6 A long cylindrical capacitor. The linear charge density l is assumed to be positive in this figure. The magnitude of charge in a length L of either cylinder is lL.

(Section 3.3) to find the potential difference Vab between the cylinders, and find the charge Q on a length L of the cylinders from the linear charge density. We then find the corresponding capacitance C using Eq. (4.1). Our target variable is this capacitance divided by L. EXECUTE: As in Example 4.3, the potential V between the cylinders is not affected by the presence of the charged outer cylinder. Hence our result in Example 3.10 for the potential outside a charged conducting cylinder also holds in this example for potential in the space between the cylinders: V =

2l

r0 l ln r 2pP0

Here r0 is the arbitrary, finite radius at which V = 0. We take r0 = rb, the radius of the inner surface of the outer cylinder. Then the potential at the outer surface of the inner cylinder (at which r = ra) is just the potential Vab of the inner (positive) cylinder a with respect to the outer (negative) cylinder b:

1l

r rb a

Vab = L

rb l ln 2pP0 ra

If l is positive as in Fig. 4.6, then Vab is positive as well: The inner cylinder is at higher potential than the outer.

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4.2 Capacitors in Series and Parallel The total charge Q in a length L is Q = lL, so from Eq. (4.1) the capacitance C of a length L is Q C = = Vab

Substituting P0 = 8.85 * 10-12 F>m = 8.85 pF>m, we get 55.6 pF>m C = L ln1rb>ra2

2pP0L = rb l ln1rb>ra2 ln 2pP0 ra lL

EVALUATE: The capacitance of coaxial cylinders is determined entirely by their dimensions, just as for parallel-plate and spherical capacitors. Ordinary coaxial cables are made like this but with an insulating material instead of vacuum between the conductors. A typical cable used for connecting a television to a cable TV feed has a capacitance per unit length of 69 pF>m.

The capacitance per unit length is 2pP0 C = L ln1rb>ra2

Test Your Understanding of Section 4.1 A capacitor has vacuum in the space between the conductors. If you double the amount of charge on each conductor, what happens to the capacitance? (i) It increases; (ii) it decreases; (iii) it remains the same; (iv) the answer depends on the size or shape of the conductors.

4.2

111

y

4.7 An assortment of commercially available capacitors.

Capacitors in Series and Parallel

Capacitors are manufactured with certain standard capacitances and working voltages (Fig. 4.7). However, these standard values may not be the ones you actually need in a particular application. You can obtain the values you need by combining capacitors; many combinations are possible, but the simplest combinations are a series connection and a parallel connection.

Capacitors in Series Figure 4.8a is a schematic diagram of a series connection. Two capacitors are connected in series (one after the other) by conducting wires between points a and b. Both capacitors are initially uncharged. When a constant positive potential difference Vab is applied between points a and b, the capacitors become charged; the figure shows that the charge on all conducting plates has the same magnitude. To see why, note first that the top plate of C1 acquires a positive charge Q. The electric field of this positive charge pulls negative charge up to the bottom plate of C1 until all of the field lines that begin on the top plate end on the bottom plate. This requires that the bottom plate have charge -Q. These negative charges had to come from the top plate of C2 , which becomes positively charged with charge +Q. This positive charge then pulls negative charge -Q from the connection at point b onto the bottom plate of C2 . The total charge on the lower plate of C1 and the upper plate of C2 together must always be zero because these plates aren’t connected to anything except each other. Thus in a series connection the magnitude of charge on all plates is the same. Referring to Fig. 4.8a, we can write the potential differences between points a and c, c and b, and a and b as '

4.8 A series connection of two capacitors. (a) Two capacitors in series Capacitors in series: • The capacitors have the same charge Q. • Their potential differences add: Vac 1 Vcb 5 Vab. a

'

Vac = V1 =

Q C1

Vcb = V2 =

Vab = V = V1 + V2 = Qa

Q C2

1Q + + + + 2Q – – – – C1 Vac 5 V1 Vab 5 V

1Q + + + + C Vcb 5 V2 2Q – – – – 2 b (b) The equivalent single capacitor

1 1 + b C1 C2

a Equivalent capacitance is less than the individual capacitances:

and so V 1 1 = + Q C1 C2

(4.3)

Following a common convention, we use the symbols V1 , V2 , and V to denote the potential differences Vac (across the first capacitor), Vcb (across the second capacitor), and Vab (across the entire combination of capacitors), respectively. '

c

'

Charge is 1Q the same ++ ++ Q V as for the C 5 – – – – eq V individual 2Q 1 1 1 capacitors. 5 1 C1 C2 Ceq b

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CHAPTER 4 Capacitance and Dielectrics

Touch Screens and Capacitance

Application

The touch screen on a mobile phone, an MP3 player, or (as shown here) a medical device uses the physics of capacitors. Behind the screen are two parallel layers, one behind the other, of thin strips of a transparent conductor such as indium tin oxide. A voltage is maintained between the two layers. The strips in one layer are oriented perpendicular to those in the other layer; the points where two strips overlap act as a grid of capacitors. When you bring your finger (a conductor) up to a point on the screen, your finger and the front conducting layer act like a second capacitor in series at that point. The circuitry attached to the conducting layers detects the location of the capacitance change, and so detects where you touched the screen.

The equivalent capacitance Ceq of the series combination is defined as the capacitance of a single capacitor for which the charge Q is the same as for the combination, when the potential difference V is the same. In other words, the combination can be replaced by an equivalent capacitor of capacitance Ceq . For such a capacitor, shown in Fig. 4.8b, '

Ceq =

Q 1 V or = V Ceq Q

(4.4)

Combining Eqs. (4.3) and (4.4), we find 1 1 1 = + Ceq C1 C2 We can extend this analysis to any number of capacitors in series. We find the following result for the reciprocal of the equivalent capacitance: 1 1 1 1 = + + + Á Ceq C1 C2 C3

(capacitors in series)

(4.5)

The reciprocal of the equivalent capacitance of a series combination equals the sum of the reciprocals of the individual capacitances. In a series connection the equivalent capacitance is always less than any individual capacitance. CAUTION Capacitors in series The magnitude of charge is the same on all plates of all the capacitors in a series combination; however, the potential differences of the individual capacitors are not the same unless their individual capacitances are the same. The potential differences of the individual capacitors add to give the total potential difference across the series combination: Vtotal = V1 + V2 + V3 + Á . y

Capacitors in Parallel 4.9 A parallel connection of two capacitors.

The arrangement shown in Fig. 4.9a is called a parallel connection. Two capacitors are connected in parallel between points a and b. In this case the upper plates of the two capacitors are connected by conducting wires to form an equipotential surface, and the lower plates form another. Hence in a parallel connection the potential difference for all individual capacitors is the same and is equal to Vab = V. The charges Q1 and Q2 are not necessarily equal, however, since charges can reach each capacitor independently from the source (such as a battery) of the voltage Vab . The charges are '

Q1 = C1V and Q2 = C2V The total charge Q of the combination, and thus the total charge on the equivalent capacitor, is Q = Q1 + Q2 = 1C1 + C22V so Q = C1 + C2 V

(4.6)

The parallel combination is equivalent to a single capacitor with the same total charge Q = Q1 + Q2 and potential difference V as the combination (Fig. 4.9b). The equivalent capacitance of the combination, Ceq , is the same as the capacitance Q>V of this single equivalent capacitor. So from Eq. (4.6), '

Ceq = C1 + C2

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4.2 Capacitors in Series and Parallel

113

In the same way we can show that for any number of capacitors in parallel, Ceq = C1 + C2 + C3 + Á

(capacitors in parallel)

(4.7)

The equivalent capacitance of a parallel combination equals the sum of the individual capacitances. In a parallel connection the equivalent capacitance is always greater than any individual capacitance. CAUTION Capacitors in parallel The potential differences are the same for all the capacitors in a parallel combination; however, the charges on individual capacitors are not the same unless their individual capacitances are the same. The charges on the individual capacitors add to give the total charge on the parallel combination: Qtotal = Q1 + Q2 + Q3 + Á . [Compare these statements to those in the “Caution” paragraph following Eq. (4.5).] y

Problem-Solving Strategy 4.1

Equivalent Capacitance

IDENTIFY the relevant concepts: The concept of equivalent capacitance is useful whenever two or more capacitors are connected. SET UP the problem using the following steps: 1. Make a drawing of the capacitor arrangement. 2. Identify all groups of capacitors that are connected in series or in parallel. 3. Keep in mind that when we say a capacitor “has charge Q,” we mean that the plate at higher potential has charge + Q and the other plate has charge - Q. EXECUTE the solution as follows: 1. Use Eq. (4.5) to find the equivalent capacitance of capacitors connected in series, as in Fig. 4.8. Such capacitors each have the same charge if they were uncharged before they were connected; that charge is the same as that on the equivalent capacitor. The potential difference across the combination is the sum of the potential differences across the individual capacitors.

Example 4.5

2. Use Eq. (4.7) to find the equivalent capacitance of capacitors connected in parallel, as in Fig. 4.9. Such capacitors all have the same potential difference across them; that potential difference is the same as that across the equivalent capacitor. The total charge on the combination is the sum of the charges on the individual capacitors. 3. After replacing all the series or parallel groups you initially identified, you may find that more such groups reveal themselves. Replace those groups using the same procedure as above until no more replacements are possible. If you then need to find the charge or potential difference for an individual original capacitor, you may have to retrace your steps. EVALUATE your answer: Check whether your result makes sense. If the capacitors are connected in series, the equivalent capacitance Ceq must be smaller than any of the individual capacitances. If the capacitors are connected in parallel, Ceq must be greater than any of the individual capacitances.

Capacitors in series and in parallel

In Figs. 4.8 and 4.9, let C1 = 6.0 mF, C2 = 3.0 mF, and Vab = 18 V. Find the equivalent capacitance and the charge and potential difference for each capacitor when the capacitors are connected (a) in series (see Fig. 4.8) and (b) in parallel (see Fig. 4.9). S O L U T IO N

EXECUTE: (a) From Eq. (4.5) for a series combination, Ceq = 2.0 mF

The charge Q on each capacitor in series is the same as that on the equivalent capacitor: Q = CeqV = 12.0 mF2118 V2 = 36 mC

36 mC Q = = 6.0 V C1 6.0 mF 36 mC Q = V2 = = = 12.0 V C2 3.0 mF

Vac = V1 = Vcb

IDENTIFY and SET UP: In both parts of this example a target variable is the equivalent capacitance Ceq, which is given by Eq. (4.5) for the series combination in part (a) and by Eq. (4.7) for the parallel combination in part (b). In each part we find the charge and potential difference using the definition of capacitance, Eq. (4.1), and the rules outlined in Problem-Solving Strategy 4.1.

1 1 1 1 1 = + = + Ceq C1 C2 6.0 mF 3.0 mF

The potential difference across each capacitor is inversely proportional to its capacitance:

(b) From Eq. (4.7) for a parallel combination, Ceq = C1 + C2 = 6.0 mF + 3.0 mF = 9.0 mF The potential difference across each of the capacitors is the same as that across the equivalent capacitor, 18 V. The charge on each capacitor is directly proportional to its capacitance: Q1 = C1V = 16.0 mF2118 V2 = 108 mC

Q2 = C2V = 13.0 mF2118 V2 = 54 mC

EVALUATE: As expected, the equivalent capacitance Ceq for the series combination in part (a) is less than either C1 or C2, while

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Continued

114

CHAPTER 4 Capacitance and Dielectrics

that for the parallel combination in part (b) is greater than either C1 or C2. For two capacitors in series, as in part (a), the charge is the same on either capacitor and the larger potential difference appears across the capacitor with the smaller capacitance. Furthermore, the sum of the potential differences across the individual capacitors in series equals the potential difference across the

Example 4.6

equivalent capacitor: Vac + Vcb = Vab = 18 V. By contrast, for two capacitors in parallel, as in part (b), each capacitor has the same potential difference and the larger charge appears on the capacitor with the larger capacitance. Can you show that the total charge Q1 + Q2 on the parallel combination is equal to the charge Q = CeqV on the equivalent capacitor?

A capacitor network

Find the equivalent capacitance of the five-capacitor network shown in Fig. 4.10a.

This gives us the equivalent combination of Fig. 4.10b. Now we see three capacitors in parallel, and we use Eq. (4.7) to replace them with their equivalent capacitance C–:

SOLUTION

C– = 3 mF + 11 mF + 4 mF = 18 mF

IDENTIFY and SET UP: These capacitors are neither all in series nor all in parallel. We can, however, identify portions of the arrangement that are either in series or parallel. We combine these as described in Problem-Solving Strategy 4.1 to find the net equivalent capacitance, using Eq. (4.5) for series connections and Eq. (4.7) for parallel connections. EXECUTE: The caption of Fig. 4.10 outlines our procedure. We first use Eq. (4.5) to replace the 12-mF and 6-mF series combination by its equivalent capacitance C¿: 1 1 1 = + C¿ 12 mF 6 mF

C¿ = 4 mF

This gives us the equivalent combination of Fig. 4.10c, which has two capacitors in series. We use Eq. (4.5) to replace them with their equivalent capacitance Ceq, which is our target variable (Fig. 4.10d): 1 1 1 + = Ceq 18 mF 9 mF

Ceq = 6 mF

EVALUATE: If the potential difference across the entire network in Fig. 4.10a is Vab = 9.0 V, the net charge on the network is Q = CeqVab = 16 mF219.0 V2 = 54 mC. Can you find the charge on, and the voltage across, each of the five individual capacitors?

4.10 (a) A capacitor network between points a and b. (b) The 12-mF and 6-mF capacitors in series in (a) are replaced by an equivalent 4-mF capacitor. (c) The 3-mF, 11-mF, and 4-mF capacitors in parallel in (b) are replaced by an equivalent 18-mF capacitor. (d) Finally, the 18-mF and 9-mF capacitors in series in (c) are replaced by an equivalent 6-mF capacitor. (a)

(b)

a

(d) a

(c) a

a

12 mF 3 mF

11 mF

3 mF

11 mF

4 mF

18 mF

... replace these series capacitors by an equivalent capacitor.

6 mF 9 mF

6 mF

Replace these series capacitors by an equivalent capacitor ...

9 mF b

b

... replace these parallel capacitors by an equivalent capacitor ...

9 mF b

b

Test Your Understanding of Section 4.2 You want to connect a 4-mF capacitor and an 8-mF capacitor. (a) With which type of connection will the 4-mF capacitor have a greater potential difference across it than the 8-mF capacitor? (i) series; (ii) parallel; (iii) either series or parallel; (iv) neither series nor parallel. (b) With which type of connection will the 4-mF capacitor have a greater charge than the 8-mF capacitor? (i) series; (ii) parallel; (iii) either series or parallel; (iv) neither series nor parallel. y

4.3

Energy Storage in Capacitors and Electric-Field Energy

Many of the most important applications of capacitors depend on their ability to store energy. The electric potential energy stored in a charged capacitor is just equal to the amount of work required to charge it—that is, to separate opposite charges and place them on different conductors. When the capacitor is discharged, this stored energy is recovered as work done by electrical forces.

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4.3 Energy Storage in Capacitors and Electric-Field Energy

We can calculate the potential energy U of a charged capacitor by calculating the work W required to charge it. Suppose that when we are done charging the capacitor, the final charge is Q and the final potential difference is V. From Eq. (4.1) these quantities are related by V =

Q C

Let q and v be the charge and potential difference, respectively, at an intermediate stage during the charging process; then v = q>C. At this stage the work dW required to transfer an additional element of charge dq is dW = v dq =

q dq C

The total work W needed to increase the capacitor charge q from zero to a final value Q is W

W =

20

Q

dW =

Q2 1 q dq = C 20 2C

(work to charge a capacitor)

(4.8)

This is also equal to the total work done by the electric field on the charge when the capacitor discharges. Then q decreases from an initial value Q to zero as the elements of charge dq “fall” through potential differences v that vary from V down to zero. If we define the potential energy of an uncharged capacitor to be zero, then W in Eq. (4.8) is equal to the potential energy U of the charged capacitor. The final stored charge is Q = CV, so we can express U (which is equal to W) as

U =

Q2 = 12 CV2 = 12 QV 2C

(potential energy stored in a capacitor)

(4.9)

When Q is in coulombs, C in farads (coulombs per volt), and V in volts (joules per coulomb), U is in joules. The last form of Eq. (4.9), U = 12 QV, shows that the total work W required to charge the capacitor is equal to the total charge Q multiplied by the average 1 potential difference 2 V during the charging process. The expression U = 121Q2>C2 in Eq. (4.9) shows that a charged capacitor is the electrical analog of a stretched spring with elastic potential energy U = 12 kx2. The charge Q is analogous to the elongation x, and the reciprocal of the capacitance, 1>C, is analogous to the force constant k. The energy supplied to a capacitor in the charging process is analogous to the work we do on a spring when we stretch it. Equations (4.8) and (4.9) tell us that capacitance measures the ability of a capacitor to store both energy and charge. If a capacitor is charged by connecting it to a battery or other source that provides a fixed potential difference V, then increasing the value of C gives a greater charge Q = CV and a greater amount of stored energy U = 12 CV2. If instead the goal is to transfer a given quantity of charge Q from one conductor to another, Eq. (4.8) shows that the work W required is inversely proportional to C; the greater the capacitance, the easier it is to give a capacitor a fixed amount of charge.

Applications of Capacitors: Energy Storage

?

Most practical applications of capacitors take advantage of their ability to store and release energy. In electronic flash units used by photographers, the energy stored in a capacitor (see Fig. 4.4) is released by depressing the camera’s shutter button. This provides a conducting path from one capacitor plate to the other through the flash tube. Once this path is established, the stored energy is rapidly converted into a brief but intense flash of light. An extreme example of the same principle is the Z machine at Sandia National Laboratories in New

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116

CHAPTER 4 Capacitance and Dielectrics

4.11 The Z machine uses a large number of capacitors in parallel to give a tremendous equivalent capacitance C (see Section 4.2). Hence a large amount of energy U = 12 CV2 can be stored with even a modest potential difference V. The arcs shown here are produced when the capacitors discharge their energy into a target, which is no larger than a spool of thread. This heats the target to a temperature higher than 2 * 109 K.

Mexico, which is used in experiments in controlled nuclear fusion (Fig. 4.11). A bank of charged capacitors releases more than a million joules of energy in just a few billionths of a second. For that brief space of time, the power output of the Z machine is 2.9 * 1014 W, or about 80 times the power output of all the electric power plants on earth combined! In other applications, the energy is released more slowly. Springs in the suspension of an automobile help smooth out the ride by absorbing the energy from sudden jolts and releasing that energy gradually; in an analogous way, a capacitor in an electronic circuit can smooth out unwanted variations in voltage due to power surges. We’ll discuss these circuits in detail in Chapter 6.

Electric-Field Energy We can charge a capacitor by moving electrons directly from one plate to another. This requires doing work against the electric field between the plates. Thus we can think of the energy as being stored in the field in the region between the plates. To develop this relationship, let’s find the energy per unit volume in the space between the plates of a parallel-plate capacitor with plate area A and separation d. We call this the energy density, denoted by u. From Eq. (4.9) the total stored potential energy is 12 CV2 and the volume between the plates is just Ad; hence the energy density is u = Energy density =

1 2 2 CV

(4.10)

Ad

From Eq. (4.2) the capacitance C is given by C = P0 A>d. The potential difference V is related to the electric-field magnitude E by V = Ed. If we use these expressions in Eq. (4.10), the geometric factors A and d cancel, and we find '

u = 12 P0 E2 '

(electric energy density in a vacuum)

(4.11)

Although we have derived this relationship only for a parallel-plate capacitor, it turns out to be valid for any capacitor in vacuum and indeed for any electric field configuration in vacuum. This result has an interesting implication. We think of vacuum as space with no matter in it, but vacuum can nevertheless have electric fields and therefore energy. Thus “empty” space need not be truly empty after all. We will use this idea and Eq. (4.11) in Chapter 12 in connection with the energy transported by electromagnetic waves. CAUTION Electric-field energy is electric potential energy It’s a common misconception that electric-field energy is a new kind of energy, different from the electric potential energy described before. This is not the case; it is simply a different way of interpreting electric potential energy. We can regard the energy of a given system of charges as being a shared property of all the charges, or we can think of the energy as being a property of the electric field that the charges create. Either interpretation leads to the same value of the potential energy. y

Example 4.7

Transferring charge and energy between capacitors

We connect a capacitor C1 = 8.0 mF to a power supply, charge it to a potential difference V0 = 120 V, and disconnect the power supply (Fig. 4.12). Switch S is open. (a) What is the charge Q0 on C1? (b) What is the energy stored in C1? (c) Capacitor C2 = 4.0 mF is initially uncharged. We close switch S. After charge no longer flows, what is the potential difference across each capacitor, and what is the charge on each capacitor? (d) What is the final energy of the system?

4.12 When the switch S is closed, the charged capacitor C1 is connected to an uncharged capacitor C2 . The center part of the switch is an insulating handle; charge can flow only between the two upper terminals and between the two lower terminals. '

Q0 ++ ++ –– ––

C1 5 8.0 mF

V0 5 120 V S

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C2 5 4.0 mF

4.3 Energy Storage in Capacitors and Electric-Field Energy

117

S O L U T IO N IDENTIFY and SET UP: In parts (a) and (b) we find the charge Q0 and stored energy Uinitial for the single charged capacitor C1 using Eqs. (4.1) and (4.9), respectively. After we close switch S, one wire connects the upper plates of the two capacitors and another wire connects the lower plates; the capacitors are now connected in parallel. In part (c) we use the character of the parallel connection to determine how Q0 is shared between the two capacitors. In part (d) we again use Eq. (4.9) to find the energy stored in capacitors C1 and C2; the energy of the system is the sum of these values. EXECUTE: (a) The initial charge Q0 on C1 is

Q0 = C1V0 = 18.0 mF21120 V2 = 960 mC

of charge requires that Q1 + Q2 = Q0. The potential difference V between the plates is the same for both capacitors because they are connected in parallel, so the charges are Q1 = C1V and Q2 = C2V. We now have three independent equations relating the three unknowns Q1, Q2, and V. Solving these, we find V =

Q1 = 640 mC

Uinitial =

=

1 2 1960

* 10

-6

Ufinal = 12 Q1V + 12 Q2V = 12 Q0V = 12 1960 * 10-6 C2180 V2 = 0.038 J

C21120 V2 = 0.058 J

(c) When we close the switch, the positive charge Q0 is distributed over the upper plates of both capacitors and the negative charge -Q0 is distributed over the lower plates. Let Q1 and Q2 be the magnitudes of the final charges on the capacitors. Conservation

Example 4.8

EVALUATE: The final energy is less than the initial energy; the difference was converted to energy of some other form. The conductors become a little warmer because of their resistance, and some energy is radiated as electromagnetic waves. We’ll study the behavior of capacitors in more detail in Chapters 6 and 11.

Electric-field energy

(a) What is the magnitude of the electric field required to store 1.00 J of electric potential energy in a volume of 1.00 m3 in vacuum? (b) If the field magnitude is 10 times larger than that, how much energy is stored per cubic meter? S O L U T IO N IDENTIFY and SET UP: We use the relationship between the electricfield magnitude E and the energy density u. In part (a) we use the given information to find u; then we use Eq. (4.11) to find the corresponding value of E. In part (b), Eq. (4.11) tells us how u varies with E. EXECUTE: (a) The desired energy density is u = 1.00 J>m3. Then from Eq. (4.11),

Example 4.9

Q2 = 320 mC

(d) The final energy of the system is

(b) The energy initially stored in C1 is 1 2 Q0V0

Q0 960 mC = = 80 V C1 + C2 8.0 mF + 4.0 mF

E =

211.00 J>m32 2u = B P0 B 8.85 * 10-12 C2>N # m2

= 4.75 * 105 N>C = 4.75 * 105 V>m (b) Equation (4.11) shows that u is proportional to E2. If E increases by a factor of 10, u increases by a factor of 102 = 100, so the energy density becomes u = 100 J>m3. EVALUATE: Dry air can sustain an electric field of about 3 * 106 V>m without experiencing dielectric breakdown, which we will discuss in Section 4.4. There we will see that field magnitudes in practical insulators can be as great as this or even larger.

Two ways to calculate energy stored in a capacitor

The spherical capacitor described in Example 4.3 (Section 4.1) has charges + Q and - Q on its inner and outer conductors. Find the electric potential energy stored in the capacitor (a) by using the capacitance C found in Example 4.3 and (b) by integrating the electric-field energy density u. S O L U T IO N IDENTIFY and SET UP: We can determine the energy U stored in a capacitor in two ways: in terms of the work done to put the charges on the two conductors, and in terms of the energy in the electric field between the conductors. The descriptions are equivalent, so they must give us the same result. In Example 4.3 we found the capacitance C and the field magnitude E in the space between the conductors. (The electric field is zero inside the inner sphere and is also zero outside the inner surface of the outer sphere, because a Gaussian surface with radius r 6 ra or r 7 rb encloses zero net

charge. Hence the energy density is nonzero only in the space between the spheres, ra 6 r 6 rb.) In part (a) we use Eq. (4.9) to find U. In part (b) we use Eq. (4.11) to find u, which we integrate over the volume between the spheres to find U. EXECUTE: (a) From Example 4.3, the spherical capacitor has capacitance rarb C = 4pP0 rb - ra where ra and rb are the radii of the inner and outer conducting spheres, respectively. From Eq. (4.9) the energy stored in this capacitor is U =

Q2 Q2 rb - ra = 2C 8pP0 rarb

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Continued

118

CHAPTER 4 Capacitance and Dielectrics rb

(b) The electric field in the region ra 6 r 6 rb between the two conducting spheres has magnitude E = Q>4pP0r2. The energy density in this region is u = 12 P0E2 = 12 P0 a

2

b = 2

Q 4pP0r

U =

2r

a

a

Q2 32p2P0r4

b4pr2 dr

rb Q2 Q2 dr 1 1 = a+ b = 2 2 r r 8pP0 ra r 8pP0 b a

Q2 32p2P0r4

= The energy density is not uniform; it decreases rapidly with increasing distance from the center of the capacitor. To find the total electric-field energy, we integrate u (the energy per unit volume) over the region ra 6 r 6 rb. We divide this region into spherical shells of radius r, surface area 4pr2, thickness dr, and volume dV = 4pr2 dr. Then

2

u dV =

Q2 rb - ra 8pP0 rarb

EVALUATE: Electric potential energy can be regarded as being associated with either the charges, as in part (a), or the field, as in part (b); the calculated amount of stored energy is the same in either case.

Test Your Understanding of Section 4.3 You want to connect a 4-mF capacitor and an 8-mF capacitor. With which type of connection will the 4-mF capacitor have a greater amount of stored energy than the 8-mF capacitor? (i) series; (ii) parallel; (iii) either series or parallel; (iv) neither series nor parallel.

4.4 4.13 A common type of capacitor uses dielectric sheets to separate the conductors.

Conductor (metal foil)

Conductor (metal foil)

Dielectric (plastic sheets)

y

Dielectrics

Most capacitors have a nonconducting material, or dielectric, between their conducting plates. A common type of capacitor uses long strips of metal foil for the plates, separated by strips of plastic sheet such as Mylar. A sandwich of these materials is rolled up, forming a unit that can provide a capacitance of several microfarads in a compact package (Fig. 4.13). Placing a solid dielectric between the plates of a capacitor serves three functions. First, it solves the mechanical problem of maintaining two large metal sheets at a very small separation without actual contact. Second, using a dielectric increases the maximum possible potential difference between the capacitor plates. As we described in Section 3.3, any insulating material, when subjected to a sufficiently large electric field, experiences a partial ionization that permits conduction through it. This is called dielectric breakdown. Many dielectric materials can tolerate stronger electric fields without breakdown than can air. Thus using a dielectric allows a capacitor to sustain a higher potential difference V and so store greater amounts of charge and energy. Third, the capacitance of a capacitor of given dimensions is greater when there is a dielectric material between the plates than when there is vacuum. We can demonstrate this effect with the aid of a sensitive electrometer, a device that measures the potential difference between two conductors without letting any appreciable charge flow from one to the other. Figure 4.14a shows an electrometer connected across a charged capacitor, with magnitude of charge Q on each plate and potential difference V0 . When we insert an uncharged sheet of dielectric, such as glass, paraffin, or polystyrene, between the plates, experiment shows that the potential difference decreases to a smaller value V (Fig. 4.14b). When we remove the dielectric, the potential difference returns to its original value V0 , showing that the original charges on the plates have not changed. The original capacitance C0 is given by C0 = Q>V0 , and the capacitance C with the dielectric present is C = Q>V. The charge Q is the same in both cases, and V is less than V0 , so we conclude that the capacitance C with the dielectric present is greater than C0 . When the space between plates is completely filled by the dielectric, the ratio of C to C0 (equal to the ratio of V0 to V ) is called the dielectric constant of the material, K: '

'

'

'

'

K =

C C0

(definition of dielectric constant)

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(4.12)

4.4 Dielectrics

When the charge is constant, Q = C0 V0 = CV and C>C0 = V0>V. In this case, Eq. (4.12) can be rewritten as V0 V = 1when Q is constant2 (4.13) K With the dielectric present, the potential difference for a given charge Q is reduced by a factor K. The dielectric constant K is a pure number. Because C is always greater than C0 , K is always greater than unity. Some representative values of K are given in Table 4.1. For vacuum, K = 1 by definition. For air at ordinary temperatures and pressures, K is about 1.0006; this is so nearly equal to 1 that for most purposes an air capacitor is equivalent to one in vacuum. Note that while water has a very large value of K, it is usually not a very practical dielectric for use in capacitors. The reason is that while pure water is a very poor conductor, it is also an excellent ionic solvent. Any ions that are dissolved in the water will cause charge to flow between the capacitor plates, so the capacitor discharges. '

119

4.14 Effect of a dielectric between the plates of a parallel-plate capacitor. (a) With a given charge, the potential difference is V0 . (b) With the same charge but with a dielectric between the plates, the potential difference V is smaller than V0 . '

'

(a) Vacuum

'

Q

2Q

V0

Table 4.1 Values of Dielectric Constant K at 20°C K

Material Vacuum

1

Material

K

Polyvinyl chloride

3.18

®

Air (1 atm)

1.00059

Plexiglas

Air (100 atm)

1.0548

Glass

Teflon

2.1

Neoprene

Polyethylene

2.25

Germanium

16

Benzene

2.28

Glycerin

42.5

Mica Mylar

3–6 3.1

Water Strontium titanate

3.40 5–10

(b) Dielectric

6.70 2Q

Q

80.4 310

No real dielectric is a perfect insulator. Hence there is always some leakage current between the charged plates of a capacitor with a dielectric. We tacitly ignored this effect in Section 4.2 when we derived expressions for the equivalent capacitances of capacitors in series, Eq. (4.5), and in parallel, Eq. (4.7). But if a leakage current flows for a long enough time to substantially change the charges from the values we used to derive Eqs. (4.5) and (4.7), those equations may no longer be accurate.

Induced Charge and Polarization When a dielectric material is inserted between the plates while the charge is kept constant, the potential difference between the plates decreases by a factor K. Therefore the electric field between the plates must decrease by the same factor. If E0 is the vacuum value and E is the value with the dielectric, then E0 1when Q is constant2 (4.14) K Since the electric-field magnitude is smaller when the dielectric is present, the surface charge density (which causes the field) must be smaller as well. The surface charge on the conducting plates does not change, but an induced charge of the opposite sign appears on each surface of the dielectric (Fig. 4.15). The dielectric was originally electrically neutral and is still neutral; the induced surface charges arise as a result of redistribution of positive and negative charge within the dielectric material, a phenomenon called polarization. We first encountered polarization in Section 1.2, and we suggest that you reread the discussion of Fig. 1.8. We will assume that the induced surface charge is directly proportional to the electric-field magnitude E in the material; this is indeed the case for many common dielectrics. (This direct proportionality is analogous to Hooke’s law for a E =

–

+

Electrometer (measures potential difference across plates)

V

–

+

Adding the dielectric reduces the potential difference across the capacitor.

4.15 Electric field lines with (a) vacuum between the plates and (b) dielectric between the plates. (a)

(b) Dielectric

Vacuum

s + + + + + + + + + + + + + + s

S

E0

2s – – – – – – – – – – – – – – 2s

s 2s 2si si – + – +– + S + E – +– +– – + +– +– – + Induced +– +– charges – + +– +– – + +– +– – + +– +– 2si si s 2s

For a given charge density s, the induced charges on the dielectric’s surfaces reduce the electric field between the plates.

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120

CHAPTER 4 Capacitance and Dielectrics

spring.) In that case, K is a constant for any particular material. When the electric field is very strong or if the dielectric is made of certain crystalline materials, the relationship between induced charge and the electric field can be more complex; we won’t consider such cases here. We can derive a relationship between this induced surface charge and the charge on the plates. Let’s denote the magnitude of the charge per unit area induced on the surfaces of the dielectric (the induced surface charge density) by si . The magnitude of the surface charge density on the capacitor plates is s, as usual. Then the net surface charge on each side of the capacitor has magnitude 1s - si2 , as shown in Fig. 4.15b. As we found in Example 1.12 (Section 1.5) and in Example 2.8 (Section 2.4), the field between the plates is related to the net surface charge density by E = snet>P0 . Without and with the dielectric, respectively, we have '

'

'

E0 =

s P0

E =

s - si P0

(4.15)

Using these expressions in Eq. (4.14) and rearranging the result, we find si = s a1 -

1 b K

(induced surface charge density)

(4.16)

This equation shows that when K is very large, si is nearly as large as s. In this case, si nearly cancels s, and the field and potential difference are much smaller than their values in vacuum. The product KP0 is called the permittivity of the dielectric, denoted by P: P = KP0

1definition of permittivity2

(4.17)

In terms of P we can express the electric field within the dielectric as E =

s P

(4.18)

The capacitance when the dielectric is present is given by

C = KC0 = KP0

A A = P d d

(parallel-plate capacitor, dielectric between plates)

(4.19)

We can repeat the derivation of Eq. (4.11) for the energy density u in an electric field for the case in which a dielectric is present. The result is u = 12 KP0 E2 = 12 PE2 '

(electric energy density in a dielectric)

(4.20)

In empty space, where K = 1, P = P0 and Eqs. (4.19) and (4.20) reduce to Eqs. (4.2) and (4.11), respectively, for a parallel-plate capacitor in vacuum. For this reason, P0 is sometimes called the “permittivity of free space” or the “permittivity of vacuum.” Because K is a pure number, P and P0 have the same units, C2>N # m2 or F>m. Equation (4.19) shows that extremely high capacitances can be obtained with plates that have a large surface area A and are separated by a small distance d by a dielectric with a large value of K. In an electrolytic double-layer capacitor, tiny carbon granules adhere to each plate: The value of A is the combined surface area of the granules, which can be tremendous. The plates with granules attached are separated by a very thin dielectric sheet. A capacitor of this kind can have a capacitance of 5000 farads yet fit in the palm of your hand (compare Example 4.1 in Section 4.1). Several practical devices make use of the way in which a capacitor responds to a change in dielectric constant. One example is an electric stud finder, used by

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4.4 Dielectrics

121

home repair workers to locate metal studs hidden behind a wall’s surface. It consists of a metal plate with associated circuitry. The plate acts as one half of a capacitor, with the wall acting as the other half. If the stud finder moves over a metal stud, the effective dielectric constant for the capacitor changes, changing the capacitance and triggering a signal.

Problem-Solving Strategy 4.2

Dielectrics

IDENTIFY the relevant concepts: The relationships in this section are useful whenever there is an electric field in a dielectric, such as a dielectric between charged capacitor plates. Typically you must relate the potential difference Vab between the plates, the electric field magnitude E in the capacitor, the charge density s on the capacitor plates, and the induced charge density si on the surfaces of the capacitor. SET UP the problem using the following steps: 1. Make a drawing of the situation. 2. Identify the target variables, and choose which equations from this section will help you solve for those variables. EXECUTE the solution as follows: 1. In problems such as the next example, it is easy to get lost in a blizzard of formulas. Ask yourself at each step what kind of

Example 4.10

EVALUATE your answer: With a dielectric present, (a) the capacitance is greater than without a dielectric; (b) for a given charge on the capacitor, the electric field and potential difference are less than without a dielectric; and (c) the magnitude of the induced surface charge density si on the dielectric is less than that of the charge density s on the capacitor plates.

A capacitor with and without a dielectric

Suppose the parallel plates in Fig. 4.15 each have an area of 2000 cm2 (2.00 * 10-1 m2) and are 1.00 cm (1.00 * 10-2 m) apart. We connect the capacitor to a power supply, charge it to a potential difference V0 = 3.00 kV, and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00 kV while the charge on each capacitor plate remains constant. Find (a) the original capacitance C0; (b) the magnitude of charge Q on each plate; (c) the capacitance C after the dielectric is inserted; (d) the dielectric constant K of the dielectric; (e) the permittivity P of the dielectric; (f ) the magnitude of the induced charge Qi on each face of the dielectric; (g) the original electric field E0 between the plates; and (h) the electric field E after the dielectric is inserted. S O L U T IO N

Q = C0V0 = 11.77 * 10-10 F213.00 * 103 V2 = 5.31 * 10-7 C = 0.531 mC

(c) When the dielectric is inserted, Q is unchanged but the potential difference decreases to V = 1.00 kV. Hence from Eq. (4.1), the new capacitance is C =

Q 5.31 * 10-7 C = = 5.31 * 10-10 F = 531 pF V 1.00 * 103 V

(d) From Eq. (4.12), the dielectric constant is K =

531 pF 5.31 * 10-10 F C = 3.00 = = -10 C0 177 pF 1.77 * 10 F

Alternatively, from Eq. (4.13), K =

IDENTIFY and SET UP: This problem uses most of the relationships we have discussed for capacitors and dielectrics. (Energy relationships are treated in Example 4.11.) Most of the target variables can be obtained in several ways. The methods used below are a sample; we encourage you to think of others and compare your results. EXECUTE: (a) With vacuum between the plates, we use Eq. (4.19) with K = 1 : '

C0 = P0

quantity each symbol represents. For example, distinguish clearly between charges and charge densities, and between electric fields and electric potential differences. 2. Check for consistency of units. Distances must be in meters. A microfarad is 10-6 farad, and so on. Don’t confuse the numerical value of P0 with the value of 1>4pP0. Electric-field magnitude can be expressed in both N>C and V>m. The units of P0 are C2>N # m2 or F>m.

A 2.00 * 10-1 m2 = 18.85 * 10-12 F>m2 d 1.00 * 10-2 m

= 1.77 * 10-10 F = 177 pF

V0 3000 V = = 3.00 V 1000 V

(e) Using K from part (d) in Eq. (4.17), the permittivity is P = KP0 = 13.00218.85 * 10-12 C2>N # m22 = 2.66 * 10-11 C2>N # m2

(f) Multiplying both sides of Eq. (4.16) by the plate area A gives the induced charge Qi = si A in terms of the charge Q = sA on each plate: Qi = Qa 1 -

1 1 b = 15.31 * 10-7 C2 a1 b K 3.00

= 3.54 * 10-7 C

(b) From the definition of capacitance, Eq. (4.1),

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Continued

122

CHAPTER 4 Capacitance and Dielectrics

(g) Since the electric field between the plates is uniform, its magnitude is the potential difference divided by the plate separation:

or, from Eq. (4.14),

1000 V V = = 1.00 * 105 V>m E = d 1.00 * 10-2 m

E =

or, from Eq. (4.18), Q 5.31 * 10-7 C s = = -11 P PA 12.66 * 10 C2>N # m2212.00 * 10-1 m22

= 1.00 * 105 V>m or, from Eq. (4.15),

SOLUTION

EXECUTE: From Eq. (4.9), the stored energies U0 and U without and with the dielectric in place are U0 = 12 C0V02 = 12 11.77 * 10-10 F213000 V22 = 7.97 * 10-4 J U = 12 CV2 = 12 15.31 * 10-10 F211000 V22 = 2.66 * 10-4 J

The final energy is one-third of the original energy. Equation (4.20) gives the energy densities without and with the dielectric: u0 = 12 P0 E02 = 12 18.85 * 10-12 C2>N # m2213.00 * 105 N>C22 = 0.398 J>m3 =

EVALUATE: Inserting the dielectric increased the capacitance by a factor of K = 3.00 and reduced the electric field between the plates by a factor of 1>K = 1>3.00. It did so by developing induced charges on the faces of the dielectric of magnitude Q11 - 1>K2 = Q11 - 1>3.002 = 0.667Q.

0.00200 m3. Since the electric field between the plates is uniform, u0 is uniform as well and the energy density is just the stored energy divided by the volume: U0 7.97 * 10-4 J = = 0.398 J>m3 V 0.00200 m3 This agrees with our earlier answer. You can use the same approach to check our result for u. In general, when a dielectric is inserted into a capacitor while the charge on each plate remains the same, the permittivity P increases by a factor of K (the dielectric constant), and the electric field E and the energy density u = 12 PE2 decrease by a factor of 1>K. Where does the energy go? The answer lies in the fringing field at the edges of a real parallel-plate capacitor. As Fig. 4.16 shows, that field tends to pull the dielectric into the space between the plates, doing work on it as it does so. We could attach a spring to the left end of the dielectric in Fig. 4.16 and use this force to stretch the spring. Because work is done by the field, the field energy density decreases. u0 =

IDENTIFY and SET UP: We now consider the ideas of energy stored in a capacitor and of electric-field energy density. We use Eq. (4.9) to find the stored energy and Eq. (4.20) to find the energy density.

u =

3.00 * 105 V>m E0 = = 1.00 * 105 V>m K 3.00

Energy storage with and without a dielectric

Find the energy stored in the electric field of the capacitor in Example 4.10 and the energy density, both before and after the dielectric sheet is inserted.

1 2 2 PE

18.85 * 10-12 C2>N # m2212.00 * 10-1 m22

= 1.00 * 105 V>m

(h) After the dielectric is inserted,

Example 4.11

15.31 - 3.542 * 10-7 C

=

V0 3000 V = = 3.00 * 105 V>m E0 = d 1.00 * 10-2 m

E =

s - si Q - Qi = P0 P0 A

E =

1 2 12.66

* 10

-11

C >N # m2211.00 * 2

4.16 The fringing field at the edges of the capacitor exerts forces S S F-i and F+i on the negative and positive induced surface charges of a dielectric, pulling the dielectric into the capacitor. S

5

F2i

2

10 N>C2

= 0.133 J>m3

+ + + + + + + + + + + + + + Dielectric S

E

The energy density with the dielectric is one-third of the original energy density. EVALUATE: We can check our answer for u0 by noting that the volume between the plates is V = 10.200 m2210.0100 m2 =

–

– – – – – – – – – – – – –

S

F1i

Dielectric Breakdown We mentioned earlier that when a dielectric is subjected to a sufficiently strong electric field, dielectric breakdown takes place and the dielectric becomes a conductor. This occurs when the electric field is so strong that electrons are ripped loose from their molecules and crash into other molecules, liberating even more

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4.5 Molecular Model of Induced Charge

electrons. This avalanche of moving charge forms a spark or arc discharge. Lightning is a dramatic example of dielectric breakdown in air. Because of dielectric breakdown, capacitors always have maximum voltage ratings. When a capacitor is subjected to excessive voltage, an arc may form through a layer of dielectric, burning or melting a hole in it. This arc creates a conducting path (a short circuit) between the conductors. If a conducting path remains after the arc is extinguished, the device is rendered permanently useless as a capacitor. The maximum electric-field magnitude that a material can withstand without the occurrence of breakdown is called its dielectric strength. This quantity is affected significantly by temperature, trace impurities, small irregularities in the metal electrodes, and other factors that are difficult to control. For this reason we can give only approximate figures for dielectric strengths. The dielectric strength of dry air is about 3 * 106 V>m. Table 4.2 lists the dielectric strengths of a few common insulating materials. Note that the values are all substantially greater than the value for air. For example, a layer of polycarbonate 0.01 mm thick (about the smallest practical thickness) has 10 times the dielectric strength of air and can withstand a maximum voltage of about 13 * 107 V>m211 * 10-5 m2 = 300 V.

123

Dielectric Cell Membrane

Application

The membrane of a living cell behaves like a dielectric between the plates of a capacitor. The membrane is made of two sheets of lipid molecules, with their water-insoluble ends in the middle and their water-soluble ends (shown in red) on the surfaces of the membrane. The conductive fluids on either side of the membrane (water with negative ions inside the cell, water with positive ions outside) act as charged capacitor plates, and the nonconducting membrane acts as a dielectric with K of about 10. The potential difference V across the membrane is about 0.07 V and the membrane thickness d is about 7 * 10-9 m, so the electric field E 5 V/d in the membrane is about 107 V/m—close to the dielectric strength of the membrane. If the membrane were made of air, V and E would be larger by a factor of K L 10 and dielectric breakdown would occur.

Table 4.2 Dielectric Constant and Dielectric Strength of Some Insulating Materials

3.3

6 * 107

Polypropylene

2.2

7 * 107

Polystyrene

2.6

2 * 107

Pyrex glass

4.7

1 * 107

Test Your Understanding of Section 4.4 The space between the plates of an isolated parallel-plate capacitor is filled by a slab of dielectric with dielectric constant K. The two plates of the capacitor have charges Q and -Q. You pull out the dielectric slab. If the charges do not change, how does the energy in the capacitor change when you remove the slab? (i) It increases; (ii) it decreases; (iii) it remains the same.

4.5

y

Molecular Model of Induced Charge

– +

'

+

'

(a) –

In the absence of an electric field, polar molecules orient randomly.

(b) S

–

–

+

+

– +

– +

In Section 4.4 we discussed induced surface charges on a dielectric in an electric field. Now let’s look at how these surface charges can arise. If the material were a conductor, the answer would be simple. Conductors contain charge that is free to move, and when an electric field is present, some of the charge redistributes itself on the surface so that there is no electric field inside the conductor. But an ideal dielectric has no charges that are free to move, so how can a surface charge occur? To understand this, we have to look again at rearrangement of charge at the molecular level. Some molecules, such as H2 O and N2 O, have equal amounts of positive and negative charges but a lopsided distribution, with excess positive charge concentrated on one side of the molecule and negative charge on the other. As we described in Section 1.7, such an arrangement is called an electric dipole, and the molecule is called a polar molecule. When no electric field is present in a gas or liquid with polar molecules, the molecules are oriented randomly (Fig. 4.17a). When they are placed in an electric field, however, they tend to orient themselves as in Fig. 4.17b, as a result of the electric-field torques described inSSection 1.7. Because of thermal agitation, the alignment of the molecules with E is not perfect.

4.17 Polar molecules (a) without and (b) S with an applied electric field E.

– +

Polyester

+

3 * 107

–

2.8

– +

Polycarbonate

+

Dielectric Strength, Em (V/m)

–

Dielectric Constant, K

Material

– +

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– +

E

When an electric field is applied, the molecules tend to align with it.

124

CHAPTER 4 Capacitance and Dielectrics S

4.18 Nonpolar molecules (a) without and (b) with an applied electric field E. PhET: Molecular Motors PhET: Optical Tweezers and Applications PhET: Stretching DNA

(a)

(b) S

In the absence of an electric field, nonpolar molecules are not electric dipoles.

4.19 Polarization of a dielectric in an S electric field E gives rise to thin layers of bound charges on the surfaces, creating surface charge densities si and - si . The sizes of the molecules are greatly exaggerated for clarity. '

2si

si

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

– +

2si

si d

S

E

–

–

–

+

+

–

+

–

E

+

–

+

+

An electric field causes the molecules’ positive and negative charges to separate slightly, making the molecule effectively polar.

Even a molecule that is not ordinarily polar becomes a dipole when it is placed in an electric field because the field pushes the positive charges in the molecules in the direction of the field and pushes the negative charges in the opposite direction. This causes a redistribution of charge within the molecule (Fig. 4.18). Such dipoles are called induced dipoles. With either polar or nonpolar molecules, the redistribution of charge caused by the field leads to the formation of a layer of charge on each surface of the dielectric material (Fig. 4.19). These layers are the surface charges described in Section 4.4; their surface charge density is denoted by si. The charges are not free to move indefinitely, as they would be in a conductor, because each charge is bound to a molecule. They are in fact called bound charges to distinguish them from the free charges that are added to and removed from the conducting capacitor plates. In the interior of the material the net charge per unit volume remains zero. As we have seen, this redistribution of charge is called polarization, and we say that the material is polarized. The four parts of Fig. 4.20 show the behavior of a slab of dielectric when it is inserted in the field between a pair of oppositely charged capacitor plates. Figure 4.20a shows the original field. Figure 4.20b is the situation after the dielectric has been inserted but before any rearrangement of charges has occurred. Figure 4.20c shows by thinner arrows the additional field set up in the dielectric by its induced surface charges. This field is opposite to the original field, but it is not great enough to cancel the original field completely because the charges in the dielectric are not free to move indefinitely. The 4.20 (a) Electric field of magnitude E0 between two charged plates. (b) Introduction of a dielectric of dielectric constant K. (c) The induced surface charges and their field. (d) Resultant field of magnitude E0 /K. (a) No dielectric s + + + + + + + + + + + + + + s

S

E0

2s – – – – – – – – – – – – – – 2s

(b) Dielectric just inserted + + + + + + + + + + + + + +

– – – – – – – – – – – – – –

(c) Induced charges create electric field 2si + +– + +– + +– + +– + +– + +– + +– 2si Original electric field

si

– +– – +– – +– – +– – +– – +– – +–

si

(d) Resultant field s 2si + +– + +– + +– + +– + +– + +– + +– 2si s

2s – +– – +– – +– – +– – +– – +– – +–

si

si

2s

Weaker field in dielectric due to induced (bound) charges

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125

4.6 Gauss’s Law in Dielectrics

resultant field in the dielectric, shown in Fig. 4.20d, is therefore decreased in magnitude. In the field-line representation, some of the field lines leaving the positive plate go through the dielectric, while others terminate on the induced charges on the faces of the dielectric. As we discussed in Section 1.2, polarization is also the reason a charged body, such as an electrified plastic rod, can exert a force on an uncharged body such as a bit of paper or a pith ball. Figure 4.21 shows an uncharged dielectric sphere B in the radial field of a positively charged body A. The induced positive charges on B experience a force toward the right, while the force on the induced negative charges is toward the left. The negative charges are closer to A, and thus are in a stronger field, than are the positive charges. The force toward the left is stronger than that toward the right, and B is attracted toward A, even though its net charge is zero. The attraction occurs whether the sign of A’s charge is positive or negative (see Fig. 1.8). Furthermore, the effect is not limited to dielectrics; an uncharged conducting body would be attracted in the same way.

4.21 A neutral sphere B in the radial electric field of a positively charged sphere A is attracted to the charge because of polarization.

+

4.6

+ + + +

A +

Test Your Understanding of Section 4.5 A parallel-plate capacitor has charges Q and -Q on its two plates. A dielectric slab with K = 3 is then inserted into the space between the plates as shown in Fig. 4.20. Rank the following electric-field magnitudes in order from largest to smallest. (i) the field before the slab is inserted; (ii) the resultant field after the slab is inserted; (iii) the field due to the bound charges. y

+ + +

+ + +

+ + +

– + –B+ – +

S

E

+

Gauss’s Law in Dielectrics

We can extend the analysis of Section 4.4 to reformulate Gauss’s law in a form that is particularly useful for dielectrics. Figure 4.22 is a close-up view of the left capacitor plate and left surface of the dielectric in Fig. 4.15b. Let’s apply Gauss’s law to the rectangular box shown in cross section by the purple line; the surface area of the left and right sides is A. The left side is embedded in the conductor that forms the left capacitor plate, and so the electric field everywhere on that surface is zero. The right side is embedded in the dielectric, where the electric field has magnitude E, and E' = 0 everywhere on the other four sides. The total charge enclosed, including both the charge on the capacitor plate and the induced charge on the dielectric surface, is Qencl = 1s - si2A, so Gauss’s law gives EA =

1s - si2A P0

(4.21)

4.22 Gauss’s law with a dielectric. This figure shows a close-up of the left-hand capacitor plate in Fig. 4.15b. The Gaussian surface is a rectangular box that lies half in the conductor and half in the dielectric. S

S

E50

E

+ Conductor Dielectric

+ – +

Side view

σ

–σi

+ –

This equation is not very illuminating as it stands because it relates two unknown quantities: E inside the dielectric and the induced surface charge density si . But now we can use Eq. (4.16), developed for this same situation, to simplify this equation by eliminating si . Equation (4.16) is '

+ Gaussian surface

'

si = sa 1 -

1 s b or s - si = K K

Conductor

Combining this with Eq. (4.21), we get EA =

Perspective view

sA sA or KEA = P0 KP0 S

(4.22)

S

Equation (4.22) says that the flux of KE, not E, through the Gaussian surface in Fig. 4.22 is equal to the enclosed free charge sA divided by P0 . It turns out that for any Gaussian surface, whenever the induced charge is proportional to the electric field in the material, we can rewrite Gauss’s law as '

S

B

#

S

KE dA =

Qencl-free P0

(Gauss’s law in a dielectric)

(4.23)

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A

A

Dielectric

126

CHAPTER 4 Capacitance and Dielectrics

where Qencl-free is the total free charge (not bound charge) enclosed by the Gaussian surface. The significance of these results is that the right sides contain only the free charge on the conductor, not the bound (induced) charge. In fact, although we have not proved it, Eq. (4.23) remains valid even when different parts of the Gaussian surface are embedded in dielectrics having different values of K, provided that the value of K in each dielectric is independent of the electric field (usually the case for electric fields that are not too strong) and that we use the appropriate value of K for each point on the Gaussian surface.

Example 4.12

A spherical capacitor with dielectric

Use Gauss’s law to find the capacitance of the spherical capacitor of Example 4.3 (Section 4.1) if the volume between the shells is filled with an insulating oil with dielectric constant K. SOLUTION IDENTIFY and SET UP: The spherical symmetry of the problem is not changed by the presence of the dielectric, so as in Example 4.3, we use a concentric spherical Gaussian surface of radius r between the shells. Since a dielectric is present, we use Gauss’s law in the form of Eq. (4.23). EXECUTE: From Eq. (4.23), S

B

#

S

KE dA = E =

B

KE dA = KE Q

4pKP0r2

=

B Q

dA = 1KE214pr22 =

Q P0

where P = KP0. Compared to the case in which there is vacuum between the shells, the electric field is reduced by a factor of 1>K. The potential difference Vab between the shells is reduced by the same factor, and so the capacitance C = Q>Vab is increased by a factor of K, just as for a parallel-plate capacitor when a dielectric is inserted. Using the result of Example 4.3, we find that the capacitance with the dielectric is C =

4pKP0rarb 4pPrarb = rb - ra rb - ra

EVALUATE: If the dielectric fills the volume between the two conductors, the capacitance is just K times the value with no dielectric. The result is more complicated if the dielectric only partially fills this volume (see Challenge Problem 4.78).

4pPr2

Test Your Understanding of Section 4.6 A single point charge q is imbedded in a dielectric of dielectric constant K. At a point inside the dielectric a distance r from the point charge, what is the magnitude of the electric field? (i) q>4pP0r2; (ii) Kq>4pP0r2; (iii) q>4pKP0r2; (iv) none of these. y

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4

CHAPTER

SUMMARY Q Vab

Capacitors and capacitance: A capacitor is any pair of conductors separated by an insulating material. When the capacitor is charged, there are charges of equal magnitude Q and opposite sign on the two conductors, and the potential Vab of the positively charged conductor with respect to the negatively charged conductor is proportional to Q. The capacitance C is defined as the ratio of Q to Vab . The SI unit of capacitance is the farad (F): 1 F = 1 C>V. A parallel-plate capacitor consists of two parallel conducting plates, each with area A, separated by a distance d. If they are separated by vacuum, the capacitance depends only on A and d. For other geometries, the capacitance can be found by using the definition C = Q>Vab . (See Examples 4.1–4.4.)

C =

Capacitors in series and parallel: When capacitors with capacitances C1 , C2 , C3 , Á are connected in series, the reciprocal of the equivalent capacitance Ceq equals the sum of the reciprocals of the individual capacitances. When capacitors are connected in parallel, the equivalent capacitance Ceq equals the sum of the individual capacitances. (See Examples 4.5 and 4.6.)

1 1 1 1 = + + + Á Ceq C1 C2 C3 (capacitors in series)

Wire Plate a, area A

(4.1) 1Q

Q A C = = P0 Vab d

(4.2)

2Q

d

Potential Wire Plate b, area A difference 5 Vab

'

'

'

'

'

a 1Q 2Q

(4.5)

Vab 5 V

C1 Vac 5 V1

c 1Q 2Q

Ceq = C1 + C2 + C3 + Á (capacitors in parallel)

++ ++

++ ++

C2 Vcb 5 V2

b

(4.7)

a ++ Vab 5 V C1

++

Q1 C2

+ +

Q2

b

Energy in a capacitor: The energy U required to charge a capacitor C to a potential difference V and a charge Q is equal to the energy stored in the capacitor. This energy can be thought of as residing in the electric field between the conductors; the energy density u (energy per unit volume) is proportional to the square of the electric-field magnitude. (See Examples 4.7–4.9.)

Dielectrics: When the space between the conductors is filled with a dielectric material, the capacitance increases by a factor K, called the dielectric constant of the material. The quantity P = KP0 is called the permittivity of the dielectric. For a fixed amount of charge on the capacitor plates, induced charges on the surface of the dielectric decrease the electric field and potential difference between the plates by the same factor K. The surface charge results from polarization, a microscopic rearrangement of charge in the dielectric. (See Example 4.10.) Under sufficiently strong fields, dielectrics become conductors, a situation called dielectric breakdown. The maximum field that a material can withstand without breakdown is called its dielectric strength. In a dielectric, the expression for the energy density is the same as in vacuum but with P0 replaced by P = KP. (See Example 4.11.) Gauss’s law in a dielectric has almostSthe same form asSin vacuum, with two key differences: E is replaced by KE and Qencl is replaced by Qencl-free, which includes only the free charge (not bound charge) enclosed by the Gaussian surface. (See Example 4.12.)

U =

Q2 = 12 CV2 = 12 QV 2C

1Q

+

(4.9)

+

+

+

+

+

–

–

–

–

S

E V

u = 12 P0 E2

(4.11)

'

A A = P d d (parallel-plate capacitor filled with dielectric)

(4.19)

u = 12 KP0 E2 = 12 PE2

(4.20)

S

B

#

S

KE dA =

Qencl-free P0

–

–

Dielectric between plates

C = KC0 = KP0

'

2Q

(4.23)

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s 2si + +– + +– + +– + +– 2si s

2s – +– – +– – +– – +–

si

si

2s

127

128

CHAPTER 4 Capacitance and Dielectrics

BRIDGING PROBLEM

Electric-Field Energy and Capacitance of a Conducting Sphere

A solid conducting sphere of radius R carries a charge Q. Calculate the electric-field energy density at a point a distance r from the center of the sphere for (a) r , R and (b) r . R. (c) Calculate the total electric-field energy associated with the charged sphere. (d) How much work is required to assemble the charge Q on the sphere? (e) Use the result of part (c) to find the capacitance of the sphere. (You can think of the second conductor as a hollow conducting shell of infinite radius.) SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution

IDENTIFY and SET UP 1. You know the electric field for this situation at all values of r from Example 2.5 (Section 2.4). You’ll use this to find the electric-field energy density u and the total electric-field energy U. You can then find the capacitance from the relationship U = Q2>2C. 2. To find U, consider a spherical shell of radius r and thickness dr that has volume dV = 4pr2dr. The energy stored in this volume

Problems

is u dV, and the total energy is the integral of u dV from r 5 0 to r S q . Set up this integral. EXECUTE 3. Find u for r , R and for r . R. 4. Substitute your results from step 3 into the expression from step 2. Then calculate the integral to find the total electric-field energy U. 5. Use your understanding of the energy stored in a charge distribution to find the work required to assemble the charge Q. 6. Find the capacitance of the sphere. EVALUATE 7. Where is the electric-field energy density greatest? Where is it least? 8. How would the results be affected if the solid sphere were replaced by a hollow conducting sphere of the same radius R? 9. You can find the potential difference between the sphere and infinity from C = Q>V. Does this agree with the result of Example 3.8 (Section 3.3)?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q4.1 Equation (4.2) shows that the capacitance of a parallel-plate capacitor becomes larger as the plate separation d decreases. However, there is a practical limit to how small d can be made, which places limits on how large C can be. Explain what sets the limit on d. (Hint: What happens to the magnitude of the electric field as d S 0?) Q4.2 Suppose several different parallel-plate capacitors are charged up by a constant-voltage source. Thinking of the actual movement and position of the charges on an atomic level, why does it make sense that the capacitances are proportional to the surface areas of the plates? Why does it make sense that the capacitances are inversely proportional to the distance between the plates? Q4.3 Suppose the two plates of a capacitor have different areas. When the capacitor is charged by connecting it to a battery, do the charges on the two plates have equal magnitude, or may they be different? Explain your reasoning. Q4.4 In the parallel-plate capacitor of Fig. 4.2, suppose the plates are pulled apart so that the separation d is much larger than the size of the plates. (a) Is it still accurate to say that the electric field between the plates is uniform? Why or why not? (b) In the situation shown in Fig. 4.2, the potential difference between the plates is Vab = Qd>P0A. If the plates are pulled apart as described above, is Vab more or less than this formula would indicate? Explain your

reasoning. (c) With the plates pulled apart as described above, is the capacitance more than, less than, or the same as that given by Eq. (4.2)? Explain your reasoning. Q4.5 A parallel-plate capacitor is charged by being connected to a battery and is kept connected to the battery. The separation between the plates is then doubled. How does the electric field change? The charge on the plates? The total energy? Explain your reasoning. Q4.6 A parallel-plate capacitor is charged by being connected to a battery and is then disconnected from the battery. The separation between the plates is then doubled. How does the electric field change? The potential difference? The total energy? Explain your reasoning. Q4.7 Two parallel-plate capacitors, identical except that one has twice the plate separation of the other, are charged by the same voltage source. Which capacitor has a stronger electric field between the plates? Which capacitor has a greater charge? Which has greater energy density? Explain your reasoning. Q4.8 The charged plates of a capacitor attract each other, so to pull the plates farther apart requires work by some external force. What becomes of the energy added by this work? Explain your reasoning. Q4.9 The two plates of a capacitor are given charges ;Q. The capacitor is then disconnected from the charging device so that the

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Single Correct

charges on the plates can’t change, and the capacitor is immersed in a tank of oil. Does the electric field between the plates increase, decrease, or stay the same? Explain your reasoning. How can this field be measured? Q4.10 As shown in Table 4.1, water has a very large dielectric constant K = 80.4. Why do you think water is not commonly used as a dielectric in capacitors? Q4.11 Is dielectric strength the same thing as dielectric constant? Explain any differences between the two quantities. Is there a simple relationship between dielectric strength and dielectric constant (see Table 4.2)? Q4.12 A capacitor made of aluminum foil strips separated by Mylar film was subjected to excessive voltage, and the resulting dielectric breakdown melted holes in the Mylar. After this, the capacitance was found to be about the same as before, but the breakdown voltage was much less. Why? Q4.13 Suppose you bring a slab of dielectric close to the gap between the plates of a charged capacitor, preparing to slide it between the plates. What force will you feel? What does this force tell you about the energy stored between the plates once the dielectric is in place, compared to before the dielectric is in place? Q4.14 The freshness of fish can be measured by placing a fish between the plates of a capacitor and measuring the capacitance. How does this work? (Hint: As time passes, the fish dries out. See Table 4.1.) Q4.15 Electrolytic capacitors use as their dielectric an extremely thin layer of nonconducting oxide between a metal plate and a conducting solution. Discuss the advantage of such a capacitor over one constructed using a solid dielectric between the metal plates. Q4.16 In terms of the dielectric constant K, what happens to the electric flux through the Gaussian surface shown in Fig. 4.22 when the dielectric is inserted into the previously empty space between the plates? Explain. Q4.17 A parallel-plate capacitor is connected to a power supply that maintains a fixed potential difference between the plates. (a) If a sheet of dielectric is then slid between the plates, what happens to (i) the electric field between the plates, (ii) the magnitude of charge on each plate, and (iii) the energy stored in the capacitor? (b) Now suppose that before the dielectric is inserted, the charged capacitor is disconnected from the power supply. In this case, what happens to (i) the electric field between the plates, (ii) the magnitude of charge on each plate, and (iii) the energy stored in the capacitor? Explain any differences between the two situations. Q4.18 Liquid dielectrics that have polar molecules (such as water) always have dielectric constants that decrease with increasing temperature. Why? Q4.19 A conductor is an extreme case of a dielectric, since if an electric field is applied to a conductor, charges are free to move within the conductor to set up “induced charges.” What is the dielectric constant of a perfect conductor? Is it K = 0, K S q , or something in between? Explain your reasoning.

SINGLE CORRECT Q4.1 Two parallel metal plates carry opposite electrical charges each with a magnitude of Q. The plates are separated by a distance d and each plate has an area A. Consider the folowing : (i) increasing Q (ii) increasing d (iii) increasing A Which of the following would have the effect of reducing the potential difference between the plates?

129

(a) II and III (b) II only (c) III only (d) I and III Q4.2 In four options below, all the four circuits are arranged in order of equivalent capactance. Select the correct order. Assume all capacitors are of equal capacitance.

(1)

(2)

(3)

(4)

(a) C1> C2> C3> C4 (b) C1> C3> C2> C4 (c) C1< C2< C3< C4 (d) C1< C3< C2< C4 Q4.3 Which of following graph correctly F A represents the force between plates of an isolated charged parallel plate capacitor with B distance x between them. (Assume x to be C small). D Q4.4 Three capacitors C1, C2, and C3 are x connected to a battery as shown in the figure. The three capacitors have equal capacitances. Which capacitor C2 stores the most energy? (a) C2 or C3 as they store the same + C1 C3 V amount of energy – (b) C2 (c) C1 (d) All three capacitors store the same amount of energy Q4.5 Statement-1 If t emperature is increased, the dielectric constant of a polar dielectric decreases whereas that of a non-polar dielectric does not change significantly. Statement-2 The magnitude of dipole moment of individual polar molecule decreases significantly with increase in temperature. (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true.

ONE OR MORE CORRECT Q4.6 A parallel plate capacitor is connected to a battery and a thick uncharged metal sheet is inserted parallel to the plates of the capacitor. Select the correct statement(s) : (a) The battery will supply more charge (b) The capacitance will decrease (c) The potential difference between the plates will increase. (d) Equal and opposite charges will appear on the two faces of the metal plate. Q4.7 A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plate is now increased (a) the force of attraction between the plates will decrease (b) The field in the region between the plates will not change (c) The energy stored in the capacitor will increase (d) The potential difference between the plates will decreases.

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130

CHAPTER 4 Capacitance and Dielectrics

Q4.8 A dielectric slab of thickness d is inserted in the parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3d The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d. (a) the magnitude of the electric field remains the same. (b) the direction of the electric field remains the same. (c) the electric potential increases continuously. (d) the electric potential increases as first, then decreases and again increases. Q4.9 Two identical capacitors are connected in series as shown in the figure. A dielectric slab (k > 1 ) is placed between the plates of the capacitor B and the battery remains connected. Which of the following statement(s) is/are correct following the insertion of the Before After dielectric? (a) The charge supplied by the battery A C C increases. V V (b) The capacitance of the system increases. B C C (c) The electric field in the capacitor B increases. (d) The electrostatic potential energy decreases. Q4.10 The figure shows a capacitor having three layers between its plates. Layer x is x y z vacuum, y is conductor and z is a dielectric. Which of the following change(s) will result in increase in capacitance? (a) Replace x by conductor (b) Replace y by dielectric (c) Replace z by conductor (d) Replace x by dielectric Q4.11 When two capacitors are charged to different potentials and then connected in parallel to each other, then (a) Final charge is equal to sum of initial charges (b) Final potential difference is equal to average of the initial potential difference (c) Final potential difference is different from the sum of initial potential difference (d) Final energy stored is less then the sum of initial stored energy Q4.12 Figure shows a capacitor having three layers between its plates. Layer x is x y z vacuum, y is conductor, and z is a dielectric. Which of the following change(s) will result in increase in capacitance? (a) Replace x by conductor (b) Replace y by dielectric (c) Replace z by conductor (d) Replace x by dielectric Q4.13 The plates of a parallel-plate capacitor are connected across the terminals of a battery. Now the plates of the capacitor are pulled slightly farther apart. When equilibrium is restored in the circuit, (a) The potential difference across the plates has increased (b) The energy stored on the capacitor has decreased (c) The capacitance of the capacitor has increased (d) The battery would have gained energy

MATCH THE COLUMNS Q4.14 Some events related to a capacitor are listed in column-I. Match these events with their effect(s) in column-II. V

Column-I (Events) Column-II (Effects) (a) Insertion of dielectric while (P) Electric field between battery remain attached plates changes (b) Removal of dielectric while (Q) Charge present on plates battery is not present changes (c) Slow decrease in separation (R) Energy stored in capacitor between plates while batincreases tery is attached (d) Slow increase of separation (S) Work done by external between plates while batagent is positive tery is not present (T) Potential difference between the plates changes Q4.15 On a capacitor of capacitance C0 following steps are performed in the order as given in column-I (a) Capacitor is charged by connecting it across a battery of EMF V0 (b) Dielectric of dielectric constant k and thickness d is inserted (c) Capacitor is disconnected from battery (d) Separation between plates is doubled Column-I (Steps performed)

(a) (a) (d) (c) (b) (b) (d) (a) (c) (b) (c) (b) (a) (c) (d) (d) (a) (b) (d) (c)

Column-II Final value of Quantity (Symbols have usual meaning) C0V0 (P) Q = 2 KC0V0 (Q) Q = K + 1 KC0 (R) C = K + 1 V0(K + 1) (S) V = 2K (T) Potential energy, K(K + 1)C0V20 U = 2

EXERCISES Section 4.1 Capacitors and Capacitance

4.1 . The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00 * 106 V>m. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the capacitance? 4.2 .. Capacitance of an Oscilloscope. Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates. Typically, they are squares 3.0 cm on a side and separated by 5.0 mm, with vacuum in between. What is the capacitance of these deflecting plates and hence of the oscilloscope? (Note: This capacitance can sometimes have an effect on the circuit you are trying to study and must be taken into consideration in your calculations.) 4.3 . A 10.0-mF parallel-plate capacitor with circular plates is connected to a 12.0-V battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0-V battery after the radius of each plate was doubled without changing their separation?

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Exercises

4.4 . A parallel-plate air capacitor is to store charge of magnitude 210 pC on each plate when the potential difference between the plates is 42.0 V. (a) If the area of each plate is 7 cm2, what is the separation between the plates? (b) If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 210 pC on each plate? 4.5 . A cylindrical capacitor consists of a solid inner conducting core with radius r1 surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is L The capacitance is C (a) Calculate the outer radius r2 of the hollow tube. (b) When the capacitor is charged to voltage V what is the charge per unit length l on the capacitor? 4.6 . A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

is 50.0 V. (a) Find the equivalent Figure E4.11 capacitance of this system 5.0 mF between a and b. (b) How much charge is stored by this combi- 10.0 mF nation of capacitors? (c) How a much charge is stored in each of the 10.0-mF and the 9.0-mF capacitors?

131

9.0 mF b

8.0 mF

Section 4.3 Energy Storage in Capacitors and Electric-Field Energy

4.12 . A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 2.55 mC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation? 4.13 . A parallel-plate vacuum capacitor with plate area A and separation x has charges + Q and -Q on its plates. The capacitor is disconnected from the source of charge, so the charge on each Section 4.2 Capacitors in Series and Parallel 4.7 . BIO Electric Eels. Electric eels and electric fish generate plate remains fixed. (a) What is the total energy stored in the large potential differences that are used to stun enemies and prey. capacitor? (b) The plates are pulled apart an additional distance dx. These potentials are produced by cells that each can generate 0.10 What is the change in the stored energy? (c) If F is the force with V. We can plausibly model such cells as charged capacitors. (a) which the plates attract each other, then the change in the stored How should these cells be connected (in series or in parallel) to energy must equal the work dW = Fdx done in pulling the plates produce a total potential of more than 0.10 V? (b) Using the con- apart. Find an expression for F. (d) Explain why F is not equal to nection in part (a), how many cells must be connected together to QE, where E is the electric field between the plates. 4.14 .. A parallel-plate vacuum capacitor has 8.38 J of energy produce the 500-V surge of the electric eel? stored in it. The separation between the plates is 2.30 mm. If the 4.8 . In Fig. E4.8, each capacitor Figure E4.8 separation is decreased to 1.15 mm, what is the energy stored (a) if has C = 4.00 mF and Vab = the capacitor is disconnected from the potential source so the C C 1 2 + 28.0 V. Calculate (a) the charge charge on the plates remains constant, and (b) if the capacitor on each capacitor; (b) the potential remains connected to the potential source so the potential differdifference across each capacitor; (c) a ence between the plates remains constant? the potential difference between C3 4.15 . You have two identical capacitors and an external potential points a and d. source. (a) Compare the total energy stored in the capacitors when d 4.9 . In Fig. E4.9, C1 = 6.00 mF, they are connected to the applied potential in series and in parallel. C2 = 3.00 mF, and C3 = 5.00 mF. b (b) Compare the maximum amount of charge stored in individual The capacitor network is concapacitors in each case. (c) Energy storage in a capacitor can be limnected to an applied potential C4 ited by the maximum electric field between the plates. What is the Vab. After the charges on the ratio of the electric field for the series and parallel combinations? capacitors have reached their Figure E4.9 4.16 . For the capacitor network shown in Fig. E4.16, the potenfinal values, the charge on C2 is tial difference across ab is 220 C1 40.0 mC. (a) What are the Figure E4.16 V. Find (a) the total charge charges on capacitors C1 and stored in this network; (b) the 35 nF C3? (b) What is the applied volta charge on each capacitor; (c) age Vab? C2 the total energy stored in the 4.10 .. For the system of ca b network; (d) the energy stored a d pacitors shown in Fig. E4.10, a in each capacitor; (e) the potential difference of 25 V is potential difference across b 75 nF maintained across ab. (a) What is each capacitor. the equivalent capacitance of this C3 4.17 . A 0.350-m-long cylindrical capacitor consists of a solid system between a and b? (b) conducting core with a radius of 1.20 mm and an outer hollow How much charge is stored by Figure E4.10 conducting tube with an inner radius of 2.00 mm. The two conducthis system? (c) How much tors are separated by air and charged to a potential difference of 7.5 nF charge does the 6.5-nF capacitor 6.00 V. Calculate (a) the charge per length for the capacitor; (b) store? (d) What is the potential the total charge on the capacitor; (c) the capacitance; (d) the energy 18 nF 30 nF 10 nF difference across the 7.5-nF stored in the capacitor when fully charged. capacitor? a b 6.5nF 5 . 4.11 Figure E4.11 shows a Given aln = 0.51b 3 system of four capacitors, where the potential difference across ab '

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132

CHAPTER 4 Capacitance and Dielectrics

Section 4.4 Dielectrics

.

4.18 A 12.5-mF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 4 is placed between the plates, completely filling the space between them. (a) How much energy is stored in the capacitor before and after the dielectric is inserted? (b) By how much did the energy change during the insertion? Did it increase or decrease? 4.19 .. BIO Potential in Human Cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is 60.50 * 10-3 C>m2, the cell wall is 5.0 nm thick, and the cellS wall material is air. (a) Find the magnitude of E in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case. 4.20 . When a 360-nF air capacitor 11 nF = 10-9 F2 is connected to a power supply, the energy stored in the capacitor is 1.8 * 10-5 J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.4 * 10-5 J. (a) What is the potential difference between the capacitor plates? (b) What is the dielectric constant of the slab? 4.21 . A parallel-plate capacitor has capacitance C = 12.5 pF when the volume between the plates is filled with air. The plates are circular, and distance between them is 2 mm. The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 pC. (a) What is the dielectric constant K of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

Section 4.6 Gauss’s Law in Dielectrics

4.22 . A parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant K. The magnitude of the charge on each plate is Q. Each plate has area A, and the distance between the plates is d. (a) Use Gauss’s law as stated in Eq. (4.23) to calculate the magnitude of the electric field in the dielectric. (b) Use the electric field determined in part (a) to calculate the potential difference between the two plates. (c) Use the result of part (b) to determine the capacitance of the capacitor. Compare your result to Eq. (4.12).

PROBLEMS 4.23 . Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the 1 flash lasts for 675 s with an average light power output of 5 2.70 * 10 W. (a) If the conversion of electrical energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value calculated in part (a). What is the capacitance?

4.24 ... In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, airfilled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 42.0 mm2, and the separation between the plates is 0.700 mm before the key is depressed. (a) Calculate the capacitance before the key is depressed. (b) If the circuitry can detect a change in capacitance of 0.250 pF, how far must the key be depressed before the circuitry detects its depression? 4.25 .. A 20.0-mF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to those of an uncharged 10.0-mF capacitor. Compute (a) the original charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected. 4.26 . In Fig. P4.26, C1 = Figure P4.26 C5 = 8.4 mF and C2 = C3 = C3 C1 C4 = 4.2 mF. The applied potential is Vab = 220 V. (a) What is a the equivalent capacitance of the network between points a and C2 C5 C4 b? (b) Calculate the charge on b each capacitor and the potential difference across each capacitor. 4.27 .. You are working on an electronics project requiring a variety of capacitors, but you have only a large supply of 100-nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) 50 nF; (b) 450 nF; (c) 25 nF; (d) 75 nF. 4.28 . The capacitors in Fig. Figure P4.28 P4.28 are initially uncharged 3.00 mF 6.00 mF d and are connected, as in the diagram, with switch S open. The applied potential difference is a b S Vab = + 210 V. (a) What is the potential difference Vcd? (b) c What is the potential difference 6.00 mF 3.00 mF across each capacitor after switch S is closed? (c) How much charge flowed through the switch when it was closed? 4.29 .. Three capacitors having capacitances of 8.4, 8.4, and 4.2 mF are connected in series across a 36-V potential difference. (a) What is the charge on the 4.2-mF capacitor? (b) What is the total energy stored in all three capacitors? (c) The capacitors are disconnected from the potential difference without allowing them to discharge. They are then reconnected in parallel with each other, with the positively charged plates connected together. What is the voltage across each capacitor in the parallel combination? (d) What is the total energy now stored in the capacitors? 4.30 . Capacitance of a Thundercloud. The charge center of a thundercloud, drifting 3.0 km above the earth’s surface, contains 20 C of negative charge. Assuming the charge center has a radius of 1.0 km, and modeling the charge center and the earth’s surface as parallel plates, calculate: (a) the capacitance of the system; (b) the potential difference between charge center and ground; (c) the average strength of the electric field between cloud and ground; (d) the electrical energy stored in the system.

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133

Exercises

4.31 . Each combination of Figure P4.31 capacitors between points a (a) and b in Fig. P4.31 is first connected across a 120-V battery, a charging the combination to 10.0 20.0 30.0 S mF mF 120 V. These combinations are mF Signal b then connected to make the cirdevice cuits shown. When the switch S is thrown, a surge of charge for the discharging capacitors flows (b) to trigger the signal device. How a much charge flows through the 1 10.0 mF signal device in each case? 2 4.32 . A parallel-plate capaciS 1 20.0 mF tor with only air between the 2 1 plates is charged by connecting 30.0 mF 2 it to a battery. The capacitor is Signal b device then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only onethird of the space between the plates? 4.33 .. An air capacitor is made by Figure P4.33 using two flat plates, each with area A, separated by a distance d. Then a metal slab having thickness a (less a d than d) and the same shape and size as the plates is inserted between them, parallel to the plates and not touching either plate (Fig. P4.33). (a) What is the capacitance of this arrangement? (b) Express the capacitance as a multiple of the capacitance C0 when the metal slab is not present. (c) Discuss what happens to the capacitance in the limits a S 0 and a S d. 4.34 .. Capacitance of the Earth. Consider a spherical capacitor with one conductor being a solid conducting sphere of radius R and the other conductor being at infinity. (a) Use Eq. (4.1) and what you know about the potential at the surface of a conducting sphere with charge Q to derive an expression for the capacitance of the charged sphere. (b) Use your result in part (a) to calculate the capacitance of the earth. The earth is a good conductor and has a radius of 6380 km. Compare your results to the capacitance of typical capacitors used in electronic circuits, which ranges from 10 pF to 100 pF. 4.35 . CALC The inner cylinder of a long, cylindrical capacitor has radius ra and linear charge density +l. It is surrounded by a coaxial cylindrical conducting shell with inner radius rb and linear charge density - l (see Fig. 4.6). (a) What is the energy density in the region between the conductors at a distance r from the axis? (b) Integrate the energy density calculated in part (a) over the volume between the conductors in a length L of the capacitor to obtain the total electric-field energy per unit length. (c) Use Eq. (4.9) and the capacitance per unit length calculated in Example 4.4 (Section 4.1) to calculate U>L. Does your result agree with that obtained in part (b)? 4.36 .. A fuel gauge uses a capacitor to determine the height of the fuel in a tank. The effective dielectric constant Keff changes from a value of 1 when the tank is empty to a value of K, the

dielectric constant of the fuel, Figure P4.36 when the tank is full. The approV priate electronic circuitry can determine the effective dielectric Battery constant of the combined air and fuel between the capacitor plates. Air L Each of the two rectangular plates has a width w and a length L (Fig. P4.36). The height of the fuel w between the plates is h. You can Fuel h ignore any fringing effects. (a) Derive an expression for Keff as a function of h. (b) What is the effective dielectric constant for a tank 12 full, if the fuel is gasoline 1K = 1.952? (c) Repeat part (b) for methanol 1K = 33.02. (d) For which fuel is this fuel gauge more practical? 4.37 Figure shows a parallel plate capacitor with plate slab width b & length l. The separation between the plates is d . The plates are rigidly clamped & connected to a battery of e.m.f. V. A diV K electric slab of thickness d & di-electric d constant K is slowly inserted between the plates. x (i) Calculate the energy of the system when a length x of the slab is introduced 1 into the capacitor. (ii) What force should be applied to the slab to ensure that it goes slowly into the capacitor ?Neglect any effect of friction or gravity ? 4.38 The gap between the plates of a parallel plate capacitor is filled with an isotropic dielectric whose dielectric constant K varies linearly from K1 to K2(>K1) in the direction perpendicular to the plates.The area of each plate is A and the separation between the plates is d. Find the expression for the capacity per unit area of the plates. C 2 4.39 Find the energy dissipated in the cir1 C 2C cuit after the switch is flipped from posi- V tion 1 to 2. 4.40 Capacitor C3 in the circuit is a variable capacitor (its capacitance can be varied). Graph is plotted between potential difference V1 (across capacitor C1) versus C3. Electric potential V1 approaches an asymptote of 10 V as C3 S q (a) Find the electric potential V across the battery. (b) Find the ratio C1/C2. 12

V1(V)

C1

V C2

C3

10 8 6 4 2 2

4.41 In the network shown we have three identical capacitors. Each of them can withstand a maximum100 V p.d. A What maximum voltage can be applied across A and B so that no capacitor gets spoiled? 4.42 There are six plates of equal area A and the plates are arranged as shown in figure. The equivalent capacitance between points A and B is

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4

6 8 10 12 C3(µF) C C B C

A

1 2

d 2d

3

3d 2d d

B 5

4 6

134

CHAPTER 4 Capacitance and Dielectrics

4.43 Find the effective capacitance between points A and B in the networks shown below. All caps = 1 µF. 4.44 Two long coaxial conducting cylinders of radii a and 3a constitute a cylindrical capacitor. It has been charged to a potential difference V. Find the electric field at r = 2a from axis. 4.45 Find the equivalent capacitance of the combination shown in the figure. Each capacitor is of capacitance C.

into the space between the plates, show that the change in stored energy is

A

B

CHALLENGE PROBLEMS 4.46 ... CP The parallel-plate air capacitor in Fig. P4.46 consists of two horizontal conducting plates of equal area A. The bottom plate rests on a fixed support, and the top plate is suspended by four springs with spring constant k, positioned at each of the four corners of the top plate as shown in the figure. When uncharged, the plates are separated by a distance z0. A battery is connected to the plates and produces a potential difference V between them. This causes the plate separation to decrease to z. Neglect any fringing effects. (a) Show that the electrostatic force between the charged plates has a magnitude P0 AV2>2z2. (Hint: See Exercise 4.34.) (b) Obtain an expression that relates the plate separation z to the potential difference V. The resulting equation will be cubic in z. Figure P4.46 k z

k k

k

1K - 12P0V2L dx 2D (c) Suppose that before the slab is moved by dx, the plates are discon- Figure P4.47 nected from the battery, so that the charges on the plates remain conL stant. Determine the magnitude of the charge on each plate, and then show L that when the slab is moved dx farther into the space between the plates, the stored energy changes by an x amount that is the negative of the Dielectric expression for dU given in part (b). slab, constant K (d) If F is the force exerted on the slab by the charges on the plates, then dU should equal the work done D against this force to move the slab a distance dx. Thus dU = - F dx. Show that applying this expression to the result of part (b) suggests that the electric force on the slab pushes it out of the capacitor, while the result of part (c) suggests that the force pulls the slab into the capacitor. (e) Figure 4.16 shows that the force in fact pulls the slab into the capacitor. Explain why the result of part (b) gives an incorrect answer for the direction of this force, and calculate the magnitude of the force. (This method does not require knowledge of the nature of the fringing field.) 4.48 ... An isolated spherical Figure P4.48 capacitor has charge +Q on its inner conductor (radius ra) and charge - Q rb on its outer conductor (radius rb). Half of the volume between the two conductors is then filled with a liquid dielectric ra of constant K, as shown in cross section in Fig. P4.48. (a) Find the capacitance of the half-filled capacitor. (b) Find the K S magnitude of E in the volume between the two conductors as a function of the distance r from the center of the capacitor. Give answers for both the upper and lower halves of this volume. dU = +

V

A A

4.47 ... Two square conducting plates with sides of length L are separated by a distance D. A dielectric slab with constant K with dimensions L * L * D is inserted a distance x into the space between the plates, as shown in Fig. P4.47. (a) Find the capacitance C of this system. (b) Suppose that the capacitor is connected to a battery that maintains a constant potential difference V between the plates. If the dielectric slab is inserted an additional distance dx

Answers Chapter Opening Question

?

Equation (4.9) shows that the energy stored in a capacitor with capacitance C and charge Q is U = Q2>2C. If the charge Q is doubled, the stored energy increases by a factor of 22 = 4. Note that if the value of Q is too great, the electric-field magnitude inside the capacitor will exceed the dielectric strength of the material between the plates and dielectric breakdown will occur (see Section 4.4). This puts a practical limit on the amount of energy that can be stored.

Test Your Understanding Questions 4.1 (iii) The capacitance does not depend on the value of the charge Q. Doubling the value of Q causes the potential difference Vab to double, so the capacitance C = Q>Vab remains the same.

These statements are true no matter what the geometry of the capacitor. 4.2 (a) (i), (b) (iv) In a series connection the two capacitors carry the same charge Q but have different potential differences Vab = Q>C; the capacitor with the smaller capacitance C has the greater potential difference. In a parallel connection the two capacitors have the same potential difference Vab but carry different charges Q = CVab; the capacitor with the larger capacitance C has the greater charge. Hence a 4-mF capacitor will have a greater potential difference than an 8-mF capacitor if the two are connected in series. The 4-mF capacitor cannot carry more charge than the 8-mF capacitor no matter how they are connected: In a series connection they will carry the same charge, and in a parallel connection the 8-mF capacitor will carry more charge. 4.3 (i) Capacitors connected in series carry the same charge Q. To compare the amount of energy stored, we use the expression

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Answers

U = Q2>2C from Eq. (4.9); it shows that the capacitor with the smaller capacitance 1C = 4 mF2 has more stored energy in a series combination. By contrast, capacitors in parallel have the same potential difference V, so to compare them we use U = 12 CV2 from Eq. (4.9). It shows that in a parallel combination, the capacitor with the larger capacitance 1C = 8 mF2 has more stored energy. (If we had instead used U = 12 CV2 to analyze the series combination, we would have to account for the different potential differences across the two capacitors. Likewise, using U = Q2>2C to study the parallel combination would require us to account for the different charges on the capacitors.) 4.4 (i) Here Q remains the same, so we use U = Q2>2C from Eq. (4.9) for the stored energy. Removing the dielectric lowers the capacitance by a factor of 1>K; since U is inversely proportional to C, the stored energy increases by a factor of K. It takes work to pull the dielectric slab out of the capacitor because the fringing field tries to pull the slab back in (Fig. 4.16). The work that you do goes into the energy stored in the capacitor. 4.5 (i), (iii), (ii) Equation (4.14) says that if E0 is the initial electric-field magnitude (before the dielectric slab is inserted), then the resultant field magnitude after the slab is inserted is E0>K = E0>3. The magnitude of the resultant field equals the difference between the initial field magnitude and the magnitude Ei of the field due to the bound charges (see Fig. 4.20). Hence E0 - Ei = E0>3 and Ei = 2E0>3. 4.6 (iii) Equation (4.23) shows that this situation is the same as an S S isolated point charge in vacuum but with E replaced by KE. Hence KE at the point of interest is equal to q>4pP0 r2, and so E = q>4pKP0 r2. As in Example 4.12, filling the space with a dielectric reduces the electric field by a factor of 1>K. '

'

Bridging Problem (a) 0 (b) Q2/32p2E0r4 (c) Q2/8pE0R (d) Q2/8pE0R (e) C = 4pE0R

Discussion Questions

¢Vrated Ebreakdown Q4.2 Metal plates contribute in generating capacitance due to surface charges which are responsible for creating electric field. The charges on atomic level do not interact with surface charge (given by cell) when charging gets completed. Greater the distance between plates greater will be the volume between plates and hence lesser energy density. Q4.3 Capacitance of capacitor depends on effective overlapping area. Due to different area fringe loss will be there and this loss will be remarkable. Induced change will be same. Q4.4 (a) No. (b) More as Q would be constant. (c) Less Q4.5 As the battery remains connected so the potential difference between the plates do not change. Due to increase in separation the capacitance will decrease so some charge will go back to battery. 1 1C 2 Ui = CV2 ; Uf = V 2 2 2 Q Due to decrease in charge E1 = will decrease. 2Ae0 Q4.1 dmin =

Q4.6 After disconnection with battery the charge on plates can not change by changing the separation. Capacitance will decrease because of increase in potential difference.

E1 =

135

Q by one plate 2Ae0

¢V = E.d Ae0 Ae0 and C2 = d 2d ‹ Q1 = C1V and Q2 = C2V Obviously Q1 > Q2 E1 > E2 1 Energy density = e0E2 . So, energy density q E2 2 Q4.8 It becomes elactrostatic potential energy of the capacitance. Q 1 Q2 d 1 Q2 F = Q. ; W = F.d = . 1W = 2e0A 2 e0 A 2 C Q4.9 Polarization effect decreases the field Q Q ;E = so, Ef < Ei Ei = Ae0 f Ae0K Due to insersion of dielectric bound charges are produced on the dielectric surface facing capacitor plates.The field produced by these bound charges is opposite to the original field and the superposition of fields decreases the net effect than the initial condition. Q4.10 One of the reasons is it can ionize certain materials. So some ions may enter into water increasing the conductivity of water, making water a bad dielectric. Q4.11 Dielectric are materials with high polarizability this quality is signified by the number called dielectric constant.Dielectric strength is related with the voltage a material can withstand before breakdown occurs. The value of dielectric strength is influenced by electrode configuration geometry, rate of application of test voltage. Q4.12 After dielectric breakdown the capacitor turns out to be capacitor with no dielectric between the plates. Q4.13 The dielectric is pulled due to attraction between charges on plates and dielectric surfaces.This force in the shown diagram is given by Q2d (K - 1) F = . 2e0b [a + (K - 1)x]2 d b Obviously it depends on x. But if battery remains connected with plates. X 1 e0b a F = (K - 1)V2 2 d Q4.7 C1 =

Q4.14 Capacitance depends on permittivity water is adielectric. Water is largest component of fish fins and tissues.After death of fish dielectric properties of fish body change. By correlating with known pattern of for capacitance change the freshness can be judge. Q4.15 Electrolytic capacitors have a much higher capacitance to volume ratio in comparison to metallic ones. They can be effectively used for communicating through low frequency signals due to their large capacitance. It can store large amount of energy also. f Q4.16 f¿ = k Q4.17 Refer answers 4.8 and 4.9 Q4.18 When the temperature is increased, the molecules of materials have higher K.E. and the mechanical amplitudes of motion are increased. Therefore an electric fields ability to allign the dipoles in the material will decrease. Q4.19 Infact a perfect conductor has infinite conductivity KS q 1 Induced charge Q¿ = Q a1 - b K

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136

CHAPTER 4 Capacitance and Dielectrics

If K S q ; Q¿ = Q (for perfect conductor)

Single correct Q4.1 (c) Q4.2 (b) Q4.3 (a) Q4.4 (c) Q4.5 (c)

4.12 (a) 2770 V (b) 5540 V (c) 3.53×10 - 3 J xQ2 4.13 (a) 2P0 A (b) a

One or More Correct Q4.6 (a), (d) Q4.7 (b), (c) Q4.8 (b), (c) Q4.9 (a), (b) Q4.10 (a), (c), (d) Q4.11 (a), (c), (d) Q4.12 (a), (c), (d) Q4.13 (b), (d)

Match the column Q4.14 (a) Q, R (b) P, R, S, T (c) P, Q, R (d) R, S, T Q4.15 (a) P, R, S (b) P, R, S (c) R, T (d) Q, R

Exercise 4.1 (a) 1.00×104V (b) 22.6 cm2 (c) 8 pF 4.2 1.59 pF 4.3 (a) 120 µC (b) 60 µC (d) 480 µC 4.4 (a) 1.24 mm (b) 84 V 4.5 (a) r2 == r1 . e( 2 pP0 L/C) (b) l = CV/ L 4.6 (a) 0.175 m (b) 2.55×10 - 8C 4.7 (a) connect the capacitors in series so their voltages will add. (b) 5000 4.8 (a) Q1 = 22.4 µC Q2 = 22.4 µC Q3 = 44.8 µC Q4 = 67.2 µC (b) V1 = 5.6 V V2 = 5.6 V V3 = 11.2 V V4 = 16.8 V (c) 11.2 V 4.9 (a) Q1 = 80×10 - 6 C Q3 = 120×10 - 6 C (b) 37.4 V 4.10 (a) 19.3 nF (b) 482 nC (c) 162 nC (d) 25 V 4.11 (a) 3.47 µF (b) 174 µC (c) 174 µC in each

(c)

Q2 b dx 2P0 A

Q2 2P0 A

E (d) E is the field due to both plates, field by one plate = 2 so E F = a b Q 2 Q2 1 2P0 A 4.14 (a) 4.19 J (b) 16.8 J 4.15 (a) Uparallel = 4 Useries (b) Qparallel = 2 Qseries (c) Eparallel = 2 Eseries 4.16 (a) 24.2 µC (b) Q35 = 7.7 µC Q75 = 16.5 µC (c) 2.66 mJ (d) U35 = 0.85 mJ U75 = 1.81 mJ (e) 220 V for each capacitor 4.17 (a) 6.53×10−10 c>m (b) 2.29×10−10 C (c) 3.81×10−11 F (d) 6.85×10−10 J 4.18 (a) before: 3.6 mJ After: 13.05 mJ (b) 9.9 mJ. The energy increased. 4.19 (a) 5.6×107V/m (b) 0.28 V. The outer wall of the cell is at higher potential, since it has positive charge. (c) E = 1.0×107v/m V = 0.052 V. 4.20 (a) 10.1 V or 10 V (b) 2.25 4.21 (a) K = 1.8 (b) 2 V; same before and After (c) before: 1000 N/C After: 1000 N/C Q 4.22 (a) E = P0 AK Qd (b) V = P0 AK K P0 A (c) C = d 4.23 (a) 421 J (b) 0.05 µF 4.24 (a) 5.31×10−13 (b) 0.224 mm 4.25 (a) 0.0160 C (b) 533 V (c) 4.26 J (d) 2.14 J

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Answers

4.26 (a) 2.5 µF (b) Q1 = 550 µc, V1 Q2 = 370 µc, V2 Q3 = 180 µc, V3 Q4 = 180 µc, V4 Q5 = 550 µc, V5 4.27

= = = = =

65 V 87 V 43 V 43 V 65 V

4.35 (a)

l2 up2P0r2

(b)

l2 ln(rb/ra) 4pP0

(c)

l2L ln(rb/ra)) 4pP0

50 nF

4.36 (a) Keff = 1 +

(a) 450 nF

h Kh L L

(b) 1.48 (c) 17 (d) methanal. P0bv2(k - 1) P0bv2 4.37 (i) U = [l + x(k - 1)] , (ii) F = 2d 2d P0(K2 - K1) 4.38 d ln (K2/K1)

(b) 25 nF

CV2 J 12 4.40 (a) 10 V, (b) 4 4.39 ¢H =

(c) 75nF

(d) 4.28 (a) Vcd = 70V (b) same p.d 105 V (c) 315 µC 4.29 (a) 76 µC (b) 1.4×10 - 3 J (c) 11 V (d) 1.3×10 - 3 J 4.30 (a) 9.3×10 - 9F (b) 2.2×109V (c) 7.3×105V/m (d) 2.2×1010J 4.31 7200 µC, 654 µC 4.32 (a) k = 3.91 (b) 22.8 V P0 A 4.33 (a) d - a C0 d (b) (d - a) (c) As a S 0 c S C0 As a S d cS q

4.41 150 V e0A 4.42 d 4.43 2µF V 4.44 2a ln (3) 4.45

11C 15

Challenge Problems 4.46 (b) 2z3 - 2z2z0 +

P0 AV2 = 0 4k

P0L [L + (K - 1)x] D P0LV [L + (K - 1)x] (c) Q = D (K - 1)P0V2L (e) F = 2D 2pP0(1 + K) (rarb) 4.48 (a) C = (rb - ra) Q 2 (b) EL = 1 + K 4pP0r2 Q 2 EU = 1 + K 4pP0r2 4.47 (a)

4.34 (a) 4pP0R (b) 710 µF

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CURRENT, RESISTANCE, AND ELECTROMOTIVE FORCE

5 LEARNING GOALS

By studying this chapter, you will learn: • The meaning of electric current, and how charges move in a conductor. • What is meant by the resistivity and conductivity of a substance. • How to calculate the resistance of a conductor from its dimensions and its resistivity. • How an electromotive force (emf) makes it possible for current to flow in a circuit. • How to do calculations involving energy and power in circuits.

?

In a flashlight, is the amount of current that flows out of the bulb less than, greater than, or equal to the amount of current that flows into the bulb?

n the past four chapters we studied the interactions of electric charges at rest; now we’re ready to study charges in motion. An electric current consists of charges in motion from one region to another. If the charges follow a conducting path that forms a closed loop, the path is called an electric circuit. Fundamentally, electric circuits are a means for conveying energy from one place to another. As charged particles move within a circuit, electric potential energy is transferred from a source (such as a battery or generator) to a device in which that energy is either stored or converted to another form: into sound in a stereo system or into heat and light in a toaster or light bulb. Electric circuits are useful because they allow energy to be transported without any moving parts (other than the moving charged particles themselves). They are at the heart of flashlights, computers, radio and television transmitters and receivers, and household and industrial power distribution systems. Your nervous system is a specialized electric circuit that carries vital signals from one part of your body to another. In Chapter 6 we will see how to analyze electric circuits and will examine some practical applications of circuits. Before we can do so, however, you must understand the basic properties of electric currents. These properties are the subject of this chapter. We’ll begin by describing the nature of electric conductors and considering how they are affected by temperature. We’ll learn why a short, fat, cold copper wire is a better conductor than a long, skinny, hot steel wire. We’ll study the properties of batteries and see how they cause current and energy transfer in a circuit. In this analysis we will use the concepts of current, potential difference (or voltage), resistance, and electromotive force. Finally, we’ll look at electric current in a material from a microscopic viewpoint.

I

138

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139

5.1 Current

5.1

Current

A current is any motion of charge from one region to another. In this section we’ll discuss currents in conducting materials. The vast majority of technological applications of charges in motion involve currents of this kind. In electrostatic situations (discussed in Chapters 1 through 4) the electric field is zero everywhere within the conductor, and there is no current. However, this does not mean that all charges within the conductor are at rest. In an ordinary metal such as copper or alumium, some of the electrons are free to move within the conducting material. These free electrons move randomly in all directions, somewhat like the molecules of a gas but with much greater speeds, of the order of 106 m/s. The electrons nonetheless do not escape from the conducting material, because they are attracted to the positive ions of the material. The motion of the electrons is random, so there is no net flow of charge in any direction and hence no current. S Now consider what happens if a constant, steady electric field E is established inside a conductor. (We’ll see later how this can be done.) A charged particle (such as a free electron) inside the conducting material is then subjected to a S S steady force F 5 qE. If the charged particle were moving in vacuum, this steady S force would cause a steady acceleration in the direction of F, and after a time the charged particle would be moving in that direction at high speed. But a charged particle moving in a conductor undergoes frequent collisions with the massive, nearly stationary ions of the material. In each such collision the particle’s direcS tion of motion undergoes a random change. The net effect of the electric field E is that in addition to the random motion of the charged particles within the conductor, there is also a very slow net motion or drift of the moving charged partiS S cles as a group in the direction of the electric force F 5 qE (Fig. 5.1). This S motion is described in terms of the drift velocity vd of the particles. As a result, there is a net current in the conductor. While the random motion of the electrons has a very fast average speed of about 106 m/s, the drift speed is very slow, often on the order of 10-4 m/s. Given that the electrons move so slowly, you may wonder why the light comes on immediately when you turn on the switch of a flashlight. The reason is that the electric field is set up in the wire with a speed approaching the speed of light, and electrons start to move all along the wire at very nearly the same time. The time that it takes any individual electron to get from the switch to the light bulb isn’t really relevant. A good analogy is a group of soldiers standing at attention when the sergeant orders them to start marching; the order reaches the soldiers’ ears at the speed of sound, which is much faster than their marching speed, so all the soldiers start to march essentially in unison.

The Direction of Current Flow The drift of moving charges through Sa conductor can be interpreted in terms of work and energy. The electric field E does work on the moving charges. The resulting kinetic energy is transferred to the material of the conductor by means of collisions with the ions, which vibrate about their equilibrium positions in the crystalline structure of the conductor. This energy transfer increases the average vibrational energy of the ions and therefore the temperature of the material. Thus much of the work done by the electric field goes into heating the conductor, not into making the moving charges move ever faster and faster. This heating is sometimes useful, as in an electric toaster, but in many situations is simply an unavoidable by-product of current flow. In different current-carrying materials, the charges of the moving particles may be positive or negative. In metals the moving charges are always (negative) electrons, while in an ionized gas (plasma) or an ionic solution the moving charges may include both electrons and positively charged ions. In a semiconductor

5.1 If there is no electric field inside a conductor, an electron moves randomly from point P1 Sto point P2 in a time ¢t. If an electric field E is present, the electric force S S F 5 qE imposes a small drift (greatly exaggerated here) that takes the electron to point P¿2 , a distance vd ¢t from P2 in the direction of the force. '

'

S

Conductor without internal E field

S

Path of electron without E field. Electron moves randomly. Path of electron S with E field. The motion is mostly P1 random, but ...

P2 vdDt

P29 S

... the E field results in a net displacement along the wire. S

Conductor with internal E field S

E

S

S

F 5 qE

S

E

An electron has a negative charge q, S so the force on it due to the S E field is in the direction opposite to E.

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CHAPTER 5 Current, Resistance, and Electromotive Force

5.2 The same current can be produced by (a) positive charges moving in the direcS tion of the electric field E or (b) the same number of negative charges moving at the S same speed in the direction opposite to E. (a)

S

S

vd

S

vd

E

S

vd

S

vd

S

vd

S

S

vd S

vd

vd

S

vd I

A conventional current is treated as a flow of positive charges, regardless of whether the free charges in the conductor are positive, negative, or both. (b)

S

vd

S

vd

S

vd

S

E

material such as germanium or silicon, conduction is partly by electrons and partly by motion of vacancies, also known as holes; these are sites of missing electrons and act like positive charges. Figure 5.2 shows segments of two different current-carrying materials. In Fig. 5.2a the moving charges are positive, the electric force is in the same direction as S S E, and the drift velocity vd is from leftS to right. In Fig. 5.2b the charges are negaS tive, the electric force is opposite to E, and the drift velocity vd is from right to left. In both cases there is a net flow of positive charge from left to right, and positive charges end up to the right of negative ones. We define the current, denoted by I, to be in the direction in which there is a flow of positive charge. Thus we describe currents as though they consisted entirely of positive charge flow, even in cases in which we know that the actual current is due to electrons. Hence the current is to the right in both Figs. 5.2a and 5.2b. This choice or convention for the direction of current flow is called conventional current. While the direction of the conventional current is not necessarily the same as the direction in which charged particles are actually moving, we’ll find that the sign of the moving charges is of little importance in analyzing electric circuits. Figure 5.3 shows a segment of a conductor in which a current is flowing. We consider the moving charges to be positive, so they are moving in the same direction as the current. We define the current through the cross-sectional area A to be the net charge flowing through the area per unit time. Thus, if a net charge dQ flows through an area in a time dt, the current I through the area is

S

vd

S

vd

S

vd S

vd

I = S

vd

S

vd I

In a metallic conductor, the moving charges are electrons — but the current still points in the direction positive charges would flow.

5.3 The current I is the time rate of charge transfer through the cross-sectional area A. The random component of each moving charged particle’s motion averages to zero, and the current is in the same S direction as E whether the moving charges are positive (as shown here) or negative (see Fig. 5.2b).

+

vd dt r v d

vrd A

vrd vrd

vrd Current I 5

(definition of current)

(5.1)

CAUTION Current is not a vector Although we refer to the direction of a current, current as defined by Eq. (5.1) is not a vector quantity. In a current-carrying wire, the current is always along the length of the wire, regardless of whether the wire is straight or curved. No single vector could describe motion along a curved path, which is why current is not a vector. We’ll usually describe the direction of current either in words (as in “the current flows clockwise around the circuit”) or by choosing a current to be positive if it flows in one direction along a conductor and negative if it flows in the other direction. y

The SI unit of current is the ampere; one ampere is defined to be one coulomb per second 11 A = 1 C/s2. This unit is named in honor of the French scientist André Marie Ampère (1775–1836). When an ordinary flashlight (D-cell size) is turned on, the current in the flashlight is about 0.5 to 1 A; the current in the wires of a car engine’s starter motor is around 200 A. Currents in radio and television circuits are usually expressed in milliamperes 11 mA = 10-3 A2 or microamperes 11 mA = 10-6 A2, and currents in computer circuits are expressed in nanoamperes 11 nA = 10-9 A2 or picoamperes 11 pA = 10-12 A2.

Current, Drift Velocity, and Current Density

I

vrd

dQ dt

dQ dt

S

E

We can express current in terms of the drift velocity of the moving charges. Let’s consider again the situation of Fig. 5.3 of a conductor with cross-sectional area A S and an electric field E directed from left to right. To begin with, we’ll assume that the free charges in the conductor are positive; then the drift velocity is in the same direction as the field. Suppose there are n moving charged particles per unit volume. We call n the concentration of particles; its SI unit is m-3. Assume that all the particles move with the same drift velocity with magnitude vd. In a time interval dt, each particle moves a distance vd dt. The particles that flow out of the right end of the shaded cylinder with length vd dt during dt are the particles that were within this cylinder at the beginning of the interval dt . The volume of the cylinder is Avd dt, and the number of particles within it is n Avd dt. If each

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5.1 Current

141

particle has a charge q , the charge dQ that flows out of the end of the cylinder during time dt is dQ = q1n Avd dt2 = nqvd A dt and the current is I =

dQ = nqvd A dt

The current per unit cross-sectional area is called the current density J: J =

I = nqvd A

The units of current density are amperes per square meter 1A/m22. If the moving charges are negative rather than positive, as in Fig. 5.2b,Sthe S drift velocity is opposite to E. But the current is still in the same direction as E at each point in the conductor. Hence the current I and current density J don’t depend on the sign of the charge, and so in the above expressions for I and J we replace the charge q by its absolute value ƒ q ƒ : I =

J =

dQ = n ƒ q ƒ vd A dt

I = n ƒ q ƒ vd A

(general expression for current)

(general expression for current density)

(5.2)

(5.3)

The current in a conductor is the product of the concentration of moving charged particles, the magnitude of charge of each such particle, the magnitude of the drift velocity, and the cross-sectional area of the conductor. S We can also define a vector current density J that includes the direction of the drift velocity: S

S

J 5 nqvd

(vector current density)

(5.4) S

There are no Sabsolute value signs in Eq. (5.4). If q Sis positive, vd is inS the same S direction as E; if q is negative, vd is opposite to E. In either case, J is in the S same direction as E. Equation (5.3) gives the magnitude J of the vector current S density J . S

CAUTION Current density vs. current Note that current density J is a vector, but current S I is not. The difference is that the current density J describes how charges flow at a certain point, and the vector’s direction tells you about the direction of the flow at that point. By contrast, the current I describes how charges flow through an extended object such as a S wire. For example, I has the same value at all points in the circuit of Fig. 5.3, but J does not: The current density is directed downward in the left-hand side of the loop and upward S in the right-hand side. The magnitude of J can also vary around a circuit. In Fig. 5.3 the current density magnitude J = I/A is less in the battery (which has a large cross-sectional area A) than in the wires (which have a small cross-sectional area). y

5.4 Part of the electric circuit that includes this light bulb passes through a beaker with a solution of sodium chloride. The current in the solution is carried by both positive charges (Na+ ions) and negative charges (Cl- ions).

In general, a conductor may contain several different kinds of moving charged particles having charges q1 , q2 , Á , concentrations n1 , n2 , Á , and drift veloci ties with magnitudes vd1 , vd2 , Á . An example is current flow in an ionic solution (Fig. 5.4). In a sodium chloride solution, current can be carried by both positive sodium ions and negative chlorine ions; the total current I is found by adding up the currents due to each kind of charged particle, using Eq. (5.2). Likewise, the S total vector current density J is found by using Eq. (5.4) for each kind of charged particle and adding the results. We will see in Section 5.4 that it is possible to have a current that is steady (that is, one that is constant in time) only if the conducting material forms a '

'

'

'

'

'

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142

CHAPTER 5 Current, Resistance, and Electromotive Force

closed loop, called a complete circuit. In such a steady situation, the total charge in every segment of the conductor is constant. Hence the rate of flow of charge out at one end of a segment at any instant equals the rate of flow of charge in at the other end of the segment, and the current is the same at all cross sections of the circuit. We’ll make use of this observation when we analyze electric circuits later in this chapter. In many simple circuits, such as flashlights or cordless electric drills, the direction of the current is always the same; this is called direct current. But home appliances such as toasters, refrigerators, and televisions use alternating current, in which the current continuously changes direction. In this chapter we’ll consider direct current only. Alternating current has many special features worthy of detailed study, which we’ll examine in Chapter 11.

Example 5.1

Current density and drift velocity in a wire

An 18-gauge copper wire (the size usually used for lamp cords), with a diameter of 1.02 mm, carries a constant current of 1.67 A to a 200-W lamp. The free-electron density in the wire is 8.5 * 1028 per cubic meter. Find (a) the current density and (b) the drift speed.

The magnitude of the current density is then J =

vd =

EXECUTE: (a) The cross-sectional area is 2

A =

p11.02 * 10 pd = 4 4

-3

J

= 8.17 * 10-7 m2

'

nƒqƒ

2.04 * 106 A > m2 '

=

18.5 * 10

'

m 2 ƒ -1.60 * 10-19 C ƒ -3

28

= 1.5 * 10-4 m > s = 0.15 mm > s '

'

'

'

EVALUATE: At this speed an electron would require 6700 s (almost 2 h) to travel 1 m along this wire. The speeds of random motion of the electrons are roughly 106 m > s, around 1010 times the drift speed. Picture the electrons as bouncing around frantically, with a very slow drift! '

2

m2

'

(b) From Eq. (5.3) for the drift velocity magnitude vd, we find

SOLUTION IDENTIFY and SET UP: This problem uses the relationships among current I, current density J, and drift speed vd. We are given I and the wire diameter d, so we use Eq. (5.3) to find J. We use Eq. (5.3) again to find vd from J and the known electron density n.

1.67 A I = = 2.04 * 106 A > m2 A 8.17 * 10-7 m2

'

Test Your Understanding of Section 5.1 Suppose we replaced the wire in Example 5.1 with 12-gauge copper wire, which has twice the diameter of 18-gauge wire. If the current remains the same, what effect would this have on the magnitude of the drift velocity vd? (i) none—vd would be unchanged; (ii) vd would be twice as great; (iii) vd would be four times greater; (iv) vd would be half as great; (v) vd would be one-fourth as great.

5.2

y

Resistivity S

S

The current density J in a conductor depends on the electric field E and on the properties of the material. In general, this dependence can be quite complex. But S for some materials, especially metals, at a given temperature, J is nearly directly S proportional to E, and the ratio of the magnitudes of E and J is constant. This relationship, called Ohm’s law, was discovered in 1826 by the German physicist Georg Simon Ohm (1787–1854). The word “law” should actually be in quotation marks, since Ohm’s law, like the ideal-gas equation and Hooke’s law, is an idealized model that describes the behavior of some materials quite well but is not a general description of all matter. In the following discussion we’ll assume that Ohm’s law is valid, even though there are many situations in which it is not. The situation is comparable to our representation of the behavior of the static and kinetic friction forces; we treated these friction forces as being directly proportional to the normal force, even though we knew that this was at best an approximate description.

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5.2 Resistivity

143

Table 5.1 Resistivities at Room Temperature (20°C)

#

R 1æ m2

Substance Conductors Metals

1011 –1015 Alloys

* * * * * * * *

10-8 10-8 10-8 10-8 10-8 10-8 10-8 10-8

Silver Copper Gold Aluminum Tungsten Steel Lead Mercury

1.47 1.72 2.44 2.75 5.25 20 22 95

Manganin (Cu 84%, Mn 12%, Ni 4%) Constantan (Cu 60%, Ni 40%) Nichrome

44 * 10-8 49 * 10-8 100 * 10-8

#

R 1æ m2

Substance Semiconductors Pure carbon (graphite) Pure germanium Pure silicon Insulators Amber Glass Lucite Mica

3.5 * 10-5 0.60 2300 5 * 1014 1010 – 1014 7 1013 75 * 1016 1015 7 1013

Quartz (fused) Sulfur Teflon

108 – 1011

Wood

We define the resistivity r of a material as the ratio of the magnitudes of electric field and current density: r =

E J

(definition of resistivity)

(5.5)

The greater the resistivity, the greater the field needed to cause a given current density, or the smaller the current density caused by a given field. From Eq. (5.5) the units of r are 1V>m2>1A>m22 = V # m>A. As we will discuss in the next section, 1 V>A is called one ohm (1 Æ; we use the Greek letter Æ, or omega, which is alliterative with “ohm”). So the SI units for r are Æ # m (ohm-meters). Table 5.1 lists some representative values of resistivity. A perfect conductor would have zero resistivity, and a perfect insulator would have an infinite resistivity. Metals and alloys have the smallest resistivities and are the best conductors. The resistivities of insulators are greater than those of the metals by an enormous factor, on the order of 1022. The reciprocal of resistivity is conductivity. Its units are 1Æ # m2-1. Good conductors of electricity have larger conductivity than insulators. Conductivity is the direct electrical analog of thermal conductivity. Comparing Table 5.1 with Table 17.5 (Thermal Conductivities), we note that good electrical conductors, such as metals, are usually also good conductors of heat. Poor electrical conductors, such as ceramic and plastic materials, are also poor thermal conductors. In a metal the free electrons that carry charge in electrical conduction also provide the principal mechanism for heat conduction, so we should expect a correlation between electrical and thermal conductivity. Because of the enormous difference in conductivity between electrical conductors and insulators, it is easy to confine electric currents to well-defined paths or circuits (Fig. 5.5). The variation in thermal conductivity is much less, only a factor of 103 or so, and it is usually impossible to confine heat currents to that extent. Semiconductors have resistivities intermediate between those of metals and those of insulators. These materials are important because of the way their resistivities are affected by temperature and by small amounts of impurities. A material that obeys Ohm’s law reasonably well is called an ohmic conductor or a linear conductor. For such materials, at a given temperature, r is a constant that does not depend on the value of E. Many materials show substantial departures from Ohm’s-law behavior; they are nonohmic, or nonlinear. In these materials, J depends on E in a more complicated manner. Analogies with fluid flow can be a big help in developing intuition about electric current and circuits. For example, in the making of wine or maple syrup, the product is sometimes filtered to remove sediments. A pump forces the fluid through the filter under pressure; if the flow rate (analogous to J) is proportional to the pressure difference between the upstream and downstream sides (analogous to E), the behavior is analogous to Ohm’s law.

5.5 The copper “wires,” or traces, on this circuit board are printed directly onto the surface of the dark-colored insulating board. Even though the traces are very close to each other (only about a millimeter apart), the board has such a high resistivity (and low conductivity) that no current can flow between the traces.

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Conducting paths (traces)

144

CHAPTER 5 Current, Resistance, and Electromotive Force

Resistivity and Nerve Conduction

Application

This false-color image from an electron microscope shows a cross section through a nerve fiber about 1 µm (10 - 6 m) in diameter. A layer of an insulating fatty substance called myelin is wrapped around the conductive material of the axon. The resistivity of myelin is much greater than that of the axon, so an electric signal traveling along the nerve fiber remains confined to the axon. This makes it possible for a signal to travel much more rapidly than if the myelin were absent.

Resistivity and Temperature The resistivity of a metallic conductor nearly always increases with increasing temperature, as shown in Fig. 5.6a. As temperature increases, the ions of the conductor vibrate with greater amplitude, making it more likely that a moving electron will collide with an ion as in Fig. 5.1; this impedes the drift of electrons through the conductor and hence reduces the current. Over a small temperature range (up to 100 C° or so), the resistivity of a metal can be represented approximately by the equation r1T2 = r031 + a1T - T024

'

Table 5.2 Temperature Coefficients of Resistivity (Approximate Values Near Room Temperature)

Axon

Material

5.6 Variation of resistivity r with absolute temperature T for (a) a normal metal, (b) a semiconductor, and (c) a superconductor. In (a) the linear approximation to r as a function of T is shown as a green line; the approximation agrees exactly at T = T0 , where r = r0 . '

'

r Metal: Resistivity increases with increasing temperature. r0

Slope 5 r0a

O

(b)

T0

T

r Semiconductor: Resistivity decreases with increasing temperature.

T

O

(c)

r

O

(5.6)

where r0 is the resistivity at a reference temperature T0 (often taken as 0°C or 20°C2 and r1T2 is the resistivity at temperature T, which may be higher or lower than T0 . The factor a is called the temperature coefficient of resistivity. Some representative values are given in Table 5.2. The resistivity of the alloy manganin is practically independent of temperature.

Myelin

(a)

(temperature dependence of resistivity)

Superconductor: At temperatures below Tc, the resistivity is zero.

Tc

T

Aluminum Brass Carbon (graphite) Constantan Copper Iron

A 31°C2214

Material

A 31°C2214

0.0039 0.0020 - 0.0005 0.00001 0.00393 0.0050

Lead Manganin Mercury Nichrome Silver Tungsten

0.0043 0.00000 0.00088 0.0004 0.0038 0.0045

The resistivity of graphite (a nonmetal) decreases with increasing temperature, since at higher temperatures, more electrons are “shaken loose” from the atoms and become mobile; hence the temperature coefficient of resistivity of graphite is negative. This same behavior occurs for semiconductors (Fig. 5.6b). Measuring the resistivity of a small semiconductor crystal is therefore a sensitive measure of temperature; this is the principle of a type of thermometer called a thermistor. Some materials, including several metallic alloys and oxides, show a phenomenon called superconductivity. As the temperature decreases, the resistivity at first decreases smoothly, like that of any metal. But then at a certain critical temperature Tc a phase transition occurs and the resistivity suddenly drops to zero, as shown in Fig. 5.6c. Once a current has been established in a superconducting ring, it continues indefinitely without the presence of any driving field. Superconductivity was discovered in 1911 by the Dutch physicist Heike Kamerlingh Onnes (1853–1926). He discovered that at very low temperatures, below 4.2 K, the resistivity of mercury suddenly dropped to zero. For the next 75 years, the highest Tc attained was about 20 K. This meant that superconductivity occurred only when the material was cooled using expensive liquid helium, with a boiling-point temperature of 4.2 K, or explosive liquid hydrogen, with a boiling point of 20.3 K. But in 1986 Karl Müller and Johannes Bednorz discovered an oxide of barium, lanthanum, and copper with a Tc of nearly 40 K, and the race was on to develop “high-temperature” superconducting materials. By 1987 a complex oxide of yttrium, copper, and barium had been found that has a value of Tc well above the 77 K boiling temperature of liquid nitrogen, a refrigerant that is both inexpensive and safe. The current (2010) record for Tc at atmospheric pressure is 138 K, and materials that are superconductors at room temperature may become a reality. The implications of these discoveries for power-distribution systems, computer design, and transportation are enormous. Meanwhile, superconducting electromagnets cooled by liquid helium are used in particle accelerators and some experimental magnetic-levitation railroads.

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5.3 Resistance

145

Superconductors have other exotic properties that require an understanding of magnetism to explore; we will discuss these further in Chapter 29. Test Your Understanding of Section 5.2 You maintain a constant electric field inside a piece of semiconductor while lowering the semiconductor’s temperature. What happens to the current density in the semiconductor? (i) It increases; (ii) it decreases; (iii) it remains the same.

5.3

y

Resistance PhET: Resistance in a Wire S

For a conductor with resistivity r, the current density J at a point where the elecS tric field is E is given by Eq. (5.5), which we can write as S

S

E 5 rJ

(5.7)

When Ohm’s law is obeyed, r is constant and independent of the magnitude of S S the electric field, so E is directly proportional to J . Often, however, we are more S interested in the total current in a conductor than in J and more interested in the S potential difference between the ends of the conductor than in E. This is so largely because current and potential difference are much easier to measure than S S are J and E. Suppose our conductor is a wire with uniform cross-sectional area A and length L, as shown in Fig. 5.7. Let V be the potential difference between the higher-potential and lower-potential ends of the conductor, so that V is positive. The direction of the current is always from the higher-potential end to the lowerS potential end. That’s because current in a conductor flows in the direction of E, S no matter what the sign of the moving charges (Fig. 5.2), and because E points in the direction of decreasing electric potential (see Section 3.2). As the current flows through the potential difference, electric potential energy is lost; this energy is transferred to the ions of the conducting material during collisions. We can also relate the value of the current I to the potential difference between S the ends of the conductor. If the magnitudes of the current density and the elecJ S tric field E are uniform throughout the conductor, the total current I is given by I = JA, and the potential difference V between the ends is V = EL. When we solve these equations for J and E, respectively, and substitute the results in Eq. (5.7), we obtain

5.7 A conductor with uniform cross section. The current density is uniform over any cross section, and the electric field is constant along the length.

Higher potential

V I

rL A

(relationship between resistance and resistivity)

I

J A

(5.9)

(5.10)

If r is constant, as is the case for ohmic materials, then so is R. The equation V = IR (relationship among voltage, current, and resistance)

S

S

Comparing this definition of R to Eq. (5.8), we see that the resistance R of a particular conductor is related to the resistivity r of its material by R =

L E

rI rL V = or V = I (5.8) L A A This shows that when r is constant, the total current I is proportional to the potential difference V. The ratio of V to I for a particular conductor is called its resistance R: R =

Lower potential

Current flows from higher to lower electric potential.

(5.11)

is often called Ohm’s law, but it is important to understand that the real content of Ohm’s law is the direct proportionality (for some materials) of V to I or of J to E.

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I

V 5 potential difference between ends

146

CHAPTER 5 Current, Resistance, and Electromotive Force

Equation (5.9) or (5.11) defines resistance R for any conductor, whether or not it obeys Ohm’s law, but only when R is constant can we correctly call this relationship Ohm’s law.

Interpreting Resistance 5.8 A long fire hose offers substantial resistance to water flow. To make water pass through the hose rapidly, the upstream end of the hose must be at much higher pressure than the end where the water emerges. In an analogous way, there must be a large potential difference between the ends of a long wire in order to cause a substantial electric current through the wire.

5.9 This resistor has a resistance of 5.7 kÆ with a precision (tolerance) of 610%. Second digit

Multiplier Tolerance

Equation (5.10) shows that the resistance of a wire or other conductor of uniform cross section is directly proportional to its length and inversely proportional to its cross-sectional area. It is also proportional to the resistivity of the material of which the conductor is made. The flowing-fluid analogy is again useful. In analogy to Eq. (5.10), a narrow water hose offers more resistance to flow than a fat one, and a long hose has more resistance than a short one (Fig. 5.8). We can increase the resistance to flow by stuffing the hose with cotton or sand; this corresponds to increasing the resistivity. The flow rate is approximately proportional to the pressure difference between the ends. Flow rate is analogous to current, and pressure difference is analogous to potential difference (“voltage”). Let’s not stretch this analogy too far, though; the water flow rate in a pipe is usually not proportional to its crosssectional area (see Section 14.6). The SI unit of resistance is the ohm, equal to one volt per ampere 11 Æ = 1 V>A2. The kilohm 11 kÆ = 103 Æ2 and the megohm 11 MÆ = 106 Æ2 are also in common use. A 100-m length of 12-gauge copper wire, the size usually used in household wiring, has a resistance at room temperature of about 0.5 Æ. A 100-W, 120-V light bulb has a resistance (at operating temperature) of 140 Æ. If the same current I flows in both the copper wire and the light bulb, the potential difference V = IR is much greater across the light bulb, and much more potential energy is lost per charge in the light bulb. This lost energy is converted by the light bulb filament into light and heat. You don’t want your household wiring to glow white-hot, so its resistance is kept low by using wire of low resistivity and large cross-sectional area. Because the resistivity of a material varies with temperature, the resistance of a specific conductor also varies with temperature. For temperature ranges that are not too great, this variation is approximately a linear relationship, analogous to Eq. (5.6): R1T2 = R031 + a1T - T024

In this equation, R1T2 is the resistance at temperature T and R0 is the resistance at temperature T0 , often taken to be 0°C or 20°C. The temperature coefficient of resistance a is the same constant that appears in Eq. (5.6) if the dimensions L and A in Eq. (5.10) do not change appreciably with temperature; this is indeed the case for most conducting materials (see Problem 5.67). Within the limits of validity of Eq. (5.12), the change in resistance resulting from a temperature change T - T0 is given by R0 a1T - T02. A circuit device made to have a specific value of resistance between its ends is called a resistor. Resistors in the range 0.01 to 107 Æ can be bought off the shelf. Individual resistors used in electronic circuitry are often cylindrical, a few millimeters in diameter and length, with wires coming out of the ends. The resistance may be marked with a standard code using three or four color bands near one end (Fig. 5.9), according to the scheme shown in Table 5.3. The first two bands (starting with the band nearest an end) are digits, and the third is a power-of-10 multiplier, as shown in Fig. 5.9. For example, green–violet–red means 57 * 102 Æ, or 5.7 kÆ. The fourth band, if present, indicates the precision (tolerance) of the value; no band means 620%, a silver band 610%, and a gold band 65%. Another important characteristic of a resistor is the maximum power it can dissipate without damage. We’ll return to this point in Section 5.5. '

First digit

'

Table 5.3 Color Codes for Resistors Color

Value as Digit

Black Brown Red Orange Yellow Green Blue Violet Gray White

0 1 2 3 4 5 6 7 8 9

Value as Multiplier 1 10 102 103 104 105 106 107 108 109

(5.12)

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5.3 Resistance

147

5.10 Current–voltage relationships for two devices. Only for a resistor that obeys Ohm’s law as in (a) is current I proportional to voltage V. (a)

(b)

Ohmic resistor (e.g., typical metal wire): At a given temperature, current is proportional to voltage. I

Semiconductor diode: a nonohmic resistor

Slope 5

I

1 R V

O

O

In the direction of positive current and voltage, I increases nonlinearly with V. V

In the direction of negative current and voltage, little current flows.

For a resistor that obeys Ohm’s law, a graph of current as a function of potential difference (voltage) is a straight line (Fig. 5.10a). The slope of the line is 1>R. If the sign of the potential difference changes, so does the sign of the current produced; in Fig. 5.7 this corresponds to interchanging the higher- and lower-potential ends of the conductor, so the electric field, current density, and current all reverse direction. In devices that do not obey Ohm’s law, the relationship of voltage to current may not be a direct proportion, and it may be different for the two directions of current. Figure 5.10b shows the behavior of a semiconductor diode, a device used to convert alternating current to direct current and to perform a wide variety of logic functions in computer circuitry. For positive potentials V of the anode (one of two terminals of the diode) with respect to the cathode (the other terminal), I increases exponentially with increasing V; for negative potentials the current is extremely small. Thus a positive V causes a current to flow in the positive direction, but a potential difference of the other sign causes little or no current. Hence a diode acts like a one-way valve in a circuit. Example 5.2

Electric field, potential difference, and resistance in a wire

The 18-gauge copper wire of Example 5.1 has a cross-sectional area of 8.20 * 10 -7 m2. It carries a current of 1.67 A. Find (a) the electric-field magnitude in the wire; (b) the potential difference between two points in the wire 50.0 m apart; (c) the resistance of a 50.0-m length of this wire.

(b) The potential difference is

V = EL = 10.0350 V > m2150.0 m2 = 1.75 V '

(c) From Eq. (5.10) the resistance of 50.0 m of this wire is R =

S O L U T IO N IDENTIFY and SET UP: We are given the cross-sectional area A and current I. Our target variables are the electric-field magnitude E, potential difference V, and resistance R. The current density is J = I > A. We find E from Eq. (5.5), E = rJ (Table 5.1 gives the resistivity r for copper). The potential difference is then the product of E and the length of the wire. We can use either Eq. (5.10) or Eq. (5.11) to find R. '

'

rL 11.72 * 10-8 Æ # m2150.0 m2 = = 1.05 Æ A 8.20 * 10-7 m2

Alternatively, we can find R using Eq. (5.11): R =

1.75 V V = = 1.05 Æ I 1.67 A

'

EXECUTE: (a) From Table 5.1, r 5 1.72 * 10-8 Æ # m. Hence, using Eq. (5.5), rI 11.72 * 10-8 Æ # m211.67 A2 = A 8.20 * 10-7 m2 = 0.0350 V > m

EVALUATE: We emphasize that the resistance of the wire is defined to be the ratio of voltage to current. If the wire is made of nonohmic material, then R is different for different values of V but is always given by R = V > I. Resistance is also always given by R = rL > A; if the material is nonohmic, r is not constant but depends on E (or, equivalently, on V = EL). '

'

'

'

E = rJ =

'

'

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148

CHAPTER 5 Current, Resistance, and Electromotive Force

Example 5.3

Temperature dependence of resistance

Suppose the resistance of a copper wire is 1.05 Æ at 20°C. Find the resistance at 0°C and 100°C. SOLUTION IDENTIFY and SET UP: We are given the resistance R0 = 1.05 Æ at a reference temperature T0 = 20°C. We use Eq. (5.12) to find the resistances at T = 0°C and T = 100°C (our target variables), taking the temperature coefficient of resistivity from Table 5.2. EXECUTE: From Table 5.2, a = 0.00393 1C°2-1 for copper. Then from Eq. (5.12),

R = 11.05 Æ251 + 30.00393 1C°2-143100°C - 20°C46 = 1.38 Æ at T = 100°C

EVALUATE: The resistance at 100°C is greater than that at 0°C by a factor of 11.38 Æ2 > 10.97 Æ2 = 1.42: Raising the temperature of copper wire from 0°C to 100°C increases its resistance by 42%. From Eq. (5.11), V = IR, this means that 42% more voltage is required to produce the same current at 100°C than at 0°C. Designers of electric circuits that must operate over a wide temperature range must take this substantial effect into account. '

'

R = R031 + a1T - T024

= 11.05 Æ251 + 30.00393 1C°2-1430°C - 20°C46 = 0.97 Æ at T = 0°C

Test Your Understanding of Section 5.3 Suppose you increase the voltage across the copper wire in Examples 5.2 and 5.3. The increased voltage causes more current to flow, which makes the temperature of the wire increase. (The same thing happens to the coils of an electric oven or a toaster when a voltage is applied to them. We’ll explore this issue in more depth in Section 5.5.) If you double the voltage across the wire, the current in the wire increases. By what factor does it increase? (i) 2; (ii) greater than 2; (iii) less than 2. y

5.4 5.11 If an electric field is produced inside a conductor that is not part of a complete circuit, current flows for only a very short time. S

(a) An electric field E1 produced inside an isolated conductor causes a current. S

E1

I J

(b) The current causes charge to build up at the ends. S

E1

I

S

S

S

E2 I

Etotal

J

+ + +

The charge buildup produces an opposing S field E2, thus reducing the current.

S

(c) After a very short time E2 has the same S S magnitude as E1; then the total field is Etotal 5 0 and the current stops completely. – – – – – –

I50 S

J50

For a conductor to have a steady current, it must be part of a path that formsSa closed loop or complete circuit. Here’s why. If you establish an electric field E1 inside an isolated conductor with resistivity r that is not part of a complete cirS S cuit, a current begins to flow with current density J 5 E1>r (Fig. 5.11a). As a result a net positive charge quickly accumulates at one end of the conductor and a net negative charge accumulates at S the other end (Fig. 5.11b). These charges S themselves produce an electric field E2 in the direction opposite to E1 , causing the total electric field and hence the current to decrease. Within a very small fraction of a second, enough charge builds up on the conductorSends that the total S S S electric field E 5 E1 1 E2 5 0 inside the conductor. Then J 5 0 as well, and the current stops altogether (Fig. 5.11c). So there can be no steady motion of charge in such an incomplete circuit. To see how to maintain a steady current in a complete circuit, we recall a basic fact about electric potential energy: If a charge q goes around a complete circuit and returns to its starting point, the potential energy must be the same at the end of the round trip as at the beginning. As described in Section 5.3, there is always a decrease in potential energy when charges move through an ordinary conducting material with resistance. So there must be some part of the circuit in which the potential energy increases. The problem is analogous to an ornamental water fountain that recycles its water. The water pours out of openings at the top, cascades down over the terraces and spouts (moving in the direction of decreasing gravitational potential energy), and collects in a basin in the bottom. A pump then lifts it back to the top (increasing the potential energy) for another trip. Without the pump, the water would just fall to the bottom and stay there. '

I

S

– – –

Electromotive Force and Circuits

S

E1

S

S

Etotal 5 0

E2

+ + + + +

Electromotive Force In an electric circuit there must be a device somewhere in the loop that acts like the water pump in a water fountain (Fig. 5.12). In this device a charge travels

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5.4 Electromotive Force and Circuits

“uphill,” from lower to higher potential energy, even though the electrostatic force is trying to push it from higher to lower potential energy. The direction of current in such a device is from lower to higher potential, just the opposite of what happens in an ordinary conductor. The influence that makes current flow from lower to higher potential is called electromotive force (abbreviated emf and pronounced “ee-em-eff”). This is a poor term because emf is not a force but an energy-per-unit-charge quantity, like potential. The SI unit of emf is the same as that for potential, the volt 11 V = 1 J>C2. A typical flashlight battery has an emf of 1.5 V; this means that the battery does 1.5 J of work on every coulomb of charge that passes through it. We’ll use the symbol E (a script capital E) for emf. Every complete circuit with a steady current must include some device that provides emf. Such a device is called a source of emf. Batteries, electric generators, solar cells, thermocouples, and fuel cells are all examples of sources of emf. All such devices convert energy of some form (mechanical, chemical, thermal, and so on) into electric potential energy and transfer it into the circuit to which the device is connected. An ideal source of emf maintains a constant potential difference between its terminals, independent of the current through it. We define electromotive force quantitatively as the magnitude of this potential difference. As we will see, such an ideal source is a mythical beast, like the frictionless plane and the massless rope. We will discuss later how real-life sources of emf differ in their behavior from this idealized model. Figure 5.13 is a schematic diagram of an ideal source of emf that maintains a potential difference between conductors a and b, called the terminals of the device. Terminal a, marked +, is maintained at higher potential than terminal b, S marked -. Associated with this potential difference is an electric field E in the region around the terminals, both inside and outside the source. The electric field inside the device is directed from a to b, as shown. A charge q within the source S S experiences an electric force Fe 5 qE. But the source also provides an additional S influence, which we represent as a nonelectrostatic force Fn . This force, operating inside the device, pushes charge from b to a in an “uphill” direction against S S the electricSforce Fe . Thus Fn maintains the potential difference between the terminals. If Fn were not present, charge would flow between the terminals until the S potential difference was zero. The origin of the additional influence Fn depends on the kind of source. In a generator it results from magnetic-field forces on moving charges. In a battery or fuel cell it is associated with diffusion processes and varying electrolyte concentrations resulting from chemical reactions. In an electrostatic machine such as a Van de Graaff generator (see Fig. 22.26), an actual mechanical force is applied by a moving belt or wheel. If a positive charge q is moved from b to a inside the source, the nonelectroS static force Fn does a positive amount of work Wn = qE on the charge. This disS placement is opposite to the electrostatic force Fe , so the potential energy associated with the charge increases by an amount equal to qVab , where Vab = Va - Vb is the (positive) potential of point a with respect to point b. For S S the ideal source of emf that we’ve described, Fe and Fn are equal in magnitude but opposite in direction, so the total work done on the charge q is zero; there is an increase in potential energy but no change in the kinetic energy of the charge. It’s like lifting a book from the floor to a high shelf at constant speed. The increase in potential energy is just equal to the nonelectrostatic work Wn , so qE = qVab , or '

5.12 Just as a water fountain requires a pump, an electric circuit requires a source of electromotive force to sustain a steady current.

5.13 Schematic diagram of a source of emf in an “open-circuit” situation. The S S electric-field force Fe 5 qE and the S nonelectrostatic force Fn are shown for a positive charge q. Terminal at higher potential

Ideal emf source + a

Va

'

'

Nonelectrostatic force tending to move charge to higher potential

S

S

Vab 5 E

E

Fn q S

S

Fe 5 qE

Force due to electric field Vb

b Terminal at lower potential

When the emf source is not part of a closed circuit, Fn 5 Fe and there is no net motion of charge between the terminals.

'

'

PhET: Battery Voltage PhET: Signal Circuit ActivPhysics 12.1: DC Series Circuits (Qualitative)

'

Vab = E

(ideal source of emf)

149

(5.13)

Now let’s make a complete circuit by connecting a wire with resistance R to the terminals of a source (Fig. 5.14). The potential difference between terminals a and b sets up an electric field within the wire; this causes current to flow around the loop from a toward b, from higher to lower potential. Where the wire bends, equal amounts of positive and negative charge persist on the “inside” and “outside”

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150

CHAPTER 5 Current, Resistance, and Electromotive Force

5.14 Schematic diagram of an ideal source of emf in a complete circuit. The S S electric-field force Fe 5 qE and the nonS electrostatic force Fn are shown for a positive charge q. The current is in the direction from a to b in the external circuit and from b to a within the source. Potential across terminals creates electric field in circuit, causing charges to move. I Ideal emf source

S

S

E

E

I

S

Fe

Vb

(ideal source of emf)

(5.14)

That is, when a positive charge q flows around the circuit, the potential rise E as it passes through the ideal source is numerically equal to the potential drop Vab = IR as it passes through the remainder of the circuit. Once E and R are known, this relationship determines the current in the circuit.

?

Fn S

Vab 5 E

E = Vab = IR

S

E + a

Va

of the bend. These charges exert the forces that cause the current to follow the bends in the wire. From Eq. (5.11) the potential difference between the ends of the wire in Fig. 5.14 is given by Vab = IR. Combining with Eq. (5.13), we have

b S

E When a real I (as opposed to ideal) emf source is connected to a circuit, Vab and thus Fe fall, so S that Fn . Fe and Fn does work on the charges.

Application Danger: Electric Ray! Electric rays deliver electric shocks to stun their prey and to discourage predators. (In ancient Rome, physicians practiced a primitive form of electroconvulsive therapy by placing electric rays on their patients to cure headaches and gout.) The shocks are produced by specialized flattened cells called electroplaques. Such a cell moves ions across membranes to produce an emf of about 0.05 V. Thousands of electroplaques are stacked on top of each other, so their emfs add to a total of as much as 200 V. These stacks make up more than half of an electric ray’s body mass. A ray can use these to deliver an impressive current of up to 30 A for a few milliseconds.

CAUTION Current is not “used up” in a circuit It’s a common misconception that in a closed circuit, current is something that squirts out of the positive terminal of a battery and is consumed or “used up” by the time it reaches the negative terminal. In fact the current is the same at every point in a simple loop circuit like that in Fig. 5.14, even if the thickness of the wires is different at different points in the circuit. This happens because charge is conserved (that is, it can be neither created nor destroyed) and because charge cannot accumulate in the circuit devices we have described. If charge did accumulate, the potential differences would change with time. It’s like the flow of water in an ornamental fountain; water flows out of the top of the fountain at the same rate at which it reaches the bottom, no matter what the dimensions of the fountain. None of the water is “used up” along the way! y

Internal Resistance Real sources of emf in a circuit don’t behave in exactly the way we have described; the potential difference across a real source in a circuit is not equal to the emf as in Eq. (5.14). The reason is that charge moving through the material of any real source encounters resistance. We call this the internal resistance of the source, denoted by r. If this resistance behaves according to Ohm’s law, r is constant and independent of the current I. As the current moves through r, it experiences an associated drop in potential equal to Ir. Thus, when a current is flowing through a source from the negative terminal b to the positive terminal a, the potential difference Vab between the terminals is Vab = E - Ir

(terminal voltage, source with internal resistance)

(5.15)

The potential Vab , called the terminal voltage, is less than the emf E because of the term Ir representing the potential drop across the internal resistance r. Expressed another way, the increase in potential energy qVab as a charge q moves from b to a within the source is now less than the work qE done by the nonelecS trostatic force Fn , since some potential energy is lost in traversing the internal resistance. A 1.5-V battery has an emf of 1.5 V, but the terminal voltage Vab of the battery is equal to 1.5 V only if no current is flowing through it so that I = 0 in Eq. (5.15). If the battery is part of a complete circuit through which current is flowing, the terminal voltage will be less than 1.5 V. For a real source of emf, the terminal voltage equals the emf only if no current is flowing through the source (Fig. 5.15). Thus we can describe the behavior of a source in terms of two properties: an emf E, which supplies a constant potential difference independent of current, in series with an internal resistance r. The current in the external circuit connected to the source terminals a and b is still determined by Vab = IR. Combining this with Eq. (5.15), we find '

'

E - Ir = IR

or

I =

E R + r

(current, source with internal resistance)

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(5.16)

5.4 Electromotive Force and Circuits

That is, the current equals the source emf divided by the total circuit resistance 1R + r2.

151

5.15 The emf of this battery—that is, the terminal voltage when it’s not connected to anything—is 12 V. But because the battery has internal resistance, the terminal voltage of the battery is less than 12 V when it is supplying current to a light bulb.

CAUTION A battery is not a “current source” You might have thought that a battery or other source of emf always produces the same current, no matter what circuit it’s used in. Equation (5.16) shows that this isn’t so! The greater the resistance R of the external circuit, the less current the source will produce. It’s analogous to pushing an object through a thick, viscous liquid such as oil or molasses; if you exert a certain steady push (emf), you can move a small object at high speed (small R, large I) or a large object at low speed (large R, small I). y

Symbols for Circuit Diagrams An important part of analyzing any electric circuit is drawing a schematic circuit diagram. Table 5.4 shows the usual symbols used in circuit diagrams. We will use these symbols extensively in this chapter and the next. We usually assume that the wires that connect the various elements of the circuit have negligible resistance; from Eq. (5.11), V = IR, the potential difference between the ends of such a wire is zero. Table 5.4 includes two meters that are used to measure the properties of circuits. Idealized meters do not disturb the circuit in which they are connected. A voltmeter, introduced in Section 3.2, measures the potential difference between its terminals; an idealized voltmeter has infinitely large resistance and measures potential difference without having any current diverted through it. An ammeter measures the current passing through it; an idealized ammeter has zero resistance and has no potential difference between its terminals. Because meters act as part of the circuit in which they are connected, these properties are important to remember. Table 5.4 Symbols for Circuit Diagrams Conductor with negligible resistance R Resistor + E

E + or

Source of emf (longer vertical line always represents the positive terminal, usually the terminal with higher potential) Source of emf with internal resistance r (r can be placed on either side)

+ E

V

Voltmeter (measures potential difference between its terminals)

A

Ammeter (measures current through it)

Conceptual Example 5.4

A source in an open circuit

Figure 5.16 shows a source (a battery) with emf E = 12 V and internal resistance r 5 2 Æ. (For comparison, the internal resistance of a commercial 12-V lead storage battery is only a few thousandths of an ohm.) The wires to the left of a and to the right of the ammeter A are not connected to anything. Determine the respective readings Vab and I of the idealized voltmeter V and the idealized ammeter A.

5.16 A source of emf in an open circuit. Vab V +

A b

a r 5 2 V, E 5 12 V

Continued

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152

CHAPTER 5 Current, Resistance, and Electromotive Force

SOLUTION There is zero current because there is no complete circuit. (Our idealized voltmeter has an infinitely large resistance, so no current flows through it.) Hence the ammeter reads I = 0. Because there is no current through the battery, there is no potential difference across

Example 5.5

A source in a complete circuit

We add a 4-Æ resistor to the battery in Conceptual Example 5.4, forming a complete circuit (Fig. 5.17). What are the voltmeter and ammeter readings Vab and I now? SOLUTION

5.17 A source of emf in a complete circuit.

a9

A

Either way, we see that the voltmeter reading is 8 V. I

EVALUATE: With current flowing through the source, the terminal voltage Vab is less than the emf E. The smaller the internal resistance r, the less the difference between Vab and E.

b9

R54V

Conceptual Example 5.6

Vab = E - Ir = 12 V - 12 A212 Æ2 = 8 V

b

+ r 5 2 V, E 5 12 V

I

Using voltmeters and ammeters

We move the voltmeter and ammeter in Example 5.5 to different positions in the circuit. What are the readings of the ideal voltmeter and ammeter in the situations shown in (a) Fig. 5.18a and (b) Fig. 5.18b? SOLUTION (a) The voltmeter now measures the potential difference between points a¿ and b¿. As in Example 5.5, Vab = Va¿b¿, so the voltmeter reads the same as in Example 5.5: Va¿b¿ = 8 V.

5.18 Different placements of a voltmeter and an ammeter in a complete circuit. (b)

(a) a

+

b

r 5 2 V, E 5 12 V

a I

A

+

b

r 5 2 V, E 5 12 V

V Vbb9

a9

R54V V Va9b9

Our idealized conducting wires and the idealized ammeter have zero resistance, so there is no potential difference between points a and a¿ or between points b and b¿; that is, Vab = Va¿b¿. We find Vab by considering a and b as the terminals of the resistor: From Ohm’s law, Eq. (5.11), we then have

Alternatively, we can consider a and b as the terminals of the source. Then, from Eq. (5.15),

V a

E 12 V = = 2 A R + r 4 Æ + 2Æ

Va¿b¿ = IR = 12 A214 Æ2 = 8 V

Vab 5 Va9b9

A

EXECUTE: The ideal ammeter has zero resistance, so the total resistance external to the source is R = 4 Æ. From Eq. (5.16), the current through the circuit aa¿b¿b is then I =

IDENTIFY and SET UP: Our target variables are the current I through the circuit aa¿b¿b and the potential difference Vab. We first find I using Eq. (5.16). To find Vab, we can use either Eq. (5.11) or Eq. (5.15).

I

its internal resistance. From Eq. (5.15) with I = 0, the potential difference Vab across the battery terminals is equal to the emf. So the voltmeter reads Vab = E = 12 V. The terminal voltage of a real, nonideal source equals the emf only if there is no current flowing through the source, as in this example.

b9

a9

R54V

b9

CAUTION Current in a simple loop As charges move through a resistor, there is a decrease in electric potential energy, but there is no change in the current. The current in a simple loop is the same at every point; it is not “used up” as it moves through a resistor. Hence the ammeter in Fig. 5.17 (“downstream” of the 4-Æ resistor) and the ammeter in Fig. 5.18b (“upstream” of the resistor) both read I = 2 A. y (b) There is no current through the ideal voltmeter because it has infinitely large resistance. Since the voltmeter is now part of the circuit, there is no current at all in the circuit, and the ammeter reads I = 0. The voltmeter measures the potential difference Vbb¿ between points b and b¿. Since I = 0, the potential difference across the resistor is Va¿b¿ = IR = 0, and the potential difference between the ends a and a¿ of the idealized ammeter is also zero. So Vbb¿ is equal to Vab, the terminal voltage of the source. As in Conceptual Example 5.4, there is no current, so the terminal voltage equals the emf, and the voltmeter reading is Vab = E = 12 V. This example shows that ammeters and voltmeters are circuit elements, too. Moving the voltmeter from the position in Fig. 5.18a to that in Fig. 5.18b makes large changes in the current and potential differences in the circuit. If you want to measure the potential difference between two points in a circuit without disturbing the circuit, use a voltmeter as in Fig. 5.17 or 5.18a, not as in Fig. 5.18b.

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153

5.4 Electromotive Force and Circuits

Example 5.7

A source with a short circuit

In the circuit of Example 5.5 we replace the 4-Æ resistor with a zero-resistance conductor. What are the meter readings now?

EXECUTE: We must have Vab = IR = I102 = 0, no matter what the current. We can therefore find the current I from Eq. (5.15): Vab = E - Ir = 0

S O L U T IO N IDENTIFY and SET UP: Figure 5.19 shows the new circuit. Our target variables are again I and Vab. There is now a zero-resistance path between points a and b, through the lower loop, so the potential difference between these points must be zero. 5.19 Our sketch for this problem.

E 12 V = 6 A = r 2 Æ

I =

EVALUATE: The current has a different value than in Example 5.5, even though the same battery is used; the current depends on both the internal resistance r and the resistance of the external circuit. The situation here is called a short circuit. The external-circuit resistance is zero, because terminals of the battery are connected directly to each other. The short-circuit current is equal to the emf E divided by the internal resistance r. Warning: Short circuits can be dangerous! An automobile battery or a household power line has very small internal resistance (much less than in these examples), and the short-circuit current can be great enough to melt a small wire or cause a storage battery to explode.

Potential Changes Around a Circuit The net change in potential energy for a charge q making a round trip around a complete circuit must be zero. Hence the net change in potential around the circuit must also be zero; in other words, the algebraic sum of the potential differences and emfs around the loop is zero. We can see this by rewriting Eq. (5.16) in the form E - Ir - IR = 0 A potential gain of E is associated with the emf, and potential drops of Ir and IR are associated with the internal resistance of the source and the external circuit, respectively. Figure 5.20 is a graph showing how the potential varies as we go around the complete circuit of Fig. 5.17. The horizontal axis doesn’t necessarily represent actual distances, but rather various points in the loop. If we take the potential to be zero at the negative terminal of the battery, then we have a rise E and a drop Ir in the battery and an additional drop IR in the external resistor, and as we finish our trip around the loop, the potential is back where it started. In this section we have considered only situations in which the resistances are ohmic. If the circuit includes a nonlinear device such as a diode (see Fig. 5.10b), Eq. (5.16) is still valid but cannot be solved algebraically because R is not a constant. In such a situation, the current I can be found by using numerical techniques. Finally, we remark that Eq. (5.15) is not always an adequate representation of the behavior of a source. The emf may not be constant, and what we have described as an internal resistance may actually be a more complex voltage–current relationship that doesn’t obey Ohm’s law. Nevertheless, the concept of internal resistance frequently provides an adequate description of batteries, generators, and other energy converters. The principal difference between a fresh flashlight battery and an old one is not in the emf, which decreases only slightly with use, but in the internal resistance, which may increase from less than an ohm when the battery is fresh to as much as 1000 Æ or more after long use. Similarly, a car battery can deliver less current to the starter motor on a cold morning than when the battery is warm, not because the emf is appreciably less but because the internal resistance increases with decreasing temperature.

5.20 Potential rises and drops in a circuit. 2A

2A

2A

+ 12 V

2V

2A

4V

V 12 V Ir 5 4 V 8V E 5 12 V IR 5 8 V

O

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154

CHAPTER 5 Current, Resistance, and Electromotive Force Test Your Understanding of Section 5.4 Rank the following circuits in order from highest to lowest current. (i) a 1.4-Æ resistor connected to a 1.5-V battery that has an internal resistance of 0.10 Æ; (ii) a 1.8-Æ resistor connected to a 4.0-V battery that has a terminal voltage of 3.6 V but an unknown internal resistance; (iii) an unknown resistor connected to a 12.0 -V battery that has an internal resistance of 0.20 Æ and a terminal voltage of 11.0 V. y

5.5 5.21 The power input to the circuit element between a and b is P = 1Va - Vb2I = Vab I. '

Vb

Va Circuit element

I

I a

b

Energy and Power in Electric Circuits

Let’s now look at some energy and power relationships in electric circuits. The box in Fig. 5.21 represents a circuit element with potential difference Va - Vb = Vab between its terminals and current I passing through it in the direction from a toward b. This element might be a resistor, a battery, or something else; the details don’t matter. As charge passes through the circuit element, the electric field Sdoes work on the charge. In a source of emf, additional work is done by the force Fn that we mentioned in Section 5.4. As an amount of charge q passes through the circuit element, there is a change in potential energy equal to qVab . For example, if q 7 0 and Vab = Va - Vb is positive, potential energy decreases as the charge “falls” from potential Va to lower potential Vb . The moving charges don’t gain kinetic energy, because the current (the rate of charge flow) out of the circuit element must be the same as the current into the element. Instead, the quantity qVab represents energy transferred into the circuit element. This situation occurs in the coils of a toaster or electric oven, in which electrical energy is converted to thermal energy. If the potential at a is lower than at b, then Vab is negative and there is a net transfer of energy out of the circuit element. The element then acts as a source, delivering electrical energy into the circuit to which it is attached. This is the usual situation for a battery, which converts chemical energy into electrical energy and delivers it to the external circuit. Thus qVab can denote either a quantity of energy delivered to a circuit element or a quantity of energy extracted from that element. In electric circuits we are most often interested in the rate at which energy is either delivered to or extracted from a circuit element. If the current through the element is I, then in a time interval dt an amount of charge dQ = I dt passes through the element. The potential energy change for this amount of charge is Vab dQ = Vab I dt. Dividing this expression by dt, we obtain the rate at which energy is transferred either into or out of the circuit element. The time rate of energy transfer is power, denoted by P, so we write '

'

PhET: Battery-Resistor Circuit PhET: Circuit Construction Kit (AC+DC) PhET: Circuit Construction Kit (DC Only) PhET: Ohm’s Law

'

(rate at which energy is delivered to or extracted from a circuit element)

P = Vab I

(5.17)

The unit of Vab is one volt, or one joule per coulomb, and the unit of I is one ampere, or one coulomb per second. Hence the unit of P = Vab I is one watt, as it should be: 11 J>C211 C>s2 = 1 J>s = 1 W

Let’s consider a few special cases.

Power Input to a Pure Resistance If the circuit element in Fig. 5.21 is a resistor, the potential difference is Vab = IR. From Eq. (5.17) the electrical power delivered to the resistor by the circuit is P = Vab I = I2R =

Vab 2 R '

(power delivered to a resistor)

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(5.18)

155

5.5 Energy and Power in Electric Circuits

In this case the potential at a (where the current enters the resistor) is always higher than that at b (where the current exits). Current enters the higher-potential terminal of the device, and Eq. (5.18) represents the rate of transfer of electric potential energy into the circuit element. What becomes of this energy? The moving charges collide with atoms in the resistor and transfer some of their energy to these atoms, increasing the internal energy of the material. Either the temperature of the resistor increases or there is a flow of heat out of it, or both. In any of these cases we say that energy is dissipated in the resistor at a rate I2R. Every resistor has a power rating, the maximum power the device can dissipate without becoming overheated and damaged. Some devices, such as electric heaters, are designed to get hot and transfer heat to their surroundings. But if the power rating is exceeded, even such a device may melt or even explode.

5.22 Energy conversion in a simple circuit. (a) Diagrammatic circuit • The emf source converts nonelectrical to electrical energy at a rate EI. • Its internal resistance dissipates energy at a rate I 2r. • The difference EI 2 I 2r is its power output. E, r S

Fn

Power Output of a Source

S

v

+ a

The upper rectangle in Fig. 5.22a represents a source with emf E and internal resistance r, connected by ideal (resistanceless) conductors to an external circuit represented by the lower box. This could describe a car battery connected to one of the car’s headlights (Fig. 5.22b). Point a is at higher potential than point b, so Va 7 Vb and Vab is positive. Note that the current I is leaving the source at the higher-potential terminal (rather than entering there). Energy is being delivered to the external circuit, at a rate given by Eq. (5.17):

q

S

b

Fe

emf source with internal resistance r

I

External circuit

+ a

I

b

P = Vab I For a source that can be described by an emf E and an internal resistance r, we may use Eq. (5.15):

(b) A real circuit of the type shown in (a)

+

Vab = E - Ir

a

b

Multiplying this equation by I, we find P = Vab I = E I - I2r '

(5.19)

What do the terms EI and I2r mean? In Section 5.4 we defined the emf E as the work per unit charge performed on the charges by the nonelectrostatic force as the charges are pushed “uphill” from b to a in the source. In a time dt, a charge dQ = I dt flows through the source; the work done on it by this nonelectrostatic force is E dQ = EI dt. Thus EI is the rate at which work is done on the circulating charges by whatever agency causes the nonelectrostatic force in the source. This term represents the rate of conversion of nonelectrical energy to electrical energy within the source. The term I2r is the rate at which electrical energy is dissipated in the internal resistance of the source. The difference EI - I2r is the net electrical power output of the source—that is, the rate at which the source delivers electrical energy to the remainder of the circuit.

Power Input to a Source Suppose that the lower rectangle in Fig. 5.22a is itself a source, with an emf larger than that of the upper source and with its emf opposite to that of the upper source. Figure 5.23 shows a practical example, an automobile battery (the upper circuit element) being charged by the car’s alternator (the lower element). The current I in the circuit is then opposite to that shown in Fig. 5.22; the lower source is pushing current backward through the upper source. Because of this reversal of current, instead of Eq. (5.15) we have for the upper source

Battery

I

+ a

5.23 When two sources are connected in a simple loop circuit, the source with the larger emf delivers energy to the other source.

+

– a

I

b

Battery (small emf) S

a+

Fn r

–b

v

S

q

Fe

and instead of Eq. (5.19), we have 2

P = Vab I = EI + I r '

b Headlight

Vab = E + Ir

'

I

Alternator (large emf) (5.20)

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I

156

CHAPTER 5 Current, Resistance, and Electromotive Force

Work is being done on, rather than by, the agent that causes the nonelectrostatic force in the upper source. There is a conversion of electrical energy into nonelectrical energy in the upper source at a rate EI. The term I2r in Eq. (5.20) is again the rate of dissipation of energy in the internal resistance of the upper source, and the sum EI + I2r is the total electrical power input to the upper source. This is what happens when a rechargeable battery (a storage battery) is connected to a charger. The charger supplies electrical energy to the battery; part of it is converted to chemical energy, to be reconverted later, and the remainder is dissipated (wasted) in the battery’s internal resistance, warming the battery and causing a heat flow out of it. If you have a power tool or laptop computer with a rechargeable battery, you may have noticed that it gets warm while it is charging. Problem-Solving Strategy 5.1

Power and Energy in Circuits

IDENTIFY the relevant concepts: The ideas of electric power input and output can be applied to any electric circuit. Many problems will ask you to explicitly consider power or energy. SET UP the problem using the following steps: 1. Make a drawing of the circuit. 2. Identify the circuit elements, including sources of emf and resistors. We will introduce other circuit elements later, including capacitors (Chapter 6) and inductors (Chapter 10). 3. Identify the target variables. Typically they will be the power input or output for each circuit element, or the total amount of energy put into or taken out of a circuit element in a given time. EXECUTE the solution as follows: 1. A source of emf E delivers power EI into a circuit when current I flows through the source in the direction from - to + . (For example, energy is converted from chemical energy in a battery, or from mechanical energy in a generator.) In this case there is a positive power output to the circuit or, equivalently, a negative power input to the source. 2. A source of emf takes power EI from a circuit when current passes through the source from + to - . (This occurs in charging a storage battery, when electrical energy is converted to chemical energy.) In this case there is a negative power output

Example 5.8

to the circuit or, equivalently, a positive power input to the source. 3. There is always a positive power input to a resistor through which current flows, irrespective of the direction of current flow. This process removes energy from the circuit, converting it to heat at the rate VI = I2R = V2 > R, where V is the potential difference across the resistor. 4. Just as in item 3, there always is a positive power input to the internal resistance r of a source through which current flows, irrespective of the direction of current flow. This process likewise removes energy from the circuit, converting it into heat at the rate I2r. 5. If the power into or out of a circuit element is constant, the energy delivered to or extracted from that element is the product of power and elapsed time. (In Chapter 26 we will encounter situations in which the power is not constant. In such cases, calculating the total energy requires an integral over the relevant time interval.) '

'

EVALUATE your answer: Check your results; in particular, check that energy is conserved. This conservation can be expressed in either of two forms: “net power input = net power output” or “the algebraic sum of the power inputs to the circuit elements is zero.”

Power input and output in a complete circuit

For the circuit that we analyzed in Example 5.5, find the rates of energy conversion (chemical to electrical) and energy dissipation in the battery, the rate of energy dissipation in the 4- Æ resistor, and the battery’s net power output.

5.24 Our sketch for this problem.

SOLUTION IDENTIFY and SET UP: Figure 5.24 shows the circuit, gives values of quantities known from Example 5.5, and indicates how we find the target variables. We use Eq. (5.19) to find the battery’s net power output, the rate of chemical-to-electrical energy conversion, and the rate of energy dissipation in the battery’s internal resistance. We use Eq. (5.18) to find the power delivered to (and dissipated in) the 4- Æ resistor. EXECUTE: From the first term in Eq. (5.19), the rate of energy conversion in the battery is EI = 112 V212 A2 = 24 W

From the second term in Eq. (5.19), the rate of dissipation of energy in the battery is I2r = 12 A2212 Æ2 = 8 W

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5.5 Energy and Power in Electric Circuits The net electrical power output of the battery is the difference between these: EI - I2r = 16 W. From Eq. (5.18), the electrical power input to, and the equal rate of dissipation of electrical energy in, the 4- Æ resistor are Va¿b¿I = 18 V212 A2 = 16 W and I2R = 12 A22 14 Æ2 = 16 W '

Example 5.9

157

EVALUATE: The rate Va¿b¿I at which energy is supplied to the 4- Æ resistor equals the rate I2R at which energy is dissipated there. This is also equal to the battery’s net power output: P = Vab I = 18 V212 A2 = 16 W. In summary, the rate at which the source of emf supplies energy is EI = 24 W, of which I2r = 8 W is dissipated in the battery’s internal resistor and I2R = 16 W is dissipated in the external resistor.

Increasing the resistance

Suppose we replace the external 4-Æ resistor in Fig. 5.24 with an 8-Æ resistor. Hiow does this affect the electrical power dissipated in this resistor? S O L U T IO N IDENTIFY and SET UP: Our target variable is the power dissipated in the resistor to which the battery is connected. The situation is the same as in Example 5.8, but with a higher external resistance R. EXECUTE: According to Eq. (5.18), the power dissipated in the resistor is P = I2R. You might conclude that making the resistance R twice as great as in Example 5.8 should also make the power twice as great, or 2116 W2 = 32 W. If instead you used the formula P = Vab2 > R, you might conclude that the power should be one-half as great as in the preceding example, or 116 W2 > 2 = 8 W. Which answer is correct? In fact, both of these answers are incorrect. The first is wrong because changing the resistance R also changes the current in the circuit (remember, a source of emf does not generate the same current in all situations). The second answer is wrong because the potential difference Vab across the resistor changes when the current changes. To get the correct answer, we first find the current just as we did in Example 5.5: '

'

'

I =

Example 5.10

'

The greater resistance causes the current to decrease. The potential difference across the resistor is Vab = IR = 11.2 A218 Æ2 = 9.6 V which is greater than that with the 4-Æ resistor. We can then find the power dissipated in the resistor in either of two ways: P = I2R = 11.2 A2218 Æ2 = 12 W or P =

Vab2 19.6 V22 = = 12 W R 8Æ

EVALUATE: Increasing the resistance R causes a reduction in the power input to the resistor. In the expression P = I2R the decrease in current is more important than the increase in resistance; in the expression P = Vab2 > R the increase in resistance is more important than the increase in Vab. This same principle applies to ordinary light bulbs; a 50-W light bulb has a greater resistance than does a 100-W light bulb. Can you show that replacing the 4-Æ resistor with an 8-Æ resistor decreases both the rate of energy conversion (chemical to electrical) in the battery and the rate of energy dissipation in the battery? '

'

E 12 V = = 1.2 A R + r 8 Æ + 2 Æ

Power in a short circuit

For the short-circuit situation of Example 5.7, find the rates of energy conversion and energy dissipation in the battery and the net power output of the battery. S O L U T IO N IDENTIFY and SET UP: Our target variables are again the power inputs and outputs associated with the battery. Figure 5.25 shows 5.25 Our sketch for this problem.

the circuit. This is the same situation as in Example 5.8, but now the external resistance R is zero. EXECUTE: We found in Example 5.7 that the current in this situation is I = 6 A. From Eq. (5.19), the rate of energy conversion (chemical to electrical) in the battery is then EI = 112 V216 A2 = 72 W and the rate of dissipation of energy in the battery is I2r = 16 A2212 Æ2 = 72 W

The net power output of the source is EI - I2r = 0. We get this same result from the expression P = Vab I, because the terminal voltage Vab of the source is zero. EVALUATE: With ideal wires and an ideal ammeter, so that R = 0, all of the converted energy from the source is dissipated within the source. This is why a short-circuited battery is quickly ruined and may explode.

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158

CHAPTER 5 Current, Resistance, and Electromotive Force Test Your Understanding of Section 5.5 Rank the following circuits in order from highest to lowest values of the net power output of the battery. (i) a 1.4 - Æ resistor connected to a 1.5-V battery that has an internal resistance of 0.10 Æ; (ii) a 1.8-Æ resistor connected to a 4.0-V battery that has a terminal voltage of 3.6 V but an unknown internal resistance; (iii) an unknown resistor connected to a 12.0-V battery that has an internal resistance of 0.20 Æ and a terminal voltage of 11.0 V. y

5.6 5.26 Random motions of an electron in a metallic crystal (a) with zero electric field and (b) with an electric field that causes drift. The curvatures of the paths are greatly exaggerated. (a) Collision with crystal

S

Without E field: random motion END

START

S

E

(b) S

E S S

START

END

E

With E field: random motion plus drift

Theory of Metallic Conduction

We can gain additional insight into electrical conduction by looking at the microscopic origin of conductivity. We’ll consider a very simple model that treats the electrons as classical particles and ignores their quantum-mechanical behavior in solids. Using this model, we’ll derive an expression for the resistivity of a metal. Even though this model is not entirely correct, it will still help you to develop an intuitive idea of the microscopic basis of conduction. In the simplest microscopic model of conduction in a metal, each atom in the metallic crystal gives up one or more of its outer electrons. These electrons are then free to move through the crystal, colliding at intervals with the stationary positive ions. The motion of the electrons is analogous to the motion of molecules of a gas moving through a porous bed of sand. If there is no electric field, the electrons move in straight lines between collisions, the directions of their velocities are random, and on average they never get anywhere (Fig. 5.26a). But if an electric field is present, the paths curve slightly because of the acceleration caused by electric-field forces. Figure 5.26b shows a few paths of an electron in an electric field directed from right to left. As we mentioned in Section 5.1, the average speed of random motion is of the order of 106 m>s, while the average drift speed is much slower, of the order of 10-4 m>s. The average time between collisions is called the mean free time, denoted by t. Figure 5.27 shows a mechanical analog of this electron motion. We would like to derive from this model an expression for the resistivity r of a material, defined by Eq. (5.5):

Net displacement

r =

E J

(5.21)

where E and J are the magnitudes of electric field and current density, respectively. S The current density J is in turn given by Eq. (5.4): S

S

J 5 nqÁd

PhET: Conductivity

5.27 The motion of a ball rolling down an inclined plane and bouncing off pegs in its path is analogous to the motion of an electron in a metallic conductor with an electric field present.

(5.22)

where n is the number of free electrons per unit volume, q = - e is the charge of S each, and Ád is their average drift velocity. S S S We need to relate the drift velocity Ád to the electric field E. The value of Ád is determined by a steady-state condition Sin which, on average, the velocity gains of the charges due to the force of the E field are just balanced by the velocity losses due to collisions. To clarify this process, let’s imagine turning on the two effects one at a time. Suppose that before time t = 0 there is no field. The electron S motion is then completely random. A typical electron has velocity Á0 at time S t = 0, and the value of Á0 averaged over many electrons (that is, the initial velocS ity of an average electron) is zero: 1Á02av 5 0. Then at time t = 0 we turn on a S S S constant electric field E. The field exerts a force F 5 qE on each charge, and this S causes an acceleration a in the direction of the force, given by S

S

qE F a5 5 m m

S

where m is the electron mass. Every electron has this acceleration.

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5.6 Theory of Metallic Conduction

We wait for a time t, the average time between collisions, and then “turn on” S the collisions. An electron that has velocity Á0 at time t = 0 has a velocity at time t = t equal to S

S

S

Á 5 Á0 1 a t S

The velocity Áav of an average electron at this time is the sum of the averages of S the two terms on the right. As we have pointed out, the initial velocity Á0 is zero for an average electron, so S

S

Áav 5 a t 5

qt S E m

(5.23)

After time t = t, the tendency of the collisions to decrease the velocity of an averageS electron (by means of randomizing collisions) just balances the tendency of the E field to increase this velocity. Thus the velocity of an average electron, S given by Eq. (5.23), is maintained over time and is equal to the drift velocity Ád : qt S S E Ád 5 m '

S

Now we substitute this equation for the drift velocity Ád into Eq. (5.22): S

S

J 5 nqÁd 5

nq2t S E m

Comparing this with Eq. (5.21), which we can rewrite as J 5 E>r, and substituting q = -e for an electron, we see that the resistivity r is given by S

r = S

S

m ne2t

(5.24)

S

If n and t are independent of E, then the resistivity is independent of E and the conducting material obeys Ohm’s law. Turning the interactions on one at a time may seem artificial. But the derivation would come out the same if each electron had its own clock and the t = 0 times were different for different electrons. If t is the average time between colliS sions, then Ád is still the average electron drift velocity, even though the motions of the various electrons aren’t actually correlated in the way we postulated. What about the temperature dependence of resistivity? In a perfect crystal with no atoms out of place, a correct quantum-mechanical analysis would let the free electrons move through the crystal with no collisions at all. But the atoms vibrate about their equilibrium positions. As the temperature increases, the amplitudes of these vibrations increase, collisions become more frequent, and the mean free time t decreases. So this theory predicts that the resistivity of a metal increases with temperature. In a superconductor, roughly speaking, there are no inelastic collisions, t is infinite, and the resistivity r is zero. In a pure semiconductor such as silicon or germanium, the number of charge carriers per unit volume, n, is not constant but increases very rapidly with increasing temperature. This increase in n far outweighs the decrease in the mean free time, and in a semiconductor the resistivity always decreases rapidly with increasing temperature. At low temperatures, n is very small, and the resistivity becomes so large that the material can be considered an insulator. Electrons gain energy between collisions through the work done on them by the electric field. During collisions they transfer some of this energy to the atoms of the material of the conductor. This leads to an increase in the material’s internal energy and temperature; that’s why wires carrying current get warm. If the electric field in the material is large enough, an electron can gain enough energy between collisions to knock off electrons that are normally bound to atoms in the material. These can then knock off more electrons, and so on, leading to an avalanche of current. This is the basis of dielectric breakdown in insulators (see Section 4.4).

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159

160

CHAPTER 5 Current, Resistance, and Electromotive Force

Example 5.11

Mean free time in copper

Calculate the mean free time between collisions in copper at room temperature. SOLUTION IDENTIFY and SET UP: We can obtain an expression for mean free time t in terms of n, r, e, and m by rearranging Eq. (5.24). From Example 5.1 and Table 5.1, for copper n = 8.5 * 1028 m-3 and r = 1.72 * 10-8 Æ # m. In addition, e = 1.60 * 10-19 C and m = 9.11 * 10-31 kg for electrons.

EXECUTE: From Eq. (5.24), we get m t = 2 ne r =

9.11 * 10-31 kg

18.5 * 1028 m-3211.60 * 10-19 C2211.72 * 10-8 Æ # m2

= 2.4 * 10-14 s

EVALUATE: The mean free time is the average time between collisions for a given electron. Taking the reciprocal of this time, we find that each electron averages 1>t = 4.2 * 1013 collisions per second!

Test Your Understanding of Section 5.6 Which of the following factors will, if increased, make it more difficult to produce a certain amount of current in a conductor? (There may be more than one correct answer.) (i) the mass of the moving charged particles in the conductor; (ii) the number of moving charged particles per cubic meter; (iii) the amount of charge on each moving particle; (iv) the average time between collisions for a typical moving charged particle. y

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CHAPTER

5

SUMMARY

Current and current density: Current is the amount of charge flowing through a specified area, per unit time. The SI unit of current is the ampere 11 A = 1 C>s2. The current I through an area A depends on the concentration n and charge q of the charge carriers, as well as on S the magnitude of their drift velocity Ád . The current density is current per unit cross-sectional area. Current is usually described in terms of a flow of positive charge, even when the charges are actually negative or of both signs. (See Example 5.1.)

I =

dQ = n ƒ q ƒ vd A dt '

S

I

(5.2) S

S

vd

vd S

S

J 5 nqÁd

S

vd

(5.4)

vd

S

S

E

S

vd

vd

'

Resistivity: The resistivity r of a material is the ratio of the magnitudes of electric field and current density. Good conductors have small resistivity; good insulators have large resistivity. Ohm’s law, obeyed approximately by many materials, states that r is a constant independent of the value of E. Resistivity usually increases with temperature; for small temperature changes this variation is represented approximately by Eq. (5.6), where a is the temperature coefficient of resistivity.

r =

E J

r

(5.5)

r1T2 = r0 31 + a1T - T024 '

(5.6)

Slope 5 r0a

r0

O

T0 Metal: r increases with increasing T.

Resistors: The potential difference V across a sample of material that obeys Ohm’s law is proportional to the current I through the sample. The ratio V>I = R is the resistance of the sample. The SI unit of resistance is the ohm 11 Æ = 1 V>A2. The resistance of a cylindrical conductor is related to its resistivity r, length L, and cross-sectional area A. (See Examples 5.2 and 5.3.)

V = IR

(5.11)

rL A

(5.10)

Circuits and emf: A complete circuit has a continuous current-carrying path. A complete circuit carrying a steady current must contain a source of electromotive force (emf) E. The SI unit of electromotive force is the volt (1 V). Every real source of emf has some internal resistance r, so its terminal potential difference Vab depends on current. (See Examples 5.4–5.7.)

Vab = E - Ir (5.15) (source with internal resistance)

Energy and power in circuits: A circuit element with a potential difference Va - Vb = Vab and a current I puts energy into a circuit if the current direction is from lower to higher potential in the device, and it takes energy out of the circuit if the current is opposite. The power P equals the product of the potential difference and the current. A resistor always takes electrical energy out of a circuit. (See Examples 5.8–5.10.)

P = Vab I (general circuit element)

R =

Lower potential Higher potential

L S S

E A

I

Vab 5 Va9b9 V

I

+

(5.17)

(5.18)

Conduction in metals: The microscopic basis of conduction in metals is the motion of electrons that move freely through the metallic crystal, bumping into ion cores in the crystal. In a crude classical model of this motion, the resistivity of the material can be related to the electron mass, charge, speed of random motion, density, and mean free time between collisions. (See Example 5.11.)

R 5 4V

I

b9

Vb

Va Circuit element

I '

b

r 5 2 V, E 5 12 V A a9

Vab2 P = Vab I = I R = R (power into a resistor)

I

J V

a

2

T

a

I b

S

E Net displacement

161

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162

CHAPTER 5 Current, Resistance, and Electromotive Force

BRIDGING PROBLEM

Resistivity, Temperature, and Power

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 20°C, the heating element carries an initial current of 1.35 A. A few seconds later the current reaches the steady value of 1.23 A. (a) What is the final temperature of the element? The average value of the temperature coefficient of resistivity for Nichrome over the relevant temperature range is 4.5 * 10-4 1C°2-1. (b) What is the power dissipated in the heating element initially and when the current reaches 1.23 A? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. A heating element acts as a resistor that converts electrical energy into thermal energy. The resistivity r of Nichrome de pends on temperature, and hence so does the resistance R = rL>A of the heating element and the current I = V>R that it carries. 2. We are given V = 120 V and the initial and final values of I. Select an equation that will allow you to find the initial and

Problems

final values of resistance, and an equation that relates resistance to temperature [the target variable in part (a)]. 3. The power P dissipated in the heating element depends on I and V. Select an equation that will allow you to calculate the initial and final values of P. EXECUTE 4. Combine your equations from step 2 to give a relationship between the initial and final values of I and the initial and final temperatures (20°C and Tfinal). 5. Solve your expression from step 4 for Tfinal. 6. Use your equation from step 3 to find the initial and final powers. EVALUATE 7. Is the final temperature greater than or less than 20°C? Does this make sense? 8. Is the final resistance greater than or less than the initial resistance? Again, does this make sense? 9. Is the final power greater than or less than the initial power? Does this agree with your observations in step 8?

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. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS

Q5.1 The definition of resistivity 1r = E > J) implies that an electric field exists inside a conductor. Yet we saw in Chapter 1 that there can be no electric field inside a conductor. Is there a contradiction here? Explain. Q5.2 A cylindrical rod has resistance R. If we triple its length and diameter, what is its resistance, in terms of R? Q5.3 A cylindrical rod has resistivity r. If we triple its length and diameter, what is its resistivity, in terms of r? Q5.4 Two copper wires with different diameters are joined end to end. If a current flows in the wire combination, what happens to electrons when they move from the larger-diameter wire into the smaller-diameter wire? Does their drift speed increase, decrease, or stay the same? If the drift speed changes, what is the force that causes the change? Explain your reasoning. Q5.5 When is a 1.5-V AAA battery not actually a 1.5-V battery? That is, when do its terminals provide a potential difference of less than 1.5 V? Q5.6 Can the potential difference between the terminals of a battery ever be opposite in direction to the emf? If it can, give an example. If it cannot, explain why not. Q5.7 A rule of thumb used to determine the internal resistance of a source is that it is the open-circuit voltage divided by the short-circuit current. Is this correct? Why or why not? Q5.8 Batteries are always labeled with their emf; for instance, an AA flashlight battery is labeled “1.5 volts.” Would it also be appropriate to put a label on batteries stating how much current they provide? Why or why not? '

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Q5.9 We have seen that a coulomb is an enormous amount of charge; it is virtually impossible to place a charge of 1 C on an object. Yet, a current of 10 A, 10 C > s, is quite reasonable. Explain this apparent discrepancy. Q5.10 Electrons in an electric circuit pass through a resistor. The wire on either side of the resistor has the same diameter. (a) How does the drift speed of the electrons before entering the resistor compare to the speed after leaving the resistor? Explain your reasoning. (b) How does the potential energy for an electron before entering the resistor compare to the potential energy after leaving the resistor? Explain your reasoning. Q5.11 Current causes the temperature of a real resistor to increase. Why? What effect does this heating have on the resistance? Explain. Q5.12 Which of the graphs in Fig. Q5.12 best illustrates the current I in a real resistor as a function of the potential difference V across it? Explain. (Hint: See Discussion Question Q5.11.) '

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Figure Q5.12 (a)

(b)

(c)

(d)

I

I

I

I

O

V

O

V

O

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V

O

V

Single Correct

Q5.13 Why does an electric light bulb nearly always burn out just as you turn on the light, almost never while the light is shining? Q5.14 A light bulb glows because it has resistance. The brightness of a light bulb increases with the electrical power dissipated in the bulb. (a) In the circuit shown in Fig. Q5.14a, the two bulbs A and B are identical. Compared to bulb A, does bulb B glow more brightly, just as brightly, or less brightly? Explain your reasoning. (b) Bulb B is removed from the circuit and the circuit is completed as shown in Fig. Q5.14b. Compared to the brightness of bulb A in Fig. Q5.14a, does bulb A now glow more brightly, just as brightly, or less brightly? Explain your reasoning. Figure Q5.14 (a)

E

(b)

+

Bulb A

Bulb B

E +

Bulb A

Q5.15 (See Discussion Question Q5.14.) Will a light bulb glow more brightly when it is connected to a battery as shown in Fig. Q5.15a, in which an ideal ammeter A is placed in the circuit, or when it is connected as shown in Fig. 5.15b, in which an ideal voltmeter V is placed in the circuit? Explain your reasoning. Figure Q5.15 E

(a)

(b)

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E +

A Light bulb

V Light bulb

Q5.16 The energy that can be extracted from a storage battery is always less than the energy that goes into it while it is being charged. Why? Q5.17 Eight flashlight batteries in series have an emf of about 12 V, similar to that of a car battery. Could they be used to start a car with a dead battery? Why or why not? Q5.18 Small aircraft often have 24-V electrical systems rather than the 12-V systems in automobiles, even though the electrical power requirements are roughly the same in both applications. The explanation given by aircraft designers is that a 24-V system weighs less than a 12-V system because thinner wires can be used. Explain why this is so. Q5.19 Long-distance, electric-power, transmission lines always operate at very high voltage, sometimes as much as 750 kV. What are the advantages of such high voltages? What are the disadvantages? Q5.20 A fuse is a device designed to break a circuit, usually by melting when the current exceeds a certain value. What characteristics should the material of the fuse have? Q5.21 High-voltage power supplies are sometimes designed intentionally to have rather large internal resistance as a safety precaution. Why is such a power supply with a large internal resistance safer than a supply with the same voltage but lower internal resistance? Q5.22 The text states that good thermal conductors are also good electrical conductors. If so, why don’t the cords used to connect toasters, irons, and similar heat-producing appliances get hot by conduction of heat from the heating element?

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SINGLE CORRECT Q5.1 Four wires A, B, C and D are made of different materials. I They each carry the same current I. Wire B has C D twice the length of other wires. Wire C has twice A B E E 2E E the internal electric field of other wries. Wire D has twice the diameter of other wires. Other than these differences, all wires share identical characteristics. Rank the conductivities of these materials, from greatest to least. (a) sA = sB > sC > sD (b) sA = sB = sC = sD (c) sA = sB < sC = sD (d) sA = sB > sC = sD Q5.2 If M = mass, L = length, T = time and I = electric current then the dimensional formula for resistance R will be (a) [R] = [M1L2T - 3I - 2] (b) [R] = [M1L2T - 3I2] (c) [R] = [M1L2T3I - 2] (d) [R] = [M1L2T3I2] Q5.3 The total momentum of electrons in a straight wire of length 1000 m carrying a current of 70 A is closest to. (a) 40 * 10 - 8 N - sec (b) 30 * 10 - 8 N - sec (c) 50 * 10 - 8 N - sec (d) 70 * 10 - 8 N - sec Q5.4 Whice one of the following statement is false? (a) The electrons in a wire carrying an electrical current, drifts slowly ( y > z a (c) y > x > z E D (d) x > z > y Q5.8 When you flip a switch to turn on a light, the delay before the light turns on is determined by: (a) the speed of the electric field moving in the wire. (b) the drift speed of the electrons in the wire. (c) the number of electron collisions per second in the wire. (d) none on these, since the light comes on instantly. Q5.9 Current flows through a metallic conductor whose area of cross-section increases in the direction of the current. If we move in this direction (a) The current will change (b) The carrier density will change (c) The drift velocity will increase (d) The drift velocity will decrease Q5.10 Three copper rods are subjected to different potential difference. Compare the drift speed of electrons through them. Assume that all 3 are at the same temperture. Length Diameter Potential difference (1) L V 3d (2) 2L d 2V (3) 3L 2d 2V (a) v1 = v2 > v3 (b) v1 > v2 > v3 (c) v1 < v2 > v3 (d) None of these Q5.11 Salt water contains n sodium ions (Na + ) per cubic meter and n chloride ions (Cl−) per cubic meter. A battery is connected to metal rods that dip into a narrow pipe full of salt water. The cross sectional area of the pipe is A. The magnitude of the drift velocity of the sodium ions is VNa and the magnitude of the drift velocity of the chloride ions is VCl. Assume that VNa > VCl ( + e is the change of a proton). What is the magnitude of the ammeter reading? Battery −

+

Ammeter

Pipe full of salt water

(a) enAVNa - enAVCl (b) enAVNa + enAVCl

Cross-sectional area A

(c) enAVNa (d) Zero Q5.12 Statement-1 : Potential difference across the terminals of a battery is always less than its emf. Statement-2 : A real battery always has some internal resistance. (a) Statement-1 is ture, Statement-2 is ture and Statement-2 is correct explanation for Statement-1. (b) Statement-1 is ture, Statement-2 is ture and Statement-2 is NOT the correct explanation for Statement-1. (c) Statement-1 is ture, Statement-2 is false. (d) Statement-1 is false, Statement-2 is ture. Q5.13 Statement-1 : When two conducting wires of difference resistivity having same cross section area are joined in series, the electric field in them would be equal when they carry current. Statement-2 : When wires are in series they carry equal current. (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement‑1. (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement‑1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true. Q5.14 Statement-1 : When a battery is supplying power to a circuit, work done by electrostatic forces on electrolyte ions inside the battery is positive Statement-2 : Electric field is dircted from positive to negative electrode inside a battery (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement‑1. (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement‑1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true.

ONE OR MORE CORRECT Q5.15 Which of the following statements are true for a metallic conductor? (a) The electrical conductivity depends on the density of atoms (b) The electrical conductivity decreases with rise in temperature (c) The current density depends upon the drift velocity of electrons (d) The electrical conductivity increases with increase in voltage across it Q5.16 The current in a wire is doubled. Which of the following statement is/are wrong? (a) The current density is doubled (b) The conduction electron density is doubled (c) The mean time between collisions is constant (d) The electron drift speed is doubled Q5.17 When 0.4 V is applied to the ends of mercury colum contained in a thin glass tube X, 5 A current flows. Same mercury is poured into a glass tube Y which has diameter one third of the tube X and same voltage is applied to it. Then (a) the ratio of resistance of mercury in the tube X to tube Y is 1/81. (b) the ratio of resistance of mercury in the tube X to tube Y is 1/9. (c) current in the tube Y is 5/81 A. (d) current in the tube Y is 5/9 A.

MATCH THE COLUMN Q5.18 Electrons are emitted by a hot filament and are accelerated

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Exercises

by an electric field as shown in figure. The two stops at the left ensure that the electron beam has a uniform cross-section. Match the entries of column-I with column-II as electron move from A to B :

E A

Section 5.4 Electromotive Force and Circuits

B

Column-I Column-II (a) Speed of an electron (P) Increases (b) Number of free elec- (Q) Decreases trons per unit volume (c) Current density (R) Remains same (d) Electric potential

(S) any of the above is possible

EXERCISES Section 5.1 Current

5.1 . Lightning Strikes. During lightning strikes from a cloud to the ground, currents as high as 25,000 A can occur and last for about 40 ms . How much charge is transferred from the cloud to the earth during such a strike? 5.2 . A silver wire 2 mm in diameter transfers a charge of 420 C in 70 min. Silver contains 6 * 1028 free electrons per cubic meter. (a) What is the current in the wire? (b) What is the magnitude of the drift velocity of the electrons in the wire? 5.3 . CALC The current in a wire varies with time according to the relationship I = 50 A - 10.75 A > s22t2. (a) How many coulombs of charge pass a cross section of the wire in the time interval between t = 0 and t = 8.0 s ? (b) What constant current would transport the same charge in the same time interval? '

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Section 5.2 Resistivity and Section 5.3 Resistance

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5.4 .. A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 15.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature 120.0°C2 the ammeter reads 18 A, while at 90.0°C it reads 16.5 A. You can ignore any thermal expansion of the rod. Find (a) the resistivity at 20.0°C and (b) the temperature coefficient of resistivity at 20°C for the material of the rod. 5.5 . A 14-gauge copper wire of diameter 1.6 mm carries a current of 12 mA. (a) What is the potential difference across a 2.00-m length of the wire? (b) What would the potential difference in part (a) be if the wire were silver instead of copper, but all else were the same? Take rcu = 1.76 * 10−8 Æ - m, rsilver = 1.43 * 10−8 Æ - m 5.6 .. A ductile metal wire has resistance R. What will be the resistance of this wire in terms of R if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.) 5.7 . You need to produce a set of cylindrical copper wires 3.50 m long that will have a resistance of 0.125 Æ each. What will be the mass of each of these wires? Take rcu = 1.76 * 10−8 Æ - m, density = 9000 kg/m3 5.8 . An aluminum cube has sides of length 2 m. What is the resistance between two opposite faces of the cube? rAl = 2.75 * 10−8 Æ - m 5.9 . A hollow aluminum cylinder is 2.2 m long and has an inner radius of 3 cm and an outer radius of 5 cm. Treat each surface (inner, outer, and the two end faces) as an equipotential surface. At room temperature, what will an ohmmeter read if it is connected between (a) the opposite faces and (b) the inner and outer surfaces?

5.10 . Consider the circuit shown in Fig. E5.10. The terminal voltage of the 24.0-V battery is 21.2 V. What are (a) the internal Figure E5.10 resistance r of the battery and (b) r 24.0 V the resistance R of the circuit + resistor? 4.00 A 5.11 . An idealized ammeter is connected to a battery as 4.00 A R shown in Fig. E5.11. Find (a) the reading of the ammeter, (b) the current through the 4.00- Æ Figure E5.11 resistor, (c) the terminal voltage of A the battery. 2.00 V 10.0 V 5.12 . An ideal voltmeter V is + – connected to a 2.0- Æ resistor and a battery with emf 5.0 V and internal resistance 0.5 Æ as shown in Fig. 4.00 V E5.12. (a) What is the current in the 2.0- Æ resistor? (b) What is the termiFigure E5.12 nal voltage of the battery? (c) What is 0.5 V 5.0 V the reading on the voltmeter? Explain + your answers. 5.13 . The circuit shown in Fig. E5.13 contains two batteries, each V 2.0 V with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit (magnitude and direction); (b) the terminal Figure E5.13 voltage Vab of the 16.0-V bat1.6 V 16.0 V + tery; (c) the potential differa b ence Vac of point a with respect to point c. 5.0 V 8V 1.4 V 8.0 V 5.14 . When switch S in Fig. + c 5.14 is open, the voltmeter V of the battery reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery, Figure E5.14 and the circuit resistance R. Assume V that the two meters are ideal, so they r E don’t affect the circuit.

Section 5.5 Energy and Power in Electric Circuits

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R

5.15 . Light Bulbs. The power ratA S ing of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference. What is the resistance of (a) a 100-W bulb and (b) a 60-W bulb? (c) How much current does each bulb draw in normal use? 5.16 . European Light Bulb. In Europe the standard voltage in homes is 220 V instead of the 120 V used in the United States. Therefore a “100-W” European bulb would be intended for use with a 220-V potential difference. (a) If you bring a “100-W” European bulb home to the United States, what should be its U.S. power rating? (b) How much current will the 100-W European bulb draw in normal use in the United States? 5.17 . Consider a resistor with length L, uniform cross-sectional area A, and uniform resistivity r that is carrying a current with uniform current density J. Find the electrical power dissipated per unit volume, p. Express your result in terms of (a) E and J; (b) J and r; (c) E and r.

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166

CHAPTER 5 Current, Resistance, and Electromotive Force

5.18 . BIO Electric Eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to 500 V and produce currents of 80 mA (or even larger). A typical pulse lasts for 10 ms. What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current? 5.19 .. The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours 1A # h2. A 50-A # h battery can supply a current of 50 A for 1.0 h, or 25 A for 2.0 h, and so on. (a) What total energy can be supplied by a 12-V, 60-A # h battery if its internal resistance is negligible? (b) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? the density of gasoline is 900 kg > m3.) (c) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully? Latent heat of combustion of gasoline = 46 * 106 J/kg. 5.20 .. A 24.5- Æ bulb is connected across the terminals of a 12.0-V battery having 3.50 Æ of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb? 5.21 . In the circuit in Fig. E5.21, find Figure E5.21 (a) the rate of conversion of internal 1.0 V 12.0 V d (chemical) energy to electrical energy a + within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of electrical energy in the external resistor. c b '

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PROBLEMS 5.22 . An electrical

conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 11 m long. The resistance between its ends is 0.14 Æ. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.4 V > m, what is the total current? (c) If the material has 8.5 * 1028 free electrons per cubic meter, find the average drift speed under the conditions of part (b). 5.23 .. A resistor with resistance R is connected to a battery that has emf 12.0 V and internal resistance r = 0.40 Æ . For what two values of R will the power dissipated in the resistor be 80.0 W? 5.24 .. CP BIO Struck by Lightning. Lightning strikes can involve currents as high as 25,000 A that last for about 40 ms . If a person is struck by a bolt of lightning with these properties, the current will pass through his body. We shall assume that his mass is 75 kg, that he is wet (after all, he is in a rainstorm) and therefore has a resistance of 1 .0 kÆ, and that his body is all water (which is reasonable for a rough, but plausible, approximation). (a) By how many degrees Celsius would this lightning bolt increase the temperature of 75 kg of water? (b) Given that the internal body temperature is about 37°C, would the person’s temperature actually increase that much? Why not? What would happen first? 5.25 . CALC A material of resistivity r is Figure P5.25 formed into a solid, truncated cone of height h and radii r1 and r2 at either end (Fig. P5.25). r1 Calculate the resistance of the cone between the two flat end faces. (Hint: Imagine slicing the cone into very many thin disks, and calculate the resistance of one such disk.) h 5.26 . CALC The region between two concentric conducting spheres with radii a and b is r2 filled with a conducting material with resistiv'

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ity r. (a) Show that the resistance between the spheres is given by r 1 1 R = a - b 4p a b (b) Derive an expression for the current density as a function of radius, in terms of the potential difference Vab between the spheres. 5.27 . (a) What is the potential difference Vad in the circuit Figure P5.27 of Fig. P5.27? (b) What is the 0.50 V 4.00 V 9.00 V c b + terminal voltage of the 4.00-V battery? (c) A battery with emf 10 V and internal resistance d 6.00 V 0.50 V 8.00 V 1 Æ is inserted in the circuit at a + d, with its negative terminal 8.00 V connected to the negative terminal of the 8.00-V battery. What is the difference of potential Vbc between the terminals of the 4.00-V battery now? 5.28 . The potential difference across the terminals of a battery is 8.40 V when there is a current of 1.50 A in the battery from the negative to the positive terminal. When the current is 3.50 A in the reverse direction, the potential difference becomes 9.4 V. (a) What is the internal resistance of the battery? (b) What is the emf of the battery? 5.29 .. A 12.6-V car battery with negligible internal resistance is connected to a series combination of a 3.2- Æ resistor that obeys Ohm’s law and a thermistor that does not obey Ohm’s law but instead has a current–voltage relationship V = aI + bI2, with a = 3.8 Æ and b = 1.3 Æ > A. What is the current through the 3.2- Æ resistor? 5.30 .. A Nonideal Ammeter. Unlike the idealized ammeter any real ammeter has a nonzero resistance. (a) An ammeter with resistance RA is connected in series with a resistor R and a battery of emf E and internal resistance r. The current measured by the ammeter is IA. Find the current through the circuit if the ammeter is removed so that the battery and the resistor form a complete circuit. Express your answer in terms of IA, r, RA, and R. The more “ideal” the ammeter, the smaller the difference between this current and the current IA. (b) If R = 3.80 Æ, E = 7.50 V, and r = 0.45 Æ, find the maximum value of the ammeter resistance RA so that IA is within 1.0% of the current in the circuit when the ammeter is absent. (c) Explain why your answer in part (b) represents a maximum value. 5.31 . CALC A cylinder of radius r and length L is made of a complicated mixture of materials. Its resistivity depends on the distance x from the left end and obeys the formula r1x2 = a + bx2, where a and b are constants. (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries a current I ? (c) If we cut the rod into two halves, what is the resistance of each half? 5.32 . In the circuit of Fig. Figure P5.32 P5.32, find (a) the current E1 5 12.0 V r1 5 1.0 V through the 8.0- Æ resistor and + (b) the total rate of dissipation of electrical energy in the 8.0- Æ R 5 8.0 V resistor and in the internal resistance of the batteries. (c) In + one of the batteries, chemical E2 5 8.0 V r2 5 1.0 V energy is being converted into electrical energy. In which one is this happening, and at what rate? (d) In one of the batteries, electrical energy is being converted into chemical energy. In which one is this happening, and at what rate? (e) Show that the overall rate of '

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167

Answers

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production of electrical energy equals the overall rate of consumption of electrical energy in the circuit. 5.33 .. CP Consider the cirFigure P5.33 cuit shown in Fig. P5.33. The R2 emf source has negligible internal resistance. The resistors have resistances R1 = 6.00 Æ R1 C E and R2 = 4.00 Æ . The capacitor has capacitance C = 9.00 mF. When the capacitor is fully charged, the magnitude of the charge on its plates is Q = 36.0 mC. Calculate the emf E. 5.34 .. CP Consider the circuit shown in Fig. P5.34. The battery has emf 60.0 V and negligible internal resistance. R2 = 2.00 Æ , C1 = 3.00 mF, and C2 = 6.00 mF. After the capacitors have attained their final charges, the charge on C1 is Q1 = 18.0 mC . (a) What is the final charge on C2? (b) What is the resistance R1? Figure P5.34

+

R1 E

R2

C1

C2

Q5.35 Two conducting plates each of area A are separated by a distance d and they are parallel to each other. A conducting medium of varying conductivity fills the space between them. The conductivity varies linearly from s and 2s as you move from one plate to the other plate. Find the resistance of the meduim between the conducting plates. Q5.36 A conductor of resistivity r is placed in an electric field E which produces current in it. The heat energy dissipated per second in a unit volume is--------------

The experiment consisted of abruptly stopping a rapidly rotating spool of wire and measuring the potential difference that this produced between the ends of the wire. In a simplified model of this experiment, consider a metal rod of length L that is given a uniform S acceleration a to the right. Initially the free charges in the metal lag S behind the rod’s motion, thus setting up an electric field E in the rod. In the steady state this field exerts a force on the free charges that makes them accelerate along with the rod. (a) Apply S S gF 5 ma to the free charges to obtain an expression for ƒ q ƒ > m in S terms of the magnitudes of the induced electric field E and the S acceleration a . (b) If all the free charges in the metal rod have the S same acceleration, the electric field E is the same at all points in the rod. Use this fact to rewrite the expression for ƒ q ƒ > m in terms of the potential Vbc between the ends of the rod (Fig. P5.34). (c) If the free charges have negative charge, which end of the rod, b or c, is at higher potential? (d) If the rod is 0.50 m long and the free charges are electrons (charge mass q = - 1.60 * 10-19 C, -31 9.11 * 10 kg2, what magnitude of acceleration is required to produce a potential difference of 1.0 mV between the ends of the rod? (e) Discuss why the actual experiment used a rotating spool of thin wire rather than a moving bar as in our simplified analysis. 5.38 ... CALC A source with emf E and internal resistance r is connected to an external circuit. (a) Show that the power output of the source is maximum when the current in the circuit is one-half the short-circuit current of the source. (b) If the external circuit consists of a resistance R, show that the power output is maximum when R = r and that the maximum power is E 2 > 4r. 5.39 ... CALC The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length L and cross-sectional area A lies along the xaxis between x = 0 and x = L. The material obeys Ohm’s law, and its resistivity varies along the rod according to r1x2 = r0 exp1 -x > L2. The end of the rod at x = 0 is at a potential V0 greater than the end at x = L. (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude E1x2 in the rod as a function of x. (c) Find the electric potential V1x2 in the rod as a function of x. (d) Graph the functions r1x2, E1x2, and V1x2 for values of x between x = 0 and x = L. '

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CHALLENGE PROBLEMS 5.37 ... The Tolman-Stewart experiment in 1916 demonstrated that the free charges in a metal have negative charge and provided a quantitative measurement of their charge-to-mass ratio, ƒ q ƒ > m. '

Figure P5.37 a b

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Answers

Chapter Opening Question

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The current out equals the current in. In other words, charge must enter the bulb at the same rate as it exits the bulb. It is not “used up” or consumed as it flows through the bulb.

Test Your Understanding Questions 5.1 (v) Doubling the diameter increases the cross-sectional area A by a factor of 4. Hence the current-density magnitude J = I>A is

reduced to 14 of the value in Example 5.1, and the magnitude of the drift velocity vd = J>n ƒ q ƒ is reduced by the same factor. The new magnitude is yd = 10.15 mm>s2>4 = 0.038 mm>s. This behavior is the same as that of an incompressible fluid, which slows down when it moves from a narrow pipe to a broader one (see Section 14.4). 5.2 (ii) Figure 5.6b shows that the resistivity r of a semiconductor increases as the temperature decreases. From Eq. (5.5), the magnitude of the current density is J = E>r, so the current density decreases as the temperature drops and the resistivity increases.

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168

CHAPTER 5 Current, Resistance, and Electromotive Force

5.3 (iii) Solving Eq. (5.11) for the current shows that I = V>R. If the resistance R of the wire remained the same, doubling the voltage V would make the current I double as well. However, we saw in Example 5.3 that the resistance is not constant: As the current increases and the temperature increases, R increases as well. Thus doubling the voltage produces a current that is less than double the original current. An ohmic conductor is one for which R = V>I has the same value no matter what the voltage, so the wire is nonohmic. (In many practical problems the temperature change of the wire is so small that it can be ignored, so we can safely regard the wire as being ohmic. We do so in almost all examples in this book.) 5.4 (iii), (ii), (i) For circuit (i), we find the current from Eq. (5.16): I = E>1R + r2 = 11.5 V2>11.4 Æ + 0.10 Æ2 = 1.0 A. For circuit (ii), we note that the terminal voltage vab = 3.6 V equals the voltage across the resistor: so IR 1.8-Æ Vab = IR, I = Vab>R = 13.6 V2>11.8 Æ2 = 2.0 A. For circuit (iii), we use Eq. (5.15) for the terminal voltage: Vab = E - Ir, so I = 1E - Vab2>r = 112.0 V - 11.0 V2>10.20 Æ2 = 5.0 A. 5.5 (iii), (ii), (i) These are the same circuits that we analyzed in Test Your Understanding of Section 5.4. In each case the net power output of the battery is P = Vab I, where Vab is the battery terminal voltage. For circuit (i), we found that I = 1.0 A, so Vab = E - Ir = 1.5 V - 11.0 A210.10 Æ2 = 1.4 V, so P = 11.4 V211.0 A2 = 1.4 W. For circuit (ii), we have Vab = 3.6 V and found that I = 2.0 A, so P = 13.6 V212.0 A2 = 7.2 W. For circuit (iii), we have Vab = 11.0 V and found that I = 5.0 A, so P = 111.0 V215.0 A2 = 55 A. 5.6 (i) The difficulty of producing a certain amount of current increases as the resistivity r increases. From Eq. (5.24), r = m>ne2t, so increasing the mass m will increase the resistivity. That’s because a more massive charged particle will respond more sluggishly to an applied electric field and hence drift more slowly. To produce the same current, a greater electric field would be needed. (Increasing n, e, or t would decrease the resistivity and make it easier to produce a given current.)

Q5.9 During current flow charge does not accumulate. Q5.10 Drift speed remains same. VA > VB

Bridging Problem

then I = P/V

Answers: (a) 237°C

(b) 162 W initially, 148 W at 1.23 A

Discussion Questions Q5.1 This electric field is developed due to potential difference between the terminals of wire. This potential difference is developed due to external source (like cell). if a conductor is charged then no electric field due to its charge will reach inside the conductor. l 3l 1 Q5.2 R = r ; Rnew = r. ; ‹ Rnew = R 3 pR2 p(3r)2 Q5.3 Resistivity will not change as it is independent of length and radius. Q5.4 We know current i = neAvd. Current does not change, so greater cross sectional area lesser drift velocity. Q5.5 When current is drawn from the cell V = E - ir, soV < E. Q5.6 when battery is discharged V = E - ir, but when it is being charged. V = E + ir Q5.7 Infact no wire can be ideally of zero resistance, so measuring internal resistance will be slightly faulty. Q5.8 Current drawn depends on load resistance. The used instrument work properly if the voltage applied across its terminal is equal to marked voltage on it.

1 VA (−e) < VB (−e) 1 UA < UB Q5.11 Due to current in a conductor electrons move. The change in potential energy of e− takes place. That difference converts into heat with the rise in temperature the resistance increases R = R0[1 + a¢T] Q5.12 In view to answer to question 5.11 to correct graph would be b. Q5.13 It draws slightly greater power just as switch is put on due to smaller resistence of cold filament. Q5.14 Their brightness will be same but the brightness will be less than their full ablility, because the emf will be divided in potential difference across both. e e VA = & VB = 2 2 But in second figure VA = e. Q5.15 Ideal ammeter has no resistance but for ideal voltmeter resistance tends to infinity. So bulb in figure a will glow brighter but in figure b it will give almost no light. Q5.16 In case of extraction V = E - ir So, energy = VQ = V (E - ir) Q but while charging energy = (E - ir) Q Q5.17 Yes it can be used VA - VB = 8 * 1.5 A1 2 3 4 5 6 7 B = 12 Volt Q5.18 For the same power, higher voltage means smaller currents. Q5.19 This is done because of the concern that in the transmission of electricity too much of power loss can happen, dissipated as heat due to resistenace. Assuming power = P, resistance = R Voltage of transmission = V

‹ PLOSS = I2R =

P2

* R V2 Since P and R are fixed so less power will be lost if high voltage is used. Q5.20 The fuse wire should have low melting point and rate of rise in temperture with time should be approprite. Q5.21 Large internal resistance will prevent flow of excessive current in case of short cricult. Q5.22 Most of the heat produced in the applicance is transfered to the surrounding by convection and radiation. Very small fraction goes into the connecting cord due to its small cross-section.

Single correct Q5.1 (a) Q5.2 (a) Q5.3 (a) Q5.4 (c) Q5.5 (d) Q5.6 (b) Q5.7 (d) Q5.8 (a)

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Answers

Q5.9 (d) Q5.10 (a) Q5.11 (b) Q5.12 (d) Q5.13 (d) Q5.14 (d)

One or More Correct Q5.15 (b), (c) Q5.16 (b) Q5.17 (a), (c)

Match the Columns Q5.18 (a) P (b) Q (c) R (d) P

EXERCISES Section 5.1 Current 5.1 1C 5.2 (a) 0.1A, (b) 3.3 * 10−6m>s 5.3 (a) 272 C, (b) 34 A

Section 5.2 Resistivity and Section 5.3 Resistance 1 (°C)−1. 770 5.5 (a) 2.1 * 10−4 V , (b) 1.7 * 10−4 V 5.6 9 R 5.7 15.5 grams 5.8 1.375 * 10 - 8 Æ 5.9 (a) 1.2 * 10−5 Æ , (b) 10−9 Æ 5.4 (a) 1.1 * 10−5 Æ m(b)

Section 5.4 Electromotive Force and Circuits 5.10 (a) 0.7 Æ

(b) 5.3 Æ

5.11 (a) 5 A(b) 0(c) 0 5.12 (a) 0(b) 5 V(c) 5 V 5.13 0.5 A anticlockwise, (b) 15.2 V, (c) 11.2 V 5.14 −(a) e = 3.08 V, r = 0.067 Æ , R = 1.8 Æ .

169

5.23 0.2 Æ and 0.8 Æ 5.24 (a) 80°C (approx), (b) No. A part of his body will vaporize. rh 5.25 pr1r2 (Vab) ab 5.26 (b) (b−a) r2 5.27 (a)6.58 V, (b) 4.08 V, (c) 3.88 V 5.28 (a) 0.2 Æ , (b) 8.7 V 5.29 1.42 A RA 5.30 (a) IA a1 + b , (b) 0.0429 Æ R + r e RA (c) (R + r) (R + r + RA) [a + b (L/2)2] I 12aL + bL3 3aL + bL3 5.31 (a) , (b) , (c) , 2 2 3pr pr 24pr2 12aL + 7bL3 24pr2 5.32 (a) 0.4 A, (b) 1.6 W, (c) 4.8 W (in 12 V battery), (d) 3.2 W (in 8 V battery) 5.33 20/3 Volts. 5.34 (a) 36 m C, (b) 18 Æ d 5.35 c d In (2) sA 2 5.36 E /r

CHALLENGE PROBLEMS a aL 5.37 (a) , (b) , (c) point ‘c’, (d)3.5 * 108 m/s2 E Vbc f0 L V0 e−x/L V0 (e−x/L −e−1) 5.39 (a) R (x) = , (c) , (1−e−x/L), (b) −1 A L (1−e ) (1−e−1) (d) r E r0

E0

O

L

x L

V V0

L

x

Section 5.5 Energy and Power in Electric Circuits 5.15 (a) 144 Æ , (b) 240 Æ , (c) 5/6 A (for 100 W), 0.5 A (for 60 W) 5.16 (a) 29.8 W, (b) 0.248 A 5.17 (a) EJ(b) rJ2 (c) E2/r 5.18 (a) 40 W, 0.4 J 5.19 (a) 2.6 * 106 J, (b) 0.063 litres, (c) 1.6 hours 5.20 (12.5 % 5.21 (a) 24 W, (b) 4 W, (c) 20 W

PROBLEMS 5.22 (a) 6.25 * 10−8 Æ m, (b) 110 A, (c) 1.64 6mm/s

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6

DIRECT-CURRENT CIRCUITS

LEARNING GOALS By studying this chapter, you will learn: • How to analyze circuits with multiple resistors in series or parallel. • Rules that you can apply to any circuit with more than one loop. • How to use an ammeter, voltmeter, ohmmeter, or potentiometer in a circuit. • How to analyze circuits that include both a resistor and a capacitor.

?

• How electric power is distributed in the home.

In a complex circuit like the one on this circuit board, is it possible to connect several resistors with different resistances so that they all have the same potential difference? If so, will the current be the same through all of the resistors?

f you look inside your TV, your computer, or under the hood of a car, you will find circuits of much greater complexity than the simple circuits we studied in Chapter 5. Whether connected by wires or integrated in a semiconductor chip, these circuits often include several sources, resistors, and other circuit elements interconnected in a network. In this chapter we study general methods for analyzing such networks, including how to find voltages and currents of circuit elements. We’ll learn how to determine the equivalent resistance for several resistors connected in series or in parallel. For more general networks we need two rules called Kirchhoff’s rules. One is based on the principle of conservation of charge applied to a junction; the other is derived from energy conservation for a charge moving around a closed loop. We’ll discuss instruments for measuring various electrical quantities. We’ll also look at a circuit containing resistance and capacitance, in which the current varies with time. Our principal concern in this chapter is with direct-current (dc) circuits, in which the direction of the current does not change with time. Flashlights and automobile wiring systems are examples of direct-current circuits. Household electrical power is supplied in the form of alternating current (ac), in which the current oscillates back and forth. The same principles for analyzing networks apply to both kinds of circuits, and we conclude this chapter with a look at household wiring systems. We’ll discuss alternating-current circuits in detail in Chapter 31.

I

ActivPhysics 12.1: DC Series Circuits (Qualitative)

170

6.1

Resistors in Series and Parallel

Resistors turn up in all kinds of circuits, ranging from hair dryers and space heaters to circuits that limit or divide current or reduce or divide a voltage. Such circuits often contain several resistors, so it’s appropriate to consider combinations of resistors. A simple example is a string of light bulbs used for holiday decorations;

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171

6.1 Resistors in Series and Parallel

each bulb acts as a resistor, and from a circuit-analysis perspective the string of bulbs is simply a combination of resistors. Suppose we have three resistors with resistances R1 , R2 , and R3 . Figure 6.1 shows four different ways in which they might be connected between points a and b. When several circuit elements such as resistors, batteries, and motors are connected in sequence as in Fig. 6.1a, with only a single current path between the points, we say that they are connected in series. We studied capacitors in series in Section 4.2; we found that, because of conservation of charge, capacitors in series all have the same charge if they are initially uncharged. In circuits we’re often more interested in the current, which is charge flow per unit time. The resistors in Fig. 6.1b are said to be connected in parallel between points a and b. Each resistor provides an alternative path between the points. For circuit elements that are connected in parallel, the potential difference is the same across each element. We studied capacitors in parallel in Section 4.2. In Fig. 6.1c, resistors R2 and R3 are in parallel, and this combination is in series with R1 . In Fig. 6.1d, R2 and R3 are in series, and this combination is in parallel with R1 . For any combination of resistors we can always find a single resistor that could replace the combination and result in the same total current and potential difference. For example, a string of holiday light bulbs could be replaced by a single, appropriately chosen light bulb that would draw the same current and have the same potential difference between its terminals as the original string of bulbs. The resistance of this single resistor is called the equivalent resistance of the combination. If any one of the networks in Fig. 6.1 were replaced by its equivalent resistance Req, we could write '

'

'

'

'

Vab = IReq or Req =

6.1 Four different ways of connecting three resistors.

Vab I

where Vab is the potential difference between terminals a and b of the network and I is the current at point a or b. To compute an equivalent resistance, we assume a potential difference Vab across the actual network, compute the corresponding current I, and take the ratio Vab>I.

(a) R1, R2, and R3 in series R1 R2 x a y

I

(b) R1, R2, and R3 in parallel R1

a I

b

I

I

R3

(d) R1 in parallel with series combination of R2 and R3 R3 R2

a I

Vyb = IR3

Vab = R1 + R2 + R3 I The ratio Vab>I is, by definition, the equivalent resistance Req. Therefore Req = R1 + R2 + R3

It is easy to generalize this to any number of resistors: (resistors in series)

I

R1

a

Vab = Vax + Vxy + Vyb = I1R1 + R2 + R32

Req = R1 + R2 + R3 + Á

b

(c) R1 in series with parallel combination of R2 and R3 R2

The potential differences across each resistor need not be the same (except for the special case in which all three resistances are equal). The potential difference Vab across the entire combination is the sum of these individual potential differences:

and so

R2

R3

We can derive general equations for the equivalent resistance of a series or parallel combination of resistors. If the resistors are in series, as in Fig. 6.1a, the current I must be the same in all of them. (As we discussed in Section 5.4, current is not “used up” as it passes through a circuit.) Applying V = IR to each resistor, we have Vxy = IR2

b

I

Resistors in Series

Vax = IR1

R3

(6.1)

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b R1

I

172

CHAPTER 6 Direct-Current Circuits

The equivalent resistance of any number of resistors in series equals the sum of their individual resistances.

The equivalent resistance is greater than any individual resistance. Let’s compare this result with Eq. (4.5) for capacitors in series. Resistors in series add directly because the voltage across each is directly proportional to its resistance and to the common current. Capacitors in series add reciprocally because the voltage across each is directly proportional to the common charge but inversely proportional to the individual capacitance.

Resistors in Parallel 6.2 A car’s headlights and taillights are connected in parallel. Hence each light is exposed to the full potential difference supplied by the car’s electrical system, giving maximum brightness. Another advantage is that if one headlight or taillight burns out, the other one keeps shining (see Example 6.2).

?

If the resistors are in parallel, as in Fig. 6.1b, the current through each resistor need not be the same. But the potential difference between the terminals of each resistor must be the same and equal to Vab (Fig. 6.2). (Remember that the potential difference between any two points does not depend on the path taken between the points.) Let’s call the currents in the three resistors I1 , I2 , and I3 . Then from I = V>R, '

I1 =

Vab R1

I2 =

Vab R2

I3 =

'

'

Vab R3

In general, the current is different through each resistor. Because charge is not accumulating or draining out of point a, the total current I must equal the sum of the three currents in the resistors: I = I1 + I2 + I3 = Vab a '

1 1 1 + + b or R1 R2 R3

I 1 1 1 = + + Vab R1 R2 R3 But by the definition of the equivalent resistance Req, I>Vab = 1>Req, so 1 1 1 1 = + + Req R1 R2 R3 Again it is easy to generalize to any number of resistors in parallel: ActivPhysics 12.2: DC Parallel Circuits

1 1 1 1 = + + + Á Req R1 R2 R3

(resistors in parallel)

(6.2)

For any number of resistors in parallel, the reciprocal of the equivalent resistance equals the sum of the reciprocals of their individual resistances.

The equivalent resistance is always less than any individual resistance. Compare this with Eq. (4.7) for capacitors in parallel. Resistors in parallel add reciprocally because the current in each is proportional to the common voltage across them and inversely proportional to the resistance of each. Capacitors in parallel add directly because the charge on each is proportional to the common voltage across them and directly proportional to the capacitance of each. For the special case of two resistors in parallel, R1 + R2 1 1 1 = + = and Req R1 R2 R1 R2 '

R1 R2 R1 + R2 '

Req =

(two resistors in parallel)

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(6.3)

6.1 Resistors in Series and Parallel

173

Because Vab = I1 R1 = I2 R2 , it follows that '

'

'

I1 R2 = I2 R1

(two resistors in parallel)

(6.4)

This shows that the currents carried by two resistors in parallel are inversely proportional to their resistances. More current goes through the path of least resistance.

Problem-Solving Strategy 6.1

Resistors in Series and Parallel

IDENTIFY the relevant concepts: As in Fig. 6.1, many resistor networks are made up of resistors in series, in parallel, or a combination thereof. Such networks can be replaced by a single equivalent resistor. The logic is similar to that of Problem-Solving Strategy 4.1 for networks of capacitors. SET UP the problem using the following steps: 1. Make a drawing of the resistor network. 2. Identify groups of resistors connected in series or parallel. 3. Identify the target variables. They could include the equivalent resistance of the network, the potential difference across each resistor, or the current through each resistor. EXECUTE the solution as follows: 1. Use Eq. (6.1) or (6.2), respectively, to find the equivalent resistance for series or parallel combinations. 2. If the network is more complex, try reducing it to series and parallel combinations. For example, in Fig. 6.1c we first replace the parallel combination of R2 and R3 with its equivalent resistance;

Example 6.1

this then forms a series combination with R1. In Fig. 6.1d, the combination of R2 and R3 in series forms a parallel combination with R1. 3. Keep in mind that the total potential difference across resistors connected in series is the sum of the individual potential differences. The potential difference across resistors connected in parallel is the same for every resistor and equals the potential difference across the combination. 4. The current through resistors connected in series is the same through every resistor and equals the current through the combination. The total current through resistors connected in parallel is the sum of the currents through the individual resistors. EVALUATE your answer: Check whether your results are consistent. The equivalent resistance of resistors connected in series should be greater than that of any individual resistor; that of resistors in parallel should be less than that of any individual resistor.

Equivalent resistance

Find the equivalent resistance of the network in Fig. 6.3a and the current in each resistor. The source of emf has negligible internal resistance.

SOLUTION IDENTIFY and SET UP: This network of three resistors is a combination of series and parallel resistances, as in Fig. 6.1c. We determine the

6.3 Steps in reducing a combination of resistors to a single equivalent resistor and finding the current in each resistor.

+

(a) E 5 18 V, r 5 0 6V a 4V

c

b 3V

(b)

(c)

(d)

(e)

(f )

Continued

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174

CHAPTER 6 Direct-Current Circuits

equivalent resistance of the parallel 6-Æ and 3-Æ resistors, and then that of their series combination with the 4-Æ resistor: This is the equivalent resistance Req of the network as a whole. We then find the current in the emf, which is the same as that in the 4-Æ resistor. The potential difference is the same across each of the parallel 6-Æ and 3-Æ resistors; we use this to determine how the current is divided between these.

We reverse these steps to find the current in each resistor of the original network. In the circuit shown in Fig. 6.3d (identical to Fig. 6.3c), the current is I = Vab > R = 118 V2 > 16 Æ2 = 3 A. So the current in the 4-Æ and 2-Æ resistors in Fig. 6.3e (identical to Fig. 6.3b) is also 3 A. The potential difference Vcb across the 2-Æ resistor is therefore Vcb = IR = 13 A212 Æ2 = 6 V. This potential difference must also be 6 V in Fig. 6.3f (identical to Fig. 6.3a). From I = Vcb > R, the currents in the 6-Æ and 3-Æ resistors in Fig. 6.3f are respectively 16 V2>16 Æ2 = 1 A and 16 V2>13 Æ2 = 2 A. '

'

EXECUTE: Figures 6.3b and 6.3c show successive steps in reducing the network to a single equivalent resistance Req. From Eq. (6.2), the 6-Æ and 3-Æ resistors in parallel in Fig. 6.3a are equivalent to the single 2-Æ resistor in Fig. 6.3b: 1 1 1 1 = + = R6 Æ + 3 Æ 6 Æ 3 Æ 2 Æ [Equation (6.3) gives the same result.] From Eq. (6.1) the series combination of this 2-Æ resistor with the 4-Æ resistor is equivalent to the single 6-Æ resistor in Fig. 6.3c.

Example 6.2

'

'

'

'

EVALUATE: Note that for the two resistors in parallel between points c and b in Fig. 6.3f, there is twice as much current through the 3-Æ resistor as through the 6-Æ resistor; more current goes through the path of least resistance, in accordance with Eq. (6.4). Note also that the total current through these two resistors is 3 A, the same as it is through the 4-Æ resistor between points a and c.

Series versus parallel combinations

Two identical light bulbs, each with resistance R = 2 Æ, are connected to a source with E = 8 V and negligible internal resistance. Find the current through each bulb, the potential difference across each bulb, and the power delivered to each bulb and to the entire network if the bulbs are connected (a) in series and (b) in parallel. (c) Suppose one of the bulbs burns out; that is, its filament breaks and current can no longer flow through it. What happens to the other bulb in the series case? In the parallel case? SOLUTION IDENTIFY and SET UP: The light bulbs are just resistors in simple series and parallel connections (Figs. 6.4a and 6.4b). Once we find the current I through each bulb, we can find the power delivered to each bulb using Eq. (5.18), P = I2R = V2 > R. '

I =

Vde 8 V = = 4 A R 2 Æ

and the power delivered to each bulb is P = I2R = 14 A2212 Æ2 = 32 W P =

2

Both the potential difference across each bulb and the current through each bulb are twice as great as in the series case. Hence the power delivered to each bulb is four times greater, and each bulb is brighter. The total power delivered to the parallel network is Ptotal = 2P = 64 W, four times greater than in the series case. The

'

Vac 8 V I = = = 2 A Req 4 Æ

6.4 Our sketches for this problem. (a) Light bulbs in series

Since the bulbs have the same resistance, the potential difference is the same across each bulb: Vab = Vbc = IR = 12 A212 Æ2 = 4 V From Eq. (5.18), the power delivered to each bulb is P = I2R = 12 A2212 Æ2 = 8 W

or

or

Vde 18 V2 = = 32 W R 2 Æ 2

'

EXECUTE: (a) From Eq. (6.1) the equivalent resistance of the two bulbs between points a and c in Fig. 6.4a is Req = 2R = 2 12 Æ2 = 4 Æ . In series, the current is the same through each bulb:

P =

'

(b) Light bulbs in parallel

Vab2 Vbc2 14 V22 = = = 8 W R R 2 Æ

The total power delivered to both bulbs is Ptot = 2P = 16 W. (b) If the bulbs are in parallel, as in Fig. 6.4b, the potential difference Vde across each bulb is the same and equal to 8 V, the terminal voltage of the source. Hence the current through each light bulb is

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175

6.2 Kirchhoff’s Rules 6.5 When connected to the same source, two light bulbs in series (shown at top) draw less power and glow less brightly than when they are in parallel (shown at bottom).

increased power compared to the series case isn’t obtained “for free”; energy is extracted from the source four times more rapidly in the parallel case than in the series case. If the source is a battery, it will be used up four times as fast. (c) In the series case the same current flows through both bulbs. If one bulb burns out, there will be no current in the circuit, and neither bulb will glow. In the parallel case the potential difference across either bulb is unchanged if a bulb burns out. The current through the functional bulb and the power delivered to it are unchanged. EVALUATE: Our calculation isn’t completely accurate, because the resistance R = V > I of real light bulbs depends on the potential difference V across the bulb. That’s because the filament resistance increases with increasing operating temperature and therefore with increasing V. But bulbs connected in series across a source do in fact glow less brightly than when connected in parallel across the same source (Fig. 6.5). '

Test Your Understanding of Section 6.1 Suppose all three of the resistors shown in Fig. 6.1 have the same resistance, so R1 = R2 = R3 = R. Rank the four arrangements shown in parts (a)–(d) of Fig. 6.1 in order of their equivalent resistance, from highest to lowest.

6.2

'

y

Kirchhoff’s Rules

Many practical resistor networks cannot be reduced to simple series-parallel combinations. Figure 6.6a shows a dc power supply with emf E1 charging a battery with a smaller emf E2 and feeding current to a light bulb with resistance R. Figure 6.6b is a “bridge” circuit, used in many different types of measurement and control systems. (Problem 6.81 describes one important application of a “bridge” circuit.) To compute the currents in these networks, we’ll use the techniques developed by the German physicist Gustav Robert Kirchhoff (1824–1887). First, here are two terms that we will use often. A junction in a circuit is a point where three or more conductors meet. A loop is any closed conducting path. In Fig. 6.6a points a and b are junctions, but points c and d are not; in Fig. 6.6b the points a, b, c, and d are junctions, but points e and f are not. The blue lines in Figs. 6.6a and 6.6b show some possible loops in these circuits. Kirchhoff’s rules are the following two statements: Kirchhoff’s junction rule: The algebraic sum of the currents into any junction is zero. That is,

6.6 Two networks that cannot be reduced to simple series-parallel combinations of resistors. (a)

Junction Loop 1 a

r2

r1

Loop 3 R

Loop 2 E1

+

E2

+

c

b

Not a junction

Junction

(junction rule, valid at any junction)

(6.5)

Kirchhoff’s loop rule: The algebraic sum of the potential differences in any loop, including those associated with emfs and those of resistive elements, must equal zero. That is,

f

r E

(2)

R1

b

(loop rule, valid for any closed loop)

(6.6)

a (3)

R2

Rm

c

+ R3

aV = 0

Not a junction

(b) (1)

aI = 0

d

e

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(4) d

R4

176

CHAPTER 6 Direct-Current Circuits

6.7 Kirchhoff’s junction rule states that as much current flows into a junction as flows out of it.

(a) Kirchhoff’s junction rule

(b) Water-pipe analogy

Junction I1 I1 1 I2

I2 The current leaving a junction equals the current entering it.

The flow rate of water leaving the pipe equals the flow rate entering it.

The junction rule is based on conservation of electric charge. No charge can accumulate at a junction, so the total charge entering the junction per unit time must equal the total charge leaving per unit time (Fig. 6.7a). Charge per unit time is current, so if we consider the currents entering a junction to be positive and those leaving to be negative, the algebraic sum of currents into a junction must be zero. It’s like a T branch in a water pipe (Fig. 6.7b); if you have a total of 1 liter per minute coming in the two pipes, you can’t have 3 liters per minute going out the third pipe. We may as well confess that we used the junction rule (without saying so) in Section 6.1 in the derivation of Eq. (6.2) for resistors in parallel. The loop rule is a statement that the electrostatic force is conservative. Suppose we go around a loop, measuring potential differences across successive circuit elements as we go. When we return to the starting point, we must find that the algebraic sum of these differences is zero; otherwise, we could not say that the potential at this point has a definite value.

Sign Conventions for the Loop Rule In applying the loop rule, we need some sign conventions. Problem-Solving Strategy 6.2 describes in detail how to use these, but here’s a quick overview. We first assume a direction for the current in each branch of the circuit and mark it on a diagram of the circuit. Then, starting at any point in the circuit, we imagine traveling around a loop, adding emfs and IR terms as we come to them. When we travel through a source in the direction from - to + , the emf is considered to be positive; when we travel from + to -, the emf is considered to be negative (Fig. 6.8a). When we travel through a resistor in the same direction as the assumed current, the IR term is negative because the current goes in the direction of decreasing potential. When we travel through a resistor in the direction opposite to the assumed current, the IR term is positive because this represents a rise of potential (Fig. 6.8b). Kirchhoff’s two rules are all we need to solve a wide variety of network problems. Usually, some of the emfs, currents, and resistances are known, and others are unknown. We must always obtain from Kirchhoff’s rules a number of independent equations equal to the number of unknowns so that we can solve the equations simultaneously. Often the hardest part of the solution is not understanding the basic principles but keeping track of algebraic signs!

6.8 Use these sign conventions when you apply Kirchhoff’s loop rule. In each part of the figure “Travel” is the direction that we imagine going around the loop, which is not necessarily the direction of the current.

(a) Sign conventions for emfs

(b) Sign conventions for resistors

1E: Travel direction from – to +:

2E: Travel direction from + to –:

1IR: Travel opposite to current direction:

Travel – +

Travel – +

E

E

2IR: Travel in current direction:

Travel I –

Travel +

R

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I –

+ R

6.2 Kirchhoff’s Rules

Problem-Solving Strategy 6.2

177

Kirchhoff’s Rules

IDENTIFY the relevant concepts: Kirchhoff’s rules are useful for analyzing any electric circuit. SET UP the problem using the following steps: 1. Draw a circuit diagram, leaving room to label all quantities, known and unknown. Indicate an assumed direction for each unknown current and emf. (Kirchhoff’s rules will yield the magnitudes and directions of unknown currents and emfs. If the actual direction of a quantity is opposite to your assumption, the resulting quantity will have a negative sign.) 2. As you label currents, it helpful to use Kirchhoff’s junction rule, as in Fig. 6.9, so as to express the currents in terms of as few quantities as possible. 3. Identify the target variables. EXECUTE the solution as follows: 1. Choose any loop in the network and choose a direction (clockwise or counterclockwise) to travel around the loop as you apply Kirchhoff’s loop rule. The direction need not be the same as any assumed current direction.

2. Travel around the loop in the chosen direction, adding potential differences algebraically as you cross them. Use the sign conventions of Fig. 6.8. 3. Equate the sum obtained in step 2 to zero in accordance with the loop rule. 4. If you need more independent equations, choose another loop and repeat steps 1–3; continue until you have as many independent equations as unknowns or until every circuit element has been included in at least one loop. 5. Solve the equations simultaneously to determine the unknowns. 6. You can use the loop-rule bookkeeping system to find the potential Vab of any point a with respect to any other point b. Start at b and add the potential changes you encounter in going from b to a, using the same sign rules as in step 2. The algebraic sum of these changes is Vab = Va - Vb. EVALUATE your answer: Check all the steps in your algebra. Apply steps 1 and 2 to a loop you have not yet considered; if the sum of potential drops isn’t zero, you’ve made an error somewhere.

6.9 Applying the junction rule to point a reduces the number of unknown currents from three to two. (a) Three unknown currents: I1, I2, I3 r1 E1 r2 E2 + + I1 I1 R1

Example 6.3

I3

I2

R3

I2

a

R2

(b) Applying the junction rule to point a eliminates I3. r1 E1 r2 E2 + + I1

I1 1 I2

I2

R3

I2

I1 R1

a

R2

A single-loop circuit

The circuit shown in Fig. 6.10a contains two batteries, each with an emf and an internal resistance, and two resistors. Find (a) the current in the circuit, (b) the potential difference Vab, and (c) the power output of the emf of each battery.

(b) To find Vab, the potential at a with respect to b, we start at b and add potential changes as we go toward a. There are two paths from b to a; taking the lower one, we find Vab = 10.5 A217 Æ2 + 4 V + 10.5 A214 Æ2 = 9.5 V

S O L U T IO N IDENTIFY and SET UP: There are no junctions in this single-loop circuit, so we don’t need Kirchhoff’s junction rule. To apply Kirchhoff’s loop rule, we first assume a direction for the current; let’s assume a counterclockwise direction as shown in Fig. 6.10a.

Point a is at 9.5 V higher potential than b. All the terms in this sum, including the IR terms, are positive because each represents an increase in potential as we go from b to a. Taking the upper path, we find

EXECUTE: (a) Starting at a and traveling counterclockwise with the current, we add potential increases and decreases and equate the sum to zero as in Eq. (6.6):

Vab = 12 V - 10.5 A212 Æ2 - 10.5 A213 Æ2

- I14 Æ2 - 4 V - I17 Æ2 + 12 V - I12 Æ2 - I13 Æ2 = 0

Here the IR terms are negative because our path goes in the direction of the current, with potential decreases through the resistors. The results for Vab are the same for both paths, as they must be in order for the total potential change around the loop to be zero. Continued

Collecting like terms and solving for I, we find 8 V = I 116 Æ2 '

and

I = 0.5 A

The positive result for I shows that our assumed current direction is correct.

= 9.5 V

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178

CHAPTER 6 Direct-Current Circuits

(c) The power outputs of the emf of the 12-V and 4-V batteries are P12V = EI = 112 V210.5 A2 = 6 W

P4V = EI = 1- 4 V210.5 A2 = - 2 W

The negative sign in E for the 4-V battery appears because the current actually runs from the higher-potential side of the battery to the lower-potential side. The negative value of P means that we are storing energy in that battery; the 12-V battery is recharging it (if it is in fact rechargeable; otherwise, we’re destroying it). EVALUATE: By applying the expression P = I2R to each of the four resistors in Fig. 6.10a, you can show that the total power dis-

sipated in all four resistors is 4 W. Of the 6 W provided by the emf of the 12-V battery, 2 W goes into storing energy in the 4-V battery and 4 W is dissipated in the resistances. The circuit shown in Fig. 6.10a is much like that used when a fully charged 12-V storage battery (in a car with its engine running) is used to “jump-start” a car with a run-down battery (Fig. 6.10b). The run-down battery is slightly recharged in the process. The 3-Æ and 7-Æ resistors in Fig. 6.10a represent the resistances of the jumper cables and of the conducting path through the automobile with the run-down battery. (The values of the resistances in actual automobiles and jumper cables are considerably lower.)

6.10 (a) In this example we travel around the loop in the same direction as the assumed current, so all the IR terms are negative. The potential decreases as we travel from + to - through the bottom emf but increases as we travel from - to + through the top emf. (b) A real-life example of a circuit of this kind. (a)

(b) Dead battery

2 V 12 V +

b

Live battery

I Travel 3V

I

7V

I I +

a

4V 4V

Example 6.4

Charging a battery

In the circuit shown in Fig. 6.11, a 12-V power supply with unknown internal resistance r is connected to a run-down rechargeable battery with unknown emf E and internal resistance 1 Æ and to an indicator light bulb of resistance 3 Æ carrying a current of 2 A. The current through the run-down battery is 1 A in the direction shown. Find r, E, and the current I through the power supply.

EXECUTE: We apply the junction rule, Eq. (6.5), to point a: -I + 1 A + 2 A = 0

so

I = 3 A

To determine r, we apply the loop rule, Eq. (6.6), to the large, outer loop (1): 12 V - 13 A2r - 12 A213 Æ2 = 0

so

r = 2 Æ

To determine E, we apply the loop rule to the left-hand loop (2): SOLUTION IDENTIFY and SET UP: This circuit has more than one loop, so we must apply both the junction and loop rules. We assume the direction of the current through the 12-V power supply, and the polarity of the run-down battery, to be as shown in Fig. 6.11. There are three target variables, so we need three equations. 6.11 In this circuit a power supply charges a run-down battery and lights a bulb. An assumption has been made about the polarity of the emf E of the run-down battery. Is this assumption correct? (1) a (3)

+

(2)

12 V

3V

1A

1V

b

+

E 2A

I

r

- E + 11 A211 Æ2 - 12 A213 Æ2 = 0

so

E = -5 V

The negative value for E shows that the actual polarity of this emf is opposite to that shown in Fig. 6.11. As in Example 6.3, the battery is being recharged. EVALUATE: Try applying the junction rule at point b instead of point a, and try applying the loop rule by traveling counterclockwise rather than clockwise around loop (1). You’ll get the same results for I and r. We can check our result for E by using the righthand loop (3): 12 V - 13 A212 Æ2 - 11 A211 Æ2 + E = 0

which again gives us E = - 5 V. As an additional check, we note that Vba = Vb - Va equals the voltage across the 3-Æ resistance, which is 12 A213 Æ2 = 6 V. Going from a to b by the top branch, we encounter potential differences +12 V - 13 A212 Æ2 = +6 V, and going by the middle branch, we find -1 -5 V2 + 11 A211 Æ2 = + 6 V. The three ways of getting Vba give the same results.

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179

6.2 Kirchhoff’s Rules

Example 6.5

Power in a battery-charging circuit

In the circuit of Example 6.4 (shown in Fig. 6.11), find the power The power output of the emf E of the battery being charged is delivered by the 12-V power supply and by the battery being Pemf = EIbattery = 1-5 V211 A2 = - 5 W recharged, and find the power dissipated in each resistor. S O L U T IO N IDENTIFY and SET UP: We use the results of Section 5.5, in which we found that the power delivered from an emf to a circuit is E and the power delivered to a resistor from a circuit is Vab I = I2R. We know the values of all relevant quantities from Example 6.4. EXECUTE: The power output Ps from the emf of the power supply is Psupply = Esupply Isupply = 112 V213 A2 = 36 W The power dissipated in the power supply’s internal resistance r is Pr - supply = Isupply2rsupply = 13 A2212 Æ2 = 18 W

This is negative because the 1-A current runs through the battery from the higher-potential side to the lower-potential side. (As we mentioned in Example 6.4, the polarity assumed for this battery in Fig. 6.11 was wrong.) We are storing energy in the battery as we charge it. Additional power is dissipated in the battery’s internal resistance; this power is Pr-battery = Ibattery2 rbattery = 11 A2211 Æ2 = 1 W The total power input to the battery is thus 1 W + ƒ - 5 W ƒ = 6 W. Of this, 5 W represents useful energy stored in the battery; the remainder is wasted in its internal resistance. The power dissipated in the light bulb is Pbulb = Ibulb2 Rbulb = 12 A2213 Æ2 = 12 W

so the power supply’s net power output is Pnet = 36 W 18 W = 18 W. Alternatively, from Example 6.4 the terminal volt- EVALUATE: As a check, note that all of the power from the supply age of the battery is Vba = 6 V, so the net power output is is accounted for. Of the 18 W of net power from the power supply, 5 W goes to recharge the battery, 1 W is dissipated in the battery’s Pnet = Vba Isupply = 16 V213 A2 = 18 W internal resistance, and 12 W is dissipated in the light bulb.

Example 6.6

A complex network

Figure 6.12 shows a “bridge” circuit of the type described at the beginning of this section (see Fig. 6.6b). Find the current in each resistor and the equivalent resistance of the network of five resistors.

EXECUTE: We apply the loop rule to the three loops shown: 13 V - I1 11 Æ2 - 1I1 - I3211 Æ2 = 0 '

-I2 11 Æ2 - 1I2 + I3212 Æ2 + 13 V = 0 '

-I1 11 Æ2 - I3 11 Æ2 + I2 11 Æ2 = 0 '

S O L U T IO N IDENTIFY and SET UP: This network is neither a series combination nor a parallel combination. Hence we must use Kirchhoff’s rules to find the values of the target variables. There are five unknown currents, but by applying the junction rule to junctions a and b, we can represent them in terms of three unknown currents I1, I2, and I3, as shown in Fig. 6.12.

c (3)

I2

(1) 1V

1V 1V

+ 13 V

a

b 1V

13 V = I1 12 Æ2 - I3 11 Æ2 '

'

I3

'

'

13 V = I1 13 Æ2 + I3 15 Æ2

I1 + I2

(1 ¿ ) (2 ¿ )

I1 = 6 A

We substitute this result into Eq. (1¿) to obtain I3 = - 1 A, and from Eq. (3) we find I2 = 5 A. The negative value of I3 tells us that its direction is opposite to the direction we assumed. The total current through the network is I1 + I2 = 11 A, and the potential drop across it is equal to the battery emf, 13 V. The equivalent resistance of the network is therefore Req =

2V

(3)

Now we can eliminate I3 by multiplying Eq. (1¿) by 5 and adding the two equations. We obtain '

(2)

'

(2)

One way to solve these simultaneous equations is to solve Eq. (3) for I2, obtaining I2 = I1 + I3, and then substitute this expression into Eq. (2) to eliminate I2. We then have

78 V = I1 113 Æ2

6.12 A network circuit with several resistors.

I1

'

(1)

13 V = 1.2 Æ 11 A

I2 + I3

I1 – I3 d

EVALUATE: You can check our results for I1, I2, and I3 by substituting them back into Eqs. (1)–(3). What do you find?

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180

CHAPTER 6 Direct-Current Circuits

Example 6.7

A potential difference in a complex network

In the circuit of Example 6.6 (Fig. 6.12), find the potential difference Vab. SOLUTION IDENTIFY and SET UP: Our target variable Vab = Va - Vb is the potential at point a with respect to point b. To find it, we start at point b and follow a path to point a, adding potential rises and drops as we go. We can follow any of several paths from b to a; the result must be the same for all such paths, which gives us a way to check our result. EXECUTE: The simplest path is through the center 1-Æ resistor. In Example 6.6 we found I3 = - 1 A, showing that the actual cur-

6.13 This ammeter (top) and voltmeter (bottom) are both d’Arsonval galvanometers. The difference has to do with their internal connections (see Fig. 6.15).

Magnetic-field torque tends to push pointer away from zero.

Spring torque tends to push pointer toward zero.

5

EVALUATE: To check our result, let’s try a path from b to a that goes through the lower two resistors. The currents through these are I2 + I3 = 5 A + 1- 1 A2 = 4 A and

and so

I1 - I3 = 6 A - 1-1 A2 = 7 A

Vab = -14 A212 Æ2 + 17 A211 Æ2 = - 1 V

You can confirm this result using some other paths from b to a.

Test Your Understanding of Section 6.2 Subtract Eq. (1) from Eq. (2) in Example 6.6. To which loop in Fig. 6.12 does this equation correspond? Would this equation have simplified the solution of Example 6.6? y

6.3

6.14 A d’Arsonval galvanometer, showing a pivoted coil with attached pointer, a permanent magnet supplying a magnetic field that is uniform in magnitude, and a spring to provide restoring torque, which opposes magnetic-field torque.

rent direction through this resistor is from right to left. Thus, as we go from b to a, there is a drop of potential with magnitude ƒ I3 ƒ R = 11 A211 Æ2 = 1 V. Hence Vab = -1 V, and the potential at a is 1 V less than at point b.

Electrical Measuring Instruments

10

We’ve been talking about potential difference, current, and resistance for two chapters, so it’s about time we said something about how to measure these quantities. Many common devices, including car instrument panels, battery chargers, and inexpensive electrical instruments, measure potential difference (voltage), current, or resistance using a d’Arsonval galvanometer (Fig. 6.13). In the following discussion we’ll often call it just a meter. A pivoted coil of fine wire is placed in the magnetic field of a permanent magnet (Fig. 6.14). Attached to the coil is a spring, similar to the hairspring on the balance wheel of a watch. In the equilibrium position, with no current in the coil, the pointer is at zero. When there is a current in the coil, the magnetic field exerts a torque on the coil that is proportional to the current. (We’ll discuss this magnetic interaction in detail in Chapter 7.) As the coil turns, the spring exerts a restoring torque that is proportional to the angular displacement. Thus the angular deflection of the coil and pointer is directly proportional to the coil current, and the device can be calibrated to measure current. The maximum deflection, typically 90° or so, is called full-scale deflection. The essential electrical characteristics of the meter are the current Ifs required for full-scale deflection (typically on the order of 10 mA to 10 mA) and the resistance Rc of the coil (typically on the order of 10 to 1000 Æ ). The meter deflection is proportional to the current in the coil. If the coil obeys Ohm’s law, the current is proportional to the potential difference between the terminals of the coil, and the deflection is also proportional to this potential difference. For example, consider a meter whose coil has a resistance Rc = 20.0 Æ and that deflects full scale when the current in its coil is Ifs = 1.00 mA. The corresponding potential difference for full-scale deflection is

0

V = Ifs Rc = 11.00 * 10-3 A2120.0 Æ2 = 0.0200 V '

Spring

Ammeters Magnetic field Permanent magnet

Soft-iron core

Pivoted coil

A current-measuring instrument is usually called an ammeter (or milliammeter, microammeter, and so forth, depending on the range). An ammeter always measures the current passing through it. An ideal ammeter, discussed in Section 5.4, would have zero resistance, so including it in a branch of a circuit would not

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181

6.3 Electrical Measuring Instruments

'

Example 6.8

||

||

|

||

Rc

I

– + a Rsh b

|||

||| |||| ||||||

||

|

|

|||||||||||||

||

|||

|

||

(b) Moving-coil voltmeter

Rc Rs + a

I Va

– b

Circuit element

I

(for an ammeter)

Vb I

(6.7)

Designing an ammeter

What shunt resistance is required to make the 1.00-mA, 20.0-Æ meter described above into an ammeter with a range of 0 to 50.0 mA? S O L U T IO N IDENTIFY and SET UP: Since the meter is being used as an ammeter, its internal connections are as shown in Fig. 6.15a. Our target variable is the shunt resistance Rsh, which we will find using Eq. (6.7). The ammeter must handle a maximum current Ia = 50.0 * 10-3 A. The coil resistance is Rc = 20.0 Æ, and the meter shows full-scale deflection when the current through the coil is Ifs = 1.00 * 10-3 A. EXECUTE: Solving Eq. (6.7) for Rsh, we find Rsh =

(a) Moving-coil ammeter

||

Ifs Rc = 1Ia - Ifs2Rsh

6.15 Using the same meter to measure (a) current and (b) voltage.

||

affect the current in that branch. Real ammeters always have some finite resistance, but it is always desirable for an ammeter to have as little resistance as possible. We can adapt any meter to measure currents that are larger than its full-scale reading by connecting a resistor in parallel with it (Fig. 6.15a) so that some of the current bypasses the meter coil. The parallel resistor is called a shunt resistor or simply a shunt, denoted as Rsh. Suppose we want to make a meter with full-scale current Ifs and coil resistance Rc into an ammeter with full-scale reading Ia. To determine the shunt resistance Rsh needed, note that at full-scale deflection the total current through the parallel combination is Ia, the current through the coil of the meter is Ifs, and the current through the shunt is the difference Ia - Ifs. The potential difference Vab is the same for both paths, so

IfsRc 11.00 * 10-3 A2120.0 Æ2 = Ia - Ifs 50.0 * 10-3 A - 1.00 * 10-3 A

EVALUATE: It’s useful to consider the equivalent resistance Req of the ammeter as a whole. From Eq. (6.2), Req = a

-1 1 1 -1 1 1 + b = a + b Rc Rsh 20.0 Æ 0.408 Æ

= 0.400 Æ The shunt resistance is so small in comparison to the coil resistance that the equivalent resistance is very nearly equal to the shunt resistance. The result is an ammeter with a low equivalent resistance and the desired 0–50.0-mA range. At full-scale deflection, I = Ia = 50.0 mA, the current through the galvanometer is 1.00 mA, the current through the shunt resistor is 49.0 mA, and Vab = 0.0200 V. If the current I is less than 50.0 mA, the coil current and the deflection are proportionally less.

= 0.408 Æ

Voltmeters This same basic meter may also be used to measure potential difference or voltage. A voltage-measuring device is called a voltmeter. A voltmeter always measures the potential difference between two points, and its terminals must be connected to these points. (Example 5.6 in Section 5.4 described what can happen if a voltmeter is connected incorrectly.) As we discussed in Section 5.4, an ideal voltmeter would have infinite resistance, so connecting it between two points in a circuit would not alter any of the currents. Real voltmeters always have finite resistance, but a voltmeter should have large enough resistance that connecting it in a circuit does not change the other currents appreciably. For the meter described in Example 6.8 the voltage across the meter coil at full-scale deflection is only Ifs Rc = 11.00 * 10-3 A2120.0 Æ2 = 0.0200 V. We can extend this range by connecting a resistor Rs in series with the coil (Fig. 6.15b). Then only a fraction of the total potential difference appears across the coil itself, and the remainder appears across Rs. For a voltmeter with full-scale reading VV, we need a series resistor Rs in Fig. 6.15b such that

Application Electromyography A fine needle containing two electrodes is being inserted into a muscle in this patient’s hand. By using a sensitive voltmeter to measure the potential difference between these electrodes, a physician can probe the muscle’s electrical activity. This is an important technique for diagnosing neurological and neuromuscular diseases.

'

VV = Ifs1Rc + Rs2

(for a voltmeter)

(6.8)

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182

CHAPTER 6 Direct-Current Circuits

Example 6.9

Designing a voltmeter

What series resistance is required to make the 1.00-mA, 20.0-Æ meter described above into a voltmeter with a range of 0 to 10.0 V?

EVALUATE: At full-scale deflection, Vab = 10.0 V, the voltage across the meter is 0.0200 V, the voltage across Rs is 9.98 V, and the current through the voltmeter is 0.00100 A. Most of the voltage appears across the series resistor. The meter’s equivalent resistance is a desirably high Req 5 20.0 Æ + 9980 Æ 5 10,000 Æ. Such a meter is called a “1000 ohms-per-volt” meter, referring to the ratio of resistance to full-scale deflection. In normal operation the current through the circuit element being measured (I in Fig. 6.15b) is much greater than 0.00100 A, and the resistance between points a and b in the circuit is much less than 10,000 Æ. The voltmeter draws off only a small fraction of the current and thus disturbs the circuit being measured only slightly.

SOLUTION IDENTIFY and SET UP: Since this meter is being used as a voltmeter, its internal connections are as shown in Fig. 6.15b. Our target variable is the series resistance Rs. The maximum allowable voltage across the voltmeter is VV = 10.0 V. We want this to occur when the current through the coil is Ifs = 1.00 * 10-3 A. Our target variable is the series resistance Rs, which we find using Eq. (6.8). EXECUTE: From Eq. (6.8), Rs =

VV 10.0 V - Rc = - 20.0 Æ = 9980 Æ Ifs 0.00100 A

Ammeters and Voltmeters in Combination ActivPhysics 12.4: Using Ammeters and Voltmeters

A voltmeter and an ammeter can be used together to measure resistance and power. The resistance R of a resistor equals the potential difference Vab between its terminals divided by the current I; that is, R = Vab>I. The power input P to any circuit element is the product of the potential difference across it and the current through it: P = Vab I. In principle, the most straightforward way to measure R or P is to measure Vab and I simultaneously. With practical ammeters and voltmeters this isn’t quite as simple as it seems. In Fig. 6.16a, ammeter A reads the current I in the resistor R. Voltmeter V, however, reads the sum of the potential difference Vab across the resistor and the potential difference Vbc across the ammeter. If we transfer the voltmeter terminal from c to b, as in Fig. 6.16b, then the voltmeter reads the potential difference Vab correctly, but the ammeter now reads the sum of the current I in the resistor and the current IV in the voltmeter. Either way, we have to correct the reading of one instrument or the other unless the corrections are small enough to be negligible. 6.16 Ammeter–voltmeter method for measuring resistance. (a) a

(b) RA

R

b

A

R

a

c

b

A

c

I

I

IV

Example 6.10

V

V

RV

RV

Measuring resistance I

The voltmeter in the circuit of Fig. 6.16a reads 12.0 V and the ammeter reads 0.100 A. The meter resistances are RV = 10,000 Æ (for the voltmeter) and RA = 2.00 Æ (for the ammeter). What are the resistance R and the power dissipated in the resistor?

between a and c. If the ammeter were ideal (that is, if RA = 0), there would be zero potential difference between b and c, the voltmeter reading V = 12.0 V would be equal to the potential difference Vab across the resistor, and the resistance would simply be equal to R = V > I = 112.0 V2 > 10.100 A2 = 120 Æ. The ammeter is not ideal, however (its resistance is RA = 2.00 Æ ), so the voltmeter reading V is actually the sum of the potential differences Vbc (across the ammeter) and Vab (across the resistor). We use Ohm’s law to find the voltage Vbc from the known current and '

SOLUTION IDENTIFY and SET UP: The ammeter reads the current I = 0.100 A through the resistor, and the voltmeter reads the potential difference

'

'

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6.3 Electrical Measuring Instruments ammeter resistance. Then we solve for Vab and the resistance R. Given these, we are able to calculate the power P into the resistor. EXECUTE: From Ohm’s law, Vbc = IRA = 10.100 A212.00 Æ2 = 0.200 V and Vab = IR. The sum of these is V = 12.0 V, so the potential difference across the resistor is Vab = V - Vbc = 112.0 V2 - 10.200 V2 = 11.8 V. Hence the resistance is

Example 6.11

R =

183

Vab 11.8 V = = 118 Æ I 0.100 A

The power dissipated in this resistor is P = Vab I = 111.8 V210.100 A2 = 1.18 W EVALUATE: You can confirm this result for the power by using the alternative formula P = I2R. Do you get the same answer?

Measuring resistance II

Suppose the meters of Example 6.10 are connected to a different resistor as shown in Fig. 6.16b, and the readings obtained on the meters are the same as in Example 6.10. What is the value of this new resistance R, and what is the power dissipated in the resistor?

EXECUTE: We have IV = V > RV = 112.0 V2 > 110,000 Æ2 = 1.20 mA. The actual current I in the resistor is I = IA - IV = 0.100 A - 0.0012 A = 0.0988 A, and the resistance is '

R =

'

'

'

Vab 12.0 V = = 121 Æ I 0.0988 A

S O L U T IO N IDENTIFY and SET UP: In Example 6.10 the ammeter read the actual current through the resistor, but the voltmeter reading was not the same as the potential difference across the resistor. Now the situation is reversed: The voltmeter reading V = 12.0 V shows the actual potential difference Vab across the resistor, but the ammeter reading IA = 0.100 A is not equal to the current I through the resistor. Applying the junction rule at b in Fig. 6.16b shows that IA = I + IV, where IV is the current through the voltmeter. We find IV from the given values of V and the voltmeter resistance RV, and we use this value to find the resistor current I. We then determine the resistance R from I and the voltmeter reading, and calculate the power as in Example 6.10.

The power dissipated in the resistor is P = Vab I = 112.0 V210.0988 A2 = 1.19 W EVALUATE: Had the meters been ideal, our results would have been R = 12.0 V>0.100 A = 120 Æ and P = VI = 112.0 V2 * 10.100 A2 = 1.2 W both here and in Example 6.10. The actual (correct) results are not too different in either case. That’s because the ammeter and voltmeter are nearly ideal: Compared with the resistance R under test, the ammeter resistance RA is very small and the voltmeter resistance RV is very large. Under these conditions, treating the meters as ideal yields pretty good results; accurate work requires calculations as in these two examples.

Ohmmeters 6.17 Ohmmeter circuit. The resistor Rs has a variable resistance, as is indicated by the arrow through the resistor symbol. To use the ohmmeter, first connect x directly to y and adjust Rs until the meter reads zero. Then connect x and y across the resistor R and read the scale.

The Potentiometer The potentiometer is an instrument that can be used to measure the emf of a source without drawing any current from the source; it also has a number of other useful applications. Essentially, it balances an unknown potential difference against an adjustable, measurable potential difference.

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|||

| | | | | | |

|

| ||

||

|

`

0

E +

An alternative method for measuring resistance is to use a d’Arsonval meter in an arrangement called an ohmmeter. It consists of a meter, a resistor, and a source (often a flashlight battery) connected in series (Fig. 6.17). The resistance R to be measured is connected between terminals x and y. The series resistance Rs is variable; it is adjusted so that when terminals x and y are short-circuited (that is, when R = 0), the meter deflects full scale. When nothing is connected to terminals x and y, so that the circuit between x and y is open (that is, when R S q ), there is no current and hence no deflection. For any intermediate value of R the meter deflection depends on the value of R, and the meter scale can be calibrated to read the resistance R directly. Larger currents correspond to smaller resistances, so this scale reads backward compared to the scale showing the current. In situations in which high precision is required, instruments containing d’Arsonval meters have been supplanted by electronic instruments with direct digital readouts. Digital voltmeters can be made with extremely high internal resistance, of the order of 100 MÆ. Figure 6.18 shows a digital multimeter, an instrument that can measure voltage, current, or resistance over a wide range.

Rs x

y

R

184

CHAPTER 6 Direct-Current Circuits

6.18 This digital multimeter can be used as a voltmeter (red arc), ammeter (yellow arc), or ohmmeter (green arc).

The principle of the potentiometer is shown schematically in Fig. 6.19a. A resistance wire ab of total resistance Rab is permanently connected to the terminals of a source of known emf E1 . A sliding contact c is connected through the galvanometer G to a second source whose emf E2 is to be measured. As contact c is moved along the resistance wire, the resistance Rcb between points c and b varies; if the resistance wire is uniform, Rcb is proportional to the length of wire between c and b. To determine the value of E2 , contact c is moved until a position is found at which the galvanometer shows no deflection; this corresponds to zero current passing through E2 . With I2 = 0, Kirchhoff’s loop rule gives '

'

'

E2 = IRcb With I2 = 0, the current I produced by the emf E1 has the same value no matter what the value of the emf E2 . We calibrate the device by replacing E2 by a source of known emf; then any unknown emf E2 can be found by measuring the length of wire cb for which I2 = 0. Note that for this to work, Vab must be greater than E2 . The term potentiometer is also used for any variable resistor, usually having a circular resistance element and a sliding contact controlled by a rotating shaft and knob. The circuit symbol for a potentiometer is shown in Fig. 6.19b. '

6.19 A potentiometer.

'

(a) Potentiometer circuit E1 + I

I

I

I a

b c

I2 5 0 G

+

rG

E2, r

(b) Circuit symbol for potentiometer (variable resistor)

Test Your Understanding of Section 6.3 You want to measure the current through and the potential difference across the 2-Æ resistor shown in Fig. 6.12 (Example 6.6 in Section 6.2). (a) How should you connect an ammeter and a voltmeter to do this? (i) ammeter and voltmeter both in series with the 2-Æ resistor; (ii) ammeter in series with the 2-Æ resistor and voltmeter connected between points b and d; (iii) ammeter connected between points b and d and voltmeter in series with the 2-Æ resistor; (iv) ammeter and voltmeter both connected between points b and d. (b) What resistances should these meters have? (i) Ammeter and voltmeter resistances should both be much greater than 2 Æ; (ii) ammeter resistance should be much greater than 2 Æ and voltmeter resistance should be much less than 2 Æ; (iii) ammeter resistance should be much less than 2 Æ and voltmeter resistance should be much greater than 2 Æ; (iv) ammeter and voltmeter resistances should both be much less than y 2 Æ.

6.4

R-C Circuits

In the circuits we have analyzed up to this point, we have assumed that all the emfs and resistances are constant (time independent) so that all the potentials, currents, and powers are also independent of time. But in the simple act of charging or discharging a capacitor we find a situation in which the currents, voltages, and powers do change with time. Many devices incorporate circuits in which a capacitor is alternately charged and discharged. These include flashing traffic lights, automobile turn signals, and electronic flash units. Understanding what happens in such circuits is thus of great practical importance.

Charging a Capacitor Figure 6.20 shows a simple circuit for charging a capacitor. A circuit such as this that has a resistor and a capacitor in series is called an R-C circuit. We idealize the battery (or power supply) to have a constant emf E and zero internal resistance 1r = 02, and we neglect the resistance of all the connecting conductors. We begin with the capacitor initially uncharged (Fig. 6.20a); then at some initial time t = 0 we close the switch, completing the circuit and permitting current around the loop to begin charging the capacitor (Fig. 6.20b). For all practical purposes, the current begins at the same instant in every conducting part of the circuit, and at each instant the current is the same in every part.

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185

6.4 R-C Circuits CAUTION Lowercase means time-varying Up to this point we have been working with constant potential differences (voltages), currents, and charges, and we have used capital letters V, I, and Q, respectively, to denote these quantities. To distinguish between quantities that vary with time and those that are constant, we will use lowercase letters v, i, and q for time-varying voltages, currents, and charges, respectively. We suggest that you follow this same convention in your own work. y

6.20 Charging a capacitor. (a) Just before the switch is closed, the charge q is zero. (b) When the switch closes (at t = 0), the current jumps from zero to E>R. As time passes, q approaches Qf and the current i approaches zero. (a) Capacitor initially uncharged

Because the capacitor in Fig. 6.20 is initially uncharged, the potential difference vbc across it is zero at t = 0. At this time, from Kirchhoff’s loop law, the voltage vab across the resistor R is equal to the battery emf E The initial 1t = 02 current through the resistor, which we will call I0 , is given by Ohm’s law: I0 = vab>R = E>R. As the capacitor charges, its voltage vbc increases and the potential difference vab across the resistor decreases, corresponding to a decrease in current. The sum of these two voltages is constant and equal to E. After a long time the capacitor becomes fully charged, the current decreases to zero, and the potential difference vab across the resistor becomes zero. Then the entire battery emf E appears across the capacitor and vbc = E. Let q represent the charge on the capacitor and i the current in the circuit at some time t after the switch has been closed. We choose the positive direction for the current to correspond to positive charge flowing onto the left-hand capacitor plate, as in Fig. 6.20b. The instantaneous potential differences vab and vbc are

E +

Switch open

'

vab = iR

vbc =

q C

q50

i50 a

R

b

c C

(b) Charging the capacitor Switch E closed + i

a

i

1q 2q

R

b

When the switch is closed, the charge on the capacitor increases over time while the current decreases.

c C

Using these in Kirchhoff’s loop rule, we find q E - iR = 0 C

(6.9)

The potential drops by an amount iR as we travel from a to b and by q>C as we travel from b to c. Solving Eq. (6.9) for i, we find

6.21 Current i and capacitor charge q as functions of time for the circuit of Fig. 6.20. The initial current is I0 and the initial capacitor charge is zero. The current asymptotically approaches zero, and the capacitor charge asymptotically approaches a final value of Qf . '

q E i = R RC

(6.10)

At time t = 0, when the switch is first closed, the capacitor is uncharged, and so q = 0. Substituting q = 0 into Eq. (6.10), we find that the initial current I0 is given by I0 = E>R, as we have already noted. If the capacitor were not in the circuit, the last term in Eq. (6.10) would not be present; then the current would be constant and equal to E>R. As the charge q increases, the term q>RC becomes larger and the capacitor charge approaches its final value, which we will call Qf. The current decreases and eventually becomes zero. When i = 0, Eq. (6.10) gives Qf E = R RC

Qf = CE

(6.11)

Note that the final charge Qf does not depend on R. Figure 6.21 shows the current and capacitor charge as functions of time. At the instant the switch is closed 1t = 02, the current jumps from zero to its initial value I0 = E>R; after that, it gradually approaches zero. The capacitor charge starts at zero and gradually approaches the final value given by Eq. (6.11), Qf = CE. We can derive general expressions for the charge q and current i as functions of time. With our choice of the positive direction for current (Fig. 6.20b), i equals the rate at which positive charge arrives at the left-hand (positive)

(a) Graph of current versus time for a charging capacitor i I0

/ /

I0 2 I0 e

The current decreases exponentially with time as the capacitor charges.

RC

O

t

(b) Graph of capacitor charge versus time for a charging capacitor q Qf

/

Qf e

/

Qf 2

O

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RC

The charge on the capacitor increases exponentially with time toward the final value Qf. t

186

CHAPTER 6 Direct-Current Circuits

plate of the capacitor, so i = dq>dt. Making this substitution in Eq. (6.10), we have dq q E 1 = = 1q - CE2 dt R RC RC We can rearrange this to dq dt = q - CE RC and then integrate both sides. We change the integration variables to q¿ and t¿ so that we can use q and t for the upper limits. The lower limits are q¿ = 0 and t¿ = 0: q

t dq¿ dt¿ = 20 q¿ - CE 20 RC

When we carry out the integration, we get lna

q - CE t b = -CE RC

Exponentiating both sides (that is, taking the inverse logarithm) and solving for q, we find q - CE = e-t/RC -CE PhET: Circuit Construction Kit (AC+DC) PhET: Circuit Construction Kit (DC Only) ActivPhysics 12.6: Capacitance ActivPhysics 12.7: Series and Parallel Capacitors ActivPhysics 12.8: Circuit Time Constants

(R-C circuit, charging q = CE11 - e-t/RC2 = Qf11 - e-t/RC2 capacitor) The instantaneous current i is just the time derivative of Eq. (6.12): i =

Pacemakers and Capacitors

(6.12)

dq E = e-t/RC = I0 e-t/RC dt R '

(R-C circuit, charging capacitor)

(6.13)

Application

This x-ray image shows a pacemaker implanted in a patient with a malfunctioning sinoatrial node, the part of the heart that generates the electrical signal to trigger heartbeats. The pacemaker circuit contains a battery, a capacitor, and a computer-controlled switch. To maintain regular beating, once per second the switch discharges the capacitor and sends an electrical pulse along the lead to the heart. The switch then flips to allow the capacitor to recharge for the next pulse.

Pacemaker

Electrical lead

Lung

Lung

Heart

The charge and current are both exponential functions of time. Figure 6.21a is a graph of Eq. (6.13) and Fig. 6.21b is a graph of Eq. (6.12).

Time Constant After a time equal to RC, the current in the R-C circuit has decreased to 1>e (about 0.368) of its initial value. At this time, the capacitor charge has reached 11 - 1>e2 = 0.632 of its final value Qf = CE. The product RC is therefore a measure of how quickly the capacitor charges. We call RC the time constant, or the relaxation time, of the circuit, denoted by t: t = RC

(time constant for R-C circuit)

(6.14)

When t is small, the capacitor charges quickly; when it is larger, the charging takes more time. If the resistance is small, it’s easier for current to flow, and the capacitor charges more quickly. If R is in ohms and C in farads, t is in seconds. In Fig. 6.21a the horizontal axis is an asymptote for the curve. Strictly speaking, i never becomes exactly zero. But the longer we wait, the closer it gets. After a time equal to 10RC, the current has decreased to 0.000045 of its initial value. Similarly, the curve in Fig. 6.21b approaches the horizontal dashed line labeled Qf as an asymptote. The charge q never attains exactly this value, but after a time equal to 10RC, the difference between q and Qf is only 0.000045 of Qf. We invite you to verify that the product RC has units of time.

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187

6.4 R-C Circuits

Discharging a Capacitor Now suppose that after the capacitor in Fig. 6.21b has acquired a charge Q0 , we remove the battery from our R-C circuit and connect points a and c to an open switch (Fig. 6.22a). We then close the switch and at the same instant reset our stopwatch to t = 0; at that time, q = Q0 . The capacitor then discharges through the resistor, and its charge eventually decreases to zero. Again let i and q represent the time-varying current and charge at some instant after the connection is made. In Fig. 6.22b we make the same choice of the positive direction for current as in Fig. 6.20b. Then Kirchhoff’s loop rule gives Eq. (6.10) but with E = 0; that is, '

'

i =

dq q = dt RC

6.22 Discharging a capacitor. (a) Before the switch is closed at time t = 0, the capacitor charge is Q0 and the current is zero. (b) At time t after the switch is closed, the capacitor charge is q and the current is i. The actual current direction is opposite to the direction shown; i is negative. After a long time, q and i both approach zero. (a) Capacitor initially charged Switch open

(6.15)

The current i is now negative; this is because positive charge q is leaving the lefthand capacitor plate in Fig. 6.22b, so the current is in the direction opposite to that shown in the figure. At time t = 0, when q = Q0 , the initial current is I0 = -Q0>RC. To find q as a function of time, we rearrange Eq. (6.15), again change the names of the variables to q¿ and t¿, and integrate. This time the limits for q¿ are Q0 to q. We get

i50

+ Q0 –Q0

'

q

t

dq¿ 1 = dt¿ 2Q0 q¿ RC 20 ln

a

R

b

(b) Discharging the capacitor Switch closed

q = Q0 e '

When the switch is closed, the charge on the capacitor and the current both decrease over time.

i

q t = Q0 RC

+q –q

i -t/RC

c C

(R-C circuit, discharging capacitor)

(6.16)

a

R

b

c C

The instantaneous current i is the derivative of this with respect to time: i =

Q0 -t/RC dq (R-C circuit, = e = I0 e-t/RC discharging capacitor) dt RC '

(6.17)

We graph the current and the charge in Fig. 6.23; both quantities approach zero exponentially with time. Comparing these results with Eqs. (6.12) and (6.13), we note that the expressions for the current are identical, apart from the sign of I0 . The capacitor charge approaches zero asymptotically in Eq. (6.16), while the difference between q and Q approaches zero asymptotically in Eq. (6.12). Energy considerations give us additional insight into the behavior of an R-C circuit. While the capacitor is charging, the instantaneous rate at which the bat tery delivers energy to the circuit is P = Ei. The instantaneous rate at which electrical energy is dissipated in the resistor is i2R, and the rate at which energy is stored in the capacitor is ivbc = iq>C. Multiplying Eq. (6.9) by i, we find

6.23 Current i and capacitor charge q as functions of time for the circuit of Fig. 6.22. The initial current is I0 and the initial capacitor charge is Q0. Both i and q asymptotically approach zero. '

'

Ei = i2R +

iq C

(6.18)

This means that of the power Ei supplied by the battery, part 1i2R2 is dissipated in the resistor and part 1iq>C2 is stored in the capacitor. The total energy supplied by the battery during charging of the capacitor equals the battery emf E multiplied by the total charge Qf, or EQf. The total energy stored in the capacitor, from Eq. (4.9), is Qf E>2. Thus, of the energy supplied by the battery, exactly half is stored in the capacitor, and the other half is dissipated in the resistor. This half-and-half division of energy doesn’t depend on C, R, or E. You can verify this result by taking the integral over time of each of the power quantities in Eq. (6.18) (see Problem 6.88).

(a) Graph of current versus time for a discharging capacitor i RC t O

/ /

I0 e I0 2

I0

The current decreases exponentially as the capacitor discharges. (The current is negative because its direction is opposite to that in Fig. 26.22.)

(b) Graph of capacitor charge versus time for a discharging capacitor q Q0

'

/ /

Q0 2 Q0 e O

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The charge on the capacitor decreases exponentially as the capacitor discharges.

RC

t

188

CHAPTER 6 Direct-Current Circuits

Example 6.12

Charging a capacitor

A 10-MÆ resistor is connected in series with a 1.0-mF capacitor and a battery with emf 12.0 V. Before the switch is closed at time t = 0, the capacitor is uncharged. (a) What is the time constant? (b) What fraction of the final charge Qf is on the capacitor at t = 46 s ? (c) What fraction of the initial current I0 is still flowing at t = 46 s ? '

'

t = RC = 110 * 106 Æ211.0 * 10-6 F2 = 10 s (b) From Eq. (6.12), q = 1 - e-t > RC = 1 - e-146 Qf '

'

s2 > 110 s2 '

'

= 0.99

(c) From Eq. (6.13),

SOLUTION IDENTIFY and SET UP: This is the same situation as shown in Fig. 6.20, with R = 10 MÆ, C = 1.0 mF, and E = 12.0 V. The charge q and current i vary with time as shown in Fig. 6.21. Our target variables are (a) the time constant t, (b) the ratio q>Qf at t = 46 s, and (c) the ratio i>I0 at t = 46 s. Equation (6.14) gives t. For a capacitor being charged, Eq. (6.12) gives q and Eq. (6.13) gives i.

Example 6.13

EXECUTE: (a) From Eq. (6.14),

i = e-t > RC = e-146 I0 '

'

s2 > 110 s2 '

'

= 0.010

EVALUATE: After 4.6 time constants the capacitor is 99% charged and the charging current has decreased to 1.0% of its initial value. The circuit would charge more rapidly if we reduced the time constant by using a smaller resistance.

Discharging a capacitor

The resistor and capacitor of Example 6.12 are reconnected as shown in Fig. 6.22. The capacitor has an initial charge of 5.0 mC and is discharged by closing the switch at t = 0. (a) At what time will the charge be equal to 0.50 mC? (b) What is the current at this time?

EXECUTE: (a) Solving Eq. (6.16) for the time t gives t = -RC ln

0.50 mC q = -110 s2 ln = 23 s = 2.3t Q0 5.0 mC

(b) From Eq. (6.17), with Q0 = 5.0 mC = 5.0 * 10-6 C, SOLUTION IDENTIFY and SET UP: Now the capacitor is being discharged, so q and i vary with time as in Fig. 6.23, with Q0 = 5.0 * 10-6 C . Again we have RC 5 t = 10 s. Our target variables are (a) the value of t at which q = 0.50 mC and (b) the value of i at this time. We first solve Eq. (6.16) for t, and then solve Eq. (6.17) for i.

i = -

Q0 -t > RC 5.0 * 10-6 C -2.3 e = e = -5.0 * 10-8 A RC 10 s '

'

EVALUATE: The current in part (b) is negative because i has the opposite sign when the capacitor is discharging than when it is charging. Note that we could have avoided evaluating e-t>RC by noticing that at the time in question, q = 0.10Q0; from Eq. (6.16) this means that e-t>RC = 0.10.

Test Your Understanding of Section 6.4 The energy stored in a capacitor is equal to q2/2C. When a capacitor is discharged, what fraction of the initial energy remains after an elapsed time of one time constant? (i) 1/e; (ii) 1/e2; (iii) 1 - 1/e; (iv) 11 - 1/e22; (v) answer depends on how much energy was stored initially.

6.5

y

Power Distribution Systems

We conclude this chapter with a brief discussion of practical household and automotive electric-power distribution systems. Automobiles use direct-current (dc) systems, while nearly all household, commercial, and industrial systems use alternating current (ac) because of the ease of stepping voltage up and down with transformers. Most of the same basic wiring concepts apply to both. We’ll talk about alternating-current circuits in greater detail in Chapter 11. The various lamps, motors, and other appliances to be operated are always connected in parallel to the power source (the wires from the power company for houses, or from the battery and alternator for a car). If appliances were connected in series, shutting one appliance off would shut them all off (see Example 6.2 in Section 6.1). Figure 6.24 shows the basic idea of house wiring. One side of the “line,” as the pair of conductors is called, is called the neutral side; it is always connected to

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6.5 Power Distribution Systems

189

6.24 Schematic diagram of part of a house wiring system. Only two branch circuits are shown; an actual system might have four to thirty branch circuits. Lamps and appliances may be plugged into the outlets. The grounding wires, which normally carry no current, are not shown. Fuse From power company

Light

Hot line

Outlets Switch Main fuse

Neutral line

Fuse

Light

Hot line

Outlets Switch Ground

Meter

“ground” at the entrance panel. For houses, ground is an actual electrode driven into the earth (which is usually a good conductor) or sometimes connected to the household water pipes. Electricians speak of the “hot” side and the “neutral” side of the line. Most modern house wiring systems have two hot lines with opposite polarity with respect to the neutral. We’ll return to this detail later. Household voltage is nominally 120 V in the United States and Canada, and often 240 V in Europe. (For alternating current, which varies sinusoidally with time, these numbers represent the root-mean-square voltage, which is 1> !2 times the peak voltage. We’ll discuss this further in Section 11.1.) The amount of current I drawn by a given device is determined by its power input P, given by Eq. (5.17): P = VI. Hence I = P>V. For example, the current in a 100-W light bulb is I =

P 100 W = = 0.83 A V 120 V

The power input to this bulb is actually determined by its resistance R. Using Eq. (25.18), which states that P = VI = I2R = V2>R for a resistor, the resistance of this bulb at operating temperature is R =

V 120 V = = 144 Æ I 0.83 A

or

R =

V2 1120 V22 = = 144 Æ P 100 W

Similarly, a 1500-W waffle iron draws a current of 11500 W2>1120 V2 = 12.5 A and has a resistance, at operating temperature, of 9.6 Æ. Because of the temperature dependence of resistivity, the resistances of these devices are considerably less when they are cold. If you measure the resistance of a 100-W light bulb with an ohmmeter (whose small current causes very little temperature rise), you will probably get a value of about 10 Æ. When a light bulb is turned on, this low resistance causes an initial surge of current until the filament heats up. That’s why a light bulb that’s ready to burn out nearly always does so just when you turn it on.

Circuit Overloads and Short Circuits The maximum current available from an individual circuit is limited by the resistance of the wires. As we discussed in Section 5.5, the I2R power loss in the wires causes them to become hot, and in extreme cases this can cause a fire or melt the wires. Ordinary lighting and outlet wiring in houses usually uses 12-gauge wire. This has a diameter of 2.05 mm and can carry a maximum current of 20 A safely (without overheating). Larger-diameter wires of the same length have lower resistance [see Eq. (5.10)]. Hence 8-gauge (3.26 mm) or 6-gauge (4.11 mm) are used for high-current appliances such as clothes dryers, and 2-gauge (6.54 mm) or larger is used for the main power lines entering a house.

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Neutral line

190

CHAPTER 6 Direct-Current Circuits

6.25 (a) Excess current will melt the thin wire of lead–tin alloy that runs along the length of a fuse, inside the transparent housing. (b) The switch on this circuit breaker will flip if the maximum allowable current is exceeded. (a)

(b)

Protection against overloading and overheating of circuits is provided by fuses or circuit breakers. A fuse contains a link of lead–tin alloy with a very low melting temperature; the link melts and breaks the circuit when its rated current is exceeded (Fig. 6.25a). A circuit breaker is an electromechanical device that performs the same function, using an electromagnet or a bimetallic strip to “trip” the breaker and interrupt the circuit when the current exceeds a specified value (Fig. 6.25b). Circuit breakers have the advantage that they can be reset after they are tripped, while a blown fuse must be replaced. If your system has fuses and you plug too many high-current appliances into the same outlet, the fuse blows. Do not replace the fuse with one of larger rating; if you do, you risk overheating the wires and starting a fire. The only safe solution is to distribute the appliances among several circuits. Modern kitchens often have three or four separate 20-A circuits. Contact between the hot and neutral sides of the line causes a short circuit. Such a situation, which can be caused by faulty insulation or by any of a variety of mechanical malfunctions, provides a very low-resistance current path, permitting a very large current that would quickly melt the wires and ignite their insulation if the current were not interrupted by a fuse or circuit breaker (see Example 5.10 in Section 5.5). An equally dangerous situation is a broken wire that interrupts the current path, creating an open circuit. This is hazardous because of the sparking that can occur at the point of intermittent contact. In approved wiring practice, a fuse or breaker is placed only in the hot side of the line, never in the neutral side. Otherwise, if a short circuit should develop because of faulty insulation or other malfunction, the ground-side fuse could blow. The hot side would still be live and would pose a shock hazard if you touched the live conductor and a grounded object such as a water pipe. For similar reasons the wall switch for a light fixture is always in the hot side of the line, never the neutral side. Further protection against shock hazard is provided by a third conductor called the grounding wire, included in all present-day wiring. This conductor corresponds to the long round or U-shaped prong of the three-prong connector plug on an appliance or power tool. It is connected to the neutral side of the line at the entrance panel. The grounding wire normally carries no current, but it connects the metal case or frame of the device to ground. If a conductor on the hot side of the line accidentally contacts the frame or case, the grounding conductor provides a current path, and the fuse blows. Without the ground wire, the frame could become “live”—that is, at a potential 120 V above ground. Then if you touched it and a water pipe (or even a damp basement floor) at the same time, you could get a dangerous shock (Fig. 6.26). In some situations, especially outlets located outdoors or near a sink or other water pipes, a special kind of circuit breaker called a ground-fault interrupter (GFI or GFCI) is used. This device senses the difference in current between the hot and neutral conductors (which is normally zero) and trips when this difference exceeds some very small value, typically 5 mA.

Household and Automotive Wiring Most modern household wiring systems actually use a slight elaboration of the system described above. The power company provides three conductors. One is neutral; the other two are both at 120 V with respect to the neutral but with opposite polarity, giving a voltage between them of 240 V. The power company calls this a three-wire line, in contrast to the 120-V two-wire (plus ground wire) line described above. With a three-wire line, 120-V lamps and appliances can be connected between neutral and either hot conductor, and high-power devices requiring 240 V, such as electric ranges and clothes dryers, are connected between the two hot lines. All of the above discussion can be applied directly to automobile wiring. The voltage is about 13 V (direct current); the power is supplied by the battery and by

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6.5 Power Distribution Systems (b) Three-prong plug

(a) Two-prong plug

191

6.26 (a) If a malfunctioning electric drill is connected to a wall socket via a two-prong plug, a person may receive a shock. (b) When the drill malfunctions when connected via a three-prong plug, a person touching it receives no shock, because electric charge flows through the ground wire (shown in green) to the third prong and into the ground rather than into the person’s body. If the ground current is appreciable, the fuse blows.

the alternator, which charges the battery when the engine is running. The neutral side of each circuit is connected to the body and frame of the vehicle. For this low voltage a separate grounding conductor is not required for safety. The fuse or circuit breaker arrangement is the same in principle as in household wiring. Because of the lower voltage (less energy per charge), more current (a greater number of charges per second) is required for the same power; a 100-W headlight bulb requires a current of about 1100 W2>113 V2 = 8 A. Although we spoke of power in the above discussion, what we buy from the power company is energy. Power is energy transferred per unit time, so energy is average power multiplied by time. The usual unit of energy sold by the power company is the kilowatt-hour 11 kW # h2: 1 kW # h = 1103 W213600 s2 = 3.6 * 106 W # s = 3.6 * 106 J

In the United States, one kilowatt-hour typically costs 8 to 27 cents, depending on the location and quantity of energy purchased. To operate a 1500-W (1.5-kW) waffle iron continuously for 1 hour requires 1.5 kW # h of energy; at 10 cents per kilowatt-hour, the energy cost is 15 cents. The cost of operating any lamp or appliance for a specified time can be calculated in the same way if the power rating is known. However, many electric cooking utensils (including waffle irons) cycle on and off to maintain a constant temperature, so the average power may be less than the power rating marked on the device.

Example 6.14

A kitchen circuit

An 1800-W toaster, a 1.3-kW electric frying pan, and a 100-W lamp are plugged into the same 20-A, 120-V circuit. (a) What current is drawn by each device, and what is the resistance of each device? (b) Will this combination trip the circuit breaker?

EXECUTE: (a) To simplify the calculation of current and resistance, we note that I = P > V and R = V2 > P. Hence '

'

'

'

'

Itoaster =

1800 W = 15 A 120 V

Rtoaster =

Ifrying pan =

1300 W = 11 A 120 V

Rfrying pan =

Ilamp =

100 W = 0.83 A 120 V

Rlamp =

S O L U T IO N IDENTIFY and SET UP: When plugged into the same circuit, the three devices are connected in parallel, so the voltage across each appliance is V = 120 V. We find the current I drawn by each device using the relationship P = VI, where P is the power input of the device. To find the resistance R of each device we use the relationship P = V2 > R.

'

1120 V22 = 8 Æ 1800 W

1120 V22 = 11 Æ 1300 W 1120 V22 = 144 Æ 100 W

For constant voltage the device with the least resistance (in this case the toaster) draws the most current and receives the most power. Continued

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192

CHAPTER 6 Direct-Current Circuits

(b) The total current through the line is the sum of the currents drawn by the three devices:

A third way to determine I is to use I = V > Req, where Req is the equivalent resistance of the three devices in parallel: '

I = Itoaster + Ifrying pan + Ilamp I =

= 15 A + 11 A + 0.83 A = 27 A This exceeds the 20-A rating of the line, and the circuit breaker will indeed trip. EVALUATE: We could also find the total current by using I = P > V and dividing the total power P delivered to all three devices by the voltage: '

I = =

'

'

1 1 V 1 = 1120 V2a + + b = 27 A Req 8 Æ 11 Æ 144 Æ

Appliances with such current demands are common, so modern kitchens have more than one 20-A circuit. To keep currents safely below 20 A, the toaster and frying pan should be plugged into different circuits.

Ptoaster + Pfrying pan + Plamp V 1800 W + 1300 W + 100 W = 27 A 120 V

Test Your Understanding of Section 6.5 To prevent the circuit breaker in Example 6.14 from blowing, a home electrician replaces the circuit breaker with one rated at 40 A. Is this a reasonable thing to do?

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y

SUMMARY

Resistors in series and parallel: When several resistors R1 , R2 , R3 , Á are connected in series, the equivalent resistance Req is the sum of the individual resistances. The same current flows through all the resistors in a series connection. When several resistors are connected in parallel, the reciprocal of the equivalent resistance Req is the sum of the reciprocals of the individual resistances. All resistors in a parallel connection have the same potential difference between their terminals. (See Examples 6.1 and 6.2.)

Req = R1 + R2 + R3 + Á (resistors in series)

Kirchhoff’s rules: Kirchhoff’s junction rule is based on conservation of charge. It states that the algebraic sum of the currents into any junction must be zero. Kirchhoff’s loop rule is based on conservation of energy and the conservative nature of electrostatic fields. It states that the algebraic sum of potential differences around any loop must be zero. Careful use of consistent sign rules is essential in applying Kirchhoff’s rules. (See Examples 6.3–6.7.)

aI = 0 aV = 0

'

1 1 1 1 = + + + Á Req R1 R2 R3 (resistors in parallel)

(6.1)

Resistors in series R1

a

R2

x

I

(6.2)

(loop rule)

I

R2

b

R3

I

Junction

(6.5) (6.6)

b

R1

Resistors in parallel a I

(junction rule)

R3

y

I2

I1

At any junction: SI 5 0

I1 1 I2 Loop 1

+

+ Loop 2

R

Loop 3

E

E Around any loop: SV 5 0

Ammeter ||

|||

|||||||||||||

||

Voltmeter ||

|

I

||| ||||||||||

||

|

Rc Rs

Rc +

|||

||

||

Electrical measuring instruments: In a d’Arsonval galvanometer, the deflection is proportional to the current in the coil. For a larger current range, a shunt resistor is added, so some of the current bypasses the meter coil. Such an instrument is called an ammeter. If the coil and any additional series resistance included obey Ohm’s law, the meter can also be calibrated to read potential difference or voltage. The instrument is then called a voltmeter. A good ammeter has very low resistance; a good voltmeter has very high resistance. (See Examples 6.8–6.11.)

|

'

||

'

|

6

||

CHAPTER

–

a R sh b

I

+

–

a

b

Va Circuit Vb element I

R-C circuits: When a capacitor is charged by a battery in series with a resistor, the current and capacitor charge are not constant. The charge approaches its final value asymptotically and the current approaches zero asymptotically. The charge and current in the circuit are given by Eqs. (6.12) and (6.13). After a time t = RC, the charge has approached within 1>e of its final value. This time is called the time constant or relaxation time of the circuit. When the capacitor discharges, the charge and current are given as functions of time by Eqs. (6.16) and (6.17). The time constant is the same for charging and discharging. (See Examples 6.12 and 6.13.)

Capacitor charging: q = CE A 1 - e

-t/RC

B

= Qf A 1 - e-t/RC B

+ E i

(6.12)

i

'

dq E = e-t/RC dt R = I0 e-t/RC

i =

I

R

(6.13)

i, q

1q

2q C q versus t

'

Capacitor discharging: q = Q0 e-t/RC '

Q0 -t/RC dq = e dt R -t/RC = I0 e

i =

O

i versus t t

(6.16)

(6.17)

'

Household wiring: In household wiring systems, the various electrical devices are connected in parallel across the power line, which consists of a pair of conductors, one “hot” and the other “neutral.” An additional “ground” wire is included for safety. The maximum permissible current in a circuit is determined by the size of the wires and the maximum temperature they can tolerate. Protection against excessive current and the resulting fire hazard is provided by fuses or circuit breakers. (See Example 6.14.)

193

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194

CHAPTER 6 Direct-Current Circuits

BRIDGING PROBLEM

Two Capacitors and Two Resistors

A 2.40-mF capacitor and a 3.60-mF capacitor are connected in series. (a) A charge of 5.20 mC is placed on each capacitor. What is the energy stored in the capacitors? (b) A 655- Æ resistor is connected to the terminals of the capacitor combination, and a voltmeter with resistance 4.58 * 104 Æ is connected across the resistor. What is the rate of change of the energy stored in the capacitors just after the connection is made? (c) How long after the connection is made has the energy stored in the capacitors decreased to 1> e of its initial value? (d) At the instant calculated in part (c), what is the rate of change of the energy stored in the capacitors? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. The two capacitors act as a single equivalent capacitor (see Section 4.2), and the resistor and voltmeter act as a single equivalent resistor. Select equations that will allow you to calculate the values of these equivalent circuit elements.

Problems

2. Equation (4.9) gives the energy stored in a capacitor. Equations (6.16) and (6.17) give the capacitor charge and current as functions of time. Use these to set up the solutions to the various parts of this problem. (Hint: The rate at which energy is lost by the capacitors equals the rate at which energy is dissipated in the resistances.) EXECUTE 3. Find the stored energy at t 5 0. 4. Find the rate of change of the stored energy at t 5 0. 5. Find the value of t at which the stored energy has 1> e of the value you found in step 3. 6. Find the rate of change of the stored energy at the time you found in step 5. EVALUATE 7. Check your results from steps 4 and 6 by calculating the rate of change in a different way. (Hint: The rate of change of the stored energy U is dU> dt.)

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems. DISCUSSION QUESTIONS Q6.1 In which 120-V light bulb does the filament have greater resistance: a 60-W bulb or a 120-W bulb? If the two bulbs are connected to a 120-V line in series, through which bulb will there be the greater voltage drop? What if they are connected in parallel? Explain your reasoning. Q6.2 Two 120-V light bulbs, one 25-W and one 200-W, were connected in series across a 240-V line. It seemed like a good idea at the time, but one bulb burned out almost immediately. Which one burned out, and why? Q6.3 You connect a number of identical light bulbs to a flashlight battery. (a) What happens to the brightness of each bulb as more and more bulbs are added to the circuit if you connect them (i) in series and (ii) in parallel? (b) Will the battery last longer if the bulbs are in series or in parallel? Explain your reasoning. Q6.4 In the circuit shown in Fig. Q6.4, three Figure Q6.4 identical light bulbs are connected to a flashlight + E battery. How do the brightnesses of the bulbs compare? Which light bulb has the greatest curA rent passing through it? Which light bulb has the greatest potential difference between its terB C minals? What happens if bulb A is unscrewed? Bulb B? Bulb C? Explain your reasoning. Q6.5 If two resistors R1 and R2 1R2 7 Figure Q6.5 R12 are connected in series as shown in I1 R1 I2 R2 I3 Fig. Q6.5, which of the following must a b c be true? In each case justify your answer. (a) I1 = I2 = I3. (b) The current is greater in R1 than in R2. (c) The electrical power consumption is the same for both resistors. (d) The electrical power consumption is greater in R2 than in R1.

(e) The potential drop is the same across both resistors. (f) The potential at point a is the same as at point c. (g) The potential at point b is lower than at point c. (h) The potential at point c is lower than at point b. Q6.6 If two resistors R1 and R2 1R2 7 Figure Q6.6 R12 are connected in parallel as shown I1 R1 in Fig. Q6.6, which of the following I I3 a c b 4 must be true? In each case justify your I2 R2 d answer. (a) I1 = I2. (b) I3 = I4. (c) The e f current is greater in R1 than in R2. (d) The rate of electrical energy consumption is the same for both resistors. (e) The rate of electrical energy consumption is greater in R2 than in R1. (f) Vcd = Vef = Vab. (g) Point c is at higher potential than point d. (h) Point f is at higher potential than point e. (i) Point c is at higher potential than point e. Q6.7 Why do the lights on a car become dimmer when the starter Figure Q6.8 is operated? A Q6.8 A resistor consists of three identical metal strips connected as shown in Fig. Q6.8. If one of the + strips is cut out, does the ammeter reading increase, decrease, or stay Figure Q6.9 the same? Why? S Q6.9 A light bulb is connected in the circuit shown in Fig. Q6.9. If we close the switch S, does the bulb’s brightness increase, decrease, or + remain the same? Explain why.

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Single Correct

Q6.1 Two identical resistors with resist-I 12V II R ance R are connected in the two circuits 12V R drawn. The battery in both circuits is a 12 R R volt ideal battery. Which statement is correct? (a) More current will flow through each R in circuit I than in circuit II. (b) More total power will be delivered by the battery in circuit II than in circuit I. (c) The potential drop across each R will be greater in circuit I than in circuit II. (d) The power dissipated in each R will be greater in circuit I than in circuit II. Q6.2 Four identical bulbs K, L, M, N are connected as shown in

()

()

(a)

()

(b)

()

() () () () (c) (d) R Q6.5 A circuit consists of a battery, a resistor R and two light bulbs A and B as shown: If the A B filament in lightbulb A burns out, then the following is true for light bulb B: (a) it is turned off (b) its brightness does not change (c) it gets dimmer (d) it gets brighter Q6.6 The following diagram represents Wire Y Wire X an electrical circuit containing two uniform resistance wires connected to a sin- 1.56 V gle flashlight cell. Both wires have the same length, but the thickness of wire X is twice that of wire Y. Which of the following would best represent the dependence of electric potential on position (distance from top) along the length of the two wires? Wire X

Wire X

Voltage

SINGLE CORRECT

the circuit below are glowing. The bulb K is L now taken out. Which of the following K statements is correct? M (a) All the bulbs will now blow off (b) Bulb N glows more brightly N (c) Bulb N becomes dimmer (d) Bulb N glows same as it were earlier Q6.3 Circuit A is made of resistors connected in series to an ideal battery; circuit B is made of resistors connected in parallel to an ideal battery. Let P be the power drawn from the batteries. As the number of resistors in each circuit is increased (a) PA increases and PB decreases (b) both PA and PB increases (c) PA decreases and PB increases (d) PA and PB remain the same Q6.4 In the house of a person who is weak of hearing, a light bulb is also lit when somebody rings the doorbell. The ring can be operated both from the garden gate and from the door of the house. Select the correct possible circuit(s) required.

Voltage

Q6.10 A real battery, having nonnegligible Figure Q6.10 internal resistance, is connected across a light bulb as shown in Fig. Q6.10. When the switch S is closed, what happens to the brightness of the bulb? Why? S Q6.11 If the battery in Discussion Question + Q6.10 is ideal with no internal resistance, what will happen to the brightness of the bulb when S is closed? Why? Q6.12 For the circuit shown in Fig. Q6.12 Figure Q6.12 what happens to the brightness of the bulbs when the switch S is closed if the battery (a) has no internal resistance and (b) has S nonnegligible internal resistance? Explain why. + Q6.13 Is it possible to connect resistors together in a way that cannot be reduced to some combination of series and parallel combinations? If so, give examples. If not, state why not. Q6.14 The direction of current in a battery can be reversed by connecting it to a second battery of greater emf with the positive terminals of the two batteries together. When the direction of current is reversed in a battery, does its emf also reverse? Why or why not? Q6.15 In a two-cell flashlight, the batteries are usually connected in series. Why not connect them in parallel? What possible advantage could there be in connecting several identical batteries in parallel? Q6.16 The greater the diameter of the wire used in household wiring, the greater the maximum current that can safely be carried by the wire. Why is this? Does the maximum permissible current depend on the length of the wire? Does it depend on what the wire is made of? Explain your reasoning. Q6.17 The emf of a flashlight battery is roughly constant with time, but its internal resistance increases with age and use. What sort of meter should be used to test the freshness of a battery? Q6.18 Is it possible to have a circuit in which the potential difference across the terminals of a battery in the circuit is zero? If so, give an example. If not, explain why not. Q6.19 Verify that the time constant RC has units of time. Q6.20 For very large resistances it is easy to construct R-C circuits that have time constants of several seconds or minutes. How might this fact be used to measure very large resistances, those that are too large to measure by more conventional means? Q6.21 Whan a capacitor, battery, and resistor are connected in series, does the resistor affect the maximum charge stored on the capacitor? Why or why not? What purpose does the resistor serve?

195

Wire Y

(a)

(0, 0)

Position

(b)

(0, 0)

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Wire Y Position

196

Wire Y

Voltage

Voltage

CHAPTER 6 Direct-Current Circuits

Wire X and Wire Y

resistance R, and all capacitors are identical with capacitance C and are initially uncharged. R

C

R

C

Wire X (0, 0)

Position Position (c) (d) (0, 0) Q6.7 The diagram below shows a junction I3 with currents labeled I1 to I6. Which of the following statements is correct? (a) I1 + I3 = I6 + I4 I6 I4 (b) I1 + I2 = I6 + I4 (c) I4 + I3 = I6 I1 I5 I2 (d) I2 = I6 + I4 Q6.8 An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. If a resistance is now joined in parallel with the voltmeter, (a) both A and V will increase (b) both A and V will decrease (c) A will decrease, V will increase (d) A will increase, V will decrease Q6.9 A potentiometer is to be calibrated with a standard cell using R the circuit shown in the diagram. The balance point is found to be near L. To improve accuracy the L M balance point should be nearer M. Galvanometer This may be achieved by P (a) Replacing the galvanometer Standard Cell with one of lower resistance (b) Replacing the potentiometer wire with one of higher resistance per unit length (c) Putting a shunt resistance in parallel with the galvanometer (d) Increasing the resistance R Q6.10 A capacitor of capacitance C is charged to a potential difference V0 and is ln I then discharged through a resistance R. The 1 discharge current gradually decreases, with a 2 straight line 1 corresponding to this process, 0 t as shown in figure where time is along x axis and the logarithm of the current on y-axis. Later on, one of the three parameters V0, R or C, is changed (keeping the other two unchanged) in such a manner that the ln I v/s t dependence is represented by the straight line 2. Which option correctly represents the change? (a) V0 is decreased (b) R is decreased (c) R is increased (d) C is decreased Q6.11 A graph between current & time during charging of a capacitor by a battery in series with a resistor is shown. The graphs are drawn for two cir- Log I (2) cuits. R1, C1 and V1 are the values of resist(1) ance, capacitance and EMF of the cell in one circuit and R2, C2 and V2 for another circuit t 0 respectively. If only two parameters (out of resistance, capacitance, EMF) are different in the two circuits. What is the correct option? (a) V1 = V2 , R1 > R2, C1 > C2 (b) V1 > V2, R1 > R2, C1 = C2 (c) V1 < V2, R1< R2, C1 = C2 (d) V1 < V2, C1< C2, R1 = R2 Q6.12 For the circuits shown below, which will take the least time to charge the capacitor(s) to 90% of the full charge? All batteries are ideal and identical with emf j, all resistors are identical with

x

(a)

(b)

R

C

x

x R

C

R x x (c) (d) V Q6.13 A capacitor is charged upto V a potential V0. It is then connected V0 to a resistance R and a battery of V0 emf E. Two possible graphs of t t potentials across capacitor vs time are shown. What is the most reasonable explanation of these graphs? (a) The first graph shows what happens when the capacitor has a less than E potential initially and the second shows what happens when it has a greater than E potential initially. (b) The first graph shows what happens when the capacitor has a greater than E potential initially and the second shows what happens when it has a less than E potential initially. (c) The first graph is the correct qualitative shape for any initial potential, but the second is not possible. (d) The second graph is the correct qualitative shape for any initial potential, but the first is not possible. Q6.14 In an RC circuit, the time required for the charge on a capacitor to build up to given fraction of its equilibrium value, is independent of (a) the value of the applied emf to the circuit (b) the value of C (c) the value of R (d) none of the above Q6.15 In the circuit shown, the 2µF charge on the 3mF capacitor at 3µF 2V 1W steady state will be 2W (a) 6 mC 1V (b) 4 mC (c) 2/3mC (d) 3 mC Q6.16 In the circuit shown, the switch 1 2 is shifted from position 1 S 2 at t = 0. The switch was initially in C 2R position 1 since a long time. The graph R between charge on upper plate of capacitor C and time 't' is e 2e

e

q q

2Ce

2Ce t

Ce

– Ce

(a)

t

(b) q q

3Ce

2Ce Ce –Ce

t

(c)

t

(d)

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One or More Correct

PARAGRAPH Paragraph for Question (6.17 to 6.19) G

The diagram shows a circuit of simple P ohm-meter. The galvanometer G gives a R R P full-scale deflection when the terminals P and Q are short circuited. The resistance R1 is very large. Q6.17 What would a graph of galvanometer deflection u against resistance R between the terminals PQ look like?

θ rad

θ rad

1

O

(a)

R Ω

O

(b)

R Ω

197

Q6.20 The value of emf E1 is (a) 8 V (b) 6 V (c) 4 V (d) 2 V Q6.21 The resistance R1 has value (a) 10Æ (b) 20Æ (c) 30Æ (d) 40Æ Q6.22 The resistance R2 is equal to (a) 10Æ (b) 20Æ (c) 30Æ (d) 40Æ

θ rad

θ rad

ONE OR MORE CORRECT

R Ω R Ω (c) O (d) O Q6.18 The scale which would be put on the galvanometer to read directly the resistance between P and Q would look like 300

200 100

(a)

0

400

0

(b)

200

300 400

100

300

200 100

400

300

200

100

0 (d) 0 (c) 400 Q6.19 The graph of potential difference accross PQ vs R is

V

V R

(a)

R

(b)

V

V R

(c)

R

(c)

Paragraph for Question (6.20 to 6.22) In the circuit shown, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1 V and 10 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite to the direction of that battery's emf. (direction from negative to positive)

E2

+ –

R1 + – E1

R2

Current (A)

0.4

0.2 0

–0.2

5

10 E2(V)

Q6.23 Two light bulbs shown in the circuit have ratings A (24 V, 24 W) and B (24 V and 36 W) as shown. 12V A When the switch is closed. (24 V, 24 W) (a) the intensity of light bulb A increases 12V B (b) the intensity of light bulb A (24 V, 36 W) decreases (c) the intensity of light bulb B increases (d) the intensity of light bulb B decreases C Q6.24 In the connection shown in the figure the switch K is open and the capacitor is G uncharged. Then we close the switch and let R2 R1 the capacitor charge upto the maximum and open the switch again. The values indicated K by the galvanometer. (a) increases after closing the switch and then decreases after opening the switch (b) direction of current is same through the galvanometer before and after opening the switch again (c) after opening the switch again energy dissipation in R1 is zero (d) after opening the switch again energy dissipation in R2 is zero R Q6.25 Referring to the circuit diagram, the C tap key is pressed at time t = 0. After suffiVR VC ciently long time (a) VC = 0 V K (b) VR = 0 (c) VC = V (d) VR = V Q6.26 An ideal battery is connected to a circuit of four resistors as shown in the fig. If the resistance of R4 is R2 increased, then e R1 R3 R4 (a) the current through R1 does not change (b) the potential difference across R3 increases (c) the potential difference across R2 decreases (d) current through the battery decreases Q6.27 The capacitor in the circuit shown above is initially charged. After closing the C R switch, (choose the correct option(s)) (a) Energy of the capacitor becomes half of its initial value in RC(ln2) time 2 (b) Charge on the capacitor becomes half of its initial value in time (RC) (ln2)

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198

CHAPTER 6 Direct-Current Circuits

(c) Current in the circuit becomes half of its initial value in time (RC) (ln2) (d) Current in the circuit becomes half of its initial value in time RC(ln2) 2

MATCH THE COLUMN Q6.28 After the switch shown in figure A is closed, there is current i through resistance R. Figure B indicates current variation curves a, b, c and d for four sets of values of R and capacitance C: Given in column I. Which set goes with which curve given in column II? Column-I

Column-II

R1 and R2 are connected in circuit as shown in figure. Voltmeter and ammeter connected in the circuit are ideal. Compari-S son of electric field, drift speed and power 1 dissipated per unit volume should be done only if R1 carries current. Column-I (a) S1 is open S2 is closed

(p)

(b) S1 is closed S2 is open

(q)

(c) S1 and S2 both are closed (r) (d) S1 and S2 both are open than in R1

(t)

i

(a) R0 and C0 (b) 2R0 and C0

R

(c) R0 and 2C0

t

Q6.29 Column I has four circuits each having an ammeter. Column II has four values of current in the ammeter. The ammeter has zero resistance. The voltmeter, in (B) has infinite resistance and a reading 8V. The resistance R has not been specified. Match the circuit with its correct ammeter reading. Column II 20 V

A

(a)

(p) 0

4W 6W 20 V

(q) 2 Ampere

.0

15

4W

8W A

(r) 4 Ampere

12W 20 V A

6W

6W

V

20 V 6W

6.1 .. A uniform wire of resistance R is Figure E6.1 cut into three equal lengths. One of these is formed into a circle and connected a b between the other two (Fig. E6.1). What is the resistance between the opposite ends a and b? 6.2 .. A machine part has a Figure E6.2 resistor X protruding from an a opening in the side. This resis15.0 5.0 10.0 tor is connected to three other X V V V resistors, shown in Fig. E6.2. b An ohmmeter connected across a and b reads 2.00Æ . What is the resistance of X? 6.3 . A triangular array of resistors is Figure E6.3 shown in Fig. E6.3. What current will b this array draw from a 35.0-V battery having negligible internal resistance if we connect it across (a) ab; (b) bc; (c) ac? c (d) If the battery has an internal resistance a 20.0 V of 3.00 Æ, what current will the array draw if the battery is connected Figure E6.4 across bc? 25.0 V 6.4 .. For the circuit shown in V A 15.0 V Fig. E6.4 both meters are idealized, the battery has no appre45.0 V 10.0 V 15.0 V ciable internal resistance, and the ammeter reads 1.25 A. (a) What does the voltmeter read? 35.0 V E 5 ? (b) What is the emf e of the battery? 6.5 .. For the circuit shown in Figure E6.5 Fig. E6.5 find the reading of the 45.0 V idealized ammeter if the battery has an internal resistance of 25.0 V 3.75 Æ. 10 V + A 6.6 . In Fig. E6.6, R1 = 3.00 Æ , R2 = 6.00 Æ , and R3 = 5.00 Æ . The battery has negligible internal 15.0 V resistance. The current I2 through .0

3W

(c)

+

(d)

6W 6W

(s) 5 Ampere (t) 7 Ampere

Q6.30 Figure shows two cylindrical uniform specimen of electrical resistances with their characteristics. p1 = Resistivity : Length : l1 = Area of cross-section : A1 = No. of electron per unit volume : n1 =

V

Column-II Reading of voltmeter is maximum Reading of ammeter is maximum Magnitude of electric field in R2 is greater than that in R1 Drift speed of electrons is greater in R2 than in R1 Power dissipated per unit volume is greater in R2 than in R1

V

V

R

A

R2

S2

10

(b)

R2

Section 6.1 Resistors in Series and Parallel

Q

Column I

A

R1

EXERCISES

S

P

(d) 2R0 and 2C0

(s)

e

R1

p p2 = 2p l l2 = 2l A A2 = A/2 n n2 = 2n

R2

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199

Excercises

R2 is 4.00 A. (a) What are the currents I1 and I3? (b) What is the emf of the battery? Figure E6.6

E

+

R1 R2

power dissipated in each bulb. (c) One bulb burns out very quickly. Which one? Why? 6.13 .. CP In the circuit in Fig. Figure E6.13 E6.13, a 20.0-Æ resistor is inside 10.0 V 10.0 V 100 g of pure water that is sur20.0 V 10.0 V 10.0 V rounded by insulating styrofoam. If the water is initially at 10.0°C, how 5.0 V 5.0 V Water long will it take for its temperature 30.0 V 5.0 V to rise to 58.0°C? +

R3

+

'

30.0 6.14 .. The batteries shown in 20.0 V V the circuit in Fig. E6.14 have neg- 10.0 V 5.00 V ligibly small internal resistances. Find the current through (a) the 30.0-Æ resistor; (b) the 20.0-Æ resis- Figure E6.15 tor; (c) the 10.0-V battery. 28.0 V R + 6.15 . In the circuit shown in Fig. E6.15 find (a) the current in resistor R; (b) the 4.00 A E resistance R; (c) the unknown emf E. (d) If + the circuit is broken at point x, what is the x 6.00 V current in resistor R? 6.00 A 6.16 . Find the emfs E1 and E2 in the cir3.00 V cuit of Fig. E6.16, and find the potential difference of point b relative to point a. +

'

Section 6.2 Kirchhoff’s Rules

+

6.7 . In the circuit of Fig. E6.7, each Figure E6.7 resistor represents a light bulb. Let R1 = R1 R2 = R3 = R4 = 4.50 Æ and e = 9.00 V. R3 (a) Find the current in each bulb. (b) Find E R2 R4 the power dissipated in each bulb. Which bulb or bulbs glow the brightest? (c) Bulb R4 is now removed from the circuit, leaving a break in the wire at its position. Now what is the current in each of the remaining bulbs R1, R2, and R3 ? (d) With bulb R4 removed, what is the power dissipated in each of the remaining bulbs? (e) Which light bulb(s) glow brighter as a result of removing R4 ? Which bulb(s) glow less brightly? Discuss why there are different effects on different bulbs. 6.8 . Consider the circuit shown in Fig. Figure E6.8 E6.8. The current through the 6.00-Æ 4.00 A 6.00 V resistor is 4.00 A, in the direction shown. 25.0 V What are the currents through the 8.00 V 25.0-Æ and 20.0-Æ resistors? 6.9 . A Three-Way Light Bulb. A 20.0 V three-way light bulb has three brightness + settings (low, medium, and high) but only two filaments. (a) A particular E three-way light bulb connected across a 120-V line can dissipate 60 W, 120 W, or 180 W. Describe how the two filaments are arranged in the bulb, and calculate the resistance of each filament. (b) Suppose the filament with the higher resistance burns out. How much power will the bulb dissipate on each of the three brightness settings? What will be the brightness (low, medium, or high) on each setting? (c) Repeat part (b) for the situation in which the filament with the lower resistance burns out. 6.10 .. Working Late! You are working late in your electronics shop and find that you need various resistors for a project. But alas, all you have is a big box of 10 .0-Æ resistors. Show how you can make each of the following equivalent resistances by a combination of your 10 .0-Æ resistors: (a) 35 Æ, (b) 1 .0 Æ, (c) 3 .33 Æ, (d) 7.5 Æ. 6.11 . In the circuit shown in Fig. Figure E6.11 E6.11, the rate at which R1 is dissipat3.50 A ing electrical energy is 20.0 W. (a) Find R1 and R2. (b) What is the emf of + 10.0 V R2 R1 the battery? (c) Find the current E through both R2 and the 10.0-Æ resis2.00 A tor. (d) Calculate the total electrical power consumption in all the resistors and the electrical power delivered by the battery. Show that your results are consistent with conservation of energy. 6.12 . Light Bulbs in Series. A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance of each bulb does not vary with current. (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.) (a) Find the current through the bulbs. (b) Find the

Figure E6.14

Figure E6.16 1.00 V 20.0 V +

1.00 A

1.00 V E1

4.00 V

a 2.00 A

+

1.00 V E2 +

6.00 V

b

2.00 V

6.17 . In the circuit shown in Fig. E6.17, find (a) the current in the 3.00-Æ resistor; (b) the unknown emfs E1 and E2; (c) the resistance R. Note that three currents are given. Figure E6.17 2.00 A +

R

E1

4.00 V 3.00 A

E2 + 3.00 V

6.00 V 5.00 A

6.18 .. In the circuit Figure E6.18 shown in Fig. E6.18 30.0 V the batteries have negliA + 20.0 gible internal resistance 25.0 V 75.0 V and the meters are both V + S E5? idealized. With the switch S open, the voltmeter reads 15.0 V. (a) Find the Figure E6.19 emf E of the battery. (b) What 12.0 V will the ammeter read when the switch is closed? 48.0 + 6.19 .. In the circuit shown in V 75.0 V Fig. E6.19 both batteries have A insignificant internal resistance

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50.0 V V

+

E5?

15.0 V

200

CHAPTER 6 Direct-Current Circuits

and the idealized ammeter reads 1.50 A in the direction shown. Find the emf E of the battery. Is the polarity shown correct? 6.20 .. In the circuit shown in Fig. E6.20, the 6.0-Æ resistor is consuming energy at a rate of 24 J/s when the current through it flows as shown. (a) Find the current through the ammeter A. (b) What are the polarity and emf E of the battery, assuming it has negligible internal resistance?

Figure E6.26 25.0 V 75.0 V 100.0 V

Figure E6.20

15.0 mF

+

50.0 V

S 20.0 V

19 V

+

A

1.0 V

+

6.21 . The resistance of a galvanometer coil is 25.0 Æ, and the current required for full-scale deflection is 500 mA. (a) Show in a diagram how to convert the galvanometer to an ammeter reading 20.0 mA full scale, and compute the shunt resistance. (b) Show how to convert the galvanometer to a voltmeter reading 500 mV full scale, and compute the series resistance. 6.22 . A galvanometer having a resistance of 25.0 Æ has a 1.00-Æ shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a circuit consisting of a 15.0-Æ resistor connected across the terminals of a 25.0-V battery having no appreciable internal resistance. (a) What current does the ammeter measure? (b) What should be the true current in the circuit (that is, the current without the ammeter present)? (c) By what percentage is the ammeter reading in error from the true current? 6.23 . In the ohmmeter in Fig. E6.23 M is a Figure E6.23 2.50-mA meter of resistance 65.0 Æ. (A M 2.50-mA meter deflects full scale when the a b B current through it is 2.50 mA.) The battery Rx B has an emf of 1.52 V and negligible interR nal resistance. R is chosen so that when the terminals a and b are shorted 1Rx = 02, the meter reads full scale. When a and b are open 1Rx = q 2, the meter reads zero. (a) What is the resistance of the resistor R? (b) What current indicates a resistance Rx of 200 Æ? (c) What values of Rx correspond to meter deflections of 14, 1 3 2 , and 4 of full scale if the deflection is proportional to the current through the galvanometer?

Section 6.4 R-C Circuits

6.24 . A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3 MÆ. After a time of 4.00 s the voltmeter reads 3.0 V. What are (a) the capacitance and (b) the time constant of the circuit? 6.25 .. CP In the circuit shown in Figure E6.25 Fig. E6.25 each capacitor initially 10.0 pF has a charge of magnitude 3.50 nC – + on its plates. After the switch S is S closed, what will be the current in 20.0 + 25.0 V the circuit at the instant that the pF – capacitors have lost 80.0% of their + – initial stored energy? 15.0 pF 6.26 . In the circuit in Fig. E6.26 the capacitors are all initially uncharged, the battery has no internal resistance, and the ammeter is idealized. Find the reading of the

+

Section 6.3 Electrical Measuring Instruments

ammeter (a) just after the switch S is closed and (b) after the switch has been closed for a very long time. 6.27 . In the circuit shown in Figure E6.27 Fig. E6.27, C = 6 mF, E = 30 V, Switch S Switch S and the emf has negligible resistin position 1 in position 2 ance. Initially the capacitor is uncharged and the switch S is in S position 1. The switch is then moved C E to position 2, so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long R time after the switch is moved to position 2? (b) After the switch has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 120 mC. What is the value of the resistance R? (c) How long after the switch is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

Section 6.5 Power Distribution Systems

6.28 . A 1500-W electric heater is plugged into the outlet of a 120-V circuit that has a 20-A circuit breaker. You plug an electric hair dryer into the same outlet. The hair dryer has power settings of 600 W, 900 W, 1200 W, and 1500 W. You start with the hair dryer on the 600-W setting and increase the power setting until the circuit breaker trips. What power setting caused the breaker to trip?

PROBLEMS 6.29 .. In Fig. P6.29, the bat-

Figure P6.29 tery has negligible internal resistance and E = 50 V. R1 = R1 R2 R2 = 4.00 Æ and R4 = 3.00 Æ . What must the resistance R3 be E R3 for the resistor network to dissipate electrical energy at a rate of R4 250 W? 6.30 .. Each of the three resistors in Fig. P6.30 has a resistance of Figure P6.30 2.4 Æ and can dissipate a maximum of 48 W without becoming excessively heated. What is the maximum power the circuit can dissipate? 6.31 . If an ohmmeter is connected between points a and b in each of the circuits shown in Fig. P6.31, what will it read? +

+

3.0 V 13 V

25.0 V

15.0 V 6.0 V 20.0 V

25 V

10.0 mF

A

17 V E5?

25.0 V

20.0 mF

Figure P6.31 (a)

a 100.0 V b

(b)

50.0 V 75.0 V 25.0 V

7.00 V a 10.0 V b 60.0 V 20.0 V

40.0 V

50.0 V

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30.0 V 45.0 V

Problems

+

Figure P6.32 5.00 V

8.00 V

I1 I2 + 12.00 1.00 I3 1.00 V V V 10.00 V

+ 9.00 V

7.00 V

3.00 V

2.00 V

R4 5 2.00 V

R5 5 1.00 V

Figure P6.35 1.00 V 12.0 V + 10.0 V b1.00 V + 1.00 V 8.0 V 3.00 V +

a

2.00 V

1.00 V 2.00 V

6.36 .. Consider the circuit shown in Fig. P6.36. (a) What must the emf E of the battery be in order for a current of 2.00 A to flow through the 5.00-V battery as shown? Is the polarity of the battery correct as shown? (b) How long does it take for 60.0 J of thermal energy to be produced in the 10.0-Æ resistor? Figure P6.36 10.0 V 20.0 V 60.0 V 2.00 A +

60.0 V 30.0 V

5.0 V

5.0 V 5.0 V

15.0 V E +

10.0 V

+

20.0 V

6.37 ... In the circuit shown in Fig. P6.37 all the resistors are rated at a maximum power of 2.00 W. What is the maximum emf E that the battery can have without burning up any of the resistors? Figure P6.37 25.0 V 25.0 V 30.0 V

15.0 V 10.0 V 20.0 V 20.0 V

50.0 V E 50.0 V

R

30.0 V

R

18.0 V 20.0 V

8.00 V X

5.00 A

6.35 .. (a) Find the potential of point a with respect to point b in Fig. P6.35. (b) If points a and b are connected by a wire with negligible resistance, find the current in the 12.0-V battery.

2.00 V

20.0 V

+

E

24.0 V

R1 5 1.00 V R 5 2.00 V R3 5 1.00 V 2

14.0 V

6.38 . In the circuit shown in Fig. P6.38, the current in the 20.0-V battery is 5.00 A in the direction shown and the voltage across the 8.00-Æ resistor is 16.0 V, with the lower end of the resistor at higher potential. Find (a) the emf (including its polarity) of the battery X; (b) the current I through the 200.0-V battery (including its direction); (c) the resistance R. Figure P6.38

Figure P6.33 +

6.32 . Calculate the three currents I1, I2, and I3 indicated in the circuit diagram shown in Fig. P6.32. 6.33 ... What must the emf E in Fig. P6.33 be in order for the current through the 7.00-Æ resistor to be 1.80 A? Each emf source has negligible internal resistance. 6.34 . (a) Find the current through the battery and each resistor in the circuit shown in Fig. P6.34. (b) What is the equivalent resistance of the resistor network? Figure P6.34

201

40.0 V

200.0 V I

6.39 .. Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 36 W. What power would be dissipated if the three resistors were connected in parallel across the same potential difference? 6.40 . A resistor R1 consumes electrical power P1 when con nected to an emf E. When resistor R2 is connected to the same emf, it consumes electrical power P2. In terms of P1 and P2, what is the total electrical power consumed when they are both connected to this emf source (a) in parallel and (b) in series? 6.41 . The capacitor in Fig. Figure P6.41 P6.41 is initially uncharged. R1 5 8.00 V The switch is closed at t = 0. (a) Immediately after the switch E 5 42.0 V R3 5 3.00 V is closed, what is the current + R2 5 through each resistor? (b) What 6.00 V C 5 4.00 mF is the final charge on the capacitor? 6.42 . Figure P6.42 employs a con Figure P6.42 vention often used in circuit diaV 5 36.0 V grams. The battery (or other power supply) is not shown explicitly. It is 3.00 understood that the point at the top, 3.00 V 6.00 V V labeled “36.0 V,” is connected to the a b S positive terminal of a 36.0-V battery 6.00 V 3.00 V having negligible internal resistance, and that the “ground” symbol at the bottom is connected to the negative terminal of the battery. The circuit is completed through the battery, even though it is not shown on the diagram. (a) What is the potential difference Vab, the potential of point a relative to point b, when the switch S is open? (b) What is the current through switch S when it is closed? (c) What is the equivalent resistance when switch S is closed? 6.43 . (See Problem 6.42.) (a) What Figure P6.43 is the potential of point a with respect V 5 18.0 V to point b in Fig. P6.43 when switch S is open? (b) Which point, a or b, is at 6.00 mF 6.00 V the higher potential? (c) What is the a b final potential of point b with respect S to ground when switch S is closed? 3.00 mF 3.00 V (d) How much does the charge on each capacitor change when S is closed? 6.44 . Point a in Fig. P6.44 is main- Figure P6.44 tained at a constant potential of 400 V 100 kV 200 kV above ground. (See Problem 6.42.) (a) a b What is the reading of a voltmeter with

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202

CHAPTER 6 Direct-Current Circuits

+

the proper range and with resistance 5.00 * 104 Æ when connected between point b and ground? (b) What is the reading of a voltmeter with resistance 5.00 * 106 Æ? (c) What is the reading of a voltmeter with infinite resistance? 6.45 .. The Wheatstone Bridge. The circuit shown in Fig. P6.45, called a Wheatstone bridge, Figure P6.45 is used to determine the value of an a unknown resistor X by comparison P with three resistors M, N, and P N whose resistances can be varied. b c G For each setting, the resistance of K2 E each resistor is precisely known. M X With switches K1 and K2 closed, these resistors are varied until the K1 d current in the galvanometer G is zero; the bridge is then said to be balanced. (a) Show that under this condition the unknown resistance is given by X = MP > N. (This method permits very high precision in comparing resistors.) (b) If the galvanometer G shows zero deflection when M = 850.0 Æ, N = 15.00 Æ, and P = 30 Æ, what is the unknown resistance X? 6.46 . A 2.36-mF capacitor that is initially uncharged is connected in series with a 6 Æ resistor and an emf source with E = 120 V and negligible internal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value. 6.47 ... An R-C circuit has a time constant RC. (a) If the circuit is discharging, how long will it take for its stored energy to be reduced to 1 > e of its initial value? (b) If it is charging, how long will it take for the stored energy to reach 1 > e of its maximum value? 6.48 . CALC The current in a charging capacitor is given by Eq. (6.13). (a) The instantaneous power supplied by the battery is Ei. Integrate this to find the total energy supplied by the battery. (b) The instantaneous power dissipated in the resistor is i2R. Integrate this to find the total energy dissipated in the resistor. (c) Find the final energy stored in the capacitor, and show that this equals the total energy supplied by the battery less the energy dissipated in the resistor, as obtained in parts (a) and (b). (d) What fraction of the energy supplied by the battery is stored in the capacitor? How does this fraction depend on R? 6.49 ... CALC (a) Using Eq. (6.17) for the current in a discharging capacitor, derive an expression for the instantaneous power P = i2R dissipated in the resistor. (b) Integrate the expression for P to find the total energy dissipated in the resistor, and show that this is equal to the total energy initially stored in the capacitor. 6.50 In a potentiometer experiment it is found that no current passes through the galvanometer when the terminals of the cell are connected across 0.52 m of the potentiometer wire. If the cell is shunted by a resistance of 5Æ , a balance is obtained when the cell is connected across 0.4m of the wire. Find the internal resistance of the cell. 6.51 A battery has an emf of E = 6V. The battery is connected in series with an ammeter and a voltmeter. If a certain resistor is connected in parallel with the voltmeter, the voltmeter reading decreases by a factor of 3, and the ammeter reading increases by '

'

'

0.4

'

E2

+ –

R1 + – E1

R2

Current (A)

'

'

the same factor. What is the initial reading V Figure P6.52 of the voltmeter? All elements of the circuit have unknown internal resistances. 6.52 A frame made of thin homogeneous wire is shown in Fig. P6.52. Assume that the number of successively embedded equilateral triangle with B sides decreasing by half tends to infinity. The side A AB has a resistance R0. The equivalent Figure P6.53 resistance between A and B is x A B 6.53 For the circuit shown in the 5 11W Fig. P6.53, find potential difference 2 11 5W VB - VA 2 11W 4 6.54 For given circuit, Fig. P6.54 heat pro11 2W duced by a current in resistance of 5Æ is 6 10 Cal/sec. What is the heat produced in resistance of 4Æ Figure P6.54 6.55 When a current of 2 A flows in a bat4W 6W tery from negative to positive terminal, the p.d. across it is 12 V. If a current of 3 A flows 5W in the opposite direction p.d. across the terminals of the battery is 15 V. Find the emf of the battery 6.56 In the circuit shown the variable resistance is so 1 adjusted that the ammeter reading is same in both the position 1 and 2 of the 2E X key. The reading of ammeter is 2A. If A Ideal E = 10V , then find x. 6.57 In the circuit shown, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1 V and 10 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (direction from negative to positive)

0.2 0

5

10 E2(V)

–0.2

(a) Find the value of (a) emf E1 (b) Find the resistance R1 (c) The resistance R2 6.58 A car battery with a 12 V emf and an internal resistance of 0.04Æ is being charged wtih a current of 50 A (a) Find the potential difference V across the terminals of the battery. (b) Find the rate at which energy is being dissipated as heat inside the battery. (c) Find the rate of energy conversion from electrical to chemical. R 6.59 Consider the circuit in the figure. The switch is initially in position A for a suffiC A B ciently long time to establish E 2C 2R a steady state. At t = 0, the switch is turned to position B. The capacitor 2C did not hold any charge before t = 0. (a) Find the magnitude of charge which each plate of capacitor C holds just before t = 0. (b) Alt t S q when a new steady state has been established. What are the charges held by capacitors C and 2C.

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203

Answers

(c) The current decreases exponentially between t = 0 to t = q . Find the time constant. (d) Find the energy lost between t = 0 to t = q . (e) Find the time, at which charge in the capacitor C is reduced to half of the value it had at t = 0. O 6.60 A window grill-like network of conductors is fitted in a ring of zero resistance. The interstices of the network are squares of side a and are made of uniform wire C of resistance r per unit length. What is the effective resistance between the ring G and the centre O. E B 6.61 Twelve identical resistors each A F of value 1Æ are connected as shown. H Find the net resistance between C and D. D 6.62 A potential difference of 220 V is b maintained across a 12000 ohm rheostat, as shown in the figure. The voltmeter has a resistance of 6000 ohm and point c is at one-fourth of the distance from a to b. Find the reading of 220 V the voltmeter. c 6.63 A galvanometer of resistance 100Æ V contains 100 division. It gives a deflection of one division on passing a current of 10 - 4 a A. Find the resistance in ohms to be connected to it, so that it 2W 2W becomes a voltmeter of range 10V. 1W 1W 6.64 In the circuit shown in 1W 4V figure find the steady state 6V 2W C = 10µF charge on capacitor C.

6.65 Two parallel plate capacitor of capacitances 2C and C are charged to the potentials 2V and V respectively and are connected in a circuit along with a resistance R as shown in the diagram. The switch k is closed at t = 0 . Find the current in the circuit as a function of time and total heat produced in the circuit. 4W 6.66 Find the charges on 4mF 4µF and 2mF capacitors in steady state.

CHALLENGE PROBLEMS 6.67 ... An Infinite Network. work of resistors of resistances R1 and R2 extends to infinity toward the right. Prove that the total resistance RT of the infinite network is equal to RT = R1 + 2R12 + 2R1R2

2C 2V + −

R

+ − C, V

k 5V

2W 2W

4W 2W

2µF

10V

As shown in Fig. P6.67, a netFigure P6.67 a

b

R1 c

R1

R1

R2

R2

R2

R1 d R1

and so on

R1 Dx

(Hint: Since the network is infinite, the resistance of the network to the right of points c and d is also equal to RT.) 6.68 ... Suppose a resistor R lies along Figure P6.68 each edge of a cube (12 resistors in all) b with connections at the corners. Find the equivalent resistance between two diagonally opposite corners of the cube (points a and b in Fig. P6.68). a

Answers

Chapter Opening Question

?

The potential difference V is the same across resistors connected in parallel. However, there is a different current I through each resistor if the resistances R are different: I = V>R.

Test Your Understanding Questions 6.1 Answer: (a), (c), (d), (b) Here’s why: The three resistors in Fig. 6.1a are in series, so Req = R + R + R = 3R. In Fig. 6.1b the three resistors are in parallel, so 1>Req = 1>R + 1>R + 1>R = 3>R and Req = R>3. In Fig. 6.1c the second and third resistors are in parallel, so their equivalent resistance R23 is given by 1>R23 = 1>R + 1>R = 2>R; hence R23 = R>2. This combination is in series with the first resistor, so the three resistors together have equivalent resistance Req = R + R>2 = 3R>2. In Fig. 6.1d the second and third resistors are in series, so their equivalent resistance is R23 = R + R = 2R. This combination is in parallel with the first resistor, so the equivalent resistance of the three-resistor combination is given by 1>Req = 1>R + 1>2R = 3>2R. Hence Req = 2R>3. 6.2 Answer: loop cbdac Equation (2) minus Eq. (1) gives

- I2 11 Æ2 - 1I2 + I3212 Æ2 + 1I1 - I3211 Æ2 + I1 11 Æ2 = 0. We can obtain this equation by applying the loop rule around the path from c to b to d to a to c in Fig. 6.12. This isn’t a new equation, so it would not have helped with the solution of Example 6.6. 6.3 Answers: (a) (ii), (b) (iii) An ammeter must always be placed in series with the circuit element of interest, and a voltmeter must always be placed in parallel. Ideally the ammeter would have zero resistance and the voltmeter would have infinite resistance so that their presence would have no effect on either the resistor current or the voltage. Neither of these idealizations is possible, but the ammeter resistance should be much less than 2 Æ and the voltmeter resistance should be much greater than 2 Æ. 6.4 Answer: (ii) After one time constant, and the initial charge Q0 has decreased to Q0 e-t/RC = Q0 e-RC/RC = Q0 e-1 = Q0>e. Hence the stored energy has decreased from Q02>2C to a fraction of its initial value. This result doesn’t depend on the initial value of the energy. 6.5 Answer: no This is a very dangerous thing to do. The circuit breaker will allow currents up to 40 A, double the rated value of the wiring. The amount of power P = I2R dissipated in a section of wire can therefore be up to four times the rated value, so the wires could get very warm and start a fire. '

'

'

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'

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204

CHAPTER 6 Direct-Current Circuits

Bridging Problem Answers: (a) 9.39 J 4 (b) 2.02 * 10 W (c) 4.65 * 10-4s 3 (d) 7.43 * 10 W

Discussion Questions V2 P So, 60 W bulb will have greater resistance. In series, voltage drop is proportional to resisance. But in parallel the voltage drop is same. P1P2 In series Peff = P1 + P 2 In parallel Peff = P1 + P2 Q6.2 25W bulb will get burned out,because potential drop across it will exceed its voltage (120 V). Q6.3 In veiw of answer to Q-6.1, if they are connected in series the bulbs will give different brightness, but identical bulbs connected in parallel will give same brightness. Q6.4 Branch of (B + C) will have greater resistance so branch of A will draw greater current. A will get potential drop E but B and C will get E and E each so brightest will be A. 2 2 If A is disconnected, B and C will get equal brightness. If B or C is disconnected (B + C) branch will go out of circuit and A will give brightness in its full potential. Q6.5 The resistances are in series so I1 = I2 = I3. (a) True (b) False Power consumption will be different Since, P = i2R and ¢V = iR (across a resistor) (d) is true, (e) is false, (f) is false, (g) false, (h) true Q6.1 Resistance R =

Q6.6 In parallel current through resistors will be in reverse ratio of I1 R1 1 = their resistances i.e I a or but the potential drop across I2 R2 R them will be same. V2 (a) is false (b) true (c) true (d) false as P = (e) false (f) true R (g) true (h) false (i) false Q6.7 At the moment of start the starter draws greater power instantaneously. R Q6.8 In case 1, Req = 3 R But in case 2, Req = 2 So, current in case 2 will decrease as resistance is increased. Q6.9 When S is closed net resistance will decrease, and hence greater current will be drawn from cell, bulb will glow brighter. Q6.10 When switch was not on, potential drop across bulb = E - ir E V1 = E r R + r E when switch is put on potential drop V2 = E .r R¿ + r as R > R¿ , so V1 > V2 so brightness will decrease. Q6.11 In case of ideal battery potential drop across bulb will remain unchanged so, brightness will remain same. R1 R3 Q6.12 (a) Incase of no internal resistance it will remain unaffected. R5 (b) Potential drop across each parallel branch will decrease so current in them will decrease than initial condiR2 R4 tion, so brightness will decrease.

Q6.13 Yes it is possible. One of the example can be unbalanced wheatstone bridge. Q6.14 The potential difference across the terminals will change during charging and discharging. V = E + ir (during charging) V = E - ir (during discharging) but E is a characteristic quantity, it will be unaltered. Q6.15 When cells are connected in series effective emf becomes greater E = E1 + E2 + Á but when connected in parallel, the effective emf E = emf of any one cell. Q6.16 Greater the diameter lesser will be resistance. Hence lesser will be heating effect in wire. Length of the wire increases the resistance also heating effect depends on material because resistance depends on nature of material also. Q6.17 The internal resistance is measured by potentiometer which works on null deflection principle. E Q6.18 V = E - ir = E r R + r At V = 0 Er = 0 E R + r 1 R + r = r 1 R = 0 Q6.19 Unit of RC = Unit of R * unit of C Coulomb Volt. * = ampere Volt Volt coulomb = * Volt coulomb a b time Q6.20 If resistance is very small then time constant will be very small for observation in experimentation. In case of large resistance time constant will be large t = RC so takQ0 ing reading during experimental observation will be easier and calculation of R can be done. Q6.21 We have Q = Q0 a1 - e - RC b t

Theoretically we say maximum charge stored in Q0 but practically it is not realizable. So on theoretical ground R does not effect the maximum charge stored but it does effect time constant.

Single Correct Q6.1 (b) Q6.2 (c) Q6.3 (c) Q6.4 (d) Q6.5 (d) Q6.6 (d) Q6.7 (b) Q6.8 (d) Q6.9 (d) Q6.10 (d) Q6.11 (c) Q6.12 (d) Q6.13 (b) Q6.14 (a) Q6.15 (b) Q6.16 (d) Q6.17 (c) Q6.18 (d)

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Answers

Q6.19 (b) Q6.20 (b) Q6.21 (b) Q6.22 (d)

205

6.22 (a) 1.57 A (b) 1.67 A (c) 6% 6.23 (a) 543Æ (b) 1.88 mA (c) 1824Æ , 608Æ , 203Æ

Section 6.4 R-C Circuits 6.24 (a) 0.96 mF (b) 2.88 sec

One or More Correct Q6.23 (b), (c) Q6.24 (a), (b), (c) Q6.25 (b), (c) Q6.26 (a), (c), (d) Q6.27 (a), (b), (c)

Match the Columns Q6.28 (a) R (b) P (c) S (d) Q Q6.29 (a) S S (b) S R (c) S P (d) S Q Q6.30 (a) S R, T (b) S P (c) S P, Q (d) S R, T

EXERCISES Section 6.1 Resistors in Series and Parallel 6.1

3R 4

6.2 7.5Æ 6.3 (a) 3.5 A (b) 4.5 A (c) 3.15 A (d) 3.25 A 6.4 (a) 206 V (b) 398 V 6.5 1 A 6.6 (a) I1 = 8 A, I3 = 12 A (b) 84 V. 6.7 (a) 1.5 A in R1 and 0.5 A in R2,R3 and R4 each 4 2 2 (d) 8, 2, 2 (e) R2, R3, R1 (b) R1 (c) , , 3 3 3 6.8 7 A, 9.95 A 6.9 (a) 240Æ,120Æ (b) 120 W: stay same, 60 W: does not work, 180 W: goes to 120 W (c) 60 W: stay same, 120 W:does not work, 180 W:goes to 60 W. 6.10 (a) (b) Ten in parallel (c) Three in parallel.

(d) 6.11 (a) R1 = 5Æ , R2 = 20Æ (b) 10 V (c) 0.5A, 1 A (d) P = 20 W, P2 = 5 W and P10 = 10 W. Pbattery = 35 W. 6.12 (a) 0.769 A (b) P60 = 142 W,P200 = 42.6 W (c) 60 W bulb. 6.13 1.01 * 103Sec,

Section 6.2 Kirchhoff’s Rules 7 1 1 A (b) A (c) A 3 4 12 6.15 (a) 2A (b) 5Æ (c) 42V (d) 3.5A 6.16 E1 = 18 V, E2 = 7 V, Vb - Va = - 13 V 6.17 (a) 8 A (b) 36 V, 54 V (c) 9Æ . 6.18 (a) 36.4 V (b) 0.5 A 6.19 52.3 V, correct 7 6.20 (a) A (b) E = - 46.1 V 30 6.14 (a)

Section 6.3 Electrical Measuring Instruments

6.25 13.6 A 20 15 A (b) A. 6.26 (a) 16 33 -4 6.27 (a) 1.8 * 10 s (b) 455Æ , (c) 0.0125 sec.

Section 6.5 Power Distribution Systems 6.28 900 W.

PROBLEMS 6.29 56Æ . 6.30 72 W 6.31 (a) 18.7Æ (b) 7.5Æ 179 452 36 A, I2 = A , I3 = A. 6.32 I1 = 211 211 211 6.33 8.6 V 6.34 (a) IR1 = 6 A, IR2 = 4 A, IR3 = 2 A, IR4 = 4 A, IR5 = 6 A (b) 1.4Æ 13 6.35 (a) 0.22 V (b) A 28 6.36 (a) - 109 V (b) 13.5 sec. 6.37 28.2 V 6.38 (a) 186 V with upper terminal positive (b) 3 A anticlock (c) 20Æ . 6.39 324 W P1P2 6.40 (a) P1 + P2 (b) . P1 + P2 6.41 (a) 2.8 A through 3Æ , 1.4 A through 6Æ and 4.2 A through 8Æ (b) 72mC. 6.42 (a) - 12 V (b) 1.71 A (c) 4.21Æ 6.43 (a) 18 V (b) a (c) 6 V (d) Both lose 36mC. 6.44 (a) 114.4 V (b) 263 V (c) 266 V 6.45 1700 Æ 6.46 (a) (i) 2400 W (ii) 0 (iii) 2400 W (b) all are zero (c) (i) 600 W (ii) 600 W (iii) 1200 W. 1 RC 6.47 (a) (b) - RC lna 1 b. 2 !e

1 6.48 (a) CE2 (b) CE2/2 (c) CE2/2 (d) ,independent of R. 2 Q20 - 2t/RC Q0 2 6.49 (a) (b) . e 2C RC2 6.50 1.5Æ 6.51 4.5 V !7 - 1 6.52 c = a bR0 3 6.53 - 4V 6.54 2 Cal/ sec 6.55 13.2 V 6.56 5Æ 6.57 (a) 6 V (b) 20Æ (c) 40Æ 6.58 (a) 14 V (b) 100 W (c) 600 W

6.21 (a) 0.64Æ (b) 975Æ

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206

CHAPTER 6 Direct-Current Circuits

6.59 (a) CE (b) CE/3 and 2CE/3 (c) 2RC (d) CE2/3 (e) 4RC ln2 21 6.60 ar 40 6.61 R = 3/4Æ 6.62 40 V 6.63 900Æ 6.64 15mC v 3t 1 6.65 i = a 1e b CV2 R 2RC 3 6.66 q4mF = 10mC, q2mF = 5mC

CHALLENGE PROBLEMS 6.68 5R/6.

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7

MAGNETIC FIELD AND MAGNETIC FORCES

LEARNING GOALS By studying this chapter, you will learn: • The properties of magnets, and how magnets interact with each other. • The nature of the force that a moving charged particle experiences in a magnetic field.

?

Magnetic resonance imaging (MRI) makes it possible to see details of soft tissue (such as in the foot shown here) that aren’t visible in x-ray images. Yet soft tissue isn’t a magnetic material (it’s not attracted to a magnet). How does MRI work?

verybody uses magnetic forces. They are at the heart of electric motors, microwave ovens, loudspeakers, computer printers, and disk drives. The most familiar examples of magnetism are permanent magnets, which attract unmagnetized iron objects and can also attract or repel other magnets. A compass needle aligning itself with the earth’s magnetism is an example of this interaction. But the fundamental nature of magnetism is the interaction of moving electric charges. Unlike electric forces, which act on electric charges whether they are moving or not, magnetic forces act only on moving charges. We saw in Chapter 1 that the electric force arises in two stages: (1) a charge produces an electric field in the space around it, and (2) a second charge responds to this field. Magnetic forces also arise in two stages. First, a moving charge or a collection of moving charges (that is, an electric current) produces a magnetic field. Next, a second current or moving charge responds to this magnetic field, and so experiences a magnetic force. In this chapter we study the second stage in the magnetic interaction—that is, how moving charges and currents respond to magnetic fields. In particular, we will see how to calculate magnetic forces and torques, and we will discover why magnets can pick up iron objects like paper clips. In Chapter 8 we will complete our picture of the magnetic interaction by examining how moving charges and currents produce magnetic fields.

E

7.1

• How magnetic field lines are different from electric field lines. • How to analyze the motion of a charged particle in a magnetic field. • Some practical applications of magnetic fields in chemistry and physics. • How to analyze magnetic forces on current-carrying conductors. • How current loops behave when placed in a magnetic field.

Magnetism

Magnetic phenomena were first observed at least 2500 years ago in fragments of magnetized iron ore found near the ancient city of Magnesia (now Manisa, in western Turkey). These fragments were examples of what are now called permanent magnets; you probably have several permanent magnets on your refrigerator 207

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208

CHAPTER 7 Magnetic Field and Magnetic Forces

7.1 (a) Two bar magnets attract when opposite poles (N and S, or S and N) are next to each other. (b) The bar magnets repel when like poles (N and N, or S and S) are next to each other. (a) Opposite poles attract. S

N

N

S

F F F F

S

N

N

S

(b) Like poles repel. F F

S

N

N

S

N

S

S

N

F F

7.2 (a) Either pole of a bar magnet attracts an unmagnetized object that contains iron, such as a nail. (b) A real-life (a) S

N

N

S

F

F

F

F

door at home. Permanent magnets were found to exert forces on each other as well as on pieces of iron that were not magnetized. It was discovered that when an iron rod is brought in contact with a natural magnet, the rod also becomes magnetized. When such a rod is floated on water or suspended by a string from its center, it tends to line itself up in a north-south direction. The needle of an ordinary compass is just such a piece of magnetized iron. Before the relationship of magnetic interactions to moving charges was understood, the interactions of permanent magnets and compass needles were described in terms of magnetic poles. If a bar-shaped permanent magnet, or bar magnet, is free to rotate, one end points north. This end is called a north pole or N pole; the other end is a south pole or S pole. Opposite poles attract each other, and like poles repel each other (Fig. 7.1). An object that contains iron but is not itself magnetized (that is, it shows no tendency to point north or south) is attracted by either pole of a permanent magnet (Fig. 7.2). This is the attraction that acts between a magnet and the unmagnetized steel door of a refrigerator. By analogy to electric interactions, we describe the interactions in Figs. 7.1 and 7.2 by saying that a bar magnet sets up a magnetic field in the space around it and a second body responds to that field. A compass needle tends to align with the magnetic field at the needle’s position. The earth itself is a magnet. Its north geographic pole is close to a magnetic south pole, which is why the north pole of a compass needle points north. The earth’s magnetic axis is not quite parallel to its geographic axis (the axis of rotation), so a compass reading deviates somewhat from geographic north. This deviation, which varies with location, is called magnetic declination or magnetic variation. Also, the magnetic field is not horizontal at most points on the earth’s surface; its angle up or down is called magnetic inclination. At the magnetic poles the magnetic field is vertical. Figure 7.3 is a sketch of the earth’s magnetic field. The lines, called magnetic field lines, show the direction that a compass would point at each location; they are discussed in detail in Section 7.3. The direction of the field at any point can be defined as the direction of the force that the field would exert on a magnetic north

7.3 A sketch of the earth’s magnetic field. The field, which is caused by currents in the earth’s molten core, changes with time; geologic evidence shows that it reverses direction entirely at irregular intervals of 10 4 to 10 6 years. (b)

North geographic pole (earth’s rotation axis)

The geomagnetic north pole is actually a magnetic south (S) pole—it attracts the N pole of a compass. Compass Magnetic field lines show the direction a compass would point at a given location.

S

The earth’s magnetic field has a shape similar to that produced by a simple bar magnet (although actually it is caused by electric currents in the core).

N

The earth’s magnetic axis is offset from its geographic axis.

The geomagnetic south pole is actually a magnetic north (N) pole.

South geographic pole

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209

7.2 Magnetic Field

pole. In Section 7.2 we’ll describe a more fundamental way to define the direction and magnitude of a magnetic field.

Magnetic Poles Versus Electric Charge The concept of magnetic poles may appear similar to that of electric charge, and north and south poles may seem analogous to positive and negative charge. But the analogy can be misleading. While isolated positive and negative charges exist, there is no experimental evidence that a single isolated magnetic pole exists; poles always appear in pairs. If a bar magnet is broken in two, each broken end becomes a pole (Fig. 7.4). The existence of an isolated magnetic pole, or magnetic monopole, would have sweeping implications for theoretical physics. Extensive searches for magnetic monopoles have been carried out, but so far without success. The first evidence of the relationship of magnetism to moving charges was discovered in 1820 by the Danish scientist Hans Christian Oersted. He found that a compass needle was deflected by a current-carrying wire, as shown in Fig. 7.5 Similar investigations were carried out in France by André Ampère. A few years later, Michael Faraday in England and Joseph Henry in the United States discovered that moving a magnet near a conducting loop can cause a current in the loop. We now know that the magnetic forces between two bodies shown in Figs. 7.1 and 7.2 are fundamentally due to interactions between moving electrons in the atoms of the bodies. (There are also electric interactions between the two bodies, but these are far weaker than the magnetic interactions because the two bodies are electrically neutral.) Inside a magnetized body such as a permanent magnet, there is a coordinated motion of certain of the atomic electrons; in an unmagnetized body these motions are not coordinated. (We’ll describe these motions further in Section 7.7, and see how the interactions shown in Figs. 7.1 and 7.2 come about.) Electric and magnetic interactions prove to be intimately connected. Over the next several chapters we will develop the unifying principles of electromagnetism, culminating in the expression of these principles in Maxwell’s equations. These equations represent the synthesis of electromagnetism, just as Newton’s laws of motion are the synthesis of mechanics, and like Newton’s laws they represent a towering achievement of the human intellect.

7.4 Breaking a bar magnet. Each piece has a north and south pole, even if the pieces are different sizes. (The smaller the piece, the weaker its magnetism.) In contrast to electric charges, magnetic poles always come in pairs and can't be isolated. Breaking a magnet in two ... N

N

S

S

N

... yields two magnets, not two isolated poles.

7.5 In Oersted’s experiment, a compass is placed directly over a horizontal wire (here viewed from above). When the compass is placed directly under the wire, the compass deflection is reversed. (a) N

When the wire carries no E current, the compass needle points north.

W S

I50 (b) When the wire carries a current, the compass needle deflects. The direction of deflection depends on the direction of the current. I

Test Your Understanding of Section 7.1 Suppose you cut off the part of the compass needle shown in Fig. 7.5a that is painted gray. You discard this part, drill a hole in the remaining red part, and place the red part on the pivot at the center of the compass. Will the red part still swing east and west when a current is applied as in Fig. 7.5b? y

S

I

N

N

W

E

W

E

S

7.2

S I

Magnetic Field

To introduce the concept of magnetic field properly, let’s review our formulation of electric interactions in Chapter 1, where we introduced the concept of electric field. We represented electric interactions in two steps: S

1. A distribution of electric charge at rest creates an electric field E in the surrounding space. S S 2. The electric field exerts a force F 5 qE on any other charge q that is present in the field. We can describe magnetic interactions in a similar way: 1. A moving charge or a current creates a magnetic field in the surrounding space (in addition to its electric field). S 2. The magnetic field exerts a force F on any other moving charge or current that is present in the field.

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I

210

CHAPTER 7 Magnetic Field and Magnetic Forces

Spiny Lobsters and Magnetic Compasses

Application

Although the Caribbean spiny lobster (Panulirus argus) has a relatively simple nervous system, it is remarkably sensitive to magnetic fields. It has an internal magnetic “compass” that allows it to distinguish north, east, south, and west. This lobster can also sense small differences in the earth’s magnetic field from one location to another, and may use these differences to help it navigate.

In this chapter we’ll concentrate on the second aspect of the interaction: Given the presence of a magnetic field, what force does it exert on a moving charge or a current? In Chapter 8 we will come back to the problem of how magnetic fields are created by moving charges and currents. Like electric field, magnetic field is a vector field—that Sis, a vector quantity associated with each point in space. We will use the symbol B for magnetic field. S At any position the direction of B is defined as the direction in which the north pole of a compass needle tends to point. The arrowsSin Fig. 7.3 suggest the direction of the earth’s magnetic field; for any magnet, B points out of its north pole and into its south pole.

Magnetic Forces on Moving Charges

S

7.6 The magnetic force F acting on a S positive charge q moving with velocity v is S perpendicular to both and the magnetic v S field B. For given values of the speed v and magnetic field strength B, the force is S S greatest when v and B are perpendicular. (a) A charge moving parallel to a magnetic field experiences zero q S magnetic v force. S B q S v + (b) A charge moving at an angle f to a magnetic field experiences a magnetic force with magnitude F 5 0 q 0 v'B 5 0 q 0 vB sin f. S

F

S

F is perpendicular to the plane q containing S S v and B. v'

S

B S

B

fS v

(c) A charge moving perpendicular to a magnetic field experiences a maximal magnetic force with magnitude S Fmax 5 qvB. Fmax

There are four key characteristics of the magnetic force on a moving charge. First, its magnitude is proportional to the magnitude of the charge. If a 1-mC charge and a 2-mC charge move through a given magnetic field with the same velocity, experiments show that the force on the 2-mC charge is twice as great as the force on the 1-mC charge. Second, the magnitude of the force is also proportional to the magnitude, or “strength,” of the field; if we double the magnitude of the field (for example, by using two identical bar magnets instead of one) without changing the charge or its velocity, the force doubles. A third characteristic is that the magnetic force depends on the particle’s velocity. This is quite different from the electric-field force, which is the same whether the charge is moving or not. A charged particle at rest experiences no S magnetic force. And fourth, we find by experiment thatS the magnetic force F does not have the sameSdirection as the magnetic field B but instead is always S perpendicular to both B and the velocity v. The magnitude F of the force is S v perpendicular to the field; when found to be proportional to the component of S S that component is zero (that is, when v and B are parallel or antiparallel), the force is zero. S Figure 7.6 shows these relationships. The direction of F is always perpendicuS S lar to the plane containing v and B. Its magnitude is given by F = ƒ q ƒ y' B = ƒ q ƒ yB sin f

(7.1)

where ƒ q ƒ is the magnitude of theS charge and f is the angle measured from the S direction of v to the direction of B, as shown in the figure. S This description does not specify the direction of F completely; there are always two directions, opposite to each other, that are both perpendicular to the S S plane of v and B. To complete the description, we use the same right-hand rule that we used to define the vector product in Section 1.10. (It would be a good idea S S to review that section before you go on.) Draw the vectors v and B with their tails S S together, as in Fig. 7.7a. Imagine turning v until it points in the direction of B (turning through the smaller of the two possible angles). Wrap the fingers of your S S right hand around the line perpendicular toSthe plane of v and B so that they curl S around with the sense of rotation from v to B. Your thumb then points in the direcS tion of the force F on a positive charge. (Alternatively, the direction of the force S F on a positive charge is the direction in which a right-hand-thread screw would advance if turned the same way.) S This discussion shows that the force on a charge q moving with velocity v in a S magnetic field B is given, both in magnitude and in direction, by S

S

S

F 5 qv 3 B

(magnetic force on a moving charged particle)

(7.2)

S

q S

v

+

B

This is the first of several vector products we will encounter in our study of magnetic-field relationships. It’s important to note that Eq. (7.2) was not deduced theoretically; it is an observation based on experiment.

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7.2 Magnetic Field

211

7.7 Finding the direction of the magnetic force on a moving charged particle. (a)

(b)

Right-hand rule for the direction of magnetic force on a positive charge moving in a magnetic field: S S

1 Place the v and B vectors tail to tail. S

2 Imagine turning v toward B in the v-B

B

1

3

q

plane (through the smaller angle).

If the charge is negative, the direction of the force is opposite to that given by the right-hand rule.

S

1

S S

S

S

S

F 5 qv 3 B

S

S

v 3 The force acts along a line perpenS S

dicular to the v-B plane. Curl the fingers of your right hand around this line in the same direction you S rotated v. Your thumb now points in the direction the force acts.

2

1

S

S

F 5 qv 3 B S

S

S

F 5 qv 3 B S

q

S

v

3

S

B S

S

Right hand!

2

S

2

S

v

Force acts along this line.

S S

v-B plane v

S

B

q

q

S

B

S

F 5 qv 3 B

Equation (7.2) is valid for bothS positive and negative charges.SWhen q is negaS tive, the direction of the force F is opposite to that of v 3 B (Fig. S 7.7b). If two charges with equal magnitude and opposite sign move in the same B field with the same velocity (Fig. 7.8), the forces have equal magnitude and opposite direction. FiguresS 7.6, 7.7, and 7.8 show several examples of the relationships S S of the directions of F, v, and B for both positive and negative charges. Be sure you understand the relationships shown in these figures. S Equation (7.1) gives the magnitude of the magnetic force F in Eq. (7.2). We can express this magnitude in a differentSbut equivalent way. Since f is the angle S between the directions of vectors v and B, we may interpret B sin f S S as the component of B perpendicular to v—that is, B' . With this notation the force magnitude is

7.8 Two charges of the same magnitude but opposite sign moving with the same velocity in the same magnetic field. The magnetic forces on the charges are equal in magnitude but opposite in direction. Positive and negative charges moving in the same direction through a magnetic field experience magnetic forces in opposite S S S directions. F 5 qv 3 B

'

F = ƒ q ƒ yB'

S

1 tesla = 1 T = 1 N>A # m Another unit of B, the gauss 11 G = 10 -4 T2, is also in common use. The magnetic field of the earth is of the order of 10 -4 T or 1 G. Magnetic fields of the order of 10 T occur in the interior of atoms and are important in the analysis of atomic spectra. The largest steady magnetic field that can be produced at present in the laboratory is about 45 T. Some pulsed-current electromagnets can produce fields of the order of 120 T for millisecond time intervals.

f

S

(7.3)

This form is sometimes more convenient, especially in problems involving currents rather than individual particles. We will discuss forces on currents later in this chapter. From Eq. (7.1) the units of B must be the same as the units of F>qy. Therefore the SI unit of B is equivalent to 1 N # s>C # m, or, since one ampere is one coulomb per second 11 A = 1 C>s2, 1 N>A # m. This unit is called the tesla (abbreviated T), in honor of Nikola Tesla (1856–1943), the prominent SerbianAmerican scientist and inventor:

B

q1 5 q . 0

q2 5 2q , 0

S

v

S

S

v

B f

S

S

F 5 (2q)v 3 B

Application Magnetic Fields of the Body All living cells are electrically active, and the feeble electric currents within the body produce weak but measurable magnetic fields. The fields produced by skeletal muscles have magnitudes less than 10-10 T, about onemillionth as strong as the earth’s magnetic field. The brain produces magnetic fields that are far weaker, only about 10-12 T.

Measuring Magnetic Fields with Test Charges To explore an unknown magnetic field, we can measure the magnitude and direcS tion of the force on a moving test charge and then use Eq. (7.2) to determine B. The electron beam in a cathode-ray tube, such as that in an older television set (not a flat screen), is a convenient device for this. The electron gun shoots out a narrow beam of electrons at a known speed. If there is no force to deflect the beam, it strikes the center of the screen.

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212

CHAPTER 7 Magnetic Field and Magnetic Forces

If a magnetic field is present, in general the electron beam is deflected. But if the beam is parallel or antiparallel to the field, then f = 0 or p in Eq. (7.1) and F = 0; there is no force and hence no deflection. If we find that the electron beam is not deflected when its direction is parallel to a certain axis as in Fig. 7.9a, S the B vector must point either up or down along that axis. If we then turn the tube 90° (Fig. 7.9b), f = p>2 in Eq. (7.1) and the magnetic force is maximum; the beam is deflected in a direction perpendicular to the S S plane of B and v. The direction and magnitude of the deflection determine the S direction and magnitude of B . We can perform additional experiments in which S S the angle between B and v is between zero and 90° to confirm Eq. (7.1). We note that the electron has a negative charge; the force in Fig. 7.9b is opposite in direction to the force on a positive charge. When a charged particle moves through a region of space where both electric and magnetic fields are present, both fields exert forces on the particle. The total S force F is the vector sum of the electric and magnetic forces:

ActivPhysics 13.4: Magnetic Force on a Particle

F 5 q1E 1 v 3 B2 S

7.9 Determining the direction of a magnetic field using a cathode-ray tube. Because electrons have a negative charge, S S S the magnetic force F 5 qv 3 B in part (b) points opposite to the direction given by the right-hand rule (see Fig. 7.7b).

(a) If the tube axis is parallel to the y-axis, the beamSis undeflected, so B is in either the 1y- or the 2y-direction.

S

S

S

y

(7.4)

(b) If the tube axis is parallel to the x-axis,Sthe beam is deflected in the 2z-direction, so B is in the 1y-direction. y

S

B

S

v

S S

F

B

x

S

v z Electron beam

Problem-Solving Strategy 7.1

z

Magnetic Forces S

S

S

IDENTIFY the relevant concepts: The equation F 5 qv 3 B allows you to determine the magnetic force on a moving charged particle. SET UP the problem using the following steps: S S 1. Draw the velocity v and magnetic field B with their tails together so that you can visualize the plane that contains them. S S 2. Determine the angle f between v and B. 3. Identify the target variables. EXECUTE the solution as follows: S S S 1. Express the magnetic force using Eq. (7.2), F 5 qv 3 B. Equation (7.1) gives the magnitude of the force, F = qyB sin f.

Example 7.1

x

S

S

2. Remember that F is perpendicular to the plane containing v S and BS (see Fig. 7.7) givesSthe direction of . The right-hand rule S S S v 3 B. If q is negative, F is opposite to v 3 B. EVALUATE your answer: Whenever possible, solve the problem in two ways to confirm that the results agree. Do it directly from the geometric definition of the vector product. Then find the components of the vectors in some convenient coordinate system and calculate the vector product from the components. Verify that the results agree.

Magnetic force on a proton

A beam of protons 1q = 1.6 * 10-19 C2 moves at 3.0 * 105 m> s through a uniform 2.0-T magnetic field directed along the positive z-axis, as in Fig. 7.10. The velocity of each proton lies in the '

'

xz-plane and is directed at 30° to the + z-axis. Find the force on a proton.

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7.3 Magnetic Field Lines and Magnetic Flux

213

F = qyB sin f

S O L U T IO N

= 11.6 * 10-19 C213.0 * 105 m> s212.0 T21sin 30°2

S

IDENTIFY and SET UP: This problem uses the expression F 5 S S S F qv 3 B for the magnetic force on a moving charged particle. S The target variable is F. EXECUTE: The charge is positive,Sso the force is in the same direcS tion as the vector product v 3 B. From the right-hand rule, this direction is along the negative y-axis. The magnitude of the force, from Eq. (7.1), is

'

'

= 4.8 * 10-14 N

EVALUATE: We check our result by evaluating the force using vector language and Eq. (7.2). We have v 5 13.0 * 105 m> s21sin 30°2≤n 1 13.0 * 105 m> s21cos 30°2kN

S

'

'

'

'

B 5 12.0 T2kN S S

S

S

F 5 qv 3 B S

S

7.10 Directions of v and B for a proton in a magnetic field. y

q S

B

S 30° v

z

5 11.6 * 10-19 C213.0 * 105 m> s212.0 T2 * 1sin 30°≤n 1 cos 30°kN 2 3 kN '

'

5 1 -4.8 * 10-14 N2≥n

x

(Recall that n≤ 3 kN 5 - n≥ and kN 3 kN 5 0.2 We again find that the force is in the negative y-direction with magnitude 4.8 * 10 -14 N. If the beam consists of electrons rather than protons, the charge is negative 1q = - 1.6 * 10-19 C2 and the direction of the force is reversed. The force is now directed along the positive y-axis, but the magnitude is the same as before, F = 4.8 * 10-14 N.

Test Your Understanding of Section 7.2 The figure at right shows a S uniform magnetic field B directed into the plane of the paper (shown by the blue * ’s). A particle with a negative charge moves in the plane. Which of the three paths—1, 2, or 3—does the particle follow?

Path 1

S

B

S

v

Path 2

Path 3

y

7.3

Magnetic Field Lines and Magnetic Flux

We can represent any magnetic field by magnetic field lines, just as we did for the earth’s magnetic field in Fig. 7.3. The idea is the same as for the electric field lines we introduced in Section 1.6. We draw the lines so that the line through any S point is tangent to the magnetic field vector B at that point (Fig. 7.11). Just as with electric field lines, we draw only a few representative lines; otherwise, the lines would fill up all of space. Where adjacent field lines are close together, the field magnitude is large; where these field lines are far apart, the field magnitude S is small. Also, because the direction of B at each point is unique, field lines never intersect. CAUTION Magnetic field lines are not “lines of force” Magnetic field lines are sometimes called “magnetic lines of force,” but that’s not a good name for them; unlike electric field lines, they do not point in the direction of the force on a charge (Fig. 7.12). Equation (7.2) shows that the force on a moving charged particle is always perpendicular to the magnetic field, and hence to the magnetic field line that passes through the particle’s position. The direction of the force depends on the particle’s velocity and the sign of its charge, so just looking at magnetic field lines cannot in itself tell you the direction of the force on an arbitrary moving charged particle. Magnetic field lines do have the direction that a compass needle would point at each location; this may help you to visualize them. y

7.11 The magnetic field lines of a permanent magnet. Note that the field lines pass through the interior of the magnet. At each point, the field line is tangent to the magnetic S field vector B.

The more densely the field lines are packed, the stronger the field is at that point. S

B S

B

S

At each point, the field lines point in the same direction a compass would . . .

Figures 7.11 and 7.13 show magnetic field lines produced by several common sources of magnetic field. In the gap between the poles of the magnet shown in Fig. 7.13a, the field lines are approximately straight, parallel, and equally spaced, showing that the magnetic field in this region is approximately uniform (that is, constant in magnitude and direction).

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N

. . . therefore, magnetic field lines point away from N poles and toward S poles.

214

CHAPTER 7 Magnetic Field and Magnetic Forces

Because magnetic-field patterns are three-dimensional, it’s often necessary to draw magnetic field lines that point into or out of the plane of a drawing. To do this we use a dot 1 # 2 to represent a vector directed out of the plane and a cross 1*2 to represent a vector directed into the plane (Fig. 7.13b). To remember these, think of a dot as the head of an arrow coming directly toward you, and think of a cross as the feathers of an arrow flying directly away from you. Iron filings, like compass needles, tend to align with magnetic field lines. Hence they provide an easy way to visualize field lines (Fig. 7.14).

7.12 Magnetic field lines are not “lines of force.” S

B S

F WRONG

Magnetic field lines are not “lines of force.” The force on a charged particle is not along the direction of a field line.

We define the magnetic flux £ B through a surface just as we defined electric flux in connection with Gauss’s law in Section 2.2. We can divide any surface into elements of area dA (Fig. 7.15). For each element we determine B ' , the component S of B normal to the surface at the position of that element, as shown. From the figS ure, B' = B cos f, where f is the angle between the direction of B and a line perpendicular to the surface. (Be careful not to confuse f with £ B.2 In general, this

S

S

F

Magnetic Flux and Gauss’s Law for Magnetism

RIGHT!

B

'

S

v

'

The direction of the magnetic force depends S on the velocity v, asSexpressed by the S S magnetic force law F 5 qv 3 B.

7.13 Magnetic field lines produced by some common sources of magnetic field. (a) Magnetic field of a C-shaped magnet

(b) Magnetic field of a straight current-carrying wire

Between flat, parallel magnetic poles, the magnetic field is nearly uniform. S

To represent a field coming out of or going into the plane of the paper, we use dots and crosses, respectively.

B

S

B directed out of plane I

S

B

I

I

I

S

B directed into plane

S

B Perspective view

Wire in plane of paper

(c) Magnetic fields of a current-carrying loop and a current-carrying coil (solenoid) I I

S

B Notice that the field of the loop and, especially, that of the coil look like the field of a bar magnet (see Fig. 7.11).

I I

S

B

7.14 (a) Like little compass needles, iron filings line up tangent to magnetic field lines. (b) Drawing of the field lines for the situation shown in (a).

(a)

(b)

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S

B

215

7.3 Magnetic Field Lines and Magnetic Flux

component varies from point to point on the surface. We define the magnetic flux d£ B through this area as S

#

7.15 The magnetic flux through an area element dA is defined to be d£ B = B' dA. '

S

d£ B = B' dA = B cos f dA = B dA '

(7.5) S

B' f

The total magnetic flux through the surface is the sum of the contributions from the individual area elements: £B =

2

B' dA = '

2

B cos f dA =

S

2

#

S

B dA

(magnetic flux through a surface)

B

S

dA Bi

(7.6)

dA

(This equation uses the concepts of vector area and surface integral that we introduced in Section 2.2; you may want to review that discussion.) S Magnetic flux is a scalar quantity. If B is uniform over a plane surface with total area A, then B' and f are the same at all points on the surface, and £ B = B' A = BA cos f '

(7.7)

S

If B happens to be perpendicular to the surface, then cos f = 1 and Eq. (7.7) reduces to £ B = BA. We will use the concept of magnetic flux extensively during our study of electromagnetic induction in Chapter 9. The SI unit of magnetic flux is equal to the unit of magnetic field (1 T) times the unit of area 11 m2). This unit is called the weber (1 Wb), in honor of the German physicist Wilhelm Weber (1804–1891):

PhET: Magnet and Compass PhET: Magnets and Electromagnets

1 Wb = 1 T # m2 Also, 1 T = 1 N>A # m, so 1 Wb = 1 T # m2 = 1 N # m>A In Gauss’s law the total electric flux through a closed surface is proportional to the total electric charge enclosed by the surface. For example, if the closed surface encloses an electric dipole, the total electric flux is zero because the total charge is zero. (You may want to review Section 2.3 on Gauss’s law.) By analogy, if there were such a thing as a single magnetic charge (magnetic monopole), the total magnetic flux through a closed surface would be proportional to the total magnetic charge enclosed. But we have mentioned that no magnetic monopole has ever been observed, despite intensive searches. We conclude: The total magnetic flux through a closed surface is always zero.

Symbolically, S

B

#

S

B dA = 0

(magnetic flux through any closed surface)

(7.8)

This equation is sometimes called Gauss’s law for magnetism. You can verify it by examining Figs. 7.11 and 7.13; if you draw a closed surface anywhere in any of the field maps shown in those figures, you will see that every field line that enters the surface also exits from it; the net flux through the surface is zero. It also follows from Eq. (7.8) that magnetic field lines always form closed loops. CAUTION Magnetic field lines have no ends Unlike electric field lines that begin and end on electric charges, magnetic field lines never have end points; such a point would indicate the presence of a monopole. You might be tempted to draw magnetic field lines that begin at the north pole of a magnet and end at a south pole. But as Fig. 7.11 shows, the field lines of a magnet actually continue through the interior of the magnet. Like all other magnetic field lines, they form closed loops. y

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216

CHAPTER 7 Magnetic Field and Magnetic Forces

For Gauss’s law, which always deals with closed surfaces, the vector area eleS ment dA in Eq. (7.6) always points out of the surface. However, some applications of magnetic flux involve an open surface with a boundary line; there is then an ambiguity of sign in Eq. (7.6) because of the two possible choices of direction S for dA. In these cases we choose one of the two sides of the surface to be the “positive” side and use that choice consistently. If the element of area dA in Eq. (7.5) is at right angles to the field lines, then B' = B; calling the area dA' , we have '

B =

d£ B dA'

(7.9)

That is, the magnitude of magnetic field is equal to flux per unit area acrossS an area at right angles to the magnetic field. For this reason, magnetic field B is sometimes called magnetic flux density.

Example 7.2

Magnetic flux calculations

Figure 7.16a is a perspective viewSof a flat surface with area 3.0 cm2 in a uniform magnetic field B. The magnetic flux through this surface is +0.90 mWb. Find the magnitude of the magnetic S field and the direction of the area vector A. S

7.16 (a) AS flat area A in a uniform magnetic field B. (b) The S S area vector A makes a 60° angle with B. (If we had chosen A to point in the opposite direction, f would have been 120° and the magnetic flux £ B would have been negative.) (b) Our sketch of the problem (edge-on view)

(a) Perspective view

SOLUTION IDENTIFY and SET UP: Our target variables are Sthe field magnitude B and the direction of the area vector. Because B is uniform, B and f are the same at all points on the surface. Hence we can use Eq. (7.7), £ B = BA cos f. S

EXECUTE: The area A is 3.0 * 10-4 m2; the direction of A is perpendicular to the surface, so f could be either 60° or 120°. But £ B, B, and A are all positive, so cos f must also be positive. This rules out 120°, so f = 60° (Fig. 7.16b). Hence we find B =

S

B 30°

£B 0.90 * 10-3 Wb = = 6.0 T A cos f 13.0 * 10-4 m221cos 60°2

EVALUATE: In many problems we are asked to calculate the flux of a given magnetic field through a given area. This example is somewhat different: It tests your understanding of the definition of magnetic flux.

A

Test Your Understanding of Section 7.3 Imagine moving along the axis of the current-carrying loop in Fig. 7.13c, starting at a point well to the left of the loop and ending at a point well to the right of the loop. (a) How would the magnetic field strength vary as you moved along this path? (i) It would be the same at all points along the path; (ii) it would increase and then decrease; (iii) it would decrease and then increase. (b) Would the magnetic field direction vary as you moved along the path?

7.4

y

Motion of Charged Particles in a Magnetic Field

When a charged particle moves in a magnetic field, it is acted on by the magnetic force given by Eq. (7.2), and the motion is determined by Newton’s laws. Figure 7.17a shows a simple example. A particle with positive charge q is at point O, S S moving with velocity v in a uniform magnetic field B directed into the plane of S S the figure. The vectors and B are perpendicular, so the magnetic force v S S S F 5 qv 3 B has magnitude F = qyB and a direction as shown in the figure. S The force is always perpendicular to v, so it cannot change the magnitude of the velocity, only its direction. To put it differently, the magnetic force never has a

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7.4 Motion of Charged Particles in a Magnetic Field

component parallel to the particle’s motion, so the magnetic force can never do work on the particle. This is true even if the magnetic field is not uniform. Motion of a charged particle under the action of a magnetic field alone is always motion with constant speed.

Using this principle, we see that in the situation shown in Fig. 7.17 the magniS S tudes of both F and v are constant. At points such as P and S the directions of force and velocity have changed as shown, but their magnitudes are the same. The particle therefore moves under the influence of a constant-magnitude force that is always at right angles to the velocity of the particle. Comparing the discussion of circular motion in Sections 3.4 and 5.4, we see that the particle’s path is a circle, traced out with constant speed y. The centripetal acceleration is y2>R and only the magnetic force acts, so from Newton’s second law,

7.17 A charged particle moves in a plane perpendicular to a uniform S magnetic field B. S

A charge moving at right angles to a uniform B in a circle at constant speed field moves S S because F and v are always perpendicular to each other. S

v

S S

F R

S

S

v

F S

y2 F = ƒ q ƒ yB = m R

217

P

F (7.10)

S

O

S

B

v

where m is the mass of the particle. Solving Eq. (7.10) for the radius R of the circular path, we find R =

my ƒqƒB

(radius of a circular orbit in a magnetic field)

(7.11)

We can also write this as R = p> ƒ q ƒ B, where p = my is the magnitude of the particle’s momentum. If the charge q is negative, the particle moves clockwise around the orbit in Fig. 7.17. The angular speed v of the particle can be found from Eq. (9.13), y = Rv. Combining this with Eq. (7.11), we get v =

ƒqƒB ƒqƒB y = = y my m R

(7.12)

The number of revolutions per unit time is ƒ = v>2p. This frequency ƒ is independent of the radius R of the path. It is called the cyclotron frequency; in a particle accelerator called a cyclotron, particles moving in nearly circular paths are given a boost twice each revolution, increasing their energy and their orbital radii but not their angular speed or frequency. Similarly, one type of magnetron, a common source of microwave radiation for microwave ovens and radar systems, emits radiation with a frequency equal to the frequency of circular motion of electrons in a vacuum chamber between the poles of a magnet. If the direction of the initial velocity is not perpendicular to the field, the velocity component parallel to the field is constant because there is no force parallel to the field. Then the particle moves in a helix (Fig. 7.18). The radius of the helix is given by Eq. (7.11), where y is now the component of velocity perpendiS cular to the B field. Motion of a charged particle in a nonuniform magnetic field is more complex. Figure 7.19 shows a field produced by two circular coils separated by some distance. Particles near either coil experience a magnetic force toward the center of the region; particles with appropriate speeds spiral repeatedly from one end of the region to the other and back. Because charged particles can be trapped in such a magnetic field, it is called a magnetic bottle. This technique is used to confine very hot plasmas with temperatures of the order of 106 K. In a similar way the earth’s nonuniform magnetic field traps charged particles coming from the sun in doughnut-shaped regions around the earth, as shown in Fig. 7.20. These regions, called the Van Allen radiation belts, were discovered in 1958 using data obtained by instruments aboard the Explorer I satellite.

7.18 The general case of a charged partiS cle moving in a uniform magnetic field B. The magnetic field does no work on the particle, so its speed and kinetic energy remain constant. This particle’s motion has components both parallel (vi) and perpendicular (v') to the magnetic field, so it moves in a helical path. y S

v'

v

vi

z

q

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S

B

x

218

CHAPTER 7 Magnetic Field and Magnetic Forces S

7.19 A magnetic bottle. Particles near either end of the region experience a magnetic force toward the center of the region. This is one way of containing an ionized gas that has a temperature of the order of 10 6 K, which would vaporize any material container.

S

v

B

S

v S

B S

F

+ I

S

F

I

S

F Coil 1

Coil 2 S

B

S

v

7.20 (a) The Van Allen radiation belts around the earth. Near the poles, charged particles from these belts can enter the atmosphere, producing the aurora borealis (“northern lights”) and aurora australis (“southern lights”). (b) A photograph of the aurora borealis. 7.21 This bubble chamber image shows the result of a high-energy gamma ray (which does not leave a track) that collides with an electron in a hydrogen atom. This electron flies off to the right at high speed. Some of the energy in the collision is transformed into a second electron and a positron (a positively charged electron). A magnetic field is directed into the plane of the image, which makes the positive and negative particles curve off in different directions. Slow-moving positron (q . 0)

Path of incoming gamma ray

S

B

Hydrogen atom

Slow-moving electron (q , 0)

Fast-moving electron (q , 0)

Problem-Solving Strategy 7.2

(b)

(a) Charged particles from sun enter earth’s magnetic field

Protons trapped in inner radiation belts

North Pole

South Pole

Magnetic forces on charged particles play an important role in studies of elementary particles. Figure 7.21 shows a chamber filled with liquid hydrogen and with a magnetic field directed into the plane of the photograph. A high-energy gamma ray dislodges an electron from a hydrogen atom, sending it off at high speed and creating a visible track in the liquid hydrogen. The track shows the electron curving downward due to the magnetic force. The energy of the collision also produces another electron and a positron (a positively charged electron). Because of their opposite charges, the trajectories of the electron and the positron curve in opposite directions. As these particles plow through the liquid hydrogen, they collide with other charged particles, losing energy and speed. As a result, the radius of curvature decreases as suggested by Eq. (7.11). (The electron’s speed is comparable to the speed of light, so Eq. (7.11) isn’t directly applicable here.) Similar experiments allow physicists to determine the mass and charge of newly discovered particles.

Motion in Magnetic Fields

IDENTIFY the relevant concepts: In analyzing the motion of a charged particle in electric and magnetic fields, you will apply S S Newton’s second law of motion, a F 5 ma , with the net force S S S S given by a F 5 q1E 1 v 3 B2. Often other forces such as gravity can be neglected. Many of the problems are similar to the trajectory and circular-motion problems in Sections 3.3, 3.4, and 5.4; it would be a good idea to review those sections. SET UP the problem using the following steps: 1. Determine the target variable(s). 2. Often the use of components is the most efficient approach. Choose a coordinate system and then express all vector quanti-

ties (including , and ) in terms of their components in this system. EXECUTE the solution as follows: 1. If the particle moves perpendicular to a uniform magnetic field, the trajectory is a circle with a radius and angular speed given by Eqs. (7.11) and (7.12), respectively. 2. If Syour calculation involves a more complex trajectory, use S gF 5 ma in component form: a Fx = max, and so forth. This approach is particularly useful when both electric and magnetic fields are present. EVALUATE your answer: Check whether your results are reasonable.

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7.4 Motion of Charged Particles in a Magnetic Field

Example 7.3

Electron motion in a magnetron

A magnetron in a microwave oven emits electromagnetic waves with frequency ƒ = 2450 MHz. What magnetic field strength is required for electrons to move in circular paths with this frequency?

EXECUTE: The angular speed that corresponds to the frequency ƒ is v = 2pƒ = 12p212450 * 106 s-12 = 1.54 * 1010 s-1. Then from Eq. (7.12), B =

S O L U T IO N IDENTIFY and SET UP: The problem refers to circular motion as shown in Fig. 7.17a. We use Eq. (7.12) to solve for the field magnitude B.

Example 7.4

mv

19.11 * 10-31 kg211.54 * 1010 s-12

=

1.60 * 10-19 C

ƒqƒ

'

EVALUATE: This is a moderate field strength, easily produced with a permanent magnet. Incidentally, 2450-MHz electromagnetic waves are useful for heating and cooking food because they are strongly absorbed by water molecules.

R =

S

S

IDENTIFY and SET UP: The magnetic force is F 5 qv 3 B; NewS ton’s second law gives the resulting acceleration. Because F is perS pendicular to v, the proton’s speed does not change. Hence Eq. (7.11) gives the radius of the helical trajectory if we replace y with S the velocity component perpendicular to B. Equation (7.12) gives the angular speed v, which yields the time T for one revolution (the period). Given the velocity component parallel to the magnetic field, we can then determine the pitch. S S EXECUTE: (a) With B 5 B≤n and v 5 yxn≤ 1 yz kN , Eq. (7.2) yields S S S F 5 qv 3 B 5 q1yxn≤ 1 yz kN 2 3 B≤n 5 qyzB≥n

5 11.60 * 10-19 C212.00 * 105 m> s210.500 T2n≥ '

-14

'

'

N2≥n

myz

11.67 * 10-27 kg212.00 * 105 m> s2 '

=

ƒqƒB

= 4.18 * 10

-3

11.60 * 10

'

C210.500 T2

m = 4.18 mm

ƒqƒB m

=

11.60 * 10-19 C210.500 T2 1.67 * 10-27 kg

= 4.79 * 107 rad> s '

'

The period is T = 2p>v = 2p>14.79 * 107 s-12 = 1.31 * 10-7 s. The pitch is the distance traveled along the x-axis in this time, or yxT = 11.50 * 105 m> s211.31 * 10-7 s2 '

'

= 0.0197 m = 19.7 mm

EVALUATE: Although the magnetic force has a tiny magnitude, it produces an immense acceleration because the proton mass is so small. Note that the pitch of the helix is almost five times greater than the radius R, so this helix is much more “stretched out” than that shown in Fig. 7.18.

(Recall that that n≤ 3 n≤ 5 0 and kN 3 n≤ 5 n≥ .) The resulting acceleration is S

F 1.60 * 10-14 N n≥ 5 19.58 * 1012 m> s22≥n 5 m 1.67 * 10-27 kg '

-19

From Eq. (7.12) the angular speed is v =

S

5 11.60 * 10

S

(b) Since yy = 0, the component of velocity perpendicular to B is yz; then from Eq. (7.11),

'

S O L U T IO N

a5

= 0.0877 T

Helical particle motion in a magnetic field

In a situation like that shown in Fig. 7.18, the charged particle is a proton 1q = 1.60 * 10-19 C, m = 1.67 * 10-27 kg2 and the uniform, 0.500-T magnetic field is directed along the x-axis. At t 5 0 the proton has velocity components yx = 1.50 * 105 m> s, yy = 0, and yz = 2.00 * 105 m>s. Only the magnetic force acts on the proton. (a) At t = 0, find the force on the proton and its acceleration. (b) Find the radius of the resulting helical path, the angular speed of the proton, and the pitch of the helix (the distance traveled along the helix axis per revolution).

S

219

'

Test Your Understanding of Section 7.4 (a) If you double the speed of the charged particle in Fig. 7.17a while keeping the magnetic field the same (as well as the charge and the mass), how does this affect the radius of the trajectory? (i) The radius is unchanged; (ii) the radius is twice as large; (iii) the radius is four times as large; (iv) the radius is 12 as large; (v) the radius is 41 as large. (b) How does this affect the time required for one complete circular orbit? (i) The time is unchanged; (ii) the time is twice as long; (iii) the time is four times as long; (iv) the time is 12 as long; (v) the time is 14 as long. y

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220

CHAPTER 7 Magnetic Field and Magnetic Forces

7.5

Applications of Motion of Charged Particles

This section describes several applications of the principles introduced in this chapter. Study them carefully, watching for applications of Problem-Solving Strategy 7.2 (Section 7.4).

Velocity Selector 7.22 (a) A velocity selector for charged S S particles uses perpendicular E and B fields. Only charged particles with y = E>B move through undeflected. (b) The electric and magnetic forces on a positive charge. The forces are reversed if the charge is negative. (a) Schematic diagram of velocity selector Source of charged particles q S

v

+ S

+

B

By the right-hand rule, S the force of the B field on the charge points to the right.

S S

+

E

+

The force of the E field on the charge points to the left.

In a beam of charged particles produced by a heated cathode or a radioactive material, not all particles move with the same speed. Many applications, however, require a beam in which all the particle speeds are the same. Particles of a specific speed can be selected from the beam using an arrangement of electric and magnetic fields called a velocity selector. In Fig. 7.22a a charged particle with mass m, charge q, and speed y enters a region of space where the electric and magnetic fields are perpendicular to the particle’s velocity and to each other. S S The electric field E is to the left, and the magnetic field B is into the plane of the figure. If q is positive, the electric force is to the left, with magnitude qE, and the magnetic force is to the right, with magnitude qyB. For given field magnitudes E and B, for a particular value of y the electric and magnetic forces will be equal in magnitude; the total force is then zero, and the particle travels in a straight line with constant velocity. For zero total force, gFy = 0, we need -qE + qyB = 0; solving for the speed y for which there is no deflection, we find y =

+ + +

For a negative charge, the directions of both forces are reversed.

(b) Free-body diagram for a positive particle FE 5 qE

Only if a charged FB 5 qvB particle has v 5 E B do the electric and magnetic forces cancel. All other v particles are deflected.

/

7.23 Thomson’s apparatus for measuring the ratio e/m for the electron.

E B

(7.13)

Only particles with speeds equal to E>B can pass through without being deflected by the fields (Fig. 7.22b). By adjusting E and B appropriately, we can select particles having a particular speed for use in other experiments. Because q divides out in Eq. (7.13), a velocity selector for positively charged particles also works for electrons or other negatively charged particles.

Thomson’s e/m Experiment In one of the landmark experiments in physics at the end of the 19th century, J. J. Thomson (1856–1940) used the idea just described to measure the ratio of charge to mass for the electron. For this experiment, carried out in 1897 at the Cavendish Laboratory in Cambridge, England, Thomson used the apparatus shown in Fig. 7.23. In a highly evacuated glass container, electrons from the hot cathode are accelerated and formed into a beam by a potential difference V between the two anodes A and A¿. The speed y of the electrons is determined by the accelerating

+ V

Electrons travel from the cathode to the screen. +

Electron beam Screen

S

B A Cathode

A9

P

Anodes S

E P9 Between plates P and P9 there are mutually perpendicular, S S uniform E and B fields.

–

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221

7.5 Applications of Motion of Charged Particles

potential V. The gained kinetic energy 12 my2 equals the lost electric potential energy eV, where e is the magnitude of the electron charge: 1 2 2 my

= eV

or

y =

2eV Ç m

(7.14)

The electrons pass between the plates P and P¿ and strike the screen at the end of the tube, which is coated with a material that fluoresces (glows) at the point of impact. The electrons pass straight through the plates when Eq. (7.13) is satisfied; combining this with Eq. (7.14), we get E 2eV = B Ç m

so

E2 e = m 2VB2

(7.15)

All the quantities on the right side can be measured, so the ratio e>m of charge to mass can be determined. It is not possible to measure e or m separately by this method, only their ratio. The most significant aspect of Thomson’s e>m measurements was that he found a single value for this quantity. It did not depend on the cathode material, the residual gas in the tube, or anything else about the experiment. This independence showed that the particles in the beam, which we now call electrons, are a common constituent of all matter. Thus Thomson is credited with the first discovery of a subatomic particle, the electron. The most precise value of e>m available as of this writing is ActivPhysics 13.8: Velocity Selector

e>m = 1.7588201501442 * 1011 C>kg In this expression, (44) indicates the likely uncertainty in the last two digits, 50. Fifteen years after Thomson’s experiments, the American physicist Robert Millikan succeeded in measuring the charge of the electron precisely (see Challenge Problem 23.91). This value, together with the value of e>m, enables us to determine the mass of the electron. The most precise value available at present is m = 9.109382151452 * 10-31 kg

Mass Spectrometers Techniques similar to Thomson’s e>m experiment can be used to measure masses of ions and thus measure atomic and molecular masses. In 1919, Francis Aston (1877–1945), a student of Thomson’s, built the first of a family of instruments called mass spectrometers. A variation built by Bainbridge is shown in Fig. 7.24. Positive ions from a source pass through the slits S1 and S2, forming a narrow S S beam. Then the ions pass through a velocity selector with crossed E and B fields, as we have described, to block all ions except those withSspeeds y equal to E>B. Finally, the ions pass into a region with a magnetic field B ¿ perpendicular to the figure, where they move in circular arcs with radius R determined by Eq. (7.11): R = my>qB¿. Ions with different masses strike the detector (in Bainbridge’s design, a photographic plate) at different points, and the values of R can be measured. We assume that each ion has lost one electron, so the net charge of each ion is just +e. With everything known in this equation except m, we can compute the mass m of the ion. One of the earliest results from this work was the discovery that neon has two species of atoms, with atomic masses 20 and 22 g>mol. We now call these species isotopes of the element. Later experiments have shown that many elements have several isotopes, atoms that are identical in their chemical behavior but different in mass owing to differing numbers of neutrons in their nuclei. This is just one of the many applications of mass spectrometers in chemistry and physics.

7.24 Bainbridge’s mass spectrometer utilizes a velocity selector to produce particles with uniform speed y. In the region of magnetic field B¿, particles with greater mass 1m2 7 m12 travel in paths with larger radius 1R2 7 R12. '

S1 S2

'

+ + + +

Velocity selector selects particles with speed v.

+ S

+ +

Particle detector

S

E, B

S3 m2 m1

R1

R2

S

B9 Magnetic field separates particles by mass; the greater a particle’s mass, the larger is the radius of its path.

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222

CHAPTER 7 Magnetic Field and Magnetic Forces

An e>m demonstration experiment

Example 7.5

You set out to reproduce Thomson’s e> m experiment with an accelerating potential of 150 V and a deflecting electric field of magnitude 6.0 * 106 N> C. (a) At what fraction of the speed of light do the electrons move? (b) What magnetic-field magnitude will yield zero beam deflection? (c) With this magnetic field, how will the electron beam behave if you increase the accelerating potential above 150 V? '

'

'

IDENTIFY and SET UP: This is the situation shown in Fig. 7.23. We use Eq. (7.14) to determine the electron speed and Eq. (7.13) to determine the required magnetic field B. EXECUTE: (a) From Eq. (7.14), the electron speed y is y = !21e> m2V = !211.76 * 1011 C> kg21150 V2 '

'

= 7.27 * 106 m> s = 0.024c '

Example 7.6

'

'

'

(c) Increasing the accelerating potential V increases the electron speed y. In Fig. 7.23 this doesn’t change the upward electric force eE, but it increases the downward magnetic force eyB. Therefore the electron beam will turn downward and will hit the end of the tube below the undeflected position. EVALUATE: The required magnetic field is relatively large because the electrons are moving fairly rapidly (2.4% of the speed of light). If the maximum available magnetic field is less than 0.83 T, the electric field strength E would have to be reduced to maintain the desired ratio E> B in Eq. (7.15). '

'

6.00 * 106 N> C E = 0.83 T = y 7.27 * 106 m> s '

B =

'

SOLUTION

'

(b) From Eq. (7.13), the required field strength is

'

'

Finding leaks in a vacuum system

There is almost no helium in ordinary air, so helium sprayed near a leak in a vacuum system will quickly show up in the output of a vacuum pump connected to such a system. You are designing a leak detector that uses a mass spectrometer to detect He+ ions (charge + e = +1.60 * 10-19 C, mass 6.65 * 10-27 kg2. The ions emerge from the velocity selector with a speed of 1.00 * 105 m> s. They are curved in a semicircular path by a magnetic field B¿ and are detected at a distance of 10.16 cm from the slit S3 in Fig. 7.24. Calculate the magnitude of the magnetic field B¿. '

'

SOLUTION

EXECUTE: The distance given is the diameter of the semicircular path shown in Fig. 7.24, so the radius is R = 12 110.16 * 10-2 m2. From Eq. (7.11), R = my > qB¿, we get '

'

16.65 * 10-27 kg211.00 * 105 m> s2 my = qR 11.60 * 10-19 C215.08 * 10-2 m2 '

B¿ =

'

= 0.0818 T

EVALUATE: Helium leak detectors are widely used with highvacuum systems. Our result shows that only a small magnetic field is required, so leak detectors can be relatively compact.

IDENTIFY and SET UP: After it passes through the slit, the ion follows a circular path as described in Section 7.4 (see Fig. 7.17). We solve Eq. (7.11) for B¿ .

ActivPhysics 13.7: Mass Spectrometer

7.25 Forces on a moving positive charge in a current-carrying conductor.

7.6

S

J Drift velocity A of charge carriers S

vd l

S

F

q

S

S

J

Test Your Understanding of Section 7.5 In Example 7.6 He+ ions with charge + e move at 1.00 * 105 m/s in a straight line through a velocity selector. Suppose the He+ ions were replaced with He2+ ions, in which both electrons have been removed from the helium atom and the ion charge is +2e. At what speed must the He2+ ions travel through the same velocity selector in order to move in a straight line? (i) about 4.00 * 105 m/s; (ii) about 2.00 * 105 m/s; (iii) 1.00 * 105 m/s; (iv) about 0.50 * 105 m/s; (v) about 0.25 * 105 m/s. y

B

Magnetic Force on a Current-Carrying Conductor

What makes an electric motor work? Within the motor are conductors that carry currents (that is, whose charges are in motion), as well as magnets that exert forces on the moving charges. Hence there is a magnetic force along the length of each current-carrying conductor, and these forces make the motor turn. The moving-coil galvanometer that we described in Section 6.3 also uses magnetic forces on conductors. We can compute the force on a current-carrying conductor starting with the magS S S netic force F 5 qv 3 B on a single moving charge. Figure 7.25 shows a straight

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7.6 Magnetic Force on a Current-Carrying Conductor

segment of a conducting wire, with length l and cross-sectionalSarea A; the current is from bottom to top. The wire is in a uniform magnetic field B, perpendicular to the plane of the diagram and directed into the plane. Let’s assume first that the moving charges are positive. Later we’ll see what happensSwhen they are negative. S The drift velocity vdSis upward, perpendicular to B. The average force on each S S S charge is F 5 qvd 3 B, directed to the left as shown in the figure; since vd and S B are perpendicular, the magnitude of the force is F = qydB. We can derive an expression for the total force on all the moving charges in a length l of conductor with cross-sectional area A using the same language we used in Eqs. (5.2) and (5.3) of Section 5.1. The number of charges per unit volume is n; a segment of conductor with length l Shas volume Al and contains a number of charges equal to nAl. The total force F on all the moving charges in this segment has magnitude F = 1nAl21qydB2 = 1nqydA21lB2 '

7.26 A straight wire segment of length S I in the l carries a current S direction of l . The magnetic forceS on this segment is perpendicular to both l and S the magnetic field B. S

Force F on a straight wire carrying a positive current and oriented at an angle f to a S magnetic field B: • Magnitude is F 5 IlB' 5 IlB sin f. S • Direction of F is given by the right-hand rule. S

(7.16)

F

From Eq. (5.3) the current density is J = nqyd. The product JA is the total current I, so we can rewrite Eq. (7.16) as F = IlB

B' 5 B sin f

(7.17)

S

F = IlB' = IlB sin f

(7.18)

The force is always perpendicular to both the conductor and the field, with the direction determined by the same right-hand rule we used for a moving positive charge (Fig. 7.26). Hence this force can be expressed as a vector product, just like the force on a single moving charge. We represent the segmentSof wire with a vector S l along the wire in the direction of the current; then the force F on this segment is

B

f Bi

S

If the B field is not perpendicular to the wire but makes an angle f with it, we handle the situationSthe same way we did in Section 7.2 for a single charge. Only the component of B perpendicular to the wire (and to the drift velocities of the charges) exerts a force; this component is B' = B sin f. The magnetic force on the wire segment is then

223

S

l

I

S

S

7.27S Magnetic field B, length l , and force F vectors for a straight wire carrying a current I. (a)

y S

F S

B S

S

S

F 5 Il 3 B

(magnetic force on a straight wire segment)

f

(7.19) S

S S

z (b) y

S

S

S

S

dF 5 I d l 3 B

(magnetic force on an infinitesimal wire section)

I

l

S

Figure 7.27 illustrates the directions of B, l , and F for several cases. S If the conductor is not straight, we can divide it into infinitesimal segments d l . S The force dF on each segment is (7.20)

Reversing B reverses the force direction.

Then we can integrate this expression along the wire to find the total force on a conductor of any shape. The integral is a line integral, the same mathematical operation we have used to define work (Section 6.3) and electric potential (Section 3.2). CAUTION Current is not a vector Recall from Section 5.1 that the current I is not a vecS tor. The direction of current flow is described by d l , not I. If the conductor is curved, the S current I is the same at all points along its length, but d l changes direction so that it is always tangent to the conductor. y

Finally, what happens when the moving charges are negative, such as electrons in a metal? Then in Fig. 7.25 an upward current corresponds to a downward S drift velocity. But because q is now negative, the direction of the force F is the same as before. Thus Eqs. (7.17) through (7.20) are valid for both positive and negative charges and even when both signs of charge are present at once. This happens in some semiconductor materials and in ionic solutions. A common application of the magnetic forces on a current-carrying wire is found in loudspeakers (Fig. 7.28). The radial magnetic field created by the permanent

x

S

B x

S

l

f

I

z

S

F

(c) Reversing the current [relative to (b)] y reverses the force direction.

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S

F S

l f x

S

B

I z

224

CHAPTER 7 Magnetic Field and Magnetic Forces

7.28 (a) Components of a loudspeaker. (b) The permanent magnet creates a magnetic field that exerts forces on the current in the voice coil; for a current I in the direction shown, the force is to the right. If the electric current in the voice coil oscillates, the speaker cone attached to the voice coil oscillates at the same frequency. (b)

(a)

S

Magnets Basket

Signal from amplifier

ActivPhysics 13.5: Magnetic Force on Wire

Example 7.7

B field of permanent magnet

Rigid speaker cone

Voice coil

Direction of motion of voice coil and speaker cone

Flexible suspension ring

I

Current in voice coil

magnet exerts a force on the voice coil that is proportional to the current in the coil; the direction of the force is either to the left or to the right, depending on the direction of the current. The signal from the amplifier causes the current to oscillate in direction and magnitude. The coil and the speaker cone to which it is attached respond by oscillating with an amplitude proportional to the amplitude of the current in the coil. Turning up the volume knob on the amplifier increases the current amplitude and hence the amplitudes of the cone’s oscillation and of the sound wave produced by the moving cone.

Magnetic force on a straight conductor

A straight horizontal copper rod carries a current of 50.0 A from west to east in a region between the poles of a large electromagnet. In this region there is a horizontal magnetic field toward the northeast (that is, 45° north of east) with magnitude 1.20 T. (a) Find the magnitude and direction of the force on a 1.00-m section of rod. (b) While keeping the rod horizontal, how should it be oriented to maximize the magnitude of the force? What is the force magnitude in this case? SOLUTION IDENTIFY and SET UP: Figure 7.29 shows the situation. This is a straight wire segment in a uniform magnetic field, as in Fig. 7.26. S Our target variables are the force F on the segment and the angle f for which the force magnitude F is greatest. We find the magnitude of the magnetic force using Eq. (7.18) and the direction from the right-hand rule. EXECUTE: (a) The angle f between the directions of current and field is 45°. From Eq. (7.18) we obtain 7.29 Our sketch of the copper rod as seen from overhead.

F = IlB sin f = 150.0 A211.00 m211.20 T21sin 45°2 = 42.4 N

The direction of the force is perpendicular to the plane of the current and the field, both of which lie in the horizontal plane. Thus the force must be vertical; the right-hand rule shows that it is vertically upward (out of the plane of the figure). S (b) From F = IlB sin f, F is maximum for f = 90°, so that l S S S S and B are perpendicular. To keep F 5 I l 3 B upward, we rotate the rod clockwise by 45° from its orientation in Fig. 7.29, so that the current runs toward the southeast. Then F = IlB = 150.0 A211.00 m211.20 T2 = 60.0 N. EVALUATE: You can check the result in part (a) by using Eq. (7.19) to calculate the force vector. If we use a coordinate system with the x-axis S pointing east, the Sy-axis north, and the z-axis upward, we have l 5 11.00 m2≤n, B 5 11.20 T231cos 45°2≤n 1 1sin 45°2≥n4, and S

S

S

F 5 Il 3 B

5 150 A211.00 m2≤n 3 11.20 T231cos 45°2≤n 1 1sin 45°2≥n4 5 142.4 N2kN

Note that the maximum upward force of 60.0 N can hold the conductor in midair against the force of gravity—that is, magnetically levitate the conductor—if its weight is 60.0 N and its mass is m = w> g = 160.0 N2> 19.8 m> s22 = 6.12 kg. Magnetic levitation is used in some high-speed trains to suspend the train over the tracks. Eliminating rolling friction in this way allows the train to achieve speeds of over 400 km>h. '

'

'

'

'

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225

7.7 Force and Torque on a Current Loop

Example 7.8

Magnetic force on a curved conductor S

In Fig. 7.30 the magnetic field B is uniform and perpendicular to the plane of the figure, pointing out of the page. The conductor, carrying current I to the left, has three segments: (1) a straight segment with length L perpendicular to the plane of the figure, (2) a semicircle with radius R, and (3) another straight segment with length L parallel to the x-axis. Find the total magnetic force on this conductor. S O L U T IO N S IDENTIFY and SET SUP: TheS magnetic field B 5 BkN is uniform, so we find the forces F1 and F3 on the straight segments (1) and (3) using Eq. (7.19). We divide the curved segment (2) into infinitesiS mal straight segments and find the corresponding force dF2 on each straight segment using Eq. (7.20). We then integrate toSfind S total magnetic force on the conductor is then F 5 F . The S2 S S F1 1 F2 1 F3.

S EXECUTE: For segment (1), L 5 -LkNS. Hence fromS Eq. (7.19), S S S S F 5 IL 3 B 5 0. For segment (3), L 5 - L≤n, so F3 5 IL 3 S1 B 5 I1-L≤n2 3 1BkN 2 5 ILB≥n. S For the curved segment (2), Fig. 7.20 shows a segment d l with length dl = RSdu, atS angle u. The right-hand rule shows that the direction of d l 3 B is radially outward from the center; make S S sure you can verify this. Because d l and B are perpendicular, the S magnitude dF2 of the force on the segment d l is just dF2 = I dl B = I1R du2B. The components of the force on this segment are

dF2x = IR du B cos u

dF2y = IR du B sin u

To find the components of the total force, we integrate these expressions with respect to θ from u = 0 to u = p to take in the whole semicircle. The results are p

F2x = IRB

cos u du = 0

20 p

7.30 What is the total magnetic force on the conductor?

F2y = IRB

y

20

sin u du = 2IRB

S

dFy

Hence F2 5 2IRB≥n. Finally, adding the forces on all three segments, we find that the total force is in the positive y-direction:

S

F 5 F1 1 F2 1 F3 5 0 1 2IRB≥n 1 ILB≥n 5 IB12R + L2≥n

dF

S

S

B (out)

dl

I

u dl

S

F

R L L

u O

dFx I

du

S

I

S

S

S

'

S

x I (in)

EVALUATE: We Scould have predicted from symmetry that the x-component of F2 would be zero: On the right half of the semicircle the x-component of the force is positive (to the right) and on the left half it is negative (to the left); the positive and negative contriS butions to the integral cancel. The result is that F2 is the force that would be exerted if we replaced the semicircle with a straight segment of length 2R along the x-axis. Do you see why?

Test Your Understanding of Section 7.6 The figure at right shows a top view of two conducting rails on which a conducting bar can slide. A uniform magnetic field is directed perpendicular to the plane of the figure as shown. A battery is to be connected to the two rails so that when the switch is closed, current will flow through the bar and cause a magnetic force to push the bar to the right. In which orientation, A or B, should the battery be placed in the circuit?

Which orientation? A B

Switch

Conducting bar

S

Conducting rails

F

S

B

y

7.7

Force and Torque on a Current Loop

Current-carrying conductors usually form closed loops, so it is worthwhile to use the results of Section 7.6 to find the total magnetic force and torque on a conductor in the form of a loop. Many practical devices make use of the magnetic force or torque on a conducting loop, including loudspeakers (see Fig. 7.28) and galvanometers (see Section 6.3). Hence the results of this section are of substantial practical importance. These results will also help us understand the behavior of bar magnets described in Section 7.1. As an example, let’s look at a rectangular current loop in a uniform magnetic field. We can represent the loop as a series of straight line segments. We will find

ActivPhysics 13.6: Magnetic Torque on a Loop

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226

CHAPTER 7 Magnetic Field and Magnetic Forces

7.31 Finding the torque on a current-carrying loop in a uniform magnetic field. (a)

(b) z

The two pairs of forces acting on the loop cancel, so no net force acts on the loop. S

S

B I

y I

S

B

S

m

x

I

F

B 90° 2 f f f

S

2F (c)

b sin f S

B

S

f

S

S

y a

z (direction normal to loop)

m

I

S

2F

x (direction normal to loop)

S

S

S

S

B

S

B

m

S

F

y S f is the angle F9 between a vector normal to the loop and the magnetic field. I

The torque is maximal S when f 5 90° (so B is in the plane of the loop).

S

However, the forces on the a sides of the loop (F and 2F ) produce a torque z t 5 (IBa)(b sin f) on the loop.

F9

B

S

S

B

B

S

F

I I

S

b

2F9

I

S

2F

S

2F9

The torque is zero when f 5 0° (as shown here) or fS 5 180°. In both cases, x B is perpendicular to the plane of the loop. The loop is in stable equilibrium when f 5 0; it is in unstable equilibrium when f 5 180°.

that the total force on the loop is zero but that there can be a net torque acting on the loop, with some interesting properties. Figure 7.31a shows a rectangular loop of wire with side lengths a and b. A line perpendicular to the plane of the loop (i.e.,Sa normal to the plane) makes an angle f with the direction of the magnetic field B, and the loop carries a current I. The wires leading the current into and out of the loop and the source of emf are omitted to keep theSdiagram simple. The force F on the right side of Sthe loop (length a) is to the right, in the +x-direction as shown. On this side, B is perpendicular to the current direction, and the force on this side has magnitude F = IaB

(7.21)

S

A force 2F with the same magnitude but opposite direction acts on the opposite side of the loop, as shown in the figure. S The sides with length b make an angle 190° - f2 with the direction of B. The S S forces on these sides are the vectors F ¿ and 2F ¿; their magnitude F¿ is given by F¿ = IbB sin190° - f2 = IbB cos f The lines of action of both forces lie along the y-axis. The total force on the loop is zero because the forces on opposite sides cancel out in pairs. The net force on a current loop in a uniform magnetic field is zero. However, the net torque is not in general equal to zero.

(You may find it helpfulS at this point to review the discussion of torque in Section S 10.1.) The two forces F ¿ and 2F ¿ in Fig. 7.31a lie along the same line and so S S give rise to zero net torque with respect to any point. The two forces F and 2F lie along different lines, and each gives rise to a torque about the y-axis. According to the right-hand ruleS for determining the direction of torques, the vector S torques due to F and 2F are both in the +y-direction; hence the net vector S torque T is in the + y-direction as well. The moment arm for each of these forces

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227

7.7 Force and Torque on a Current Loop

(equal to the perpendicular distance from the rotation axis to the line of action of the force) is 1b>22 sin f, so the torque due to each force has magnitude F1b>22 sin f. If we use Eq. (7.21) for F, the magnitude of the net torque is t = 2F1b>22 sin f = 1IBa21b sin f2

(7.22)

S

The torque is greatest when f = 90°, B is in the plane of the loop, and the norS mal to this plane is perpendicular to B (Fig. 7.31b). The torque is zero when f is 0° or 180° and the normal to the loop is parallel or antiparallel to the field (Fig. 7.31c). The value f = 0° is a stable equilibrium position because the torque is zero there, and when the loop is rotated slightly from this position, the resulting torque tends to rotate it back toward f = 0°. The position f = 180° is an unstable equilibrium position; if displaced slightly from this position, the loop tends to move farther away from f = 180°. Figure 7.31 shows rotation about the y-axis, but because the net force on the loop is zero, Eq. (7.22) for the torque is valid for any choice of axis. The area A of the loop is equal to ab, so we can rewrite Eq. (7.22) as t = IBA sin f

(magnitude of torque on a current loop)

(7.23)

The product IA is called the magnetic dipole moment or magnetic moment of the loop, for which we use the symbol m (the Greek letter mu): m = IA

(7.24)

It is analogous to the electric dipole moment introduced in Section 1.7. In terms of m, the magnitude of the torque on a current loop is t = mB sin f

(7.25)

whereS f is the angle between the normal to the loop (the direction of the vector S area A2 and B. The torque tends to rotate the loop in the direction of decreasing f—that is, toward its stable equilibrium position in which the loop lies in the S xy-plane perpendicular to the direction of the field B (Fig. 7.31c). A current loop, or any other body that experiences a magnetic torque given by Eq. (7.25), is also called a magnetic dipole.

Magnetic Torque: Vector Form S

We can also define a vector magnetic moment M with magnitude IA: This is shown S in Fig. 7.31. The direction of M is defined to be perpendicular to the plane of the loop, with a sense determined by a right-hand rule, as shown in Fig. 7.32. Wrap the fingers of your right hand around the perimeter of the loop in the direction of the current. Then extend your thumb so that it is perpendicularSto the plane of the S loop; its direction is the direction of M (and of the vector area A of the loop). The S S torque is greatest when M and B are perpendicular and is zero when they are parS S allel or antiparallel. In the stable equilibrium position, M and B are parallel. S Finally, we can express this interaction in terms of the torque vector T, which we used for electric-dipole interactions in Section 1.7. From Eq. (7.25) the magS S S nitude of T is equal to the magnitude of M 3 B, and reference to Fig. 7.31 shows that the directions are also the same. So we have S

S

S

T5M 3 B

(vector torque on a current loop)

7.32 The right-hand rule determines the direction of the magnetic moment of a current-carrying loop. This is also the direcS S S tion of the loop’s area vector A; M 5 IA is a vector equation.

(7.26)

This result is directly analogous to the result we found in Section 1.7 for the S S torque exerted by an electric field E on an electric dipole with dipole moment p .

Potential Energy for a Magnetic Dipole When a magnetic dipole changes orientation in a magnetic field, the field does work on it. In an infinitesimal angular displacement df, the work dW is given by

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I I S

m S

A I I

228

CHAPTER 7 Magnetic Field and Magnetic Forces

t df, and there is a corresponding change in potential energy.SAs the above disS cussion suggests, the potential energy is least when M and B are parallel and greatest when they are antiparallel. To find an expression for the potential energy U as a function of orientation, we can make use of the beautiful symmetry between the electric and magnetic dipole interactions. The torque on an electric S S S dipole in an electric field is T 5 p 3 E; we found in Section 1.7 that the correS S sponding potential energy is SU = 2 p E. The torque on a magnetic dipole in a S S magnetic field is T 5 M 3 B, so we can conclude immediately that the corresponding potential energy is

#

S

#

S

U = - M B = - mB cos f

(potential energy for a magnetic dipole)

(7.27)

With this definition, U is zero when the magnetic dipole moment is perpendicular to the magnetic field.

Magnetic Torque: Loops and Coils 7.33 The collection of rectangles exactly matches the irregular plane loop in the limit as the number of rectangles approaches infinity and the width of each rectangle approaches zero. A planar current loop of any shape can be approximated by a set of rectangular loops.

I

I

I

I S

S

S

7.34 The torque T 5 M 3 B on this solenoid in a uniform magnetic field is directed straight into the page. An actual solenoid has many more turns, wrapped closely together. S

m

f I

S

t

I

S

B

The torque tends to make the solenoid rotate clockwise in the plane of the page, aligning S S magnetic moment m with field B.

Although we have derived Eqs. (7.21) through (7.27) for a rectangular current loop, all these relationships are valid for a plane loop of any shape at all. Any planar loop may be approximated as closely as we wish by a very large number of rectangular loops, as shown in Fig. 7.33. If these loops all carry equal currents in the same clockwise sense, then the forces and torques on the sides of two loops adjacent to each other cancel, and the only forces and torques that do not cancel are due to currents around the boundary. Thus all the above relationships are S valid for aS plane current loop of any shape, with the magnetic moment M given S by M 5 IA. We can also generalize this whole formulation to a coil consisting of N planar loops close together; the effect is simply to multiply each force, the magnetic moment, the torque, and the potential energy by a factor of N. An arrangement of particular interest is the solenoid, a helical winding of wire, such as a coil wound on a circular cylinder (Fig. 7.34). If the windings are closely spaced, the solenoid can be approximated by a number of circular loops lying in planes at right angles to its long axis. The total torque on a solenoid in a magnetic field is simply the sum of the torques on the individual turns. For a solenoid with N turns in a uniform field B, the magnetic moment is m = NIA and t = NIAB sin f

(7.28)

where f is the angle between the axis of the solenoid and the direction of the S field. The magnetic moment vector M is along the solenoid axis. The torque is greatest when the solenoid axis is perpendicular to the magnetic field and zero when they are parallel. The effect of this torque is to tend to rotate the solenoid into a position where its axis is parallel to the field. Solenoids are also useful as sources of magnetic field, as we’ll discuss in Chapter 8. The d’Arsonval galvanometer, described in Section 6.3, makes use of a magnetic torque on a coil carrying a current. As Fig. 6.14 shows, the magnetic field is not uniform but is radial, so the side thrusts on the coil are always perpendicular to its plane. Thus the angle f in Eq. (7.28) is always 90°, and the magnetic torque is directly proportional to the current, no matter what the orientation of the coil. A restoring torque proportional to the angular displacement of the coil is provided by two hairsprings, which also serve as current leads to the coil. When current is supplied to the coil, it rotates along with its attached pointer until the restoring spring torque just balances the magnetic torque. Thus the pointer deflection is proportional to the current. An important medical application of the torque on a magnetic dipole is magnetic resonance imaging (MRI). A patient is placed in a magnetic field of about 1.5 T, more than 10 4 times stronger than the earth’s field. The nucleus of each hydrogen atom in the tissue to be imaged has a magnetic dipole moment,

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?

7.7 Force and Torque on a Current Loop

229

which experiences a torque that aligns it with the applied field. The tissue is then illuminated with radio waves of just the right frequency to flip these magnetic moments out of alignment. The extent to which these radio waves are absorbed in the tissue is proportional to the amount of hydrogen present. Hence hydrogenrich soft tissue looks quite different from hydrogen-deficient bone, which makes MRI ideal for analyzing details in soft tissue that cannot be seen in x-ray images (see the image that opens this chapter).

Example 7.9

Magnetic torque on a circular coil S

A circular coil 0.0500 m in radius, with 30 turns of wire, lies in a horizontal plane. It carries a counterclockwise (as viewed from above) current of 5.00 A. The coil is in a uniform 1.20-T magnetic field directed toward the right. Find the magnitudes of the magnetic moment and the torque on the coil. S O L U T IO N IDENTIFY and SET UP: This problem uses the definition of magnetic moment and the expression for the torque on a magnetic dipole in a magnetic field. Figure 7.35 shows the situation. Equation (7.24) gives the magnitude m of the magnetic moment of a single turn of wire; for N turns, the magnetic moment is N times greater. Equation (7.25) gives the magnitude t of the torque.

S

The angle f between the direction of B and the direction of M (which is along the normal to the plane of the coil) is 90°. From Eq. (7.25), the torque on the coil is t = mtotalB sin f = 11.18 A # m2211.20 T21sin 90°2 = 1.41 N # m

EVALUATE: The torque tends to rotate the right side of the coil down and the left side up, into a position where the normal to its S plane is parallel to B. 7.35 Our sketch for this problem.

EXECUTE: The area of the coil is A = pr2. From Eq. (7.24), the total magnetic moment of all 30 turns is mtotal = NIA = 3015.00 A2p10.0500 m22 = 1.18 A # m2

Example 7.10

Potential energy for a coil in a magnetic field

If the coil in Example 7.9 rotates from its initial orientation to one S S in which its magnetic moment M is parallel to B, what is the change in potential energy? S O L U T IO N IDENTIFY and SET UP: Equation (7.27) gives the potential energy for each orientation. The initial position is as shown in Fig. 7.35, with f1 = 90°. In the final orientation, the coil has been rotated S S 90° clockwise so that M and B are parallel, so the angle between these vectors is f2 = 0.

EXECUTE: From Eq. (7.27), the potential energy change is ¢U = U2 - U1 = - mB cos f2 - 1-mB cos f12 = -mB1cos f2 - cos f12

= - 11.18 A # m2211.20 T21cos 0° - cos 90°2 = - 1.41 J EVALUATE: The potential energy decreases because the rotation is in the direction of the magnetic torque that we found in Example 7.9.

Magnetic Dipole in a Nonuniform Magnetic Field We have seen that a current loop (that is, a magnetic dipole) experiences zero net force in a uniform magnetic field. Figure 7.36 shows two current loops in the S nonuniform B field of a bar magnet; in both cases the net force on the loop is not S zero. In Fig. 7.36a the magnetic moment M is in the direction opposite to the S S S field, and the force dF 5 I d l 3 B on a segment of the loop has both a radial component and a component to the right. When these forces are summed to find S the net force F on the loop, the radial components cancel so that the net force is to the right, away from the magnet. Note that in this case the force is toward the region where the field lines are farther apart and the field magnitude B is less. The S S polarity of the bar magnet is reversed in Fig. 7.36b, so M and B are parallel; now

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230

CHAPTER 7 Magnetic Field and Magnetic Forces

the net force on the loop is to the left, toward the region of greater field magnitude near the magnet. Later in this section we’ll use these observations to explain why bar magnets can pick up unmagnetized iron objects.

7.36 Forces on current loops in a S nonuniform B field. In each case the axis of the bar magnet is perpendicular to the plane of the loop and passes through the center of the loop.

Magnetic Dipoles and How Magnets Work

(a) Net force on this coil is away from north pole of magnet. S net F S

B

S

dF S dF dF I S dF S

S

N

S

m

(b) Net force on same coil is toward south pole of magnet. S net F S dF N

S

S

S

dF

dF

S

m

S

B

I S

dF

7.37 (a) An unmagnetized piece of iron. (Only a few representative atomic moments are shown.) (b) A magnetized piece of iron (bar magnet). The net magnetic moment of the bar magnet points from its south pole to its north pole. (c) A bar magnet in a magnetic field. (a) Unmagnetized iron: magnetic moments are oriented randomly.

S

matom (b) In a bar magnet, the magnetic moments are aligned. S

m

N

S (c) A magnetic field creates a torque on the bar magnet that tends to align its dipole S moment with the B field. N S

t

S

B S

S

m

The behavior of a solenoid in a magnetic field (see Fig. 7.34) resembles that of a bar magnet or compass needle; if free to turn,Sboth the solenoid and the magnet orient themselves with their axes parallel to the B field. In both cases this is due to the interaction of moving electric charges with a magnetic field; the difference is that in a bar magnet the motion of charge occurs on the microscopic scale of the atom. Think of an electron as being like a spinning ball of charge. In this analogy the circulation of charge around the spin axis is like a current loop, and so the electron has a net magnetic moment. (This analogy, while helpful, is inexact; an electron isn’t really a spinning sphere. A full explanation of the origin of an electron’s magnetic moment involves quantum mechanics, which is beyond our scope here.) In an iron atom a substantial fraction of the electron magnetic moments align with each other, and the atom has a nonzero magnetic moment. (By contrast, the atoms of most elements have little or no net magnetic moment.) In an unmagnetized piece of iron there is no overall alignment of the magnetic moments of the atoms; their vector sum is zero, and the net magnetic moment is zero (Fig. 7.37a). But in an iron bar magnet the magnetic moments of many of S the atoms are parallel, and there is a substantial netSmagnetic moment M (Fig. 7.37b). If the magnet is placed in a magneticSfield B, the field exerts a torque S given by Eq. (7.26) that tends to align M with B (Fig. 7.37c). A bar magnet tends S to align with a B field so that Sa line from the south pole to the north pole of the magnet is in the direction of B; hence the real significance of a magnet’s north and south poles is that they represent the head and tail, respectively, of the magS net’s dipole moment M . The torque experienced by a current loop in a magnetic field also explains how an unmagnetized iron object like that in Fig. 7.37a becomes magnetized. If an unmagnetized iron paper clip is placed next to a powerful Smagnet, the magnetic moments of the paper clip’s atoms tend to align with the B field of the magnet. When the paper clip is removed, its atomic dipoles tend to remain aligned, and the paper clip has a net magnetic moment. The paper clip can be demagnetized by being dropped on the floor or heated; the added internal energy jostles and rerandomizes the atomic dipoles. The magnetic-dipole picture of a bar magnet explains the attractive and repulS sive forces between bar magnets shown in Fig. 7.1. The magnetic moment M of a bar magnet points from its south pole to its north pole, so the current loops in Figs. 7.36a and 7.36b are both equivalent to a magnet with its north pole on the left. Hence the situation in Fig. 7.36a is equivalent to two bar magnets with their north poles next to each other; the resultant force is repulsive, just as in Fig. 7.1b. In Fig. 7.36b we again have the equivalent of two bar magnets end to end, but with the south pole of the left-hand magnet next to the north pole of the righthand magnet. The resultant force is attractive, as in Fig. 7.1a. Finally, we can explain how a magnet can attract an unmagnetized iron object (see Fig. 7.2). It’s a two-step process. First, the atomic magnetic moments of the S iron tend to align with the B field of the magnet, so the iron acquires a net magS netic dipole moment M parallel to the field. Second, the nonuniform field of the magnet attracts the magnetic dipole. Figure 7.38a shows an example. The north pole of the magnet is closer to the nail (which contains iron), and the magnetic dipole produced in the nail is equivalent to a loop with a current that circulates in a direction opposite to that shown in Fig. 7.36a. Hence the net magnetic force on the nail is opposite to the force on the loop in Fig. 7.36a, and the nail is attracted toward the magnet. Changing the polarity of the magnet, as in Fig. 7.38b, reverses S S the directions of both B and M . The situation is now equivalent to that shown in

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7.8 The Direct-Current Motor

Fig. 7.36b; like the loop in that figure, the nail is attracted toward the magnet. Hence a previously unmagnetized object containing iron is attracted to either pole of a magnet. By contrast, objects made of brass, aluminum, or wood hardly respond at all to a magnet; the atomic magnetic dipoles of these materials, if present at all, have less tendency to align with an external field. Our discussion of how magnets and pieces of iron interact has just scratched the surface of a diverse subject known as magnetic properties of materials. We’ll discuss these properties in more depth in Section 8.8.

231

7.38 A bar magnet attracts an unmagnetized iron nail in two steps. First, the S B field of the bar magnet gives rise to a net magnetic moment in the nail. Second, because the field of the bar magnet is not uniform, this magnetic dipole is attracted toward the magnet. The attraction is the same whether the nail is closer to (a) the magnet’s north pole or (b) the magnet’s south pole.

Test Your Understanding of Section 7.7 Figure 7.13c depicts the magnetic field lines due to a circular current-carrying loop. (a) What is the direction of the magnetic moment of this loop? (b) Which side of the loop is equivalent to the north pole of a magnet, and which side is equivalent to the south pole? y

(a) r m

S

N

r

B

The Direct-Current Motor

7.8

Electric motors play an important role in contemporary society. In a motor a magnetic torque acts on a current-carrying conductor, and electric energy is converted to mechanical energy. As an example, let’s look at a simple type of directcurrent (dc) motor, shown in Fig. 7.39. The moving part of the motor is the rotor, a length of wire formed into an open-ended loop and free to rotate about an axis. The ends of the rotor wires are attached to circular conducting segments that form a commutator. In Fig. 7.39a, each of the two commutator segments makes contact with one of the terminals, or brushes, of an external circuit that includes a source of emf. This causes a current to flow into the rotor on one side, shown in red, and out of the rotor on the other S side, shown in blue. Hence the rotor is a current loop with a magnetic moment M . The rotor lies between opposing poles of a permanent magnet, so there is a magS S S S netic field B that exerts a torque T 5 M 3 B on the rotor. For the rotor orientation shown in Fig. 7.39a the torqueScauses the rotor to turn counterclockwise, in S the direction that will align M with B. In Fig. 7.39b the rotor has rotated by 90° from its orientation in Fig. 7.39a. If the current through the rotor were constant, the rotor would now be in its equilibrium orientation; it would simply oscillate around this orientation. But here’s where the commutator comes into play; each brush is now in contact with both

(b) N

r m

S

r

B

7.39 Schematic diagram of a simple dc motor. The rotor is a wire loop that is free to rotate about an axis; the rotor ends are attached to the two curved conductors that form the commutator. (The rotor halves are colored red and blue for clarity.) The commutator segments are insulated from one another. (a) Brushes are aligned with commutator segments.

(c) Rotor has turned 180°.

(b) Rotor has turned 90°. v

Rotation axis v Rotor

I50 v S

S

BI

N

IS t Brush

S

N

I S

BI

I S t

S

S

m

m

S

m50 S t50

Commutator I

N

S

S

B

I +

I

I

I

+

• Current flows into the red side of the rotor and out of the blue side. • Therefore the magnetic torque causes the rotor to spin counterclockwise.

• Each brush is in contact with both commutator segments, so the current bypasses the rotor altogether. • No magnetic torque acts on the rotor.

I

I +

• The brushes are again aligned with commutator segments. This time the current flows into the blue side of the rotor and out of the red side. • Therefore the magnetic torque again causes the rotor to spin counterclockwise.

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232

CHAPTER 7 Magnetic Field and Magnetic Forces

7.40 This motor from a computer disk drive has 12 current-carrying coils. They interact with permanent magnets on the turntable (not shown) to make the turntable rotate. (This design is the reverse of the design in Fig. 7.39, in which the permanent magnets are stationary and the coil rotates.) Because there are multiple coils, the magnetic torque is very nearly constant and the turntable spins at a very constant rate.

segments of the commutator. There is no potential difference between the commutators, so at this instant no current flows through the rotor, and the magnetic moment is zero. The rotor continues to rotate counterclockwise because of its inertia, and current again flows through the rotor as in Fig. 7.39c. But now current enters on the blue side of the rotor and exits on the red side, just the opposite of the situation in Fig. 7.39a. While the direction of the current has reversed with S respect to the rotor, the rotor itself has rotated 180° and the magnetic moment M is in the same direction with respect to the magnetic field. Hence the magnetic S torque T is in the same direction in Fig. 7.39c as in Fig. 7.39a. Thanks to the commutator, the current reverses after every 180° of rotation, so the torque is always in the direction to rotate the rotor counterclockwise. When the motor has come “up to speed,” the average magnetic torque is just balanced by an opposing torque due to air resistance, friction in the rotor bearings, and friction between the commutator and brushes. The simple motor shown in Fig. 7.39 has only a single turn of wire in its rotor. In practical motors, the rotor has many turns; this increases the magnetic moment and the torque so that the motor can spin larger loads. The torque can also be increased by using a stronger magnetic field, which is why many motor designs use electromagnets instead of a permanent magnet. Another drawback of the simple design in Fig. 7.39 is that the magnitude of the torque rises and falls as the rotor spins. This can be remedied by having the rotor include several independent coils of wire oriented at different angles (Fig. 7.40).

Power for Electric Motors Because a motor converts electric energy to mechanical energy or work, it requires electric energy input. If the potential difference between its terminals is Vab and the current is I, then the power input is P = VabI. Even if the motor coils have negligible resistance, there must be a potential difference between the terminals if P is to be different from zero. This potential difference results principally from magnetic forces exerted on the currents in the conductors of the rotor as they rotate through the magnetic field. The associated electromotive force E is called an induced emf; it is also called a back emf because its sense is opposite to that of the current. In Chapter 9 we will study induced emfs resulting from motion of conductors in magnetic fields. In a series motor the rotor is connected in series with the electromagnet that produces the magnetic field; in a shunt motor they are connected in parallel. In a series motor with internal resistance r, Vab is greater than E, and the difference is '

the potential drop Ir across the internal resistance. That is,

Coils

Vab = E + Ir

(7.29)

Because the magnetic force is proportional to velocity, E is not constant but is proportional to the speed of rotation of the rotor. Example 7.11

A series dc motor

A dc motor with its rotor and field coils connected in series has an internal resistance of 2.00 Æ. When running at full load on a 120-V line, it draws a 4.00-A current. (a) What is the emf in the rotor? (b) What is the power delivered to the motor? (c) What is the rate of dissipation of energy in the internal resistance? (d) What is the mechanical power developed? (e) What is the motor’s efficiency? (f) What happens if the machine being driven by the motor jams, so that the rotor suddenly stops turning?

resistance r = 2.00 Æ, the voltage Vab = 120 V across the motor, and the current I = 4.00 A through the motor. We use Eq. (7.29) to determine the emf E from these quantities. The power delivered to the motor is Vab I, the rate of energy dissipation is I2r, and the power output by the motor is the difference between the power input and the power dissipated. The efficiency e is the ratio of mechanical power output to electric power input. EXECUTE: (a) From Eq. (7.29), Vab = E + Ir, we have

SOLUTION IDENTIFY and SET UP: This problem uses the ideas of power and potential drop in a series dc motor. We are given the internal

120 V = E + 14.00 A212.00 Æ2

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and so

E = 112 V

7.9 The Hall Effect

(f) With the rotor stalled, the back emf E (which is proportional to rotor speed) goes to zero. From Eq. (7.29) the current becomes

(b) The power delivered to the motor from the source is Pinput = VabI = 1120 V214.00 A2 = 480 W (c) The power dissipated in the resistance r is

I =

Pdissipated = I2r = 14.00 A2212.00 Æ2 = 32 W

Poutput = Pinput - Pdissipated = 480 W - 32 W = 448 W (e) The efficiency e is the ratio of mechanical power output to electric power input: Poutput Pinput

=

Vab 120 V = 60 A = r 2.00 Æ

and the power dissipated in the resistance r becomes

(d) The mechanical power output is the electric power input minus the rate of dissipation of energy in the motor’s resistance (assuming that there are no other power losses):

e =

233

448 W = 0.93 = 93% 480 W

Pdissipated = I2r = 160 A2212.00 Æ2 = 7200 W EVALUATE: If this massive overload doesn’t blow a fuse or trip a circuit breaker, the coils will quickly melt. When the motor is first turned on, there’s a momentary surge of current until the motor picks up speed. This surge causes greater-than-usual voltage drops 1V = IR2 in the power lines supplying the current. Similar effects are responsible for the momentary dimming of lights in a house when an air conditioner or dishwasher motor starts.

Test Your Understanding of Section 7.8 In the circuit shown in Fig. 7.39, you add a switch in series with the source of emf so that the current can be turned on and off. When you close the switch and allow current to flow, will the rotor begin to turn no matter what its original orientation? y

7.9

The Hall Effect

The reality of the forces acting on the moving charges in a conductor in a magnetic field is strikingly demonstrated by the Hall effect, an effect analogous to the transverse deflection of an electron beam in a magnetic field in vacuum. (The effect was discovered by the American physicist Edwin Hall in 1879 while he was still a graduate student.) To describe this effect, let’s consider a conductor in the form of a flat strip, as shown in Fig. 7.41. The current is in the direction of the + x-axis and there S is a uniform magnetic field B perpendicular to the plane of the strip, in the +y-direction. The drift velocity of the moving charges (charge magnitude ƒ q ƒ ) has magnitude yd. Figure 7.41a shows the case of negative charges, such as electrons in a metal, and Fig. 7.41b shows positive charges. In both cases the magnetic force is upward, just as the magnetic force on a conductor is the same whether the moving charges are positive or negative. In either case a moving charge is driven toward the upper edge of the strip by the magnetic force Fz = ƒ q ƒ ydB. If the charge carriers are electrons, as in Fig. 7.41a, an excess negative charge accumulates at the upper edge of the strip, leaving an excess positive charge at its lower edge.SThis accumulation continues until the resulting transverse electrostatic field Ee becomes large enough to cause a force (magnitude ƒ q ƒ Ee) that is equal and opposite to the magnetic force (magnitude ƒ q ƒ ydB). After that, there is no longer any net transverse force to deflect the moving charges. This electric field causes a transverse potential difference between opposite edges of the strip, called the Hall voltage or the Hall emf. The polarity depends on whether the moving charges are positive or negative. Experiments show that for metals the upper edge of the strip in Fig. 7.41a does become negatively charged, showing that the charge carriers in a metal are indeed negative electrons. However, if the charge carriers are positive, as in Fig. 7.41b, then positive charge accumulates at the upper edge, and the potential difference is opposite to the situation with negative charges. Soon after the discovery of the Hall effect in 1879, it was observed that some materials, particularly some semiconductors, show a Hall emf opposite to that of the metals, as if their charge carriers were positively charged. We now know that these materials conduct by a process known as hole conduction. Within such a material there are locations, called holes, that would normally be occupied by an electron but are actually empty. A missing negative charge is equivalent to a positive charge. When an electron moves in one direction to fill a hole, it leaves another hole behind it. The hole migrates in the direction opposite to that of the electron.

7.41 Forces on charge carriers in a conductor in a magnetic field. (a) Negative charge carriers (electrons) The charge carriers are pushed toward the top of the strip ... z y b – – – – – – By – – – – Fz vd Jx Ee q + + + + + + + + + By + a

Jx

'

'

'

x

... so point a is at a higher potential than point b.

(b) Positive charge carriers The charge carriers are again pushed toward the top of the strip ... z

Jx

y b + + + + + + By + + + Ee + q Fz vd Jx – – – – – – – – – By – a

... so the polarity of the potential difference is opposite to that for negative charge carriers.

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x

234

CHAPTER 7 Magnetic Field and Magnetic Forces S

In terms of the coordinate axes in Fig. 7.41b, the electrostatic field Ee for the positive-q case is in the -z-direction; its z-component Ez is negative. The magnetic field is in the + y-direction, and we write it as By. The magnetic force (in the +z-direction) is qydBy. The current density Jx is in the +x-direction. In the steady state, when the forces qEz and qydBy are equal in magnitude and opposite in direction, '

'

'

'

qEz + qydBy = 0 '

or

Ez = - ydBy '

This confirms that when q is positive, Ez is negative. The current density Jx is Jx = nqyd Eliminating yd between these equations, we find - Jx By '

nq =

(Hall effect)

Ez

(7.30)

Note that this result (as well as the entire derivation) is valid for both positive and negative q. When q is negative, Ez is positive, and conversely. We can measure Jx, By, and Ez, so we can compute the product nq. In both metals and semiconductors, q is equal in magnitude to the electron charge, so the Hall effect permits a direct measurement of n, the concentration of current-carrying charges in the material. The sign of the charges is determined by the polarity of the Hall emf, as we have described. The Hall effect can also be used for a direct measurement of electron drift speed yd in metals. As we saw in Chapter 5, these speeds are very small, often of the order of 1 mm>s or less. If we move the entire conductor in the opposite direction to the current with a speed equal to the drift speed, then the electrons are at rest with respect to the magnetic field, and the Hall emf disappears. Thus the conductor speed needed to make the Hall emf vanish is equal to the drift speed. '

Example 7.12

SOLUTION IDENTIFY and SET UP: This problem describes a Hall-effect experiment. We use Eq. (7.30) to determine the mobile electron concentration n. EXECUTE: First we find the current density Jx and the electric field Ez: '

Ez =

'

A Hall-effect measurement

You place a strip of copper, 2.0 mm thick and 1.50 cm wide, in a uniform 0.40-T magnetic field as shown in Fig. 7.41a. When you run a 75-A current in the + x-direction, you find that the potential at the bottom of the slab is 0.81mV higher than at the top. From this measurement, determine the concentration of mobile electrons in copper.

Jx =

'

I 75 A = = 2.5 * 106 A> m2 A 12.0 * 10-3 m211.50 * 10-2 m2 '

Then, from Eq. (7.30), n =

-JxBy qEz

-12.5 * 106 A> m2210.40 T2 '

=

1- 1.60 * 10 28

= 11.6 * 10

m

-3

-19

'

C215.4 * 10-5 V> m2 '

'

EVALUATE: The actual value of n for copper is 8.5 * 1028 m-3. The difference shows that our simple model of the Hall effect, which ignores quantum effects and electron interactions with the ions, must be used with caution. This example also shows that with good conductors, the Hall emf is very small even with large current densities. In practice, Hall-effect devices for magnetic-field measurements use semiconductor materials, for which moderate current densities give much larger Hall emfs.

'

V 0.81 * 10-6 V = = 5.4 * 10-5 V> m d 1.5 * 10-2 m '

'

Test Your Understanding of Section 7.9 A copper wire of square cross section is oriented vertically. The four sides of the wire face north, south, east, and west. There is a uniform magnetic field directed from east to west, and the wire carries current downward. Which side of the wire is at the highest electric potential? (i) north side; (ii) south side; (iii) east side; (iv) west side. y

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CHAPTER

7

SUMMARY S

S

S

Magnetic forces: Magnetic interactions are fundamentally interactions between moving charged particles. These interactionsSare described by the vector magnetic field, denoted by B. A particle withScharge q moving S with velocity v in a magnetic field B experiences a force S S S F that is perpendicular to both v and B. The SI unit of magnetic field is the tesla 11 T = 1 N>A # m2. (See Example 7.1.)

F 5 qv 3 B

Magnetic field lines and flux: A magnetic field can be represented graphically by magnetic field lines. At each point a magnetic field line is tangent to the direction of S B at that point. Where field lines are close together the field magnitude is large, and vice versa. Magnetic flux £ B through an area is defined in an analogous way to electric flux. The SI unit of magnetic flux is the weber 11 Wb = 1 T # m22. The net magnetic flux through any closed surface is zero (Gauss’s law for magnetism). As a result, magnetic field lines always close on themselves. (See Example 7.2.)

£B =

S

(7.2)

F S

S

q fS v

v'

2

B

B

B' dA

S

B' f B

'

S

dA

=

S

#

B cos f dA

2

B dA

S

=

B

2

#

(7.6)

Bi

dA S

S

B dA = 0 (closed surface)

my

Motion in a magnetic field: The magnetic force is always S perpendicular to v; a particle moving under the action of a magnetic field alone moves with constant speed. In a uniform field, a particle with initial velocity perpendicular to the field moves in a circle with radius R that depends on the magnetic field strength B and the particle mass m, speed y, and charge q. (See Examples 7.3 and 7.4.) Crossed electric and magnetic fields can be used as a velocity selector. The electric and magnetic forces exactly cancel when y = E>B. (See Examples 7.5 and 7.6.)

R =

Magnetic force on a conductor: A straight segment of a conductor carrying current I in a uniform magnetic field S S S B experiences aS force F that is perpendicular to both B and the vector l , which points in the direction of the current and has magnitude equal to the length of Sthe segment. A similar relationship gives theSforce dF on an infinitesimal current-carrying segment d l . (See Examples 7.7 and 7.8.)

F 5 Il 3 B

Magnetic torque: A current loop with area A and current S I in a uniform magnetic field B experiences no net magnetic force, but does experience a magnetic torque of S magnitude t. The vector torque T can Sbe expressed in S terms of the magnetic moment M 5 IA of the loop, as can the potential energy U of a magnetic moment in a S magnetic field B. The magnetic moment of a loop depends only on the current and the area; it is independent of the shape of the loop. (See Examples 7.9 and 7.10.)

t = IBA sin f

(7.8)

S

(7.11)

ƒqƒB

v S

F

R

S

F

S

v

S

F S S

B

S

S

S

v

S

S

(7.19)

S

F B'

S

dF 5 I d l 3 B

(7.20)

f

S

B

Bi

S

l

I

z

(7.23 ) S

S

B

S

S

T5M 3 B S

#

(7.26)

S

U = 2M B = - mB cos f

(7.27)

I

y

S

F

S

B

x

S S

m

B I S

2F

235

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236

CHAPTER 7 Magnetic Field and Magnetic Forces

Electric motors: In a dc motor a magnetic field exerts a torque on a current in the rotor. Motion of the rotor through the magnetic field causes an induced emf called a back emf. For a series motor, in which the rotor coil is in parallel with coils that produce the magnetic field, the terminal voltage is the sum of the back emf and the drop Ir across the internal resistance. (See Example 7.11.)

Rotation axis v Rotor S

N

B I

I S t

S

S

m

Brush

Commutator I

I +

The Hall effect: The Hall effect is a potential difference perpendicular to the direction of current in a conductor, when the conductor is placed in a magnetic field. The Hall potential is determined by the requirement that the associated electric field must just balance the magnetic force on a moving charge. Hall-effect measurements can be used to determine the sign of charge carriers and their concentration n. (See Example 7.12.)

BRIDGING PROBLEM

- JxBy

z

'

nq =

Ez

(7.30) Jx

– – – b – – – – – F

y By

z

vd

E

e + + + q + + + + + B y

Jx x

a

Magnetic Torque on a Current-Carrying Ring

A circular ring with area 4.45 cm2 is carrying a current of 12.5 A. The ring, initially at rest, is immersed in a region of uniform magS netic field given by B 5 11.15 * 10-2 T2112≤n 1 3≥n 2 4kN 2. The ring is positioned initially such that its magnetic moment is given S by M i 5 m1 - 0.800≤n 1 0.600≥n2, where m is the (positive) magnitude of the magnetic moment. (a) Find the initial magnetic torque on the ring. (b) The ring (which is free to rotate around one diameter) is released and turns through an angle of 90.0°, at which point S its magnetic moment is given by M f 5 - mkN . Determine the decrease in potential energy. (c) If the moment of inertia of the ring about a diameter is 8.50 * 10-7 kg # m2, determine the angular speed of the ring as it passes through the second position. SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. The current-carrying ring acts as a magnetic dipole, so you can use the equations for a magnetic dipole in a uniform magnetic field.

Problems

2. There are no nonconservative forces acting on the ring as it rotates, so the sum of its rotational kinetic energy (discussed in Section 9.4) and the potential energy is conserved. EXECUTE 3. Use the vector expression for the torque on a magnetic dipole to find the answer to part (a). (Hint: You may want to review Section 1.10.) 4. Find the change in potential energy from the first orientation of the ring to the second orientation. 5. Use your result from step 4 to find the rotational kinetic energy of the ring when it is in the second orientation. 6. Use your result from step 5 to find the ring’s angular speed when it is in the second orientation. EVALUATE 7. If the ring were free to rotate around any diameter, in what direction would the magnetic moment point when the ring is in a state of stable equilibrium?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems. DISCUSSION QUESTIONS Q7.1 Can a charged particle move through a magnetic field without experiencing any force? If so, how? If not,Swhy not? Q7.2 At any point in space, the electric field E is defined to be in the direction of the electric force on a positively charged particle at S that point. Why don’t we similarly define the magnetic field B to

be in the direction of the magnetic force on a moving, positively charged particle? Q7.3 Section 7.2 describes a procedure for finding the direction of the magnetic force using your right hand. If you use the same procedure, but with your left hand, will you get the correct direction for the force? Explain.

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Single Correct

Q7.4 The magnetic force on a moving charged particle is always S perpendicular to the magnetic field B. Is the trajectory of a moving charged particle always perpendicular to the magnetic field lines? Explain your reasoning. Q7.5 A charged particle is fired into a cubical region of space where there is a uniform magnetic field. Outside this region, there is no magnetic field. Is it possible that the particle will remain inside the cubical region? Why or why not? Q7.6 If the magnetic force does no work on a charged particle, how can it have any effect on the particle’s motion? Are there other examples of forces that do no work but have a significant effect on a particle’s motion? Q7.7 A charged particle moves through a region of space with constant velocity (magnitude and direction). If the external magnetic field is zero in this region, can you conclude that the external electric field in the region is also zero? Explain. (By “external” we mean fields other than those produced by the charged particle.) If the external electric field is zero in the region, can you conclude that the external magnetic field in the region is also zero? Q7.8 How might a loop of wire carrying a current be used as a compass? Could such a compass distinguish between north and south? Why or why not? Q7.9 How could the direction of a magnetic field be determined by making only qualitative observations of the magnetic force on a straight wire carrying a current? Q7.10 A loose, floppy loop of wire is carrying current I. The loop S of wire is placed on a horizontal table in a uniform magnetic field B perpendicular to the plane of the table. This causes the loop of wire to expand into a circular shape while still lying on the table. In a diagram, show all possible orientations of the current I and magS netic field B that could cause this to occur. Explain your reasoning. Q7.11 Several charges enter a uniform magnetic field directed into the page. (a) What path would a positive charge q moving with a velocity of magnitude v follow through the field? (b) What path would a positive charge q moving with a velocity of magnitude 2v follow through the field? (c) What path would a negative charge -q moving with a velocity of magnitude v follow through the field? (d) What path would a neutral particle follow through the field? Q7.12 Each of the lettered Figure Q7.12 points at the corners of the cube y in Fig. Q7.12 represents a posiS B tive charge q moving with a b velocity of magnitude v in the direction indicated. The region d in the figure is in a uniform c S magnetic field B, parallel to the x-axis and directed toward the a right. Which charges experiS x ence a force due to B ? What is the direction of the force on z each charge? e Q7.13 A student claims that if lightning strikes a metal flagpole, the force exerted by the earth’s magnetic field on the current in the pole can be large enough to bend it. Typical lightning currents are of the order of 104 to 105 A. Is the student’s opinion justified? Explain your reasoning. Q7.14 Could an accelerator be built in which all the forces on the particles, for steering and for increasing speed, are magnetic forces? Why or why not? Q7.16 The magnetic force acting on a charged particle can never do work because at every instant the force is perpendicular to the velocity. The torque exerted by a magnetic field can do work on a '

237

current loop when the loop rotates. Explain how these seemingly contradictory statements can be reconciled. Q7.17 If an emf is produced in a dc motor, would it be possible to use the motor somehow as a generator or source, taking power out of it rather than putting power into it? How might this be done? Q7.18 When the polarity of the voltage applied to a dc motor is reversed, the direction of motion does not reverse. Why not? How could the direction of motion be reversed? Q7.19 In a Hall-effect experiment, is it possible that no transverse potential difference will be observed? Under what circumstances might this happen? Q7.20 Hall-effect voltages are much greater for relatively poor conductors (such as germanium) than for good conductors (such as copper), for comparable currents, fields, and dimensions. Why?

SINGLE CORRECT Q7.1 A charged particle is projected in air horizontally near the earth’s surface at a point where horizontal componenet of magnetic field is absent. Considering the effect of vertical component of the earth’s magnetic field and the gravitational field due to earth, the path of the particle will be (a) circular (b) helical (c) straight line (d) parabolic Q7.2 A negative electrical charge is made to move clockwise in a circular path due to an external magnetic field. What would be the direction of the magnitude field? (a) towards the top of the page c (b) towards the bottom of the page T (c) out of the page } (d) into the page {

−

Q7.3 An electrostatic field E and a magnetic field B act over the same region. Which one of the combination shown in the figure can cause the electrons to pass undeflected? E B (b) E (c) e− E (a) B (d) − e−

e

B

E

e−

B

Q7.4 A proton moves in the + z -direction after being accelerated from rest through a ptential difference V. The proton then passes through a region with a uniform electric field E in the + xdirection and a uniform magnetic feild B in the + y-direction, but the proton’s trajectory is not affacted. If the experiment were repeated using a potential difference of 2V, The proton would then be (a) deflected in the + x-direction (b) deflected in the - x-direction (c) deflected in the + y-direction (d) deflected in the - y-direction z Q7.5 A positively charged particle has a velocity in the negative z direction, as shown in the figure. The Lorentz force on the particle is in the negative y direction. From this observation alone, what can be F y said about the magnetic feild at this point? B v x (a) Bx is positive (b) Bx is negative (c) By is positive (d) By is negative

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238

CHAPTER 7 Magnetic Field and Magnetic Forces

1 Q7.6 Figure shows the path of an electron in a 2 region of uniform magnetic fields. The path consists of two straight sections, each between a pair 3 4 of uniformly charged plates, and two half-circles. The plates are named 1, 2, 3 & 4. Then (a) 1 and 3 at higher (positive) potential and 2 and 4 at lower (negative) potential (b) 1 and 3 at lower potential and 2 and 4 at higher potential (c) 1 and 4 at higher potential and 2 and 3 at lower potential (d) 1 and 4 at lower potential and 2 and 3 at higher potential Q7.7 At t = 0 a charge q is at the origin y and moving in the y-direction with velocity S

^

v = vj . The charge moves in a magnetic field that is for y > 0 out of page and given ^ by B1 z and for y < 0 into the page and ^

Z

B1 ×

B2

given - B2 z. The charge’s subsequent trajectory is shown in the sketch. From this information, we can deduce that (a) q > 0 and |B1| < |B2| (b) q < 0 and |B1| < |B2| (c) q > 0 and |B1| > |B2| (d) q < 0 and |B1| > |B2| S Q7.8 A chargedS particle is moving in the Spresence of electric (E) S and magnetic (B) fields. The directions of E and B are such that the charged particle movesS inS a straight lineS and its speed increases. E, B and velocity V must be such that The relations amongst S S S (a) E , is arbitrary V .B = 0 S S S (b) E , B and VS are all parallel to each other S S S S S (c) E , but, .V = 0 B .V = 0 E .B Z 0 S S S (d) V is parallel to E and perpendicular to B +y Q7.9 The figure shows a particle (carry⊗ ⊗ ⊗ ⊗ ⊗ ing charge + q) at the origin. Auniform ⊗ ⊗ ⊗ ⊗ ⊗ magnetic field is directed into the plane of the paper. The particle can be projected −x ⊗ ⊗ +q ⊗ ⊗ +x only in the plane of paper and along posi⊗ ⊗ ⊗ ⊗ ⊗ tive or negative x- or y-axis. The particle Target ⊗ ⊗ ⊗ ⊗ ⊗ moves with constant speed and has to hit a −y target located in the third quadrant. There are two direction of projections, which can make it possible, these are (a) + x and + y (b) + x and - y (c) - x and + y (d) - x and - y Q7.10 A proton, projected perpendicularly into a magnetic field with a certain velocity, follows a circular path. An electron is then projected into the same region with the same initial velocity as the proton. If the electron is to follow the exact same circular path as the proton, should the direction of the magnetic field be kept the same or reversed, and should the magnitude of the magnetic field be increased or decreased? (a) keep the same field direction and increase its magnitude (b) keep the same field direction and decrease its magnitude (c) reverse the field direction and increase its magnitude (d) reverse the field direction and decrease its magnitude (e) none of the above Q7.11 The given figure shows a set-up for accelerating protons from rest, through a potential (fixed) by the battery, and then through a velocity selector. The electric field E and magnetic field B are carefully adjusted so that the accelerated protons go straight through the slit after exiting the selector. You want to modify this

B so that electrons instead of protons are × × × × × used. You reverse the leads on the accel+ + + + + × × × × × erating battery. What do you need to do × × × × × E × × × × × to the potential difference between the × × × × × plates that create the electric field in the × × × × × × × × × × selector? × × × × × Slit − − − − − (a) Nothing Selector (b) Reduce its magnitude (c) Increase its magnitude (d) Reverse the polarity, and reduce its magnitude Q7.12 The coil below has current flowing clockwise as seen from above. It sits in an B external magnetic field shown by the field lines below. The coil will feel a magnetic force that is (a) upwards (b) downwards (c) zero (d) can’t be predicted Q7.13 Statement-1: Net force on a current carrying loop in a nonuniform magnetic field must be non-zero. S Statement-2: Force on a current carrying wire of length dl placed S in magnetic field B is given by dlS = idlS * BS

(a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true.

ONE OR MORE CORRECT Q7.14 A charged particles shoots through a space and continues its path undeflected. Then in that space. (a) the uniform electric and magnetic fields, perpendicular to each other may be present. (b) electric field alone may be present (c) uniform electric field and magnetic fields, parallel to each other may be present. (d) uniform electric fields alone may be present. (e) uniform electric field and magnetic field inclined to each other may be present. Q7.15 An electron moves in the plane of the Region II page through two regions of space along the dotted-line trajectory shown in the figure. There is a uniform electric field in Region I Region I directed into the plane of the page (as × × × × × × × shown). There is no electric field in Region × × × × × × × II. What is a necessary direction of the magnetic field in regions I and II? Ignore gravitational forces (a) Downwards along the plane of the page in region I (b) Upwards along the plane of the page in region I (c) In to the plane of the page in region II (d) Out of the plane of the page in region II S Q7.16 A particle of mass m and charge q moving withS velocity v enters a region of uniform magnetic field of induction B. Then (a) its path in the region of the field is always circular S S (b) its path in the region of the field is circular if v . B =S 0 S (c) its path in the region of the field is a straight line if v × B = 0 (d) distance travelled by the Sparticle in time T does not depend on S the angle between v and B.

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Exercises N Q7.17 Figure shows the essential parts Q of an apparatus to demonstrate the Hall effect. Which of the following Constant statement(s) is/are correct? P current (a) In the arrangement above, the Hall potential difference is developed S across PQ (b) The magnitude of the Hall potential is greater if the applied magnetic flux density is increased (c) The magnitude of the Hall potential is less if the width PQ of the specimen is decreased (d) Hall potential is independent of material used for conductor Q7.18 Two charged particles A and × × B each of charge + e and masses × × Velocity 12 amu and 13 amu respectively × × selector × × follow a circular trajectory in × × chamber X after the velocity selec- ×× ×× ×× ×× × × Y tor as shown in the figure. Both × × × × × × particles enter the velocity selector × × × × B × × X Z × × with speed 1.5 × 106 ms–1. A uni- ×× ×× ×× ×× × × X form magnetic field of strength 1.0 × × × × × × T is maintained within the chamber X and in the velocity selector. (a) Electric field across the conducting plates of the velocity ^ selector is - 1.5 × 106 NC–1 i (b) Electric field across the ^conducting plates of the velocity selector is 1.5 × 106 NC–1 i rA (c) The ratio of the radii of the circular orbits for the two partirB 12 cles is 13 r (d) The ratio A of the radii of the circular orbits for the two rB 13 particles is 12 Q7.19 A horizontal cathode ray tube (CRT) is set so that its electron beam produces a spot of light at the centre of the screen when no external electromagnetic field is present. When you look straight at the screen, however, you discover that the spot, instead of being at the centre as it should, is shifted a bit to the right. Suspecting what the cause of this deflection may be, you rotate the CRT by 180° around its vertical axis. Facing the screen, you find that the spot of light is still shifted to right of centre by the same distance as before. You conclude that the CRT is immersed in (a) an electric field directed horizontally to the left with respect to the screen’s initial position (b) an electric field directed horizontally to the right with respect to the screen’s initial position (c) a magnetic field directed vertically upward with respect to the screen’s initial position (d) a magnetic field directed vertically downward with respect to the screen’s initial position

239

7.2 . In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50 mC and initially moving northward at 4.75 km > s is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle. 7.3 . An electron moves at 2.50 * 106 m > s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 * 10-2 T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field? 7.4 .. CP A particle with charge 7.80 mC is moving with velocity S y 5 213.80 * 103 m > s2≥n. The magnetic force on the particle is S measured to be F 5 117.60 * 10-3 N2≤n 2 15.20 * 10-3 N2kN . (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? S S # Explain. S (c) Calculate the scalar product . What is the angle B F S between B and F ? 7.5 .. A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km > s in the + x-direction experiences a force of 2.25 * 10-16 N in the + y-direction, and an electron moving at 4.75 km > s in the -z-direction experiences a force of 8.50 * 10-16 N in the + y-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the - y-direction at 3.20 km > s? '

'

'

'

'

'

'

'

'

'

'

'

'

Section 7.3 Magnetic Field Lines and Magnetic Flux

7.6 . A flat, square surface with side length 3.40 cm is in the xy-plane at z = 0. Calculate the magnitude Sof the flux through this surface produced by a magnetic field B 5 10 .200 T2≤n 1 10.300 T2≥n 2 10.500 T2kN . 7.7 .. An open plastic soda bottle with an opening diameter of 2.5 cm is placed on a table. A uniform 1.75-T magnetic field directed upward and oriented 30° from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle? S 7.8 .. The magnetic field B in a Figure E7.8 certain region is 0.128 T, and its y direction is that of the +z-axis in Fig. E7.8. (a) What is b 30.0 cm 40.0 cm the magnetic flux across the sure face abcd in the figure? a 30.0 cm (b) What is the magnetic flux c across the surface befc? (c) What f is the magnetic flux across the x 50.0 cm d surface aefd? (d) What is the net flux through all five surfaces that z enclose the shaded volume?

EXERCISES Section 7.2 Magnetic Field

.

7.1 A particle of mass 0.195 g carries a charge of -2.50 * 10-8 C. The particle is given an initial horizontal velocity that is due north and has magnitude 4.00 * 104 m > s. What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth’s gravitational field in the same horizontal, northward direction? '

'

Section 7.4 Motion of Charged Particles in a Magnetic Field

7.9 .. An electron at point A in Fig. E7.9 has a speed v0 of 1.41 * 106 m > s. Find (a) the magnitude and direction of the magnetic field that will cause '

Figure E7.9 v0

'

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A

B 10.0 cm

CHAPTER 7 Magnetic Field and Magnetic Forces

the electron to follow the semicircular path from A to B, and (b) the time required for the electron to move from A to B. 7.10 . CP A 150-g ball containing 4.00 * 108 excess electrons is dropped into a 125-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field. 7.11 . An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64 * 10-27 kg) traveling horizontally at 35.6 km > s enters a uniform, vertical, 1.10-T magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it. 7.12 . (a) An 16O nucleus (charge + 8e) moving horizontally from west to east with a speed of 500 km > s experiences a magnetic force of 0.00320 nN vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, 2.10-T magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be 4.60 pN, vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction. 7.13 . A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 * 10-27 kg and a charge of + e. The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed? 7.14 . In an experiment with cosmic Figure E7.14 rays, a vertical beam of particles that have charge of magnitude 3e and 95.0 cm mass 12 times the proton mass enters a uniform horizontal magnetic field of 0.250 T and is bent in a semicircle of S diameter 95.0 cm, as shown in Fig. B E7.14. (a) Find the speed of the particles and the sign of their charge. (b) Is it reasonable to ignore the gravity force on the particles? (c) How does the speed of the particles as they enter the field compare to their speed Figure E7.15 as they exit the field? .. 7.15 A beam of protons traveling at 1.20 km>s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its S original direction (Fig. E7.15). The beam B travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field? 7.16 .. A proton 1q = 1.60 * 10-19 C, m = 1.67 * 10-27 kg2 '

'

'

'

moves in a uniform magnetic field B 5 10.500 T2≤n. At t = 0 the proton has velocity components vx = 1.50 * 105 m > s, vy = 0, S

'

'

and vz = 2.00 * 105 m > s (see Example 7.4). (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric S field in the +x-direction, E 5 1+2.00 * 104 V > m2≤n. (b) At t = T > 2, where T is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its position at t = 0? '

'

'

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Section 7.5 Applications of Motion of Charged Particles

7.17 . (a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56 * 104 V > m and a magnetic field of 4.62 * 10-3 T, with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) InS a diagram, show the relative orientation of the vectors S S y, E, and B. (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit? 7.18 . In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge +5e that are traveling perpendicular to the fields at 8.75 km> s. The magnetic field available to you has a magnitude of 0.550 T. (a) What magnitude of electric field do you need? (b) Show how the two fields should be oriented relative to each other and to the velocity of the ions. (c) Will your velocity selector also allow the following ions (having the same velocity as the + 5e ions) to pass through undeflected: (i) negative ions of charge - 5e, (ii) positive ions of charge different from +5e? 7.19 .. A 150-V battery is Figure E7.19 connected across two parallel metal plates of area 28.5 cm2 v and separation 8.20 mm. A 150 V beam of alpha particles (charge -27 + 2e, mass 6.64 * 10 kg2 is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. E7.19. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates? 7.20 . Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V> m and the magnetic field is 0.031 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass? '

'

+

240

Section 7.6 Magnetic Force on a Current-Carrying Conductor

7.21 . A straight, 2.00-m, 150-g wire carries a current in a region where the earth’s magnetic field is horizontal with a magnitude of 0.5 gauss. (a) What is the minimum value of the current in this wire so that its weight is completely supported by the magnetic force due to earth’s field, assuming that no other forces except gravity act on it? Does it seem likely that such a wire could support this size of current? (b) Show how the wire would have to be oriented relative to the earth’s magnetic field to be supported in this way. 7.22 .. A long wire carrying 4 A of current makes two 90° bends, as shown in Fig. E7.22. The bent part of the wire passes through a uniform 0.25-T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and

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Exercises

direction of the force that the Figure E7.22 magnetic field exerts on the Magnetic field region wire. 7.23 . A thin, 50.0-cm-long metal bar with mass 750 g rests on, but is not attached to, two metallic supports in a uniform 30.0 cm 60.0 cm 4.50 A 0.450-T magnetic field, as S shown in Fig. E7.23. A battery B and a 25.0- Æ resistor in series 60.0 cm are connected to the supports. (a) What is the highest voltage the Figure E7.23 battery can have without breaking V R the circuit at the supports? (b) The battery voltage has the maximum value calculated in part S (a). If the resistor suddenly gets B partially short-circuited, decreasing its resistance to 2.0 Æ, find the initial acceleration of the bar. 7.24 . Magnetic Balance. The circuit shown in Fig. E7.24 is used to make a magnetic balance to weigh objects. The mass m to be measured is hung from the center of the bar that is in a uniform magnetic field of 1.50 T, directed into the plane of the figure. The battery voltage can be adjusted to vary the current in the circuit. The horizontal bar is 60.0 cm long and is made of extremely light-weight material. It is connected to the battery by thin vertical wires that can support no appreciable tension; all the weight of the suspended mass m is supported by the magnetic force Figure E7.24 on the bar. A resistor with R = 5.00 Æ is in series with the a b 5.00 V Battery bar; the resistance of the rest of the circuit is much less than this. S (a) Which point, a or b, should be the B Bar positive terminal of the battery? (b) If the maximum terminal voltage of the battery is 175 V, what is the greatest mass m that this instrument m can measure?

241

calculate the torque that the magnetic field exerts on the circuit about the hinge axis ab. 7.27 . A rectangular coil of Figure E7.27 wire, 22.0 cm by 35.0 cm and S Axis B carrying a current of 1.40 A, is oriented with the plane of its loop perpendicular to a uniform I 22.0 cm 1.50-T magnetic field, as shown in Fig. E7.27. (a) Calculate the 35.0 cm net force and torque that the magnetic field exerts on the coil. (b) The coil is rotated through a 30.0° angle about the axis shown, with the left side coming out of the plane of the figure and the right side going into the plane. Calculate the net force and torque that the magnetic field now exerts on the coil. (Hint: In order to help visualize this three-dimensional problem, make a careful drawing of the coil as viewed along the rotation axis.) 7.28 . CP A uniform rectangular coil of total mass 212 g Figure E7.28 and dimensions 0.500 m * A1 1.00 m is oriented with its S B plane parallel to a uniform 3.00T magnetic field (Fig. E7.28). A current of 2.00 A is suddenly A2 0.500 m started in the coil. (a) About which axis 1A1 or A22 will the 1.00 m coil begin to rotate? Why? (b) Find the initial angular acceleration of the coil just after the current is started. 7.29 . A circular coil with area A and N turns is free to Figure E7.29 y z rotate about a diameter that y z I I I coincides with the x-axis. x x x x Current I is circulating in the I coil. There Sis a uniform mag(a) (b) (c) (d) netic field B in the positive yS direction. Calculate the magnitude and direction of the torque T and the value of the potential energy U, as given in Eq. (7.27), when the coil is oriented as shown in parts (a) through (d) of Fig. E7.29.

Section 7.7 Force and Torque on a Current Loop

7.25 .. The plane of a 5.0 cm * 8.0 cm rectangular loop of wire is parallel to a 0.19-T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field? 7.26 . The 20.0 cm * 35.0 Figure E7.26 cm rectangular circuit shown in Fig. E7.26 is hinged along S B side ab. It carries a clockwise b c 5.00-A current and is located in a uniform 1.20-T magnetic Hinge 20.0 cm field oriented perpendicular to two of its sides, as shown. (a) 5.00 A Draw a clear diagram showing a d 35.0 cm the direction of the force that the magnetic field exerts on each segment of the circuit (ab, bc, etc.). (b) Of the four forces you drew in part (a), decide which ones exert a torque about the hinge ab. Then calculate only those forces that exert this torque. (c) Use your results from part (b) to

Section 7.8 The Direct-Current Motor

7.30 . A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 Æ. When the motor is running at full load on a 120-V line, the emf in the rotor is 105 V. (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor? 7.31 .. In a shunt-wound dc Figure E7.31 motor with the field coils and + rotor connected in parallel (Fig. E7.31), the resistance Rf of Rf the field coils is 106 Æ, and the 120 V E, Rr resistance Rr of the rotor is 5.9 Æ. When a potential differ– ence of 120 V is applied to the brushes and the motor is running at full speed delivering mechanical power, the current supplied to it is 4.82 A. (a) What is the current in the field coils? (b) What is the current in the rotor? (c) What is the induced emf developed by the motor? (d) How much mechanical power is developed by this motor?

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242

CHAPTER 7 Magnetic Field and Magnetic Forces

PROBLEMS 7.32 . When a particle of charge q

7 0 moves with a velocity of y1 at 45.0° from the + x-axis in the xy-plane, a uniform magnetic

S

S

field exerts a force F1 along the - z-axis (Fig. P7.32). When the S same particle moves with a velocity y2 with the same magnitude as S

S

y1 but along the + z-axis, a force F2 of magnitude F2 is exerted on it along the + x-axis. (a) What are the magnitude (in terms of q, y1, and F2) and direction of the magnetic field? (b) What is the magniS

tude of F1 in terms of F2 ? '

Figure P7.32

'

y

S

v1

S

45.0°

F1

x S

v2 z

S

F2

7.33 .. Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 8.5), in the lowest energy state the electron orbits the proton at a speed of 2.2 * 106 m > s in a circular orbit of radius 5.3 * 10-11 m. (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current I? (c) What is the magnetic moment of the atom due to the motion of the electron? 7.34 .. You wish to hit a target from several meters away with a charged coin having a mass of 4.25 g and a charge of + 2500 mC. The coin is given an initial velocity of 12.8 m > s, and a downward, uniform electric field with field strength 27.5 N > C exists throughout the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target? (take g = 9.8 m/s2) 7.35 .. A particle with charge q is moving with speed v in the -y S -direction. It is moving in a uniform magnetic field B 5S N Bx n≤ 1 By n≥ 1 Bz k. (a) What are the components of the force F exerted on the particle by the magnetic field? (b) If q 7 0, what S S B must the signs of the components of be if the components of F are all nonnegative? (c) If q 6 0 and Bx =S By = Bz 7 0, find the S direction of F and find the magnitude of F in terms of ƒ q ƒ , v, and Bx. 7.36 .. A particle with negative charge q and mass m = 2.58 * 10-15 kg is traveling through a region containing a uniform magnetic S field B 5 2 10.120 T2kN . At a particular instant of time the velocity S 6 of the particle is v 5 11.05 * 10 m > s21- 3≤n 1 4≥n 1 12kN 2 and S the force F on the particle has a magnitude of 2.45 N. (a) DeterS mine the charge q. (b) Determine the acceleration a of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature R of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the x- and y-coordinates do vary in a periodic way. If the coordinates of the particle at t = 0 are 1x, y, z2 = 1R, 0, 02, determine its coordinates at a time t = 2T, where T is the period of the motion in the xy-plane. '

7.37 .. BIO Medical Uses of Cyclotrons. The largest cyclotron in the United States is the Tevatron at Fermilab, near Chicago, Illinois. It is called a Tevatron because it can accelerate particles to energies in the TeV range: 1 tera-eV = 1012 eV . Its circumference is 6.4 km, and it currently can produce a maximum energy of 2.0 TeV. In a certain medical experiment, protons will be accelerated to energies of 1.25 MeV and aimed at a tumor to destroy its cells. (a) How fast are these protons moving when they hit the tumor? (b) How strong must the magnetic field be to bend the protons in the circle indicated? 7.38 .. A particle of charge q 7 0 is moving at speed v in the + z S -direction through a region ofS uniform magnetic field B. The magnetic force on the particle is F 5 F0 13≤n 1 4≥n2, where F0 is a positive constant. (a) Determine the components Bx, By, and Bz, or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude 6F0 > qv, determine as much as you can about S the remaining components of B. 7.39 .. A straight piece of con- Figure P7.39 ducting wire with mass M and S B (vertical) length L is placed on a frictionless incline tilted at an angle u from the Wire, mass M horizontal (Fig. P7.39). There is Sa uniform, vertical magnetic field B at all points (produced by an u arrangement of magnets not L shown in the figure). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to remain at rest. Copy the figure and draw the direction of the current on your copy. In addition, show in a free-body diagram all the forces that act on the wire. 7.40 .. CP A 2.60-N metal bar, 1.50 m long and having a resistance of 10.0 Æ, rests horizontally on conducting wires connecting it to the circuit shown in Fig. P7.40. The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit. What is the acceleration of the bar just after the switch S is closed?

'

'

'

'

'

'

'

'

Figure P7.40 25.0 V 120.0 V

+

S S

10.0 V

B

'

7.41 .. CP An Electromagnetic Rail Gun. A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical magS netic field B fills the region between the rails (Fig. P7.41). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass m, find the distance d that the bar must move along the rails from rest to attain speed v. (c) It has been sug-

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Problems

Figure P7.41 S

B

I

gested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth 111.2 km > s2. Let B = 0.80 T, I = 2.0 * 103 A, m = 25 kg, and L = 50 cm. For simplicity asssume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space. 7.42 . A long wire carrying a Figure P7.42 6.00-A current reverses direction Magnetic field by means of two right-angle region bends, as shown in Fig. P7.42. 6.00 A The part of the wire where the S B bend occurs is in a magnetic field of 0.666 T confined to the circu- 45.0 cm lar region of diameter 75 cm, as shown. Find the magnitude and 75 cm direction of the net force that the magnetic field exerts on this wire. 7.43 .. CP The rectangular loop of Figure P7.43 wire shown in Fig. P7.43 has a y mass of 0.15 g per centimeter of length and is pivoted about side ab on a frictionless axis. The cura rent in the wire is 8.2 A in the 6.00 cm x direction shown. Find the magnib tude and direction of the magnetic field parallel to the y-axis that will cause the loop to swing up until z its plane makes an angle of 30.0° 30.0° with the yz-plane in equilibrium. 8.00 7.44 .. The rectangular loop cm shown in Fig. P7.44 is pivoted about the y-axis and carries a current of 15.0 A in the direction Figure P7.44 indicated. (a) If the loop is in a y uniform magnetic field with magnitude 0.48 T in the + xdirection, find the magnitude and 15.0 A direction of the torque required to hold the loop in the position 8.00 cm shown. (b) Repeat part (a) for the case in which the field is in the - z-direction. (c) For each of the above magnetic fields, what torque would be required if the 30.0° loop were pivoted about an axis x through its center, parallel to the 6.00 cm y-axis? '

7.45 .. CP CALC A thin, uniform rod Figure P7.45 with negligible mass and length 0.200 m is attached to the floor by a frictionless hinge at point P (Fig. P7.45). A k horizontal spring with force constant S B k = 4.80 N > m connects the other end I of the rod to a vertical wall. The rod is in a uniform magnetic field B = 0.340 T directed into the plane of the 53.0° figure. There is current I = 6.50 A in P the rod, in the direction shown. (a) Calculate the torque due to the magnetic force on the rod, for an axis at P. Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? Explain. (b) When the rod is in equilibrium and makes an angle of 53.0° with the floor, is the spring stretched or compressed? (c) How much energy is stored in the spring when the rod is in equilibrium? 7.46 .. The loop of wire shown Figure P7.46 in Fig. P7.46 forms a right triangle Q and carries a current I = 5.00 A S in the direction shown. The loop is B in a uniform magnetic field that has magnitude B = 3.00 T and 0.600 m I I the same direction as the current in I side PQ of the loop. (a) Find the force exerted by the magnetic field P R 0.800 m on each side of the triangle. If the force is not zero, specify its direction. (b) What is the net force on the loop? (c) The loop is pivoted about an axis that lies along side PR. Use the forces calculated in part (a) to calculate the torque on each side of the loop (see Problem 7.79). (d) What is the magnitude of the net torque on the loop? Calculate the net torque from the torques calculated in part (c) and also from Eq. (7.28). Do these two results agree? (e) Is the net torque directed to rotate point Q into the plane of the figure or out of the plane of the figure? 7.47 .. CP A uniform, 458-g metal bar 75.0 Figure P7.47 cm long carries a current I in a uniform, horizontal 1.25-T magnetic field asSshown in Fig. S a B P7.47. The directions of I and B are shown in I the figure. The bar is free to rotate about a frictionless hinge at point b. The other end of the 60.0° bar rests on a conducting support at point a but b is not attached there. The bar rests at an angle of 60.0° above the horizontal. What is the largest value the current I can have without breaking the electrical contact at a? (See Problem 7.77.) 7.48 .. CALC A Voice Coil. It was shown in Section 7.7 that the net force on a current loop in a uniform magnetic field is zero. The mag- Figure P7.48 netic force on the voice coil of a y loudspeaker (see Fig. 7.28) is nonzero because the magnetic field at the coil is not uniform. A voice coil in a loud60.0° 60.0° speaker has 50 turns of wire and a S S diameter of 1.56 cm, and the current B B in the coil is 0.950 A. Assume that the I magnetic field at each point of the x coil has a constant magnitude of '

L

'

z

243

'

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CHAPTER 7 Magnetic Field and Magnetic Forces

(L, 0)

CHALLENGE PROBLEMS 7.56 ... CP A Cycloidal Path. x

7.50 . . CALC Torque on a Current Loop in a Nonuniform Magnetic Field. In Section 7.7 the expression for the torque on a S current loop was derived assuming that the magnetic field was B S uniform. But what if B is not uniform? Figure P7.49 shows a square loop of wire that lies in the xy-plane. The loop has corners at 10, 02, 10, L2, 1L, 02, and 1L, L2 and carries a constant current I in the clockwise direction. The magnetic field has no z-component but has S both x- and y-components: B 5 1B0y > L2≤n 1 1B0x > L2≥n, where B0 is a positive constant. (a) Sketch the magnetic field lines in the xyplane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating Eq. (7.20). (c) If the loop is free to rotate about the x-axis, find the magnitude and direction of the magnetic torque on the loop. (d) Repeat part (c) for the case in which the loop is free to rotate about the y-axis. (e) Is Eq. S S S (7.26), T 5 M : B, an appropriate description of the torque on this loop? Why or why not? 7.51 .. A circular loop of wire with area A lies in the xy-plane. As viewed along the z-axis looking in the - z-direction toward the origin, a current I is circulating clockwise around Sthe loop. The torque produced by an external magnetic field B is given by S T 5 D14≤n 2 3≥n), where D is a positive constant, and for this oriS S entation of the loop the magnetic potential energy U = -M # B is negative. The magnitude of the magnetic field is B0 = 13D > IA. (a) Determine the vector magnetic moment of theS current loop. (b) Determine the components Bx, By, and Bz of B. x 7.52 In a certain region uniform electric field E and magnetic field B are present in the B opposite direction. At the instant t = 0, a particle of mass m carrying a charge q is E q z '

×

(0, 0)

×

(L, L)

×

I

×

(0, L)

×

y

×

Figure P7.49

×

'

×

'

×

'

×

'

given velocity v0 at an angle θ, with the y axis, in the yz plane. Find the time after which the speed of the particle would be minimum. 7.53 A particle having mass m, charge q × × × enters a cylinder region having uniform × × × × × × × × × magnetic field B in the inward direction as shown. If the particle is deviated by q, m 60° as it emerges out of the field then what is the time spent by it in the field. 7.54 In a coordinate system an electric field E and a magnetic field B exist in X-direction and particle of charge q and mass m is projected with a velocity v from origin along Y-direction. A target is located at point (x0, 0, 0). Find the magnitude of magnetic induction B, so that the particle will hit the target when it is crossing the X-axis third time. 7.55 A non conducting uniform rod of length m q l and mass m is uniformly charged with l B charge q and released from horizontal position shown. A uniform magnetic field is existing in space perpendicular to plane of motion. Find (a) angular velocity of rod when it becomes vertical (b) normal reaction on the hinge at that instant. ×

0.220 T and is directed at an angle of 60.0° outward from the normal to the plane of the coil (Fig. P7.48). Let the axis of the coil be in the y-direction. The current in the coil is in the direction shown (counterclockwise as viewed from a point above the coil on the y-axis). Calculate the magnitude and direction of the net magnetic force on the coil. 7.49 .. CALC Force on a Current Loop in a Nonuniform Magnetic Field. It was shown in Section 7.7 that the net forceSon a current loop in a uniform magnetic field is zero. But what if B is not uniform? Figure P7.85 shows a square loop of wire that lies in the xy-plane. The loop has corners at 10, 02, 10, L2, 1L, 02, and constantI current I in the direction. L, 1L, 02, and 1L,carries L2 anda carries constanta current in the clockwise L2 and x The magnetic Sfield has no x-component but has both y-and both -and zz-component: -components: B 5 1 B0z > L2n≥ 1 1 B0y > L2kn, where B0 is a positive constant. (a) Sketch the magnetic field lines inyzthe - yz-plane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating Eq. (7.20). (c) Find the magnitude and direction of the net magnetic force on the loop. Figure

×

244

'

'

'

'

y

'

A particle with mass m and positive charge q starts from rest at Sthe origin shown in Fig. P7.56. There is a uniform electric field E in the +y-direction and a uniS form magnetic field B directed out of the page. It is shown in more advanced books that the path is a cycloid whose radius of curvature at the top points is twice the y-coordinate at that level. (a) Explain why the path has this general shape and why it is repetitive. (b) Prove that the speed at any point is equal to (Hint: Use energy conservation.) (c) Applying Newton’s second law at the top point and taking as given that the radius of curvature here equals 2y, prove that the speed at this point is 2E/B Figure P7.56 y S

E S

B

x

7.57 A particle of mass m and charge + q Y starts from rest at origin. There is a uniform E electric field E in the positive y-direction & a uniform magnetic field B directed towards the reader. The path of the particle is cycloid X as shown. (a) Obtain the speed of the particle at any point as a function of y. (b) Find the speed of the particle at the top point of the path.

V0

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Answers

245

Answers Chapter Opening Question

?

In MRI the nuclei of hydrogen atoms within soft tissue act like miniature current loops whose magnetic moments align with an applied field. See Section 7.7 for details.

Test Your Understanding Questions 7.1 yes When a magnet is cut apart, each part has a north and south pole (see Fig. 7.4). Hence the small red part behaves much like the original, full-sized compass needle. 7.2 path 3 Applying the right-hand rule to the vectors (which points to the right) and (which points into the plane of the figure) says that the force on a positive charge would point upward. Since the charge is negative, the force points downward and the particle follows a trajectory that curves downward. 7.3 (a) (ii), (b) no The magnitude of would increase as you moved to the right, reaching a maximum as you pass through the plane of the loop. As you moved beyond the plane of the loop, the field magnitude would decrease. You can tell this from the spacing of the field lines: The closer the field lines, the stronger the field. The direction of the field would be to the right at all points along the path, since the path is along a field line and the direction of at any point is tangent to the field line through that point. 7.4 (a) (ii), (b) (i) The radius of the orbit as given by Eq. (7.11) is directly proportional to the speed, so doubling the particle speed causes the radius to double as well. The particle has twice as far to travel to complete one orbit but is traveling at double the speed, so the time for one orbit is unchanged. This result also follows from Eq. (7.12), which states that the angular speed is independent of the linear speed Hence the time per orbit, likewise does not depend on 7.5 (iii) From Eq. (7.13), the speed v = E>B at which particles travel straight through the velocity selector does not depend on the magnitude or sign of the charge or the mass of the particle. All that is required is that the particles (in this case, ions) have a nonzero charge. 7.6 A This orientation will cause current to flow clockwise around the circuit and hence through the conducting bar in the direction from the top to the bottom of the figure. From the right-hand rule, S S S the magnetic force F 5 I l : B on the bar will then point to the right. 7.7 (a) to the right; (b) north pole on the right, south pole on the left If you wrap the fingers of your right hand around the coil in the direction of the current, your right thumb points to the right (perpendicular to the plane of the coil). This is the direction of the S magnetic moment M . The magnetic moment points from the south pole to the north pole, so the right side of the loop is equivalent to a north pole and the left side is equivalent to a south pole. 7.8 no The rotor will not begin to turn when the switch is closed if the rotor is initially oriented as shown in Fig. 7.39b. In this case there is no current through the rotor and hence no magnetic torque. This situation can be remedied by using multiple rotor coils oriented at different angles around the rotation axis. With this arrangement, there is always a magnetic torque no matter what the orientation.

7.9 (ii) The mobile charge carriers in copper are negatively charged electrons, which move upward through the wire to give a downward current. From the right-hand rule, the force on a positively charged particle moving upward in a westward-pointing magnetic field would be to the south; hence the force on a negatively charged particle is to the north. The result is an excess of negative charge on the north side of the wire, leaving an excess of positive charge— and hence a higher electric potential—on the south side.

Bridging Problem

(a) tx = -1.54 * 10-4 N # m,

ty = - 2.05 * 10-4 N # m, tz = -6.14 * 10-4 N # m

(c) 42.1 rad> s

(b) -7.55 * 10-4 J

Discussion Questions 7.1 Yes, if its velocity is parallel or anti-parallel to magnetic field. 7.2 Magnetic force depends on the magnitude and direction of velocity of charged particle. It is variable. 7.3 No, direction will be opposite. 7.4 No, the component of velocity parallel to magnetic field is not affected by magnetic field. 7.5 No, in order to complete the circular path it has to come out of the cubical region. 7.6 It can change the direction of velocity without changing its magnitude. Yes, centripetal force in circular motions. 7.7 If the external magnetic field is zero then external electric field also must be zero otherwise it will exert a force on charged particle and its velocity will not be constant. If external electric field is zero, and velocity of particle is parallel to magnetic field, velocity will be constant. 7.8 S

N

We can attach the loop to a paper wick marked with N, S. But N (north pole) and S (south pole) will depend on the direction of current. 7.9 Maximum force and maximum acceleration of the wire when current in it is perpendicular to magnetic field and zero force and acceleration when current is parallel or antiparallel to the magnetic field. 7.10 dF

dF I

B

I B

S

perpendicular to plane of paper inwards B S B perpendicular to plane of paper outwards 7.11

(a)

(b)

(c)

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(d)

246

CHAPTER 7 Magnetic Field and Magnetic Forces

^ ^ 7.12 Force on a is along ( - k); Force on b is along (j ); Force on d ^ is along ( - j ) ^ ^ Force on e is along vector ( - k - j ) ; No force on c. 7.13 Horizontal component of earth’s magnetic field L 10 - 4T = B L Torque about base = (iBL) (L S height of pole) 2 2

2

2

iBL 10L L = to (in Newton – m) 2 2 2 7.14 Magnetic force can never increase the speed of a charged particle. 7.15 Magnet of Loudspeaker disturbs the electron beam. 7.16 In a current carrying wire, all electrons move with a net drift velocity. So all electrons experience magnetic force. Since electrons cannot be detached from the wire, this force is shifted onto wire. If this force displaces the wire, work is done. 7.17 No, even 100% efficiency is not possible for a DC motor as in this case heat lost in armature must be zero, or current in armature is zero. But with no current in armature, motor cannot work. 7.18 Due to commutator and brushes, the commutator segments being insulated from each other. 7.19 If we move the entire conductor in the opposite direction to the current with a speed equal to drift speed, then the electrons are at rest with respect to magnetic field and the Hall emf disappears. 7.20 Hall-effect voltage is inversely proportional to number of charge carriers per unit volume. It is lesser in Germanium. =

Single Correct Q7.1 (b) Q7.2 (d) Q7.3 (c) Q7.4 (b) Q7.5 (a) Q7.6 (c) Q7.7 (a) Q7.8 (b) Q7.9 (c) Q7.10 (d) Q7.11 (c) Q7.12 (b) Q7.13 (d)

1 7.3 (a) 0, 3.25 * 1016 m/s2 (b) sin- 1 a b 4 7.4 (a) Bx = - 0.175 T, Bz = - 0.256 T (b) By is not determined (c) 0, 90° 7.5 (a) 1.46 T, in the xz plane directed at 40° from towards ( - z) axis

+ x axis

(b) Force = c a - 4.8î - 5.72kN b * 10 - 16 dN 7.6 5.78 * 10 - 4 Wb 7.7 - 7.4 * 10 - 4 Wb 7.8 (a) Zero (b) - 0.0115 Wb (c) + 0.0115 Wb (d) Zero 7.9 (a) 1.6 * 10 - 4 T, perpendicular to the page of paper inwards (b) 1.11 * 10 - 7 S 7.10 8 * 10 - 10 N, south 7.11 (a) 1.35 mm (c) 1.88 * 1012 m/s2 7.12 (a) 5 T, south (b) 1.37 * 107 m/s, south 7.13 (a) 8.35 * 105 m/s (b) 2.62 * 10 - 8 s (c) 7.27 * 103 Volt 7.14 (a) 2.84 * 106 m/s (b) 2 * 10 - 25 N, Yes (c) speed remains same 7.15 1.67 * 10 - 3 T

7.16 (a) a 1.60 * 10 - 14 N b jN (b) 1.4 cm 7.17 (a) 3.38 * 106 m/s (c) 4.17 * 10 - 3 m, 7.74 * 10 - 9 s

One or More Correct Q7.14 (a), (b), (c), (d) Q7.15 (b), (d) Q7.16 (b), (c), (d) Q7.17 (a), (b), (c) Q7.18 (a), (c) Q7.19 (c)

Exercises 7.1 1.91 T, eastward 7.2 (a) + ive (b) 0.0505 N

7.18 (a) 4.81 * 103 N/C (c) (i) Yes (ii) Yes 7.19 0.0445 T, perpendicular to electric field 7.20 (a) 5 * 103 m/s (b) 10 - 25 kg 7.21 (a) 1.5 * 104 A, No (b) Wire must be horizontal and perpendicular to the earth’s magnetic field 7.22 0.67 N at an angle tan - 1(2) clockwise from rightward direction 7.23 (a) 833 V

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Answers

(b) 115 m/s2 (b) Bz = ;

7.24 (a) a (b) 3.15 kg 7.25 (a) 4.7 * 10 - 3 N-m (b) 0.025 A-m2 (c) 6.34 * 10 - 3 N-m 7.26 (b) force on cd, 1.2 N (c) 0.42 N-m

7.39 current = 7.40 a

247

211 F0 qv

Mg(tanu) LB

, from right to left

240 bm/s2 13

7.41 (a) Force = ILB, rightwards v2m 2ILB (c) d = 1960 km (b) d =

7.27 (a) zero, zero (b) 0.0808 N-m, anticlockwise 7.28 (a) about A2

7.42 1.8 N, leftwards 7.43 0.024 T, + y direction 7.44 (a) 0.03 N # m, + Nj direction

(b) 290 rad/s2 S

7.29 (a) t = - (NIAB)î, U = 0

(b) 0.017 N # m, - Nj direction

S

(b) t = zero, U = - NIAB S

(c) t = (NIAB)î, U = 0 S

(d) t = zero, U = NIAB

(b) 0.05765 m

7.46 (a) FPQ = zero, FRP = 12 N (into the page), FQR = 12 N (out of the page) (b) zero

(b) 564 W (c) 493 W 7.31 (a) 1.13 A

(c) tPQ = tRP = zero, tQR = 3.6 N # m (d) 3.6 N # m

(b) 3.69 A (c) 98.2 V

(e) out of the plane

(d) 362 W 7.32 (a) magnitude

F2 , in - y direction qv1

(b) F2/ 22 7.33 (a) 1.5 * 10 - 16s (b) 1.1 mA (c) 9.3 * 10 - 24 A-m2 7.34 3.45 T, direction should be perpendicular to the initial velocity of the coin. S 7.35 (a) F = qvBxkN - qvBzî (b) Bx>0, Bzs; in much of the rest of the world, ƒ = 50 Hz 1v = 314 rad>s2 is used. Similarly, a sinusoidal current might be described as i = I cos vt

(11.2)

t

O

where i (lowercase) is the instantaneous current and I (uppercase) is the maximum current or current amplitude.

Phasor Diagrams 11.2 A phasor diagram. Length of phasor equals maximum current I.

Phasor rotates with frequency f and angular speed v 5 2pf. v I

Phasor

vt O

Projection of phasor onto horizontal axis at time t equals current i at that instant: i 5 I cos vt.

i 5 I cos vt

To represent sinusoidally varying voltages and currents, we will use rotating vector diagrams similar to those we used in the study of simple harmonic motion in Section 14.2 (see Figs. 14.5b and 14.6). In these diagrams the instantaneous value of a quantity that varies sinusoidally with time is represented by the projection onto a horizontal axis of a vector with a length equal to the amplitude of the quantity. The vector rotates counterclockwise with constant angular speed v. These rotating vectors are called phasors, and diagrams containing them are called phasor diagrams. Figure 11.2 shows a phasor diagram for the sinusoidal current described by Eq. (11.2). The projection of the phasor onto the horizontal axis at time t is I cos vt; this is why we chose to use the cosine function rather than the sine in Eq. (11.2). CAUTION Just what is a phasor? A phasor is not a real physical quantity with a direction in space, such as velocity, momentum, or electric field. Rather, it is a geometric entity that helps us to describe and analyze physical quantities that vary sinusoidally with time. In Section 14.2 we used a single phasor to represent the position of a point mass undergoing simple harmonic motion. In this chapter we will use phasors to add sinusoidal voltages and currents. Combining sinusoidal quantities with phase differences then becomes a matter of vector addition. We will find a similar use for phasors in Chapters 35 and 36 in our study of interference effects with light. y

Rectified Alternating Current How do we measure a sinusoidally varying current? In Section 6.3 we used a d’Arsonval galvanometer to measure steady currents. But if we pass a sinusoidal current through a d’Arsonval meter, the torque on the moving coil varies sinusoidally, with one direction half the time and the opposite direction the other half. The needle may wiggle a little if the frequency is low enough, but its average deflection is zero. Hence a d’Arsonval meter by itself isn’t very useful for measuring alternating currents.

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11.1 Phasors and Alternating Currents

To get a measurable one-way current through the meter, we can use diodes, which we described in Section 5.3. A diode is a device that conducts better in one direction than in the other; an ideal diode has zero resistance for one direction of current and infinite resistance for the other. Figure 11.3a shows one possible arrangement, called a full-wave rectifier circuit. The current through the galvanometer G is always upward, regardless of the direction of the current from the ac source (i.e., which part of the cycle the source is in). The graph in Fig. 11.3b shows the current through G: It pulsates but always has the same direction, and the average meter deflection is not zero. The rectified average current Irav is defined so that during any whole number of cycles, the total charge that flows is the same as though the current were constant with a value equal to Irav . The notation Irav and the name rectified average current emphasize that this is not the average of the original sinusoidal current. In Fig. 11.3b the total charge that flows in time t corresponds to the area under the curve of i versus t (recall that i = dq>dt, so q is the integral of t); this area must equal the rectangular area with height Irav . We see that Irav is less than the maximum current I; the two are related by

355

11.3 (a) A full-wave rectifier circuit. (b) Graph of the resulting current through the galvanometer G. (a) A full-wave rectifier circuit Source of alternating current

Alternating current

'

G

'

Irav =

2 I = 0.637I p

(rectified average value of a sinusoidal current)

(11.3)

(The factor of 2>p is the average value of ƒ cos vt ƒ or of ƒ sin vt ƒ ; see Example 29.4 in Section 29.2.) The galvanometer deflection is proportional to Irav . The galvanometer scale can be calibrated to read I, Irav , or, most commonly, Irms (discussed below).

Diode (arrowhead and bar indicate the directions in which current can and cannot pass) (b) Graph of the full-wave rectified current and its average value, the rectified average current Irav

'

Rectified current through galvanometer G

i

'

I Irav

Root-Mean-Square (rms) Values A more useful way to describe a quantity that can be either positive or negative is the root-mean-square (rms) value. We used rms values in Section 18.3 in connection with the speeds of molecules in a gas. We square the instantaneous current i, take the average (mean) value of i2, and finally take the square root of that average. This procedure defines the root-mean-square current, denoted as Irms (Fig. 11.4). Even when i is negative, i2 is always positive, so Irms is never zero (unless i is zero at every instant). Here’s how we obtain Irms for a sinusoidal current, like that shown in Fig. 11.4. If the instantaneous current is given by i = I cos vt, then

cos A =

11.4 Calculating the root-mean-square (rms) value of an alternating current. Meaning of the rms value of a sinusoidal quantity (here, ac current with I 5 3 A): 2 Square the instantaneous current i. 3 Take the average (mean) value of i 2.

Using a double-angle formula from trigonometry, 1 2 11

Area under curve 5 total charge that flows through galvanometer in time t.

1 Graph current i versus time.

i2 = I2 cos2 vt

2

t

O

4 Take the square root of that average.

+ cos 2A2

i, i 2 I2 i 2 5 I 2 cos2 vt (i 2)av 5 2 I 2 5 9A2

we find i2 = I2 12 11 + cos 2vt2 = 12 I2 + 12 I2 cos 2vt

2 3

The average of cos 2vt is zero because it is positive half the time and negative half the time. Thus the average of i2 is simply I2>2. The square root of this is Irms :

I 5 3A 4

'

Irms =

I 22

(root-mean-square value of a sinusoidal current)

(11.4)

O 2I

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t 1

i 5 I cos vt Irms 5 (i 2)av 5

I 2

356

CHAPTER 11 Alternating Current

11.5 This wall socket delivers a rootmean-square voltage of 120 V. Sixty times per second, the instantaneous voltage across its terminals varies from 11221120 V2 = 170 V to - 170 V and back again.

In the same way, the root-mean-square value of a sinusoidal voltage with amplitude (maximum value) V is Vrms =

V 22

(root-mean-square value of a sinusoidal voltage)

(11.5)

We can convert a rectifying ammeter into a voltmeter by adding a series resistor, just as for the dc case discussed in Section 26.3. Meters used for ac voltage and current measurements are nearly always calibrated to read rms values, not maximum or rectified average. Voltages and currents in power distribution systems are always described in terms of their rms values. The usual household power supply, “120-volt ac,” has an rms voltage of 120 V (Fig. 11.5). The voltage amplitude is V = 22 Vrms = 22 1120 V2 = 170 V Example 11.1

Current in a personal computer

The plate on the back of a personal computer says that it draws 2.7 A from a 120-V, 60-Hz line. For this computer, what are (a) the average current, (b) the average of the square of the current, and (c) the current amplitude?

11.6 Our graphs of the current i and the square of the current i2 versus time t.

SOLUTION IDENTIFY and SET UP: This example is about alternating current. In part (a) we find the average, over a complete cycle, of the alternating current. In part (b) we recognize that the 2.7-A current draw of the computer is the rms value Irms—that is, the square root of the mean (average) of the square of the current, 1i22av. In part (c) we use Eq. (11.4) to relate Irms to the current amplitude. EXECUTE: (a) The average of any sinusoidally varying quantity, over any whole number of cycles, is zero. (b) We are given Irms = 2.7 A. From the definition of rms value, Irms = 21i22av so 1i22av = 1Irms22 = 12.7 A22 = 7.3 A2 (c) From Eq. (11.4), the current amplitude I is

I = 12Irms = 1212.7 A2 = 3.8 A Figure 11.6 shows graphs of i and i2 versus time t.

B

A

I

I v

C

I

EVALUATE: Why would we be interested in the average of the square of the current? Recall that the rate at which energy is dissipated in a resistor R equals i2R. This rate varies if the current is alternating, so it is best described by its average value 1i22avR = Irms 2R. We’ll use this idea in Section 11.4.

Test Your Understanding of Section 11.1 The figure at left shows four different current phasors with the same angular frequency v. At the time shown, which phasor corresponds to (a) a positive current that is becoming more positive; (b) a positive current that is decreasing toward zero; (c) a negative current that is becoming more negative; (d) a negative current that is decreasing in magnitude toward zero?

I D

y

11.2

Resistance and Reactance

In this section we will derive voltage–current relationships for individual circuit elements carrying a sinusoidal current. We’ll consider resistors, inductors, and capacitors.

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357

11.2 Resistance and Reactance

Resistor in an ac Circuit 11.7 Resistance R connected across an ac source.

First let’s consider a resistor with resistance R through which there is a sinusoidal current given by Eq. (11.2): i = I cos vt. The positive direction of current is counterclockwise around the circuit, as in Fig. 11.7a. The current amplitude (maximum current) is I. From Ohm’s law the instantaneous potential vR of point a with respect to point b (that is, the instantaneous voltage across the resistor) is

(a) Circuit with ac source and resistor

vR = iR = 1IR2 cos vt

i

(11.6)

The maximum voltage VR , the voltage amplitude, is the coefficient of the cosine function:

a

'

VR = IR

(amplitude of voltage across a resistor, ac circuit)

(11.7)

b

R vR

(b) Graphs of current and voltage versus time i, v

Hence we can also write

i 5 I cos vt

I

vR = VR cos vt

(11.8)

The current i and voltage vR are both proportional to cos vt, so the current is in phase with the voltage. Equation (11.7) shows that the current and voltage amplitudes are related in the same way as in a dc circuit. Figure 11.7b shows graphs of i and vR as functions of time. The vertical scales for current and voltage are different, so the relative heights of the two curves are not significant. The corresponding phasor diagram is given in Fig. 11.7c. Because i and vR are in phase and have the same frequency, the current and voltage phasors rotate together; they are parallel at each instant. Their projections on the horizontal axis represent the instantaneous current and voltage, respectively.

vR 5 IR cos vt 5 VR cos vt VR t O

Amplitudes are in the same relationship as for a dc circuit: VR 5 IR. (c) Phasor diagram

Inductor in an ac Circuit Next, we replace the resistor in Fig. 11.7 with a pure inductor with self-inductance L and zero resistance (Fig. 11.8a). Again we assume that the current is i = I cos vt, with the positive direction of current taken as counterclockwise around the circuit. Although there is no resistance, there is a potential difference vL between the inductor terminals a and b because the current varies with time, giving rise to a self-induced emf. The induced emf in the direction of i is given by Eq. (30.7), E = -L di>dt; however, the voltage vL is not simply equal to E. To see why, notice that if the current in the inductor is in the positive (counterclockwise) direction from a to b and is increasing, then di>dt is positive and the induced emf is directed to the left to oppose the increase in current; hence point a is at higher potential than is point b. Thus the potential of point a with respect to point b is positive and is given by vL = +L di>dt, the negative of the induced emf. (You

Current is in phase with voltage: crests and troughs occur together.

I

Current phasor

Voltage phasor

VR

vt

vR

Current and voltage phasors are in phase: they rotate together. i

O Instantaneous voltage

Instantaneous current

11.8 Inductance L connected across an ac source. (a) Circuit with ac source and inductor

(b) Graphs of current and voltage versus time i, v I

i a

L vL

b

VL

(c) Phasor diagram Voltage phasor leads current phasor by f 5 p 2 rad 5 90°.

/

i 5 I cos vt vL 5 IvL cos 1 vt 1 90°2

I t

O

p T, rad 5 90° 2 Voltage curve leads current curve by a quartercycle (corresponding to f 5 p/2 rad 5 90°).

Voltage phasor VL vL

1 4

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O

Phase angle f Current phasor vt

i

358

CHAPTER 11 Alternating Current

ActivPhysics 14.3: AC Circuits: The Driven Oscillator (Questions 1–5)

should convince yourself that this expression gives the correct sign of vL in all cases, including i counterclockwise and decreasing, i clockwise and increasing, and i clockwise and decreasing; you should also review Section 30.2.) So we have vL = L

di d = L 1I cos vt2 = -IvL sin vt dt dt

(11.9)

The voltage vL across the inductor at any instant is proportional to the rate of change of the current. The points of maximum voltage on the graph correspond to maximum steepness of the current curve, and the points of zero voltage are the points where the current curve instantaneously levels off at its maximum and minimum values (Fig. 11.8b). The voltage and current are “out of step” or out of phase by a quarter-cycle. Since the voltage peaks occur a quarter-cycle earlier than the current peaks, we say that the voltage leads the current by 90°. The phasor diagram in Fig. 11.8c also shows this relationship; the voltage phasor is ahead of the current phasor by 90°. We can also obtain this phase relationship by rewriting Eq. (11.9) using the identity cos1A + 90°2 = - sin A: vL = IvL cos1vt + 90°2

(11.10)

This result shows that the voltage can be viewed as a cosine function with a “head start” of 90° relative to the current. As we have done in Eq. (11.10), we will usually describe the phase of the voltage relative to the current, not the reverse. Thus if the current i in a circuit is i = I cos vt and the voltage v of one point with respect to another is v = V cos1vt + f2 we call f the phase angle; it gives the phase of the voltage relative to the current. For a pure resistor, f = 0, and for a pure inductor, f = 90°. From Eq. (11.9) or (11.10) the amplitude VL of the inductor voltage is VL = IvL

(11.11)

We define the inductive reactance XL of an inductor as XL = vL

(inductive reactance)

(11.12)

Using XL , we can write Eq. (11.11) in a form similar to Eq. (11.7) for a resistor 1VR = IR2: '

VL = IXL

(amplitude of voltage across an inductor, ac circuit)

(11.13)

Because XL is the ratio of a voltage and a current, its SI unit is the ohm, the same as for resistance. CAUTION Inductor voltage and current are not in phase Keep in mind that Eq. (11.13) is a relationship between the amplitudes of the oscillating voltage and current for the inductor in Fig. 11.8a. It does not say that the voltage at any instant is equal to the current at that instant multiplied by XL . As Fig. 11.8b shows, the voltage and current are 90° out of phase. Voltage and current are in phase only for resistors, as in Eq. (11.6). y '

The Meaning of Inductive Reactance The inductive reactance XL is really a description of the self-induced emf that opposes any change in the current through the inductor. From Eq. (11.13), for a given current amplitude I the voltage vL = + L di>dt across the inductor and the self-induced emf E = -L di>dt both have an amplitude VL that is directly proportional to XL . According to Eq. (11.12), the inductive reactance and self-induced emf increase with more rapid variation in current (that is, increasing angular frequency v) and increasing inductance L. '

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11.2 Resistance and Reactance

359

If an oscillating voltage of a given amplitude VL is applied across the inductor terminals, the resulting current will have a smaller amplitude I for larger values of XL . Since XL is proportional to frequency, a high-frequency voltage applied to the inductor gives only a small current, while a lower-frequency voltage of the same amplitude gives rise to a larger current. Inductors are used in some circuit applications, such as power supplies and radio-interference filters, to block high frequencies while permitting lower frequencies or dc to pass through. A circuit device that uses an inductor for this purpose is called a low-pass filter (see Problem 11.52). '

Example 11.2

An inductor in an ac circuit

The current amplitude in a pure inductor in a radio receiver is to be 250 mA when the voltage amplitude is 3.60 V at a frequency of 1.60 MHz (at the upper end of the AM broadcast band). (a) What inductive reactance is needed? What inductance? (b) If the voltage amplitude is kept constant, what will be the current amplitude through this inductor at 16.0 MHz? At 160 kHz? S O L U T IO N IDENTIFY and SET UP: There may be other elements of this circuit, but in this example we don’t care: All they do is provide the inductor with an oscillating voltage, so the other elements are lumped into the ac source shown in Fig. 11.8a. We are given the current amplitude I and the voltage amplitude V. Our target variables in part (a) are the inductive reactance XL at 1.60 MHz and the inductance L, which we find using Eqs. (11.13) and (11.12). Knowing L, we use these equations in part (b) to find XL and I at any frequency.

From Eq. (11.12), with v = 2pƒ, L =

XL 1.44 * 104 Æ = 2pƒ 2p11.60 * 106 Hz2

= 1.43 * 10-3 H = 1.43 mH (b) Combining Eqs. (11.12) and (11.13), we find I = VL>XL = VL>vL = VL>2pƒL. Thus the current amplitude is inversely proportional to the frequency ƒ. Since I = 250 mA at ƒ = 1.60 MHz, the current amplitudes at 16.0 MHz 110ƒ2 and 160 kHz = 0.160 MHz 1ƒ>102 will be, respectively, one-tenth as great 125.0 mA2 and ten times as great 12500 mA = 2.50 mA2. EVALUATE: In general, the lower the frequency of an oscillating voltage applied across an inductor, the greater the amplitude of the resulting oscillating current.

EXECUTE: (a) From Eq. (11.13), XL =

VL 3.60 V = = 1.44 * 104 Æ = 14.4 kÆ I 250 * 10-6 A

Capacitor in an ac Circuit Finally, we connect a capacitor with capacitance C to the source, as in Fig. 11.9a, producing a current i = I cos vt through the capacitor. Again, the positive direction of current is counterclockwise around the circuit. CAUTION Alternating current through a capacitor You may object that charge can’t really move through the capacitor because its two plates are insulated from each other. True enough, but as the capacitor charges and discharges, there is at each instant a current i into one plate, an equal current out of the other plate, and an equal displacement current between the plates just as though the charge were being conducted through the capacitor. (You may want to review the discussion of displacement current in Section 29.7.) Thus we often speak about alternating current through a capacitor. y

To find the instantaneous voltage vC across the capacitor—that is, the potential of point a with respect to point b—we first let q denote the charge on the lefthand plate of the capacitor in Fig. 11.9a (so - q is the charge on the right-hand plate). The current i is related to q by i = dq>dt; with this definition, positive current corresponds to an increasing charge on the left-hand capacitor plate. Then i =

dq = I cos vt dt

Integrating this, we get q =

I sin vt v

(11.14)

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360

CHAPTER 11 Alternating Current

Also, from Eq. (24.1) the charge q equals the voltage vC multiplied by the capacitance, q = CvC . Using this in Eq. (11.14), we find

11.9 Capacitor C connected across an ac source.

'

(a) Circuit with ac source and capacitor

vC = 2q

q

i

i

I sin vt vC

(11.15)

The instantaneous current i is equal to the rate of change dq>dt of the capacitor charge q; since q = CvC , i is also proportional to the rate of change of voltage. (Compare to an inductor, for which the situation is reversed and vL is proportional to the rate of change of i.) Figure 11.9b shows vC and i as functions of t. Because i = dq>dt = C dvC>dt, the current has its greatest magnitude when the vC curve is rising or falling most steeply and is zero when the vC curve instantaneously levels off at its maximum and minimum values. The capacitor voltage and current are out of phase by a quarter-cycle. The peaks of voltage occur a quarter-cycle after the corresponding current peaks, and we say that the voltage lags the current by 90°. The phasor diagram in Fig. 11.9c shows this relationship; the voltage phasor is behind the current phasor by a quartercycle, or 90°. We can also derive this phase difference by rewriting Eq. (11.15) using the identity cos1A - 90°2 = sin A: '

a

b C vC

(b) Graphs of current and voltage versus time i, v I VC

i 5 I cos vt I vC 5 cos 1vt 2 90°2 vC t

O

p rad 5 90° 2 Voltage curve lags current curve by a quartercycle (corresponding to f 5 2p/2 rad 5 290°). 1 4

T,

vC =

(11.16)

This corresponds to a phase angle f = - 90°. This cosine function has a “late start” of 90° compared with the current i = I cos vt. Equations (11.15) and (11.16) show that the maximum voltage VC (the voltage amplitude) is

(c) Phasor diagram Current phasor I

VC =

Phase angle f vt

I cos1vt - 90°2 vC

I vC

(13.17)

To put this expression in a form similar to Eq. (11.7) for a resistor, VR = IR, we define a quantity XC , called the capacitive reactance of the capacitor, as

i

'

O Voltage phasor

vC VC

Voltage phasor lags current phasor by f 5 2p 2 rad 5 290°.

XC =

/

1 vC

(capacitive reactance)

(11.18)

Then VC = IXC

(amplitude of voltage across a capacitor, ac circuit)

(11.19)

The SI unit of XC is the ohm, the same as for resistance and inductive reactance, because XC is the ratio of a voltage and a current. CAUTION Capacitor voltage and current are not in phase Remember that Eq. (11.19) for a capacitor, like Eq. (11.13) for an inductor, is not a statement about the instantaneous values of voltage and current. The instantaneous values are actually 90° out of phase, as Fig. 11.9b shows. Rather, Eq. (11.19) relates the amplitudes of the voltage and current. y

The Meaning of Capacitive Reactance The capacitive reactance of a capacitor is inversely proportional both to the capacitance C and to the angular frequency v; the greater the capacitance and the higher the frequency, the smaller the capacitive reactance XC . Capacitors tend to pass high-frequency current and to block low-frequency currents and dc, just the opposite of inductors. A device that preferentially passes signals of high frequency is called a high-pass filter (see Problem 11.51). '

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361

11.2 Resistance and Reactance

Example 11.3

A resistor and a capacitor in an ac circuit

A 200-Æ resistor is connected in series with a 5.0-mF capacitor. The voltage across the resistor is vR = 11.20 V2 cos12500 rad>s2t (Fig. 11.10). (a) Derive an expression for the circuit current. (b) Determine the capacitive reactance of the capacitor. (c) Derive an expression for the voltage across the capacitor. S O L U T IO N IDENTIFY and SET UP: Since this is a series circuit, the current is the same through the capacitor as through the resistor. Our target variables are the current i, the capacitive reactance XC, and the capacitor voltage vC. We use Eq. (11.6) to find an expression for i in terms of the angular frequency v = 2500 rad>s, Eq. (11.18) to find XC, Eq. (11.19) to find the capacitor voltage amplitude VC, and Eq. (11.16) to write an expression for vC.

11.10 Our sketch for this problem.

EXECUTE: (a) From Eq. (11.6), vR = iR, we find i =

11.20 V2 cos12500 rad>s2t vR = R 200 Æ

= 16.0 * 10-3 A2 cos12500 rad>s2t (b) From Eq. (11.18), the capacitive reactance at v = 2500 rad>s is XC =

1 1 = = 80 Æ vC 12500 rad>s215.0 * 10-6 F2

(c) From Eq. (11.19), the capacitor voltage amplitude is VC = IXC = 16.0 * 10-3 A2180 Æ2 = 0.48 V (The 80-Æ reactance of the capacitor is 40% of the resistor’s 200-Æ resistance, so VC is 40% of VR.) The instantaneous capacitor voltage is given by Eq. (11.16): vC = VC cos1vt - 90°2

= 10.48 V2 cos312500 rad>s2t - p>2 rad4

EVALUATE: Although the same current passes through both the capacitor and the resistor, the voltages across them are different in both amplitude and phase. Note that in the expression for vC we converted the 90° to p>2 rad so that all the angular quantities have the same units. In ac circuit analysis, phase angles are often given in degrees, so be careful to convert to radians when necessary.

Comparing ac Circuit Elements Table 11.1 summarizes the relationships of voltage and current amplitudes for the three circuit elements we have discussed. Note again that instantaneous voltage and current are proportional in a resistor, where there is zero phase difference between vR and i (see Fig. 11.7b). The instantaneous voltage and current are not proportional in an inductor or capacitor, because there is a 90° phase difference in both cases (see Figs. 11.8b and 11.9b). Figure 11.11 shows how the resistance of a resistor and the reactances of an inductor and a capacitor vary with angular frequency v. Resistance R is independent of frequency, while the reactances XL and XC are not. If v = 0, corresponding to a dc circuit, there is no current through a capacitor because XC S q , and there is no inductive effect because XL = 0. In the limit v S q , XL also approaches infinity, and the current through an inductor becomes vanishingly small; recall that the self-induced emf opposes rapid changes in current. In this same limit, XC and the voltage across a capacitor both approach zero; the current changes direction so rapidly that no charge can build up on either plate. Figure 11.12 shows an application of the above discussion to a loudspeaker system. Low-frequency sounds are produced by the woofer, which is a speaker with large diameter; the tweeter, a speaker with smaller diameter, produces highfrequency sounds. In order to route signals of different frequency to the appropriate speaker, the woofer and tweeter are connected in parallel across the amplifier

11.11 Graphs of R, XL , and XC as functions of angular frequency v. '

R, X XC

R

O

Table 11.1 Circuit Elements with Alternating Current Circuit Element

Amplitude Relationship

Circuit Quantity

Phase of v

Resistor Inductor Capacitor

VR = IR VL = IXL VC = IXC

R XL = vL XC = 1>vC

In phase with i Leads i by 90° Lags i by 90°

XL

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v

362

CHAPTER 11 Alternating Current

11.12 (a) The two speakers in this loudspeaker system are connected in parallel to the amplifier. (b) Graphs of current amplitude in the tweeter and woofer as functions of frequency for a given amplifier voltage amplitude. (a) A crossover network in a loudspeaker system From amplifier

Tweeter C

V

output. The capacitor in the tweeter branch blocks the low-frequency components of sound but passes the higher frequencies; the inductor in the woofer branch does the opposite.

Test Your Understanding of Section 11.2 An oscillating voltage of fixed amplitude is applied across a circuit element. If the frequency of this voltage is increased, will the amplitude of the current through the element (i) increase, (ii) decrease, or (iii) remain the same if it is (a) a resistor, (b) an inductor, or (c) a capacitor?

11.3 R A L B

R Woofer

(b) Graphs of rms current as functions of frequency for a given amplifier voltage The inductor and capacitor feed low Irms frequencies mainly to the woofer and high frequencies mainly to the tweeter. Tweeter

The L-R-C Series Circuit

Many ac circuits used in practical electronic systems involve resistance, induc tive reactance, and capacitive reactance. Figure 11.13a shows a simple example: A series circuit containing a resistor, an inductor, a capacitor, and an ac source. (In Section 30.6 we considered the behavior of the current in an L-R-C series circuit without a source.) To analyze this and similar circuits, we will use a phasor diagram that includes the voltage and current phasors for each of the components. In this circuit, because of Kirchhoff’s loop rule, the instantaneous total voltage vad across all three components is equal to the source voltage at that instant. We will show that the phasor representing this total voltage is the vector sum of the phasors for the individual voltages. Figures 11.13b and 11.13c show complete phasor diagrams for the circuit of Fig. 11.13a. We assume that the source supplies a current i given by i = I cos vt. Because the circuit elements are connected in series, the current at any instant is the same at every point in the circuit. Thus a single phasor I, with length proportional to the current amplitude, represents the current in all circuit elements. As in Section 11.2, we use the symbols vR , vL , and vC for the instantaneous voltages across R, L, and C, and the symbols VR , VL , and VC for the maximum voltages. We denote the instantaneous and maximum source voltages by v and V. Then, in Fig. 11.13a, v = vad , vR = vab , vL = vbc , and vC = vcd . We have shown that the potential difference between the terminals of a resistor is in phase with the current in the resistor and that its maximum value VR is given by Eq. (11.7): '

'

'

Crossover point

'

Woofer f

O

y

'

'

'

'

VR = IR 11.13 An L-R-C series circuit with an ac source. (a) L-R-C series circuit

(b) Phasor diagram for the case XL . XC Source voltage phasor is the vector sum of the VR, VL, and VC phasors.

i d

a 2q

C

R q b

Inductor voltage phasor leads current V 5 IX L L phasor by 90°.

All circuit elements have V 5 IZ the same current phasor.

(c) Phasor diagram for the case XL , XC If XL , XC, the source voltage phasor lags the current phasor, X , 0, and f is a negative angle between 0 and 290°. I VR 5 IR VL 5 IXL

c

I

L

O

f VL 2 VC

vt VR 5 IR

O

Capacitor voltage phasor lags current phasor VC 5 IXC by 90°. It is thus always antiparallel to the VL phasor.

Resistor voltage phasor is in phase with current phasor.

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f

V 5 IZ

vt VL 2 VC VC 5 IXC

11.3 The L-R-C Series Circuit

The phasor VR in Fig. 11.13b, in phase with the current phasor I, represents the voltage across the resistor. Its projection onto the horizontal axis at any instant gives the instantaneous potential difference vR . The voltage across an inductor leads the current by 90°. Its voltage amplitude is given by Eq. (11.13): '

363

PhET: Circuit Construction Kit (AC+DC) PhET: Faraday’s Electromagnetic Lab ActivPhysics 14.3: AC Circuits: The Driven Oscillator (Questions 6, 7, and 10)

VL = IXL The phasor VL in Fig. 11.13b represents the voltage across the inductor, and its projection onto the horizontal axis at any instant equals vL . The voltage across a capacitor lags the current by 90°. Its voltage amplitude is given by Eq. (11.19): '

VC = IXC The phasor VC in Fig. 11.13b represents the voltage across the capacitor, and its projection onto the horizontal axis at any instant equals vC . The instantaneous potential difference v between terminals a and d is equal at every instant to the (algebraic) sum of the potential differences vR , vL , and vC . That is, it equals the sum of the projections of the phasors VR , VL , and VC . But the sum of the projections of these phasors is equal to the projection of their vector sum. So the vector sum V must be the phasor that represents the source voltage v and the instantaneous total voltage vad across the series of elements. To form this vector sum, we first subtract the phasor VC from the phasor VL . (These two phasors always lie along the same line, with opposite directions.) This gives the phasor VL - VC . This is always at right angles to the phasor VR , so from the Pythagorean theorem the magnitude of the phasor V is '

'

'

'

'

'

'

'

'

'

V = 2VR2 + 1VL - VC22 = 21IR22 + 1IXL - IXC22 or

V = I 2R2 + 1XL - XC22

(11.20)

We define the impedance Z of an ac circuit as the ratio of the voltage amplitude across the circuit to the current amplitude in the circuit. From Eq. (11.20) the impedance of the L-R-C series circuit is Z = 2R2 + 1XL - XC22

(11.21)

so we can rewrite Eq. (11.20) as V = IZ

(amplitude of voltage across an ac circuit)

11.14 This gas-filled glass sphere has an alternating voltage between its surface and the electrode at its center. The glowing streamers show the resulting alternating current that passes through the gas. When you touch the outside of the sphere, your fingertips and the inner surface of the sphere act as the plates of a capacitor, and the sphere and your body together form an L-R-C series circuit. The current (which is low enough to be harmless) is drawn to your fingers because the path through your body has a low impedance.

(11.22)

While Eq. (11.21) is valid only for an L-R-C series circuit, we can use Eq. (11.22) to define the impedance of any network of resistors, inductors, and capacitors as the ratio of the amplitude of the voltage across the network to the current amplitude. The SI unit of impedance is the ohm.

The Meaning of Impedance and Phase Angle Equation (11.22) has a form similar to V = IR, with impedance Z in an ac circuit playing the role of resistance R in a dc circuit. Just as direct current tends to follow the path of least resistance, so alternating current tends to follow the path of lowest impedance (Fig. 11.14). Note, however, that impedance is actually a function of R, L, and C, as well as of the angular frequency v. We can see this by substituting Eq. (11.12) for XL and Eq. (11.18) for XC into Eq. (11.21), giving the following complete expression for Z for a series circuit: Z = 2R2 + 1XL - XC22

= 2R2 + 3vL - 11>vC242

(impedance of an L-R-C series circuit)

(11.23)

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364

CHAPTER 11 Alternating Current

Measuring Body Fat by Bioelectric Impedance Analysis

Application

The electrodes attached to this overweight patient’s chest are applying a small ac voltage of frequency 50 kHz. The attached instrumentation measures the amplitude and phase angle of the resulting current through the patient’s body. These depend on the relative amounts of water and fat along the path followed by the current, and so provide a sensitive measure of body composition.

Hence for a given amplitude V of the source voltage applied to the circuit, the amplitude I = V>Z of the resulting current will be different at different frequencies. We’ll explore this frequency dependence in detail in Section 11.5. In the phasor diagram shown in Fig. 11.13b, the angle f between the voltage and current phasors is the phase angle of the source voltage v with respect to the current i; that is, it is the angle by which the source voltage leads the current. From the diagram, tan f =

tan f =

V L - VC I1XL - XC2 XL - XC = = VR IR R

vL - 1>vC R

(phase angle of an L-R-C series circuit2

(11.24)

If the current is i = I cos vt, then the source voltage v is v = V cos1vt + f2

(11.25)

Figure 11.13b shows the behavior of a circuit in which XL 7 XC . Figure 11.13c shows the behavior when XL 6 XC ; the voltage phasor V lies on the opposite side of the current phasor I and the voltage lags the current. In this case, XL - XC is negative, tan f is negative, and f is a negative angle between 0 and -90°. Since XL and XC depend on frequency, the phase angle f depends on frequency as well. We’ll examine the consequences of this in Section 11.5. All of the expressions that we’ve developed for an L-R-C series circuit are still valid if one of the circuit elements is missing. If the resistor is missing, we set R = 0; if the inductor is missing, we set L = 0. But if the capacitor is missing, we set C = q , corresponding to the absence of any potential difference 1vC = q>C = 02 or any capacitive reactance 1XC = 1>vC = 02. In this entire discussion we have described magnitudes of voltages and currents in terms of their maximum values, the voltage and current amplitudes. But we remarked at the end of Section 11.1 that these quantities are usually described in terms of rms values, not amplitudes. For any sinusoidally varying quantity, the rms value is always 1> 12 times the amplitude. All the relationships between voltage and current that we have derived in this and the preceding sections are still valid if we use rms quantities throughout instead of amplitudes. For example, if we divide Eq. (11.22) by 12 , we get '

'

V I = Z 12 12 which we can rewrite as Vrms = Irms Z '

(11.26)

We can translate Eqs. (11.7), (11.13), and (11.19) in exactly the same way. We have considered only ac circuits in which an inductor, a resistor, and a capacitor are in series. You can do a similar analysis for an L-R-C parallel circuit; see Problem 11.56. Finally, we remark that in this section we have been describing the steadystate condition of a circuit, the state that exists after the circuit has been connected to the source for a long time. When the source is first connected, there may be additional voltages and currents, called transients, whose nature depends on the time in the cycle when the circuit is initially completed. A detailed analysis of transients is beyond our scope. They always die out after a sufficiently long time, and they do not affect the steady-state behavior of the circuit. But they can cause dangerous and damaging surges in power lines, which is why delicate electronic systems such as computers are often provided with power-line surge protectors.

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11.3 The L-R-C Series Circuit

Problem-Solving Strategy 11.1

Alternating-Current Circuits

IDENTIFY the relevant concepts: In analyzing ac circuits, we can apply all of the concepts used to analyze direct-current circuits, particularly those in Problem-Solving Strategies 6.1 and 6.2. But now we must distinguish between the amplitudes of alternating currents and voltages and their instantaneous values, and among resistance (for resistors), reactance (for inductors or capacitors), and impedance (for composite circuits). SET UP the problem using the following steps: 1. Draw a diagram of the circuit and label all known and unknown quantities. 2. Identify the target variables. EXECUTE the solution as follows: 1. Use the relationships derived in Sections 11.2 and 11.3 to solve for the target variables, using the following hints. 2. It’s almost always easiest to work with angular frequency v = 2pƒ rather than ordinary frequency ƒ. 3. Keep in mind the following phase relationships: For a resistor, voltage and current are in phase, so the corresponding phasors always point in the same direction. For an inductor, the voltage leads the current by 90° (i.e., f = +90° = p>2 radians), so the voltage phasor points 90° counterclockwise from the current phasor. For a capacitor, the voltage lags the current by 90° (i.e., f = - 90° = - p>2 radians), so the voltage phasor points 90° clockwise from the current phasor.

Example 11.4

365

4. Kirchhoff’s rules hold at each instant. For example, in a series circuit, the instantaneous current is the same in all circuit elements; in a parallel circuit, the instantaneous potential difference is the same across all circuit elements. 5. Inductive reactance, capacitive reactance, and impedance are analogous to resistance; each represents the ratio of voltage amplitude V to current amplitude I in a circuit element or combination of elements. However, phase relationships are crucial. In applying Kirchhoff’s loop rule, you must combine the effects of resistance and reactance by vector addition of the corresponding voltage phasors, as in Figs. 11.13b and 11.13c. When you have several circuit elements in series, for example, you can’t just add all the numerical values of resistance and reactance to get the impedance; that would ignore the phase relationships. EVALUATE your answer: When working with an L-R-C series circuit, you can check your results by comparing the values of the inductive and capacitive reactances XL and XC. If XL 7 XC, then the voltage amplitude across the inductor is greater than that across the capacitor and the phase angle f is positive (between 0 and 90°). If XL 6 XC, then the voltage amplitude across the inductor is less than that across the capacitor and the phase angle f is negative (between 0 and -90°).

An L-R-C series circuit I

In the series circuit of Fig. 11.13a, suppose R = 300 Æ, L = 60 mH, With source voltage amplitude V = 50 V, the current amplitude I C = 0.50 mF, V = 50 V, and v = 10,000 rad>s. Find the reactances and phase angle f are XL and XC, the impedance Z, the current amplitude I, the phase angle V 50 V f, and the voltage amplitude across each circuit element. I = = = 0.10 A Z 500 Æ XL - XC 400 Æ = arctan = 53° f = arctan S O L U T IO N R 300 Æ IDENTIFY and SET UP: This problem uses the ideas developed in From Table 11.1, the voltage amplitudes VR, VL, and VC across Section 11.2 and this section about the behavior of circuit elements in the resistor, inductor, and capacitor, respectively, are an ac circuit. We use Eqs. (11.12) and (11.18) to determine XL and XC, and Eq. (11.23) to find Z. We then use Eq. (11.22) to find VR = IR = 10.10 A21300 Æ2 = 30 V the current amplitude and Eq. (11.24) to find the phase angle. The VL = IXL = 10.10 A21600 Æ2 = 60 V relationships in Table 11.1 then yield the voltage amplitudes. VC = IXC = 10.10 A21200 Æ2 = 20 V EXECUTE: The inductive and capacitive reactances are EVALUATE: As in Fig. 11.13b, XL 7 XC; hence the voltage ampliXL = vL = 110,000 rad>s2160 mH2 = 600 Æ tude across the inductor is greater than that across the capacitor 1 1 and f is positive. The value f = 53° means that the voltage XC = = = 200 Æ vC leads the current by 53°. 110,000 rad>s210.50 * 10-6 F2 Note that the source voltage amplitude V = 50 V is not equal The impedance Z of the circuit is then to the sum of the voltage amplitudes across the separate circuit elements: 50 V Z 30 V + 60 V + 20 V. Instead, V is the vector Z = 2R2 + 1XL - XC22 = 21300 Æ22 + 1600 Æ - 200 Æ22 sum of the VR, VL, and VC phasors. If you draw the phasor dia= 500 Æ gram like Fig. 11.13b for this particular situation, you’ll see that VR, VL - VC, and V constitute a 3-4-5 right triangle.

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366

CHAPTER 11 Alternating Current

Example 11.5

An L-R-C series circuit II

For the L-R-C series circuit of Example 11.4, find expressions for the time dependence of the instantaneous current i and the instantaneous voltages across the resistor (vR), inductor (vL), capacitor (vC), and ac source (v,). SOLUTION IDENTIFY and SET UP: We describe the current using Eq. (11.2), which assumes that the current is maximum at t = 0. The voltages are then given by Eq. (11.8) for the resistor, Eq. (11.10) for the inductor, Eq. (11.16) for the capacitor, and Eq. (11.25) for the source. EXECUTE: The current and the voltages all oscillate with the same angular frequency, v = 10,000 rad>s, and hence with the same period, 2p>v = 2p>110,000 rad>s2 = 6.3 * 10-4 s = 0.63 ms. From Eq. (11.2), the current is i = I cos vt = 10.10 A2 cos110,000 rad>s2t

v = V cos1vt + f2 = 150 V2 cos c110,000 rad>s2t + a

2p rad b 153°2 d 360°

= 150 V2 cos3110,000 rad>s2t + 0.93 rad4

EVALUATE: Figure 11.15 graphs the four voltages versus time. The inductor voltage has a larger amplitude than the capacitor voltage because XL 7 XC. The instantaneous source voltage v is always equal to the sum of the instantaneous voltages vR, vL, and vC. You should verify this by measuring the values of the voltages shown in the graph at different values of the time t.

11.15 Graphs of the source voltage v, resistor voltage vR , inductor voltage vL , and capacitor voltage vC as functions of time for the situation of Example 11.4. The current, which is not shown, is in phase with the resistor voltage. '

'

The resistor voltage is in phase with the current, so

vR = VR cos vt = 130 V2 cos110,000 rad>s2t

v (V)

The inductor voltage leads the current by 90°, so 60

vL = VL cos1vt + 90°2 = - VL sin vt

40

= -160 V2 sin110,000 rad>s2t

20

VL 5 60 V V 5 50 V VR 5 30 V

The capacitor voltage lags the current by 90°, so –0.2

vC = VC cos1vt - 90°2 = VC sin vt = 120 V2 sin110,000 rad>s2t

0 –20

0.2

0.4

VC 5 20 V 0.6

t (ms)

–40

We found in Example 11.4 that the source voltage (equal to the voltage across the entire combination of resistor, inductor, and capacitor) leads the current by f = 53°, so

–60 KEY:

v

vR

vL

vC

Test Your Understanding of Section 11.3 Rank the following ac circuits in order of their current amplitude, from highest to lowest value. (i) the circuit in Example 11.4; (ii) the circuit in Example 11.4 with the capacitor and inductor both removed; (iii) the circuit in Example 11.4 with the resistor and capacitor both removed; (iv) the circuit in Example 11.4 with the resistor and inductor both removed.

11.4

y

Power in Alternating-Current Circuits

Alternating currents play a central role in systems for distributing, converting, and using electrical energy, so it’s important to look at power relationships in ac circuits. For an ac circuit with instantaneous current i and current amplitude I, we’ll consider an element of that circuit across which the instantaneous potential difference is v with voltage amplitude V. The instantaneous power p delivered to this circuit element is p = vi Let’s first see what this means for individual circuit elements. We’ll assume in each case that i = I cos vt.

Power in a Resistor Suppose first that the circuit element is a pure resistor R, as in Fig. 11.7a; then v = vR and i are in phase. We obtain the graph representing p by multiplying the

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11.4 Power in Alternating-Current Circuits

367

heights of the graphs of v and i in Fig. 11.7b at each instant. This graph is shown by the black curve in Fig. 11.16a. The product vi is always positive because v and i are always either both positive or both negative. Hence energy is supplied to the resistor at every instant for both directions of i, although the power is not constant. The power curve for a pure resistor is symmetrical about a value equal to onehalf its maximum value VI, so the average power Pav is Pav = 12 VI

1for a pure resistor)

(11.27)

An equivalent expression is Pav =

V I = Vrms Irms 12 12

(for a pure resistor)

'

(11.28)

Also, Vrms = Irms R, so we can express Pav by any of the equivalent forms '

Pav = Irms 2R =

Vrms 2 = Vrms Irms R '

(for a pure resistor)

(11.29)

Note that the expressions in Eq. (11.29) have the same form as the corresponding relationships for a dc circuit, Eq. (5.18). Also note that they are valid only for pure resistors, not for more complicated combinations of circuit elements.

Power in an Inductor Next we connect the source to a pure inductor L, as in Fig. 11.8a. The voltage v = vL leads the current i by 90°. When we multiply the curves of v and i, the product vi is negative during the half of the cycle when v and i have opposite signs. The power curve, shown in Fig. 11.16b, is symmetrical about the horizontal axis; it is positive half the time and negative the other half, and the average power is zero. When p is positive, energy is being supplied to set up the magnetic field in the inductor; when p is negative, the field is collapsing and the inductor is returning energy to the source. The net energy transfer over one cycle is zero.

Power in a Capacitor Finally, we connect the source to a pure capacitor C, as in Fig. 11.9a. The voltage v = vC lags the current i by 90°. Figure 11.16c shows the power curve; the average power is again zero. Energy is supplied to charge the capacitor and is returned

11.16 Graphs of current, voltage, and power as functions of time for (a) a pure resistor, (b) a pure inductor, (c) a pure capacitor, and (d) an arbitrary ac circuit that can have resistance, inductance, and capacitance. (b) Pure inductor

(a) Pure resistor For a resistor, p 5 vi is always positive because v and i are either both positive or both negative at any instant. v, i, p VI

v, i, p

p

I

Instantaneous current, i

v, i, p 1 Pav 5 2 VI cos f p

p

p v t v i

KEY:

v, i, p

V O

For an arbitrary combination of resistors, inductors, and capacitors, the average power is positive.

For an inductor or capacitor, p 5 vi is alternately positive and negative, and the average power is zero.

1

Pav 5 2 VI

(d) Arbitrary ac circuit

(c) Pure capacitor

t

O

O

i Instantaneous voltage across device, v

t

v i

O

t v i

Instantaneous power input to device, p

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368

CHAPTER 11 Alternating Current

to the source when the capacitor discharges. The net energy transfer over one cycle is again zero.

Power in a General ac Circuit In any ac circuit, with any combination of resistors, capacitors, and inductors, the voltage v across the entire circuit has some phase angle f with respect to the current i. Then the instantaneous power p is given by p = vi = 3V cos1vt + f243I cos vt4

(11.30)

The instantaneous power curve has the form shown in Fig. 11.16d. The area between the positive loops and the horizontal axis is greater than the area between the negative loops and the horizontal axis, and the average power is positive. We can derive from Eq. (11.30) an expression for the average power Pav by using the identity for the cosine of the sum of two angles: p = 3V1cos vt cos f - sin vt sin f243I cos vt4 = VI cos f cos2 vt - VI sin f cos vt sin vt

From the discussion in Section 11.1 that led to Eq. (11.4), we see that the average value of cos2 vt (over one cycle) is 12 . The average value of cos vt sin vt is zero because this product is equal to 12 sin 2vt, whose average over a cycle is zero. So the average power Pav is (average power into a general ac circuit)

Pav = 12 VI cos f = Vrms Irms cos f '

11.17 Using phasors to calculate the average power for an arbitrary ac circuit. 1

Average power 5 2 I(V cos f), where V cos f is the component of V in phase with I. I

V f V cos f vt

When v and i are in phase, so f = 0, the average power equals 12 VI = Vrms Irms ; when v and i are 90° out of phase, the average power is zero. In the general case, when v has a phase angle f with respect to i, the average power equals 12 I multiplied by V cos f, the component of the voltage phasor that is in phase with the current phasor. Figure 11.17 shows the general relationship of the current and voltage phasors. For the L-R-C series circuit, Figs. 11.13b and 11.13c show that V cos f equals the voltage amplitude VR for the resistor; hence Eq. (11.31) is the average power dissipated in the resistor. On average there is no energy flow into or out of the inductor or capacitor, so none of Pav goes into either of these circuit elements. The factor cos f is called the power factor of the circuit. For a pure resistance, f = 0, cos f = 1, and Pav = Vrms Irms . For a pure inductor or capacitor, f = 6 90°, cos f = 0, and Pav = 0. For an L-R-C series circuit the power factor is equal to R>Z; we leave the proof of this statement to you (see Exercise 11.21). A low power factor (large angle f of lag or lead) is usually undesirable in power circuits. The reason is that for a given potential difference, a large current is needed to supply a given amount of power. This results in large i2R losses in the transmission lines. Your electric power company may charge a higher rate to a client with a low power factor. Many types of ac machinery draw a lagging current; that is, the current drawn by the machinery lags the applied voltage. Hence the voltage leads the current, so f 7 0 and cos f 6 1. The power factor can be corrected toward the ideal value of 1 by connecting a capacitor in parallel with the load. The current drawn by the capacitor leads the voltage (that is, the voltage across the capacitor lags the current), which compensates for the lagging current in the other branch of the circuit. The capacitor itself absorbs no net power from the line. '

'

O

Example 11.6

(11.31)

'

'

Power in a hair dryer

An electric hair dryer is rated at 1500 W (the average power) at 120 V (the rms voltage). Calculate (a) the resistance, (b) the rms

current, and (c) the maximum instantaneous power. Assume that the dryer is a pure resistor. (The heating element acts as a resistor.) Continued

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11.5 Resonance in Alternating-Current Circuits S O L U T IO N IDENTIFY and SET UP: We are given Pav = 1500 W and Vrms = 120 V. Our target variables are the resistance R, the rms current Irms, and the maximum value p max of the instantaneous power p. We solve Eq. (11.29) to find R, Eq. (11.28) to find Irms from Vrms and Pav, and Eq. (11.30) to find p max . EXECUTE: (a) From Eq. (11.29), the resistance is Vrms 2 1120 V22 R = = = 9.6 Æ Pav 1500 W

369

(c) For a pure resistor, the voltage and current are in phase and the phase angle f is zero. Hence from Eq. (11.30), the instantaneous power is p = VI cos2 vt and the maximum instantaneous power is pmax = VI. From Eq. (11.27), this is twice the average power Pav, so pmax = VI = 2Pav = 211500 W2 = 3000 W EVALUATE: We can confirm our result in part (b) by using Eq. (11.7): Irms 5 Vrms>R 5 1120 V2>19.6 Æ2 = 12.5 A. Note that some unscrupulous manufacturers of stereo amplifiers advertise the peak power output rather than the lower average value.

(b) From Eq. (11.28), Irms =

Example 11.7

Pav 1500 W = = 12.5 A Vrms 120 V

Power in an L-R-C series circuit

For the L-R-C series circuit of Example 11.4, (a) calculate the power factor and (b) calculate the average power delivered to the entire circuit and to each circuit element. S O L U T IO N IDENTIFY and SET UP: We can use the results of Example 11.4. The power factor is the cosine of the phase angle f, and we use Eq. (11.31) to find the average power delivered in terms of f and the amplitudes of voltage and current.

EXECUTE: (a) The power factor is cos f = cos 53° = 0.60. (b) From Eq. (11.31), Pav = 12 VI cos f = 12 150 V210.10 A210.602 = 1.5 W EVALUATE: Although Pav is the average power delivered to the L-R-C combination, all of this power is dissipated in the resistor. As Figs. 11.16b and 11.16c show, the average power delivered to a pure inductor or pure capacitor is always zero.

Test Your Understanding of Section 11.4 Figure 11.16d shows that during part of a cycle of oscillation, the instantaneous power delivered to the circuit is negative. This means that energy is being extracted from the circuit. (a) Where is the energy extracted from? (i) the resistor; (ii) the inductor; (iii) the capacitor; (iv) the ac source; (v) more than one of these. (b) Where does the energy go? (i) the resistor; (ii) the inductor; (iii) the capacitor; (iv) the ac source; (v) more than one of these. y

11.5

Resonance in Alternating-Current Circuits

?

Much of the practical importance of L-R-C series circuits arises from the way in which such circuits respond to sources of different angular frequency v. For example, one type of tuning circuit used in radio receivers is simply an L-R-C series circuit. A radio signal of any given frequency produces a current of the same frequency in the receiver circuit, but the amplitude of the current is greatest if the signal frequency equals the particular frequency to which the receiver circuit is “tuned.” This effect is called resonance. The circuit is designed so that signals at other than the tuned frequency produce currents that are too small to make an audible sound come out of the radio’s speakers. To see how an L-R-C series circuit can be used in this way, suppose we connect an ac source with constant voltage amplitude V but adjustable angular frequency v across an L-R-C series circuit. The current that appears in the circuit has the same angular frequency as the source and a current amplitude I = V>Z, where Z is the impedance of the L-R-C series circuit. This impedance depends on the frequency, as Eq. (11.23) shows. Figure 11.18a shows graphs of R, XL , XC , and Z as functions of v. We have used a logarithmic angular frequency scale so that we can cover a wide range of frequencies. As the frequency increases, XL increases and XC decreases; hence there is always one frequency at which '

ActivPhysics 14.3: AC Circuits: The Driven Oscillator (Questions 8, 9, and 11)

'

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370

CHAPTER 11 Alternating Current

11.18 How variations in the angular frequency of an ac circuit affect (a) reactance, resistance, and impedance, and (b) impedance, current amplitude, and phase angle. (a) Reactance, resistance, and impedance as functions of angular frequency Impedance Z is least at the angular frequency at which XC 5 XL. Z 5 !R2 1 (XL 2 XC)2

R, X, Z

XL and XC are equal and XL - XC is zero. At this frequency the impedance Z = 2R2 + 1XL - XC22 has its smallest value, equal simply to the resistance R.

Circuit Behavior at Resonance As we vary the angular frequency v of the source, the current amplitude I = V>Z varies as shown in Fig. 11.18b; the maximum value of I occurs at the frequency at which the impedance Z is minimum. This peaking of the current amplitude at a certain frequency is called resonance. The angular frequency v0 at which the resonance peak occurs is called the resonance angular frequency. This is the angular frequency at which the inductive and capacitive reactances are equal, so at resonance,

XC

XL = XC XL v0

log v

XL 2 XC

Logarithmic scale

(b) Impedance, current, and phase angle as functions of angular frequency Current peaks at the angular frequency at which impedance is least. This is the resonance angular frequency v0. I, Z I

'

1 v0 C '

R

O

v0 L =

Z

f 90°

v0 =

(L-R-C series circuit at resonance)

1 1LC

(11.32)

Note that this is equal to the natural angular frequency of oscillation of an L-C circuit, which we derived in Section 30.5, Eq. (30.22). The resonance frequency f0 is v0>2p. This is the frequency at which the greatest current appears in the circuit for a given source voltage amplitude; in other words, f0 is the frequency to which the circuit is “tuned.” It’s instructive to look at what happens to the voltages in an L-R-C series circuit at resonance. The current at any instant is the same in L and C. The voltage across an inductor always leads the current by 90°, or 14 cycle, and the voltage across a capacitor always lags the current by 90°. Therefore the instantaneous voltages across L and C always differ in phase by 180°, or 12 cycle; they have opposite signs at each instant. At the resonance frequency, and only at the resonance frequency, XL = XC and the voltage amplitudes VL = IXL and VC = IXC are equal; then the instantaneous voltages across L and C add to zero at each instant, and the total voltage vbd across the L-C combination in Fig. 11.13a is exactly zero. The voltage across the resistor is then equal to the source voltage. So at the resonance frequency the circuit behaves as if the inductor and capacitor weren’t there at all! The phase of the voltage relative to the current is given by Eq. (11.24). At frequencies below resonance, XC is greater than XL ; the capacitive reactance dominates, the voltage lags the current, and the phase angle f is between zero and -90°. Above resonance, the inductive reactance dominates, the voltage leads the current, and the phase angle f is between zero and +90°. Figure 11.18b shows this variation of f with angular frequency. '

O

v0

log v

f Logarithmic scale

290° f

Tailoring an ac Circuit 11.19 Graph of current amplitude I as a function of angular frequency v for an L-R-C series circuit with V = 100 V, L = 2.0 H, C = 0.50 mF, and three different values of the resistance R. I (A) 200 V

0.5

The lower a circuit’s resistance, the higher and sharper is the resonance peak in the current near the 500 V resonance angular frequency v0.

0.4 0.3 0.2 0.1

2000 V

O

500

1000

1500

2000

/

v (rad s)

If we can vary the inductance L or the capacitance C of a circuit, we can also vary the resonance frequency. This is exactly how a radio or television receiving set is “tuned” to receive a particular station. In the early days of radio this was accomplished by the use of capacitors with movable metal plates whose overlap could be varied to change C. (This is what is being done with the radio tuning knob shown in the photograph that opens this chapter.) A more modern approach is to vary L by using a coil with a ferrite core that slides in or out. In an L-R-C series circuit the impedance reaches its minimum value and the current its maximum value at the resonance frequency. The middle curve in Fig. 11.19 is a graph of current as a function of frequency for such a circuit, with source voltage amplitude V = 100 V, L = 2.0 H, C = 0.50 mF, and R = 500 Æ. This curve is called a response curve or a resonance curve. The resonance angular frequency is v0 = 1LC2-1>2 = 1000 rad>s. As we expect, the curve has a peak at this angular frequency. The resonance frequency is determined by L and C; what happens when we change R? Figure 11.19 also shows graphs of I as a function of v for R = 200 Æ and for R = 2000 Æ. The curves are similar for frequencies far away

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11.5 Resonance in Alternating-Current Circuits

371

from resonance, where the impedance is dominated by XL or XC . But near resonance, where XL and XC nearly cancel each other, the curve is higher and more sharply peaked for small values of R and broader and flatter for large values of R. At resonance, Z = R and I = V>R, so the maximum height of the curve is inversely proportional to R. The shape of the response curve is important in the design of radio and television receiving circuits. The sharply peaked curve is what makes it possible to discriminate between two stations broadcasting on adjacent frequency bands. But if the peak is too sharp, some of the information in the received signal is lost, such as the high-frequency sounds in music. The shape of the resonance curve is also related to the overdamped and underdamped oscillations that we described in Section 30.6. A sharply peaked resonance curve corresponds to a small value of R and a lightly damped oscillating system; a broad, flat curve goes with a large value of R and a heavily damped system. In this section we have discussed resonance in an L-R-C series circuit. Resonance can also occur in an ac circuit in which the inductor, resistor, and capacitor are connected in parallel. We leave the details to you (see Problem 11.57). Resonance phenomena occur not just in ac circuits, but in all areas of physics. We discussed examples of resonance in mechanical systems in Sections 13.8 and 16.5. The amplitude of a mechanical oscillation peaks when the driving-force frequency is close to a natural frequency of the system; this is analogous to the peaking of the current in an L-R-C series circuit. We suggest that you review the sections on mechanical resonance now, looking for the analogies. '

Example 11.8

Tuning a radio

The series circuit in Fig. 11.20 is similar to some radio tuning circuits. It is connected to a variable-frequency ac source with an rms terminal voltage of 1.0 V. (a) Find the resonance frequency. At the resonance frequency, find (b) the inductive reactance XL, the capacitive reactance XC, and the impedance Z; (c) the rms current I rms; (d) the rms voltage across each circuit element.

have Z = R. We use Eqs. (11.7), (11.13), and (11.19) to find the voltages across the circuit elements. EXECUTE: (a) The values of v0 and ƒ0 are v0 =

1 2LC

1

= 210.40 * 10

-3

H21100 * 10-12 F2

= 5.0 * 10 6 rad>s

S O L U T IO N IDENTIFY and SET UP: Figure 11.20 shows an L-R-C series circuit, with ideal meters inserted to measure the rms current and voltages, our target variables. Equations (11.32) include the formula for the resonance angular frequency v0, from which we find the resonance frequency ƒ0. We use Eqs. (11.12) and (11.18) to find XL and XC, which are equal at resonance; at resonance, from Eq. (11.23), we

11.20 A radio tuning circuit at resonance. The circles denote rms current and voltages.

ƒ0 = 8.0 * 10 5 Hz = 800 kHz This frequency is in the lower part of the AM radio band. (b) At this frequency, XL = vL = 15.0 * 106 rad>s210.40 * 10-3 H2 = 2000 Æ XC =

1 1 = = 2000 Æ vC 15.0 * 106 rad>s21100 * 10-12 F2

Since XL = XC at resonance as stated above, Z = R = 500 Æ. (c) From Eq. (11.26) the rms current at resonance is Vrms Vrms 1.0 V Irms = = = = 0.0020 A = 2.0 mA Z R 500 Æ (d) The rms potential difference across the resistor is

VR-rms = IrmsR = 10.0020 A21500 Æ2 = 1.0 V

1.0 V 2.0 mA

The rms potential differences across the inductor and capacitor are

R 5 500 V L 5 0.40 mH

a

b 1.0 V

C 5 100 pF

c

d

4.0 V

4.0 V 0 V

VL-rms = IrmsXL = 10.0020 A212000 Æ2 = 4.0 V VC-rms = IrmsXC = 10.0020 A212000 Æ2 = 4.0 V

EVALUATE: The potential differences across the inductor and the capacitor have equal rms values and amplitudes, but are 180° out of phase and so add to zero at each instant. Note also that at resonance, VR-rms is equal to the source voltage Vrms, while in this example, VL-rms and VC-rms are both considerably larger than Vrms.

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372

CHAPTER 11 Alternating Current Test Your Understanding of Section 11.5 How does the resonance frequency of an L-R-C series circuit change if the plates of the capacitor are brought closer together? (i) It increases; (ii) it decreases; (iii) it is unaffected. y

11.6

11.21 Schematic diagram of an idealized step-up transformer. The primary is connected to an ac source; the secondary is connected to a device with resistance R. The induced emf per turn is the same in both coils, so we adjust the ratio of terminal voltages by adjusting the ratio of turns: N V2 5 2 N1 V1 Source of alternating current I1

Iron core

Transformers

One of the great advantages of ac over dc for electric-power distribution is that it is much easier to step voltage levels up and down with ac than with dc. For longdistance power transmission it is desirable to use as high a voltage and as small a current as possible; this reduces i2R losses in the transmission lines, and smaller wires can be used, saving on material costs. Present-day transmission lines routinely operate at rms voltages of the order of 500 kV. On the other hand, safety considerations and insulation requirements dictate relatively low voltages in generating equipment and in household and industrial power distribution. The standard voltage for household wiring is 120 V in the United States and Canada and 240 V in many other countries. The necessary voltage conversion is accomplished by the use of transformers.

How Transformers Work Figure 11.21 shows an idealized transformer. The key components of the transformer are two coils or windings, electrically insulated from each other but wound on the same core. The core is typically made of a material, such as iron, with a very large relative permeability Km . This keeps the magnetic field lines due to a current in one winding almost completely within the core. Hence almost all of these field lines pass through the other winding, maximizing the mutual inductance of the two windings (see Section 30.1). The winding to which power is supplied is called the primary; the winding from which power is delivered is called the secondary. The circuit symbol for a transformer with an iron core, such as those used in power distribution systems, is '

V1 N1

N2 V2

Primary winding

R

FB Secondary winding

Dangers of ac Versus dc Voltages

Application

Alternating current at high voltage (above 500 V) is more dangerous than direct current at the same voltage. When a person touches a high-voltage dc source, it usually causes a single muscle contraction that can be strong enough to push the person away from the source. By contrast, touching a high-voltage ac source can cause a continuing muscle contraction that prevents the victim from letting go of the source. Lowering the ac voltage with a transformer reduces the risk of injury.

Here’s how a transformer works. The ac source causes an alternating current in the primary, which sets up an alternating flux in the core; this induces an emf in each winding, in accordance with Faraday’s law. The induced emf in the secondary gives rise to an alternating current in the secondary, and this delivers energy to the device to which the secondary is connected. All currents and emfs have the same frequency as the ac source. Let’s see how the voltage across the secondary can be made larger or smaller in amplitude than the voltage across the primary. We neglect the resistance of the windings and assume that all the magnetic field lines are confined to the iron core, so at any instant the magnetic flux £ B is the same in each turn of the primary and secondary windings. The primary winding has N1 turns and the secondary winding has N2 turns. When the magnetic flux changes because of changing currents in the two coils, the resulting induced emfs are E1 = - N1

d£ B d£ B and E2 = - N2 dt dt

(11.33)

The flux per turn £ B is the same in both the primary and the secondary, so Eqs. (11.33) show that the induced emf per turn is the same in each. The ratio of the secondary emf E2 to the primary emf E1 is therefore equal at any instant to the ratio of secondary to primary turns: N2 E2 = E1 N1

(11.34)

Since E1 and E2 both oscillate with the same frequency as the ac source, Eq. (11.34) also gives the ratio of the amplitudes or of the rms values of the induced

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11.6 Transformers

emfs. If the windings have zero resistance, the induced emfs E1 and E2 are equal to the terminal voltages across the primary and the secondary, respectively; hence N2 V2 = V1 N1

(terminal voltages of transformer primary and secondary)

(11.35)

where V1 and V2 are either the amplitudes or the rms values of the terminal voltages. By choosing the appropriate turns ratio N2>N1 , we may obtain any desired secondary voltage from a given primary voltage. If N2 7 N1 , as in Fig. 11.21, then V2 7 V1 and we have a step-up transformer; if N2 6 N1 , then V2 6 V1 and we have a step-down transformer. At a power generating station, step-up transformers are used; the primary is connected to the power source and the secondary is connected to the transmission lines, giving the desired high voltage for transmission. Near the consumer, step-down transformers lower the voltage to a value suitable for use in home or industry (Fig. 11.22). Even the relatively low voltage provided by a household wall socket is too high for many electronic devices, so a further step-down transformer is necessary. This is the role of an “ac adapter” such as those used to recharge a mobile phone or laptop computer from line voltage. Such adapters contain a step-down transformer that converts line voltage to a lower value, typically 3 to 12 volts, as well as diodes to convert alternating current to the direct current that small electronic devices require (Fig. 11.23).

373

11.22 The cylindrical can near the top of this power pole is a step-down transformer. It converts the high-voltage ac in the power lines to low-voltage (120 V) ac, which is then distributed to the surrounding homes and businesses.

'

'

'

11.23 An ac adapter like this one converts household ac into low-voltage dc for use in electronic devices. It contains a step-down transformer to lower the voltage and diodes to rectify the output current (see Fig. 11.3).

Energy Considerations for Transformers If the secondary circuit is completed by a resistance R, then the amplitude or rms value of the current in the secondary circuit is I2 = V2>R. From energy considerations, the power delivered to the primary equals that taken out of the secondary (since there is no resistance in the windings), so V1 I1 = V2 I2 '

'

(currents in transformer primary and secondary)

(11.36)

We can combine Eqs. (11.35) and (11.36) and the relationship I2 = V2>R to eliminate V2 and I2 ; we obtain '

V1 R = I1 1N2>N122

(11.37)

This shows that when the secondary circuit is completed through a resistance R, the result is the same as if the source had been connected directly to a resistance equal to R divided by the square of the turns ratio, 1N2>N122. In other words, the transformer “transforms” not only voltages and currents, but resistances as well. More generally, we can regard a transformer as “transforming” the impedance of the network to which the secondary circuit is completed. Equation (11.37) has many practical consequences. The power supplied by a source to a resistor depends on the resistances of both the resistor and the source. It can be shown that the power transfer is greatest when the two resistances are equal. The same principle applies in both dc and ac circuits. When a high-impedance ac source must be connected to a low-impedance circuit, such as an audio amplifier connected to a loudspeaker, the source impedance can be matched to that of the circuit by the use of a transformer with an appropriate turns ratio N2>N1 . Real transformers always have some energy losses. (That’s why an ac adapter like the one shown in Fig. 11.23 feels warm to the touch after it’s been in use for a while; the transformer is heated by the dissipated energy.) The windings have some resistance, leading to i2R losses. There are also energy losses through hysteresis in the core (see Section 28.8). Hysteresis losses are minimized by the use of soft iron with a narrow hysteresis loop. Another important mechanism for energy loss in a transformer core involves eddy currents (see Section 29.6). Consider a section AA through an iron transformer core (Fig. 11.24a). Since iron is a conductor, any such section can be pictured as '

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374

CHAPTER 11 Alternating Current

11.24 (a) Primary and secondary windings in a transformer. (b) Eddy currents in the iron core, shown in the cross section at AA. (c) Using a laminated core reduces the eddy currents. (a) Schematic transformer

(b) Large eddy currents in solid core A

Solid core

(c) Smaller eddy currents in laminated core Laminated core

A Primary winding Secondary winding

Eddy currents Section at AA

Eddy currents Section at AA

several conducting circuits, one within the other (Fig. 11.24b). The flux through each of these circuits is continually changing, so eddy currents circulate in the entire volume of the core, with lines of flow that form planes perpendicular to the flux. These eddy currents are very undesirable; they waste energy through i2R heating and themselves set up an opposing flux. The effects of eddy currents can be minimized by the use of a laminated core— that is, one built up of thin sheets or laminae. The large electrical surface resistance of each lamina, due either to a natural coating of oxide or to an insulating varnish, effectively confines the eddy currents to individual laminae (Fig. 11.24c). The possible eddy-current paths are narrower, the induced emf in each path is smaller, and the eddy currents are greatly reduced. The alternating magnetic field exerts forces on the current-carrying laminae that cause them to vibrate back and forth; this vibration causes the characteristic “hum” of an operating transformer. You can hear this same “hum” from the magnetic ballast of a fluorescent light fixture (see Section 30.2). Thanks to the use of soft iron cores and lamination, transformer efficiencies are usually well over 90%; in large installations they may reach 99%. Example 11.9

“Wake up and smell the (transformer)!”

A friend returns to the United States from Europe with a 960-W coffeemaker, designed to operate from a 240-V line. (a) What can she do to operate it at the USA-standard 120 V? (b) What current will the coffeemaker draw from the 120-V line? (c) What is the resistance of the coffeemaker? (The voltages are rms values.) SOLUTION IDENTIFY and SET UP: Our friend needs a step-up transformer to convert 120-V ac to the 240-V ac that the coffeemaker requires. We use Eq. (11.35) to determine the transformer turns ratio N2>N1, Pav = VrmsIrms for a resistor to find the current draw, and Eq. (11.37) to calculate the resistance. EXECUTE: (a) To get V2 = 240 V from V1 = 120 V, the required turns ratio is N2>N1 = V2>V1 = 1240 V2>1120 V2 = 2. That is, the secondary coil (connected to the coffeemaker) should have twice as many turns as the primary coil (connected to the 120-V line).

(b) We find the rms current I1 in the 120-V primary by using Pav = V1I1, where Pav is the average power drawn by the coffeemaker and hence the power supplied by the 120-V line. (We’re assuming that no energy is lost in the transformer.) Hence I1 = Pav>V1 = 1960 W2>1120 V2 = 8.0 A. The secondary current is then I2 = Pav>V2 = 1960 W2>1240 V2 = 4.0 A. (c) We have V1 = 120 V, I1 = 8.0 A, and N2>N1 = 2, so V1 120 V = = 15 Æ I1 8.0 A

From Eq. (11.37),

R = 22115 Æ2 = 60 Æ

EVALUATE: As a check, V2>R = 1240 V2>160 Æ2 = 4.0 A = I2, the same value obtained previously. You can also check this result for R by using the expression Pav = V22>R for the power drawn by the coffeemaker.

Test Your Understanding of Section 11.6 Each of the following four transformers has 1000 turns in its primary coil. Rank the transformers from largest to smallest number of turns in the secondary coil. (i) converts 120-V ac into 6.0-V ac; (ii) converts 120-V ac into 240-V ac; (iii) converts 240-V ac into 6.0-V ac; (iv) converts 240-V ac into 120-V ac. y

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CHAPTER

11

SUMMARY

Phasors and alternating current: An alternator or ac source produces an emf that varies sinusoidally with time. A sinusoidal voltage or current can be represented by a phasor, a vector that rotates counterclockwise with constant angular velocity v equal to the angular frequency of the sinusoidal quantity. Its projection on the horizontal axis at any instant represents the instantaneous value of the quantity. For a sinusoidal current, the rectified average and rms (root-mean-square) currents are proportional to the current amplitude I. Similarly, the rms value of a sinusoidal voltage is proportional to the voltage amplitude V. (See Example 11.1.)

Irav =

2 I = 0.637I p

Irms =

I 12

v

(11.3)

I

(11.4)

vt O

Vrms =

V 12

i 5 I cos vt

(11.5)

Voltage, current, and phase angle: In general, the instantaneous voltage between two points in an ac circuit is not in phase with the instantaneous current passing through those points. The quantity f is called the phase angle of the voltage relative to the current.

i = I cos vt

Resistance and reactance: The voltage across a resistor R is in phase with the current. The voltage across an inductor L leads the current by 90° 1f = +90°2, while the voltage across a capacitor C lags the current by 90° 1f = - 90°2. The voltage amplitude across each type of device is proportional to the current amplitude I. An inductor has inductive reactance XL = vL, and a capacitor has capacitive reactance XC = 1>vC. (See Examples 11.2 and 11.3.)

VR = IR

Impedance and the L-R-C series circuit: In a general ac circuit, the voltage and current amplitudes are related by the circuit impedance Z. In an L-R-C series circuit, the values of L, R, C, and the angular frequency v determine the impedance and the phase angle f of the voltage relative to the current. (See Examples 11.4 and 11.5.)

V = IZ

Power in ac circuits: The average power input Pav to an ac circuit depends on the voltage and current amplitudes (or, equivalently, their rms values) and the phase angle f of the voltage relative to the current. The quantity cos f is called the power factor. (See Examples 11.6 and 11.7.)

Pav = 12 VI cos f

(11.2)

I

v = V cos1vt + f2

V f

V cos f vt

O

(11.7) R

a

VL = IXL

(11.13)

VC = IXC

(11.19)

L

a

i

C

a

Z = 2R + 1XL - XC2

b

i

2q

q

(11.22)

b

i

i

2

V 5 IZ

VL 5 IXL

2

= 2R2 + 3vL - 11>vC242 (11.23)

tan f =

b

vL - 1>vC R

I f VL 2 VC O

(11.24)

vt VR 5 IR

VC 5 IXC

v, i, p Pav 5 12 VI cos f

(11.31)

p

= Vrms Irms cos f '

f v

t v i

Resonance in ac circuits: In an L-R-C series circuit, the current becomes maximum and the impedance becomes minimum at an angular frequency called the resonance angular frequency. This phenomenon is called resonance. At resonance the voltage and current are in phase, and the impedance Z is equal to the resistance R. (See Example 11.8.)

v0 =

1 2LC

(11.32)

I (A) 0.5 0.4 0.3 0.2 0.1 O

200 V 500 V 2000 V 1000 2000

/

v (rad s)

375

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376

CHAPTER 11 Alternating Current

Transformers: A transformer is used to transform the voltage and current levels in an ac circuit. In an ideal transformer with no energy losses, if the primary winding has N1 turns and the secondary winding has N2 turns, the amplitudes (or rms values) of the two voltages are related by Eq. (11.35). The amplitudes (or rms values) of the primary and secondary voltages and currents are related by Eq. (11.36). (See Example 11.9.)

BRIDGING PROBLEM

V2 N2 = V1 N1

(11.35)

I1 V1

V1 I1 = V2 I2 '

'

(11.36)

N2 V2

N1 Primary

FB

R

Secondary

An Alternating-Current Circuit

A series circuit consists of a 1.50-mH inductor, a 125-Æ resistor, and a 25.0-nF capacitor connected across an ac source having an rms voltage of 35.0 V and variable frequency. (a) At what angular frequencies will the current amplitude be equal to 13 of its maximum possible value? (b) At the frequencies in part (a), what are the current amplitude and the voltage amplitude across each circuit element (including the ac source)? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. The maximum current amplitude occurs at the resonance angular frequency. This problem concerns the angular frequencies at which the current amplitude is one-third of that maximum. 2. Choose the equation that will allow you to find the angular frequencies in question, and choose the equations that you will

Problems

then use to find the current and voltage amplitudes at each angular frequency. EXECUTE 3. Find the impedance at the angular frequencies in part (a); then solve for the values of angular frequency. 4. Find the voltage amplitude across the source and the current amplitude for each of the angular frequencies in part (a). (Hint: Be careful to distinguish between amplitude and rms value.) 5. Use the results of steps 3 and 4 to find the reactances at each angular frequency. Then calculate the voltage amplitudes for the resistor, inductor, and capacitor. EVALUATE 6. Are there any voltage amplitudes that are greater than the voltage amplitude of the source? If so, does this mean your results are in error?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q11.1 Household electric power in most of western Europe is supplied at 240 V, rather than the 120 V that is standard in the United States and Canada. What are the advantages and disadvantages of each system? Q11.2 The current in an ac power line changes direction 120 times per second, and its average value is zero. Explain how it is possible for power to be transmitted in such a system. Q11.3 In an ac circuit, why is the average power for an inductor and a capacitor zero, but not for a resistor? Q11.4 Equation (11.14) was derived by using the relationship i = dq>dt between the current and the charge on the capacitor. In Fig. 11.9a the positive counterclockwise current increases the charge on the capacitor. When the charge on the left plate is positive but decreasing in time, is i = dq>dt still correct or should it be i = -dq>dt ? Is i = dq>dt still correct when the right-hand plate has positive charge that is increasing or decreasing in magnitude? Explain. '

Q11.5 Fluorescent lights often use an inductor, called a ballast, to limit the current through the tubes. Why is it better to use an inductor rather than a resistor for this purpose? Q11.6 Equation (11.9) says that vab = L di>dt (see Fig. 11.8a). Using Faraday’s law, explain why point a is at higher potential than point b when i is in the direction shown in Fig. 11.8a and is increasing in magnitude. When i is counterclockwise and decreasing in magnitude, is vab = L di>dt still correct, or should it be vab = -L di>dt? Is vab = L di>dt still correct when i is clockwise and increasing or decreasing in magnitude? Explain. Q11.7 Is it possible for the power factor of an L-R-C series ac circuit to be zero? Justify your answer on physical grounds. Q11.8 In an L-R-C series circuit, can the instantaneous voltage across the capacitor exceed the source voltage at that same instant? Can this be true for the instantaneous voltage across the inductor? Across the resistor? Explain. Q11.9 In an L-R-C series circuit, what are the phase angle f and power factor cos f when the resistance is much smaller than the

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Single Correct

inductive or capacitive reactance and the circuit is operated far from resonance? Explain. Q11.10 When an L-R-C series circuit is connected across a 120-V ac line, the voltage rating of the capacitor may be exceeded even if it is rated at 200 or 400 V. How can this be? Q11.11 In Example 11.6 (Section 11.4), a hair dryer is treated as a pure resistor. But because there are coils in the heating element and in the motor that drives the blower fan, a hair dryer also has inductance. Qualitatively, does including an inductance increase or decrease the values of R, Irms, and P? Q11.12 A light bulb and a parallel-plate capacitor with air between the plates are connected in series to an ac source. What happens to the brightness of the bulb when a dielectric is inserted between the plates of the capacitor? Explain. Q11.13 A coil of wire wrapped on a hollow tube and a light bulb are connected in series to an ac source. What happens to the brightness of the bulb when an iron rod is inserted in the tube? Q11.14 A circuit consists of a light bulb, a capacitor, and an inductor connected in series to an ac source. What happens to the brightness of the bulb when the inductor is removed? When the inductor is left in the circuit but the capacitor is removed? Explain. Q11.15 A circuit consists of a light bulb, a capacitor, and an inductor connected in series to an ac source. Is it possible for both the capacitor and the inductor to be removed and the brightness of the bulb to remain the same? Explain. Q11.16 Can a transformer be used with dc? Explain. What happens if a transformer designed for 120-V ac is connected to a 120-V dc line? Q11.17 An ideal transformer has N1 windings in the primary and N2 windings in its secondary. If you double only the number of secondary windings, by what factor does (a) the voltage amplitude in the secondary change, and (b) the effective resistance of the secondary circuit change? Q11.18 Some electrical appliances operate equally well on ac or dc, and others work only on ac or only on dc. Give examples of each, and explain the differences.

SINGLE CORRECT Q11.1 A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then (a) bulb will give more ontense light (b) bulb will give less intense light (c) bulb will give light of same intensity as before (d) bulb will stop radiating light. Q11.2 The graph shows the current versus the voltage in a driven RLC circuit at a fixed frequency. The arrow indicates the direction that Voltage this curve is drawn as time progresses. In this plot, the (a) Current lags the voltage by about 90 degrees (b) Current leads the voltage by about 90 degrees (c) Current and voltage are in phase (d) Current and voltage are 180 degrees out of phase Q11.3 The curve shows data taken in a driven RLC circuit experiment. The horizontal axis is the time axis. The solid curve represents the voltage. The dotted curve represents the current. Then

377

(a) this circuit acts inductively (b) this circuit acts capacitively (c) this circuits acts resistively (d) There is not enough information to decide nature of circuit Q11.4 Statement-1: Peak voltage across the resistance can be greater than the peak voltage of the source in a series LCR circuit. Statement-2: Peak voltage across the induction can be greater then the peak voltage of the source in an series LCR circuit. (a) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true.

Paragraph Question (2 question) In the RLC series circuit shown, the readings of voltmeters are V1 = 150 V, V2 = 50 V and the source has emf 130 V Q11.5 Find the power factor of the V1 circuit. (a) 3/5 (b) 4/5 (c) 12/13 V2 (d) 5/13 ~ Q11.6 Values of VL and VC and V respectively (a) 90 V and 40 V (b) 100 V and 50 V (c) 70 V and 20 V (d) 10 V and 40 V

ONE OR MORE CORRECT Q11.7 A series LCR circuit is opertaed at resonance. Then (a) voltage across R is minimum (b) impendance is minimum (c) power transferred is maximum (d) current amplitude is minimum Q11.8 When a resistance R is connected in series with an element A, the electric current is found to be lagging behing the voltage by u1. When the same resistance is connected in series with element B,current leads voltage by u2. When R, A, B are connected in series, the current now leads voltage by u. Assume same AC source is used in all cases, then : (a) u = u2 - u1 (b) tanu = tanu2 - tanu1 u1 + u2 (c) u = 2 (d) tanu = 2tanu2 - tanu1

MATCH THE COLUMN Q11.9 Four different circuit components are given in each situation of column-I and all the components are connected across an ac source of same angular frequency v = 200 rad/sec. The information of phase difference between the current and source voltage in each situation of column-I is given in column-II. Match the circuit components in column-I with corresponding results in column-II.

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378

CHAPTER 11 Alternating Current

Column-I (a)

Column-II

10W

500µF

(b) 5H

(c) 4H

3µF

(d) 1kW

5H

(p) the magnitude of require phase p difference is 2 (q) the magnitude of required phase p difference is 4 (r) the current leads in phase to source voltage (s) the current lags in phase to source voltage (t) power factor of the circuit is greater than 1 / 2

EXERCISES Section 11.1 Phasors and Alternating Currents

11.1 . You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 2A, even for an instant. What is the largest root-mean-square current you can run through this bulb? 11.2 . The voltage across the terminals of an ac power supply varies with time according to Eq. (11.1). The voltage amplitude is V = 50 V. What are (a) the root-mean-square potential difference Vrms and (b) the average potential difference Vav between the two terminals of the power supply?

Section 11.2 Resistance and Reactance 11.3

.

A capacitor is connected across an ac source that has volt80 age amplitude 50 V and frequency Hz (a) What is the phase p angle f for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What is the capacitance C of the capacitor if the current amplitude is 5 A? 11.4 . An inductor with L = 10 mH is connected across an ac source that has voltage amplitude 44 V. (a) What is the phase angle f for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of 4 A? 11.5 . A capacitance C and an inductance L are operated at the same angular frequency. (a) At what angular frequency will they have the same reactance? (b) If L = 5.00 mH and C = 3.50 mF, what is the numerical value of the angular frequency in part (a), and what is the reactance of each element? 11.6 .. A 0.2-H inductor is connected in series with a 100.0-Æ resistor and an ac source. The voltage across the inductor is vL = -112.0 V2 sin31500 rad>s2t4. (a) Derive an expression for the voltage vR across the resistor. (b) What is vR at t = 2.00 ms? 11.7 .. A 250-Æ resistor is connected in series with a 5-mF capacitor and an ac source. The voltage across the capacitor is vC = 17.5 V2 sin31120 rad>s2t4. (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage vR across the resistor.

Section 11.3 The L-R-C Series Circuit

11.8 . You have a 200-Æ resistor, a 0.400-H inductor, and a 6.00-mF capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 V and an angular frequency of 250 rad>s. (a) What is the

impedance of the circuit? (b) What is the current amplitude? (c) What are the voltage amplitudes across the resistor and across the inductor? (d) What is the phase angle f of the source voltage with respect to the current? Does the source voltage lag or lead the current? (e) Construct the phasor diagram. 11.9 . In an L-R-C series circuit, the rms voltage across the resistor is 30.0 V, across the capacitor it is 90.0 V, and across the inductor it is 50.0 V. What is the rms voltage of the source?

Section 11.4 Power in Alternating-Current Circuits

11.10 . The power of a certain CD player operating at 120 V rms is 20.0 W. Assuming that the CD player behaves like a pure resistor, find (a) the maximum instantaneous power; (b) the rms current; (c) the resistance of this player. 11.11 . (a) Show that for an L-R-C series circuit the power factor is equal to R>Z. (b) An L-R-C series circuit has phase angle - 37o. The voltage amplitude of the source is 90.0 V. What is the voltage amplitude across the resistor? 11.12 .. An L-R-C series circuit is connected to a 120-Hz ac source that has Vrms = 80.0 V. The circuit has a resistance of 75.0 Æ and an impedance at this frequency of 105 Æ. What average power is delivered to the circuit by the source?

Section 11.5 Resonance in Alternating-Current Circuits

11.13 .. In an L-R-C series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance XC of the capacitor is 200 Æ and the voltage amplitude across the capacitor is 600 V. The circuit has R = 300 Æ . What is the voltage amplitude of the source? 11.14 . An L-R-C series circuit is constructed using a 200-Æ resistor, a 20-mF capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V. (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source? 11.15 . An L-R-C series circuit consists of a source with voltage amplitude 120 V and angular frequency 50.0 rad>s, a resistor with R = 400 Æ , an inductor with L = 9.00 H, and a capacitor with capacitance C. (a) For what value of C will the current amplitude in the circuit be a maximum? (b) When C has the value calculated in part (a), what is the amplitude of the voltage across the inductor? 11.16 . In an L-R-C series circuit, R = 400 Æ, L = 0.3 H, and C = 0.0120 mF. (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 550 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

Section 11.6 Transformers

11.17 . A Step-Down Transformer. A transformer connected to a 120-V (rms) ac line is to supply 12.0 V (rms) to a portable electronic device. The load resistance in the secondary is 5.00 Æ. (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply?

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Problems

(c) What average power is delivered to the load? (d) What resistance connected directly across the 120-V line would draw the same power as the transformer? Show that this is equal to 5.00 Æ times the square of the ratio of primary to secondary turns. 11.18 . A Step-Up Transformer. A transformer connected to a 120-V (rms) ac line is to supply 12,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit exceeds 8.50 mA. (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is 8.50 mA? (c) What current rating should the fuse in the primary circuit have? 11.19 . Off to Europe! You plan to take your hair dryer to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The dryer puts out 1600 W at 120 V. (a) What could you do to operate your dryer via the 240-V line in Europe? (b) What current will your dryer draw from a European outlet? (c) What resistance will your dryer appear to have when operated at 240 V?

PROBLEMS 11.20 . A coil has a resistance of 48.0 Æ. At a frequency of 80.0 Hz the voltage across the coil leads the current in it by 53°. Determine the inductance of the coil. 11.21 .. Five infinite-impedance voltmeters, calibrated to read rms values, are connected as shown in Fig. P11.21. Let R = 600 Æ, L = 1 H, C = 5 mF, and V = 30.0 V. What is the reading of each voltmeter if (a) v = 200 rad>s and (b) v = 1000 rad>s?

379

changed to v3 = v1>3, what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (c) If the capacitor and inductor are placed in series with a resistor of resistance R to form an L-R-C series circuit, what will be the resonance angular frequency of the circuit? 11.26 .. A High-Pass Filter. Figure P11.26 One application of L-R-C series circuits is to high-pass or lowC Vs pass filters, which filter out R L either the low- or high-frequency components of a signal. Vout A high-pass filter is shown in Fig. P11.26, where the output voltage is taken across the L-R combination. (The L-R combination represents an inductive coil that also has resistance due to the large length of wire in the coil.) Derive an expression for Vout>Vs, the ratio of the output and source voltage amplitudes, as a function of the angular frequency v of the source. Show that when v is small, this ratio is proportional to v and thus is small, and show that the ratio approaches unity in the limit of large frequency. 11.27 .. A Low-Pass Filter. Figure P11.27 shows a low-pass filter (see Problem 11.26); the output voltage is taken across the capacitor in an L-R-C series circuit. Derive an expression for Vout>Vs, the ratio of the output and source voltage amplitudes, as a function of the angular frequency v of the source. Show that when v is large, this ratio is proportional to v-2 and thus is very small, and show that the ratio approaches unity in the limit of small frequency. Figure P11.27

Figure P11.21 C

Vs

a

c V1

R

C

L

R

d

b

V2

V3

L

11.28 ... An L-R-C series circuit is connected to an ac source of constant voltage amplitude V and variable angular frequency v. (a) Show that the current amplitude, as a function of v, is

V4

I = V5

11.22 .. An L-R-C series circuit has C = 5 mF, L = 0.8 H,and source voltage amplitude V = 56.0 V. The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude 80.0 V, what is the value of R for the resistor in the circuit? 11.23 .. A series circuit has an impedance of 60.0 Æ and a power factor of 0.8 at 50.0 Hz. The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity? 11.24 .. In an L-R-C series circuit, R = 300 Æ, XC = 100 Æ, and XL = 500 Æ . The average power consumed in the resistor is 48 W. (a) What is the power factor of the circuit? (b) What is the rms voltage of the source? 11.25 . At a frequency v1 the reactance of a certain capacitor equals that of a certain inductor. (a) If the frequency is changed to v2 = 2v1, what is the ratio of the reactance of the inductor to that of the capacitor? Which reactance is larger? (b) If the frequency is

Vout

V

2R

2

+ 1vL - 1>vC22

(b) Show that the average power dissipated in the resistor is P =

V2R>2

R2 + 1vL - 1>vC22

(c) Show that I and P are both maximum when v = 1> 2LC, the resonance frequency of the circuit. 11.29 .. The L-R-C Parallel Circuit. A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude V and angular frequency v. Let the source voltage be given by v = V cos vt. (a) Show that the instantaneous voltages vR, vL, and vC at any instant are each equal to v and that i = iR + iL + iC, where i is the current through the source and iR, iL, and iC are the currents through the resistor, the inductor, and the capacitor, respectively. (b) What are the phases of iR, iL, and iC with respect to v? Use current phasors to represent i, iR, iL, and iC. In a phasor diagram, show the phases of these four currents with respect to v. (c) Use the phasor diagram of part (b) to show that the current amplitude I for the current i through the source is given by

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380

CHAPTER 11 Alternating Current

I = ÄIR2 + 1IC - IL22. (d) Show that the result of part (c) can be

written as I = V>Z, with 1>Z = Ä1>R2 + 1vC - 1>vL22 . 11.30 . You want to double the resonance angular frequency of an L-R-C series circuit by changing only the pertinent circuit elements all by the same factor. (a) Which ones should you change? (b) By what factor should you change them? 11.31 .. In an L-R-C series circuit, the source has a voltage amplitude of 120 V, R = 80.0 Æ, and the reactance of the capacitor is 480 Æ. The voltage amplitude across the capacitor is 360 V. (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain. 11.32 .. CALC The current in a certain circuit varies with time Figure P11.32 as shown in Fig. P11.32. Find i the average current and the rms I0 current in terms of I0. 11.33 .. An inductor, a capacitor, t O t 2t and a resistor are all connected in 2I 0 series across an ac source. If the resistance, inductance, and capacitance are all doubled, by what factor does each of the following quantities change? Indicate whether they increase or decrease: (a) the resonance angular frequency; (b) the inductive reactance; (c) the capacitive reactance. (d) Does the impedance double? 11.34 . A resistance R, capacitance C, and inductance L are connected in series to a voltage source with amplitude V and variable angular frequency v. If v = v0, the resonance angular frequency, find (a) the maximum current in the resistor; (b) the maximum voltage across the capacitor; (c) the maximum voltage across the inductor; (d) the maximum energy stored in the capacitor; (e) the maximum energy stored in the inductor. Give your answers in terms of R, C, L, and V.

11.35 . Repeat Problem 11.34 for the case v = v0>2. 11.36 . Repeat Problem 11.34 for the case v = 2v0. 11.37 Two circuit element in a series connection have current and total voltage i = 4 cos (2000 t) A and V = 200 sin (2000 t + 37°) V Identify the two elements. 11.38 A series RC circuit with R = 10 Æ has an impendance with phase angle of 45° at frequency f = 500 Hz.Find the frequency for which the magnitude of the impedance is (a) twice that off, (b) one-half that at f. 11.39 An LCR series circuit with 100 Æ resistance is connected to an AC source of 200 V and angular frequency 300 radians/sec . When only the capacitance is removed The current lags behind the voltage by 60°. When only the inductance is removed the current leads the voltage 60°. Calculate the current and the power 130 V dissipated in the LCR circuit. V 11.40 The rms readings of voltmeters V1, V2, V3 are 150V, 130V and 120 V V 120 V V respectively.Find the applied ac rms 150 V voltage and power factor of the circuit. ~ 11.41 An A.C. source of angular frequency v is fed across a resistor R and a capacitor C in series, The current registered is I. If now the frequency of source is changed to v/3 (but maintanining the same voltage), the current in the circuit is found to be halved. Calculate the ratio ofreactance to resistance at the original frequency v . R L 11.42 The A.C. circuit shown in figure. Find the frequency v0 of the AC voltage source so C that current through the source will be same phase as of voltage of source. ~ 2

3

1

CHALLENGE PROBLEMS 11.43 ... CALC (a) At what angular frequency is the voltage amplitude across the resistor in an L-R-C series circuit at maximum value? (b) At what angular frequency is the voltage amplitude across the inductor at maximum value? (c) At what angular frequency is the voltage amplitude across the capacitor at maximum value?

Answers Chapter Opening Question

?

Yes. In fact, the radio simultaneously detects transmissions at all frequencies. However, a radio is an L-R-C series circuit, and at any given time it is tuned to have a resonance at just one frequency. Hence the response of the radio to that frequency is much greater than its response to any other frequency, which is why you hear only one broadcasting station through the radio’s speaker. (You can sometimes hear a second station if its frequency is sufficiently close to the tuned frequency.)

Test Your Understanding Questions 11.1 (a) D; (b) A; (c) B; (d) C For each phasor, the actual current is represented by the projection of that phasor onto the horizontal axis. The phasors all rotate counterclockwise around the origin with angular speed v, so at the instant shown the projection of phasor A is positive but trending toward zero; the projection of phasor B is negative and becoming more negative; the projection

of phasor C is negative but trending toward zero; and the projection of phasor D is positive and becoming more positive. 11.2 (a) (iii); (b) (ii); (c) (i) For a resistor, VR = IR, so I = VR>R. The voltage amplitude VR and resistance R do not change with frequency, so the current amplitude I remains constant. For an inductor, VL = IXL = IvL, so I = VL>vL. The voltage amplitude VL and inductance L are constant, so the current amplitude I decreases as the frequency increases. For a capacitor, VC = IXC = I>vC, so I = VCvC. The voltage amplitude VC and capacitance C are constant, so the current amplitude I increases as the frequency increases. 11.3 (iv), (ii), (i), (iii) For the circuit in Example 11.4, I = V>Z = 150 V2>1500 Æ2 = 0.10 A. If the capacitor and inductor are removed so that only the ac source and resistor remain, the circuit is like that shown in Fig. 11.7a; then I = V>R = 150 V2> 1300 Æ2 = 0.17 A. If the resistor and capacitor are removed so that only the ac source and inductor remain, the circuit is like that shown in Fig. 11.8a; then I = V>XL = 150 V2>1600 Æ2 = 0.083 A. Finally, if the resistor and inductor are removed so that only the ac source and capacitor remain, the circuit is like that '

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Bridging Problems

shown in Fig. 11.9a; then I = V>XC = 150 V2>1200 Æ2 = 0.25 A. 11.4 (a) (v); (b) (iv) The energy cannot be extracted from the resistor, since energy is dissipated in a resistor and cannot be recovered. Instead, the energy must be extracted from either the inductor (which stores magnetic-field energy) or the capacitor (which stores electric-field energy). Positive power means that energy is being transferred from the ac source to the circuit, so negative power implies that energy is being transferred back into the source. 11.5 (ii) The capacitance C increases if the plate spacing is decreased (see Section 24.1). Hence the resonance frequency ƒ0 = v0>2p = 1>2p 2LC decreases.

11.6 (ii), (iv), (i), (iii) From Eq. (11.35) the turns ratio is N2>N1 = V2>V1, so the number of turns in the secondary is N2 = N1V2>V1. Hence for the four cases we have (i) N2 = 11000216.0 V2>1120 V2 = 50 turns; (ii) N2 = 1100021240 V2> 1120 V2 = 2000 turns; (iii) N2 = 11000216.0 V2>1240 V2 = 25 turns; and (iv) N2 = 1100021120 V2>1240 V2 = 500 turns. Note that (i), (iii), and (iv) are step-down transformers with fewer turns in the secondary than in the primary, while (ii) is a step-up transformer with more turns in the secondary than in the primary. '

'

Bridging Problem

(a) 8.35 * 104 rad> s and 3.19 * 105 rad> s (b) At 8.35 * 104 rad> s: Vsource = 49.5 V, I = 0.132 A, VR = 16.5 V, VL = 16.5 V, VC = 63.2 V. At 3.19 * 105 rad> s: Vsource = 49.5 V, I = 0.132 A, VR = 16.5 V, VL = 63.2 V, VC = 16.5 V.

Discussion Questions 11.1 Efficiency is more at 240 V due to smaller losses but it is less safe. 11.2 Power is given to a system when instantaneous voltage and instantaneous current are in same direction. 11.3 Capacitor and inductor state and release energy whereas resistor dissipates energy. dq 11.4 Yes, as is negative and current is in opposite direction dt dq When right hand plate has +ve charge i = is still correct if we dt take q to be negative. 11.5 No power loss in inductor. 11.6 When current i is increasing induced EMF will be in such direction as to oppose this increase. So this EMF will produce some current in the external circuit in the direction opposite to i. di For this Vab (Va - Vb) must be + ve. And is position so dt di Vab = L . When i is counter clockwise and decreasing we have dt di di Vba = L and so Vab = - L . dt dt 11.7 Power factor is zero, when resistance of circuit is zero. Net power loss in inductor and capactior is zero. 11.8 Yes, Yes, Yes. For example source voltage may be zero at an instance but none of the three (i.e. VC, VL ande VR )is zero. R 11.9 cosf = L 0 Q f = 90° (±) 2 2 2R + (XC - XL) net power consumed will be almost zero.

381

11.10 Instantaneous voltage across capacitor may exceed source voltage as explained in question 11.8. 11.11 R increases as these inductance wires also have some resistVrms ance : Z = 2R2 + (XL)2, Irms = so Irms decreases Z (Vrms)2R (Vrms)2 P= cosf = Z Z2 If the resistance of inductance wire is almost zero, R is equal to its previous value and (Vrms)2R (Vrms)2 (Vrms)2 so P decreases P = 2 = < R XL 2 R + (XL)2 R + c d R 11.12 (Vrms)2cosf (Vrms)2R 2 (Vrms)2R (Vrms)2 = = 2 < Z Z R R + (Xc)2 1 When dielectric is inverted, Xc = decreases as capacitance C vC increases, so power P decreases and brightness also. 11.13 Insertion of iron rod increases inductance L. Power P =

(Vrms)2cosf (Vrms)2R (Vrms)2R (Vrms)2 = = 2 < 2 2 Z R Z R + (Xc)

Power P =

P decreases and brightness also (Vrms)2R

11.14 P =

Z2

= power

Z2 = R2 + (XC - XL)2 when inductor is removed Z2 = R2 + (XC)2 (XC - XL)2 may be greater than, equal to or less than (XC)2 So may be the relation between Z and Z' So P may increase, decrease or remain same 11.15 Yes, if XC = XL Z =

2R

2

+ (XC - XL)2 = R

11.16 No current will flow only in primary circuit and change of magnetic flux will not occur. 11.17 (a) Two times increases 1 (b) It become a b 4 11.18 Heat producing appliances can work both with AC or DC. AC motor work and DC motor respectively with AC and DC. Slip ring are used. AC is converted to DC in some appliance by using diode.

Single correct Q11.1 (a) Q11.2 (b) Q11.3 (a) Q11.4 (d) Q11.5 (c) Q11.6 (a)

One or More Correct Q11.7 (b), (c) Q11.8 (b)

Match the Column Q11.9 a S q, r, t; b S p, s; c S p, r; d S q, s, t

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382

CHAPTER 11 Alternating Current 11.26 c

Exercise 11.1 1 22 2A 11.2 (a) 25 22V, (b) zero 11.3 (a) f = - 90°, V lags i by 90°, (b) c = 625 mF 11.4 (a) f = 90° V leads i by 90°, (b) 175 HZ 100 1 25000 11.5 (a) , (b) rad/sec, Æ 3 3 LC

11.27

R2 + W2L2 R2 + a WL -

1

VL = 10 25 V, 1 (d) tan - 1 a b with voltage leading the current 2 (e) V VL

ø

I

VR

11.9 (a) 50 V 11.10 (a) 40 W, (b)

1 A, (c) 720Æ 6

11.11 (a) 72 V 11.12 48 W 11.13 900 V 11.14 (a) 2500 rad/sec, 200 Æ , (b) 0.125 A (c) Vsource = 12.5 V, VR = 12.5 V, VC =

5 23

V , VL =

- 5 23

Problems 7 11.20 (a) H 55 11.21 (a) 12.72 V, 4.24 V, 21.2 V, 16.96 V, 21.2 V, (b) 12.72 V, 21.2 V, 4.24 V, 16.96 V, 21.2 V 11.22 280 Æ 63 11.23 (a) Inductor, (b) H 550 11.24 (a) 0.6, (b) 200 V 1 11.25 (a) 4, XL > XC, (b) , XC > XL. 9

1/2

11.29 (b) IC

I V IR

IL

11.30 (a) L and C, (b) 1/2 11.31 (a) 0.75 A, (b) 160 Æ , (c) 619 Æ , 341 Æ , (d) for 341 Æ I0 11.32 23 1 11.33 (a) , decreases, (b) 2, increases, (c) 1/2, decreases, 2 (d) No V L 1 LV 2 V 1 LV2 V L 11.34 (a) , (b) , (c) , (d) , (e) 2 Ä R ÇC R 2 R 2 R2 R C L V 2V a b 11.35 (a) , (b) , ÇC 9L 9L 2 2 R + R + Ç 4C Ç 4C (c)

L a ÇC

V/2 9L R2 + Ç 4C V

11.36 (a)

b , (d)

, (b)

9L R + Ç 4C 2

4 4 (d) VR + VL + VC = Vsource 11.15 (a) 44.4mF (b) 135 V 5 11.16 (a) * 104 rad/sec, (b) 44 V 3 11.17 (a) 10, (b) 2.4 A, (c) 28.8 W, (d) 500 Æ 11.18 (a) 100, (b) 102 W, (c) 0.85 A 11.19 (a) use step down transformer with N2/N1 = 1/2, (b) 6.67 A, (c) 36 Æ

d

1 2 1/2 b d WC

WC cR2 + a WL -

2

11.6 (a) 12 cos [(500 rad/s) t], (b) 6.5 V 5000 11.7 (a) Æ , (b) 1.125 cos [(120 rad/s)t] 3 1 11.8 (a) 100 25 Æ , (b) A, (c) VR = 20 25 V, 2 25

1 b WC

2

(c)

L a ÇC

2V

LV2 2 L V2 1 b , (e) a 9L 2 2 9L R2 + R + 4C 4C

L a ÇC

V/2 9L R + Ç 4C 2

LV2

b , (d)

b

9L 9L R2 + 8 R2 + Ç 4C Ç 4C 11.37 R = 30Æ , C = 12.5 mF 11.38 (a) 189 Hz, (b) not possible 11.39 2 amp, 400 W 3 11.40 (i) 40 210 V, (ii) 210 11.41 xci /R = 11.42 v0 =

LV2

, (e) 2

9L R2 + Ç 4C

23/5

1 R2 - 2 ÇLC L

Challenge Problems 11.43 (a)

1

2LC

, (b) ,

1 2 2 2LC - R C /2

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(c)

1 R2 Ä a LC - 2 L2 b

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12

ELECTROMAGNETIC WAVES

LEARNING GOALS By studying this chapter, you will learn: • Why there are both electric and magnetic fields in a light wave. • How the speed of light is related to the fundamental constants of electricity and magnetism. • How to describe the propagation of a sinusoidal electromagnetic wave.

?

Metal objects reflect not only visible light but also radio waves. What aspect of metals makes them so reflective?

hat is light? This question has been asked by humans for centuries, but there was no answer until electricity and magnetism were unified into electromagnetism, as described by Maxwell’s equations. These equations show that a time-varying magnetic field acts as a source of electric field and that Sa time-varying electric field acts as a source of magnetic field. S These E and B fields can sustain each other, forming an electromagnetic wave that propagates through space. Visible light emitted by the glowing filament of a light bulb is one example of an electromagnetic wave; other kinds of electromagnetic waves are produced by TV and radio stations, x-ray machines, and radioactive nuclei. In this chapter we’ll use Maxwell’s equations as the theoretical basis for understanding electromagnetic waves. We’ll find that these waves carry both S S energy and momentum. In sinusoidal electromagnetic waves, the E and B fields are sinusoidal functions of time and position, with a definite frequency and wavelength. Visible light, radio, x rays, and other types of electromagnetic waves differ only in their frequency and wavelength. Our study of optics in the following chapters will be based in part on the electromagnetic nature of light. Unlike waves on a string or sound waves in a fluid, electromagnetic waves do not require a material medium; the light that you see coming from the stars at night has traveled without difficulty across tens or hundreds of light-years of (nearly) empty space. Nonetheless, electromagnetic waves and mechanical waves have much in common and are described in much the same language. Before reading further in this chapter, you should review the properties of mechanical waves as discussed in Chapters 15 and 16.

W

• What determines the amount of power carried by an electromagnetic wave. • How to describe standing electromagnetic waves.

383

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384

CHAPTER 12 Electromagnetic Waves

12.1

Maxwell’s Equations and Electromagnetic Waves

In the last several chapters we studied various aspects of electric and magnetic fields. We learned that when the fields don’t vary with time, such as an electric field produced by charges at rest or the magnetic field of a steady current, we can analyze the electric and magnetic fields independently without considering interactions between them. But when the fields vary with time, they are no longer independent. Faraday’s law (see Section 19.2) tells us that a time-varying magnetic field acts as a source of electric field, as shown by induced emfs in inductors and transformers. Ampere’s law, including the displacement current discovered by Maxwell (see Section 19.7), shows that a time-varying electric field acts as a source of magnetic field. This mutual interaction between the two fields is summarized in Maxwell’s equations, presented in Section 19.7. Thus, when either an electric or a magnetic field is changing with time, a field of the other kind is induced in adjacent regions of space. We are led (as Maxwell was) to consider the possibility of an electromagnetic disturbance, consisting of time-varying electric and magnetic fields, that can propagate through space from one region to another, even when there is no matter in the intervening region. Such a disturbance, if it exists, will have the properties of a wave, and an appropriate term is electromagnetic wave. Such waves do exist; radio and television transmission, light, x rays, and many other kinds of radiation are examples of electromagnetic waves. Our goal in this chapter is to see how such waves are explained by the principles of electromagnetism that we have studied thus far and to examine the properties of these waves.

Electricity, Magnetism, and Light

12.1 James Clerk Maxwell (1831–1879) was the first person to truly understand the fundamental nature of light. He also made major contributions to thermodynamics, optics, astronomy, and color photography. Albert Einstein described Maxwell’s accomplishments as “the most profound and the most fruitful that physics has experienced since the time of Newton.”

As often happens in the development of science, the theoretical understanding of electromagnetic waves evolved along a considerably more devious path than the one just outlined. In the early days of electromagnetic theory (the early 19th century), two different units of electric charge were used: one for electrostatics and the other for magnetic phenomena involving currents. In the system of units used at that time, these two units of charge had different physical dimensions. Their ratio had units of velocity, and measurements showed that the ratio had a numerical value that was precisely equal to the speed of light, 3.00 * 108 m>s. At the time, physicists regarded this as an extraordinary coincidence and had no idea how to explain it. In searching to understand this result, Maxwell (Fig. 12.1) proved in 1865 that an electromagnetic disturbance should propagate in free space with a speed equal to that of light and hence that light waves were likely to be electromagnetic in nature. At the same time, he discovered that the basic principles of electromagnetism can be expressed in terms of the four equations that we now call Maxwell’s equations, which we discussed in Section 19.7. These four equations are (1) Gauss’s law for electric fields; (2) Gauss’s law for magnetic fields, show ing the absence of magnetic monopoles; (3) Ampere’s law, including displacement current; and (4) Faraday’s law:

# QP (Gauss’s law) B # dA = 0 (Gauss’s law for magnetism) B d£ B # d l = m ai + P (Ampere’s law) b B dt d£ E # dl = (Faraday’s law) B dt S

B

S

E dA =

encl

(9.18)

0

S

S

S

E

S

0

'

C

0

(9.19)

(9.20)

encl

S

S

B

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(9.21)

12.1 Maxwell’s Equations and Electromagnetic Waves

These equations apply to electric and magnetic fields in vacuum. If a material is present, the permittivity P0 and permeability m0 of free space are replaced by the permittivity P and permeability m of the material. If the values of P and m are different at different points in the regions of integration, then P and m have to be transferred to the left sides of Eqs. (9.18) and (9.20), respectively, and placed inside the integrals. The P in Eq. (9.20) also has to be included in the integral that gives d£ E>dt. S AccordingS to Maxwell’s equations, a point charge at rest produces a static E field but no B field; Sa pointScharge moving with a constant velocity (see Section 18.1) produces both E and B fields. Maxwell’s equations can also be used to show that in order for a point charge to produce electromagnetic waves, the charge must accelerate. In fact, it’s a general result of Maxwell’s equations that every accelerated charge radiates electromagnetic energy (Fig. 12.2).

385

12.2 (Top) Every mobile phone, wireless modem, or radio transmitter emits signals in the form of electromagnetic waves that are made by accelerating charges. (Bottom) Power lines carry a strong alternating current, which means that a substantial amount of charge is accelerating back and forth and generating electromagnetic waves. These waves can produce a buzzing sound from your car radio when you drive near the lines.

Generating Electromagnetic Radiation One way in which a point charge can be made to emit electromagnetic waves is by making it oscillate in simple harmonic motion, so that it has an acceleration at almost every instant (the exception is when the charge is passing through its equilibrium position). Figure 12.3 shows some of the electric field lines produced by such an oscillating point charge. Field lines are not material objects, but you may nonetheless find it helpful to think of them as behaving somewhat like strings that extend from the point charge off to infinity. Oscillating the charge up and down makes waves that propagate outward from the charge along these “strings.” Note that the charge does not emit waves equally in all directions; the waves are strongest at 90° to the axis of motion of the charge, while there are no waves along this axis. This is just what the “string” picture would lead you to conclude. There is also a magnetic disturbance that spreads outward from the charge; this is not shown in Fig. 12.3. Because the electric and magnetic disturbances spread or radiate away from the source, the name electromagnetic radiation is used interchangeably with the phrase “electromagnetic waves.” Electromagnetic waves with macroscopic wavelengths were first produced in the laboratory in 1887 by the German physicist Heinrich Hertz. As a source of waves, he used charges oscillating in L-C circuits of the sort discussed in Section 10.5; he detected the resulting electromagnetic waves with other circuits tuned to the same frequency. Hertz also produced electromagnetic standing waves and measured the distance between adjacent nodes (one half-wavelength) to determine the wavelength. Knowing the resonant frequency of his circuits, he then found the speed of the waves from the wavelength–frequency relationship v = lƒ. He established that their speed was the same as that of light; this verified Maxwell’s theoretical prediction directly. The SI unit of frequency is named in honor of Hertz: One hertz (1 Hz) equals one cycle per second.

12.3 Electric field lines of a point charge oscillating in simple harmonic motion, seen at five instants during an oscillation period T. The charge’s trajectory is in the plane of the drawings. At t = 0 the point charge is at its maximum upward displacement. The arrow S shows one “kink” in the lines of E as it propagates outward from the point charge. The magnetic field (not shown) comprises circles that lie in planes perpendicular to these figures and concentric with the axis of oscillation. (a) t 5 0

(b) t 5 T/4

q

S

E

(c) t 5 T/2

q

S

E

(d) t 5 3T/4

q

S

E

(e) t 5 T

q

S

E

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q

S

E

386

CHAPTER 12 Electromagnetic Waves

The modern value of the speed of light, which we denote by the symbol c, is 299,792,458 m>s. (Recall from Section 1.3 that this value is the basis of our standard of length: One meter is defined to be the distance that light travels in 1>299,792,458 second.) For our purposes, c = 3.00 * 108 m>s is sufficiently accurate. The possible use of electromagnetic waves for long-distance communication does not seem to have occurred to Hertz. It was left to Marconi and others to make radio communication a familiar household experience. In a radio transmitter, electric charges are made to oscillate along the length of the conducting antenna, producing oscillating field disturbances like those shown in Fig. 12.3. Since many charges oscillate together in the antenna, the disturbances are much stronger than those of a single oscillating charge and can be detected at a much greater distance. In a radio receiver the antenna is also a conductor; the fields of the wave emanating from a distant transmitter exert forces on free charges within the receiver antenna, producing an oscillating current that is detected and amplified by the receiver circuitry. For the remainder of this chapter our concern will be with electromagnetic waves themselves, not with the rather complex problem of how they are produced.

The Electromagnetic Spectrum The electromagnetic spectrum encompasses electromagnetic waves of all frequencies and wavelengths. Figure 12.4 shows approximate wavelength and frequency ranges for the most commonly encountered portion of the spectrum. Despite vast differences in their uses and means of production, these are all electromagnetic waves with the same propagation speed (in vacuum) c = 299,792,458 m>s. Electromagnetic waves may differ in frequency ƒ and wavelength l, but the relationship c = lƒ in vacuum holds for each. We can detect only a very small segment of this spectrum directly through our sense of sight. We call this range visible light. Its wavelengths range from about 380 to 750 nm 1380 to 750 * 10-9 m2, with corresponding frequencies from about 790 to 400 THz 17.9 to 4.0 * 1014 Hz2. Different parts of the visible spectrum evoke in humans the sensations of different colors. Table 12.1 gives the approximate wavelengths for colors in the visible spectrum. Ordinary white light includes all visible wavelengths. However, by using special sources or filters, we can select a narrow band of wavelengths within a range of a few nm. Such light is approximately monochromatic (single-color) light. Absolutely monochromatic light with only a single wavelength is an unattainable idealization. When we use the expression “monochromatic light with l = 550 nm” with reference to a laboratory experiment, we really mean a small band

Table 12.1 Wavelengths of Visible Light 380–450 nm 450–495 nm 495–570 nm 570–590 nm 590–620 nm 620–750 nm

Violet Blue Green Yellow Orange Red

12.4 The electromagnetic spectrum. The frequencies and wavelengths found in nature extend over such a wide range that we have to use a logarithmic scale to show all important bands. The boundaries between bands are somewhat arbitrary.

Wavelengths in m 10

1

1021

1023

1024

1025

1026

1027

1028

1029

Infrared

Radio, TV 108

1022

X rays Ultraviolet

Microwave 109

1010

10210 10211 10212 10213

1011

1012

1013

1014

1015

1016

Gamma rays 1017

1018

Visible light 700 nm

RED

650

ORANGE

600

YELLOW

550

GREEN

1019

1020

1021

Frequencies in Hz

500

450

BLUE

400 nm

VIOLET

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1022

12.2 Plane Electromagnetic Waves and the Speed of Light

of wavelengths around 550 nm. Light from a laser is much more nearly monochromatic than is light obtainable in any other way. Invisible forms of electromagnetic radiation are no less important than visible light. Our system of global communication, for example, depends on radio waves: AM radio uses waves with frequencies from 5.4 * 105 Hz to 1.6 * 106 Hz, while FM radio broadcasts are at frequencies from 8.8 * 107 Hz to 1.08 * 108 Hz. (Television broadcasts use frequencies that bracket the FM band.) Microwaves are also used for communication (for example, by cellular phones and wireless networks) and for weather radar (at frequencies near 3 * 109 Hz). Many cameras have a device that emits a beam of infrared radiation; by analyz ing the properties of the infrared radiation reflected from the subject, the camera determines the distance to the subject and automatically adjusts the focus. X rays are able to penetrate through flesh, which makes them invaluable in dentistry and medicine. Gamma rays, the shortest-wavelength type of electromagnetic radiation, are used in medicine to destroy cancer cells. Test Your Understanding of Section 12.1 (a) Is it possible to have a purely electric wave propagate through empty space—-that is, a wave made up of an electric field but no magnetic field? (b) What about a purely magnetic wave, with a magnetic field but no electric field?

12.2

387

Application Ultraviolet Vision Many insects and birds can see ultraviolet wavelengths that humans cannot. As an example, the left-hand photo shows how black-eyed Susans (genus Rudbeck ia) look to us. The right-hand photo (in false color), taken with an ultraviolet-sensitive camera, shows how these same flowers appear to the bees that pollinate them. Note the prominent central spot that is invisible to humans. Similarly, many birds with ultraviolet vision—-including budgies, parrots, and peacocks—-have ultraviolet patterns on their bodies that make them even more vivid to each other than they appear to us.

y

Plane Electromagnetic Waves and the Speed of Light

We are now ready to develop the basic ideas of electromagnetic waves and their relationship to the principles of electromagnetism. Our procedure will be to postulate a simple field configuration that has wavelike behavior. We’ll assume an S S electric field E that has only a y-component and a magnetic field B with only a z-component, and we’ll assume that both fields move together in the +x-direction with a speed c that is initially unknown. (As we go along, it will become clear S S why we choose E and B to be perpendicular to the direction of propagation as well as to each other.) Then we will test whether these fields are physically possible by asking whether they are consistent with Maxwell’s equations, particularly Ampere’s law and Faraday’s law. We’ll find that the answer is yes, provided that c has a particular value. We’ll also show that the wave equation, which we encountered during our study of mechanical waves in Chapter 15, can be derived from Maxwell’s equations.

A Simple Plane Electromagnetic Wave Using an xyz-coordinate system (Fig. 12.5), we imagine that all space is divided into two regions by a plane perpendicular to the x-axis (parallel to the yz-plane). S At every point to the left of this plane thereS are a uniform electric field E in the +y-direction and a uniform magnetic field B in the +z-direction, as shown. Furthermore, we suppose that the boundary plane, which we call the wave front, moves to the right in the +x-direction with aSconstant speed c, the value of which S we’ll leave undetermined for now. Thus the E and B fields travel to the right into previously field-free regions with a definite speed. This is a rudimentary electromagnetic wave. A wave such as this, in which at any instant the fields are uniform over any plane perpendicular to the direction of propagation, is called a plane wave. In the case shown in Fig. 12.5, the fields are zero for planes to the right of the wave front and have the same values on all planes to the left of the wave front; later we will consider more complex plane waves. We won’t concern ourselves with the problem of actually producing such a field configuration. Instead, we simply ask whether it is consistent with the laws of electromagnetism—-that is, with Maxwell’s equations. We’ll consider each of these four equations in turn.

12.5 An electromagnetic wave front. The plane representing the wave front moves to the right (in the positive x-direction) with speed c. y Planar wave front

S

E

S

E

S

B

S

B

S

E S

S

B

E50 c

S

E

O E

S

B z

S

B50

S S

E

S

B

S

B

x

The electric and magnetic fields are uniform behind the advancing wave front and zero in front of it.

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388

CHAPTER 12 Electromagnetic Waves

12.6 Gaussian surface for a transverse plane electromagnetic wave. The electric field is the same on the top and bottom sides of the Gaussian surface, so the total electric flux through the surface is zero. y

S

E

S

E

S

B

S

E

S

B

S

B

S

E

S

E

S

B

S

B x

z

Let us first verify that our wave satisfies Maxwell’s first and second equations—-that is, Gauss’s laws for electric and magnetic fields. To do this, we take as our Gaussian surface a rectangular box with sides parallel to the xy, xz, and yz coordinate planes (Fig. 12.6). The box encloses no electric charge. The total electric flux and magnetic flux through the box are both zero, even if part S of the box is in the region where E = B = 0. This would not be the case if E S or B had an x-component, parallel to the direction of propagation; if the wave front were inside the box, there would be flux through the left-hand side of the box (at x = 0) but not the right-hand side (at x 7 0). Thus to satisfy Maxwell’s first and second equations, the electric and magnetic fields must be perpendicular to the direction of propagation; that is, the wave must be transverse. The next of Maxwell’s equations to be considered is Faraday’s law: S

The magnetic field is the same on the left and right sides of the Gaussian surface, so the total magnetic flux through the surface is zero.

12.7 (a) Applying Faraday’s law to a plane wave. (b) In a time dt, the magnetic flux through the rectangle in the xy-plane increases by an amount d£ B . This increase equals the flux through the shaded rectan gle with area ac dt; that is, d£ B = Bac dt. Thus d£ B/dt = Bac. '

(a) In time dt, the wave front moves a distance c dt in the 1x-direction.

B

#

S

E dl = -

d£ B dt

(12.1)

To test whether our wave satisfies Faraday’s law, we apply this law to a rectangle eƒgh that is parallel to the xy-plane (Fig. 12.7a). As shown in Fig. 12.7b, a cross section in the xy-plane, this rectangle has height a and width ¢x. At the time S shown, the wave front has progressed partway through the rectangle, and S E is zero along the side eƒ. In applying Faraday’s law we take the vector area dA of rectangle eƒgh to be in the With this choice the right-hand rule +z-direction. S S requires that we integrate counterclockwise around the rectangle. At every E d l S S point on side eƒ, ES is zero. At every point on sides ƒg and he, E is either zero or S perpendicular to Only side gh contributes to the integral. On this side, and d l . E S d l are opposite, and we obtain

#

y S

B

#

S

E d l = - Ea

(12.2)

S

E

S

E

S

B

S

E

S

B

S

E

S

B

g

S

B

f a e

h S

B

S

E

S

B

z

S

B

x

S

B

c dt

(b) Side view of situation in (a) y

Hence, the left-hand side of Eq. (12.1) is nonzero. S To satisfy Faraday’s law, Eq. (12.1), there must be a component of B in the S z-direction (perpendicular to E2 so that there can be a nonzero magnetic flux £ B through the rectangle efgh and a nonzero derivative d£ B>dt. Indeed, in our wave, S B has only a z-component. We have assumed that this component is in the positive z-direction; let’s see whether this assumption is consistent with Faraday’s law. During a time interval dt the wave front moves a distance c dt to the right in Fig. 12.7b, sweeping out an area ac dt of the rectangle efgh. During this interval the magnetic flux £ B through the rectangle eƒgh increases by d£ B = B1ac dt2, so the rate of change of magnetic flux is d£ B = Bac dt

Dx c dt f

g

(12.3)

Now we substitute Eqs. (32.2) and (32.3) into Faraday’s law, Eq. (32.1); we get

S

dl S S S dl dl dA

a

dl

-Ea = - Bac

e

h S

E

O

S

S

B

E = cB

S

E50 S B50 x

(electromagnetic wave in vacuum)

(12.4)

This shows that our wave is consistent with Faraday’s law only if the wave speed c S S and the magnitudes of the perpendicular vectors E and B are related as in Eq. (12.4). S Note that if we had assumed that B was in the negative z-direction, there would have been an additional minus sign in Eq. (12.4); since E, c, and B are all positive magnitudes, no solution would then haveSbeen possible. Furthermore, any comS ponent of B in the y-direction (parallel to E2 would not contribute to the changing magnetic flux £ B through the rectangle eƒgh (which is parallel to the xy-plane) and so would not be part of the wave.

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12.2 Plane Electromagnetic Waves and the Speed of Light

389

Finally, we carry out a similar calculation using Ampere’s law, the remaining member of Maxwell’s equations. There is no conduction current 1iC = 02, so Ampere’s law is

#

S

B

S

B d l = m0 P0 '

d£ E dt

(12.5)

To check whether our wave is consistent with Ampere’s law, we move our rectangle so that it lies in the xz-plane, as shown in Fig. 12.8, and we again look at the situation at a time when the wave front has traveled partway through the recS + y-direction, and so the right-hand rule dA tangle. We take the vector area in the S S S requires that we integrate B d l counterclockwise around the rectangle. The B eƒ, and at each point on sides ƒg and he it is field is zero at every point along side S S S either zero or perpendicular to d l . Only side gh, where B and d l are parallel, contributes to the integral, and we find

#

S

B

#

12.8 (a) Applying Ampere’s law to a plane wave. (Compare to Fig. 12.7a.) (b) In a time dt, the electric flux through the rectangle in the xz-plane increases by an amount d£ E . This increase equals the flux through the shaded rectangle with area ac dt; that is, d£ E = Eac dt. Thus d£ E/dt = Eac. '

(a) In time dt, the wave front moves a distance c dt in the 1x-direction.

S

B d l = Ba

(12.6)

Hence, the left-hand side of Ampere’s law, Eq. (12.5), is nonzero; the right-hand S side must be nonzero as well. Thus E must have a y-component (perpendicular to S B2 so that the electric flux £ E through the rectangle and the time derivative d£ E>dt can be nonzero. We come to the same conclusion that we inferred from S S Faraday’s law: In an electromagnetic wave, E and B must be mutually perpendicular. In a time interval dt the electricS flux £ E through the rectangle increases by d£ E = E1ac dt2. Since we chose dA to be in the + y-direction, this flux change is positive; the rate of change of electric field is d£ E = Eac dt

(12.7)

Substituting Eqs. (12.6) and (12.7) into Ampere’s law, Eq. (12.5), we find

y

S

E B

S

B = P0 m0 cE '

'

B

S

B

g

B

S

E

S

B

B a e

(b) Top view of situation in (a) O

x S

S

E

S

B

E50 S B50

(12.8) f

g

Thus our assumed wave obeys Ampere’s law only if B, c, and E are related as in Eq. (12.8). Our electromagnetic wave must obey both Ampere’s law and Faraday’s law, so Eqs. (12.4) and (12.8) must both be satisfied. This can happen only if P0 m0 c = 1>c, or '

'

c =

1 2P0 m0

(speed of electromagnetic waves in vacuum)

(12.9)

'

Inserting the numerical values of these quantities, we find c =

1

218.85 * 10

-12

= 3.00 * 108 m>s

#

C >N m2214p * 10-7 N>A22 2

x f

S

c dt

'

(electromagnetic wave in vacuum)

S

S

E

S

h

z

S

E

S

B

B

Ba = P0 m0 Eac '

S

E

S

Our assumed wave is consistent with all of Maxwell’s equations, provided that the wave front moves with the speed given above, which you should recognize as the speed of light! Note that the exact value of c is defined to be 299,792,458 m>s; the modern value of P0 is defined to agree with this when used in Eq. (12.9) (see Section 1.3).

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S S dl dl S S dl dA

h

c dt Dx

z

S

dl e

a

390

CHAPTER 12 Electromagnetic Waves

Key Properties of Electromagnetic Waves 12.9 A right-hand rule for electromagS netic waves relates the directions of E and S B and the direction of propagation.

We chose a simple wave for our study in order to avoid mathematical complications, but this special case illustrates several important features of all electromagnetic waves:

Right-hand rule for an electromagnetic wave: 1 Point the thumb of your right hand in the

wave’s direction of propagation. S

2 Imagine rotating the E-field vector 90° in

the sense your fingers curl. S

That is the direction of the B field. y S

E 90° O

2 S

B z

1

c x Direction of propagation S S 5 direction of E 3 B.

S

S

1. The wave is transverse; both E and B are perpendicular to the direction of propagation of the wave. The electric and magnetic fields are also perpendicular to eachSother.S The direction of propagation is the direction of the vector product E : B (Fig. 12.9). S S 2. There is a definite ratio between the magnitudes of E and B: E = cB. 3. The wave travels in vacuum with a definite and unchanging speed. 4. Unlike mechanical waves, which need the oscillating particles of a medium such as water or air to transmit a wave, electromagnetic waves require no medium. We can generalize this discussion to a more realistic situation. Suppose we have several wave fronts in the form of parallel planes perpendicular to the S x-axis, all of which are moving to the right with speed c. Suppose that the E and S B fields are the same at all points within a single region between two planes, but that the fields differ from region to region. The overall wave is a plane wave, but one in which the fields vary in steps along the x-axis. Such a wave could be constructed by superposing several of the simple step waves we have just disS S cussed (shown in Fig. 12.5). This is possible because the E and B fields obey the superposition principle in waves just as in static situations: When two waves are S S superposed, the total E field at each point is Sthe vector sum of the E fields of the individual waves, and similarly for the total B field. We can extend the above development to show that a wave with fields that vary in steps is also consistent with Ampere’s and Faraday’s laws, provided that the wave fronts all move with the speed c given by Eq. (12.9). In the limit that we S makeS the individual steps infinitesimally small, we have a wave in which the E and B fields at any instant vary continuously along the x-axis. The entire field pattern Smoves S to the right with speed c. In Section 12.3 we will consider waves in which E and BS are sinusoidal functions of x and t. Because at each point the S magnitudes of E and B are related by E = cB, the periodic variations of the two fields in any periodic traveling wave must be in phase. Electromagnetic waves have the property of polarization. In the above disS cussion the choice of the y-direction for E was arbitrary. We could just as well S S have specified the z-axis for E ; then B would have been in the - y-direction. A S wave in which E is always parallel to a certain axis is said to be linearly polarized along that axis. More generally, any wave traveling in the x-direction can be represented as a superposition of waves linearly polarized in the y- and z-directions. We will study polarization in greater detail in Chapter 13.

Derivation of the Electromagnetic Wave Equation Here is an alternative derivation of Eq. (12.9) for the speed of electromagnetic waves. It is more mathematical than our other treatment, but it includes a derivation of the wave equation for electromagnetic waves. This part of the section can be omitted without loss of continuity in the chapter. During our discussion of mechanical waves in Section 15.3, we showed that a function y1x, t2 that represents the displacement of any point in a mechanical wave traveling along the x-axis must satisfy a differential equation, Eq. (15.12): 0 2y1x, t2 0x2

=

1 0 2y1x, t2 v2 0t2

(12.10)

This equation is called the wave equation, and v is the speed of propagation of the wave.

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391

12.2 Plane Electromagnetic Waves and the Speed of Light

To derive the corresponding equation for an electromagnetic wave, we again consider a plane wave. That is, we assume that at each instant, Ey and Bz are uniform over any plane perpendicular to the x-axis, the direction of propagation. But now we let Ey and Bz vary continuously as we go along the x-axis; then each is a function of x and t. We consider the values of Ey and Bz on two planes perpendicular to the x-axis, one at x and one at x + ¢x. Following the same procedure as previously, we apply Faraday’s law to a rectangle lying parallel to the xy-plane, as in Fig. 12.10. This figure is similar to Fig. 12.7. Let the left end gh of the rectangle be at position x, and let the right end eƒ be at position 1x + ¢x2. At time t, the values of Ey on these two sides are Ey 1x, t2 and Ey 1x + ¢x, t2, respectively. When we apply Faraday’s law to this recS S tangle, we find that instead of A E d l = - Ea as before, we have '

y

(a)

#

'

#

x Dx

E d l = -Ey 1x, t2a + Ey 1x + ¢x, t2a S

B

S

'

S

'

= a3Ey 1x + ¢x, t2 - Ey 1x, t24 '

'

E

S

B g

f

O

a z

h

S

E

e

x

S

B

'

0Bz 1x, t2 d£ B = a ¢x dt 0t We use partial-derivative notation because Bz is a function of both x and t. When we substitute this expression and Eq. (12.11) into Faraday’s law, Eq. (12.1), we get

S

E

S

B

(12.11)

To find the magnetic flux £ B through this rectangle, we assume that ¢x is small enough that Bz is nearly uniform over the rectangle. In that case, £ B = Bz 1x, t2A = Bz 1x, t2a ¢x, and '

12.10 Faraday’s law applied to a rectangle with height a and width ¢x parallel to the xy-plane.

'

a3Ey 1x + ¢x, t2 - Ey 1x, t24 = '

'

Ey 1x + ¢x, t2 - Ey 1x, t2 '

'

= -

¢x

0Bz 0t 0Bz

(b) Side view of the situation in (a) y Dx

g

a ¢x

Ey

f Ey

S

A

a

e

h

0t

x

O

Finally, imagine shrinking the rectangle down to a sliver so that ¢x approaches zero. When we take the limit of this equation as ¢x S 0, we get 0Ey 1x, t2

0Bz 1x, t2

'

'

= -

0x

(12.12)

0t

This equation shows that if there is a time-varying component Bz of magnetic field, there must also be a component Ey of electric field that varies with x, and conversely. We put this relationship on the shelf for now; we’ll return to it soon. Next weS apply Ampere’s law to the rectangle shown in Fig. 12.11. The line S integral A B d l becomes

12.11 Ampere’s law applied to a rectangle with height a and width ¢x parallel to the xz-plane. y

(a)

x Dx

#

S

B

#

B d l = -Bz 1x + ¢x, t2a + Bz 1x, t2a

S

E

S

B

S

'

'

(12.13)

Again assuming that the rectangle is narrow, we approximate the electric flux £ E through it as £ E = Ey 1x, t2A = Ey 1x, t2a ¢x. The rate of change of £ E , which we need for Ampere’s law, is then '

'

S

E

S

B

O O z

g

0Ey 1x, t2 d£ E = a ¢x dt 0t

x

f

'

h

a e

'

(b) Top view of the situation in (a)

Now we substitute this expression and Eq. (12.13) into Ampere’s law, Eq. (12.5): -Bz 1x + ¢x, t2a + Bz 1x, t2a = P0 m0

0Ey 1x, t2

O

x

'

a ¢x 0t Again we divide both sides by a ¢x and take the limit as ¢x S 0. We find '

'

0Bz 1x, t2

'

0x

0Ey 1x, t2

Bz

= P0 m0 '

0t

Dx

S

A h

'

'

-

g

(12.14)

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z

f Bz e

a

392

CHAPTER 12 Electromagnetic Waves

Now comes the final step. We take the partial derivatives with respect to x of both sides of Eq. (12.12), and we take the partial derivatives with respect to t of both sides of Eq. (12.14). The results are -

0 2Ey 1x, t2

=

-

0 Bz 1x, t2

= P0 m0

'

0 2Bz 1x, t2 '

2

0x

0 x0t

0 2Ey 1x, t2

2

'

'

0 x 0t 0t2 Combining these two equations to eliminate Bz , we finally find '

'

0 2Ey 1x, t2 '

2

0 2Ey 1x, t2 '

= P0 m 0 '

2

(electromagnetic wave equation in vacuum)

(12.15)

0x 0t This expression has the same form as the general wave equation, Eq. (12.10). Because the electric field Ey must satisfy this equation, it behaves as a wave with a pattern that travels through space with a definite speed. Furthermore, comparison of Eqs. (12.15) and (12.10) shows that the wave speed v is given by 1 1 = P0 m0 or v = 2 v 2P0 m0 '

'

This agrees with Eq. (12.9) for the speed c of electromagnetic waves. We can show that Bz also must satisfy the same wave equation as Ey , Eq. (12.15). To prove this, we take the partial derivative of Eq. (12.12) with respect to t and the partial derivative of Eq. (12.14) with respect to x and combine the results. We leave this derivation for you to carry out. '

Test Your Understanding of Section 12.2 For each of the following electromagnetic waves, state the direction of the magnetic field. (a) The wave is propagating in S the positive z-direction, and E is in the positive x-direction; (b) the wave is propagating in S the positive y-direction, and E isSin the negative z-direction; (c) the wave is propagating in the negative x-direction, and E is in the positive z-direction. y

12.3

12.12 Waves passing through a small area at a sufficiently great distance from a source can be treated as plane waves. Waves that pass through a large area propagate in different directions ...

Source of electromagnetic waves

... but waves that pass through a small area all propagate in nearly the same direction, so we can treat them as plane waves.

Sinusoidal Electromagnetic Waves

Sinusoidal electromagnetic waves are directly analogous to sinusoidal transverse mechanical waves on a stretched string, which we studied in Section 15.3. In a sinuS S soidal electromagnetic wave, E and B at any point in space are sinusoidal functions of time, and at any instant of time the spatial variation of the fields is also sinusoidal. Some sinusoidal electromagnetic waves are plane waves; they share with the waves described in Section 12.2 the property that at any instant the fields are uniform over any plane perpendicular to the direction of propagation. The entire patS tern travels in the direction of propagation with speed c. The directions of E and S B are perpendicular to the direction of propagation (and to each other), so the wave is transverse. Electromagnetic waves produced by an oscillating point charge, shown in Fig. 12.3, are an example of sinusoidal waves that are not plane waves. But if we restrict our observations to a relatively small region of space at a sufficiently great distance from the source, even these waves are well approximated by plane waves (Fig. 12.12). In the same way, the curved surface of the (nearly) spherical earth appears flat to us because of our small size relative to the earth’s radius. In this section we’ll restrict our discussion to plane waves. The frequency ƒ, the wavelength l, and the speed of propagation c of any periodic wave are related by the usual wavelength–frequency relationship c = lƒ. If the frequency ƒ is 108 Hz (100 MHz), typical of commercial FM radio broadcasts, the wavelength is 3 * 108 m>s

= 3m 108 Hz Figure 12.4 shows the inverse proportionality between wavelength and frequency. l =

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12.3 Sinusoidal Electromagnetic Waves

393

Fields of a Sinusoidal Wave Figure 12.13 shows a linearly polarized sinusoidal electromagnetic wave travelS S ing in the +x-direction. The E and B vectors are shown for only a few points on S the positive x-axis. Note that the electric and magnetic fields oscillate in phase: E S S S is maximum where B is maximum and E is zero where B is zero. Note also that S S S where E is in the +y-direction, B is in the +z-direction; where E is in the S S S -y-direction, B is in the -z-direction. At all points the vector product E : B is in the direction in which the wave is propagating (the +x-direction). We mentioned this in Section 12.2 in the list of characteristics of electromagnetic waves.

12.13 Representation of the electric and magnetic fields as functions of x for a linearly polarized sinusoidal plane electromagnetic wave. One wavelength of the wave is shown at time t = 0. The fields are shown only for points along the x-axis. The wave is traveling in the y positive x-direction, the same S S as the direction of E 3 B. S c E S

S

S

CAUTION In a plane wave, E and B are everywhere Figure 12.13 may give you the erroneous impression that the electric and magnetic fields exist only along the x-axis. In fact, in a sinusoidal plane wave there are electric and magnetic fields at all points in space. Imagine a plane perpendicular to the x-axis (that is, parallel to the yz-plane) at a particular point, at a particular time; the fields have the same values at all points in that plane. The values are different on different planes. y

B

O

S

E

S

B z

x S

E

S

B

S

E: y-component only S B: z-component only

We can describe electromagnetic waves by means of wave functions, just as we did in Section 15.3 for waves on a string. One form of the wave function for a transverse wave traveling in the +x-direction along a stretched string is Eq. (15.7): y1x, t2 = A cos1kx - vt2 where y1x, t2 is the transverse displacement from its equilibrium position at time t of a point with coordinate x on the string. The quantity A is the maximum displacement, or amplitude, of the wave; v is its angular frequency, equal to 2p times the frequency ƒ ; and k is the wave number, equal to 2p>l, where l is the wavelength. Let Ey 1x, t2 and Bz 1x, t2 represent the instantaneous values of the y-component S S of E and the z-component of B, respectively, in Fig. 12.13, and let Emax and Bmax represent the maximum values, or amplitudes, of these fields. The wave functions for the wave are then '

PhET: Radio Waves & Electromagnetic Fields ActivPhysics 10.1: Properties of Mechanical Waves

'

Ey 1x, t2 = Emax cos1kx - vt2 '

Bz 1x, t2 = Bmax cos1kx - vt2 '

(12.16)

(sinusoidal electromagnetic plane wave, propagating in + x-direction) We can also write the wave functions in vector form: E1x, t2 5 ≥NEmax cos1kx - vt2 S

S B1x, t2 5 kN Bmax cos1kx - vt2

(12.17)

CAUTION The symbol k has two meanings Note the two different k’s: the unit vector kN in the z-direction and the wave number k. Don’t get these confused! y

The sine curves in Fig. 12.13 represent instantaneous values of the electricSand S magnetic fields as functions of x at time t = 0—that is, E1x, t = 02 and B1x, t = 02. As time goes by, the wave travels to the right with speed c. Equations S S (12.16) and (12.17) show that at any point the sinusoidal oscillations of E and B are in phase. From Eq. (12.4) the amplitudes must be related by Emax = cBmax

(electromagnetic wave in vacuum)

(12.18)

These amplitude and phase relationships are also required for E1x, t2 and B1x, t2 to satisfy Eqs. (12.12) and (12.14), which came from Faraday’s law and Ampere’s law, respectively. Can you verify this statement? (See Problem 12.38.)

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394

CHAPTER 12 Electromagnetic Waves

12.14 Representation of one wave length of a linearly polarized sinusoidal plane electromagnetic wave traveling in the negative x-direction at t = 0. The fields are shown only for points along the x-axis. (Compare with Fig. 12.13.)

Figure 12.14 shows the electric andSmagnetic fields of a wave traveling in the S negative x-direction. At points where E is in the positive y-direction, B is in the S S negative z-direction; where E is in the negative y-direction, B is in the positive z-direction. The wave functions for this wave are Ey 1x, t2 = Emax cos1kx + vt2 Bz 1x, t2 = - Bmax cos1kx + vt2 '

y S

E

The wave is traveling in the S theSsame B negative x-direction, S as the direction of E 3 B.

O

c

S

E

z S

B x S

B

S

E E: y-component only S B: z-component only

(12.19)

'

(sinusoidal electromagnetic plane wave, propagating in - x-direction) As with the wave traveling in the +x-direction, at any point the sinusoidal oscilS S S S lations of the E and B fields are in phase, and the vector product E : B points in the direction of propagation. The sinusoidal waves shown in Figs. 12.13 and 12.14 are both linearly polarS ized in the y-direction; the E field is always parallel to the y-axis. Example 12.1 concerns a wave that is linearly polarized in the z-direction.

S

Problem-Solving Strategy 12.1

Electromagnetic Waves

IDENTIFY the relevant concepts: Many of the same ideas that apply to mechanical waves apply to electromagnetic waves. One difference is that electromagnetic waves are describedSby two quanS tities (in this case, electric field E and magnetic field B), rather than by a single quantity, such as the displacement of a string. SET UP the problem using the following steps: 1. Draw a diagram showing the direction of wave propagation and S S the directions of E and B. 2. Identify the target variables. EXECUTE the solution as follows: 1. Review the treatment of sinusoidal mechanical waves in Chapters 15 and 16, and particularly the four problem-solving strategies suggested there. 2. Keep in mind the basic relationships for periodic waves: v = lƒ and v = vk. For electromagnetic waves in vacuum,

Example 12.1

v = c. Distinguish between ordinary frequency ƒ, usually expressed in hertz, and angular frequency v = 2pƒ, expressed in rad > s. Remember that the wave number is k = 2p > l. S 3. Concentrate on basic relationships, such as those between E S and B (magnitude, direction, and relative phase), how the wave speed is determined, and the transverse nature of the waves. '

'

'

EVALUATE your answer: Check that your result is reasonable. For electromagnetic waves in vacuum, the magnitude of the magnetic field in teslas is much smaller (by a factor of 3.00 * 108) than the magnitude of the electric field in volts per meter. If your answer suggests otherwise, you probably made an error using the relationship E = cB. (We’ll see later in this section that the relationship between E and B is different for electromagnetic waves in a material medium.)

Electric and magnetic fields of a laser beam

A carbon dioxide laser emits a sinusoidal electromagnetic wave that travels in vacuum in the negative x-direction. The wavelength S is 10.6 mm (in the infrared; see Fig. 12.4) and the E field is parallel to the z-axis, with Emax = 1.5 MV > m. Write vector equations S S for E and B as functions of time and position. '

12.15 Our sketch for this problem.

'

SOLUTION IDENTIFY and SET UP: Equations (12.19) describe a wave travelS ing in the negative x-direction with E along the y-axis—that is, a wave that is linearly polarized along the y-axis. By contrast, the wave in this Sexample is linearly polarized Salong the z-axis. At points where E is in the positive z-direction, B must be in the posiS S tive y-direction for the vector product E : B to be in the negative x-direction (the direction of propagation). Figure 12.15 shows a wave that satisfies these requirements. EXECUTE: A possible pair of wave functions that describe the wave shown in Fig. 12.15 are

'

S E1x, t2 5 kN Emax cos 1kx + vt2 '

B1x, t2 5 n≥ Bmax cos 1kx + vt2 S

'

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395

12.3 Sinusoidal Electromagnetic Waves B1x, t2 5 n≥ 15.0 * 10-3 T2

The plus sign in the arguments of the cosine functions indicates that the wave is propagating in the negative x-direction, as it should. Faraday’s law requires that Emax = cBmax [Eq. (12.18)], so

* cos315.93 * 105 rad > m2x + 11.78 * 1014 rad > s2t4 '

1.5 * 106 V > m Emax = 5.0 * 10-3 T = c 3.0 * 108 m > s '

Bmax =

S

'

To check unit consistency, note that 1 V = 1 Wb > s and 1 Wb > m2 = 1 T. We have l = 10.6 * 10-6 m, so the wave number and angular frequency are '

'

'

'

k =

2p rad 2p = = 5.93 * 105 rad > m l 10.6 * 10-6 m '

'

v = ck = 13.00 * 108 m > s215.93 * 105 rad > m2 '

'

'

'

= 1.78 * 1014 rad > s '

'

'

EVALUATE: As we expect, the magnitude Bmax in teslas is much smaller than the magnitude Emax in volts per meter. To check the S S S S directions of E and B, note that E : B is in the direction of Nk : n≥ 5 2 n≤ . This is as it should be for a wave that propagates in the negative x-direction.S S Our expressions for E1x, t2 and B1x, t2 are not the only possible solutions. We could always add a phase angle f to the arguments of the cosine function, so that kx + vt would become kx + vt + f. To determine the value of f we would need to S S know E and B either as functions of x at a given time t or as functions of t at a given coordinate x. However, the statement of the problem doesn’t include this information.

'

'

'

'

Substituting these values into the above wave functions, we get S E1x, t2 5 kN 11.5 * 106 V > m2 '

'

* cos315.93 * 105 rad > m2x + 11.78 * 1014 rad > s2t4 '

'

'

'

Electromagnetic Waves in Matter So far, our discussion of electromagnetic waves has been restricted to waves in vacuum. But electromagnetic waves can also travel in matter; think of light traveling through air, water, or glass. In this subsection we extend our analysis to electromagnetic waves in nonconducting materials—that is, dielectrics. In a dielectric the wave speed is not the same as in vacuum, and we denote it by v instead of c. Faraday’s law is unaltered, but in Eq. (12.4), derived from Faraday’s law, the speed c is replaced by v. In Ampere’s law the displacement current S is given not by P0 d£ E>dt, where £ E is the flux of E through a surface, but by P d£ E>dt = KP0 d£ E>dt, where K is the dielectric constant and P is the permittivity of the dielectric. (We introduced these quantities in Section 4.4.) Also, the constant m0 in Ampere’s law must be replaced by m = Km m0 , where Km is the relative permeability of the dielectric and m is its permeability (see Section 8.8). Hence Eqs. (12.4) and (12.8) are replaced by '

E = vB

B = PmvE

and

'

(12.20)

Following the same procedure as for waves in vacuum, we find that the wave speed v is v =

1

=

2Pm

1

1

=

2KKm 2P0 m0 '

c 2KKm

(speed of electromagnetic (12.21) waves in a dielectric)

12.16 The dielectric constant K of water is about 1.8 for visible light, so the speed of visible light in water is slower than in vacuum by a factor of 1/1K = 1> 11.8 = 0.75.

For most dielectrics the relative permeability Km is very nearly equal to unity (except for insulating ferromagnetic materials). When Km > 1, v =

1

1

2K 2P0 m0 '

=

c 2K

Because K is always greater than unity, the speed v of electromagnetic waves in a dielectric is always less than the speed c in vacuum by a factor of 1> 2K (Fig. 12.16). The ratio of the speed c in vacuum to the speed v in a material is known in optics as the index of refraction n of the material. When Km > 1, c = n = 2KKm > 2K v

(12.22)

Usually, we can’t use the values of K in Table 24.1 in this equation because those values are measured using constant electric fields. When the fields oscillate rapidly,

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396

CHAPTER 12 Electromagnetic Waves

there is usually not time for the reorientation of electric dipoles that occurs with steady fields. Values of K with rapidly varying fields are usually much smaller than the values in the table. For example, K for water is 80.4 for steady fields but only about 1.8 in the frequency range of visible light. Thus the dielectric “constant” K is actually a function of frequency (the dielectric function).

Example 12.2

Electromagnetic waves in different materials

(a) Visiting a jewelry store one evening, you hold a diamond up to the light of a sodium-vapor street lamp. The heated sodium vapor emits yellow light with a frequency of 5.09 * 1014 Hz. Find the wavelength in vacuum and the wave speed and wavelength in diamond, for which K = 5.84 and Km = 1.00 at this frequency. (b) A 90.0-MHz radio wave (in the FM radio band) passes from vacuum into an insulating ferrite (a ferromagnetic material used in computer cables to suppress radio interference). Find the wavelength in vacuum and the wave speed and wavelength in the ferrite, for which K = 10.0 and Km = 1000 at this frequency.

EXECUTE: (a) The wavelength in vacuum of the sodium light is 3.00 * 108 m > s

c = = 5.89 * 10-7 m = 589 nm ƒ 5.09 * 1014 Hz '

lvacuum =

'

The wave speed and wavelength in diamond are

3.00 * 108 m > s '

=

2KKm

'

215.84211.002

= 1.24 * 108 m > s '

'

1.24 * 108 m > s vdiamond = ƒ 5.09 * 1014 Hz '

ldiamond =

'

= 2.44 * 10-7 m = 244 nm (b) Following the same steps as in part (a), we find 3.00 * 108 m > s c = = 3.33 m ƒ 90.0 * 106 Hz '

lvacuum =

SOLUTION IDENTIFY and SET UP: In each case we find the wavelength in vacuum using c = lƒ. To use the corresponding equation v = lƒ to find the wavelength in a material medium, we find the speed v of electromagnetic waves in the medium using Eq. (12.21), which relates v to the values of dielectric constant K and relative permeability Km for the medium.

c

vdiamond =

vferrite =

c

'

3.00 * 108 m > s '

=

2KKm

2110.02110002

= 3.00 * 106 m > s '

'

3.00 * 106 m > s vferrite = ƒ 90.0 * 106 Hz '

lferrite =

'

'

= 3.33 * 10-2 m = 3.33 cm EVALUATE: The speed of light in transparent materials is typically between 0.2c and c; our result in part (a) shows that vdiamond = 0.414c. As our results in part (b) show, the speed of electromagnetic waves in dense materials like ferrite (for which vferrite = 0.010c) can be far slower than in vacuum.

Test Your Understanding of Section 12.3 The first of Eqs. (12.17) gives the electric field for a plane wave as measured at points along the x-axis. For this plane wave, how does the electric field at points off the x-axis differ from the expression in Eqs. (12.17)? (i) The amplitude is different; (ii) the phase is different; (iii) both the amplitude and phase are different; (iv) none of these.

12.4

y

Energy and Momentum in Electromagnetic Waves

It is a familiar fact that energy is associated with electromagnetic waves; think of the energy in the sun’s radiation. Microwave ovens, radio transmitters, and lasers for eye surgery all make use of the energy that these waves carry. To understand how to utilize this energy, it’s helpful to derive detailed relationships for the energy in an electromagnetic wave. We begin with the expressions derived in Sections 4.3 and 10.3 for the energy densities in electric and magnetic fields; we suggest you review those derivations S now.SEquations (24.11) and (30.10) show that in a region of empty space where E and B fields are present, the total energy density u is given by u = 12 P0 E2 + '

1 2 B 2m0

(12.23)

where P0 and m0 are, respectively, the permittivity and permeability of free space. For electromagnetic waves in vacuum, the magnitudes E and B are related by

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397

12.4 Energy and Momentum in Electromagnetic Waves

B =

E = 2P0 m0 E c

(12.24)

'

Combining Eqs. (12.23) and (12.24), we can also express the energy density u in a simple electromagnetic wave in vacuum as u = 12 P0 E2 + '

1 1 2P0 m0 E22 = P0 E2 2m0 '

(12.25)

'

S

This shows that in vacuum, the energy density associated with the E field in our S simple wave is equal to the energy density of the B field. In general, the electricfield magnitude E is a function of position and time, as for the sinusoidal wave described by Eqs. (12.16); thus the energy density u of an electromagnetic wave, given by Eq. (12.25), also depends in general on position and time.

Electromagnetic Energy Flow and the Poynting Vector Electromagnetic waves such as those we have described are traveling waves that transport energy from one region to another. We can describe this energy transfer in terms of energy transferred per unit time per unit cross-sectional area, or power per unit area, for an area perpendicular to the direction of wave travel. To see how the energy flow is related to the fields, consider a stationary plane, perpendicular to the x-axis, that coincides with the wave front at a certain time. In a time dt after this, the wave front moves a distance dx = c dt to the right of the plane. Considering an area A on this stationary plane (Fig. 12.17), we note that the energy in the space to the right of this area must have passed through the area to reach the new location. The volume dV of the relevant region is the base area A times the length c dt, and the energy dU in this region is the energy density u times this volume:

12.17 A wave front at a time dt after it passes through the stationary plane with area A. At time dt, the volume between the stationary plane and the wave front contains an amount of electromagnetic energy dU 5 uAc dt. y

c dt S S

B

dU = u dV = 1P0 E221Ac dt2

O

'

This energy passes through the area A in time dt. The energy flow per unit time per unit area, which we will call S, is 1 dU = P0 cE2 A dt

S =

(in vacuum)

'

(12.26)

E

Poynting vector S

A

S S

z

E S

x

B Stationary plane

Wave front at time dt later

Using Eqs. (12.4) and (12.9), we can derive the alternative forms S =

P0 2P0 m0

E2 =

'

P0 2 EB E = m m0 A 0

(in vacuum)

(12.27)

We leave the derivation of Eq. (12.27) from Eq. (12.26) as an exercise for you. The units of S are energy per unit time per unit area, or power per unit area. The SI unit of S is 1 J>s # m2 or 1 W>m2. We can define a vector quantity that describes both the magnitude and direction of the energy flow rate: S

S5

S 1 S E : B m0

(Poynting vector in vacuum)

S

(12.28)

12.18 These rooftop solar panels are tilted to be face-on to the sun—that is, face-on to the Poynting vector of electromagnetic waves from the sun, so that the panels can absorb the maximum amount of wave energy.

The vector S is called the Poynting vector; it was introduced by the British physicist John Poynting (1852–1914). Its direction is in the direction of propagaS S tion of the wave (Fig. 12.18). Since E and B are perpendicular, the magnitude of S S is S = EB>m0 ; from Eqs. (12.26) and (12.27) this is the energy flow per unit area and per unit time through a cross-sectional area perpendicular to the propagation direction. The total energy flow per unit time (power, P) out of any closed S surface is the integral of S over the surface: '

P =

S

B

#

S

S dA

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398

CHAPTER 12 Electromagnetic Waves

For the sinusoidal waves studied in Section 12.3, as well as for other more complex waves, the electric and magnetic fields at any point vary with time, so the Poynting vector at any point is also a function of time. Because the frequencies of typical electromagnetic waves are very high, the time variation of the Poynting vector is so rapid that it’s mostSappropriate to look at its average value. The magnitude of the average value of S at a point is called the intensity of the radiation at that point. The SI unit of intensity is the same as for S, 1 W>m2 (watt per square meter). Let’s work out the intensity of the sinusoidal wave described by Eqs. (12.17). S S We first substitute E and B into Eq. (12.28): S1x, t2 5

S

S 1 S E1x, t2 : B1x, t2 m0

1 3≥nEmax cos1kx - vt24 : 3kN Bmax cos1kx - vt24 m0 The vector product of the unit vectors is n≥ : kN 5 n≤ and cos21kx - vt2 is never 5

negative, so S1x, t2 always points in the positive x-direction (the direction of wave propagation). The x-component of the Poynting vector is S

Application Laser Surgery Lasers are used widely in medicine as ultraprecise, bloodless “scalpels.” They can reach and remove tumors with minimal damage to neighboring healthy tissues, as in the brain surgery shown here. The power output of the laser is typically below 40 W, less than that of a typical light bulb. However, this power is concentrated into a spot from 0.1 to 2.0 mm in diameter, so the intensity of the light (equal to the average value of the Poynting vector) can be as high as 5 * 109 W/m2.

Sx 1x, t2 = '

Emax Bmax Emax Bmax cos21kx - vt2 = 31 + cos 21kx - vt24 m0 2m0 '

'

The time average value of cos 21kx - vt2 is zero because at any point, it is positive during one half-cycle and negative during the other half. So the average value S of the Poynting vector over a full cycle is Sav 5 ≤NSav , where '

Emax Bmax 2m0 '

Sav =

S

That is, the magnitude of the average value of S for a sinusoidal wave (the 1 intensity I of the wave) is 2 the maximum value. By using the relationships Emax = Bmax c and P0 m0 = 1>c2, we can express the intensity in several equivalent forms: '

'

Emax Bmax Emax 2 = 2m0 2m0 c '

I = Sav =

'

=

1 2

P0 Emax2 = 12 P0 cEmax2 m Å 0 '

(intensity of a sinusoidal wave in vacuum)

(12.29)

We invite you to verify that these expressions are all equivalent. For a wave traveling in the -x-direction, represented by Eqs. (12.19), the Poynting vector is in the - x-direction at every point, but its magnitude is the same as for a wave traveling in the + x-direction. Verifying these statements is left to you (see Exercise 12.24). CAUTION Poynting vector vs. intensity At any point x, the magnitude of the Poynting vector varies with time. Hence, the instantaneous rate at which electromagnetic energy in a sinusoidal plane wave arrives at a surface is not constant. This may seem to contradict everyday experience; the light from the sun, a light bulb, or the laser in a grocery-store scanner appears steady and unvarying in strength. In fact the Poynting vector from these sources does vary in time, but the variation isn’t noticeable because the oscillation frequency is so high (around 5 * 1014 Hz for visible light). All that you sense is the average rate at which energy reaches your eye, which is why we commonly use intensity (the average value of S) to describe the strength of electromagnetic radiation. y

Throughout this discussion we have considered only electromagnetic waves propagating in vacuum. If the waves are traveling in a dielectric medium, however,

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399

12.4 Energy and Momentum in Electromagnetic Waves

the expressions for energy density [Eq. (12.23)], the Poynting vector [Eq. (12.28)], and the intensity of a sinusoidal wave [Eq. (12.29)] must be modified. It turns out that the required modifications are quite simple: Just replace P0 with the permittivity P of the dielectric, replace m0 with the permeability m of the dielectric, and replace c with the speed v Sof electromagnetic waves in the dielectric. Remarkably, S the energy densities in the E and B fields are equal even in a dielectric. Example 12.3

Energy in a nonsinusoidal wave

For the nonsinusoidal wave described in Section 12.2, suppose that E = 100 V > m = 100 N > C. Find the value of B, the energy density u, and the rate of energy flow per unit area S. '

'

'

'

u = P0E2 = 18.85 * 10-12 C2 > N # m221100 N > C22 '

= 8.85 * 10

'

'

N > m = 8.85 * 10

-8

2

'

'

J> m

-8

3

'

'

'

The magnitude of the Poynting vector is

1100 V > m213.33 * 10-7 T2 EB = m0 4p * 10-7 T # m > A

S O L U T IO N

'

S

S =

S

IDENTIFY and SET UP: In this wave E and B are uniform behind the wave front (and zero ahead of it). Hence the target variables B, u, and S must also be uniform behind the wave front. Given the magnitude E, we use Eq. (12.4) to find B, Eq. (12.25) to find u, and Eq. (12.27) to find S. (We cannot use Eq. (12.29), which applies to sinusoidal waves only.)

'

'

'

'

'

S = P0cE2 = 18.85 * 10-12 C2 > N # m2213.00 * 108 m > s2 '

'

'

'

'

'

'

From Eq. (12.25),

Example 12.4

'

'

* 1100 N > C2 = 26.5 W > m

EXECUTE: From Eq. (12.4),

100 V > m E = 3.33 * 10-7 T = c 3.00 * 108 m > s

'

EVALUATE: We can check our result for S by using Eq. (12.26): 2

B =

'

= 26.5 V # A > m2 = 26.5 W > m2

S

2

'

'

'

S

Since E and B have the same values at all points behind the wave front, u and S likewise have the same value everywhere behind the S S wave front. In front of the wave front, E 5 0 and B 5 0, and so u = 0 and S = 0; where there are no fields, there is no field energy.

Energy in a sinusoidal wave

A radio station on the earth’s surface emits a sinusoidal wave with average total power 50 kW (Fig. 12.19). Assuming that the transmitter radiates equally in all directions above the ground (which is unlikely in real situations), find the electric-field and magnetic-field amplitudes Emax and Bmax detected by a satellite 100 km from the antenna. S O L U T IO N

EXECUTE: The surface area of a hemisphere of radius r = 100 km = 1.00 * 105 m is A = 2pR2 = 2p11.00 * 105 m22 = 6.28 * 1010 m2 All the radiated power passes through this surface, so the average power per unit area (that is, the intensity) is I =

IDENTIFY and SET UP: We are given the transmitter’s average total power P. The intensity I is just the average power per unit area, so to find I at 100 km from the transmitter we divide P by the surface area of the hemisphere shown in Fig. 12.19. For a sinusoidal wave, I is also equal to the magnitude of the average value Sav of the Poynting vector, so we can use Eqs. (12.29) to find Emax ; Eq. (12.4) then yields Bmax . 12.19 A radio station radiates waves into the hemisphere shown.

P P 5.00 * 104 W = = = 7.96 * 10-7 W > m2 A 2pR2 6.28 * 1010 m2 '

From Eqs. (12.29), I = Sav = Emax2 > 2m0c, so '

= 2214p * 10-7 T # m > A213.00 * 108 m > s217.96 * 10-7 W > m22 '

= 2.45 * 10

-2

'

'

'

'

'

V> m '

'

Then from Eq. (12.4), Bmax =

Transmitter

'

Emax = 22m0cSav

Satellite

r 5 100 km

'

Emax = 8.17 * 10-11 T c

EVALUATE: Note that Emax is comparable to fields commonly seen S in the laboratory, but Bmax is extremely small in comparison to B fields we saw in previous chapters. For this reason, most detectors of electromagnetic radiation respond to the effect of the electric field, not the magnetic field. Loop radio antennas are an exception (see the Bridging Problem at the end of this chapter).

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400

CHAPTER 12 Electromagnetic Waves

Electromagnetic Momentum Flow and Radiation Pressure By using the observation that energy is required to establish electric and magnetic fields, we have shown that electromagnetic waves transport energy. It can also be shown that electromagnetic waves carry momentum p, with a corresponding momentum density (momentum dp per volume dV) of magnitude dp EB S = = 2 2 dV m0 c c

(12.30)

'

This momentum is a property of the field; it is not associated with the mass of a moving particle in the usual sense. There is also a corresponding momentum flow rate. The volume dV occupied by an electromagnetic wave (speed c) that passes through an area A in time dt is dV = Ac dt. When we substitute this into Eq. (12.30) and rearrange, we find that the momentum flow rate per unit area is EB 1 dp S = = c m0 c A dt

(flow rate of electromagnetic momentum)

(12.31)

'

This is the momentum transferred per unit surface area per unit time. We obtain the average rate of momentum transfer per unit area by replacing S in Eq. (12.31) by Sav = I. This momentum is responsible for radiation pressure. When an electromagnetic wave is completely absorbed by a surface, the wave’s momentum is also transferred to the surface. For simplicity we’ll consider a surface perpendicular to the propagation direction. Using the ideas developed in Section 8.1, we see that the rate dp>dt at which momentum is transferred to the absorbing surface equals the force on the surface. The average force per unit area due to the wave, or radiation pressure prad , is the average value of dp>dt divided by the absorbing area A. (We use the subscript “rad” to distinguish pressure from momentum, for which the symbol p is also used.) From Eq. (12.31) the radiation pressure is '

12.20 At the center of this interstellar gas cloud is a group of intensely luminous stars that exert tremendous radiation pressure on their surroundings. Aided by a “wind” of particles emanating from the stars, over the past million years the radiation pressure has carved out a bubble within the cloud 70 light-years across.

prad =

Sav I = c c

(radiation pressure, wave totally absorbed)

(12.32)

If the wave is totally reflected, the momentum change is twice as great, and the pressure is prad =

2Sav 2I = c c

(radiation pressure, wave totally reflected)

(12.33)

For example, the value of I (or Sav) for direct sunlight, before it passes through the earth’s atmosphere, is approximately 1.4 kW>m2. From Eq. (12.32) the corresponding average pressure on a completely absorbing surface is prad =

1.4 * 103 W>m2 I = 4.7 * 10-6 Pa = c 3.0 * 108 m>s

From Eq. (12.33) the average pressure on a totally reflecting surface is twice this: 2I>c or 9.4 * 10-6 Pa. These are very small pressures, of the order of 10-10 atm, but they can be measured with sensitive instruments. The radiation pressure of sunlight is much greater inside the sun than at the earth (see Problem 12.45). Inside stars that are much more massive and luminous than the sun, radiation pressure is so great that it substantially augments the gas pressure within the star and so helps to prevent the star from collapsing under its own gravity. In some cases the radiation pressure of stars can have dramatic effects on the material surrounding them (Fig. 12.20).

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401

12.5 Standing Electromagnetic Waves

Example 12.5

Power and pressure from sunlight

An earth-orbiting satellite has solar energy–collecting panels with a total area of 4.0 m2 (Fig. 12.21). If the sun’s radiation is perpendicular to the panels and is completely absorbed, find the average solar power absorbed and the average radiation-pressure force.

points above the atmosphere, which is where the satellite orbits.) Multiplying each value by the area of the solar panels gives the average power absorbed and the net radiation force on the panels. EXECUTE: The intensity I (power per unit area) is 1.4 * 103 W > m2. Although the light from the sun is not a simple sinusoidal wave, we can still use the relationship that the average power P is the intensity I times the area A: '

S O L U T IO N IDENTIFY and SET UP: This problem uses the relationships among intensity, power, radiation pressure, and force. In the above discussion we calculated the intensity I (power per unit area) of sunlight as well as the radiation pressure prad (force per unit area) of sunlight on an absorbing surface. (We calculated these values for

P = IA = 11.4 * 103 W > m2214.0 m22 '

'

3

= 5.6 * 10 W = 5.6 kW

The radiation pressure of sunlight on an absorbing surface is prad = 4.7 * 10-6 Pa = 4.7 * 10-6 N > m2. The total force F is the pressure prad times the area A: '

12.21 Solar panels on a satellite. Sun sensor (to keep panels facing the sun)

S

S

Solar panels

S

S

'

'

F = pradA = 14.7 * 10-6 N > m2214.0 m22 = 1.9 * 10-5 N '

'

EVALUATE: The absorbed power is quite substantial. Part of it can be used to power the equipment aboard the satellite; the rest goes into heating the panels, either directly or due to inefficiencies in the photocells contained in the panels. The total radiation force is comparable to the weight (on earth) of a single grain of salt. Over time, however, this small force can have a noticeable effect on the orbit of a satellite like that in Fig. 12.21, and so radiation pressure must be taken into account.

Test Your Understanding of Section 12.4 Figure 12.13 shows one wavelength of a sinusoidal electromagnetic wave at time t = 0. For which of the following four values of x is (a) the energy density a maximum; (b) the energy density a minimum; (c) the magnitude of the instantaneous (not average) Poynting vector a maximum; (d) the magnitude of the instantaneous (not average) Poynting vector a minimum? (i) x = 0; (ii) x = l>4; (iii) x = l>2; (iv) x = 3l>4. y

12.5

Standing Electromagnetic Waves

Electromagnetic waves can be reflected; the surface of a conductor (like a pol ished sheet of metal) or of a dielectric (such as a sheet of glass) can serve as a reflector. The superposition principle holds for electromagnetic waves just as for electric and magnetic fields. The superposition of an incident wave and a reflected wave forms a standing wave. The situation is analogous to standing waves on a stretched string, discussed in Section 15.7; you should review that discussion. Suppose a sheet of a perfect conductor (zero resistivity) is placed in the yz-plane of Fig. 12.22 and a linearly polarized electromagnetic wave, travelS ing in the negative x-direction, strikes it. As we discussed in Section 3.4, E cannot have a component Sparallel to the surface of a perfect conductor. Therefore in the present situation, E must be zero everywhere in the yz-plane. The electric field of the incident electromagnetic wave is not zero at all times in the yz-plane. But this incident wave induces oscillating currents on the surface of the conductor, and these currents give rise to an additional electric field. The net electric S field, which is the vector sum of this field and the incident E, is zero everywhere inside and on the surface of the conductor. The currents induced on the surface of the conductor also produce a reflected wave that travels out from the plane in the + x-direction. Suppose the incident wave is described by the wave functions of Eqs. (12.19) (a sinusoidal wave traveling in the - x-direction) and the reflected wave by the negative of Eqs. (12.16) (a sinusoidal wave traveling in the +x-direction). We take the negative of the

?

12.22 Representation of the electric and magnetic fields of a linearly polarized electromagnetic standing wave when vt = 3p>4 rad. In any plane perpendicular to the x-axis, E is maximum (an antinode) where B is zero (a node), and vice versa. As time elapses, the pattern does not move x-axis; instead, at every point the along the S S E and B vectors simply oscillate. y

Perfect conductor x 5 l: S nodal plane of E S S antinodal plane of B B

z

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S

E

x

/

x 5 3l 4: S antinodal planeSof E nodal plane of B

402

CHAPTER 12 Electromagnetic Waves

PhET: Microwaves

wave given by Eqs. (12.16) so that the incident and reflected electric fields cancel at x = 0 (the plane of the conductor, where the total electric field must be zero). S The superposition principle states that the total E field at any point is the vector S S sum of the E fields of the incident and reflected waves, and similarly for the B field. Therefore the wave functions for the superposition of the two waves are Ey 1x, t2 = Emax3cos1kx + vt2 - cos1kx - vt24 '

Bz 1x, t2 = Bmax3-cos1kx + vt2 - cos1kx - vt24 '

We can expand and simplify these expressions, using the identities cos1A 6 B2 = cos A cos B 7 sin A sin B The results are Ey 1x, t2 = - 2Emax sin kx sin vt

(12.34)

'

Bz 1x, t2 = - 2Bmax cos kx cos vt

(12.35)

'

Equation (12.34) is analogous to Eq. (15.28) for a stretched string. We see that at x = 0 the electric field Ey1x = 0, t2 is always zero; this is required by the nature of the ideal conductor, which plays the same role as a fixed point at the end of a string. Furthermore, Ey 1x, t2 is zero at all times at points in those planes perpendicular to the x-axis for which sin kx = 0—that is, kx = 0, p, 2p, Á . Since k = 2p>l, the positions of these planes are '

x = 0,

l 3l , l, ,Á 2 2

1nodal planes of E 2 S

(12.36)

S

These planes are called the nodal planes of the E field; they are the equivalent of the nodes, or nodal points, of a standing wave on a string. Midway between any two adjacent nodal planes is a plane on which sin kx = 61; on each such plane, the magnitude of E1x, t2 equals the maximum possible value of 2Emax twice per S oscillation cycle. These are the antinodal planes of E, corresponding to the antinodes of waves on a string. The total magnetic field is zero at all times at points in planes on which cos kx = 0. This occurs where x =

l 3l 5l , , ,Á 4 4 4

1nodal planes of B2

S

S

(12.37) S

These are the nodal planes of the B field; there is an antinodal plane of B midway between any two adjacent nodal planes. Figure 12.22 shows a standing-wave pattern at one instant of time. The magnetic field is not zero at the conducting surface 1x = 02. The surface currents that S must be present to make E exactly zero at the surface cause magnetic fields at the surface. The nodal planes of each field are separated by one half-wavelength. The nodalSplanes of one field are midway between those of the other; hence the S nodes of E coincide with the antinodes of B, and conversely. Compare this sit uation to the distinction between pressure nodes and displacement nodes in Section 16.4. The total electric field is a sine function of t, and the total magnetic field is a cosine function of t. The sinusoidal variations of the two fields are therefore 90° out of phase at each point. At times when sin vt = 0, the electric field is zero everywhere, and the magnetic field is maximum. When cos vt = 0, the magnetic field is zero everywhere, and the electric field is maximum. This is in contrast to a wave traveling in one direction, as described by SEqs. (12.16) or (12.19) sepaS rately, in which the sinusoidal variations of E and B at any particular point are in phase. You can show that Eqs. (12.34) and (12.35) satisfy the wave equation, Eq. (12.15). You can also show that they satisfy Eqs. (12.12) and (12.14), the equivalents of Faraday’s and Ampere’s laws (see Exercise 12.36).

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403

12.5 Standing Electromagnetic Waves

Standing Waves in a Cavity Let’s now insert a second conducting plane, parallel to the first and a distance L from it, along the +x-axis. The cavity between the two planes is analogous to a stretched string held Sat the points x = 0 and x = L. Both conducting planes must be nodal planes for E; a standing wave can exist only when the second plane is placed at one of the positions where E1x, t2 = 0, so L must be an integer multiple of l>2. The wavelengths that satisfy this condition are 2L ln = n

1n = 1, 2, 3, Á 2

(12.38)

The corresponding frequencies are ƒn =

c c = n ln 2L

1n = 1, 2, 3, Á 2

(12.39)

12.23 A typical microwave oven sets up a standing electromagnetic wave with l = 12.2 cm, a wavelength that is strongly absorbed by the water in food. Because the wave has nodes spaced l>2 = 6.1 cm apart, the food must be rotated while cooking. Otherwise, the portion that lies at a node—where the electric-field amplitude is zero—will remain cold.

Thus there is a set of normal modes, each with a characteristic frequency, wave shape, and node pattern (Fig. 12.23). By measuring the node positions, we can measure the wavelength. If the frequency is known, the wave speed can be determined. This technique was first used by Hertz in the 1880s in his pioneering investigations of electromagnetic waves. Conducting surfaces are not the only reflectors of electromagnetic waves. Reflections also occur at an interface between two insulating materials with different dielectric or magnetic properties. The mechanical analog is a junction of two strings with equal tension but different linear mass density. In general, a wave incident on such a boundary surface is partly transmitted into the second material and partly reflected back into the first. For example, light is transmitted through a glass window, but its surfaces also reflect light.

Example 12.6

Intensity in a standing wave

Calculate the intensity of the standing wave represented by Eqs. (12.34) and (12.35).

IDENTIFY and SET UP: The intensity I of the wave is the Stimeaveraged value Sav of the magnitude of the Poynting vector S . To Sav, we first use Eq. (12.28) to find the instantaneous value of find S S and then average it over a whole number of cycles of the wave. EXECUTE: Using the wave functionsSof Eqs. (12.34) and (12.35) in Eq. (12.28) for the Poynting vector S, we find S1x, t2 5

'

Sx 1x, t2 = '

S O L U T IO N

S

Using the identity sin 2A = 2 sin A cos A, we can rewrite Sx 1x, t2 as

S 1 S E1x, t2 : B1x, t2 m0

1 3 - 2≥nEmax sin kx cos vt4 : 3 -2kN Bmax cos kx sin vt4 m0 EmaxBmax 5 n≤ 12 sin kx cos kx212 sin vt cos vt2 m0 5 n≤ Sx 1x, t2 5

EmaxBmax sin 2kx sin 2vt m0

The average value of a sine function overS any whole number of cycles is zero. Thus the time average of S at any point is zero; I = Sav = 0. EVALUATE: This result is what we should expect. The standing wave is a superposition of two waves with the same frequency and amplitude, traveling in opposite directions. All the energy transferred by one wave is cancelled by an equal amount transferred in the opposite direction by the other wave. When we use electro magnetic waves to transmit power, it is important to avoid reflections that give rise to standing waves.

'

Example 12.7

Standing waves in a cavity

Electromagnetic standing waves are set up in a cavity with two parallel, highly conducting walls 1.50 cm apart. (a) Calculate the longest wavelength l and lowest frequency ƒ of these standing

waves. (b) For a standing wave of this wavelength,S where in the S cavitySdoes E have maximum magnitude? Where is E zero? Where S does B have maximum magnitude? Where is B zero? Continued

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404

CHAPTER 12 Electromagnetic Waves

SOLUTION IDENTIFY and SET UP: Only certain normal modes are possible for electromagnetic waves in a cavity, just as only certain normal modes are possible for standing waves on a string. The longest possible wavelength and lowest possible frequency correspond to the n = 1 mode in Eqs. (12.38) and (12.39); we use these to find l and f. Equations S (12.36)Sand (12.37) then give the locations of the nodal planes of E and B. The antinodal planes of each field are midway between adjacent nodal planes. EXECUTE: (a) From Eqs. (12.38) and (12.39), the n = 1 wavelength and frequency are

(b) With n = 1 there is a single half-wavelength between the S walls. The electric field hasSnodal planes 1E 5 02 at the walls and an antinodal plane (where E has its maximum magnitude) midway between them. The magnetic field has antinodal planes at the walls and a nodal plane midway between them. EVALUATE: One application of such standing waves is to produce S an oscillating E field of definite frequency, which is used to probe the behavior of a small sample of material placed in the cavity. To subject the sample to the strongest possible field, itS should be placed near the center of the cavity, at the antinode of E.

l1 = 2L = 211.50 cm2 = 3.00 cm

3.00 * 108 m > s c = = 1.00 * 1010 Hz = 10 GHz 2L 211.50 * 10-2 m2 '

ƒ1 =

'

Test Your Understanding of Section 12.5 In the standing wave described in Example 12.7, is there any point in the cavity where the energy density is zero at all times? If so, where? If not, why not? y

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CHAPTER

12

SUMMARY

Maxwell’s equations and electromagnetic waves: Maxwell’s equations predict the existence of electromagnetic waves that propagate in vacuum at the speed of light c. The electromagnetic spectrum covers frequencies from at least 1 to 1024 Hz and a correspondingly broad range of wavelengths. Visible light, with wavelengths from 380 to 750 nm, is only a very small S S part of this spectrum. In a plane wave, E and B are uniform over any plane perpendicular to the propagation direction. Faraday’s law and Ampere’s law bothSgive S relationships between the magnitudes of E and B; requiring both of these relationships to be satisfied gives an expression for c in terms of P0 and m0 . ElectromagS S netic waves are transverse; the E and B fields are perpendicular to the direction of propagation and to each other. The direction of propagation is the direction S S of E : B.

E = cB

y

(12.4)

Planar wave front

S

B = P0 m0 cE '

c =

(12.8)

'

E

S

E

S

B

S

B

1

S

E S

(12.9)

2P0 m0

S

B

E50 c

S

'

E

S

B50

O

S

S

B

E

S

E

S

z

B

x

S

B

'

Sinusoidal electromagnetic waves: Equations (12.17) and (12.18) describe a sinusoidal plane electromagnetic wave traveling in vacuum in the +x-direction. If the wave is propagating in the -x-direction, replace kx - vt by kx + vt. (See Example 12.1.)

E1x, t2 5 n≥ Emax cos1kx - vt2 S

y

B1x, t2 5 kN Bmax cos1kx - vt2

(12.17)

Emax = cBmax

(12.18)

S

c

S

E O

S

S

B

B

S

E

z x S

E

Electromagnetic waves in matter: When an electromagnetic wave travels through a dielectric, the wave speed v is less than the speed of light in vacuum c. (See Example 12.2.)

Energy and momentum in electromagnetic waves: The energy flow rate (power per unit area) in an electromagS netic wave in vacuum is given by the Poynting vector S. The magnitude of the time-averaged value of the Poynting vector is called the intensity I of the wave. Electromagnetic waves also carry momentum. When an electromagnetic wave strikes a surface, it exerts a radiation pressure prad . If the surface is perpendicular to the wave propagation direction and is totally absorbing, prad = I>c; if the surface is a perfect reflector, prad = 2I>c. (See Examples 12.3–12.5.) '

v = =

1

1

=

2Pm c

S

B

1

2KKm 2P0m0

(12.21)

2KKm

S

S5

S 1 S E : B m0

y

(12.28)

c dt S

Emax Bmax Emax 2 = = 2m0 2m0 c

S

I = Sav

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=

1 2

1 dp S EB = = c m0 c A dt

S S

B

E

x

Stationary Wave front at plane time dt later

(12.29)

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S

A

z

P0 E 2 A m0 max

= 12 P0 cEmax2

E

B

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(12.31)

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(flow rate of electromagnetic momentum)

Standing electromagnetic waves: If a perfect reflecting surface is placed at x = 0, the incident and S reflected waves form a standing wave. Nodal planes for E occur at kx = 0, p, 2p, Á ,Sand nodal S S planes for B at kx = p>2, 3p>2, 5p>2, Á . At each point, the sinusoidal variations of E and B with time are 90° out of phase. (See Examples 12.6 and 12.7.)

y

Perfect conductor S

B z S

E

x

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CHAPTER 12 Electromagnetic Waves

Detecting Electromagnetic Waves

BRIDGING PROBLEM

A circular loop of wire can be used as a radio antenna. If an 18.0-cmdiameter antenna is located 2.50 km from a 95.0-MHz source with a total power of 55.0 kW, what is the maximum emf induced in the loop? Assume that the plane of the antenna loop is perpendicular to the direction of the radiation’s magnetic field and that the source radiates uniformly in all directions. SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP: 1. The electromagnetic wave has an oscillating magnetic field. This causes a magnetic flux through the loop antenna that varies sinusoidally with time. By Faraday’s law, this produces an emf equal in magnitude to the rate of change of the flux. The target variable is the magnitude of this emf. 2. Select the equations that you will need to find (i) the intensity of the wave at the position of the loop, a distance r 5 2.50 km

Problems

from the source of power P = 55.0 kW; (ii) the amplitude of the sinusoidally varying magnetic field at that position; (iii) the magnetic flux through the loop as a function of time; and (iv) the emf produced by the flux. EXECUTE 3. Find the wave intensity at the position of the loop. 4. Use your result from step 3 to write expressions for the timedependent magnetic field at this position and the time-dependent magnetic flux through the loop. 5. Use the results of step 4 to find the time-dependent induced emf in the loop. The amplitude of this emf is your target variable. EVALUATE 6. Is the induced emf large enough to detect? (If it is, a receiver connected to this antenna will be able to pick up signals from the source.)

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q12.1 By measuring the electric and magnetic fields at a point in space where there is an electromagnetic wave, can you determine the direction from which the wave came? Explain. Q12.2 According to Ampere’s law, is it possible to have both a conduction current and a displacement current at the same time? Is it possible for the effects of the two kinds of current to cancel each other exactly so that no magnetic field is produced? Explain. Q12.3 Give several examples of electromagnetic waves that are encountered in everyday life. How are they all alike? How do they differ? Q12.4 Sometimes neon signs located near a powerful radio station are seen to glow faintly at night, even though they are not turned on. What is happening? Q12.5 Is polarization a property of all electromagnetic waves, or is it unique to visible light? Can sound waves be polarized? What fundamental distinction in wave properties is involved? Explain. Q12.6 Suppose that a positive point charge q is initially at rest on the x-axis, in the path of the electromagnetic plane wave described in Section 12.2. Will the charge move after the wave front reaches it? If not, why not? If the charge does move, describe its motion S S qualitatively. (Remember that E and B have the same value at all points behind the wave front.) Q12.7 The light beam from a searchlight may have an electricfield magnitude of 1000 V > m, corresponding to a potential differ'

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ence of 1500 V between the head and feet of a 1.5-m-tall person on whom the light shines. Does this cause the person to feel a strong electric shock? Why or why not? Q12.8 For a certain sinusoidal wave of intensity I, the amplitude of the magnetic field is B. What would be the amplitude (in terms of B) in a similar wave of twice the intensity? Q12.9 The magnetic-field amplitude of the electromagnetic wave from the laser described in Example 12.1 (Section 12.3) is about 100 times greater than the earth’s magnetic field. If you illuminate a compass with the light from this laser, would you expect the compass to deflect? Why or why not? Q12.10 Most automobiles have vertical antennas for receiving radio broadcasts. Explain what this tells you about the direction of S polarization of E in the radio waves used in broadcasting. Q12.11 If a light beam carries momentum, should a person holding a flashlight feel a recoil analogous to the recoil of a rifle when it is fired? Why is this recoil not actually observed? Q12.12 A light source radiates a sinusoidal electromagnetic wave uniformly in all directions. This wave exerts an average pressure p on a perfectly reflecting surface a distance R away from it. What average pressure (in terms of p) would this wave exert on a perfectly absorbing surface that was twice as far from the source? Q12.13 Does an electromagnetic standing wave have energy? Does it have momentum? Are your answers to these questions the same as for a traveling wave? Why or why not?

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Exercises

EXERCISES Section 12.2 Plane Electromagnetic Waves and the Speed of Light

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Section 12.3 Sinusoidal Electromagnetic Waves

12.5 . BIO Medical X rays. Medical x rays are taken with electromagnetic waves having a wavelength of around 0.10 nm. What are the frequency, period, and wave number of such waves? 12.6 . BIO Ultraviolet Radiation. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nm to 400 nm. It is not harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 nm and 320 nm, is much more dangerous because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB? 12.7 . A sinusoidal electromagnetic wave having a magnetic field of amplitude 1.25 mT and a wavelength of 432 nm is traveling in the + x-direction through empty space. (a) What is the frequency of this wave? (b) What is the amplitude of the associated electric field? (c) Write the equations for the electric and magnetic fields as functions of x and t in the form of Eqs. (12.17). 12.8 . An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the - z-direction. The electric field has amplitude 2.70 * 10-3 V > m and is parallel to the x-axis. What are (a) the frequency and (b)Sthe magnetic-field amplitude? (c) Write the vecS tor equations for E1z, t2 and B1z, t2. 12.9 . Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) 5.0 km, (ii) 5.0 mm, (iii) 5.0 nm. (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency 6.50 * 1021 Hz and (ii) an AM station radio wave of frequency 590 kHz? 12.10 . The electric field of a sinusoidal electromagnetic wave obeys the equation E = 1375 V > m2 cos311.99 * 107 rad > m2x + 15.97 * 1015 rad > s2t4. (a) What are the amplitudes of the electric and magnetic fields of this wave? (b) What are the frequency, wavelength, and period of the wave? Is this light visible to humans? (c) What is the speed of the wave? . An electromagnetic wave has an electric field given by 12.11 S E1y, t2 5 13.10 * 105 V > m2kN cos3ky - 112.65 *1012 rad > s2t4. (a) In which direction is the wave traveling? (b) What is Sthe wavelength of the wave? (c) Write the vector equation for B1y, t2. '

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. An electromagnetic wave has a magnetic field given by 12.12 S B1x, t2 5 - 18.25 * 10-9 T2≥n cos311.38 * 104 rad > m2x + vt4. (a) In which direction is the wave traveling? (b) What Sis the frequency ƒ of the wave? (c) Write the vector equation for E1x, t2. 12.13 . Radio station WCCO in Minneapolis broadcasts at a frequency of 830 kHz. At a point some distance from the transmitter, the magnetic-field amplitude of the electromagnetic wave from WCCO is 4.82 * 10-11 T. Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude. 12.14 . An electromagnetic wave with frequency 65.0 Hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude 7.20 * 10-3 V > m. (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field? 12.15 . An electromagnetic wave with frequency 5.70 * 1014 Hz propagates with a speed of 2.17 * 108 m > s in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction n of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity. '

12.1 . (a) How much time does it take light to travel from the moon to the earth, a distance of 384,000 km? (b) Light from the star Sirius takes 8.61 years to reach the earth. What is the distance from earth to Sirius in kilometers? 12.2 . Consider each of the electric- and magnetic-field orientations given next.SIn each case, what is the direction of propagation S S of the wave? (a) E in the -direction, in the +y-direction; (b) E B + x S S in the -yS-direction, B in the + x-direction; (c) E in the +zS S direction, B in the - x-direction; (d) E in the +y-direction, B in the - z-direction. 12.3 . A sinusoidal electromagnetic wave is propagating in vacuum in the +z-direction. If at a particular instant and at a certain point in space the electric field is in the + x-direction and has magnitude 4.00 V > m, what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time? 12.4 . Consider each of the following electric- and magnetic-field orientations. InS each case, what is the Sdirection Sof propagation of S S n, B 5 - B≥n; (b) E 5 E≥n, B 5 B≤n; (c) E 5 the wave? (a) E 5 E≤ S S S n, B n. - Ek 5 - B≤n; (d) E 5 E≤n, B 5 - Bk

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Section 12.4 Energy and Momentum in Electromagnetic Waves

12.16 . BIO High-Energy Cancer Treatment. Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pulses of light that last for an extremely short time (a few nanoseconds). These short pulses scramble the interior of a cell without causing it to explode, as long pulses would do. We can model a typical such cell as a disk 5.0 mm in diameter, with the pulse lasting for 4.0 ns with an average power of 2.0 * 1012 W . We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse. (a) How much energy is given to the cell during this pulse? (b) What is the intensity 1in W> m22 delivered to the cell? (c) What are the maximum values of the electric and magnetic fields in the pulse? 12.17 .. Fields from a Light Bulb. We can reasonably model a 75-W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in W > m2) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity? 12.18 .. A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.500 m2. At the window, the electric field of the wave has rms value 0.0200 V > m. How much energy does this wave carry through the window during a 30.0-s commercial? 12.19 .. Testing a Space Radio Transmitter. You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in physics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle’s movable arm, you aim a sensitive detector at the ISS, which is 2.5 km away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is 0.090 V > m and that the frequency of the waves is 244 MHz. Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations? '

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CHAPTER 12 Electromagnetic Waves

12.20 . The intensity of a cylindrical laser beam is 0.800 W > m2. The cross-sectional area of the beam is 3.0 * 10-4 m2 and the intensity is uniform across the cross section of the beam. (a) What is the average power output of the laser? (b) What is the rms value of the electric field in the beam? 12.21 . A space probe 2.0 * 1010 m from a star measures the total intensity of electromagnetic radiation from the star to be 5.0 * 103 W > m2. If the star radiates uniformly in all directions, what is its total average power output? 12.22 . A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 cm and an electric-field amplitude of 5.40 * 10-2 V > m at a distance of 250 m from the phone. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave. 12.23 . A monochromatic light source with power output 60.0 W radiates light of wavelength 700 nm uniformly in all directions. Calculate Emax and Bmax for the 700-nm light at a distance of 5.00 m from the source. 12.24 . For the electromagnetic wave represented by Eqs. (12.19), show that the Poynting vector (a) is in the same direction as the propagation of the wave and (b) has average magnitude given by Eqs. (12.29). 12.25 .. An intense light source radiates uniformly in all directions. At a distance of 5.0 m from the source, the radiation pressure on a perfectly absorbing surface is 9.0 * 10-6 Pa. What is the total average power output of the source? 12.26 . Television Broadcasting. Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 316 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home 5.00 km away from the antenna, (a) what average pressure does this wave exert on a totally reflecting surface, (b) what are the amplitudes of the electric and magnetic fields of the wave, and (c) what is the average density of the energy this wave carries? (d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field? 12.27 .. BIO Laser Safety. If the eye receives an average intensity greater than 1.0 * 102 W>m2, damage to the retina can occur. This quantity is called the damage threshold of the retina. (a) What is the largest average power (in mW) that a laser beam 1.5 mm in diameter can have and still be considered safe to view head-on? (b) What are the maximum values of the electric and magnetic fields for the beam in part (a)? (c) How much energy would the beam in part (a) deliver per second to the retina? (d) Express the damage threshold in W> cm2. 12.28 . In the 25-ft Space Simulator facility at NASA’s Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 W > m2 at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.) Find the average radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing section of the floor and (b) a totally reflecting section of the floor. (c) Find the average momentum density (momentum per unit volume) in the light at the floor. 12.29 . Laboratory Lasers. He–Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 nm and a power of 0.500 mW spread over a cylindrical beam 1.00 mm in diameter (although these quantities can vary). (a) What is the intensity of this laser beam? (b) What are the maximum values of the electric and magnetic fields? (c) What is the average energy density in the laser beam? 12.30 .. Solar Sail 1. During 2004, Japanese scientists successfully tested two solar sails. One had a somewhat complicated shape that we shall model as a disk 9.0 m in diameter and 7.5 mm '

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thick. The intensity of solar energy at that location was about 1400 W> m2. (a) What force did the sun’s light exert on this sail, assuming that it struck perpendicular to the sail and that the sail was perfectly reflecting? (b) If the sail was made of magnesium, of density 1.74 g>cm3, what acceleration would the sun’s radiation give to the sail? (c) Does the acceleration seem large enough to be feasible for space flight? In what ways could the sail be modified to increase its acceleration?

Section 12.5 Standing Electromagnetic Waves

12.31 . Microwave Oven. The microwaves in a certain microwave oven have a wavelength of 12.2 cm. (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing-wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made 5.0 cm longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven? 12.32 . An electromagnetic standing wave in air of frequency 750 MHz is set up between two conducting planes 80.0 cm apart. At which positions between the planes could a point charge be placed at rest so that it would remain at rest? Explain. 12.33 . A standing electromagnetic wave in a certain material has S frequency 2.20 * 1010 Hz. The nodal planes of B are 3.55 mm apart. Find (a) the wavelength of the wave in this material; (b) the S distance between adjacent nodal planes of the E field; (c) the speed of propagation of the wave. 12.34 . An electromagnetic standing wave in air has frequency S 75.0 MHz. (a) What is the distance between nodal planesSof the E field? (b) What is the distance between a nodal plane of E and the S closest nodal plane of B ? 12.35 . An electromagnetic standing wave in a certain material '

has frequency 1.20 * 1010 Hz and speed of propagation 2.10 *

108 m > s. (a) What is the distance between a nodal plane of B and S

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the closest antinodal plane of B? (b) What is the distance between S

S

an antinodal plane of E and the closest antinodal plane of B? S

(c) What is the distance between a nodal plane of E and the closest S

nodal plane of B? 12.36 . CALC Show that the electric and magnetic fields for standing waves given by Eqs. (12.34) and (12.35) (a) satisfy the wave equation, Eq. (12.15), and (b) satisfy Eqs. (12.12) and (12.14).

PROBLEMS 12.37 .. BIO Laser Surgery.

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses of energy absorbed by the retina weld the detached portions back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 250 mW of power spread over a circular spot 510 mm in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34. (a) If the laser pulses are each 1.50 ms long, how much energy is delivered to the retina with each pulse? (b) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot? (c) What are the wavelength and frequency of the laser light inside the vitreous humor of the eye? (d) What are the maximum values of the electric and magnetic fields in the laser beam? 12.38 S.. CALC Consider a sinusoidal electromagnetic wave with S fields E 5 Emaxn≥ cos1kx - vt2 and B 5 BmaxkN cos1kx - vt + f2,

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Problems S

S

with - p … p … p. Show that if E and B are to satisfy Eqs. (12.12) and (12.14), thenS Emax = cBmax and p = 0. (The S result p = 0 means the E and B fields oscillate in phase.) 12.39 .. You want to support a sheet of fireproof paper horizontally, using only a vertical upward beam of light spread uniformly over the sheet. There is no other light on this paper. The sheet measures 22.0 cm by 28.0 cm and has a mass of 1.50 g. (a) If the paper is black and hence absorbs all the light that hits it, what must be the intensity of the light beam? (b) For the light in part (a), what are the amplitudes of its electric and magnetic fields? (c) If the paper is white and hence reflects all the light that hits it, what intensity of light beam is needed to support it? (d) To see if it is physically reasonable to expect to support a sheet of paper this way, calculate the intensity in a typical 0.500-mW laser beam that is 1.00 mm in diameter, and compare this value with your answer in part (a). 12.40 .. For a sinusoidal electromagnetic wave in vacuum, such as that described by Eq. (12.16), show that the average energy density in the electric field is the same as that in the magnetic field. 12.41 . A satellite 575 km above the earth’s surface transmits sinusoidal electromagnetic waves of frequency 92.4 MHz uniformly in all directions, with a power of 25.0 kW. (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring 15.0 cm by 40.0 cm oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects? 12.42 . A plane sinusoidal electromagnetic wave in air has a S wavelength of 3.84 cm and an E-field amplitude of 1.35 V> m. S (a) What is the frequency? (b) What is the B-field amplitude? (c) What is the intensity? (d) What average force does this radiation exert on a totally absorbing surface with area 0.240 m2 perpendicular to the direction of propagation? 12.43 . A small helium–neon laser emits red visible light with a power of 4.60 mW in a beam that has a diameter of 2.50 mm. (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a 1.00-m length of the beam? 12.44 .. The electric-field component of a sinusoidal electromagnetic wave traveling through a plastic cylinder is given by the equation E = 15.35 V>m2 cos311.39 * 107 rad>m2x - 13.02 * 1015 rad>s2t4. (a) Find the frequency, wavelength, and speed of this wave in the plastic. (b) What is the index of refraction of the plastic? (c) Assuming that the amplitude of the electric field does not change, write a comparable equation for the electric field if the light is traveling in air instead of in plastic. 12.45 . The sun emits energy in the form of electromagnetic waves at a rate of 3.9 * 1026 W. This energy is produced by nuclear reactions deep in the sun’s interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius r = R = 6.96 * 105 km2 and at r = R > 2, in the sun’s interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 12.4 for sunlight just before it enters the earth’s atmosphere. (b) The gas pressure at the sun’s surface is about 1.0 * 104 Pa; at r = R > 2, the gas pressure is calculated from solar models to be about 4.7 * 1013 Pa. Comparing with '

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your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not? 12.46 .. A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At 10.0 m from this source, the amplitude of the electric field is measured to be 1.50 N > C. What is the electric-field amplitude at a distance of 20.0 cm from the source? 12.47 .. CP Two square reflectors, each 1.50 cm on a side and of mass 4.00 g, are located at Figure P12.47 opposite ends of a thin, ex1.00 m tremely light, 1.00-m rod that can rotate without friction and in vacuum about an axle perAxis of rotation pendicular to it through its center (Fig. P12.47). These reflectors are small enough to be treated as point masses in moment-of-inertia calculations. Both reflectors are illuminated on one face by a sinusoidal light wave having an electric field of amplitude 1.25 N > C that falls uniformly on both surfaces and always strikes them perpendicular to the plane of their surfaces. One reflector is covered with a perfectly absorbing coating, and the other is covered with a perfectly reflecting coating. What is the angular acceleration of this device? 12.48 .. CP A circular loop of wire has radius 7.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0195 W>m2, and the wavelength of the wave is 6.90 m. What is the maximum emf induced in the loop? 12.49 . CALC CP A cylindrical conductor with a circular cross section has a radius a and a resistivity r and carries a constant current I. (a) What are the magnitude and direction of the electric-field S vector E at a point just inside the wire at a distance a from the axis? (b) What are the magnitude and direction of the magneticS field vector B at the same point?S(c) What are the magnitude and directionS of the Poynting vector S at the same point? (The direction of S is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length l of S the conductor. (Hint: Integrate S over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a currentcarrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor. 12.50 . In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0 MHz in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing-wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing-wave pattern is determined to be in its eighth harmonic, how long is the cavity? 12.51 .. CP Flashlight to the Rescue. You are the sole crew member of the interplanetary spaceship T:1339 Vorga, which makes regular cargo runs between the earth and the mining colonies in the asteroid belt. You are working outside the ship one day while at a distance of 2.0 AU from the sun. [1 AU (astronomical unit) is the average distance from the earth to the sun, 149,600,000 km.] Unfortunately, you lose contact with the ship’s hull and begin to drift away into space. You use your spacesuit’s rockets to try to push yourself back toward the ship, but they run out of fuel and stop working before you can return to the ship. You find yourself in an awkward position, floating 16.0 m from the spaceship with zero velocity relative to it. Fortunately, you are '

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CHAPTER 12 Electromagnetic Waves

carrying a 200-W flashlight. You turn on the flashlight and use its beam as a “light rocket” to push yourself back toward the ship. (a) If you, your spacesuit, and the flashlight have a combined mass of 150 kg, how long will it take you to get back to the ship? (b) Is there another way you could use the flashlight to accomplish the same job of returning you to the ship? 12.52 . The 19th-century inventor Nikola Tesla proposed to transmit electric power via sinusoidal electromagnetic waves. Suppose power is to be transmitted in a beam of cross-sectional area 100 m2. What electric- and magnetic-field amplitudes are required to transmit an amount of power comparable to that handled by modern transmission lines (that carry voltages and currents of the order of 500 kV and 1000 A)? 12.53 .. CP Global Positioning System (GPS). The GPS network consists of 24 satellites, each of which makes two orbits around the earth per day. Each satellite transmits a 50.0-W (or even less) sinusoidal electromagnetic signal at two frequencies, one of which is 1575.42 MHz. Assume that a satellite transmits half of its power at each frequency and that the waves travel uniformly in a downward hemisphere. (a) What average intensity does a GPS receiver on the ground, directly below the satellite, receive? (Hint: First use Newton’s laws to find the altitude of the satellite.) (b) What are the amplitudes of the electric and magnetic fields at the GPS receiver in part (a), and how long does it take the signal to reach the receiver? (c) If the receiver is a square panel 1.50 cm on a side that absorbs all of the beam, what average pressure does the signal exert on it? (d) What wavelength must the receiver be tuned to? 12.54 .. CP Solar Sail 2. NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, lowmass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 * 1026 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun. 12.55 .. CP Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius R and mass density r. (a) Write an expression for the gravitational force exerted on this particle by the sun (mass M) when the particle is a distance r from the sun. (b) Let L represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun’s radiation also depends on the distance r. The relevant area is the cross-sectional area of the particle, not the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about 3000 kg > m3. Find the particle radius R such that the gravitational and radiation forces acting on the particle are equal in mag nitude. The luminosity of the sun is 3.9 * 1026 W. Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. '

'

[Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

CHALLENGE PROBLEMS 12.56 ... CALC Electromagnetic

waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an S electric field E1x, t2 5 Ey 1x, t2≥n propagating in the + x-direction within a conductor is '

d 2Ey 1x, t2 dx

2

m d Ey 1x, t2 r dt '

=

where m is the permeability of the conductor and r is its resistivity. (a) A solution to this wave equation is Ey1x, t2 = Emaxe-kCx cos1kCx - vt2 where KC = #vm/2r. Verify this by substituting Ey1x, t2 into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes i2R heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude

decreases by a factor of 1 > e in a distance 1 > kC = #2r/vm, and calculate this distance for a radio wave with frequency ƒ = 1.0 MHz in copper (resistivity 1.72 * 10-8 Æ # m; permeability m = m02. Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure. 12.57 ... CP Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by '

'

'

'

q2a2 dE = dt 6pP0c3 where c is the speed of light. (a) Verify that this equation is dimensionally correct. (b) If a proton with a kinetic energy of 6.0 MeV is traveling in a particle accelerator in a circular orbit of radius 0.750 m, what fraction of its energy does it radiate per second? (c) Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second? 12.58 ... CP The Classical Hydrogen Atom. The electron in a hydrogen atom can be considered to be in a circular orbit with a radius of 0.0529 nm and a kinetic energy of 13.6 eV. If the electron behaved classically, how much energy would it radiate per second (see Challenge Problem 12.57)? What does this tell you about the use of classical physics in describing the atom?

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Answers

411

Answers

?

Chapter Opening Question

Metals are reflective because they are good conductors of electricity. When an electromagnetic wave strikes a conductor, the electric field of the wave sets up currents on the conductor surface that generate a reflected wave. For a perfect conductor, this reflected wave is just as intense as the incident wave. Tarnished metals are less shiny because their surface is oxidized and less conductive; polishing the metal removes the oxide and exposes the conducting metal.

Test Your Understanding Questions 12.1 (a) no, (b) no A purely electric wave would have a varying electric field. Such a field necessarily generates a magnetic field through Ampere’s law, Eq. (9.20), so a purely electric wave is impossible. In the same way, a purely magnetic wave is impossible: The varying magnetic field in such a wave would automatically give rise to an electric field through Faraday’s law, Eq. (9.21). 12.2 (a) positive y-direction, (b) negative x-direction, (c) positive y-direction You canS verify these answers by using the rightS hand rule to show that E : B in each case is in the direction of propagation, or by using the rule shown in Fig. 12.9. 12.3 (iv) In an ideal electromagnetic plane wave, at any instant the fields are the same anywhere in a plane perpendicular to the direction of propagation. The plane wave described by Eqs. (12.17) is propagating in the x-direction, so the fields depend on the coordinate x and time t but do not depend on the coordinates y and z. 12.4 (a) (i) and (iii), (b) (ii) and (iv), (c) (i) and (iii), (d) (ii) and (iv) Both the energy density the Poynting vector magnitude S u and S S are maximum where the E and B fields have their maximum magnitudes. (The directions of the fields doesn’t matter.) From Fig. 12.13, this occurs at x = 0 and x = l>2. Both u and S have a minS S imum value of zero; that occurs where E and B are both zero. From Fig. 12.13, this occurs at x = l>4 and x = 3l>4. S 12.5 no There are places where E 5 0 at all times (at the walls) and the electric energy density 12 P0 E2 is always zero. There are '

S

also places where B 5 0 at all times (on the plane midway between the walls) and the magnetic energy density B2>2m0 is S

Q12.2 It is possible in a case of discharging through dielectric of the capacitor. Q12.3 Heat radiation, visible light, radio waves. They differ in wave length and frequency but have always a fixed speed c = 3 * 108 m/s in vacuum. Q12.4 Non-conservative electric field across tube ends due to varyup magnetic field in air near. Radio stations act as a source of emf. Q12.5 Polarization is a property of all electromagnetic waves as they are transverse. Sound waves cannot be polarised as they are longitudinal. Q12.6 First the electric field will exert a force on the charge in its direction. As soon as the positively charged particle obtains some velocity, magnetic field will also begin to exert force on it. Force due to electric field will always be along +y direction. But force due to magnetic field will always be perpendicular to both magnetic field and velocity of particle. Force due to magnetic field will always be in x-y plane. So the particle will move in x-y plane. Q12.7 Shock is felt due to current and not due to potential difference only. Since the potential difference due to electromagnetic waves very rapidly changes its direction, current does not flow and no shock. Q12.8 (Ë2)B Q12.9 No, the rapidly changing direction of magnetic field of the electromagnetic wave cannot exert a net torque. The compass cannot move due to inertia. Q12.10 Electromagnetic waves are polarised in the vertical direction parallel to antenna. Q12.11 This recoil force is too small to observe. 1 Q12.12 Intensity I r 2 . As power radiated by the point like small r source is passed through spherical area 4pr2 at a distance r. Pres2I sure p = (for a perfectly reflecting surface at some distance C from the source) for a perfectly absorbing surface twice as for p I I' I from the distance intensity I' = pressure p' = = = 4 C 4C 8 Q12.13 Yes due to presence of electric field and magnetic field. It has momentum also as per equation 12.30.

S

always zero. However, there are no locations where both E and B are always zero. Hence the energy density at any point in the standing wave is always nonzero.

Bridging Problem 0.0368 V

Discussion Questions Q12.1 If there are many electromagnetic waves travelling in different directions, they will superpose to give a resultant electric field and magnetic field at a point in space so the propagation directions of difference waves cannot be determined. The case would be very easy to analyse if there is a simple electromagnetic wave having a unique direction of propagation.

Exercises 12.1 (a) 1.28 s (b) 8.15 × 1013 km 12.2 (a) The wave is propagating in the + z direction (b) + z direction (c) - y direction (d) - x direction 12.3 13.3 nT, + y-direction ^

^

^

^

^

^

12.4 (a) S = i * (- j ) = - k ^

^

(b) S = j * i = - k ^

^

^

^

(c) S = (- k) * (- i ) = j ^

^

^

^

(d) S = i * (- k) = - j

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411a

CHAPTER 12 Electromagnetic Waves 18

12.21 2.5 * 1025 W

10

12.22 (a) 8.47 * 108 Hz. (b) 1.80 * 10 - 10 T. (c) 3.87 * 10 - 6 W/m2. 12.23 12.0 V/m, 40.0 nT 12.25 850 kW 12.26 (a) 1.34 * 10 - 11 Pa. (b) 1.23 N/C, 4.10 * 10 - 9 T

12.5 3.0 * 10 Hz, 3.3 * 10-19 s 6.3 * 10 rad/m 12.6 (a) UVA: 7.50 * 1014 Hz to 9.38 * 1014 Hz UVB: 9.38 * 1014 Hz to 1.07 * 1015 Hz (b) UVA: 1.57 * 107 rad/m to 1.96 * 107 rad/m UVB:1.96 * 107 rad/m to 2.24 * 107 rad/m 12.7 (a) 6.94 * 1014 Hz (b) 375 V/m (c) E(x,t) = (375 V/m) cos[(1.45 * 107 rad/m)x - (4.36 * 1015 rad/s)t], B(x,t) = (1.25mT) cos[(1.45 * 107 rad/m)x - (4.36 * 1015 rad/s)t] 14 12.8 (a) 6.90 * 10 Hz (b) 9.00 * 10 - 12 T S S S (c) B (z, t) = - ^j Bmax cos(kz + vt) , so that E * B will be in the - k^ direction S E(z, t) = ^i (2.70 * 10 - 3 V/m)cos([1.44 * 107 rad/m)z + [4.34 * 1015 rad/s]t) and S B(z, t) = - ^j (9.00 * 10 - 12 T)cos([1.44 * 107 rad/m)z + [4.34 * 1015 rad/s]t) 12.9 (a) (i) 60 kHz (ii) 6.0 * 1013 Hz (iii) 6.0 * 1016Hz (b) (i) 4.62 * 10-14 m = 4.62 * 10-5 nm (ii) 508 m = 5.08 * 1011 nm 12.10 (a) Emax = 375 V/m, Bmax = 1.25 μT (b) 9.50 * 1014 Hz, 316 nm, 1.05 * 10 - 15 s. This wavelength is too short to be visible. (c) 3.00 * 108 m/s 12.11 (a) +y-direction (b) 0.149 mm S (c) B = (1.03 mT)î cos[(4.22 × 104 rad/m)y - (1.265 × 1013 rad/s)t] 12.12 (a) The wave is traveling in the - x direction. (b) 6.59 * 1011 Hz S (c) E(x, t) = - (2.48 V/m) cos((1.38 * 104 rad/m)x + (4.14 * 1012 rad/s)t)k^ . 12.13 (a) 361 m, (b) 0.0174 rad/m, (c) 5.22 * 106 rad/s, (d) 0.0144 V/m 12.14 (a) 6.91 * 107 m/s. (b) 1.06 * 106 m. (c) 1.04 * 10 - 10 T 12.15 (a) 0.381 * 10-8 m, (b) 0.526 * 10-8 m, (c) 1.38, (d) 1.90 12.16 (a) 8.0 kJ (b) 1.0 * 1021 W/m2 (c) 8.7 * 1011 V/m, 2.9 * 103 T 12.17 (a) 330 W/m2 (b) 500 V/m, 1.7 μT 12.18 15.9 μJ 12.19 (a) 11 * 10-6 W/m (b) 0.30 nT (c) 840 W 2

12.20 (a) 2.4 * 10 - 4 W (b) 17.4 V/m

(c) 6.69 * 10-12 J/m3 (d) each one has 50%. 12.27 (a) 0.18 mW (b) 274 V/m, 0.913 μT (c) 0.18 mJ/s (d) 0.010 W/cm2 12.28 (a) 8.23 * 10 - 11 atm (b) 1.65 * 10 - 10 atm (c) 2.78 * 10 - 14 kg/m2 # s 12.29 (a) 637 W/m2 (b) 693 V/m, 2.31 μT (c) 2.12 * 10-6 J/m3 12.30 (a) 5.9 * 10 - 4 N (b) 7.1 * 10 - 4 m/s2. (c) With this acceleration it would take the sail 1.4 * 106 s = 16 days to reach a speed of 1 km/s. This would be useful only in specialized applications. The acceleration could be increased by decreasing the mass of the sail, either by reducing its density or its thickness. 12.31 (a) 30.5 cm (b) 2.46 GHz (c) 2.11 GHz 12.32 at 20 cm, 40 cm, and 60 cm. 12.33 (a) 7.10 mm (b) 3.55 mm (c) 1.56 * 108 m/s 12.34 (a) 2.00 m (b) 1.00 m 12.35 (a) 4.38 mm (b) 4.38 mm (c) 4.38 mm 12.37 (a) 0.375 mJ (b) 4.08 mPa (c) 604 nm, 3.70 * 1014 Hz (d) 30.3 kV/m, 101 μT 12.39 (a) 71.6 MW/m2 (b) 232 kV/m, 774 μT (c) 35.8 MW/m2 (d) 637 W/m2 12.41 (a) 0.00602 W/m2 (b) 2.13 V/m, 7.10 nT (c) 1.20 pN, no 12.42 (a) 7.81 * 109 Hz (b) 4.50 * 10 - 9 T (c) 2.42 * 10 - 3 W/m2 (d) 1.94 * 10 - 12 N 12.43 (a) 840 V/m, 2.80 μT (b) 1.56 μ/m3 for each (c) 15.3 pJ 12.44 (a) 4.81 * 1014 Hz, 452 nm, 2.17 * 108 m/s (b) 1.38 (c) E = (535 V/m)cosc(1.01 * 107 rad/m)x - (3.02 * 1015 rad/s)t d 12.45 (a) at r = R; 64 MW/m2, 0.21 Pa; at r = R/2; 260 MW/m2, 0.85 Pa (b) no 12.46 75.0 N/C.

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12.47 3.89 * 10 - 13 rad/s2 12.48 61.7 mV 12.49 (a) rI/pa2, in the direction of the current (b) m0I/2pa, counterclockwise if the current is out of the page rI2 (c) , radially inward 2p2a3 rII2 2 (d) =I R pa2 12.50 (a) 1.364 m apart (b) 10.91 m 12.51 (a) 23.6 h (b) throw it 12.52 6.14 * 104 V/m, 2.05 * 10 - 4 T. 12.53 (a) 9.75 * 10 - 15 W/m2 (b) 2.71 μN/C, 9.03 * 10 - 15 T, 67.3 ms (c) 3.25 * 10 - 23 Pa (d) 0.190 m

411b

12.54 (a) The sail should be reflective, to produce the maximum radiation pressure. (b) 6.42 km2 (c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance from the sun, so this distance divides out when we set Frad = Fg. 4rGpMR3

LR2

(c) 0.19 μm, no 3r2 4cr2 12.57 (b) 1.39 * 10 - 11 (c) 2.54 * 10 - 8 12.58 4.64 * 10 - 8 J/s = 2.89 * 1011 eV/s. This large value of dE would mean that the electron would almost immediately dt lose all its energy! 12.55 (a)

(b)

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13

THE NATURE AND PROPAGATION OF LIGHT

LEARNING GOALS By studying this chapter, you will learn: • What light rays are, and how they are related to wave fronts. • The laws that govern the reflection and refraction of light. • The circumstances under which light is totally reflected at an interface. • How to make polarized light out of ordinary light. • How Huygens’s principle helps us analyze reflection and refraction.

?

These drafting tools are made of clear plastic, but a rainbow of colors appears when they are placed between two special filters called polarizers. How does this cause the colors?

lue lakes, ochre deserts, green forests, and multicolored rainbows can be enjoyed by anyone who has eyes with which to see them. But by studying the branch of physics called optics, which deals with the behavior of light and other electromagnetic waves, we can reach a deeper appreciation of the visible world. A knowledge of the properties of light allows us to understand the blue color of the sky and the design of optical devices such as telescopes, microscopes, cameras, eyeglasses, and the human eye. The same basic principles of optics also lie at the heart of modern developments such as the laser, optical fibers, holograms, optical computers, and new techniques in medical imaging. The importance of optics to physics, and to science and engineering in general, is so great that we will devote the next four chapters to its study. In this chapter we begin with a study of the laws of reflection and refraction and the concepts of dispersion, polarization, and scattering of light. Along the way we compare the various possible descriptions of light in terms of particles, rays, or waves, and we introduce Huygens’s principle, an important link that connects the ray and wave viewpoints. In Chapter 14 we’ll use the ray description of light to understand how mirrors and lenses work, and we’ll see how mirrors and lenses are used in optical instruments such as cameras, microscopes, and telescopes. We’ll explore the wave characteristics of light further in Chapters 15 and 16.

B

13.1

The Nature of Light

Until the time of Isaac Newton (1642–1727), most scientists thought that light consisted of streams of particles (called corpuscles) emitted by light sources. Galileo and others tried (unsuccessfully) to measure the speed of light. Around 1665, evidence of wave properties of light began to be discovered. By the early 19th century, evidence that light is a wave had grown very persuasive. 412

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13.1 The Nature of Light

413

In 1873, James Clerk Maxwell predicted the existence of electromagnetic waves and calculated their speed of propagation, as we learned in Chapter 12. This development, along with the experimental work of Heinrich Hertz starting in 1887, showed conclusively that light is indeed an electromagnetic wave.

The Two Personalities of Light The wave picture of light is not the whole story, however. Several effects associated with emission and absorption of light reveal a particle aspect, in that the energy carried by light waves is packaged in discrete bundles called photons or quanta. These apparently contradictory wave and particle properties have been reconciled since 1930 with the development of quantum electrodynamics, a comprehensive theory that includes both wave and particle properties. The propagation of light is best described by a wave model, but understanding emission and absorption requires a particle approach. The fundamental sources of all electromagnetic radiation are electric charges in accelerated motion. All bodies emit electromagnetic radiation as a result of thermal motion of their molecules; this radiation, called thermal radiation, is a mixture of different wavelengths. At sufficiently high temperatures, all matter emits enough visible light to be self-luminous; a very hot body appears “red-hot” (Fig. 13.1) or “white-hot.” Thus hot matter in any form is a light source. Familiar examples are a candle flame, hot coals in a campfire, the coils in an electric room heater, and an incandescent lamp filament (which usually operates at a temperature of about 3000°C). Light is also produced during electrical discharges through ionized gases. The bluish light of mercury-arc lamps, the orange-yellow of sodium-vapor lamps, and the various colors of “neon” signs are familiar. A variation of the mercury-arc lamp is the fluorescent lamp (see Fig. 30.7). This light source uses a material called a phosphor to convert the ultraviolet radiation from a mercury arc into visible light. This conversion makes fluorescent lamps more efficient than incandescent lamps in transforming electrical energy into light. In most light sources, light is emitted independently by different atoms within the source; in a laser, by contrast, atoms are induced to emit light in a cooperative, coherent fashion. The result is a very narrow beam of radiation that can be enormously intense and that is much more nearly monochromatic, or singlefrequency, than light from any other source. Lasers are used by physicians for microsurgery, in a DVD or Blu-ray player to scan the information recorded on a video disc, in industry to cut through steel and to fuse high-melting-point materials, and in many other applications (Fig. 13.2). No matter what its source, electromagnetic radiation travels in vacuum at the same speed. As we saw in Sections 1.3 and 12.1, the speed of light in vacuum is defined to be

13.1 An electric heating element emits primarily infrared radiation. But if its temperature is high enough, it also emits a discernible amount of visible light.

13.2 Ophthalmic surgeons use lasers for repairing detached retinas and for cauterizing blood vessels in retinopathy. Pulses of blue-green light from an argon laser are ideal for this purpose, since they pass harmlessly through the transparent part of the eye but are absorbed by red pigments in the retina.

c = 2.99792458 * 108 m>s or 3.00 * 108 m>s to three significant figures. The duration of one second is defined by the cesium clock (see Section 1.3), so one meter is defined to be the distance that light travels in 1>299,792,458 s.

Waves, Wave Fronts, and Rays We often use the concept of a wave front to describe wave propagation. We introduced this concept in Section 12.2 to describe the leading edge of a wave. More generally, we define a wave front as the locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same. That is, at any instant, all points on a wave front are at the same part of the cycle of their variation. When we drop a pebble into a calm pool, the expanding circles formed by the wave crests, as well as the circles formed by the wave troughs between

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414

CHAPTER 13 The Nature and Propagation of Light

them, are wave fronts. Similarly, when sound waves spread out in still air from a pointlike source, or when electromagnetic radiation spreads out from a pointlike emitter, any spherical surface that is concentric with the source is a wave front, as shown in Fig. 13.3. In diagrams of wave motion we usually draw only parts of a few wave fronts, often choosing consecutive wave fronts that have the same phase and thus are one wavelength apart, such as crests of water waves. Similarly, a diagram for sound waves might show only the “pressure crests,” Expanding wave y the surfaces over which the pressure is maximum, and a diagram for electrofronts magnetic waves might show only the “crests” on which the electric or magnetic field is maximum. We will often use diagrams that show the shapes of the wave fronts or their cross sections in some reference plane. For example, when electromagnetic waves are radiated by a small light source, we can represent the wave fronts as spherical surfaces concentric with the source or, as in Fig. 13.4a, by the circular x intersections of these surfaces with the plane of the diagram. Far away from the source, where the radii of the spheres have become very large, a section of a spherical surface can be considered as a plane, and we have a plane wave like z those discussed in Sections 12.2 and 12.3 (Fig, 13.4b). To describe the directions in which light propagates, it’s often convenient to represent a light wave by rays rather than by wave fronts. Rays were used to describe light long before its wave nature was firmly established. In a particle Point sound source producing spherical theory of light, rays are the paths of the particles. From the wave viewpoint a ray sound waves (alternating compressions and rarefactions of air) is an imaginary line along the direction of travel of the wave. In Fig. 13.4a the rays are the radii of the spherical wave fronts, and in Fig. 13.4b they are straight 13.4 Wave fronts (blue) and rays (purple). lines perpendicular to the wave fronts. When waves travel in a homogeneous isotropic material (a material with the same properties in all regions and in all (a) directions), the rays are always straight lines normal to the wave fronts. At a boundary surface between two materials, such as the surface of a glass plate in When wave fronts are air, the wave speed and the direction of a ray may change, but the ray segments in spherical, the rays the air and in the glass are straight lines. radiate from the Rays center of the The next several chapters will give you many opportunities to see the interplay sphere. of the ray, wave, and particle descriptions of light. The branch of optics for which the ray description is adequate is called geometric optics; the branch dealing specifically with wave behavior is called physical optics. This chapter and the Source following one are concerned mostly with geometric optics. In Chapters 15 and 16 Wave fronts we will study wave phenomena and physical optics. 13.3 Spherical wave fronts of sound spread out uniformly in all directions from a point source in a motionless medium, such as still air, that has the same properties in all regions and in all directions. Electromagnetic waves in vacuum also spread out as shown here.

(b) When wave fronts are planar, the rays are perpendicular to the wave fronts and parallel to each other. Rays

Wave fronts

ActivPhysics 15.1: Reflection and Refraction ActivPhysics 15.3: Refraction Applications

Test Your Understanding of Section 13.1 Some crystals are not isotropic: Light travels through the crystal at a higher speed in some directions than in others. In a crystal in which light travels at the same speed in the x- and z-directions but at a faster speed in the y-direction, what would be the shape of the wave fronts produced by a light source at the origin? (i) spherical, like those shown in Fig. 13.3; (ii) ellipsoidal, flattened along the y-axis; (iii) ellipsoidal, stretched out along the y-axis. y

13.2

Reflection and Refraction

In this section we’ll use the ray model of light to explore two of the most important aspects of light propagation: reflection and refraction. When a light wave strikes a smooth interface separating two transparent materials (such as air and glass or water and glass), the wave is in general partly reflected and partly refracted (transmitted) into the second material, as shown in Fig. 13.5a. For example, when you look into a restaurant window from the street, you see a reflection of the street scene, but a person inside the restaurant can look out through the window at the same scene as light reaches him by refraction.

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13.2 Reflection and Refraction

415

13.5 (a) A plane wave is in part reflected and in part refracted at the boundary between two media (in this case, air and glass). The light that reaches the inside of the coffee shop is refracted twice, once entering the glass and once exiting the glass. (b), (c) How light behaves at the interface between the air outside the coffee shop (material a) and the glass (material b). For the case shown here, material b has a larger index of refraction than material a1nb 7 na2 and the angle ub is smaller than ua . '

(b) The waves in the outside air and glass represented by rays

(a) Plane waves reflected and refracted from a window

Incident rays Hat outside window

Incident wave

Reflected image of hat

a b

Reflected rays

Woman sees reflected image of hat.

Refracted rays

Refracted wave Man sees refracted image of hat. Reflected wave

(c) The representation simplified to show just one set of rays a b

Incident ray ua ur Reflected ray

The segments of plane waves shown in Fig. 13.5a can be represented by bundles of rays forming beams of light (Fig. 13.5b). For simplicity we often draw only one ray in each beam (Fig. 13.5c). Representing these waves in terms of rays is the basis of geometric optics. We begin our study with the behavior of an individual ray. We describe the directions of the incident, reflected, and refracted (transmitted) rays at a smooth interface between two optical materials in terms of the angles they make with the normal (perpendicular) to the surface at the point of incidence, as shown in Fig. 13.5c. If the interface is rough, both the transmitted light and the reflected light are scattered in various directions, and there is no single angle of transmission or reflection. Reflection at a definite angle from a very smooth surface is called specular reflection (from the Latin word for “mirror”); scattered reflection from a rough surface is called diffuse reflection. This distinction is shown in Fig. 13.6. Both kinds of reflection can occur with either transparent materials or opaque materials that do not transmit light. The vast majority of objects in your environment (including plants, other people, and this book) are visible to you because they reflect light in a diffuse manner from their surfaces. Our primary concern, however, will be with specular reflection from a very smooth surface such as highly polished glass or metal. Unless stated otherwise, when referring to “reflection” we will always mean specular reflection. The index of refraction of an optical material (also called the refractive index), denoted by n, plays a central role in geometric optics. It is the ratio of the speed of light c in vacuum to the speed v in the material: n =

c v

(index of refraction)

Refracted ray

Normal ub

13.6 Two types of reflection. (a) Specular reflection

(b) Diffuse reflection

(13.1)

Light always travels more slowly in a material than in vacuum, so the value of n in anything other than vacuum is always greater than unity. For vacuum, n = 1.

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CHAPTER 13 The Nature and Propagation of Light

Since n is a ratio of two speeds, it is a pure number without units. (The relationship of the value of n to the electric and magnetic properties of a material is described in Section 12.3.) CAUTION Wave speed and index of refraction Keep in mind that the wave speed v is inversely proportional to the index of refraction n. The greater the index of refraction in a material, the slower the wave speed in that material. Failure to remember this point can 13.7 The laws of reflection and refraction. lead to serious confusion! y 1. The incident, reflected, and refracted rays and the normal to the surface all lie in the same plane. Angles ua, ub, and ur are measured from the normal. Incident ray ua

2. ur 5 ua

ur

Refracted ray

Reflected ray Material a

Normal ub

Material b

3. When a monochromatic light ray crosses the interface between two given materials a and b, the angles ua and ub are related to the indexes of refraction of a and b by sin ua nb 5 sin ub na

13.8 Refraction and reflection in three cases. (a) Material b has a larger index of refraction than material a. (b) Material b has a smaller index of refraction than material a. (c) The incident light ray is normal to the interface between the materials.

The Laws of Reflection and Refraction Experimental studies of the directions of the incident, reflected, and refracted rays at a smooth interface between two optical materials lead to the following conclusions (Fig. 13.7): 1. The incident, reflected, and refracted rays and the normal to the surface all lie in the same plane. The plane of the three rays and the normal, called the plane of incidence, is perpendicular to the plane of the boundary surface between the two materials. We always draw ray diagrams so that the incident, reflected, and refracted rays are in the plane of the diagram. 2. The angle of reflection Ur is equal to the angle of incidence Ua for all wavelengths and for any pair of materials. That is, in Fig. 13.5c, u r = ua

Material a

3. For monochromatic light and for a given pair of materials, a and b, on opposite sides of the interface, the ratio of the sines of the angles Ua and Ub, where both angles are measured from the normal to the surface, is equal to the inverse ratio of the two indexes of refraction: nb sin ua = na sin ub

na sin ua = nb sin ub Normal

ub Reflected

Refracted

(b) A ray entering a material of smaller index of refraction bends away from the normal. Incident

nb , na

ua

Normal

Reflected

ub

Material a

Material b

Refracted (c) A ray oriented along the normal does not bend, regardless of the materials. ua

ub

Incident Reflected

Refracted

(13.3)

or

Material b nb . na

ua

(13.2)

This relationship, together with the observation that the incident and reflected rays and the normal all lie in the same plane, is called the law of reflection.

(a) A ray entering a material of larger index of refraction bends toward the normal. Incident

(law of reflection)

Normal

(law of refraction)

(13.4)

This experimental result, together with the observation that the incident and refracted rays and the normal all lie in the same plane, is called the law of refraction or Snell’s law, after the Dutch scientist Willebrord Snell (1591–1626). There is some doubt that Snell actually discovered it. The discovery that n = c>v came much later. While these results were first observed experimentally, they can be derived theoretically from a wave description of light. We do this in Section 13.7. Equations (13.3) and (13.4) show that when a ray passes from one material (a) into another material (b) having a larger index of refraction 1nb 7 na2 and hence a slower wave speed, the angle ub with the normal is smaller in the second material than the angle ua in the first; hence the ray is bent toward the normal (Fig. 13.8a). When the second material has a smaller index of refraction than the first material 1nb 6 na2 and hence a faster wave speed, the ray is bent away from the normal (Fig. 13.8b). No matter what the materials on either side of the interface, in the case of normal incidence the transmitted ray is not bent at all (Fig. 13.8c). In this case

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13.2 Reflection and Refraction 13.9 (a) This ruler is actually straight, but it appears to bend at the surface of the water. (b) Light rays from any submerged object bend away from the normal when they emerge into the air. As seen by an observer above the surface of the water, the object appears to be much closer to the surface than it actually is. (a) AA straight ruler ruler half-immersed in water in water (a) straight half-immersed

(b) Why the ruler appears bent Observer Apparent position of end of ruler

nb (air) 5 1.00 na (water) 5 1.33

417

13.10 (a) The index of refraction of air is slightly greater than 1, so light rays from the setting sun bend downward when they enter our atmosphere. (The effect is exaggerated in this figure.) (b) Stronger refraction occurs for light coming from the lower limb of the sun (the part that appears closest to the horizon), which passes through denser air in the lower atmosphere. As a result, the setting sun appears flattened vertically. (See Problem 13.55.) (a)

Atmosphere (not to scale)

Ruler

Light ray from the sun

Actual position of end of ruler (

Earth

ua = 0 and sin ua = 0, so from Eq. (13.4) ub is also equal to zero, so the transmitted ray is also normal to the interface. Equation (13.2) shows that ur , too, is equal to zero, so the reflected ray travels back along the same path as the incident ray. The law of refraction explains why a partially submerged ruler or drinking straw appears bent; light rays coming from below the surface change in direction at the air–water interface, so the rays appear to be coming from a position above their actual point of origin (Fig. 13.9). A similar effect explains the appearance of the setting sun (Fig. 13.10). An important special case is refraction that occurs at an interface between vacuum, for which the index of refraction is unity by definition, and a material. When a ray passes from vacuum into a material (b), so that na = 1 and nb 7 1, the ray is always bent toward the normal. When a ray passes from a material into vacuum, so that na 7 1 and nb = 1, the ray is always bent away from the normal. The laws of reflection and refraction apply regardless of which side of the interface the incident ray comes from. If a ray of light approaches the interface in Fig. 13.8a or 13.8b from the right rather than from the left, there are again reflected and refracted rays; these two rays, the incident ray, and the normal to the surface again lie in the same plane. Furthermore, the path of a refracted ray is reversible; it follows the same path when going from b to a as when going from a to b. [You can verify this using Eq. (13.4).] Since reflected and incident rays make the same angle with the normal, the path of a reflected ray is also reversible. That’s why when you see someone’s eyes in a mirror, they can also see you. The intensities of the reflected and refracted rays depend on the angle of incidence, the two indexes of refraction, and the polarization (that is, the direction of the electric-field vector) of the incident ray. The fraction reflected is smallest at normal incidence 1ua = 0°2, where it is about 4% for an air–glass interface. This fraction increases with increasing angle of incidence to 100% at grazing incidence, when ua = 90°. It’s possible to use Maxwell’s equations to predict the amplitude, intensity, phase, and polarization states of the reflected and refracted waves. Such an analysis is beyond our scope, however. The index of refraction depends not only on the substance but also on the wavelength of the light. The dependence on wavelength is called dispersion; we will consider it in Section 13.4. Indexes of refraction for several solids and liquids are given in Table 13.1 for a particular wavelength of yellow light. '

(b)

Application Transparency and Index of Refraction An eel in its larval stage is nearly as transparent as the seawater in which it swims. The larva in this photo is nonetheless easy to see because its index of refraction is higher than that of seawater, so that some of the light striking it is reflected instead of transmitted. The larva appears particularly shiny around its edges because the light reaching the camera from those points struck the larva at neargrazing incidence 1ua = 90°2, resulting in almost 100% reflection.

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418

CHAPTER 13 The Nature and Propagation of Light

Table 13.1 Index of Refraction for Yellow Sodium Light, L0 5 589 nm Index of Refraction, n

Substance Solids Ice 1H2 O2 Fluorite 1CaF22 Polystyrene Rock salt (NaCl) Quartz 1SiO22 Zircon 1ZrO2 # SiO22 Diamond (C) Fabulite 1SrTiO32 Rutile 1TiO22 Glasses (typical values) Crown Light flint Medium flint Dense flint Lanthanum flint Liquids at 20°C Methanol 1CH3 OH2 Water 1H2 O2 Ethanol 1C2 H5 OH2 Carbon tetrachloride 1CCl42 Turpentine Glycerine Benzene Carbon disulfide 1CS22 '

'

'

'

'

1.309 1.434 1.49 1.544 1.544 1.923 2.417 2.409 2.62 1.52 1.58 1.62 1.66 1.80 1.329 1.333 1.36 1.460 1.472 1.473 1.501 1.628

The index of refraction of air at standard temperature and pressure is about 1.0003, and we will usually take it to be exactly unity. The index of refraction of a gas increases as its density increases. Most glasses used in optical instruments have indexes of refraction between about 1.5 and 2.0. A few substances have larger indexes; one example is diamond, with 2.417.

Index of Refraction and the Wave Aspects of Light We have discussed how the direction of a light ray changes when it passes from one material to another material with a different index of refraction. It’s also important to see what happens to the wave characteristics of the light when this happens. First, the frequency ƒ of the wave does not change when passing from one material to another. That is, the number of wave cycles arriving per unit time must equal the number leaving per unit time; this is a statement that the boundary surface cannot create or destroy waves. Second, the wavelength l of the wave is different in general in different materials. This is because in any material, v = lƒ; since ƒ is the same in any material as in vacuum and v is always less than the wave speed c in vacuum, l is also correspondingly reduced. Thus the wavelength l of light in a material is less than the wavelength l0 of the same light in vacuum. From the above discussion, ƒ = c>l0 = v>l. Combining this with Eq. (13.1), n = c>v, we find l0 n

l =

(wavelength of light in a material)

(13.5)

When a wave passes from one material into a second material with larger index of refraction, so that nb 7 na , the wave speed decreases. The wavelength lb = l0>nb in the second material is then shorter than the wavelength la = l0>na in the first material. If instead the second material has a smaller index of refraction than the first material, so that nb 6 na , then the wave speed increases. Then the wavelength lb in the second material is longer than the wavelength la in the first material. This makes intuitive sense; the waves get “squeezed” (the wavelength gets shorter) if the wave speed decreases and get “stretched” (the wavelength gets longer) if the wave speed increases. '

'

Problem-Solving Strategy 13.1

Reflection and Refraction

IDENTIFY the relevant concepts: Use geometric optics, discussed in this section, whenever light (or electromagnetic radiation of any frequency and wavelength) encounters a boundary between materials. In general, part of the light is reflected back into the first material and part is refracted into the second material. SET UP the problem using the following steps: 1. In problems involving rays and angles, start by drawing a large, neat diagram. Label all known angles and indexes of refraction. 2. Identify the target variables. EXECUTE the solution as follows: 1. Apply the laws of reflection, Eq. (13.2), and refraction, Eq. (13.4). Measure angles of incidence, reflection, and refraction with respect to the normal to the surface, never from the surface itself.

2. Apply geometry or trigonometry in working out angular relationships. Remember that the sum of the acute angles of a right triangle is 90° (they are complementary) and the sum of the interior angles in any triangle is 180°. 3. The frequency of the electromagnetic radiation does not change when it moves from one material to another; the wavelength changes in accordance with Eq. (13.5), l = l0>n. EVALUATE your answer: In problems that involve refraction, check that your results are consistent with Snell’s law 1na sin ua = nb sin ub2. If the second material has a higher index of refraction than the first, the angle of refraction must be smaller than the angle of incidence: The refracted ray bends toward the normal. If the first material has the higher index of refraction, the refracted angle must be larger than the incident angle: The refracted ray bends away from the normal. '

'

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13.2 Reflection and Refraction

Example 13.1

419

Reflection and refraction

In Fig. 13.11, material a is water and material b is glass with index of refraction 1.52. The incident ray makes an angle of 60.0° with the normal; find the directions of the reflected and refracted rays. S O L U T IO N IDENTIFY and SET UP: This is a problem in geometric optics. We are given the angle of incidence ua = 60.0° and the indexes of 13.11 Reflection and refraction of light passing from water to glass.

refraction na = 1.33 and nb = 1.52. We must find the angles of reflection and refraction ur and ub; to do this we use Eqs. (13.2) and (13.4), respectively. Figure 13.11 shows the rays and angles; nb is slightly greater than na, so by Snell’s law [Eq. (13.4)] ub is slightly smaller than ua, as the figure shows. EXECUTE: According to Eq. (13.2), the angle the reflected ray makes with the normal is the same as that of the incident ray, so ur = ua = 60.0°. To find the direction of the refracted ray we use Snell’s law, Eq. (13.4):

Normal

na sin ua = nb sin ub na 1.33 sin ub = sin ua = sin 60.0° = 0.758 nb 1.52 ub = arcsin(0.758) = 49.3° '

ua 5 60°

ur na (water) 5 1.33

a b

nb (glass) 5 1.52 ub

Example 13.2

EVALUATE: The second material has a larger refractive index than the first, as in Fig. 13.8a. Hence the refracted ray is bent toward the normal and ub 6 ua.

Index of refraction in the eye

The wavelength of the red light from a helium-neon laser is 633 nm in air but 474 nm in the aqueous humor inside your eyeball. Calculate the index of refraction of the aqueous humor and the speed and frequency of the light in it. S O L U T IO N

This is about the same index of refraction as for water. Then, using n = c > v and v = lƒ, we find '

'

3.00 * 108 m > s c = 2.25 * 108 m > s = n 1.34 2.25 * 108 m > s v = = 4.74 * 1014 Hz ƒ = l 474 * 10-9 m '

'

v =

'

'

IDENTIFY and SET UP: The key ideas here are (i) the definition of index of refraction n in terms of the wave speed v in a medium and the speed c in vacuum, and (ii) the relationship between wavelength l0 in vacuum and wavelength l in a medium of index n. We use Eq. (13.1), n = c > v; Eq. (13.5), l = l0 > n; and v = lƒ. '

'

'

'

EVALUATE: Note that while the speed and wavelength have different values in air and in the aqueous humor, the frequency in air, ƒ0, is the same as the frequency ƒ in the aqueous humor:

'

3.00 * 108 m > s c = = 4.74 * 1014 Hz l0 633 * 10-9 m '

EXECUTE: The index of refraction of air is very close to unity, so we assume that the wavelength l0 in vacuum is the same as that in air, 633 nm. Then from Eq. (13.5), l0 l = n

Example 13.3

l0 633 nm = = 1.34 n = l 474 nm

'

ƒ0 =

'

When a light wave passes from one material into another, the wave speed and wavelength both change but the wave frequency is unchanged.

A twice-reflected ray

Two mirrors are perpendicular to each other. A ray traveling in a plane perpendicular to both mirrors is reflected from one mirror, then the other, as shown in Fig. 13.12. What is the ray’s final direction relative to its original direction? S O L U T IO N IDENTIFY and SET UP: This problem involves the law of reflection, which we must apply twice (once for each mirror).

EXECUTE: For mirror 1 the angle of incidence is u1, and this equals the angle of reflection. The sum of interior angles in the triangle shown in the figure is 180°, so we see that the angles of incidence and reflection for mirror 2 are both 90° - u1. The total change in direction of the ray after both reflections is therefore 2190° - u12 + 2u1 = 180°. That is, the ray’s final direction is opposite to its original direction.

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Continued

420

CHAPTER 13 The Nature and Propagation of Light

13.12 A ray moving in the xy-plane. The first reflection changes the sign of the y-component of its velocity, and the second reflection changes the sign of the x-component. For a different ray with a z-component of velocity, a third mirror (perpendicular to the two shown) could be used to change the sign of that component. y 2u1 u1 u1 90° 2 u1 90° 2 u1

EVALUATE: An alternative viewpoint is that reflection reverses the sign of the component of light velocity perpendicular to the surface but leaves the other components unchanged. We invite you to verify this in detail. You should also be able to use this result to show that when a ray of light is successively reflected by three mirrors forming a corner of a cube (a “corner reflector”), its final direction is again opposite to its original direction. This principle is widely used in tail-light lenses and bicycle reflectors to improve their night-time visibility. Apollo astronauts placed arrays of corner reflectors on the moon. By use of laser beams reflected from these arrays, the earth–moon distance has been measured to within 0.15 m.

Mirror 2

u1

u1

90° 2 u1

180° 2 2u1

Mirror 1

x

Test Your Understanding of Section 13.2 You are standing on the shore of a lake. You spot a tasty fish swimming some distance below the lake surface. (a) If you want to spear the fish, should you aim the spear (i) above, (ii) below, or (iii) directly at the apparent position of the fish? (b) If instead you use a high-power laser to simultaneously kill and cook the fish, should you aim the laser (i) above, (ii) below, or (iii) directly at the apparent position of the fish? y

13.3 ActivPhysics 15.2: Total Internal Reflection

Total Internal Reflection

We have described how light is partially reflected and partially transmitted at an interface between two materials with different indexes of refraction. Under certain circumstances, however, all of the light can be reflected back from the interface, with none of it being transmitted, even though the second material is transparent. Figure 13.13a shows how this can occur. Several rays are shown radiating from a point source in material a with index of refraction na . The rays '

13.13 (a) Total internal reflection. The angle of incidence for which the angle of refraction is 90° is called the critical angle: This is the case for ray 3. The reflected portions of rays 1, 2, and 3 are omitted for clarity. (b) Rays of laser light enter the water in the fishbowl from above; they are reflected at the bottom by mirrors tilted at slightly different angles. One ray undergoes total internal reflection at the air–water interface. (a) Total internal reflection

(b) Total internal reflection demonstrated with a laser, mirrors, and water in a fishbowl

Total internal reflection occurs only if nb , na. b ub na ucrit ua 1

2

3

4

Refracted at interface

ub 5 90°

nb

. ucrit

4

At the critical angle of incidence, ucrit , the angle of refraction ub 5 90°.

Any ray with ua . ucrit shows total internal reflection.

Incident laser beams

a

Total internal reflection

Two mirrors at different angles

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13.3 Total Internal Reflection

421

strike the surface of a second material b with index nb , where na 7 nb . (Materials a and b could be water and air, respectively.) From Snell’s law of refraction, '

sin ub =

'

na sin ua nb

Because na>nb is greater than unity, sin ub is larger than sin ua; the ray is bent away from the normal. Thus there must be some value of ua less than 90° for which sin ub = 1 and ub = 90°. This is shown by ray 3 in the diagram, which emerges just grazing the surface at an angle of refraction of 90°. Compare Fig. 13.13a to the photograph of light rays in Fig. 13.13b. The angle of incidence for which the refracted ray emerges tangent to the surface is called the critical angle, denoted by ucrit. (A more detailed analysis using Maxwell’s equations shows that as the incident angle approaches the critical angle, the transmitted intensity approaches zero.) If the angle of incidence is larger than the critical angle, the sine of the angle of refraction, as computed by Snell’s law, would have to be greater than unity, which is impossible. Beyond the critical angle, the ray cannot pass into the upper material; it is trapped in the lower material and is completely reflected at the boundary surface. This situation, called total internal reflection, occurs only when a ray in material a is incident on a second material b whose index of refraction is smaller than that of material a (that is, nb 6 na). We can find the critical angle for two given materials a and b by setting ub = 90°1sin ub = 12 in Snell’s law. We then have

sin ucrit =

nb na

(critical angle for total internal reflection)

(13.6)

13.14 (a) Total internal reflection in a Porro prism. (b) A combination of two Porro prisms in binoculars. (a) Total internal reflection in a Porro prism

Total internal reflection will occur if the angle of incidence ua is larger than or equal to ucrit.

45°

Applications of Total Internal Reflection

45°

Total internal reflection finds numerous uses in optical technology. As an example, consider glass with index of refraction n = 1.52. If light propagating within this glass encounters a glass–air interface, the critical angle is sin ucrit =

1 = 0.658 1.52

ucrit = 41.1°

90°

If the incident beam is oriented as shown, total internal reflection occurs on the 45° faces (because, for a glass–air interface, ucrit 5 41.1). (b) Binoculars use Porro prisms to reflect the light to each eyepiece.

The light will be totally reflected if it strikes the glass–air surface at an angle of 41.1° or larger. Because the critical angle is slightly smaller than 45°, it is 4 reflecting surpossible to use a prism with angles of 45°- 45°-90° as a totally face. As reflectors, totally reflecting prisms have some advantages over metallic surfaces such as ordinary coated-glass mirrors. While no metallic surface reflects 100% of the light incident on it, light can be totally reflected by a prism. These reflecting properties of a prism are permanent and unaffected by tarnishing. A 45°– 45°– 90° prism, used as in Fig. 13.14a, is called a Porro prism. Light enters and leaves at right angles to the hypotenuse and is totally reflected at each of the shorter faces. The total change of direction of the rays is 180°. Binoculars often use combinations of two Porro prisms, as in Fig. 13.14b. When a beam of light enters at one end of a transparent rod (Fig. 13.15), the light can be totally reflected internally if the index of refraction of the rod is greater than that of the surrounding material. The light is “trapped” within the rod even if the rod is curved, provided that the curvature is not too great. Such a

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Porro prisms

422

CHAPTER 13 The Nature and Propagation of Light 13.15 A transparent rod with refractive index greater than that of the surrounding material.

a

The light is trapped in the rod if all the angles of incidence (such as a, b, and g) exceed the critical angle.

13.16 This colored x-ray image of a patient’s abdomen shows an endoscope winding through the colon.

13.17 To maximize their brilliance, diamonds are cut so that there is total internal reflection on their back surfaces.

b g

rod is sometimes called a light pipe. A bundle of fine glass or plastic fibers behaves in the same way and has the advantage of being flexible. A bundle may consist of thousands of individual fibers, each of the order of 0.002 to 0.01 mm in diameter. If the fibers are assembled in the bundle so that the relative positions of the ends are the same (or mirror images) at both ends, the bundle can transmit an image. Fiber-optic devices have found a wide range of medical applications in instruments called endoscopes, which can be inserted directly into the bronchial tubes, the bladder, the colon, and other organs for direct visual examination (Fig. 13.16). A bundle of fibers can even be enclosed in a hypodermic needle for studying tissues and blood vessels far beneath the skin. Fiber optics also have applications in communication systems, in which they are used to transmit a modulated laser beam. The rate at which information can be transmitted by a wave (light, radio, or whatever) is proportional to the frequency. To see qualitatively why this is so, consider modulating (modifying) the wave by chopping off some of the wave crests. Suppose each crest represents a binary digit, with a chopped-off crest representing a zero and an unmodified crest representing a one. The number of binary digits we can transmit per unit time is thus proportional to the frequency of the wave. Infrared and visiblelight waves have much higher frequency than do radio waves, so a modulated laser beam can transmit an enormous amount of information through a single fiber-optic cable. Another advantage of optical fibers is that they can be made thinner than conventional copper wire, so more fibers can be bundled together in a cable of a given diameter. Hence more distinct signals (for instance, different phone lines) can be sent over the same cable. Because fiber-optic cables are electrical insulators, they are immune to electrical interference from lightning and other sources, and they don’t allow unwanted currents between source and receiver. For these and other reasons, fiber-optic cables play an important role in long-distance telephone, television, and Internet communication. Total internal reflection also plays an important role in the design of jewelry. The brilliance of diamond is due in large measure to its very high index of refraction 1n = 2.4172 and correspondingly small critical angle. Light entering a cut diamond is totally internally reflected from facets on its back surface, and then emerges from its front surface (Fig. 13.17). “Imitation diamond” gems, such as cubic zirconia, are made from less expensive crystalline materials with comparable indexes of refraction.

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13.4 Dispersion

Conceptual Example 13.4

A leaky periscope

A submarine periscope uses two totally reflecting 45°- 45° - 90° prisms with total internal reflection on the sides adjacent to the 45° angles. Explain why the periscope will no longer work if it springs a leak and the bottom prism is covered with water. '

SOLUTION

'

The critical angle for water 1nb = 1.332 on glass 1na = 1.522 is

ucrit = arcsin

'

1.33 = 61.0° 1.52

The 45° angle of incidence for a totally reflecting prism is smaller than this new 61° critical angle, so total internal reflection does not occur at the glass–water interface. Most of the light is transmitted into the water, and very little is reflected back into the prism.

Test Your Understanding of Section 13.3 In which of the following situations is there total internal reflection? (i) Light propagating in water 1n = 1.332 strikes a water–air interface at an incident angle of 70°; (ii) light propagating in glass 1n = 1.522 strikes a glass–water interface at an incident angle of 70°; (iii) light propagating in water strikes a water–glass interface at an incident angle of 70°.

13.4

423

y

Dispersion

Ordinary white light is a superposition of waves with wavelengths extending throughout the visible spectrum. The speed of light in vacuum is the same for all wavelengths, but the speed in a material substance is different for different wavelengths. Therefore the index of refraction of a material depends on wavelength. The dependence of wave speed and index of refraction on wavelength is called dispersion. Figure 13.18 shows the variation of index of refraction n with wavelength for some common optical materials. Note that the horizontal axis of this figure is the wavelength of the light in vacuum, l0 ; the wavelength in the material is given by Eq. (13.5), l = l0>n. In most materials the value of n decreases with increasing wavelength and decreasing frequency, and thus n increases with decreasing wavelength and increasing frequency. In such a material, light of longer wavelength has greater speed than light of shorter wavelength. Figure 13.19 shows a ray of white light incident on a prism. The deviation (change of direction) produced by the prism increases with increasing index of refraction and frequency and decreasing wavelength. Violet light is deviated most, and red is deviated least; other colors are in intermediate positions. When it comes out of the prism, the light is spread out into a fan-shaped beam, as shown. The light is said to be dispersed into a spectrum. The amount of dispersion depends on the difference between the refractive indexes for violet light and for red light. From Fig. 13.18 we can see that for a substance such as fluorite, the difference between the indexes for red and violet is small, and the dispersion will also be small. A better choice of material for a prism whose purpose is to produce a spectrum would be silicate flint glass, for which there is a larger difference in the value of n between red and violet.

13.18 Variation of index of refraction n with wavelength for different transparent materials. The horizontal axis shows the wavelength l0 of the light in vacuum; the wavelength in the material is equal to

l = l0/n. Index of refraction (n) 1.7

'

White light

Deviation of yellow light

Silicate flint glass 1.6

Borate flint glass Quartz Silicate crown glass

1.5 Fused quartz Fluorite 1.4 400

500 600 700 Wavelength in vacuum (nm)

13.19 Dispersion of light by a prism. The band of colors is called a spectrum.

Measure of dispersion

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CHAPTER 13 The Nature and Propagation of Light

As we mentioned in Section 13.3, the brilliance of diamond is due in part to its unusually large refractive index; another important factor is its large dispersion, which causes white light entering a diamond to emerge as a multicolored spectrum. Crystals of rutile and of strontium titanate, which can be produced synthetically, have about eight times the dispersion of diamond.

Rainbows When you experience the beauty of a rainbow, as in Fig. 13.20a, you are seeing the combined effects of dispersion, refraction, and reflection. Sunlight comes from behind you, enters a water droplet, is (partially) reflected from the back surface of the droplet, and is refracted again upon exiting the droplet (Fig. 13.20b). A light ray that enters the middle of the raindrop is reflected straight back. All other rays 13.20 How rainbows form. (a) A double rainbow

(b) The paths of light rays entering the upper half of a raindrop 6 5 4 Light rays 3 from sun 2

Secondary rainbow (note reversed colors)

Down-sun point

1 Raindrop

2 Primary rainbow D 5 maximum angle of light from raindrop

The pattern of rays entering the lower half of the drop (not shown) is the same but flipped upside down.

6 3

4 5

(d) A primary rainbow is formed by rays that undergo two refractions and one internal reflection. The angle D is larger for red light than for violet. Light from sun

(c) Forming a rainbow. The sun in this illustration is directly behind the observer at P. The rays of sunlight that form the z primary rainbow refract into the droplets, undergo internal reflection, and refract out. ight hite l ent w Incid

The two refractions disperse the colors.

Water droplets in cloud

To the down-sun point y 42.5° P

Observer at P

D 5 40.8° (violet) to 42.5° (red)

(e) A secondary rainbow is formed by rays that undergo two refractions and two internal reflections. The angle D is larger for violet light than for red.

Light from sun

O

40.8° Angles are exaggerated for clarity. Only a primary rainbow is shown.

D 5 50.1° (red) to 53.2° (violet)

x

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425

13.5 Polarization

exit the raindrop within an angle ¢ of that middle ray, with many rays “piling up” at the angle ¢. What you see is a disk of light of angular radius ¢ centered on the down-sun point (the point in the sky opposite the sun); due to the “piling up” of light rays, the disk is brightest around its rim, which we see as a rainbow (Fig. 13.20c). Because no light reaches your eye from angles larger than ¢, the sky looks dark outside the rainbow (see Fig. 13.20a). The value of the angle ¢ depends on the index of refraction of the water that makes up the raindrops, which in turn depends on the wavelength (Fig. 13.20d). The bright disk of red light is slightly larger than that for orange light, which in turn is slightly larger than that for yellow light, and so on. As a result, you see the rainbow as a band of colors. In many cases you can see a second, larger rainbow. It is the result of dispersion, refraction, and two reflections from the back surface of the droplet (Fig. 13.20e). Each time a light ray hits the back surface, part of the light is refracted out of the drop (not shown in Fig. 13.20); after two such hits, relatively little light is left inside the drop, which is why the secondary rainbow is noticeably fainter than the primary rainbow. Just as a mirror held up to a book reverses the printed letters, so the second reflection reverses the sequence of colors in the secondary rainbow. You can see this effect in Fig. 13.20a.

Polarization

13.5

Polarization is a characteristic of all transverse waves. This chapter is about light, but to introduce some basic polarization concepts, let’s go back to the transverse waves on a string that we studied in Chapter 15. For a string that in equilibrium lies along the x-axis, the displacements may be along the y-direction, as in Fig. 13.21a. In this case the string always lies in the xy-plane. But the displacements might instead be along the z-axis, as in Fig. 13.21b; then the string always lies in the xz-plane. When a wave has only y-displacements, we say that it is linearly polarized in the y-direction; a wave with only z-displacements is linearly polarized in the z-direction. For mechanical waves we can build a polarizing filter, or polarizer, that permits only waves with a certain polarization direction to pass. In Fig. 13.21c the string can slide vertically in the slot without friction, but no horizontal motion is possible. This filter passes waves that are polarized in the y-direction but blocks those that are polarized in the z-direction. This same language can be applied to electromagnetic waves, which also have polarization. As we learned in Chapter 12, an electromagnetic wave is a transverse wave; the fluctuating electric and magnetic fields are perpendicular to each other and to the direction of propagation. We always define the direction of polarization of an electromagnetic wave to be the direction of the electric-field vector S E, not the magnetic field, because many common electromagnetic-wave detectors respond to the electric forces on electrons in materials, not the magnetic forces. Thus the electromagnetic wave described by Eq. (12.17),

ActivPhysics 16.9: Physical Optics: Polarization

E1x, t2 5 n≥ Emax cos1kx - vt2 S B1x, t2 5 nkBmax cos1kx - vt2 S

13.21 (a), (b) Polarized waves on a string. (c) Making a polarized wave on a string from an unpolarized one using a polarizing filter. (a) Transverse wave linearly polarized in the y-direction

(b) Transverse wave linearly polarized in the z-direction

y

O z

y

O

x z

(c) The slot functions as a polarizing filter, passing only components polarized in the y-direction. Slot y Barrier

O

x z

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x

426

CHAPTER 13 The Nature and Propagation of Light

13.22 (a) Electrons in the red and white broadcast antenna oscillate vertically, producing vertically polarized electromagnetic waves that propagate away from the antenna in the horizontal direction. (The small gray antennas are for relaying cellular phone signals.) (b) No matter how this light bulb is oriented, the random motion of electrons in the filament produces unpolarized light waves.

(a)

(b)

is said to be polarized in the y-direction because the electric field has only a y-component. CAUTION The meaning of “polarization” It’s unfortunate that the same word “polarizaS tion” that is used to describe the direction of E in an electromagnetic wave is also used to describe the shifting of electric charge within a body, such as in response to a nearby charged body; we described this latter kind of polarization in Section 1.2 (see Fig. 1.7). You should remember that while these two concepts have the same name, they do not describe the same phenomenon. y

Polarizing Filters

13.23 A Polaroid filter is illuminated by S unpolarized natural light (shown by E vectors that point in all directions perpendicular to the direction of propagation). The transmitted light is linearly polarized along S the polarizing axis (shown by E vectors along the polarization direction only). Filter only partially absorbs vertically polarized component of light. Incident unpolarized light

Polarizing axis

Polaroid filter Filter almost completely absorbs horizontally polarized component of light.

Transmitted light is linearly polarized in the vertical direction.

Waves emitted by a radio transmitter are usually linearly polarized. The vertical antennas that are used for radio broadcasting emit waves that, in a horizontal plane around the antenna, are polarized in the vertical direction (parallel to the antenna) (Fig. 13.22a). The situation is different for visible light. Light from incandescent light bulbs and fluorescent light fixtures is not polarized (Fig. 13.22b). The “antennas” that radiate light waves are the molecules that make up the sources. The waves emitted by any one molecule may be linearly polarized, like those from a radio antenna. But any actual light source contains a tremendous number of molecules with random orientations, so the emitted light is a random mixture of waves linearly polarized in all possible transverse directions. Such light is called unpolarized light or natural light. To create polarized light from unpolarized natural light requires a filter that is analogous to the slot for mechanical waves in Fig. 13.21c. Polarizing filters for electromagnetic waves have different details of construction, depending on the wavelength. For microwaves with a wavelength of a few centimeters, a good polarizer is an array of closely spaced, parallel conducting wires that are insulated from each other. (Think of a barbecue grill with the outer metal ring replaced by an insulating one.) Electrons are free to move along the S length of the conducting wires and will do so in response to a wave whose E field is parallel to the wires. The resulting currents in the wires dissipate energy by I2R heating; the dissipated energy comes from the wave, so whatever wave passes S through the grid is greatly reduced in amplitude. Waves with E oriented perpendicular to the wires pass through almost unaffected, since electrons cannot move through the air between the wires. Hence a wave that passes through such a filter will be predominantly polarized in the direction perpendicular to the wires. The most common polarizing filter for visible light is a material known by the trade name Polaroid, widely used for sunglasses and polarizing filters for camera lenses. Developed originally by the American scientist Edwin H. Land, this material incorporates substances that have dichroism, a selective absorption in which one of the polarized components is absorbed much more strongly than the other (Fig. 13.23). A Polaroid filter transmits 80% or more of the intensity of a wave that is polarized parallel to a certain axis in the material, called the polarizing

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13.5 Polarization Transmitted light, linearly polarized parallel to polarizing axis

Polarizer Incident unpolarized light

427

13.24 Unpolarized natural light is incident on the polarizing filter. The photocell measures the intensity of the transmitted linearly polarized light.

Photocell • The intensity of the transmitted light is the same for all orientations of the polarizing filter. • For an ideal polarizing filter, the transmitted intensity is half the incident intensity.

Polarizing axis

axis, but only 1% or less for waves that are polarized perpendicular to this axis. In one type of Polaroid filter, long-chain molecules within the filter are oriented with their axis perpendicular to the polarizing axis; these molecules preferentially absorb light that is polarized along their length, much like the conducting wires in a polarizing filter for microwaves.

Using Polarizing Filters An ideal polarizing filter (polarizer) passes 100% of the incident light that is polarized parallel to the filter’s polarizing axis but completely blocks all light that is polarized perpendicular to this axis. Such a device is an unattainable idealization, but the concept is useful in clarifying the basic ideas. In the following discussion we will assume that all polarizing filters are ideal. In Fig. 13.24 S unpolarized light is incident on a flat polarizing filter. The E vector of the incident wave can be represented in terms of components parallel and perpendicular S to the polarizer axis (shown in blue); only the component of E parallel to the polarizing axis is transmitted. Hence the light emerging from the polarizer is linearly polarized parallel to the polarizing axis. When unpolarized light is incident on an ideal polarizer as in Fig. 13.24, the intensity of the transmitted light is exactly half that of the incident unpolarized light,S no matter how the polarizing axis is oriented. Here’s why: We can resolve the E field of the incident wave into a component parallel to the polarizing axis and a component perpendicular to it. Because the incident light is a random mixture of all states of polarization, these two components are, on average, equal. The ideal polarizer transmits only the component that is parallel to the polarizing axis, so half the incident intensity is transmitted. What happens when the linearly polarized light emerging from a polarizer passes through a second polarizer, or analyzer, as in Fig. 13.25? Suppose the polarizing axis of the analyzer makes an angle f with the polarizing axis of the 13.25 An ideal analyzer transmits only the electric field component parallel to its transmission direction (that is, its polarizing axis).

f is the angle between the polarizing Analyzer axes of the polarizer and analyzer. Ei 5 E cos f

Polarizer Incident unpolarized light

Ei 5 E cos f S

f

f

E

Photocell E'

The intensity I of light from the analyzer is maximal (Imax ) when f 5 0. At other angles,

The linearly polarized light I 5 Imax cos2 f from the first polarizer can be resolved into components Ei and E' parallel and perpendicular, respectively, to the polarizing axis of the analyzer.

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CHAPTER 13 The Nature and Propagation of Light

13.26 These photos show the view through Polaroid sunglasses whose polarizing axes are (left) aligned 1f = 02 and (right) perpendicular 1f = 90°2. The transmitted intensity is greatest when the axes are aligned; it is zero when the axes are perpendicular.

first polarizer. We can resolve the linearly polarized light that is transmitted by the first polarizer into two components, as shown in Fig. 13.25, one parallel and the other perpendicular to the axis of the analyzer. Only the parallel component, with amplitude E cos f, is transmitted by the analyzer. The transmitted intensity is greatest when f = 0, and it is zero when the polarizer and analyzer are crossed so that f = 90° (Fig. 13.26). To determine the direction of polarization of the light transmitted by the first polarizer, rotate the analyzer until the photocell in Fig. 13.25 measures zero intensity; the polarization axis of the first polarizer is then perpendicular to that of the analyzer. To find the transmitted intensity at intermediate values of the angle f, we recall from our energy discussion in Section 12.4 that the intensity of an electromagnetic wave is proportional to the square of the amplitude of the wave [see Eq. (12.29)]. The ratio of transmitted to incident amplitude is cos f, so the ratio of transmitted to incident intensity is cos2 f. Thus the intensity of the light transmitted through the analyzer is I = Imax cos2 f

(Malus’s law, polarized light passing through an analyzer)

(13.7)

where Imax is the maximum intensity of light transmitted (at f = 0) and I is the amount transmitted at angle f. This relationship, discovered experimentally by Etienne Louis Malus in 1809, is called Malus’s law. Malus’s law applies only if the incident light passing through the analyzer is already linearly polarized. Problem-Solving Strategy 13.2

Linear Polarization

IDENTIFY the relevant concepts: In all electromagnetic waves, including light waves, the direction of polarization is the direction of S the E field and is perpendicular to the propagation direction. ProbS lems about polarizers are therefore about the components of E parallel and perpendicular to the polarizing axis. SET UP the problem using the following steps: 1. Start by drawing a large, neat diagram. Label all known angles, including the angles of all polarizing axes. 2. Identify the target variables. EXECUTE the solution as follows: 1. Remember that a polarizer lets pass only electric-field components parallel to its polarizing axis. 2. If the incident light is linearly polarized and has amplitude E and intensity Imax, the light that passes through an ideal polarizer has amplitude E cos f and intensity Imax cos2 f, where f is

the angle between the incident polarization direction and the filter’s polarizing axis. 3. Unpolarized light is a random mixture of all possible polarization states, so on the average it has equal components in any two perpendicular directions. When passed through an ideal polarizer, unpolarized light becomes linearly polarized light with half the incident intensity. Partially linearly polarized light is a superposition of linearly polarized and unpolarized light. 4. The intensity (average power per unit area) of a wave is proportional to the square of its amplitude. If you find that two waves differ in amplitude by a certain factor, their intensities differ by the square of that factor. EVALUATE your answer: Check your answer for any obvious errors. If your results say that light emerging from a polarizer has greater intensity than the incident light, something’s wrong: A polarizer can’t add energy to a light wave.

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13.5 Polarization

Example 13.5

429

Two polarizers in combination

In Fig. 13.25 the incident unpolarized light has intensity I0. Find the intensities transmitted by the first and second polarizers if the angle between the axes of the two filters is 30°.

EXECUTE: The incident light is unpolarized, so the intensity of the linearly polarized light transmitted by the first polarizer is I0 > 2. From Eq. (13.7) with f = 30°, the second polarizer reduces the intensity by a further factor of cos2 30° = 34. Thus the intensity transmitted by the second polarizer is I0 a b A 34 B = 38 I0 2 '

S O L U T IO N IDENTIFY and SET UP: This problem involves a polarizer (a polarizing filter on which unpolarized light shines, producing polarized light) and an analyzer (a second polarizing filter on which the polarized light shines). We are given the intensity I0 of the incident light and the angle f = 30° between the axes of the polarizers. We use Malus’s law, Eq. (13.7), to solve for the intensities of the light emerging from each polarizer.

'

EVALUATE: Note that the intensity decreases after each passage through a polarizer. The only situation in which the transmitted intensity does not decrease is if the polarizer is ideal (so it absorbs none of the light that passes through it) and if the incident light is linearly polarized along the polarizing axis, so f = 0.

13.27 When light is incident on a reflecting surface at the polarizing angle, the reflected light is linearly polarized. 1 If unpolarized light is incident at the polarizing angle ...

2 ... then the reflected light is 100% polarized perpendicular to the plane of incidence ...

Normal Plane of incidence

4 Alternatively, if unpolarized light is incident on the reflecting surface at an angle other than up, the reflected light is partially polarized.

na up

up

Reflecting surface nb ub 3 ... and the transmitted light is partially polarized parallel to the plane of incidence.

Polarization by Reflection Unpolarized light can be polarized, either partially or totally, by reflection. In Fig. 13.27, unpolarized natural light is incident on a reflecting surface between two transparent optical materials. For most angles of incidence, waves for which the S electric-field vector E is perpendicular to the plane of incidence (that is, parallel S to the reflecting surface) are reflected more strongly than those for which E lies in this plane. In this case the reflected light is partially polarized in the direction perpendicular to the plane of incidence. But at one particular angle of incidence, called the polarizing angle up , the S light for which E lies in the plane of incidence is not reflected at allSbut is completely refracted. At this same angle of incidence the light for which E is perpendicular to the plane of incidence is partially reflected and partially refracted. The reflected light is therefore completely polarized perpendicular to the plane of incidence, as shown in Fig. 13.27. The refracted (transmitted) light is partially polarized parallel to this plane; the refracted light is a mixture of the component parallel to the plane of incidence, all of which is refracted, and the remainder of the perpendicular component. In 1812 the British scientist Sir David Brewster discovered that when the angle of incidence is equal to the polarizing angle up , the reflected ray and the refracted ray are perpendicular to each other (Fig. 13.28). In this case the angle of refraction ub equals 90° - up. From the law of refraction, '

13.28 The significance of the polarizing angle. The open circles represent a compoS nent of E that is perpendicular to the plane of the figure (the plane of incidence) and parallel to the surface between the two materials. Note: This is a side view of the situation shown in Fig. 33.27. Component perpendicular Reflected to plane of page ray Nor mal

up

up na nb

Refracted ray ub

'

na sin up = nb sin ub

When light strikes a surface at the polarizing angle, the reflected and refracted rays are perpendicular to each other and nb tan up 5 n a

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430

CHAPTER 13 The Nature and Propagation of Light

so we find na sin up = nb sin190° - up2 = nb cos up tan up =

nb na

(Brewster’s law for the polarizing angle)

(13.8)

This relationship is known as Brewster’s law. Although discovered experimentally, it can also be derived from a wave model using Maxwell’s equations. Polarization by reflection is the reason polarizing filters are widely used in sunglasses (Fig. 13.26). When sunlight is reflected from a horizontal surface, the plane of incidence is vertical, and the reflected light contains a preponderance of light that is polarized in the horizontal direction. When the reflection occurs at a smooth asphalt road surface or the surface of a lake, it causes unwanted glare. Vision can be improved by eliminating this glare. The manufacturer makes the polarizing axis of the lens material vertical, so very little of the horizontally polarized light reflected from the road is transmitted to the eyes. The glasses also reduce the overall intensity of the transmitted light to somewhat less than 50% of the intensity of the unpolarized incident light.

Example 13.6 Reflection from a swimming pool’s surface Sunlight reflects off the smooth surface of a swimming pool. (a) For what angle of reflection is the reflected light completely polarized? (b) What is the corresponding angle of refraction? (c) At night, an underwater floodlight is turned on in the pool. Repeat parts (a) and (b) for rays from the floodlight that strike the surface from below. SOLUTION IDENTIFY and SET UP: This problem involves polarization by reflection at an air–water interface in parts (a) and (b) and at a water–air interface in part (c). Figure 13.29 shows our sketches.

For both cases our first target variable is the polarizing angle up, which we find using Brewster’s law, Eq. (13.8). For this angle of reflection, the angle of refraction ub is the complement of up 1that is, ub = 90° 2 up2. EXECUTE: (a) During the day (shown in the upper part of Fig. 13.29) the light moves in air toward water, so na = 1.00 (air) and nb = 1.33 (water). From Eq. (13.8), up = arctan

nb 1.33 = arctan = 53.1° na 1.00

(b) The incident light is at the polarizing angle, so the reflected and refracted rays are perpendicular; hence ub = 90° - up = 90° - 53.1° = 36.9°

13.29 Our sketches for this problem.

(c) At night (shown in the lower part of Fig. 13.29) the light moves in water toward air, so now na = 1.33 and nb = 1.00. Again using Eq. (13.8), we have 1.00 = 36.9° 1.33 ub = 90° - 36.9° = 53.1°

up = arctan

EVALUATE: We check our answer in part (b) by using Snell’s law, na sin ua = nb sin ub, to solve for ub: '

na sin up

1.00 sin 53.1° = = 0.600 nb 1.33 ub = arcsin10.6002 = 36.9°

sin ub =

Note that the two polarizing angles found in parts (a) and (c) add to 90°. This is not an accident; can you see why?

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13.5 Polarization

Circular and Elliptical Polarization Light and other electromagnetic radiation can also have circular or elliptical polarization. To introduce these concepts, let’s return once more to mechanical waves on a stretched string. In Fig. 13.21, suppose the two linearly polarized waves in parts (a) and (b) are in phase and have equal amplitude. When they are superposed, each point in the string has simultaneous y- and z-displacements of equal magnitude. A little thought shows that the resultant wave lies in a plane oriented at 45° to the y- and z-axes (i.e., in a plane making a 45° angle with the xyand xz-planes2. The amplitude of the resultant wave is larger by a factor of 12 than that of either component wave, and the resultant wave is linearly polarized. But now suppose the two equal-amplitude waves differ in phase by a quartercycle. Then the resultant motion of each point corresponds to a superposition of two simple harmonic motions at right angles, with a quarter-cycle phase difference. The y-displacement at a point is greatest at times when the z-displacement is zero, and vice versa. The motion of the string as a whole then no longer takes place in a single plane. It can be shown that each point on the rope moves in a circle in a plane parallel to the yz-plane. Successive points on the rope have successive phase differences, and the overall motion of the string has the appearance of a rotating helix. This is shown to the left of the polarizing filter in Fig. 13.21c. This particular superposition of two linearly polarized waves is called circular polarization. Figure 13.30 shows the analogous situation for an electromagnetic wave. Two sinusoidal waves of equal amplitude, polarized in the y- and z-directions and with a quarter-cycle phase difference, are superposed. The result is a wave in which S the E vector at each point has a constant magnitude but rotates around the direcS tion of propagation. The wave in Fig. 13.30 is propagating toward you and the E vector appears to be rotating clockwise,Sso it is called a right circularly polarized electromagnetic wave. If instead the E vector of a wave coming toward you appears to be rotating counterclockwise, it is called a left circularly polarized electromagnetic wave. If the phase difference between the two component waves is something other than a quarter-cycle, or if the two component waves have different amplitudes, then each point on the string traces out not a circle but an ellipse. The resulting wave is said to be elliptically polarized. For electromagnetic waves with radio frequencies, circular or elliptical polarization can be produced by using two antennas at right angles, fed from the same S

13.30 Circular polarization of an electromagnetic wave moving toward you parallel to the x-axis. The y-component of E lags the z-component by a quarter-cycle. This phase difference results in right circular polarization. y S

S

E (E y)max z

Ey E

x

S

E

Ez

/

t50

S

2(Ez)max

/

t5T8

t5T4

E

/

t 5 3T 8

S

E

S

(Ez)max

E

(Ey)max

S

E

S

E

/

t 5 5T 8

/

t 5 3T 4

/

t 5 7T 8

t5T

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2(Ey)max S

E

/

t5T2

Circular polarization: The S E vector of the wave has constant magnitude and rotates in a circle.

431

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CHAPTER 13 The Nature and Propagation of Light

transmitter but with a phase-shifting network that introduces the appropriate phase difference. For light, the phase shift can be introduced by use of a material that exhibits birefringence—that is, has different indexes of refraction for different directions of polarization. A common example is calcite 1CaCO32. When a calcite crystal is oriented appropriately in a beam of unpolarized light, its refractive index (for a wavelength in vacuum of 589 nm) is 1.658 for one direction of polarization and 1.486 for the perpendicular direction. When two waves with equal amplitude and with perpendicular directions of polarization enter such a material, they travel with different speeds. If they are in phase when they enter the material, then in general they are no longer in phase when they emerge. If the crystal is just thick enough to introduce a quarter-cycle phase difference, then the crystal converts linearly polarized light to circularly polarized light. Such a crystal is called a quarter-wave plate. Such a plate also converts circularly polarized light to linearly polarized light. Can you prove this?

Photoelasticity 13.31 This plastic model of an artificial hip joint was photographed between two polarizing filters (a polarizer and an analyzer) with perpendicular polarizing axes. The colored interference pattern reveals the direction and magnitude of stresses in the model. Engineers use these results to help design the metal artificial hip joint used in medicine.

?

Some optical materials that are not normally birefringent become so when they are subjected to mechanical stress. This is the basis of the science of photoelasticity. Stresses in girders, boiler plates, gear teeth, and cathedral pillars can be analyzed by constructing a transparent model of the object, usually of a plastic material, subjecting it to stress, and examining it between a polarizer and an analyzer in the crossed position. Very complicated stress distributions can be studied by these optical methods. Figure 13.31 is a photograph of a photoelastic model under stress. The polarized light that enters the model can be thought of as having a component along each of the two directions of the birefringent plastic. Since these two components travel through the plastic at different speeds, the light that emerges from the other side of the model can have a different overall direction of polarization. Hence some of this transmitted light will be able to pass through the analyzer even though its polarization axis is at a 90° angle to the polarizer’s axis, and the stressed areas in the plastic will appear as bright spots. The amount of birefringence is different for different wavelengths and hence different colors of light; the color that appears at each location in Fig. 13.31 is that for which the transmitted light is most nearly polarized along the analyzer’s polarization axis. Test Your Understanding of Section 13.5 You are taking a photograph of a sunlit high-rise office building. In order to minimize the reflections from the building’s windows, you place a polarizing filter on the camera lens. How should you orient the filter? (i) with the polarizing axis vertical; (ii) with the polarizing axis horizontal; (iii) either orientation will minimize the reflections just as well; (iv) neither orientation will have any effect.

13.6

y

Scattering of Light

The sky is blue. Sunsets are red. Skylight is partially polarized; that’s why the sky looks darker from some angles than from others when it is viewed through Polaroid sunglasses. As we will see, a single phenomenon is responsible for all of these effects. When you look at the daytime sky, the light that you see is sunlight that has been absorbed and then re-radiated in a variety of directions. This process is called scattering. (If the earth had no atmosphere, the sky would appear as black in the daytime as it does at night, just as it does to an astronaut in space or on the moon.) Figure 13.32 shows some of the details of the scattering process. Sunlight, which is unpolarized, comes from the left along the x-axis and passes over an observer looking vertically upward along the y-axis. (We are viewing the situation

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13.6 Scattering of Light

433

13.32 When the sunbathing observer on the left looks up, he sees blue, polarized sunlight that has been scattered by air molecules. The observer on the right sees reddened, unpolarized light when he looks at the sun. Incident white light, y unpolarized

z

x

O

Electric charges in air molecules at O oscillate in S the direction of the E field of the incident light from the sun, acting as antennas that produce scattered light. The scattered light that reaches the observer directly below O is polarized in the z-direction. Air molecules scatter blue light more effectively than red light; we see the sky overhead by scattered light, so it looks blue.

This observer sees reddened sunlight because most of the blue light has been scattered out.

from the side.) Consider the molecules of the earth’s atmosphere located at point O. The electric field in the beam of sunlight sets the electric charges in these molecules into vibration. Since light is a transverse wave, the direction of the electric field in any component of the sunlight lies in the yz-plane, and the motion of the charges takes place in this plane. There is no field, and hence no motion of charges, in the direction of the x-axis. An incident light wave sets the electric charges in the molecules at point O S vibrating along the line of E. We can resolve this vibration into two components, one along the y-axis and the other along the z-axis. Each component in the incident light produces the equivalent of two molecular “antennas,” oscillating with the same frequency as the incident light and lying along the y- and z-axes. We mentioned in Chapter 12 that an oscillating charge, like those in an antenna, does not radiate in the direction of its oscillation. (See Fig. 12.3 in Section 12.1.) Thus the “antenna” along the y-axis does not send any light to the observer directly below it, although it does emit light in other directions. Therefore the only light reaching this observer comes from the other molecular “antenna,” corresponding to the oscillation of charge along the z-axis. This light is linearly polarized, with its electric field along the z-axis (parallel to the “antenna”). The red vectors on the y-axis below point O in Fig. 13.32 show the direction of polarization of the light reaching the observer. As the original beam of sunlight passes though the atmosphere, its intensity decreases as its energy goes into the scattered light. Detailed analysis of the scattering process shows that the intensity of the light scattered from air molecules increases in proportion to the fourth power of the frequency (inversely to the fourth power of the wavelength). Thus the intensity ratio for the two ends of the visible spectrum is 1750 nm>380 nm24 = 15. Roughly speaking, scattered light contains 15 times as much blue light as red, and that’s why the sky is blue. Clouds contain a high concentration of water droplets or ice crystals, which also scatter light. Because of this high concentration, light passing through the cloud has many more opportunities for scattering than does light passing through a clear sky. Thus light of all wavelengths is eventually scattered out of the cloud, so the cloud looks white (Fig. 13.33). Milk looks white for the same reason; the scattering is due to fat globules in the milk. Near sunset, when sunlight has to travel a long distance through the earth’s atmosphere, a substantial fraction of the blue light is removed by scattering. White light minus blue light looks yellow or red. This explains the yellow or red hue that we so often see from the setting sun (and that is seen by the observer at the far right of Fig. 13.32).

+Application Bee Vision and Polarized Light from the Sky The eyes of a bee can detect the polarization of light. Bees use this to help them navigate between the hive and food sources. As Fig. 13.32 shows, a bee sees unpolarized light if it looks directly toward the sun and sees completely polarized light if it looks 90° away from the sun. These polarizations are unaffected by the presence of clouds, so a bee can navigate relative to the sun even on an overcast day.

Eyes

13.33 Clouds are white because they efficiently scatter sunlight of all wavelengths.

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434

CHAPTER 13 The Nature and Propagation of Light

13.7

13.34 Applying Huygens’s principle to wave front AA¿ to construct a new wave front BB¿. B Secondary wavelets

A

Huygens’s Principle

The laws of reflection and refraction of light rays that we introduced in Section 13.2 were discovered experimentally long before the wave nature of light was firmly established. However, we can derive these laws from wave considerations and show that they are consistent with the wave nature of light. We begin with a principle called Huygens’s principle. This principle, stated originally by the Dutch scientist Christiaan Huygens in 1678, is a geometrical method for finding, from the known shape of a wave front at some instant, the shape of the wave front at some later time. Huygens assumed that every point of a wave front may be considered the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the wave. The new wave front at a later time is then found by constructing a surface tangent to the secondary wavelets or, as it is called, the envelope of the wavelets. All the results that we obtain from Huygens’s principle can also be obtained from Maxwell’s equations. Thus it is not an independent principle, but it is often very convenient for calculations with wave phenomena. Huygens’s principle is shown in Fig. 13.34. The original wave front AA¿ is traveling outward from a source, as indicated by the arrows. We want to find the shape of the wave front after a time interval t. We assume that v, the speed of propagation of the wave, is the same at all points. Then in time t the wave front travels a distance vt. We construct several circles (traces of spherical wavelets) with radius r = vt, centered at points along AA¿. The trace of the envelope of these wavelets, which is the new wave front, is the curve BB¿.

r 5 vt

Reflection and Huygens’s Principle

A9 B9

13.35 Using Huygens’s principle to derive the law of reflection. (a) Successive positions of a plane wave AA9 as it is reflected from a plane surface A9 vt B9 C9 C B ua

M A O

M9

N ur

(b) Magnified portion of (a) B

A9 Q

P

To derive the law of reflection from Huygens’s principle, we consider a plane wave approaching a plane reflecting surface. In Fig. 13.35a the lines AA¿, OB¿, and NC¿ represent successive positions of a wave front approaching the surface MM¿. Point A on the wave front AA¿ has just arrived at the reflecting surface. We can use Huygens’s principle to find the position of the wave front after a time interval t. With points on AA¿ as centers, we draw several secondary wavelets with radius vt. The wavelets that originate near the upper end of AA¿ spread out unhindered, and their envelope gives the portion OB¿ of the new wave front. If the reflecting surface were not there, the wavelets originating near the lower end of AA¿ would similarly reach the positions shown by the broken circular arcs. Instead, these wavelets strike the reflecting surface. The effect of the reflecting surface is to change the direction of travel of those wavelets that strike it, so the part of a wavelet that would have penetrated the surface actually lies to the left of it, as shown by the full lines. The first such wavelet is centered at point A; the envelope of all such reflected wavelets is the portion OB of the wave front. The trace of the entire wave front at this instant is the bent line BOB¿. A similar construction gives the line CNC¿ for the wave front after another interval t. From plane geometry the angle ua between the incident wave front and the surface is the same as that between the incident ray and the normal to the surface and is therefore the angle of incidence. Similarly, ur is the angle of reflection. To find the relationship between these angles, we consider Fig. 13.35b. From O we draw OP = vt, perpendicular to AA¿. Now OB, by construction, is tangent to a circle of radius vt with center at A. If we draw AQ from A to the point of tangency, the triangles APO and OQA are congruent because they are right triangles with the side AO in common and with AQ = OP = vt. The angle ua therefore equals the angle ur , and we have the law of reflection. '

vt

vt

B9 ua A

ur O

Refraction and Huygens’s Principle We can derive the law of refraction by a similar procedure. In Fig. 13.36a we consider a wave front, represented by line AA¿, for which point A has just

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13.7 Huygens’s Principle

arrived at the boundary surface SS¿ between two transparent materials a and b, with indexes of refraction na and nb and wave speeds va and vb . (The reflected waves are not shown in the figure; they proceed as in Fig. 13.35.) We can apply Huygens’s principle to find the position of the refracted wave fronts after a time t. With points on AA¿ as centers, we draw several secondary wavelets. Those originating near the upper end of AA¿ travel with speed va and, after a time interval t, are spherical surfaces of radius vat. The wavelet originating at point A, however, is traveling in the second material b with speed vb and at time t is a spherical surface of radius vbt. The envelope of the wavelets from the original wave front is the plane whose trace is the bent line BOB¿. A similar construction leads to the trace CPC¿ after a second interval t. The angles ua and ub between the surface and the incident and refracted wave fronts are the angle of incidence and the angle of refraction, respectively. To find the relationship between these angles, refer to Fig. 13.36b. We draw OQ = va t, perpendicular to AQ, and we draw AB = vb t, perpendicular to BO. From the right triangle AOQ, '

13.36 Using Huygens’s principle to derive the law of refraction. The case vb 6 va 1nb 7 na2 is shown.

(a) Successive positions of a plane wave AA9 as it is refracted by a plane surface S9 na n . n b a C9 vb , va B9 P

A9

'

O

va t

'

va t AO

ua

'

sin ua =

A

and from the right triangle AOB,

ub vb t

C

B S Material b

Material a

vb t sin ub = AO '

(b) Magnified portion of (a)

Combining these, we find

B9

sin ua va = vb sin ub

(13.9) O

We have defined the index of refraction n of a material as the ratio of the speed of light c in vacuum to its speed v in the material: na = c>va and nb = c>vb . Thus

A9

va t

'

Q

ub

c>vb nb va = = na vb c>va and we can rewrite Eq. (13.9) as ua

sin ua nb = or na sin ub na sin ua = nb sin ub which we recognize as Snell’s law, Eq. (13.4). So we have derived Snell’s law from a wave theory. Alternatively, we may choose to regard Snell’s law as an experimental result that defines the index of refraction of a material; in that case this analysis helps to confirm the relationship v = c>n for the speed of light in a material. Mirages offer an interesting example of Huygens’s principle in action. When the surface of pavement or desert sand is heated intensely by the sun, a hot, less dense, smaller-n layer of air forms near the surface. The speed of light is slightly greater in the hotter air near the ground, the Huygens wavelets have slightly larger radii, the wave fronts tilt slightly, and rays that were headed toward the surface with a large angle of incidence (near 90°) can be bent up as shown in Fig. 13.37. Light farther from the ground is bent less and travels nearly in a straight line. The observer sees the object in its natural position, with an inverted image below it, as though seen in a horizontal reflecting surface. The mind of a thirsty traveler can interpret the apparent reflecting surface as a sheet of water.

435

A Material a

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vb t

B

Material b

436

CHAPTER 13 The Nature and Propagation of Light

13.37 How mirages are formed. Higher up, the wave fronts travel approximately straight. vt

Hot surface

vt The hot air near ground has a smaller n than cooler air higher up, so light travels fastest near the ground. Thus, the secondary wavelets nearest the ground have the largest radii vt, and the wave fronts tilt as they travel.

It is important to keep in mind that Maxwell’s equations are the fundamental relationships for electromagnetic wave propagation. But Huygens’s principle provides a convenient way to visualize this propagation. Test Your Understanding of Section 13.7 Sound travels faster in warm air than in cold air. Imagine a weather front that runs north-south, with warm air to the west of the front and cold air to the east. A sound wave traveling in a northeast direction in the warm air encounters this front. How will the direction of this sound wave change when it passes into the cold air? (i) The wave direction will deflect toward the north; (ii) the wave direction will deflect toward the east; (iii) the wave direction will be unchanged. y

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CHAPTER

13

SUMMARY

Light and its properties: Light is an electromagnetic wave. When emitted or absorbed, it also shows particle properties. It is emitted by accelerated electric charges. A wave front is a surface of constant phase; wave fronts move with a speed equal to the propagation speed of the wave. A ray is a line along the direction of propagation, perpendicular to the wave fronts. When light is transmitted from one material to another, the frequency of the light is unchanged, but the wavelength and wave speed can change. The index of refraction n of a material is the ratio of the speed of light in vacuum c to the speed v in the material. If l0 is the wavelength in vacuum, the same wave has a shorter wavelength l in a medium with index of refraction n. (See Example 13.2.)

n =

c v

(13.1)

l =

l0 n

(13.5)

Rays

Source Wave fronts

Reflection and refraction: At a smooth interface between two optical materials, the incident, reflected, and refracted rays and the normal to the interface all lie in a single plane called the plane of incidence. The law of reflection states that the angles of incidence and reflection are equal. The law of refraction relates the angles of incidence and refraction to the indexes of refraction of the materials. (See Examples 13.1 and 13.3.)

ur = ua (law of reflection)

(13.2)

na sin ua = nb sin ub (law of refraction)

(13.4)

Total internal reflection: When a ray travels in a material of greater index of refraction na toward a material of smaller index nb , total internal reflection occurs at the interface when the angle of incidence exceeds a critical angle ucrit . (See Example 13.4.)

sin ucrit =

Polarization of light: The direction of polarization of a linearlyS polarized electromagnetic wave is the direction of the E field. A polarizing filter passes waves that are linearly polarized along its polarizing axis and blocks waves polarized perpendicularly to that axis. When polarized light of intensity Imax is incident on a polarizing filter used as an analyzer, the intensity I of the light transmitted through the analyzer depends on the angle f between the polarization direction of the incident light and the polarizing axis of the analyzer. (See Example 13.5.)

I = Imax cos2f (Malus’s law)

Polarization by reflection: When unpolarized light strikes an interface between two materials, Brewster’s law states that the reflected light is completely polarized perpendicular to the plane of incidence (parallel to the interface) if the angle of incidence equals the polarizing angle up . (See Example 13.6.)

nb na (Brewster’s law)

Incident ua

Reflected

nb na

(13.6)

Normal ub Refracted

ur

na , nb

Material a

Material b

nb na ucrit

'

. ucrit

'

(13.7) Incident natural light f

E cos f

E cos f S

E f

Photocell Analyzer Polarizer

tan up =

Normal

(13.8)

up

up na nb ub

'

Huygens’s principle: Huygens’s principle states that if the position of a wave front at one instant is known, then the position of the front at a later time can be constructed by imagining the front as a source of secondary wavelets. Huygens’s principle can be used to derive the laws of reflection and refraction.

r 5 vt

B

A

A9 B9

437

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438

CHAPTER 13 The Nature and Propagation of Light

BRIDGING PROBLEM

Reflection and Refraction

Figure 13.38 shows a rectangular glass block that has a metal reflector on one face and water on an adjoining face. A light beam strikes the reflector as shown. You gradually increase the angle u of the light beam. If u Ú 59.2°, no light enters the water. What is the speed of light in this glass?

13.38 Light beam Glass

u

SOLUTION GUIDE

Water

Reflector

See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. Specular reflection occurs where the light ray in the glass strikes the reflector. If no light is to enter the water, we require that there be reflection only and no refraction where this ray strikes the glass–water interface—that is, there must be total internal reflection. 2. The target variable is the speed of light v in the glass, which you can determine from the index of refraction n of the glass. (Table 13.1 gives the index of refraction of water.) Write down the equations you will use to find n and v.

Problems

EXECUTE 3. Use the figure to find the angle of incidence of the ray at the glass–water interface. 4. Use the result of step 3 to find n. 5. Use the result of step 4 to find v. EVALUATE 6. How does the speed of light in the glass compare to the speed in water? Does this make sense?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q13.1 Light requires about 8 minutes to travel from the sun to the earth. Is it delayed appreciably by the earth’s atmosphere? Explain. Q13.2 Sunlight or starlight passing through the earth’s atmosphere is always bent toward the vertical. Why? Does this mean that a star is not really where it appears to be? Explain. Q13.3 A beam of light goes from one material into another. On physical grounds, explain why the wavelength changes but the frequency and period do not. Q13.4 A student claimed that, because of atmospheric refraction (see Discussion Question Q13.2), the sun can be seen after it has set and that the day is therefore longer than it would be if the earth had no atmosphere. First, what does she mean by saying that the sun can be seen after it has set? Second, comment on the validity of her conclusion. Q13.5 When hot air rises from a radiator or heating duct, objects behind it appear to shimmer or waver. What causes this? Q13.6 Devise straightforward experiments to measure the speed of light in a given glass using (a) Snell’s law; (b) total internal reflection; (c) Brewster’s law. Q13.7 Sometimes when looking at a window, you see two reflected images slightly displaced from each other. What causes this? Q13.8 If you look up from underneath toward the surface of the water in your aquarium, you may see an upside-down reflection of your pet fish in the surface of the water. Explain how this can happen. Q13.9 A ray of light in air strikes a glass surface. Is there a range of angles for which total reflection occurs? Explain.

Q13.10 When light is incident on an interface between two materials, the angle of the refracted ray depends on the wavelength, but the angle of the reflected ray does not. Why should this be? Q13.11 A salesperson at a bargain counter claims that a certain pair of sunglasses has Polaroid filters; you suspect that the glasses are just tinted plastic. How could you find out for sure? Q13.12 Does it make sense to talk about the polarization of a longitudinal wave, such as a sound wave? Why or why not? Q13.13 How can you determine the direction of the polarizing axis of a single polarizer? Q13.14 It has been proposed that automobile windshields and headlights should have polarizing filters to reduce the glare of oncoming lights during night driving. Would this work? How should the polarizing axes be arranged? What advantages would this scheme have? What disadvantages? Q13.15 When a sheet of plastic food wrap is placed between two crossed polarizers, no light is transmitted. When the sheet is stretched in one direction, some light passes through the crossed polarizers. What is happening? Q13.16 If you sit on the beach and look at the ocean through Polaroid sunglasses, the glasses help to reduce the glare from sunlight reflecting off the water. But if you lie on your side on the beach, there is little reduction in the glare. Explain why there is a difference. Q13.17 When unpolarized light is incident on two crossed polarizers, no light is transmitted. A student asserted that if a third polarizer is inserted between the other two, some transmission will occur. Does this make sense? How can adding a third filter increase transmission?

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Exercises

Q13.18 For the old “rabbit-ear” style TV antennas, it’s possible to alter the quality of reception considerably simply by changing the orientation of the antenna. Why? Q13.19 In Fig. 13.32, since the light that is scattered out of the incident beam is polarized, why is the transmitted beam not also partially polarized? Q13.20 You are sunbathing in the late afternoon when the sun is relatively low in the western sky. You are lying flat on your back, looking straight up through Polaroid sunglasses. To minimize the amount of sky light reaching your eyes, how should you lie: with your feet pointing north, east, south, west, or in some other direction? Explain your reasoning. Q19.21 Light scattered from blue sky is strongly polarized because of the nature of the scattering process described in Section 13.6. But light scattered from white clouds is usually not polarized. Why not? Q13.22 Atmospheric haze is due to water droplets or smoke particles (“smog”). Such haze reduces visibility by scattering light, so that the light from distant objects becomes randomized and images become indistinct. Explain why visibility through haze can be improved by wearing red-tinted sunglasses, which filter out blue light. Q13.23 The explanation given in Section 13.6 for the color of the setting sun should apply equally well to the rising sun, since sunlight travels the same distance through the atmosphere to reach your eyes at either sunrise or sunset. Typically, however, sunsets are redder than sunrises. Why? (Hint: Particles of all kinds in the atmosphere contribute to scattering.) Q13.24 Huygens’s principle also applies to sound waves. During the day, the temperature of the atmosphere decreases with increasing altitude above the ground. But at night, when the ground cools, there is a layer of air just above the surface in which the temperature increases with altitude. Use this to explain why sound waves from distant sources can be heard more clearly at night than in the daytime. (Hint: The speed of sound increases with increasing temperature. Use the ideas displayed in Fig. 13.37 for light.) Q13.25 Can water waves be reflected and refracted? Give examples. Does Huygens’s principle apply to water waves? Explain.

SINGLE CORRECT Q13.1 A plane glass slab is placed over various coloured letters. The letter which appears to be raised the least is (a) red (b) yellow (c) violet (d) green Q13.2 Dispersion occurs when (a) some material bend light more that other material (b) a material changes some frequencies more than other. (c) light has different speeds in different materials (d) a material slows down some wavelengths more than others. Q13.3 White light is dispersed by the prism, and falls on a screen to form a visible spectrum. Which of the following is/are true? (a) The frequency changes for each colour, but the speed stays the same. (b) Red wavelenghts are deviated through larger angles than green wavelengths. (c) Violet light propagates at a higher speed than green light while in the prism. (d) The speed of all colours is reduced in the prism, with maximum reduction for violet light.

439

Q13.4 A secondary rainbow is formed when light rays coming from the sun undergo the following through spherical water droplets: (IR = internal reflection) (a) a refraction, IR and the refraction (b) two refractions only (c) a refraction, IR, again IR and the refraction (d) a refraction, IR and again IR

EXERCISES Section 13.2 Reflection and Refraction

13.1 .. A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction at this wavelength is 1.47? (b) What is the wavelength of these waves in the liquid? 13.2 .. Light with a frequency of 5.80 * 1014 Hz travels in a block of glass that has an index of refraction of 1.52. What is the wavelength of the light (a) in vacuum and (b) in the glass? 13.3 .. A laser beam Figure E13.3 shines along the surface of a block of transparent Detector material (see Fig. E13.3.). n5? Half of the beam goes 2.50 m straight to a detector, while the other half travels through the block and then hits the detector. The time delay between the arrival of the two light beams at the detector is 6.25 ns. What is the index of refraction of this material? 13.4 .. As shown in Fig. Figure E13.4 E13.4, a layer of water covers a slab of material X in a beaker. A ray of light traveling upward Air follows the path indicated. Using the information on the 45° Water figure, find (a) the index of refraction of material X and (b) the angle the light makes with the normal in the air. 13.5 . In a material having an 60° X index of refraction n, a light ray has frequency ƒ, wavelength l, and speed v. What are the frequency, wavelength, and speed of this light (a) in vacuum and (b) in a material having refractive index n¿ ? In each case, express your answers in terms of only ƒ, l, v, n, and n¿. 13.6 .. A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is 775 nm and its wavelength in the glass is 544 nm. If the ray in water makes an angle of 45o with respect to the normal to the interface, what angle does the refracted ray in the glass make with respect to the normal? '

Section 13.3 Total Internal Reflection

13.7 .. A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of ua =37o , the ray refracted into the water makes an angle of 53o with the normal to the interface. What is the smallest value of the incident angle ua for which none of the ray refracts into the water?

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440

CHAPTER 13 The Nature and Propagation of Light

13.8 .. Light Pipe. Light Figure E13.8 enters a solid pipe made of A plastic having an index of u refraction of 1.60. The light travels parallel to the upper B part of the pipe (Fig. E13.8). You want to cut the face AB so that all the light will reflect back into the pipe after it first strikes that face. (a) What is the largest that u can be if the pipe is in air? (b) If the pipe is immersed in water of refractive index 1.33, what is the largest that u can be? 13.9 .. The critical angle for total internal reflection at a liquid– air interface is 45°. (a) If a ray of light traveling in the liquid has an angle of incidence at the interface of 30°, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence at the interface of 35.0°, what angle does the refracted ray in the liquid make with the normal? 13.10 . Light is incident along Figure E13.10 the normal on face AB of a glass A prism of refractive index Ë2 , as shown in Fig. E13.10. Find the largest value the angle a can Incident ray have without any light refracted a B C out of the prism at face AC if (a) the prism is immersed in air and (b) the prism is immersed in liquid 33 . of n = 2 Ç 13.11 .. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell’s law then applies to the refraction of sound waves. The speed of a sound wave is 344 m> s in air and 1320 m> s in water. (a) Which medium has the higher index of refraction for sound? (b) What is the critical angle for a sound wave incident on the surface between air and water? (c) For total internal reflection to occur, must the sound wave be traveling in the air or in the water? (d) Use your results to explain why it is possible to hear people on the opposite shore of a river or small lake extremely clearly (Take sin 15° =0.26).

Section 13.5 Polarization

13.12 . Unpolarized light with intensity I0 is incident on two polarizing filters. The axis of the first filter makes an angle of 60.0o with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter? 13.13 .. A beam of unpolarized light of intensity I0 passes through a series of ideal polarizing filters with their polarizing directions turned to various angles as shown in Fig. E13.13. (a) What is the light intensity (in terms of I0) at points A, B, and C ? (b) If we remove the middle filter, what will be the light intensity at point C ? '

'

Figure E13.13 60° 90°

I0 Unpolarized

A

B

C

13.14 . Light traveling in water strikes a glass plate at an angle of incidence of 53.0°; part of the beam is reflected and part is refracted. If the reflected and refracted portions make an angle of 90.0° with each other, what is the index of refraction of the glass? 13.15 .. Light of original intensity I0 passes through two ideal polarizing filters having their polarizing axes oriented as shown in Fig. E13.15. You want to adjust the angle f so that the intensity at point P is equal to I0 > 10. (a) If the original light is unpolarized, what should f be? (b) If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should f be? '

'

Figure E13.15 f

P

I0

13.16 .. A beam of polarized light passes through a polarizing filter. When the angle between the polarizing axis of the filter and the direction of polarization of the light is u, the intensity of the emerging beam is I. If you now want the intensity to be I > 2, what should be the angle (in terms of u) between the polarizing angle of the filter and the original direction of polarization of the light? 13.17 .. Unpolarized light of intensity 20.0 W > cm2 is incident on two polarizing filters. The axis of the first filter is at an angle of 25.0° counterclockwise from the vertical (viewed in the direction the light is traveling), and the axis of the second filter is at 62.0° counterclockwise from the vertical. What is the intensity of the light after it has passed through the second polarizer? 13.18 . Three Polarizing Filters. Three polarizing filters are stacked with the polarizing axes of the second and third at 45.0° and 90.0°, respectively, with that of the first. (a) If unpolarized light of intensity I0 is incident on the stack, find the intensity and state of polarization of light emerging from each filter. (b) If the second filter is removed, what is the intensity of the light emerging from each remaining filter? '

'

'

'

PROBLEMS 13.19 . A light beam is directed parallel to the axis of a hollow cylindrical tube. When the tube contains only air, it takes the light 8.72 ns to travel the length of the tube, but when the tube is filled with a transparent jelly, it takes the light 2.04 ns longer to travel its length. What is the refractive index of this jelly? 13.20 ... In a physics lab, light with wavelength 490 nm travels in air from a laser to a photocell in 17.0 ns. When a slab of glass 0.840 m thick is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light 21.2 ns to travel from the laser to the photocell. What is the wavelength of the light in the glass? 13.21 ... A glass plate 2.50 mm thick, with an index of refraction of 1.40, is placed between a point source of light with wavelength 540 nm (in vacuum) and a screen. The distance from source to screen is 1.80 cm. How many wavelengths are there between the source and the screen? 13.22 .. Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. This emulsion was somewhat transparent. When a bright point source is focused on the front of the plate, the developed photograph will show a halo around the image of

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Problems

the spot. If the glass plate is 1.54 mm thick and the halos have an inner radius of 5.34 mm, what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion. Some of it is then totally reflected at the back surface of the plate and returns to the front surface.) 13.23 .. A 45° - 45° - 90° prism is immersed in water. A ray of light is incident normally on one of its shorter faces. What is the minimum index of refraction that the prism must have if this ray is to be totally reflected within the glass at the long face of the prism? 13.24 . A thin layer of ice 1n = 1.3092 floats on the surface of water 1n =1.662 in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice–water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts? 13.25 .. Light is incident normally on the short face of a 30°60°- 90° prism (Fig. P13.25). A drop of liquid is placed on the Figure P13.25 hypotenuse of the prism. If the 60° 30° index of refraction of the prism is 1.62, find the maximum index that the liquid may have if the 90° light is to be totally reflected. 13.26 .. The prism shown in Fig. P13.26 has a refractive index of 1.66, and Figure P13.26 the angles A are 25°. Two light rays m and n are parallel as they enter the prism. What is the angle between them after they emerge? 13.27 .. When the sun is either rising or setting and appears to be just on the horizon, m A it is in fact below the horizon. The explanation for this seeming paradox is that light from the sun bends slightly when entering A n the earth’s atmosphere, as shown in Fig. P13.27. Since our perception is based on the idea that light travels in straight lines, we perceive the light to be coming from an apparent position that is an angle d above the sun’s true position. (a) Make the simplifying assumptions that the atmosphere has uniform density, and hence uniform index of refraction n, and extends to a height h above the earth’s surface, at which point it abruptly stops. Show that the angle d is given by '

13.28 .. Light is incident in air at an angle ua (Fig. P13.28) on the upper surface of a transparent plate, the surfaces of the plate being plane and parallel to each other. (a) Prove that ua = u¿a. (b) Show that this is true for any number of different parallel plates. (c) Prove that the lateral displacement d of the emergent beam is given by the relationship

'

d = t

Figure P13.28 ua

'

d = arcsina

n

t

u9b

n9

P

d u9a

13.29 .. Angle of Deviation. The incident angle ua shown in Fig. P13.29 is chosen so that the light passes symmetrically through the prism, which has refractive index n and apex angle A. (a) Show that the angle of deviation d (the angle between the initial and final directions of the ray) is given by sin

A + d A = n sin 2 2

(When the light passes through symmetrically, as shown, the angle of deviation is a minimum.) (b) Use the result of part (a) to find the angle of deviation for a ray of light passing symmetrically through a prism having three equal angles 1A = 60.0°2 and n = 1.52. (c) A certain glass has a refractive index of 1.61 for red light (700 nm) and 1.66 for violet light (400 nm). If both colors pass through symmetrically, as described in part (a), and if A = 60.0°, find the difference between the angles of deviation for the two colors. Figure P13.29

R nR b - arcsina b R + h R + h

Figure P13.27 Apparent position d of the sun From the sun R1h

A A 2 2 d ua

ua ub

Atmosphere

ub

n

13.30 .. A certain birefringent material has indexes of refraction n1 and n2 for the two perpendicular components of linearly polarized light passing through it. The corresponding wavelengths are l1 = l0 > n1 and l0 > n2, where l0 is the wavelength in vacuum. (a) If the crystal is to function as a quarter-wave plate, the number of wavelengths of each component within the material must differ by 1 4 . Show that the minimum thickness for a quarter-wave plate is l0 d = 41n1 - n22 '

'

'

'

'

Earth

Q

ub

n

where R = 6378 km is the radius of the earth. (b) Calculate d using n = 1.0003 and h = 20 km. How does this compare to the angular radius of the sun, which is about one quarter of a degree? (In actuality a light ray from the sun bends gradually, not abruptly, since the density and refractive index of the atmosphere change gradually with altitude.)

h

sin1ua - ub ¿2 cos ub ¿

where t is the thickness of the plate. (d) A ray of light is incident at an angle of 66.0° on one surface of a glass plate 2.40 cm thick with an index of refraction of 1.80. The medium on either side of the plate is air. Find the lateral displacement between the incident and emergent rays.

'

R

441

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442

CHAPTER 13 The Nature and Propagation of Light

(b) Find the minimum thickness of a quarter-wave plate made of siderite 1FeO # CO22 if the indexes of refraction are n1 = 1.875 and n2 = 1.635 and the wavelength in vacuum is l0 = 589 nm. y 13.31 A plane mirror of length 2m is kept along the line y = - x as shown in the figl = 2m ure. An insect having velocity of 4 iN cm/s is moving along the x-axis from far away. Find the time span for which the insect can see its x O V image. 13.32 The mirror of length 2l makes 10 revolutions per minute about the axis crossing its midpoint O and perpendicular to the plane of the figure. There is a light source in point A and an observer in point B of the circle of radius R drawn around centre O(AOB = 90°). A (a) Along what curve does the virtual image of the point-like light source A move? j (b) At what speed does the virtual image of R A move? l B O (c) What is the ratio l/R if the observer B l first sees the light source when the ange of the mirror is w = 15°? 13.33 A large Lucite cube (n = 1.5) has a ? small air bubble (a defect in the casting process) d below one surface. When a rupee coin (diameter 2 cm) is placed directly over the bubble on the outside of the cube, the bubble cannot be seen by looking down into the cube at any angle. However, when a 50 paise coin (diameter 1.5 cm) is placed directly over it, the bubble can be seen by looking down into the cube. What is the range of the possible depths d of

the air bubble beneath the surface? y 13.34 The cross-section of a prism is shown in the figure. One of its refracting C B surfaces (AC) is given by y = x2. A ray µ of light travelling parallel to x-axis is y = x2 incident normally on face AB and h x A refracted. Find the minimum distance of the incident ray from point A to get refracted ray from surface AC.

CHALLENGE PROBLEMS 13.35 An opaque sphere of radius R lies on a horizontal plane. On the perpendicular through the point of contact there is a point source of light a distance R above the sphere. (a) Show that the area of the shadow on the R plane is 3pR2. (b) A transparent liquid of refractive index R Ë3 is filled above the plane such that the sphere is just covered with the liquid. Show that the area of shadow now become 2pR2. 13.36 A circular disc of diameter d lies horizontally inside a metallic hemisphere bowl radius a. The disc is just visible to an eye looking over the edge. The bowl is now filled with a liquid of refractive index m. Now, the whole of the disc is just visible to the eye in the (m2 - 1) same position. Show that d = 2a 2 . (m + 1)

Answers Chapter Opening Question

?

the fish to you (though in the opposite direction). 13.3 Answers: (i), (ii) Total internal reflection can occur only if two conditions are met: nb must be less than na , and the critical angle ucrit (where sin ucrit = nb>na2 must be smaller than the angle of incidence ua . In the first two cases both conditions are met: The critical angles are (i) ucrit = sin-111>1.332 = 48.8° and (ii) both of which are smaller than In the third case is greater than so total internal reflection cannot occur for any incident angle. 13.5 Answer: (ii) The sunlight reflected from the windows of the high-rise building is partially polarized in the vertical direction, since each window lies in a vertical plane. The Polaroid filter in front of the lens is oriented with its polarizing axis perpendicular to the dominant direction of polarization of the reflected light. 13.7 Answer: (ii) Huygens’s principle applies to waves of all kinds, including sound waves. Hence this situation is exactly like that shown in Fig. 13.36, with material a representing the warm air, material representing the cold air in which the waves travel more slowly, and the interface between the materials representing the weather front. North is toward the top of the figure and east is toward the right, so Fig. 13.36 shows that the rays (which indicate the direction of propagation) deflect toward the east. '

This is the same effect as shown in Fig. 13.31. The drafting tools are placed between two polarizing filters whose polarizing axes are perpendicular. In places where the clear plastic is under stress, the plastic becomes birefringent; that is, light travels through it at a speed that depends on its polarization. The result is that the light that emerges from the plastic has a different polarization than the light that enters. A spot on the plastic appears bright if the emerging light has the same polarization as the second polarizing filter. The amount of birefringence depends on the wavelength of the light as well as the amount of stress on the plastic, so different colors are seen at different locations on the plastic.

Test Your Understanding Questions 13.1 Answer: (iii) The waves go farther in the y-direction in a given amount of time than in the other directions, so the wave fronts are elongated in the y-direction. 13.2 Answers: (a) (ii), (b) (iii) As shown in the figure, light rays coming from the fish bend away from the normal when they pass from the water 1n = 1.332 into the air 1n = 1.002. As a result, the fish appears to be higher in the water than it actually is. Hence you should aim a spear below the apparent position of the fish. If you use a laser beam, you should aim at the apparent position of the fish: The beam of laser light takes the same path from you to the fish as ordinary light takes from

'

Bridging Problem

Answer: 1.93 * 108 m> s

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Answers

D ISCUSSION Q UESTIONS

E XERCISES

Q13.1 Not appreciable delay due to atmosphere as distance is very small. Q13.2 As ray enter atmosphere air is more denser near surface of earth ray sufferes refraction. Due to refraction position of starts is apparently different. Q13.3 As the source is common frequency can not change. But velocity changes when light travels from one medium to another hence wavelength will change. Q13.4 Due to bending of light sun will be visible for a small time ever after it goes below the horizon. The statement is correct as in morning sun rise will be earlier. Q13.5 Due to refraction of light. Q13.6 Each experiment can help us find m of glass using which we can find speed of light in glass. Q13.7 Images formed due to reflection from 1st surfaces and reflection from 2nd surface. Q13.8 Due to image formation by partial reflection at interface. Q13.9 Light ray will not suffer total internal reflection. Q13.10 In case of reflection medium remains same therefore it does not depends upon refractive index. Q13.11 Just hold two glasses one on other and now rotate one keeping second fix intensity of light will varies with angle if glasses have polaloid. Q13.12 For polarisation wave must be a transverse wave. Q13.13 Pass PPL through polariser when axis will become perpendicular to plane of PPL intensity will become zero. Q13.14 Yes, it will reduce glare but it will also decrease intensity of light which will create difficulty in low light condition. There can be many arrangements one of the arrangement can have axis tilted by 45° with the direction of light. Q13.15 On streching wrap also starts working as poloroid and due to this plane of PPL rotates hence light will pass through. Q13.16 When we look directly light is unpolarised therefore less reduction. While light reflecting from water surface is polarised and axis of polarised is kept perpendicular this PPL. Q13.17 Yes, as inserting polaroid will change the plane & hence there will be some intensity. Q13.18 Yes, it will change the component of electric field received by antenna. Q13.19 Scattered light is coming from molecules while transmitted light is coming from sun directly. Therefore light will have oscillation in all the plane. Q13.20 Feet must point towards west. Q13.21 Light is scattered many times by water droplets in the clouds. Q13.22 Due to Haze blue light is scattered most while red color scattered least. Q13.23 Because in evening dust particle & other gases liberated in atmosphere will scatter red light therefore intensity will decrease compared to morning time. Q13.24 As sound will move up it will turn away from normal and may suffer reflection which will increase the distance travelled in horizontal. Q13.25 Yes, Yes.

13.1 (a) 2.04 * 108 m/s, (b) 442 nm 13.2 (a) 517 nm, (b) 340 nm 13.3 1.75 4 t2 2 t2 13.4 (a) a b , (b) u = sin - 1 a b 3 3 nl nv 13.5 (a) f, nl, nv, (b) f, 1 , 1 n n 13.6 30° 3 13.7 u = sin - 1 a b 4

S INGLE C ORRECT Q13.1 (a) Q13.2 (d) Q13.3 (d) Q13.4 (c)

443

5 13.8 (a) cos- 1 a b 8 5 (b) cos- 1 a b 6

13.9 (a) ub = sin - 1 a0.7 b , (b) ub = sin - 1 a0.4 b 13.10 (a) 45° (b) 30° 13.11 (a) For air, (b) uc = 15.1° (c) The sound wave must be travelling in air (d) sound waves can be totally reflected from the surface of the water. 13.12 0.375 I0 13.13(a) IA = 2 IB = 0.125I0 IC = 0.0938 I0 (b) I = 0 13.14 1.77 13.15 (a) cos - 1 a

1 b or 63.4° t5 1 (b) cos - 1 a b or 71.6° t10 cosu 13.16 cos - 1 a b t2 13.17 I = 6.38 w/m2 13.18 (a) 0.125I0, light is linearly polarized along the axis of the third polarizer (b) I = 0, No light is passed 13.19 1.23 13.20 l = 196 nm 13.21 3.52 * 104 13.22 1.16 13.23 1.89 13.24 (a) 37°, (b) 37° 13.25 1.4 13.26 39° 13.27 (b) d = 0.23° Calculated d is about the same as the angular radius of the sun 13.28 (d) 1.62 cm 13.29 (b) 38.9°, (c) 5° 13.30 (b) 6.14 * 10 - 7m 13.31 50Ë2 sec 2pR 1 13.32 (a) circle, (b) m/s, (c) 3 Ë2 13.33 dmax = 13.34 h =

3Ë5 Ë5 ,d = 2 min 8

m2 - 1 4

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14

GEOMETRIC OPTICS

LEARNING GOALS By studying this chapter, you will learn: • How a plane mirror forms an image. • Why concave and convex mirrors form different kinds of image. • How images can be formed by a curved interface between two transparent materials. • What aspects of a lens determine the type of image that it produces.

?

How do magnifying lenses work? At what distance from the object being examined do they provide the sharpest view?

• What determines the field of view of a camera lens. • What causes various defects in human vision, and how they can be corrected. • The principle of the simple magnifier. • How microscopes and telescopes work.

our reflection in the bathroom mirror, the view of the moon through a telescope, the patterns seen in a kaleidoscope—all of these are examples of images. In each case the object that you’re looking at appears to be in a different place than its actual position: Your reflection is on the other side of the mirror, the moon appears to be much closer when seen through a telescope, and objects seen in a kaleidoscope seem to be in many places at the same time. In each case, light rays that come from a point on an object are deflected by reflection or refraction (or a combination of the two), so they converge toward or appear to diverge from a point called an image point. Our goal in this chapter is to see how this is done and to explore the different kinds of images that can be made with simple optical devices. To understand images and image formation, all we need are the ray model of light, the laws of reflection and refraction, and some simple geometry and trigonometry. The key role played by geometry in our analysis is the reason for the name geometric optics that is given to the study of how light rays form images. We’ll begin our analysis with one of the simplest image-forming optical devices, a plane mirror. We’ll go on to study how images are formed by curved mirrors, by refracting surfaces, and by thin lenses. Our results will lay the foundation for understanding many familiar optical instruments, including camera lenses, magnifiers, the human eye, microscopes, and telescopes.

Y

14.1 ActivPhysics 15.4: Geometric Optics: Plane Mirrors

Reflection and Refraction at a Plane Surface

Before discussing what is meant by an image, we first need the concept of object as it is used in optics. By an object we mean anything from which light rays radiate. This light could be emitted by the object itself if it is self-luminous, like the glowing filament of a light bulb. Alternatively, the light could be emitted by 444

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445

14.1 Reflection and Refraction at a Plane Surface

another source (such as a lamp or the sun) and then reflected from the object; an example is the light you see coming from the pages of this book. Figure 14.1 shows light rays radiating in all directions from an object at a point P. For an observer to see this object directly, there must be no obstruction between the object and the observer’s eyes. Note that light rays from the object reach the observer’s left and right eyes at different angles; these differences are processed by the observer’s brain to infer the distance from the observer to the object. The object in Fig. 14.1 is a point object that has no physical extent. Real objects with length, width, and height are called extended objects. To start with, we’ll consider only an idealized point object, since we can always think of an extended object as being made up of a very large number of point objects. Suppose some of the rays from the object strike a smooth, plane reflecting surface (Fig. 14.2). This could be the surface of a material with a different index of refraction, which reflects part of the incident light, or a polished metal surface that reflects almost 100% of the light that strikes it. We will always draw the reflecting surface as a black line with a shaded area behind it, as in Fig. 14.2. Bathroom mirrors have a thin sheet of glass that lies in front of and protects the reflecting surface; we’ll ignore the effects of this thin sheet. According to the law of reflection, all rays striking the surface are reflected at an angle from the normal equal to the angle of incidence. Since the surface is plane, the normal is in the same direction at all points on the surface, and we have specular reflection. After the rays are reflected, their directions are the same as though they had come from point P¿. We call point P an object point and point P¿ the corresponding image point, and we say that the reflecting surface forms an image of point P. An observer who can see only the rays reflected from the surface, and who doesn’t know that he’s seeing a reflection, thinks that the rays originate from the image point P¿. The image point is therefore a convenient way to describe the directions of the various reflected rays, just as the object point P describes the directions of the rays arriving at the surface before reflection. If the surface in Fig. 14.2 were not smooth, the reflection would be diffuse, and rays reflected from different parts of the surface would go in uncorrelated directions (see Fig. 33.6b). In this case there would not be a definite image point P¿ from which all reflected rays seem to emanate. You can’t see your reflection in the surface of a tarnished piece of metal because its surface is rough; polishing the metal smoothes the surface so that specular reflection occurs and a reflected image becomes visible. An image is also formed by a plane refracting surface, as shown in Fig. 14.3. Rays coming from point P are refracted at the interface between two optical materials. When the angles of incidence are small, the final directions of the rays after refraction are the same as though they had come from point P¿, as shown, and again we call P¿ an image point. In Section 33.2 we described how this effect makes underwater objects appear closer to the surface than they really are (see Fig. 33.9). In both Figs. 14.2 and 14.3 the rays do not actually pass through the image point P¿. Indeed, if the mirror in Fig. 14.2 is opaque, there is no light at all on its right side. If the outgoing rays don’t actually pass through the image point, we call the image a virtual image. Later we will see cases in which the outgoing rays really do pass through an image point, and we will call the resulting image a real image. The images that are formed on a projection screen, on the photographic film in a camera, and on the retina of your eye are real images.

Image Formation by a Plane Mirror Let’s concentrate for now on images produced by reflection; we’ll return to refraction later in the chapter. To find the precise location of the virtual image P¿ that a plane mirror forms of an object at P, we use the construction shown in Fig. 14.4. The figure shows two rays diverging from an object point P at a

14.1 Light rays radiate from a point object P in all directions.

P

14.2 Light rays from the object at point P are reflected from a plane mirror. The reflected rays entering the eye look as though they had come from image point P¿.

P

P9

Image point: apparent source of reflected rays

Object point: source of rays Plane mirror

14.3 Light rays from the object at point P are refracted at the plane interface. The refracted rays entering the eye look as though they had come from image point P¿. When na . nb, P9 is closer to the surface than P; for na , nb, the reverse is true. n a . nb

nb

P P9

Object point: source of rays

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Image point: apparent source of refracted rays

446

CHAPTER 14 Geometric Optics

14.4 Construction for determining the location of the image formed by a plane mirror. The image point P¿ is as far behind the mirror as the object point P is in front of it. After reflection, all rays originating at P diverge from P. Because the u B rays do not actually pass through P, the image u is virtual. h u u P P V |u| |v| Object distance Image distance Triangles PVB and PVB are congruent, so 0 u0 5 0v 0 .

14.5 Construction for determining the height of an image formed by reflection at a plane reflecting surface. For a plane mirror, PQV and PQV are congruent, so y 5 y and the object and image are the same size (the lateral magnification is 1). Object

Image V

Q y P

u u u |u|

Q y

u V

P |v|

distance |u| to the left of a plane mirror. We call |u| the object distance. As per sign rules of section 14.2, u may be (-ive) also that is why we take distance of object from V as |u| and same case for |v|. The ray PV is incident normally on the mirror (that is, it is perpendicular to the mirror surface), and it returns along its original path. The ray PB makes an angle u with PV. It strikes the mirror at an angle of incidence u and is reflected at an equal angle with the normal. When we extend the two reflected rays backward, they intersect at point P¿, at a distance |v| behind the mirror. We call |v| the image distance. The line between P and P¿ is perpendicular to the mirror. The two triangles PVB and P¿VB are congruent, so P and P¿ are at equal distances from the mirror, and |u| and |v| have equal magnitudes. The image point P¿ is located exactly opposite the object point P as far behind the mirror as the object point is from the front of the mirror. We can repeat the construction of Fig. 14.4 for each ray diverging from P. The directions of all the outgoing reflected rays are the same as though they had originated at point P¿, confirming that P¿ is the image of P. No matter where the observer is located, she will always see the image at the point P¿.

Image of an Extended Object: Plane Mirror Next we consider an extended object with finite size. For simplicity we often consider an object that has only one dimension, like a slender arrow, oriented parallel to the reflecting surface; an example is the arrow PQ in Fig. 14.5. The distance from the head to the tail of an arrow oriented in this way is called its height; in Fig. 14.5 the height is y. The image formed by such an extended object is an extended image; to each point on the object, there corresponds a point on the image. Two of the rays from Q are shown; all the rays from Q appear to diverge from its image point Q¿ after reflection. The image of the arrow is the line P¿Q¿, with height y¿. Other points of the object PQ have image points between P¿ and Q¿. The triangles PQV and P¿Q¿V are congruent, so the object PQ and image P¿Q¿ have the same size and orientation, and y = y¿. The ratio of image height to object height, y¿>y, in any image-forming situation is called the lateral magnification m; that is,

m =

14.6 The image formed by a plane mirror is virtual, erect, and reversed. It is the same size as the object. An image made by a plane mirror is reversed back to front: the image thumb P9R9 and object thumb PR point in opposite directions (toward each other). Q9 S9 R9

P9

Q S P

Object

R

y¿ y

(lateral magnification)

(14.1)

Thus for a plane mirror the lateral magnification m is unity. When you look at yourself in a plane mirror, your image is the same size as the real you. In Fig. 14.5 the image arrow points in the same direction as the object arrow; we say that the image is erect. In this case, y and y¿ have the same sign, and the lateral magnification m is positive. The image formed by a plane mirror is always erect, so y and y¿ have both the same magnitude and the same sign; from Eq. (14.1) the lateral magnification of a plane mirror is always m = + 1. Later we will encounter situations in which the image is inverted; that is, the image arrow points in the direction opposite to that of the object arrow. For an inverted image, y and y¿ have opposite signs, and the lateral magnification m is negative. The object in Fig. 14.5 has only one dimension. Figure 14.6 shows a threedimensional object and its three-dimensional virtual image formed by a plane mirror. The object and image are related in the same way as a left hand and a right hand.

Image

CAUTION Reflections in a plane mirror At this point, you may be asking, “Why does a plane mirror reverse images left and right but not top and bottom?” This question is quite misleading! As Fig. 14.6 shows, the up-down image P¿Q¿ and the left-right image P¿S¿ are parallel to their objects and are not reversed at all. Only the front-back image P¿R¿ is reversed relative to PR. Hence it’s most correct to say that a plane mirror reverses back to

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14.2 Reflection at a Spherical Surface

HAND

The reversed image of a three-dimensional object formed by a plane mirror is the same size as the object in all its dimensions. When the transverse dimensions of object and image are in the same direction, the image is erect. Thus a plane mirror always forms an erect but reversed image. Figure 14.7 illustrates this point. An important property of all images formed by reflecting or refracting surfaces is that an image formed by one surface or optical device can serve as the object for a second surface or device. Figure 14.8 shows a simple example. Mirror 1 forms an image P1œ of the object point P, and mirror 2 forms another image P2œ , each in the way we have just discussed. But in addition, the image P1œ formed by mirror 1 serves an object for mirror 2, which then forms an image of this object at point P3œ as shown. Similarly, mirror 1 uses the image P2œ formed by mirror 2 as an object and forms an image of it. We leave it to you to show that this image point is also at P3œ . The idea that an image formed by one device can act as the object for a second device is of great importance in geometric optics. We will use it later in this chapter to locate the image formed by two successive curved-surface refractions in a lens. This idea will help us to understand image formation by combinations of lenses, as in a microscope or a refracting telescope.

14.7 The image formed by a plane mirror is reversed; the image of a right hand is a left hand, and so on. (The hand is resting on a horizontal mirror.) Are images of the letters H and A reversed?

DNAH

front. To verify this object-image relationship, point your thumbs along PR and P¿R¿, your forefingers along PQ and P¿Q¿, and your middle fingers along PS and P¿S¿. When an object and its image are related in this way, the image is said to be reversed; this means that only the front-back dimension is reversed. y

447

Mirror

'

'

14.8 Images P1œ and P2œ are formed by a single reflection of each ray from the object at P. Image P3œ , located by treating either of the other images as an object, is formed by a double reflection of each ray. '

Image of object P formed by mirror 1 Mirror 1

Test Your Understanding of Section 14.1 If you walk directly toward a plane mirror at a speed v, at what speed does your image approach you? (i) slower than v; (ii) v; (iii) faster than v but slower than 2v; (iv) 2v; (v) faster than 2v.

Image of image P19 formed by mirror 2

P19

P39

P

P29

y Image of object P formed by mirror 2

14.2

Reflection at a Spherical Surface

A plane mirror produces an image that is the same size as the object. But there are many applications for mirrors in which the image and object must be of different sizes. A magnifying mirror used when applying makeup gives an image that is larger than the object, and surveillance mirrors (used in stores to help spot shoplifters) give an image that is smaller than the object. There are also applications of mirrors in which a real image is desired, so light rays do indeed pass through the image point P¿. A plane mirror by itself cannot perform any of these tasks. Instead, curved mirrors are used.

Image of a Point Object: Spherical Mirror We’ll consider the special (and easily analyzed) case of image formation by a spherical mirror. Figure 14.9a shows a spherical mirror with magnitude of radius of curvature |R|, with its concave side facing the incident light. Note that R may be − ive also as per sign rules discussed below that is why we have taken |R| = cV. The center of curvature of the surface (the center of the sphere of which the surface is a part) is at C, and the vertex of the mirror (the center of the mirror surface) is at V. The line CV is called the optic axis. Point P is an object point that lies on the optic axis; for the moment, we assume that the distance from P to V is greater than R.

Sign Rules We take optic axis as x-axis and vertex V as origin. Now the direction in which the rays are incident on the mirror before reflection is taken as the direction of +ive x-axis and the opposite direction is taken as direction of −ve x-axis

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Mirror 2

448

CHAPTER 14 Geometric Optics

14.9 (a) A concave spherical mirror forms a real image of a point object P on the mirror’s optic axis. (b) The eye sees some of the outgoing rays and perceives them as having come from P¿. (a) Construction for finding the position P of an image formed by a concave spherical mirror For a spherical mirror, a 1 b 5 2f. Point u u object R b f a

h V

P

C

P Center of curvature

B

d

Optic axis

Vertex

u and v are both negative.

|v| |u|

(b) The paraxial approximation, which holds for rays with small a

P

Virtual Object

V

P

The x-coordinate of object-point (P) is termed as ‘u’. The x-coordinate of image-point (P¿) is termed as ‘v’. The x-coordinate of centre of curvature (C) is temed as R or radius of curvature. The x-coordinate of focal-point or focus is termed as f or focal length. Focalpoint will be explained later. In figure 14.9 (a), coordinates of object-point P = (- VP, 0) so, u = - VP . And distance (always positive) of object-point P from vertex V = VP = - u . Similarly, coordinates of image point P¿ = ( - VP¿, 0) so, v = - VP′ And distance of image point P¿ from vertex V = VP′ = - v Coordinates of centre of curvature C = (- VC,0) so, R = - VC and distance of centre of curvature = VC = - R Note that VP, VP¿, VC are distances and always +ive. In the above shown figure 14.9(b), direction of incident rays is leftwards. P is the image of object P. We have u = - VP, v = - VP, R = - VC In the above shown figure 14.9(c), P¿ is the image of object P. Direction of incident ray is rightwards. We have u = - VP v = + VP¿ R = - VC Important: For length of objects and images perpendicular to optic axis, upward direction is +y and downward -ive

In figure 14.9(d), incident rays on the concave mirror would have converged on P, if the mirror had not been there. So P is a virtual object. The rays meet at P¿ after reflection. u = + VP (Note) v = - VP¿

All rays from P that have a small angle a pass through P, forming a real image.

Virtual Image

14.9 (b) + x axis

−x axis

V

P'

C

P

14.9 (c) −x

C

+x

P

V

P'

In figure 14.9(d), after reflection, rays meet at P¿ . So P¿ is real image. But, if object P is placed in front of a plane mirror, reflected rays appear to emerge from P¿ , as shown in figure 14.9(e). So P¿ is a virtual image. In figure 14.10(f) P¿ is the virtual image of virtual object P. u = + VP (Note) v = + VP¿ Ray PV, passing through C, strikes the mirror normally and is reflected back on itself. Ray PB, at an angle a with the axis, strikes the mirror at B, where the angles of incidence and reflection are u. The reflected ray intersects the axis at point P¿. We will show shortly that all rays from P intersect the axis at the same point P¿, as in Fig. 14.9b, provided that the angle a is small. Point P¿ is therefore the image of object point P. Unlike the reflected rays in Fig. 14.1, the reflected rays in Fig. 14.9b actually do intersect at point P¿, then diverge from P¿ as if they had originated at this point. Thus P¿ is a real image. To see the usefulness of having a real image, suppose that the mirror is in a darkened room in which the only source of light is a self-luminous object at P. If you place a small piece of photographic film at P¿, all the rays of light coming from point P that reflect off the mirror will strike the same point P¿ on the film; when developed, the film will show a single bright spot, representing a sharply focused image of the object at point P. This principle is at the heart of most astronomical telescopes, which use large concave mirrors to make photographs of celestial objects. With a plane mirror like that in Fig. 14.2, placing a piece of film at the image point P¿ would be a waste of time; the light rays never actually pass

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449

14.2 Reflection at a Spherical Surface

through the image point, and the image can’t be recorded on film. Real images are essential for photography. Let’s now find the location of the real image point P¿ in Fig. 14.9a and prove the assertion that all rays from P intersect at P¿ (provided that their angle with the optic axis is small). The distance of object, measured from the vertex V, is |u|; the distance of image, also measured from V, is |v|. The signs of u, v, and the radius of curvature R are determined by the sign rules. The object point P is on the same side as the incident light, so according to sign rules, u is negative. The image point P¿ is on the same side as the reflected light, so according to sign rules, v is also negative. The center of curvature C is on the same side as the reflected light, so according to sign rules, R, too, is negative; R is always negative when reflection occurs at the concave side of a surface (Fig. 14.10). We now use the following theorem from plane geometry: An exterior angle of a triangle equals the sum of the two opposite interior angles. Applying this theorem to triangles PBC and P¿BC in Fig. 14.9a, we have f = a + u

14.9 (d)

C

V

P'

P

14.9 (e)

b = f + u P

P'

Eliminating u between these equations gives a + b = 2f

(14.2)

We may now compute the image distance v. Let h represent the height of point B above the optic axis, and let d represent the short distance from V to the foot of this vertical line. We now write expressions for the tangents of a, b, and f, remembering that u, y, and R are all negative quantities: |u| = - u , |v| = - v, h h h |R| = - R ; tan a = tan b = tan f = - u - d - v - d - R - d These trigonometric equations cannot be solved as simply as the corresponding algebraic equations for a plane mirror. However, if the angle a is small, the angles b and f are also small. The tangent of an angle that is much less than one radian is nearly equal to the angle itself (measured in radians), so we can replace tan a by a, and so on, in the equations above. Also, if a is small, we can neglect the distance d compared with |v|, |u| and )R). So for small angles we have the following approximate relationships: a =

h - u

b =

h - v

f =

h - R

Substituting these into Eq. (14.2) and dividing by h, we obtain a general relationship among u, v, and R: 1 2 1 + = v u R

(object–image relationship, spherical mirror)

14.9 (f)

P' V

P

14.10 The sign rule for the radius of a spherical mirror.

P C Outgoing

|R|

Center of curvature on same side as outgoing light: R is negative.

(14.3)

This equation does not contain the angle a. Hence all rays from P that make sufficiently small angles with the axis intersect at P¿ after they are reflected; this verifies our earlier assertion. Such rays, nearly parallel to the axis and close to it, are called paraxial rays. (The term paraxial approximation is often used for the approximations we have just described.) Since all such reflected light rays converge on the image point, a concave mirror is also called a converging mirror. Be sure you understand that Eq. (14.3), as well as many similar relationships that we will derive later in this chapter and the next, is only approximately correct. It results from a calculation containing approximations, and it is valid only for paraxial rays. If we increase the angle a that a ray makes with the optic axis, the point P¿ where the ray intersects the optic axis moves somewhat closer to the vertex than for a paraxial ray. As a result, a spherical mirror, unlike a plane mirror, does not form a precise point image of a point object; the image is “smeared out.” This property of a spherical mirror is called spherical aberration. When

P

Outgoing

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|R| C Center of curvature not on same side as outgoing light: R is positive.

450

CHAPTER 14 Geometric Optics

14.11 (a), (b) Soon after the Hubble Space Telescope (HST) was placed in orbit in 1990, it was discovered that the concave primary mirror (also called the objective 1 mirror) was too shallow by about 50 the width of a human hair, leading to spherical aberration of the star’s image. (c) After corrective optics were installed in 1993, the effects of spherical aberration were almost completely eliminated. (a) The 2.4-m-diameter primary mirror of the Hubble Space Telescope

the primary mirror of the Hubble Space Telescope (Fig. 14.11a) was manufactured, tiny errors were made in its shape that led to an unacceptable amount of spherical aberration (Fig. 14.11b). The performance of the telescope improved dramatically after the installation of corrective optics (Fig. 14.11c). If the radius of curvature becomes infinite 1R = q 2, the mirror becomes plane, and Eq. (14.3) satisfies the reflection from a plane surface.

Focal Point and Focal Length

When the object point P is very far from the spherical mirror 1s = q 2, the incoming rays are parallel. (The star shown in Fig. 14.11c is an example of such a distant object.) From Eq. (14.3) the image distance y in this case is given by 1 2 1 + = ` v R

(b) A star seen with the original mirror

v =

R 2

The situation is shown in Fig. 14.12a. The beam of incident parallel rays converges, after reflection from the mirror, to a point F at a distance |R|/2 from the vertex of the mirror. The point F at which the incident parallel rays converge is called the focal point; we say that these rays are brought to a focus. The x coordinate of focal point F is called focal length. f when optic axis is taken as x axis, vertex of the mirror is taken as origin and direction of incident ray is taken as that of +ive x-axis. We see that ƒ is related to the radius of curvature R by ƒ =

R 2

(focal length of a spherical mirror)

(14.5)

(Note that f and R are both -ive here) The opposite situation is shown in Fig. 14.12b. Now the object is placed at the focal point F, so the object distance is v = ƒ = R>2. The image distance y is again given by Eq. (14.3): 2 2 1 = + v R R

(c) The same star with corrective optics

1 = 0 v

v = q

With the object at the focal point, the reflected rays in Fig. 14.12b are parallel to the optic axis; they meet only at a point infinitely far from the mirror, so the image is at infinity. Thus the focal point F of a spherical mirror has the properties that (1) any incoming ray parallel to the optic axis is reflected through the focal point and (2) any incoming ray that passes through the focal point is reflected parallel to the optic axis. For spherical mirrors these statements are true only for paraxial rays. For parabolic mirrors these statements are exactly true; this is why parabolic mirrors are preferred for astronomical telescopes. Spherical or parabolic mirrors are used in flashlights and headlights to form the light from the bulb into a parallel beam. Some solar-power plants use an array of plane mirrors to simulate an approximately spherical concave mirror; light from the sun is collected by the mirrors and directed to the focal point, where a steam boiler is placed. (The concepts of focal point and focal length also apply to lenses, as we’ll see in Section 14.4.) We will usually express the relationship between object and image distances for a mirror, Eq. (14.4), in terms of the focal length f : 1 1 1 + = v u ƒ

(object–image relationship, spherical mirror)

(14.5)

Image of an Extended Object: Spherical Mirror Now suppose we have an object with finite size, represented by the arrow PQ in Fig. 14.13, perpendicular to the optic axis CV. The image of P formed by parax-

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14.2 Reflection at a Spherical Surface

ial rays is at P¿. The object distance for point Q is very nearly equal to that for point P, so the image P¿Q¿ is nearly straight and perpendicular to the axis. Note that the object and image arrows have different sizes, y and y¿, respectively, and that they have opposite orientation. In Eq. (14.1) we defined the lateral magnification m as the ratio of image size y¿ to object size y: m =

y¿ y

Because triangles PVQ and P¿VQ¿ in Fig. 14.13 are similar, we also have the relationship y>|u| = -y¿>|v|. The negative sign is needed because object and image are on opposite sides of the optic axis; if y is positive, y¿ is negative. Therefore )v) y¿ v m = = = (as v and u are both negative) y u )u) (lateral magnification, spherical mirror)

ActivPhysics 15.5: Spherical Mirrors: Ray Diagrams ActivPhysics 15.6: Spherical Mirrors: The Mirror Equation ActivPhysics 15.7: Spherical Mirrors: Linear Magnification ActivPhysics 15.8: Spherical Mirrors: Problems

14.12 The focal point and focal length of a concave mirror. (a) All parallel rays incident on a spherical mirror reflect through the focal point.

(14.6) ƒRƒ Focal point

If m is positive, the image is erect in comparison to the object; if m is negative, the image is inverted relative to the object, as in Fig. 14.13.

F

C

CAUTION Lateral magnification can be less than 1 Although the ratio of image size to object size is called the lateral magnification, the image formed by a mirror or lens may be larger than, smaller than, or the same size as the object. If it is smaller, then the lateral magnification is less than unity in absolute value: ƒ m ƒ 6 1. The image formed by an astronomical telescope mirror or a camera lens is usually much smaller than the object. For example, the image of the bright star shown in Fig. 14.11c is just a few millimeters across, while the star itself is hundreds of thousands of kilometers in diameter. y

In our discussion of concave mirrors we have so far considered only objects that lie outside or at the focal point, so that the object distance |u| is greater than or equal to the magnitude of focal length ƒ. In this case the image point is on the same side of the mirror as the outgoing rays, and the image is real and inverted. If an object is placed inside the focal point of a concave mirror, so that |ƒ| 6 |u|, the resulting image is virtual (that is, the image point is on the opposite side of the mirror from the object), erect, and larger than the object. Mirrors used when applying makeup (referred to at the beginning of this section) are concave mirrors; in use, the distance from the face to the mirror is less than the focal length, and an enlarged, erect image is seen. You can prove these statements about concave mirrors by applying Eqs. (14.5) and (14.6) (see Exercise 14.11). We’ll also be able to verify these results later in this section, after we’ve learned some graphical methods for relating the positions and sizes of the object and the image. The beige and blue triangles are similar, so the v v lateral magnification is m 5 y y 5

/

Q

P

C

P y Image

u u

ƒvƒ 5  Focal length

ƒuƒ 5

ƒRƒ 5f 2

(b) Rays diverging from the focal point reflect to form parallel outgoing rays. ƒRƒ

C

F Focal point

ƒvƒ 5  ƒuƒ 5

R 5f 2

14.13 Construction for determining the position, orientation, and height of an image formed by a concave spherical mirror.

v

The negative value of m means that the image is inverted.

y Object

451

V

Q

ƒvƒ ƒRƒ ƒuƒ

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452

CHAPTER 14 Geometric Optics

Example 14.1

Image formation by a concave mirror I

A concave mirror forms an image, on a wall 3.00 m in front of the mirror, of a headlamp filament 10.0 cm in front of the mirror. (a) What are the radius of curvature and focal length of the mirror? (b) What is the lateral magnification? What is the image height if the object height is 5.00 mm?

EXECUTE: (a) Both the object and the image are on the concave (reflective) side of the mirror, so both u and v are positive; we have u = - 10 cm and (v = - 300 cm). We solve Eq. (14.3) for R: 1 1 2 + = - 10.0 cm - 300 cm R

SOLUTION

- R = 2a

IDENTIFY and SET UP: Figure 14.15 shows our sketch. Our target variables are the radius of curvature R, focal length ƒ, lateral magnification m, and image height y¿. We are given the distances from the mirror to the object |u| and from the mirror to the image (|v|). We solve Eq. (14.3) for R, and then use Eq. (14.4) to find ƒ. Equation (14.6) yields both m and y¿.

The focal length of the mirror is ƒ = R > 2 = - 9.7 cm. (b) From Eq. (14.7) the lateral magnification is '

m = -

'

v - 300 cm = - 30.0 = u - 10 cm

Because m is negative, the image is inverted. The height of the image is 30.0 times the height of the object, or 130.0215.00 mm2 = 150 mm.

14.14 Our sketch for this problem.

ƒuƒ ƒRƒ ƒvƒ

Conceptual Example 14.2

-1 1 1 + b = 19.4 cm 10.0 cm 300 cm

EVALUATE: Our sketch indicates that the image is inverted; our calculations agree. Note that the object (at u = - 10.0 cm) is just outside the focal point (ƒ = - 9.7 cm). This is very similar to what is done in automobile headlights. With the filament close to the focal point, the concave mirror produces a beam of nearly parallel rays.

Image formation by a concave mirror II

In Example 14.1, suppose that the lower half of the mirror’s reflecting surface is covered with nonreflective soot. What effect will this have on the image of the filament? SOLUTION It would be natural to guess that the image would now show only half of the filament. But in fact the image will still show the entire filament. You can see why by examining Fig. 14.9b. Light rays coming from any object point P are reflected from all parts of the mirror and converge on the corresponding image point P¿. If part

of the mirror surface is made nonreflective (or is removed altogether), rays from the remaining reflective surface still form an image of every part of the object. Reducing the reflecting area reduces the light energy reaching the image point, however: The image becomes dimmer. If the area is reduced by one-half, the image will be one-half as bright. Conversely, increasing the reflective area makes the image brighter. To make reasonably bright images of faint stars, astronomical telescopes use mirrors that are up to several meters in diameter (see Fig. 14.11a).

Convex Mirrors In Fig. 14.15a the convex side of a spherical mirror faces the incident light. According to sign rule in Section 14.2, R is positive. Ray PB is reflected, with the angles of incidence and reflection both equal to u. The reflected ray, projected backward, intersects the axis at P¿. As with a concave mirror, all rays from P that are reflected by the mirror diverge from the same point P¿, provided that the angle a is small. Therefore P¿ is the image of P. u is negative, v is positive, and the radius of curvature R is positive for a convex mirror. Figure 14.15b shows two rays diverging from the head of the arrow PQ and the virtual image P¿Q¿ of this arrow. The same procedure that we used for a concave mirror can be used to show that for a convex mirror, 2 1 1 = (with sign) + u v R

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14.2 Reflection at a Spherical Surface

453

14.15 Image formation by a convex mirror. (a) Construction for finding the position of an image formed by a convex mirror

(b) Construction for finding the magnification of an image formed by a convex mirror Q

Norma

l

u

As with a concave spherical mirror, v y V m5 y 52u Q

B R is negative.

u u

h

a

b

V

u

f

P

C

P' ƒvƒ = v

ƒuƒ

y

R P

P

V

u

u is (-ive); v is (+ive).

y

u

ƒuƒ

ƒvƒ = v

C u is (-ive); v is (+ive).

14.16 The focal point and focal length of a convex mirror. (b) Rays aimed at the virtual focal point are parallel to the axis after reflection.

(a) Paraxial rays incident on a convex spherical mirror diverge from a virtual focal point.

ƒRƒ = R

ƒRƒ = R

F

C

F

C

Virtual focal point

u =u= u5 

5

R R = =f 2 2

and the lateral magnification is m =

y¿ v = y u

These expressions are exactly the same as Eqs. (14.3) and (14.6) for a concave mirror. Thus when we use our sign rules consistently, Eqs. (14.3) and (14.6) are valid for both concave and convex mirrors. When R is positive (convex mirror), incoming rays that are parallel to the optic axis are not reflected through the focal point F. Instead, they diverge as though they had come from the point F at a distance |ƒ| = f behind the mirror, as shown in Fig. 14.16a. In this case, ƒ is the focal length, and F is called a virtual focal point. The corresponding image distance v is positive, so both ƒ and R are positive, and Eq. (14.4), ƒ = R>2, holds for convex as well as concave mirrors. In Fig. 14.16b the incoming rays are converging as though they would meet at the virtual focal point F, and they are reflected parallel to the optic axis. In summary, Eqs. (14.3) through (14.6), the basic relationships for image formation by a spherical mirror, are valid for both concave and convex mirrors, provided that we use the sign rules consistently. Example 14.3

Santa’s image problem

Santa checks himself for soot, using his reflection in a silvered Christmas tree ornament 0.750 m away (Fig. 14.17a). The diameter of the ornament is 7.20 cm. Standard reference texts state that

he is a “right jolly old elf,” so we estimate his height to be 1.6 m. Where and how tall is the image of Santa formed by the ornament? Is it erect or inverted? Continued

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454

CHAPTER 14 Geometric Optics

14.17 (a) The ornament forms a virtual, reduced, erect image of Santa. (b) Our sketch of two of the rays forming the image. (a)

(b)

|v| = v

SOLUTION IDENTIFY and SET UP: Figure 14.17b shows the situation. Santa is the object, and the surface of the ornament closest to him acts as a convex mirror. The relationships among object distance, image distance, focal length, and magnification are the same as for concave mirrors, provided we use the sign rules consistently. The radius of curvature and the focal length of a convex mirror are negative. Here, u = - 0.75 cm = - 75 cm, and Santa’s height is y = 1.6 m. We solve Eq. (14.5) to v, and then use Eq. (14.6) to find the lateral magnification m and the image height y¿. The sign of m tells us whether the image is erect or inverted. EXECUTE: The radius of the mirror (half the diameter) is R = + 17.20 cm2 >2 = + 3.60 cm, and the focal length is ƒ = R > 2 = + 1.80 cm. From Eq. (14.5), '

'

'

1 1 1 1 1 = = v u ƒ + 1.80 cm -75.0 cm v = -1.76 cm

Because v is positive, the image is behind the mirror—that is, on the side opposite to the outgoing light (Fig. 14.17b)—and it is virtual. The image is about halfway between the front surface of the ornament and its center. From Eq. (14.6), the lateral magnification and the image height are m =

y¿ v -1.76 cm = 0.0234 = - = y u - 75 cm

y¿ = my = 10.0234211.6 m2 = 3.8 * 10-2 m = 3.8 cm EVALUATE: Our sketch indicates that the image is erect so both m and y¿ are positive; our calculations agree. When u is negative, a convex mirror always forms an erect, virtual, reduced, reversed image. For this reason, convex mirrors are used at blind intersections, for surveillance in stores, and as wide-angle rear-view mirrors for cars and trucks. (Many such mirrors read “Objects in mirror are closer than they appear.”)

Graphical Methods for Mirrors In Examples 14.1 and 14.3, we used Eqs. (14.5) and (14.6) to find the position and size of the image formed by a mirror. We can also determine the properties of the image by a simple graphical method. This method consists of finding the point of intersection of a few particular rays that diverge from a point of the object (such as point Q in Fig. 14.18) and are reflected by the mirror. Then (neglecting aberrations) all rays from this object point that strike the mirror will intersect at the same point. For this construction we always choose an object point that is not on the optic axis. Four rays that we can usually draw easily are shown in Fig. 14.18. These are called principal rays. 1. A ray parallel to the axis, after reflection, passes through the focal point F of a concave mirror or appears to come from the (virtual) focal point of a convex mirror. 2. A ray through (or proceeding toward) the focal point F is reflected parallel to the axis. 3. A ray along the radius through or away from the center of curvature C intersects the surface normally and is reflected back along its original path. 4. A ray to the vertex V is reflected forming equal angles with the optic axis. Once we have found the position of the image point by means of the intersection of any two of these principal rays 11, 2, 3, 42, we can draw the path of any other ray from the object point to the same image point.

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14.2 Reflection at a Spherical Surface

455

14.18 The graphical method of locating an image formed by a spherical mirror. The colors of the rays are for identification only; they do not refer to specific colors of light. (b) Principal rays for convex mirror

(a) Principal rays for concave mirror

1

Q

Q

1

1 3

3

4

2

C

P9

F

V

P

Q9

4

P

V

2

Q9

4

4

2

2

P9

F

C

4

3 1

1 Ray parallel to axis reflects through focal point.

1 Reflected parallel ray appears to come from focal point.

2 Ray through focal point reflects parallel to axis.

2 Ray toward focal point reflects parallel to axis.

3 Ray through center of curvature intersects the surface normally

3 As with concave mirror: Ray radial to center of curvature intersects

and reflects along its original path. 4 Ray to vertex reflects symmetrically around optic axis.

4 As with concave mirror: Ray to vertex reflects symmetrically around

the surface normally and reflects along its original path. optic axis.

CAUTION Principal rays are not the only rays Although we’ve emphasized the principal rays, in fact any ray from the object that strikes the mirror will pass through the image point (for a real image) or appear to originate from the image point (for a virtual image). Usually, you only need to draw the principal rays, because these are all you need to locate the image. y

Problem-Solving Strategy 14.1

Image Formation by Mirrors

IDENTIFY the relevant concepts: Problems involving image formation by mirrors can be solved in two ways: using principal-ray diagrams and using equations. A successful problem solution uses both approaches. SET UP the problem: Identify the target variables. One of them is likely to be the focal length, the object distance, or the image distance, with the other two quantities given. EXECUTE the solution as follows: 1. Draw a large, clear principal-ray diagram if you have enough information. 2. Orient your diagram so that incoming rays go from left to right. Draw only the principal rays; color-code them as in Fig. 14.18. If possible, use graph paper or quadrille-ruled paper. Use a ruler and measure distances carefully! A freehand sketch will not give good results. 3. If your principal rays don’t converge at a real image point, you may have to extend them straight backward to locate a virtual

Example 14.4

image point, as in Fig. 14.18b. We recommend drawing the extensions with broken lines. 4. Measure the resulting diagram to obtain the magnitudes of the target variables. 5. Solve for the target variables using Eq. (14.5), 1>v + 1>u = 1>ƒ, and the lateral magnification equation, Eq. (14.6), as appropriate. Apply the sign rules given in Section 14.2 to object and image distances, radii of curvature, and object and image heights. 6. Use the sign rules to interpret the results that you deduced from your ray diagram and calculations. Note that the same sign rules (given in Section 14.2) work for all four cases in this chapter: reflection and refraction from plane and spherical surfaces. EVALUATE your answer: Check that the results of your calculations agree with your ray-diagram results for image position, image size, and whether the image is real or virtual.

Concave mirror with various object distances

A concave mirror has a radius of curvature with absolute value 20 cm. Find graphically the image of an object in the form of an arrow perpendicular to the axis of the mirror at object distances (magnitude = |u|) of (a) 30 cm, (b) 20 cm, (c) 10 cm, and (d) 5 cm. Check the construction by computing the size and lateral magnification of each image.

SOLUTION IDENTIFY and SET UP: We must use graphical methods and calculations to analyze the image made by a mirror. The mirror is concave, so its radius of curvature is R = - 20 cm and its focal length is ƒ = R > 2 = - 10 cm. Our target variables are '

'

Continued

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456

CHAPTER 14 Geometric Optics

x cordinate of image v and lateral magnifications m corresponding to four cases with successively smaller object distances s. In each case we solve Eq. (14.5) for v and use m = - v > u to find m. '

1 1 + = v - 30 cm 1 1 (b) + = v - 20 cm 1 1 (c) + = v - 10 cm -

1 10 cm 1 10 cm 1 10 cm

v = + 10 cm

'

EXECUTE: Figure 14.19 shows the principal-ray diagrams for the four cases. Study each of these diagrams carefully and confirm that each numbered ray is drawn in accordance with the rules given earlier (under “Graphical Methods for Mirrors”). Several points are worth noting. First, in case (b) the object and image distances are equal. Ray 3 cannot be drawn in this case because a ray from Q through the center of curvature C does not strike the mirror. In case (c), ray 2 cannot be drawn because a ray from Q through F does not strike the mirror. In this case the outgoing rays are parallel, corresponding to an infinite image distance. In case (d), the outgoing rays diverge; they have been extended backward to the virtual image point Q¿, from which they appear to diverge. Case (d) illustrates the general observation that an object placed inside the focal point of a concave mirror produces a virtual image. Measurements of the figures, with appropriate scaling, give the following approximate image distances: (a) 15 cm; (b) 20 cm; (c) q or - q (because the outgoing rays are parallel and do not converge at any finite distance); (d) -10 cm. To compute these distances, we solve Eq. (14.6) for s¿ and insert ƒ = 10 cm : (a)

1 1 1 + = v - 5 cm - 10 cm

(d)

v = - 15 cm v = - 20 cm v = q 1or - q 2

The signs of v tell us that the image is real in cases (a) and (b) and virtual in case (d). The lateral magnifications measured from the figures are approximately (a) - 12; (b) - 1; (c) q or - q ; (d) +2. From Eq. (14.7), - 15 cm 1 (a) m = - a b = - 30 cm 2 - 20 cm (b) m = - a b = -1 - 20 cm - ` cm (c) m = - a b = - q 1or + q 2 - 10 cm + 10 cm (d) m = - a b = +2 - 5 cm The signs of m tell us that the image is inverted in cases (a) and (b) and erect in case (d). EVALUATE: Notice the trend of the results in the four cases. When the object is far from the mirror, as in Fig. 14.19a, the image is smaller than the object, inverted, and real. As the object distance s decreases, the image moves farther from the mirror and gets larger (Fig. 14.19b). When the object is at the focal point, the image is at infinity (Fig. 14.19c). When the object is inside the focal point, the image becomes larger than the object, erect, and virtual (Fig. 14.19d). You can confirm these conclusions by looking at objects reflected in the concave bowl of a shiny metal spoon.

'

14.19 Using principal-ray diagrams to locate the image P¿Q¿ made by a concave mirror. (b) Construction for u 5 220 cm

(a) Construction for u 5 230 cm

Q

All principal rays can be drawn. The image is inverted. 4 32

1 C

P 2 4

P' 3

F

V

Ray 3 (from Q through C ) cannot be drawn because it does not strike Q the mirror. 1 4 2 The image P P is inverted. C F

V

Q 2

Q

4

1

1

u and v are equal. ƒ uƒ 5 ƒ vƒ 5 20 cm

ƒvƒ ƒ uƒ 5 30 cm (c) Construction for u 5 210 cm Ray 2 (from Q through F ) cannot be drawn because it does not strike the mirror. 3 Q 4 1 C F P 3 1 4

(d) Construction for u 5 25 cm

Q

2 Q 2 3 C

V

Parallel outgoing rays correspond to an ƒ uƒ 5 10 cm infinite image distance.

F P

3 1 4

1 4 V

P The image is and erect. (N ƒ uƒ 5 v, as v> ƒ vƒ

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14.3 Refraction at a Spherical Surface

457

Test Your Understanding of Section 14.2 A cosmetics mirror is designed so that your reflection appears right-side up and enlarged. (a) Is the mirror concave or convex? (b) To see an enlarged image, what should be the distance from the mirror (of focal length f ) to your face? (i) ƒ ƒ ƒ ; (ii) less than ƒ ƒ ƒ ; (iii) greater than ƒ ƒ ƒ . y

Refraction at a Spherical Surface

14.3

As we mentioned in Section 14.1, images can be formed by refraction as well as by reflection. To begin with, let’s consider refraction at a spherical surface—that is, at a spherical interface between two optical materials with different indexes of refraction. This analysis is directly applicable to some real optical systems, such as the human eye. It also provides a stepping-stone for the analysis of lenses, which usually have two spherical (or nearly spherical) surfaces.

Image of a Point Object: Spherical Refracting Surface Sign ruler for spherical refracting surface is exactly same as that discussed in section 14.2 for spherical mirrors. In fig. 14.20, V is the vertex or central point of refracting surface. A line through vertex V and centre of curvature c is taken as x-axis. Point V is taken as origin. The direction of +ive x-axis is that of incident rays. In Fig. 14.20 a spherical surface with radius R forms an interface between two materials with different indexes of refraction na and nb . The surface forms an image P¿ of an object point P; we want to find how the x coordinates of the object and image distances (u and v) are related. We will use the same sign rules that we used for spherical mirrors. R is positive. Ray PV strikes the vertex V and is perpendicular to the surface (that is, to the plane that is tangent to the surface at the point of incidence V ). It passes into the second material without deviation. Ray PB, making an angle a with the axis, is incident at an angle ua with the normal and is refracted at an angle ub . These rays intersect at P¿, a distance |v| = v to the right of the vertex. The figure is drawn for the case na 6 nb . Here u is -ive but v is +ive. We are going to prove that if the angle a is small, all rays from P intersect at the same point P¿, so P¿ is the real image of P. We use much the same approach as we did for spherical mirrors in Section 14.2. We again use the theorem that an exterior angle of a triangle equals the sum of the two opposite interior angles; applying this to the triangles PBC and P¿BC gives '

'

'

ua = a + f

f = b + ub

(14.7)

From the law of refraction, na sin ua = nb sin ub Also, the tangents of a, b, and f are h h h h tan a = = tan b = = |u| + d -u + d |v| - d v - d h h tan f = = |R| - d R - d na , nb

nb

B

ua

u is -ive, v, R are +ive R

h P

a

(14.8)

f

V d

ub

b C

P

14.20 Construction for finding the position of the image point P¿ of a point object P formed by refraction at a spherical surface. The materials to the left and right of the interface have refractive indexes na and nb , respectively. In the case shown here, na 6 nb . '

'

|u|

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458

CHAPTER 14 Geometric Optics

Note that u is -ive, so |u| = -u, v is +ive, |v| = v. R is +ive, so |R| = R. For paraxial rays, ua and ub are both small in comparison to a radian, and we may approximate both the sine and tangent of either of these angles by the angle itself (measured in radians). The law of refraction then gives na ua = nb ub '

'

Combining this with the first of Eqs. (14.7), we obtain na ub = 1a + f2 nb When we substitute this into the second of Eqs. (14.7), we get na a + nb b = 1nb - na2f '

(14.9)

'

Now we use the approximations tan a = a, and so on, in Eqs. (14.8) and also neglect the small distance d; those equations then become h R Finally, we substitute these into Eq. (14.9) and divide out the common factor h. We obtain nb nb - na na (object–image relationship, + = (14.10) spherical refracting surface) - u v R h - u

a =

b =

h v

f =

This equation does not contain the angle a, so the image distance is the same for all paraxial rays emanating from P; this proves our assertion that P¿ is the image of P. To obtain the lateral magnification m for this situation, we use the construction in Fig. 14.21. We draw two rays from point Q, one through the center of curvature C and the other incident at the vertex V. From the triangles PQV and P¿Q¿V, tan ua =

y - u

tan ub =

- y¿ v

and from the law of refraction, na sin ua = nb sin ub For small angles, tan ua = sin ua

tan ub = sin ub

so finally na y nb y¿ = - u v '

na v y¿ = y nb u '

m =

'

'

or

(lateral magnification, spherical refracting surface)

(14.12)

Equations (14.10) and (14.11) can be applied to both convex and concave refracting surfaces, provided that you use the sign rules consistently. It doesn’t matter whether nb is greater or less than na . To verify these statements, you should construct diagrams like Figs. 14.20 and 14.21 for the following three cases: (i) R 7 0 and na 7 nb , '

'

14.21 Construction for determining the height of an image formed by refraction at a spherical surface. In the case shown here, n a 6 nb .

n a ! nb

Q

nb u is -ive and v is +ive

'

y ua P

C ub

V

P y Q

|u|

|v| = v

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14.3 Refraction at a Spherical Surface

(ii) R 6 0 and na 6 nb , and (iii) R 6 0 and na 7 nb . Then in each case, use your diagram to again derive Eqs. (14.10) and (14.11). Refraction at a curved surface is one reason gardeners avoid watering plants at midday. As sunlight enters a water drop resting on a leaf (Fig. 14.22), the light rays are refracted toward each other as in Figs. 14.20 and 14.21. The sunlight that strikes the leaf is therefore more concentrated and able to cause damage. An important special case of a spherical refracting surface is a plane surface between two optical materials. This corresponds to setting R = q in Eq. (14.10). In this case, nb na = 0 (plane refracting surface) (14.12) v u '

'

459

14.22 Light rays refract as they pass through the curved surfaces of these water droplets.

To find the lateral magnification m for this case, we combine this equation with the general relationship, Eq. (14.11), obtaining the simple result m = 1 That is, the image formed by a plane refracting surface always has the same lateral size as the object, and it is always erect. An example of image formation by a plane refracting surface is the appearance of a partly submerged drinking straw or canoe paddle. When viewed from some angles, the object appears to have a sharp bend at the water surface because the submerged part appears to be only about three-quarters of its actual distance below the surface. (We commented on the appearance of a submerged object in Section 33.2; see Fig. 33.9.) Example 14.5

Image formation by refraction I

A cylindrical glass rod (Fig. 14.23) has index of refraction 1.52. It is surrounded by air. One end is ground to a hemispherical surface with radius R = 2.00 cm. A small object is placed on the axis of the rod, 8.00 cm to the left of the vertex. Find (a) the image distance and (b) the lateral magnification.

material b is the glass of which the rod is made 1nb = 1.522. We are given |u| = 8.00 cm. The radius is positive: R = +2.00 cm . We solve Eq. (14.10) for v, and we use Eq. (14.11) to find m. EXECUTE: (a) From Eq. (14.10), 1.00 1.52 1.52 - 1.00 + = v - 8.00 cm +2.00 cm

S O L U T IO N IDENTIFY and SET UP: This problem uses the ideas of refraction at a curved surface. Our target variables are the image distance s¿ and the lateral magnification m. Here material a is air 1na = 1.002 and

14.23 The glass rod in air forms a real image. na 5 1.00 (air) P

C

|u| 5 8.00 cm

Example 14.6

nb 5 1.52

P

v = +11.3 cm (b) From Eq. (14.12), m = -

na v 11.002111.3 cm2 = - 0.929 = nb u 11.5221 - 8.00 cm2

EVALUATE: Because the image distance v is positive, the image is formed 11.3 cm to the right of the vertex, as Fig. 14.23 shows. The value of m tells us that the image is somewhat smaller than the object and that it is inverted. If the object is an arrow 1.000 mm high, pointing upward, the image is an arrow 0.929 mm high, pointing downward.

R 5 2.00 cm |v|

Image formation by refraction II

The glass rod of Example 14.5 is immersed in water, which has index of refraction n = 1.33 (Fig. 14.24). The object distance is again 8.00 cm. Find the image distance and lateral magnification. S O L U T IO N

EXECUTE: Our solution of Eq. (14.10) in Example 14.5 yields 1.52 - 1.33 1.52 1.33 = v (- 8 cm) +2.00 cm v = -21.3 cm

IDENTIFY and SET UP: The situation is the same as in Example 14.5 except that now na = 1.33. We again use Eqs. (14.10) and (14.11) to determine v and m, respectively.

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Continued

460

CHAPTER 14 Geometric Optics

14.24 When immersed in water, the glass rod forms a virtual image. na 5 1.33 (water) P

P

C

nb 5 1.52

|u| 5 8.00 cm |v|

The lateral magnification in this case is m =

1.331- 21.3 cm2 = + 2.33 1.521 - 8.00 cm2

left of the vertex. We saw a similar case in the reflection of light from a convex mirror; in both cases we call the result a virtual image. The vertical image is erect (because m is positive) and 2.33 times as large as the object.

EVALUATE: The negative value of v means that the refracted rays do not converge, but appear to diverge from a point 21.3 cm to the

Example 14.7

Apparent depth of a swimming pool

If you look straight down into a swimming pool where it is 2.00 m deep, how deep does it appear to be?

nb na 1 1.33 = 0 = v u v (- 2.00 m) v = -1.50 m

SOLUTION IDENTIFY and SET UP: Figure 14.25 shows the situation. The surface of the water acts as a plane refracting surface. To determine the pool’s apparent depth, we imagine an arrow PQ painted on the bottom. The pool’s refracting surface forms a virtual image P¿Q¿ of this arrow. We solve Eq. (14.12) to find the image depth s¿; that’s the pool’s apparent depth. EXECUTE: The object distance is the actual depth of the pool, u = - 2.00 m. Material a is water 1na = 1.332 and material b is air 1nb = 1.002. From Eq. (14.12),

14.25 Arrow P¿Q¿ is the virtual image of the underwater arrow PQ. The angles of the ray with the vertical are exaggerated for clarity.

The image distance is negative. By the sign rules in Section 14.2, this means that the image is virtual and on the incoming side of the refracting surface—that is, on the same side as the object, namely underwater. The pool’s apparent depth is 1.50 m, or just 75% of its true depth. EVALUATE: Recall that the lateral magnification for a plane refracting surface is m = 1. Hence the image P¿Q¿ of the arrow has the same horizontal length as the actual arrow PQ (Fig. 14.26). Only its depth is different from that of PQ.

14.26 The submerged portion of this straw appears to be at a shallower depth (closer to the surface) than it actually is.

Same horizontal dimension

nb 5 1.00 (air) na 5 1.33 (water)

V

|v| |u| Q

P

Q

P

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14.4 Thin Lenses

461

Test Your Understanding of Section 14.3 The water droplets in Fig. 14.22 have radius of curvature R and index of refraction n = 1.33. Can they form an image of the sun on the leaf? y

14.4

Thin Lenses

The most familiar and widely used optical device (after the plane mirror) is the lens. A lens is an optical system with two refracting surfaces. The simplest lens has two spherical surfaces close enough together that we can neglect the distance between them (the thickness of the lens); we call this a thin lens. If you wear eyeglasses or contact lenses while reading, you are viewing these words through a pair of thin lenses. We can analyze thin lenses in detail using the results of Section 14.3 for refraction by a single spherical surface. However, we postpone this analysis until later in the section so that we can first discuss the properties of thin lenses.

Properties of a Lens A lens of the shape shown in Fig. 14.27 has the property that when a beam of rays parallel to the axis passes through the lens, the rays converge to a point F2 (Fig. 14.27a) and form a real image at that point. Such a lens is called a converging lens. Similarly, rays passing through point F1 emerge from the lens as a beam of parallel rays (Fig. 14.27b). The points F1 and F2 are called the first and second focal points, and the x coordinate of F2 is called focal length f. So f is here with sign centre of the lens is taken as origin and direction in which rays are incident on the lens is taken as the direction of +ive x axis. Note the similarities between the two focal points of a converging lens and the single focal point of a concave mirror (see Fig. 14.12). As opposite to concave mirror, the focal length of a converging lens is defined to be a positive quantity, and such a lens is also called a positive lens. The central horizontal line in Fig. 14.27 is called the optic axis, as with spherical mirrors. The centers of curvature of the two spherical surfaces lie on and define the optic axis. The magnitudes of x coordinates of F1 and F2 are always equal for a thin lens, even when the two sides have different curvatures. We will derive this somewhat surprising result later in the section, when we derive the relationship of ƒ to the index of refraction of the lens and the radii of curvature of its surfaces.

Image of an Extended Object: Converging Lens Like a concave mirror, a converging lens can form an image of an extended object. Figure 14.28 shows how to find the position and lateral magnification of an image made by a thin converging lens. Using the same notation and sign rules as before, we let u and v be the object and image distances, respectively, and let y and y¿ be the object and image heights. Ray QA, parallel to the optic axis before refraction, passes through the second focal point F2 after refraction. Ray QOQ¿ passes undeflected straight through the center of the lens because at the center the two surfaces are parallel and (we have assumed) very close together. There is refraction where the ray enters and leaves the material but no net change in direction. Q

A

y P

F1

a

O

u is -ive and v is +ve; the image is inverted. F2

b a

P b

|v|-|f|=v - f |f|=f |u|

PhET: Geometric Optics ActivPhysics 15.9: Thin Lens Ray Diagram ActivPhysics 15.10: Converging Thin Lenses ActivPhysics 15.11: Diverging Thin Lenses

14.27 F1 and F2 are the first and second focal points of a converging thin lens. The numerical value of ƒ is positive. (a) Second focal point: the point to which incoming parallel rays converge

Optic axis (passes through centers of curvature of both lens surfaces)

F1

F2

|f|=f

|f|=f

Focal length • Measured from lens center • Always the same on both sides of the lens • Positive for a converging thin lens (b) First focal point: Rays diverging from this point emerge from the lens parallel F1 to the axis. |f|=f

F2

|f|=f

14.28 Construction used to find image position for a thin lens. To emphasize that the lens is assumed to be very thin, the ray QAQ¿ is shown as bent at the midplane of the lens rather than at the two surfaces and ray QOQ¿ is shown as a straight line.

y Q

|f|=f |v|

f is +ive, u is -ive, v is +ive

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462

CHAPTER 14 Geometric Optics

The two angles labeled a in Fig. 14.28 are equal. Therefore the two right triangles PQO and P¿Q¿O are similar, and ratios of corresponding sides are equal. Thus, PO = |u| = -u (as u is -ive); P'O = |v| = v (v is +ive) y y¿ = -u v

or

y¿ v = y u

(14.13)

(The reason for the negative sign is that the image is below the optic axis and y¿ is negative.) Also, the two angles labeled b are equal, and the two right triangles OAF2 and P¿Q¿F2 are similar, so y y¿ = |ƒ| |v| - |ƒ| = v - f

or

f - v y¿ = y ƒ

(14.14)

We now equate Eqs. (14.13) and (14.14), divide by v, and rearrange to obtain 1 1 1 = v u ƒ

(object– image relationship, thin lens)

(14.15)

This analysis also gives the lateral magnification m = y¿>y for the lens; from Eq. (14.13), m = -

v u

(lateral magnification, thin lens)

(14.16)

The negative sign tells us that when u and v are of opposite signs, as in Fig. 14.28, the image is inverted, and y and y¿ have opposite signs. Equations (14.15) and (14.16) are the basic equations for thin lenses. As we will see, the same sign rules that we used for spherical mirrors are also applicable to lenses. In particular, consider a lens with a positive focal length (a converging lens). When an object is outside the first focal point F1 of this lens (|u| > |f| = f), the image distance v is positive (that is, the image is on the same side as the outgoing rays); this image is real and inverted, as in Fig. 14.28. An object placed inside the first focal point of a converging lens, so that |f| = f > |u| produces an image with a negative value of v; this image is located on the same side of the lens as the object and is virtual, erect, and larger than the object. You can verify these statements algebraically using Eqs. (14.15) and (14.16); we’ll also verify them in the next section, using graphical methods analogous to those introduced for mirrors in Section 14.2. Figure 14.29 shows how a lens forms a three-dimensional image of a threedimensional object. Point R is nearer the lens than point P. From Eq. (14.15), image point R¿ is farther from the lens than is image point P¿, and the image P¿R¿ 14.29 The image S¿P¿Q¿R¿ of a threedimensional object SPQR is not reversed by a lens.

A real image made by a converging lens is inverted but not reversed back to front: the image thumb P9R9 and object thumb PR point in the same direction. R9 P9 F2 S9 Q9 Q F1 Image S R P

Object

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14.4 Thin Lenses

points in the same direction as the object PR. Arrows P¿S¿ and P¿Q¿ are reversed relative to PS and PQ. Let’s compare Fig. 14.29 with Fig. 14.6, which shows the image formed by a plane mirror. We note that the image formed by the lens is inverted, but it is not reversed front to back along the optic axis. That is, if the object is a left hand, its image is also a left hand. You can verify this by pointing your left thumb along PR, your left forefinger along PQ, and your left middle finger along PS. Then rotate your hand 180°, using your thumb as an axis; this brings the fingers into coincidence with P¿Q¿ and P¿S¿. In other words, an inverted image is equivalent to an image that has been rotated by 180° about the lens axis.

463

14.30 F2 and F1 are the second and first focal points of a diverging thin lens, respectively. The numerical value of ƒ is negative. (a) Second focal point: The point from which parallel incident rays appear to diverge F1

F2

Diverging Lenses So far we have been discussing converging lenses. Figure 14.30 shows a diverging lens; the beam of parallel rays incident on this lens diverges after refraction. The focal length of a diverging lens is a negative quantity, and the lens is also called a negative lens. The focal points of a negative lens are reversed, relative to those of a positive lens. The second focal point, F2 , of a negative lens is the point from which rays that are originally parallel to the axis appear to diverge after refraction, as in Fig. 14.30a. Incident rays converging toward the first focal point F1 , as in Fig. 14.30b, emerge from the lens parallel to its axis. Comparing with Section 14.2, you can see that a diverging lens has the same relationship to a converging lens as a convex mirror has to a concave mirror. Equations (14.15) and (14.16) apply to both positive and negative lenses. Figure 14.31 shows various types of lenses, both converging and diverging. Here’s an important observation: Any lens that is thicker at its center than at its edges is a converging lens with positive ƒ; and any lens that is thicker at its edges than at its center is a diverging lens with negative ƒ (provided that the lens has a greater index of refraction than the surrounding material). We can prove this using the lensmaker’s equation, which it is our next task to derive.

(b)

|f| |f| For a diverging thin lens, f is negative. |f| = -f First focal point: Rays converging on this point emerge from the lens parallel to the axis.

'

'

F2

F1

|f|

|f|

14.31 Various types of lenses. (a)

Converging lenses

The Lensmaker’s Equation Note that sign converntion is similar to that of spherical refracting surface. We’ll now derive Eq. (14.15) in more detail and at the same time derive the lensmaker’s equation, which is a relationship among the focal length ƒ, the index of refraction n of the lens, and the radii of curvature R1 and R2 of the lens surfaces. We use the principle that an image formed by one reflecting or refracting surface can serve as the object for a second reflecting or refracting surface. We begin with the somewhat more general problem of two spherical interfaces separating three materials with indexes of refraction na , nb , and nc , as shown in Fig. 14.32. The values of u and v are for the first surface are u and v and for 2nd surface are u2 and v2. There values are +ive or -ive. We assume that the lens is thin, so that the distance t between the two surfaces is small in comparison with the object and image distances and can therefore be neglected. This is usually the case with eyeglass lenses (Fig. 14.33). So we can say that v1 = u2 We need to use the single-surface equation, Eq. (14.10), twice, once for each surface. The two resulting equations are nb na nb - na = v1 u1 R1 nc nb nc - nb = v2 u2 R2 '

'

'

Meniscus Planoconvex (b)

Double convex

Diverging lenses

Meniscus Planoconcave Double concave

Ordinarily, the first and third materials are air or vacuum, so we set na = nc = 1. The second index nb is that of the lens, which we can call simply n. Substituting these values and the relationship v1 = u2, we get

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464

CHAPTER 14 Geometric Optics

14.32 The image formed by the first surface of a lens serves as the object for the second surface. The distances |v1| and |u2| are taken to be equal; this is a good approximation if the lens thickness t is small. na nb nc

Q

P P

|R1| |R2|

Q

t |u1|

P

C1 C 2 Q

|v1| |u2| |v2|

n 1 n - 1 = v1 u1 R1 1 n 1 - n = v2 u2 R2

14.33 These eyeglass lenses satisfy the thin-lens approximation: Their thickness is small compared to the object and image distances.

To get a relationship between the initial object position u1 and the final image n n position v2, we add these two equations. This eliminates the term = and we v1 u2 obtain 1 1 1 1 = 1n - 12a b v1 u1 R1 R2 Finally, thinking of the lens as a single unit, we call the object distance simply u instead of u1, and we call the final image distance v instead of v2. Making these substitutions, we have 1 1 1 1 = 1n - 12a b v u R1 R2

(14.17)

Now we compare this with the other thin-lens equation, Eq. (14.15). We see that the object and image distances u and v appear in exactly the same places in both equations and that the focal length ƒ is given by 1 1 1 = 1n - 12a b ƒ R1 R2

(lensmaker’s equation for a thin lens)

(14.18)

This is the lensmaker’s equation. In the process of rederiving the relationship among object distance, image distance, and focal length for a thin lens, we have also derived an expression for the focal length ƒ of a lens in terms of its index of refraction n and the radii of curvature R1 and R2 of its surfaces. This can be used to show that all the lenses in Fig. 14.31a are converging lenses with positive focal lengths and that all the lenses in Fig. 14.31b are diverging lenses with negative focal lengths. We use all our sign rules from Section 14.2 with Eqs. (14.17) and (14.18). For example, in Fig. 14.34, u is -ive, v is +ive, R1 is +ive, R2 is -ive, and R1. It is not hard to generalize Eq. (14.18) to the situation in which the lens is immersed in a material with an index of refraction greater than unity. We invite you to work out the lensmaker’s equation for this more general situation. We stress that the paraxial approximation is indeed an approximation! Rays that are at sufficiently large angles to the optic axis of a spherical lens will not be brought to the same focus as paraxial rays; this is the same problem of spherical aberration that plagues spherical mirrors (see Section 14.2). To avoid this and other limitations of thin spherical lenses, lenses of more complicated shape are used in precision optical instruments.

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14.4 Thin Lenses

465

14.34 A converging thin lens with a positive focal length ƒ. R1 is positive. (C1 has +ive x coordinate if we take centre of lens as origin and direction of incident ray as +ive x axis direction)

R2 is negative. (C2 has -ive x coordinate)

x coordinate of C2: |R2| Q

x coordinate of C1: |R1| n

y

C2

P

C1

P y v is +ive, u is -ive so m is negative. |u|

Example 14.8

Q |v|

Determining the focal length of a lens

(a) Suppose the absolute values of the radii of curvature of the lens surfaces in Fig. 14.34 are both equal to 10 cm and the index of refraction of the glass is n = 1.52. What is the focal length ƒ of the lens? (b) Suppose the lens in Fig. 14.30 also has n = 1.52 and the absolute values of the radii of curvature of its lens surfaces are also both equal to 10 cm. What is the focal length of this lens? S O L U T IO N IDENTIFY and SET UP: We are asked to find the focal length ƒ of (a) a lens that is convex on both sides (Fig. 14.34) and (b) a lens that is concave on both sides (Fig. 14.31). In both cases we solve the lensmaker’s equation, Eq. (14.18), to determine ƒ. We apply the sign rules to the radii of curvature R1 and R2 to take account of whether the surfaces are convex or concave. EXECUTE: (a) The lens in Fig. 14.34 is double convex: The center of curvature of the first surface 1C12 has +ive x coordinate, so R1 is positive, and the center of curvature of the second surface 1C22, has -ive x coordinate. Hence R1 = + 10 cm and R2 = -10 cm. Then from Eq. (14.18),

1 1 1 = 11.52 - 12 a b ƒ +10 cm -10 cm ƒ = 9.6 cm (b) The lens in Fig. 14.30 is double concave: The center of curvature of the first surface has -ive x co-ordinate, so R1 is negative, and the center of curvature of the second surface has +ive x coordinate, so R2 is positive. Hence in this case R1 = - 10 cm and R2 = +10 cm. Again using Eq. (14.19), 1 1 1 = 11.52 - 12 a b ƒ -10 cm +10 cm ƒ = - 9.6 cm EVALUATE: In part (a) the focal length is positive, so this is a converging lens; this makes sense, since the lens is thicker at its center than at its edges. In part (b) the focal length is negative, so this is a diverging lens; this also makes sense, since the lens is thicker at its edges than at its center.

Graphical Methods for Lenses We can determine the position and size of an image formed by a thin lens by using a graphical method very similar to the one we used in Section 14.2 for spherical mirrors. Again we draw a few special rays called principal rays that diverge from a point of the object that is not on the optic axis. The intersection of these rays, after they pass through the lens, determines the position and size of the image. In using this graphical method, we will consider the entire deviation of a ray as occurring at the midplane of the lens, as shown in Fig. 14.35. This is consistent with the assumption that the distance between the lens surfaces is negligible. The three principal rays whose paths are usually easy to trace for lenses are shown in Fig. 14.35: 1. A ray parallel to the axis emerges from the lens in a direction that passes through the second focal point F2 of a converging lens, or appears to come from the second focal point of a diverging lens. 2. A ray through the center of the lens is not appreciably deviated; at the center of the lens the two surfaces are parallel, so this ray emerges at essentially the same angle at which it enters and along essentially the same line. 3. A ray through (or proceeding toward) the first focal point F1 emerges parallel to the axis.

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CHAPTER 14 Geometric Optics

14.35 The graphical method of locating an image formed by a thin lens. The colors of the rays are for identification only; they do not refer to specific colors of light. (Compare Fig. 14.18 for spherical mirrors.) 1

(b) Diverging lens

(a) Converging lens Q

Q

1 2

F2

3

P

1 3 2

P9 P

F1

F2

Q9 3

P9

F1 2

3

Q9

1

2

1 Parallel incident ray refracts to pass through second focal point F2.

1 Parallel incident ray appears after refraction to have come

2 Ray through center of lens does not deviate appreciably.

from the second focal point F2.

3 Ray through the first focal point F1 emerges parallel to the axis.

2 Ray through center of lens does not deviate appreciably. 3 Ray aimed at the first focal point F1 emerges parallel to

the axis.

When the image is real, the position of the image point is determined by the intersection of any two rays 1, 2, and 3 (Fig. 14.35a). When the image is virtual, we extend the diverging outgoing rays backward to their intersection point to find the image point (Fig. 14.35b). CAUTION Principal rays are not the only rays Keep in mind that any ray from the object that strikes the lens will pass through the image point (for a real image) or appear to originate from the image point (for a virtual image). (We made a similar comment about image formation by mirrors in Section 14.2.) We’ve emphasized the principal rays because they’re the only ones you need to draw to locate the image. y

Figure 14.36 shows principal-ray diagrams for a converging lens for several object distances. We suggest you study each of these diagrams very carefully, comparing each numbered ray with the above description. 14.36 Formation of images by a thin converging lens for various object distances. The principal rays are numbered. (Compare Fig. 14.19 for a concave spherical mirror.) (a) Object O is outside focal point; image I is real. 1 2 3

O

(b) Object O is closer to focal point; image I is real and farther away.

F2

O I

F1

3

1 2

F2 I

F1

3 2

3 1 2

1

(d) Object O is at focal point; image I is at infinity.

(c) Object O is even closer to focal point; image I is real and even farther away. 1 2

O

F2

F1

O

1 2

F2 I at infinity

F1

I

1

2 1

(e) Object O is inside focal point; image I is virtual and larger than object.

I

1 2

O

2 (f) A virtual object O (light rays are converging on lens)

F2

2

F1

1

2

1 3

F1

3 O I

F2

2

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1

14.4 Thin Lenses

467

Parts (a), (b), and (c) of Fig. 14.36 help explain what happens in focusing a camera. For a photograph to be in sharp focus, the film must be at the position of the real image made by the camera’s lens. The image distance increases as the object is brought closer, so the film is moved farther behind the lens (i.e., the lens is moved farther in front of the film). In Fig. 14.36d the object is at the focal point; ray 3 can’t be drawn because it doesn’t pass through the lens. In Fig. 14.36e the object distance is less than the focal length. The outgoing rays are divergent, and the image is virtual; its position is located by extending the outgoing rays backward, so the image distance v is negative. Note also that the image is erect and larger than the object. (We’ll see the usefulness of this in Section 14.6.) Figure 14.36f corresponds to a virtual object. The incoming rays do not diverge from a real object, but are converging as though they would meet at the tip of the virtual object O on the right side; the object distance u is positive in this case. The image is real and is located between the lens and the second focal point. This situation can arise if the rays that strike the lens in Fig. 14.36f emerge from another converging lens (not shown) to the left of the figure. Problem-Solving Strategy 14.2

Image Formation by Thin Lenses

IDENTIFY the relevant concepts: Review Problem-Solving Strategy 14.1 (Section 14.2) for mirrors, which is equally applicable here. As for mirrors, you should solve problems involving image formation by lenses using both principal-ray diagrams and equations. SET UP the problem: Identify the target variables. EXECUTE the solution as follows: 1. Draw a large principal-ray diagram if you have enough information, using graph paper or quadrille-ruled paper. Orient your diagram so that incoming rays go from left to right. Draw the rays with a ruler, and measure distances carefully. 2. Draw the principal rays so they change direction at the midplane of the lens, as in Fig. 14.35. For a lens there are only three principal rays (compared to four for a mirror). Draw all three whenever possible; the intersection of any two rays determines the image location, but the third ray should pass through the same point.

Example 14.9

3. If the outgoing principal rays diverge, extend them backward to find the virtual image point on the incoming side of the lens, as in Fig. 14.36e. 4. Solve Eqs. (14.15) and (14.16), as appropriate, for the target variables. Make sure that you carefully use the sign rules given in Section 14.2. 5. The image from a first lens or mirror may serve as the object for a second lens or mirror. In finding the object and image distances for this intermediate image, be sure you include the distance between the two elements (lenses and> or mirrors) correctly. EVALUATE your answer: Your calculated results must be consistent with your ray-diagram results. Check that they give the same image position and image size, and that they agree on whether the image is real or virtual.

Image position and magnification with a converging lens

Use ray diagrams to find the image position and magnification for an object at each of the following distances from a converging lens with a focal length of 20 cm: (a) 50 cm; (b) 20 cm; (c) 15 cm; (d) Object is virtual at a distance of 40 cm. Check your results by calculating the image position and lateral magnification using Eqs. (14.15) and (14.16), respectively. S O L U T IO N IDENTIFY and SET UP: We are given the focal length ƒ = 20 cm and four object distances |u| = -u. Our target variables are the corresponding values of v and lateral magnifications m. We solve Eq. (14.15) for v, and find m from Eq. (14.16), m = v > u. '

'

EXECUTE: Figures 14.36a, d, e, and f, respectively, show the appropriate principal-ray diagrams. You should be able to reproduce these without referring to the figures. Measuring these diagrams yields the approximate results: v = 35 cm, - q , - 40 cm, +15 cm and m = - 23, + q , + 3, and + 13, respectively.

Calculating the image distances from Eq. (14.15), we find 1 v 1 (b) v 1 (c) v 1 (d) v (a)

-

(+

1 1 = v = 33.3 cm 50 cm 20 cm 1 1 = v = —q 20 cm ) 20 cm 1 1 = v = - 60 cm 15 cm 20 cm 1 1 = v = 13.3 cm 40cm 20 cm

The graphical results are fairly close to these except for part (c); the accuracy of the diagram in Fig. 14.36e is limited because the rays extended backward have nearly the same direction. From Eq. (14.16), (a) m = (c) m =

33.3 cm = - 23 - 50 cm - 60 cm = +4 - 15 cm

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6q cm = 6q - 20 cm 13.3 cm (d) m = = + 13 40 cm Continued (b) m =

468

CHAPTER 14 Geometric Optics

EVALUATE: Note that the x coordinate of image ‘v’ is positive in parts (a) and (d) but negative in part (c). This makes sense: The image is real in parts (a) and (d) but virtual in part (c). The light rays that emerge from the lens in part (b) are parallel and never converge, so the image can be regarded as being at either + q or - q.

Example 14.10

The values of magnification m tell us that the image is inverted in part (a) and erect in parts (c) and (d), in agreement with the principal-ray diagrams. The infinite value of magnification in part (b) is another way of saying that the image is formed infinitely far away.

Image formation by a diverging lens

A beam of parallel rays spreads out after passing through a thin diverging lens, as if the rays all came from a point 20.0 cm from the center of the lens. You want to use this lens to form an erect, virtual image that is 13 the height of the object. (a) Where should the object be placed? Where will the image be? (b) Draw a principalray diagram. SOLUTION IDENTIFY and SET UP: The result with parallel rays shows that the focal length is ƒ = - 20 cm. We want the lateral magnification to be m = + 13 (positive because the image is to be erect). Our target variables are u and v. In part (a), we solve the magnification equation, Eq. (14.16), for v in terms of u; we then use the object–image relationship, Eq. (14.15), to find u and v individually. 1 v u EXECUTE: (a) From Eq. (14.17), m = + = , so v = . We u 3 3 insert this result into Eq. (14.16) and solve for the object distance s: 1 3 1 2 1 1 = - = = u u u u u/3 ƒ

v =

- 40.0 cm = -13.3 cm 3

v is negative, so the object and image are on the same side of the lens. (b) Figure 14.37 is a principal-ray diagram for this problem, with the rays numbered as in Fig. 14.35b. EVALUATE: You should be able to draw a principal-ray diagram like Fig. 14.37 without referring to the figure. From your diagram, you can confirm our results in part (a) for the object and image distances. You can also check our results for u and v by substituting them back into Eq. (14.15). 14.37 Principal-ray diagram for an image formed by a thin diverging lens. 1 1 3 O

3

2 F2

u = 2ƒ = - 40.0 cm

I

F1 2

213.3 cm 220.0 cm 40.0 cm

The object should be 40.0 cm from the lens. The image distance will be

Example 14.11

u = 3

220.0 cm

An image of an image

Converging lenses A and B, of focal lengths 8.0 cm and 6.0 cm, respectively, are placed 36.0 cm apart. Both lenses have the same optic axis. An object 8.0 cm high is placed 12.0 cm to the left of lens A. Find the position, size, and orientation of the image produced by the lenses in combination. (Such combinations are used in telescopes and microscopes, to be discussed in Section 14.7.)

14.38 Principal-ray diagram for a combination of two converging lenses. The first lens (A) makes a real image of the object. This real image acts as an object for the second lens (B). Lens A

Lens B

3, 19 29 39

1

SOLUTION IDENTIFY and SET UP: Figure 14.38 shows the situation. The object O lies outside the first focal point F1 of lens A, which therefore produces a real image I. The light rays that strike lens B diverge from this real image just as if I was a material object; image I therefore acts as an object for lens B. Our goal is to determine the properties of the image I¿ made by lens B. We use both ray-diagram and computational methods to do this. EXECUTE: In Fig. 14.38 we have drawn principal rays 1, 2, and 3 from the head of the object arrow O to find the position of the image I made by lens A, and principal rays 1¿, 2¿, and 3¿ from the head of I to find the position of the image I¿ made by lens B (even though rays 2¿ and 3¿ don’t actually exist in this case). The image

O

I9

2 F19

F2

3 F1

39

F29

I

1 12.0 cm 8.0 cm

24.0 cm 8.0 cm

36.0 cm

29 3, 19 2 12.0 cm 6.0 cm

12.0 cm 6.0 cm

is inverted twice, once by each lens, so the second image I¿ has the same orientation as the original object.

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469

14.5 Cameras We first find the position and size of the first image I. Applying Eq. (14.15), 1>v + 1>u = 1>ƒ, to lens A gives 1 1 1 = vA ( - 12cm) 8.0 cm

v A = + 24.0 cm

Image I is 24.0 cm to the right of lens A. The lateral magnification is mA = + 24.0 cm> - 12.0 cm = - 2.00, so image I is inverted and twice as tall as object O. Image I is 36.0 cm - 24.0 cm = 12.0 cm to the left of lens B, so the value of u (with sign) for lens B is ( -12.0 cm) . Applying Eq. (14.15) to lens B then gives '

1 1 1 = vB - 12.0 cm 6.0 cm

vB = + 12.0 cm

The final image I¿ is 12.0 cm to the right of lens B. The magnification produced by lens B is mB = 112.0 cm2>1- 12.0 cm2 = - 1.00.

EVALUATE: The value of mB means that the final image I¿ is just as large as the first image I but has the opposite orientation. The overall magnification is mAmB = 1-2.0021 -1.002 = + 2.00. Hence the final image I¿ is 12.00218.0 cm2 = 16 cm tall and has the same orientation as the original object O, just as Fig. 14.38 shows.

Test Your Understanding of Section 14.4 A diverging lens and an object are positioned as shown in the figure at right. Which of the rays A, B, C, and D could emanate from point Q at the top of the object?

A

Q 2F2

B F2

F1

2F1 C

D

y

14.5

Cameras

The concept of image, which is so central to understanding simple mirror and lens systems, plays an equally important role in the analysis of optical instruments (also called optical devices). Among the most common optical devices are cameras, which make an image of an object and record it either electronically or on film. The basic elements of a camera are a light-tight box (“camera” is a Latin word meaning “a room or enclosure”), a converging lens, a shutter to open the lens for a prescribed length of time, and a light-sensitive recording medium (Fig. 14.39). In a digital camera this is an electronic detector called a charge-coupled device (CCD) array; in an older camera, this is photographic film. The lens forms an inverted real image on the recording medium of the object being photographed. High-quality camera lenses have several elements, permitting partial correction of various aberrations, including the dependence of index of refraction on wavelength and the limitations imposed by the paraxial approximation. When the camera is in proper focus, the position of the recording medium coincides with the position of the real image formed by the lens. The resulting photograph will then be as sharp as possible. With a converging lens, the image distance increases as the object distance decreases (see Figs. 14.40a, 14.40b, and 14.40c, and the discussion in Section 14.4). Hence in “focusing” the camera, we move the lens closer to the film for a distant object and farther from the film for a nearby object. Real image CCD array

Shutter

To adjust for different object Elements distances, the image distance of lens is changed by moving the lens in or out.

Aperture-control diaphragm

14.39 Key elements of a digital camera.

Object

The lens forms an inverted, usually reduced, real image, in the plane of the CCD array.

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470

CHAPTER 14 Geometric Optics

14.40 Three photographs are taken with the same camera from the same position using lenses with focal lengths f = 28 mm, 105 mm, and 300 mm. The larger the value of ƒ, the smaller the angle of view.

8° (300 mm)

25° (105 mm)

Camera Lenses: Focal Length The choice of the focal length ƒ for a camera lens depends on the film size and the desired angle of view. Figure 14.40 shows photographs taken on 35-mm film with the same camera at the same position, but with lenses of different focal lengths. A lens of long focal length, called a telephoto lens, gives a small angle of view and a large image of a distant object; a lens of short focal length gives a small image and a wide angle of view and is called a wide-angle lens. To understand this behavior, recall that the focal length is the distance from the lens to the image when the object is infinitely far away. In general, for any object distance, using a lens of longer focal length gives a greater image distance. This also increases the height of the image; as was discussed in Section 14.4, the ratio of the image height y¿ to the object height y (the lateral magnification) is equal in absolute value to the ratio of x coordinate of image (v) to the x coordinate of object (u) [Eq. (14.16)]:

75° (28 mm)

m =

y¿ v = y u

With a lens of short focal length, the ratio |v>u| is small, and a distant object gives only a small image. When a lens with a long focal length is used, the image of this same object may entirely cover the area of the film. Hence the longer the focal length, the narrower the angle of view.

Camera Lenses: f-Number

Application

Inverting an Inverted

Image A camera lens makes an inverted image on the camera’s light-sensitive electronic detector. The internal software of the camera then inverts the image again so it appears the right way around on the camera’s display. A similar thing happens with your vision: The image formed on the retina of your eye is inverted, but your brain’s “software” erects the image so you see the world right-side up.

For the film to record the image properly, the total light energy per unit area reaching the film (the “exposure”) must fall within certain limits. This is controlled by the shutter and the lens aperture. The shutter controls the time interval during which light enters the lens. This is usually adjustable in steps correspon1 ding to factors of about 2, often from 1 s to 1000 s. The intensity of light reaching the film is proportional to the area viewed by the camera lens and to the effective area of the lens. The size of the area that the lens “sees” is proportional to the square of the angle of view of the lens, and so is roughly proportional to 1>ƒ 2. The effective area of the lens is controlled by means of an adjustable lens aperture, or diaphragm, a nearly circular hole with variable diameter D; hence the effective area is proportional to D2. Putting these factors together, we see that the intensity of light reaching the film with a particular lens is proportional to D2>ƒ 2. The light-gathering capability of a lens is commonly expressed by photographers in terms of the ratio ƒ>D, called the ƒ-number of the lens:

ƒ-number =

Focal length ƒ = Aperture diameter D

(14.19)

For example, a lens with a focal length ƒ = 50 mm and an aperture diameter D = 25 mm is said to have an ƒ-number of 2, or “an aperture of ƒ>2.” The light intensity reaching the film is inversely proportional to the square of the ƒ-number. For a lens with a variable-diameter aperture, increasing the diameter by a factor of 12 changes the ƒ-number by 1> 12 and increases the intensity at the film by a factor of 2. Adjustable apertures usually have scales labeled with successive numbers (often called ƒ-stops) related by factors of 12, such as ƒ>2

ƒ>2.8

ƒ>4

ƒ>5.6

ƒ>8

ƒ>11

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ƒ>16

471

14.5 Cameras

and so on. The larger numbers represent smaller apertures and exposures, and each step corresponds to a factor of 2 in intensity (Fig. 14.41). The actual exposure (total amount of light reaching the film) is proportional to both the aperture area 1 1 1 and the time of exposure. Thus ƒ>4 and 500 s, ƒ>5.6 and 250 s, and ƒ>8 and 125 s all correspond to the same exposure.

14.41 A camera lens with an adjustable diaphragm. Changing the diameter by a factor of 2 changes the intensity by a factor of 2. Å

2.

8

f-stops

16

Zoom Lenses and Projectors

D

Many photographers use a zoom lens, which is not a single lens but a complex collection of several lens elements that give a continuously variable focal length, often over a range as great as 10 to 1. Figures 14.42a and 14.42b show a simple system with variable focal length, and Fig. 14.42c shows a typical zoom lens for a single-lens reflex camera. Zoom lenses give a range of image sizes of a given object. It is an enormously complex problem in optical design to keep the image in focus and maintain a constant ƒ-number while the focal length changes. When you vary the focal length of a typical zoom lens, two groups of elements move within the lens and a diaphragm opens and closes. A projector for viewing slides, digital images, or motion pictures operates very much like a camera in reverse. In a movie projector, a lamp shines on the film, which acts as an object for the projection lens. The lens forms a real, enlarged, inverted image of the film on the projection screen. Because the image is inverted, the film goes through the projector upside down so that the image on the screen appears right-side up.

/

f 4 aperture

2.

8

Adjustable diaphragm

Larger f-numbers mean a smaller aperture.

4 5.6 8 11 .3

16

Example 14.12

4 5.6 8 11 .3

/

f 8 aperture

Photographic exposures

A common telephoto lens for a 35-mm camera has a focal length of 200 mm; its ƒ-stops range from ƒ > 2.8 to ƒ > 22. (a) What is the corresponding range of aperture diameters? (b) What is the corresponding range of image intensities on the film? '

'

'

proportional to D2, the square of the aperture diameter.

'

EXECUTE: (a) From Eq. (14.19), the diameter ranges from D =

S O L U T IO N IDENTIFY and SET UP: Part (a) of this problem uses the relationship among lens focal length ƒ, aperture diameter D, and ƒ-number. Part (b) uses the relationship between intensity and aperture diameter. We use Eq. (14.19) to relate D (the target variable) to the ƒ-number and the focal length ƒ = 200 mm. The intensity of the light reaching the film is proportional to D2>ƒ2 ; since ƒ is the same in each case, we conclude that the intensity in this case is

ƒ 200 mm = = 71 mm ƒ-number 2.8

to 200 mm = 9.1 mm 22 (b) Because the intensity is proportional to D2, the ratio of the intensity at ƒ > 2.8 to the intensity at ƒ > 22 is D =

'

a

'

'

'

2

2

'

71 mm 22 b = a b = 62 9.1 mm 2.8

1about 262

14.42 A simple zoom lens uses a converging lens and a diverging lens in tandem. (a) When the two lenses are close together, the combination behaves like a single lens of long focal length. (b) If the two lenses are moved farther apart, the combination behaves like a short-focal-length lens. (c) A typical zoom lens for a single-lens reflex camera, containing twelve elements arranged in four groups. (a) Zoom lens set for long focal length Image

4 cm

24 cm

(c) A practical zoom lens

(b) Zoom lens set for short focal length Image

8 cm

6 cm

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CHAPTER 14 Geometric Optics

1 EVALUATE: If the correct exposure time at ƒ > 2.8 is 1000 s, then the 1 1 exposure at ƒ > 22 is 1622 A 1000 s B = 16 s to compensate for the lower intensity. In general, the smaller the aperture and the larger the ƒ-number, the longer the required expo'

'

sure. Nevertheless, many photographers prefer to use small apertures so that only the central part of the lens is used to make the image. This minimizes aberrations that occur near the edges of the lens and gives the sharpest possible image.

'

'

Test Your Understanding of Section 14.5 When used with 35-mm film (image area 24 mm * 36 mm), a lens with ƒ = 50 mm gives a 45° angle of view and is called a “normal lens.” When used with a CCD array that measures 5 mm * 5 mm, this same lens is (i) a wide-angle lens; (ii) a normal lens; (iii) a telephoto lens.

14.6 PhET: Color Vision

y

The Eye

The optical behavior of the eye is similar to that of a camera. The essential parts of the human eye, considered as an optical system, are shown in Fig. 14.43a. The eye is nearly spherical and about 2.5 cm in diameter. The front portion is somewhat more sharply curved and is covered by a tough, transparent membrane called the cornea. The region behind the cornea contains a liquid called the aqueous humor. Next comes the crystalline lens, a capsule containing a fibrous jelly, hard at the center and progressively softer at the outer portions. The crystalline lens is held in place by ligaments that attach it to the ciliary muscle, which encircles it. Behind the lens, the eye is filled with a thin watery jelly called the vitreous humor. The indexes of refraction of both the aqueous humor and the vitreous humor are about 1.336, nearly equal to that of water. The crystalline lens, while not homogeneous, has an average index of 1.437. This is not very different from the indexes of the aqueous and vitreous humors. As a result, most of the refraction of light entering the eye occurs at the outer surface of the cornea. Refraction at the cornea and the surfaces of the lens produces a real image of the object being viewed. This image is formed on the light-sensitive retina, lining the rear inner surface of the eye. The retina plays the same role as the film in a camera. The rods and cones in the retina act like an array of miniature photocells (Fig. 14.43b); they sense the image and transmit it via the optic nerve to the brain. Vision is most acute in a small central region called the fovea centralis, about 0.25 mm in diameter.

14.43 (a) The eye. (b) There are two types of light-sensitive cells on the retina. The rods are more sensitive to light than the cones, but only the cones are sensitive to differences in color. A typical human eye contains about 1.3 * 108 rods and about 7 * 106 cones. (a) Diagram of the eye Contraction of the ciliary Ciliary muscle causes the lens to muscle become more convex, decreasing its focal Crystalline length to allow lens near vision. Iris Pupil

(b) Scanning electron micrograph showing retinal rods and cones in different colors Rod Cone

Retina Fovea centralis Image

Optic nerve

Vitreous humor

Object Cornea

Aqueous humor

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473

14.6 The Eye

In front of the lens is the iris. It contains an aperture with variable diameter called the pupil, which opens and closes to adapt to changing light intensity. The receptors of the retina also have intensity adaptation mechanisms. For an object to be seen sharply, the image must be formed exactly at the location of the retina. The eye adjusts to different object distances s by changing the focal length ƒ of its lens; the lens-to-retina distance, corresponding to s¿, does not change. (Contrast this with focusing a camera, in which the focal length is fixed and the lens-to-film distance is changed.) For the normal eye, an object at infinity is sharply focused when the ciliary muscle is relaxed. To permit sharp imaging on the retina of closer objects, the tension in the ciliary muscle surrounding the lens increases, the ciliary muscle contracts, the lens bulges, and the radii of curvature of its surfaces decrease; this decreases the focal length. This process is called accommodation. The extremes of the range over which distinct vision is possible are known as the far point and the near point of the eye. The far point of a normal eye is at infinity. The position of the near point depends on the amount by which the ciliary muscle can increase the curvature of the crystalline lens. The range of accommodation gradually diminishes with age because the crystalline lens grows throughout a person’s life (it is about 50% larger at age 60 than at age 20) and the ciliary muscles are less able to distort a larger lens. For this reason, the near point gradually recedes as one grows older. This recession of the near point is called presbyopia. Table 14.1 shows the approximate position of the near point for an average person at various ages. For example, an average person 50 years of age cannot focus on an object that is closer than about 40 cm.

Defects of Vision Several common defects of vision result from incorrect distance relationships in the eye. A normal eye forms an image on the retina of an object at infinity when the eye is relaxed (Fig. 14.44a). In the myopic (nearsighted) eye, the eyeball is too long from front to back in comparison with the radius of curvature of the cornea (or the cornea is too sharply curved), and rays from an object at infinity are focused in front of the retina (Fig. 14.44b). The most distant object for which an image can be formed on the retina is then nearer than infinity. In the hyperopic (farsighted) eye, the eyeball is too short or the cornea is not curved enough, and the image of an infinitely distant object is behind the retina (Fig. 14.44c). The myopic eye produces too much convergence in a parallel bundle of rays for an image to be formed on the retina; the hyperopic eye, not enough convergence. All of these defects can be corrected by the use of corrective lenses (eyeglasses or contact lenses). The near point of either a presbyopic or a hyperopic eye is farther from the eye than normal. To see clearly an object at normal reading distance (often assumed to be 25 cm), we need a lens that forms a virtual image of the object at or beyond the near point. This can be accomplished by a converging (positive) lens, as shown in Fig. 14.45. In effect the lens moves the object farther away from the eye to a point where a sharp retinal image can be formed. Similarly, correcting the myopic eye involves using a diverging (negative) lens to move the image closer to the eye than the actual object, as shown in Fig. 14.47. Astigmatism is a different type of defect in which the surface of the cornea is not spherical but rather more sharply curved in one plane than in another. As a result, horizontal lines may be imaged in a different plane from vertical lines (Fig. 14.47a). Astigmatism may make it impossible, for example, to focus clearly on both the horizontal and vertical bars of a window at the same time. Astigmatism can be corrected by use of a lens with a cylindrical surface. For example, suppose the curvature of the cornea in a horizontal plane is correct to focus rays from infinity on the retina but the curvature in the vertical plane is too great to form a sharp retinal image. When a cylindrical lens with its axis horizon-

Table 14.1 Receding of Near Point with Age Age (years)

Near Point (cm)

10

7

20

10

30

14

40

22

50

40

60

200

14.44 Refractive errors for (a) a normal eye, (b) a myopic (nearsighted) eye, and (c) a hyperopic (farsighted) eye viewing a very distant object. The dashed blue curve indicates the required position of the retina. (a) Normal eye Rays from distant object

(b) Myopic (nearsighted) eye Eye too long or cornea . . . rays focus in too sharply front of the curved . . . retina.

(c) Hyperopic (farsighted) eye Eye too short or . . . rays focus cornea not curved behind the enough . . . retina.

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474

CHAPTER 14 Geometric Optics

Focusing in the Animal Kingdom

Application

The crystalline lens and ciliary muscle found in humans and other mammals are among a number of focusing mechanisms used by animals. Birds can change the shape not only of their lens but also of the corneal surface. In aquatic animals the corneal surface is not very useful for focusing because its refractive index is close to that of water. Thus, focusing is accomplished entirely by the lens, which is nearly spherical. Fish focus by using a muscle to move the lens either inward or outward. Whales and dolphins achieve the same effect by filling or emptying a fluid chamber behind the lens to move the lens in or out.

14.45 (a) An uncorrected hyperopic (farsighted) eye. (b) A positive (converging) lens gives the extra convergence needed for a hyperopic eye to focus the image on the retina. Image not focused on retina

Nearby object Farsighted people have trouble focusing on nearby objects. A converging lens creates a virtual image at or beyond the eye’s near point.

(a) Hyperopic eye

Converging lens

Image focused on retina

(b)

tal is placed before the eye, the rays in a horizontal plane are unaffected, but the additional divergence of the rays in a vertical plane causes these to be sharply imaged on the retina (Fig. 14.47b). Lenses for vision correction are usually described in terms of the power, defined as the reciprocal of the focal length expressed in meters. The unit of power is the diopter. Thus a lens with ƒ = 0.50 m has a power of 2.0 diopters, ƒ = - 0.25 m corresponds to - 4.0 diopters, and so on. The numbers on a prescription for glasses are usually powers expressed in diopters. When the correction involves both astigmatism and myopia or hyperopia, there are three numbers: one for the spherical power, one for the cylindrical power, and an angle to describe the orientation of the cylinder axis. Example 14.13

Correcting for farsightedness

The near point of a certain hyperopic eye is 100 cm in front of the eye. Find the focal length and power of the contact lens that will permit the wearer to see clearly an object that is 25 cm in front of the eye. SOLUTION

the eye, 100 cm from it. The contact lens (which we treat as having negligible thickness) is at the surface of the cornea, so the object distance is u = - 25 cm. The virtual image is on the incoming side of the contact lens, so the image distance is v = -100 cm. We use Eq. (14.15) to determine the required focal length ƒ of the contact lens; the corresponding power is 1>ƒ.

IDENTIFY and SET UP: Figure 14.48 shows the situation. We want the lens to form a virtual image of the object at the near point of 14.46 (a) An uncorrected myopic (nearsighted) eye. (b) A negative (diverging) lens spreads the rays farther apart to compensate for the excessive convergence of the myopic eye. Image not focused on retina

Distant object (a)

Myopic eye

Nearsighted people have trouble seeing distant objects. A diverging lens creates a virtual image that is inside the eye’s far point.

Diverging lens

(b)

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Image focused on retina

14.6 The Eye (a) Vertical lines are imaged in front of the retina.

(b) A cylindrical lens corrects for astigmatism.

Shape of eyeball or lens causes vertical and horizontal elements to focus at different distances.

This cylindrical lens is curved in the vertical, but not the horizontal, direction; it changes the focal length of vertical elements.

EXECUTE: From Eq. (14.16), 1 1 1 1 1 = = v u ƒ - 100 cm -25 cm

475

14.47 One type of astigmatism and how it is corrected.

14.48 Using a contact lens to correct for farsightedness. For clarity, the eye and contact lens are shown much larger than the scale of the figure; the 2.5-cm diameter of the eye is actually much smaller than the focal length ƒ of the contact lens.

ƒ = +33 cm We need a converging lens with focal length ƒ = 33 cm and power 1>10.33 m2 = + 3.0 diopters.

Converging lens Image

|v| 5 |2100 cm| = 100 cm

EVALUATE: In this example we used a contact lens to correct hyperopia. Had we used eyeglasses, we would have had to account for the separation between the eye and the eyeglass lens, and a somewhat different power would have been required (see Example 14.14).

Example 14.14

Object

|f| |u| 5|2 25 cm| = 25

Correcting for nearsightedness

The far point of a certain myopic eye is 50 cm in front of the eye. Find the focal length and power of the eyeglass lens that will permit the wearer to see clearly an object at infinity. Assume that the lens is worn 2 cm in front of the eye.

nate of image to be v = -48 cm . As in Example 14.12, we use the values of u and v to calculate the required focal length. EXECUTE: From Eq. (14.15), 1 1 1 1 = = v u ƒ -48 cm

S O L U T IO N IDENTIFY and SET UP: Figure 14.49 shows the situation. The far point of a myopic eye is nearer than infinity. To see clearly objectsbeyond the far point, we need a lens that forms a virtual image of such objects no farther from the eye than the far point. Assume that the virtual image of the object at infinity is formed at the far point, 50 cm in front of the eye (48 cm in front of the eyeglass lens). Then when the (object is at infinity) u = - q , we want the x coordi-

1 -`

ƒ = - 48 cm We need a diverging lens with focal length ƒ = -48 cm and power 1>1 -0.48 m2 = - 2.1 diopters. EVALUATE: If a contact lens were used to correct this myopia, we would need ƒ = -50 cm and a power of - 2.0 diopters. Can you see why?

14.49 Using an eyeglass lens to correct for nearsightedness. For clarity, the eye and eyeglass lens are shown much larger than the scale of the figure.

Object at infinity

When the distance of object from eye is infinity, all rays are parallel to the axis and the distance of image from lens equals |f |.

Diverging lens

|v| = |f| = |-48cm| = 48 cm |u| = q

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476

CHAPTER 14 Geometric Optics Test Your Understanding of Section 14.6 A certain eyeglass lens is thin at its center, even thinner at its top and bottom edges, and relatively thick at its left and right edges. What defects of vision is this lens intended to correct? (i) hyperopia for objects oriented both vertically and horizontally; (ii) myopia for objects oriented both vertically and horizontally; (iii) hyperopia for objects oriented vertically and myopia for objects oriented horizontally; (iv) hyperopia for objects oriented horizontally and myopia for objects oriented vertically. y

14.7

The Magnifier

The apparent size of an object is determined by the size of its image on the retina. If the eye is unaided, this size depends on the angle u subtended by the object at the eye, called its angular size (Fig. 14.50a). To look closely at a small object, such as an insect or a crystal, you bring it close to your eye, making the subtended angle and the retinal image as large as possible. But your eye cannot focus sharply on objects that are closer than the near point, so the angular size of an object is greatest (that is, it subtends the largest possible viewing angle) when it is placed at the near point. In the following discussion we will assume an average viewer for whom the near point is 25 cm from the eye. A converging lens can be used to form a virtual image that is larger and farther from the eye than the object itself, as shown in Fig. 14.50b. Then the object can be moved closer to the eye, and the angular size of the image may be substantially larger than the angular size of the object at 25 cm without the lens. A lens used in this way is called a magnifier, otherwise known as a magnifying glass or a simple magnifier. The virtual image is most comfortable to view when it is placed at infinity, so that the ciliary muscle of the eye is relaxed; this means that the object is placed at the focal point F1 of the magnifier. In the following discussion we assume that this is done. In Fig. 14.50a the object is at the near point, where it subtends an angle u at the eye. In Fig. 14.50b a magnifier in front of the eye forms an image at infinity, and the angle subtended at the magnifier is u¿. The usefulness of the magnifier is given by the ratio of the angle u¿ (with the magnifier) to the angle u (without the magnifier). This ratio is called the angular magnification M:

?

M =

u¿ u

(angular magnification)

(14.20)

CAUTION Angular magnification vs. lateral magnification Don’t confuse the angular magnification M with the lateral magnification m. Angular magnification is the ratio of the angular size of an image to the angular size of the corresponding object; lateral magnifica-

14.50 (a) The angular size u is largest when the object is at the near point. (b) The magnifier gives a virtual image at infinity. This virtual image appears to the eye to be a real object subtending a larger angle u¿ at the eye. (a)

(b) When the inchworm is at the eye’s near point, its image on the retina is as large as it can be and still be in focus. At the near point, the inchworm subtends an angle u. y

u |u|5 |225 cm| = 25cm

With a magnifier, the inchworm can be placed closer than the near point. The magnifier creates an enlarged, upright, virtual image. Parallel

/

M 5 u u u 5 y f

/

u y When the object is placed at the magnifier’s focal point, the image is at infinity. |v|5 `

F1

|v|5|f|= f

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u

14.7 The Magnifier

477

tion refers to the ratio of the height of an image to the height of the corresponding object. For the situation shown in Fig. 14.50b, the angular magnification is about 3 *, since the inchworm subtends an angle about three times larger than that in Fig. 14.50a; hence the inchworm will look about three times larger to the eye. The lateral magnification m = v>u in Fig. 14.50b is infinite because the virtual image is at infinity, but that doesn’t mean that the inchworm looks infinitely large through the magnifier! (That’s why we didn’t attempt to draw an infinitely large inchworm in Fig. 14.50b.) When dealing with a magnifier, M is useful but m is not. y

To find the value of M, we first assume that the angles are small enough that each angle (in radians) is equal to its sine and its tangent. Using Fig. 14.51a and drawing the ray in Fig. 14.51b that passes undeviated through the center of the lens, we find that u and u¿ (in radians) are y y u = u¿ = 25 cm ƒ Combining these expressions with Eq. (14.20), we find M =

y>ƒ u¿ 25 cm = = u y>25 cm ƒ

(angular magnification for a simple magnifier)

(14.21)

It may seem that we can make the angular magnification as large as we like by decreasing the focal length ƒ. In fact, the aberrations of a simple double-convex lens set a limit to M of about 3* to 4*. If these aberrations are corrected, the angular magnification may be made as great as 20* . When greater magnification than this is needed, we usually use a compound microscope, discussed in the next section. 14.51 (a) Elements of a microscope. (b) The object O is placed just outside the first focal point of the objective (the distance s1 has been exaggerated for clarity). (c) This microscope image shows single-celled organisms about 2 * 10-4 m (0.2 mm) across. Typical light microscopes can resolve features as small as 2 * 10-7 m, comparable to the wavelength of light. (a) Elements of a microscope

(b) Microscope optics

(c) Single-celled freshwater algae (Micrasterias denticulata)

Eyepiece

|f1| = f1 |f2| = f2 are both lenses are converging

F2

Real image

|f2|

Eyepiece |f2|

|v1| Object

F1 |f1|

|u1| Light source

The objective forms a real, inverted image I inside the focal point F2 of the eyepiece.

I F2

Objective lens

|f2| F1 O

The eyepiece uses the image I as an object Objective and creates an enlarged, virtual image I (still inverted). I

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478

CHAPTER 14 Geometric Optics Test Your Understanding of Section 14.7 You are examining a gem using a magnifier. If you change to a different magnifier with twice the focal length of the first one, (i) you will have to hold the object at twice the distance and the angular magnification will be twice as great; (ii) you will have to hold the object at twice the distance and the angular magnification will be 12 as great; (iii) you will have to hold the object at 12 the distance and the angular magnification will be twice as great; (iv) you will have to hold the object at 12 the distance and the angular magnification will be 21 as great.

14.8 ActivPhysics 15.12: Two-Lens Optical Systems

y

Microscopes and Telescopes

Cameras, eyeglasses, and magnifiers use a single lens to form an image. Two important optical devices that use two lenses are the microscope and the telescope. In each device a primary lens, or objective, forms a real image, and a second lens, or eyepiece, is used as a magnifier to make an enlarged, virtual image.

Microscopes When we need greater magnification than we can get with a simple magnifier, the instrument that we usually use is the microscope, sometimes called a compound microscope. The essential elements of a microscope are shown in Fig. 14.51a. To analyze this system, we use the principle that an image formed by one optical element such as a lens or mirror can serve as the object for a second element. We used this principle in Section 14.4 when we derived the thin-lens equation by repeated application of the single-surface refraction equation; we used this principle again in Example 14.11 (Section 14.4), in which the image formed by a lens was used as the object of a second lens. The object O to be viewed is placed just beyond the first focal point F1 of the objective, a converging lens that forms a real and enlarged image I (Fig. 14.51b). In a properly designed instrument this image lies just inside the first focal point F1œ of a second converging lens called the eyepiece or ocular. (The reason the image should lie just inside F1œ is left for you to discover; see Problem 14.108.) The eyepiece acts as a simple magnifier, as discussed in Section 14.7, and forms a final virtual image I¿ of I. The position of I¿ may be anywhere between the near and far points of the eye. Both the objective and the eyepiece of an actual microscope are highly corrected compound lenses with several optical elements, but for simplicity we show them here as simple thin lenses. As for a simple magnifier, what matters when viewing through a microscope is the angular magnification M. The overall angular magnification of the compound microscope is the product of two factors. The first factor is the lateral magnification m1 of the objective, which determines the linear size of the real image I; the second factor is the angular magnification M2 of the eyepiece, which relates the angular size of the virtual image seen through the eyepiece to the angular size that the real image I would have if you viewed it without the eyepiece. The first of these factors is given by m1 = -

v1 u1

(14.22)

where v1 and u1 are values of v and u (with sign), respectively, for the objective lens. Carefully note the signs of v1 and u1 and that line through the centre of lens is x axis. Ordinarily, the object is very close to the focal point, and the resulting x coordinate of image I (v1) is very great in comparison to the focal length ƒ1 of the objective lens. Thus u1 = (-|f1|) = f1 is approximately equal to |f1| = f1 and we can write m1 = ( -v1>ƒ1 ). The real image I is close to the focal point F1œ of the eyepiece, so to find the eyepiece angular magnification, we can use Eq. (14.21): M2 = 125 cm2>ƒ2 , where ƒ2 is the focal length of the eyepiece (considered as a simple lens). The '

'

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14.8 Microscopes and Telescopes

overall angular magnification M of the compound microscope (apart from a negative sign, which is customarily ignored) is the product of the two magnifications: M = m1 M2 = '

125 cm2(- v1) ƒ1 ƒ2 '

(angular magnification for a microscope)

(14.23)

where v1, ƒ1 , and ƒ2 are measured in centimeters. The final image is inverted with respect to the object. Microscope manufacturers usually specify the values of m1 and M2 for microscope components rather than the focal lengths of the objective and eyepiece. Equation (14.23) shows that the angular magnification of a microscope can be increased by using an objective of shorter focal length ƒ1 , thereby increasing m1 and the size of the real image I. Most optical microscopes have a rotating “turret” with three or more objectives of different focal lengths so that the same object can be viewed at different magnifications. The eyepiece should also have a short focal length ƒ2 to help to maximize the value of M. To take a photograph using a microscope (called a photomicrograph or micrograph), the eyepiece is removed and a camera placed so that the real image I falls on the camera’s CCD array or film. Figure 14.51c shows such a photograph. In this case what matters is the lateral magnification of the microscope as given by Eq. (14.22). '

'

Telescopes The optical system of a telescope is similar to that of a compound microscope. In both instruments the image formed by an objective is viewed through an eyepiece. The key difference is that the telescope is used to view large objects at large distances and the microscope is used to view small objects close at hand. Another difference is that many telescopes use a curved mirror, not a lens, as an objective. Figure 14.52 shows an astronomical telescope. Because this telescope uses a lens as an objective, it is called a refracting telescope or refractor. The objective lens forms a real, reduced image I of the object. This image is the object for the eyepiece lens, which forms an enlarged, virtual image of I. Objects that are viewed with a telescope are usually so far away from the instrument that the first image I is formed very nearly at the second focal point of the objective lens. If the final image I¿ formed by the eyepiece is at infinity (for most comfortable viewing by a normal eye), the first image 14.52 Optical system of an astronomical refracting telescope.

Objective lens

F1

u u

|f2| = f2 |f2| = f2

|f1| = f1

|f1| = f1

F2, F1

a

The objective forms a real, b inverted image I of the distant Objective object at its second focal point F1, which is also the first focal I at point F2 of the eyepiece. infinity

u

y

I

c u F2 d

Eyepiece The eyepiece uses image I to form a magnified, virtual image I (still inverted) at infinity.

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Eyepiece

479

480

CHAPTER 14 Geometric Optics

must also be at the first focal point of the eyepiece. The distance between objective and eyepiece, which is the length of the telescope, is therefore the sum of the focal lengths of objective and eyepiece, ƒ1 + ƒ2. Note that f1 and f2 are positive. The angular magnification M of a telescope is defined as the ratio of the angle subtended at the eye by the final image I¿ to the angle subtended at the (unaided) eye by the object. We can express this ratio in terms of the focal lengths of objective and eyepiece. In Fig. 14.52 the ray passing through F1 , the first focal point of the objective, and through F2œ , the second focal point of the eyepiece, is shown in red. The object (not shown) subtends an angle u at the objective and would subtend essentially the same angle at the unaided eye. Also, since the observer’s eye is placed just to the right of the focal point F2œ , the angle subtended at the eye by the final image is very nearly equal to the angle u¿. Because bd is parallel to the optic axis, the distances ab and cd are equal to each other and also to the height y¿ of the real image I. Because the angles u and u¿ are small, they may be approximated by their tangents. From the right triangles F1 ab and F2œ cd, '

'

u =

-y¿ ƒ1

u¿ =

'

y¿ ƒ2

and the angular magnification M is M =

y¿>ƒ2 ƒ1 u¿ = = u y¿>ƒ1 ƒ2

(angular magnification for a telescope)

(14.24)

The angular magnification M of a telescope is equal to the ratio of the focal length of the objective to that of the eyepiece. The negative sign shows that the final image is inverted. Equation (14.24) shows that to achieve good angular magnification, a telescope should have a long objective focal length ƒ1 . By contrast, Eq. (14.23) shows that a microscope should have a short objective focal length. However, a telescope objective with a long focal length should also have a large diameter D so that the ƒ-number ƒ1>D will not be too large; as described in Section 14.5, a large ƒ-number means a dim, low-intensity image. Telescopes typically do not have interchangeable objectives; instead, the magnification is varied by using different eyepieces with different focal lengths ƒ2 . Just as for a microscope, smaller values of ƒ2 give larger angular magnifications. An inverted image is no particular disadvantage for astronomical observations. When we use a telescope or binoculars—essentially a pair of telescopes mounted side by side—to view objects on the earth, though, we want the image to be rightside up. In prism binoculars, this is accomplished by reflecting the light several times along the path from the objective to the eyepiece. The combined effect of the reflections is to flip the image both horizontally and vertically. Binoculars are usually described by two numbers separated by a multiplication sign, such as 7 * 50. The first number is the angular magnification M, and the second is the diameter of the objective lenses (in millimeters). The diameter helps to determine the lightgathering capacity of the objective lenses and thus the brightness of the image. In the reflecting telescope (Fig. 14.53a) the objective lens is replaced by a concave mirror. In large telescopes this scheme has many advantages. Mirrors are inherently free of chromatic aberrations (dependence of focal length on wavelength), and spherical aberrations (associated with the paraxial approximation) are easier to correct than with a lens. The reflecting surface is sometimes parabolic rather than spherical. The material of the mirror need not be transparent, and it can be made more rigid than a lens, which has to be supported only at its edges. The largest reflecting telescopes in the world, the Keck telescopes atop Mauna Kea in Hawaii, each have an objective mirror of overall diameter 10 m made up of 36 separate hexagonal reflecting elements. One challenge in designing reflecting telescopes is that the image is formed in front of the objective mirror, in a region traversed by incoming rays. Isaac New'

'

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14.8 Microscopes and Telescopes

481

14.53 (a), (b), (c) Three designs for reflecting telescopes. (d) This photo shows the interior of the Gemini North telescope, which uses the design shown in (c). The objective mirror is 8 meters in diameter. (c)

(b)

(a) Starlight

Starlight

(d) Starlight

Planar secondary mirror

Cage

Planar secondary mirror

Eyepiece

Secondary mirror (in housing)

The cage at the focal point may contain a camera.

Concave objective mirror This is a common design for the telescopes of amateur astronomers.

Objective mirror

Concave objective mirror Eyepiece This is a common design for large modern telescopes. A camera or other instrument package is typically used instead of an eyepiece.

ton devised one solution to this problem. A flat secondary mirror oriented at 45° to the optic axis causes the image to be formed in a hole on the side of the telescope, where it can be magnified with an eyepiece (Fig. 14.53b). Another solution uses a secondary mirror that causes the focused light to pass through a hole in the objective mirror (Fig. 14.53c). Large research telescopes, as well as many amateur telescopes, use this design (Fig. 14.53d). Like a microscope, when a telescope is used for photography the eyepiece is removed and a CCD array or photographic film is placed at the position of the real image formed by the objective. (Some long-focal-length “lenses” for photography are actually reflecting telescopes used in this way.) Most telescopes used for astronomical research are never used with an eyepiece.

Test Your Understanding of Section 14.8 Which gives a lateral magnification of greater absolute value: (i) the objective lens in a microscope (Fig. 14.51); (ii) the objective lens in a refracting telescope (Fig. 14.52); or (iii) not enough information is given to decide? y

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Hole in objective mirror Reflection of secondary mirror

CHAPTER

14

SUMMARY

Reflection or refraction at a plane surface: When rays diverge from an object point P and are reflected or refracted, the directions of the outgoing rays are the same as though they had diverged from a point P¿ called the image point. If they actually converge at P¿ and diverge again beyond it, P¿ is a real image of P; if they only appear to have diverged from P¿, it is a virtual image. Images can be either erect or inverted.

Lateral magnification: The lateral magnification m in any reflecting or refracting situation is defined as the ratio of image height y¿ to object height y. When m is positive, the image is erect; when m is negative, the image is inverted.

m =

y¿ - v = y u

P

P9

Plane mirror

Q

(14.1)

y

y P

C

P

u

Q

u

|R|

V

|v|

|u|

Focal point and focal length: The focal point of a mirror is the point where parallel rays converge after reflection from a concave mirror, or the point from which they appear to diverge after reflection from a convex mirror. Rays diverging from the focal point of a concave mirror are parallel after reflection; rays converging toward the focal point of a convex mirror are parallel after reflection. The x coordinate of focal point is called focal length f, when vertex (or pole) v of mirror is taken as origin, optic axis (or principal axis) as x-axis, and direction of incident rays as that of +ive x-axis. The focal points of a lens are defined similarly.

|R|

F

C

|u| 5  |R| |v| 5 5 |f| 2

Relating object and image distances: The formulas for u and v for plane and spherical mirrors and single refracting surfaces are summarized in the table. The equation for a plane surface can be obtained from the corresponding equation for a spherical surface by setting R = q . (See Examples 14.1–14.7.)

Q

1 3

4 2 C

P9 F

2 4

P

Q9

V 3

1

Plane Mirror

Spherical Mirror

Object and image distances

1 1 + = 0 v u

1 1 2 1 + = = v u R f

Lateral magnification

m = -

v = 1 u

m = -

v u

Plane Refracting Surface

Spherical Refracting Surface

nb

nb

-

v

na u

= 0

na v

v

-

na

'

m = -

nb u

=

nb - na

u

R na v '

= 1

m = -

'

nb u '

Object–image relationships derived in this chapter are valid only for rays close to and nearly parallel to the optic axis; these are called paraxial rays. Nonparaxial rays do not converge precisely to an image point. This effect is called spherical aberration. Thin lenses: The object–image relationship, given by Eq. (14.15), is the same for a thin lens as for a spherical mirror. Equation (14.18), the lensmaker’s equation, relates the focal length of a lens to its index of refraction and the radii of curvature of its surfaces. (See Examples 14.8–14.11.)

1 1 1 = v u ƒ

(14.15)

Q

1 2

3

1 1 1 = 1n - 12 a b ƒ R1 R2

P

F2

F1

(14.18)

P9 3 Q9 1 2

Sign rules:

• Direction of incident rays is taken as that of +ive x axis

• Centre of mirror, refracting surface or lens is taken as origin

• x coordinates of object, image give the value of u and v. x coor-

• Line through this origin and perpendicular to surface is taken as x axis. In case of lens line joining the centres of curvature of two surfaces or optic axis is taken as x axis

dinates of focal point

• x coordinates of centres of curvature and focal point are equal to R, R, R2, focal length f.

482

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Summary Bridging Problem Cameras: A camera forms a real, inverted, reduced image of the object being photographed on a light-sensitive surface. The amount of light striking this surface is controlled by the shutter speed and the aperture. The intensity of this light is inversely proportional to the square of the ƒ-number of the lens. (See Example 14.12.)

ƒ-number = =

483 483

Focal length Aperture diameter ƒ D

(14.19)

Object

Inverted, real image

The eye: In the eye, refraction at the surface of the cornea forms a real image on the retina. Adjustment for various object distances is made by squeezing the lens, thereby making it bulge and decreasing its focal length. A nearsighted eye is too long for its lens; a farsighted eye is too short. The power of a corrective lens, in diopters, is the reciprocal of the focal length in meters. (See Examples 14.13 and 14.14.)

The simple magnifier: The simple magnifier creates a virtual image whose angular size u¿ is larger than the angular size u of the object itself at a distance of 25 cm, the nominal closest distance for comfortable viewing. The angular magnification M of a simple magnifier is the ratio of the angular size of the virtual image to that of the object at this distance.

M =

u¿ 25 cm = u ƒ

(14.21)

Normal eye

Parallel u9 y F1

u9

s5f s9 5 2`

Microscopes and telescopes: In a compound microscope, the objective lens forms a first image in the barrel of the instrument, and the eyepiece forms a final virtual image, often at infinity, of the first image. The telescope operates on the same principle, but the object is far away. In a reflecting telescope, the objective lens is replaced by a concave mirror, which eliminates chromatic aberrations.

BRIDGING PROBLEM

Rays from distant object

Eyepiece

Objective I F1 I O F1 F2 F2 |f 1| |f 1| |f 2| |f 2| |v1| |u1|

Image Formation by a Wine Goblet

A thick-walled wine goblet can be considered to be a hollow glass sphere with an outer radius of 4.00 cm and an inner radius of 3.40 cm. The index of refraction of the goblet glass is 1.50. (a) A beam of parallel light rays enters the side of the empty goblet along a horizontal radius. Where, if anywhere, will an image be formed? (b) The goblet is filled with white wine 1n = 1.372. Where is the image formed? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. The goblet is not a thin lens, so you cannot use the thin-lens formula. Instead, you must think of the inner and outer surfaces of the goblet walls as spherical refracting surfaces. The image formed by one surface serves as the object for the next surface. 2. Choose the appropriate equation that relates the image and object distances for a spherical refracting surface.

EXECUTE 3. For the empty goblet, each refracting surface has glass on one side and air on the other. Find the position of the image formed by the first surface, the outer wall of the goblet. Use this as the object for the second surface (the inner wall of the same side of the goblet) and find the position of the second image. (Hint: Be sure to account for the thickness of the goblet wall.) 4. Continue the process of step 3. Consider the refractions at the inner and outer surfaces of the glass on the opposite side of the goblet, and find the position of the final image. (Hint: Be sure to account for the distance between the two sides of the goblet.) 5. Repeat steps 3 and 4 for the case in which the goblet is filled with wine. EVALUATE 6. Are the images real or virtual? How can you tell?

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484

CHAPTER 14 Geometric Optics

Problems

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q14.1 A spherical mirror is cut in half horizontally. Will an image be formed by the bottom half of the mirror? If so, where will the image be formed? Q14.2 For the situation shown in Fig. 14.3, is the image distance s¿ positive or negative? Is the image real or virtual? Explain your answers. Q14.3 The laws of optics also apply to electromagnetic waves invisible to the eye. A satellite TV dish is used to detect radio waves coming from orbiting satellites. Why is a curved reflecting surface (a “dish”) used? The dish is always concave, never convex; why? The actual radio receiver is placed on an arm and suspended in front of the dish. How far in front of the dish should it be placed? Q14.4 Explain why the focal length of a plane mirror is infinite, and explain what it means for the focal point to be at infinity. Q14.5 If a spherical mirror is immersed in water, does its focal length change? Explain. Q14.6 For what range of object positions does a concave spherical mirror form a real image? What about a convex spherical mirror? Q14.7 When a room has mirrors on two opposite walls, an infinite series of reflections can be seen. Discuss this phenomenon in terms of images. Why do the distant images appear fainter? Q14.8 For a spherical mirror, if s = ƒ, then s¿ = q , and the lateral magnification m is infinite. Does this make sense? If so, what does it mean? Q14.9 You may have noticed a small convex mirror next to your bank’s ATM. Why is this mirror convex, as opposed to flat or concave? What considerations determine its radius of curvature? Q14.10 A student claims that she can start a fire on a sunny day using just the sun’s rays and a concave mirror. How is this done? Is the concept of image relevant? Can she do the same thing with a convex mirror? Explain. Q14.11 A person looks at his reflection in the concave side of a shiny spoon. Is it right side up or inverted? Does it matter how far his face is from the spoon? What if he looks in the convex side? (Try this yourself!) Q14.12 In Example 14.4 (Section 14.2), there appears to be an ambiguity for the case s = 10 cm as to whether s¿ is + q or - q and whether the image is erect or inverted. How is this resolved? Or is it? Q14.13 Suppose that in the situation of Example 14.7 of Section 14.3 (see Fig. 14.26) a vertical arrow 2.00 m tall is painted on the side of the pool beneath the water line. According to the calculations in the example, this arrow would appear to the person shown in Fig. 14.26 to be 1.50 m long. But the discussion following Eq. (14.13) states that the magnification for a plane refracting surface is m = 1, which suggests that the arrow would appear to the person to be 2.00 m long. How can you resolve this apparent contradiction? Q14.14 The bottom of the passenger-side mirror on your car notes, “Objects in mirror are closer than they appear.” Is this true? Why? Q14.15 How could you very quickly make an approximate measurement of the focal length of a converging lens? Could the same method be applied if you wished to use a diverging lens? Explain.

Q14.16 The focal length of a simple lens depends on the color (wavelength) of light passing through it. Why? Is it possible for a lens to have a positive focal length for some colors and negative for others? Explain. Q14.17 When a converging lens is immersed in water, does its focal length increase or decrease in comparison with the value in air? Explain. Q14.18 A spherical air bubble in water can function as a lens. Is it a converging or diverging lens? How is its focal length related to its radius? Q14.19 Can an image formed by one reflecting or refracting surface serve as an object for a second reflection or refraction? Does it matter whether the first image is real or virtual? Explain. Q14.20 If a piece of photographic film is placed at the location of a real image, the film will record the image. Can this be done with a virtual image? How might one record a virtual image? Q14.21 You’ve entered a survival contest that will include building a crude telescope. You are given a large box of lenses. Which two lenses do you pick? How do you quickly identify them? Q14.22 BIO You can’t see clearly underwater with the naked eye, but you can if you wear a face mask or goggles (with air between your eyes and the mask or goggles). Why is there a difference? Could you instead wear eyeglasses (with water between your eyes and the eyeglasses) in order to see underwater? If so, should the lenses be converging or diverging? Explain. Q14.23 You take a lens and mask it so that light can pass through only the bottom half of the lens. How does the image formed by the masked lens compare to the image formed before masking?

SINGLE CORRECT V Q14.1 A point source of length S is located on a wall. A plane mirror M having length l is moving l parallel to the wall with constant velocity v. The bright patch formed on the wall by reflected S Wall light will (a) move with uniform velocity v and will have a length 2l. (b) move with uniform velocity 2v and will have a length l. (c) move with uniform velocity 2v and will have a length 2l. (d) move with uniform velocity but will have a changing length. Q14.2 There are two plane mirror with reflecting surfaces facing each other. The mirrors are moving with speed v away from each other. A point object is placed between the mirrors. The velocity of the n-th image will be (a) nv (b) 2nv (c) 3nv (d) 4nv Q14.3 A ray of light strikes a corner reflector as shown in figure. As the angle of incidence u is increased, the angle between the final reflected ray and the incident ray (a) increases (b) decreases (c) remains the same (d) Information are insufficient to decide

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Single Correct Second reflected ray q'

r'

Q14.9 In the diagram shown below, a point source O is placed vertically below the center of a circular plane mirror. The light rays starting from the source are reflected from the mirror such that a circular area A on the ground receives light. Now, a large glass slab is placed between the mirror and the source O. What will the magnitude of the new area on the ground receiving light? powers, in decreasing order, are :

M" b'

q'

Incident i ray

q

q Mirror

M

Circular plane mirror Circular plane mirror

M'

b Reflected ray

o

Q14.4 A tall girl standing in front of a vertical plane mirror, is able to see hereself only upto her knees. She can see the lower part of her legs, if (a) She moves towards the mirror (b) She moves away from the mirror (c) She bends down (d) She stands on a stool Q14.5 A small object AB is being viewed by the observer through the given arrangement of mirrors. The image of AB as seen by the observer will be A B

485

o

(a) A (b) Greater than A (c) Less than A (d) Cannot tell, as the information given is insufficient Q14.10 A light ray strikes a hexagonal ice crystal floating in the air as shown in the figure. The correct path of ray may be? (A)

Incoming ray

(B)

(C)

45˚

(D)

45˚

(a) vertical, pointing upwards (b) vertical, pointing downwards (c) horizontal, pointing towards right (d) horizontal, pointing towards left Q34.6 A beam of light passes from medium 1 to medium 2 to medium 3 as shown in the diagram. What may be concluded about the three indexes of refraction, n1, n2 and n3? (a) n1 > n2 > n3 (b) n1 > n3 > n2 (c) n2 > n3 > n1 (d) n2 > n1 > n3 Q14.7 The observer at O views two closely spaced spots on a vertical wall through an angled glass slab as shown. As seen by observer, the spots appear. (a) shifted upward (b) shifted downward (c) spaced farther apart (d) spaced closer together Q14.8 A light ray strikes a prism as shown in the drawing. The angles of the prism are 90°, 45° and 45°. The critical angle of the prism material is 49°. What rays are the possible continuations of the incident ray? (a) The ray 1 only (b) The ray 2 only (c) The ray 3 only (d) Both rays 2 and 3

n1 n2 n3

O

Slab

1 Incident ray 2 3

Q14.11 When the reflection of a real object is seen in a convex mirror the image will: (a) always be real (b) always be virtual (c) may be either real or virtual (d) will always be magnified. Q14.12 A concave mirror is used to form image of the Sun on a white screen. If the lower half of the mirror were covered with an opaque card, the effect on the image on the screen would be (a) negligible (b) it will make the image less bright than before (c) it will make the upper half of the image disappear (d) it will make the lower half of the image disappear Q14.13 A point object is between the Pole and Focus of a concave mirror, and moving away from the mirror with a constant speed. Then, the velocity of the image is: (a) away from mirror and increasing in magnitude (b) towards mirror and increasing in magnitude (c) away from mirror and decreasing in magnitude (d) towards mirror and decreasing in magnitude Q14.14 The passenger side-view mirror on an automobile often has the notation. “Objects seen in mirror are closer than they appear”. Is the image really farther away than the object? (a) Yes, the image is smaller and farther away than the object (b) No, the image is smaller and closer than the object (c) No, the image is larger and closer than the object (d) Yes, the image is larger and farther away than the object Q14.15 A point object P moves towards a convex mirror with a constant speed v, along the optical axis. The speed of the image (a) Is always less than v (b) May be greater than, equal to or less than v, depending upon the position of P (c) Increases as P comes closer to the mirror (d) Decreases as P comes closer to the mirror Q14.16 A concave mirror forms a real image of an object. If the whole arrangement is immersed in water, then the image will

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486

CHAPTER 14 Geometric Optics

(a) Disappear (b) Shift towards the mirror (c) Shift away from the mirror (d) Not suffer any change in position Q14.17 Let spherical aberration and chromatic aberration be denoted by S and C respectively. In case of concave mirror, (a) S and C both may be present (b) Both S and C can not exist (c) S may be present but C cannot exist (d) C may be present but S cannot exist Q14.18 A opaque card is held over the lower half of a converging lens as shown in figure. Which picture is best shows the image that appears on the screen. (a)

(b)

(c)

(d)

F

F Card

Screen

Q14.19 A screen is placed at a distance 80 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen.If no position could be found, the focal length of the lens: (a) must be greater than 40 cm (b) must be greater than 20 cm (c) must be less than 20 cm (d) must not be greater than 40 cm Q14.20 A point object O moves from the P Q R principal axis of a converging lens in a direction OP. I is the image of O, will O move initially in the direction U S (a) IQ T (b) IR (c) IS (d) IU

Paragraph for Question Nos. 1 to 3 All objects referred to, in the subsequent problems lie on the principal axis Air

Air

Water

Surface 1

Surface 2

Q14.21 If light is incident on surface 1 from left, the image formed after first refraction is definitely (a) Real for a real object (b) Virtual for a real object (c) Real for a virtual object (d) Virtual for a virtual object Q14.22 In above question, if the object is real, then the final image formed after two refractions (a) Must be real (b) May be virtual or real (c) Must be virtual (d) None of these Q14.23 If light is incident on surface 2 from right then which of the following is true for image formed after a single refraction (a) Real object will result in a real image (b) Virtual object will result in a virtual image

(c) Real object will result in a virtual image (d) Virtual object will result in a real image

ONE OR MORE CORRECT Q14.24 Which of the following statements about a plane mirror is (are) correct. (a) When a plane mirror is displaced a distance L perpendicular to its surface, image point for any fixed object point moves a distance 2L. (b) When a plane mirror is displaced a distance L perpendicular to its surface, image point for any fixed object point moves a distance L. (c) When a plane mirror is rotated through an a angle about an axis that lies in the plane of the mirror, the reflected ray for any particular incident ray that is perpendicular to the axis rotates through anangle 2a. (d) When a plane mirror is rotated through an angle a about an axis that lies in the plane of the mirror,the reflected ray for any particular incident ray that is perpendicular to the axis rotates through an angle a. Q14.25 A ray of light is incident upon an air/water interface (it passes from air into water) at an angle of 45°.Which of the following quantities change as the light enters the water? (a) wavelength (b) frequency (c) speed of propagation (d) direction of propagation Q14.26 Figure, shows positions of a point image "I" of a self luminous point object "O" formed by a lens in air.This is possible if (a) a convex lens is placed to left of O. O I (b) a concave lens is placed to left of O. (c) a convex lens is placed between O and I (d) a concave lens is placed to right of I Q14.27 A flat mirror creates a virtual image of your face. Which of the following optical elements in combination with the flat mirror can form a real image ? (a) Convex lens (b) Concave lens µ2 (c) Concave mirror White (d) convex mirror light Q14.28 A beam of white light incident R on a combinations of thin prisms with µ1 Y V equal angles. In the out going beam the rainbow colours are observed as shown. m stands for refractive index and u for angle of dispersion. Then (a) m1 < m2 (b) m2 = m1 (c) u1 > u2 (d) m1 > m2 Q14.29 I is the image of a point object O formed by spherical mirror, then which of the following statement is/are incorrect? (a) If O and I are on the same side of the principal axis, then they have to be on opposite sides of the mirror (b) If O and I are on opposite sides of the same principal axis, then they have to be on opposite side of the mirror (c) If O and I are on opposite sides of the principal axis, then they can be on opposite sides of the mirror as well (d) If O is an principal axis, then I has to lie on principal axis only

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Exercises

Q14.30 Choose the correct option(s) (a) When we see a virtual image in plane mirror, a real image is formed on the retina of our eye (b) A plane mirror always forms a real image of virtual object (c) A bi-concave lens may form a real image (d) A bi-convex lens may form a virtual image Q14.31 Figure shows the behaviour of light rays with respect to a thin plano-convex lenswhose plane surface is silvered as shown in the figure. If the refractive index of the material of the lens is 1.5. Choose the correct option(s). (a) The focal length of the silvered lens (which acts as a mirror) is 10 cm. 20 cm (b) Radius of curvature of the curved part of the lens is 10 cm. (c) In the silvered lens, rays fall normally on the silvered surface (d) The focal length of the lens is 20 cm. Q14.32 A point object is placed at 30 cm from a convex glass lens 3 a mg = b of focal length 20 cm. The final image of object will be 2 formed at infinity, if (a) another concave lens of focal length 60 cm is placed in contact with the previous lens (b) another convex lens of focal length 60 cm is placed at a distance of 30 cm from the first lens (c) the whole system is immersed in a liquid of refractive index 4/3 (d) the whole system is immersed in a liquid of refractive index 9/8 Q14.33 The distance between a screen and an object is 120 cm. A convex lens is placed closed to the object and is moved along the line joining object and screen, towards the screen. Two sharp images of the object are found on the screen. The ratio of magnification of two real images is 1 : 9. Then, (a) focal length of the lens is 22.5 cm (b) small image is brighter than the larger one (c) larger image is brighter than the smaller one (d) brightness of both the images is same

487

Q14.35 A monochromatic point source of light is placed on the axis of a converging lens of focal length 20 cm. A screen is there on the other side at a distance of 30 cm from the lens as shown in the figure below. Match the entities of columns Screen f = 20 cm S

30 cm 30 cm

Column-I (a) The screen is shifted by 15 cm towards the lens

Column-II (P) The intensity of bright circle on the screen increases

(b)The screen is shifted by 15 cm away from the lens

(Q) The intensity of bright circle on the screen decreases

(c)The screen is shifted by 45 cm away from the lens

(R) The intensity of bright circle on the screen remains same

(d) The screen is shifted by 60 cm away from the lens

(S) Diameter of bright circle on the screen becomes half (T) Diameter of bright circle on the screen remains same

Q14.36 Consider a linear extended object that could be real or virtual with its length at right angles to the optic axis of a lens. With regard to image formation by lenses, match Column-I with Column-II :

MATCH THE COLUMN

Column-I

Column-II

Q14.34 Medium of lens in (R), (S), (T) is denser than surroundings

(a) Image of the same size as the object

(P) Concave lens in case of real object

(b) Virtual image of a size greater than the object

(Q) Convex lens in case of real object

Column-I (a) Real erect image may be formed

Column-II (P) Object placed in front of a convex mirror

(b) Virtual erect image may be formed

(Q) Converging beam incident on a convex mirror

(c) Real image of a size smaller than the object

(R) Concave lens in case of virtual object

(d) Real and erect image

(c) Real inverted image may be formed

(R) Object placed in front of a lens having theshape as shown

(S) Convex lens in case of virtual object

(d) Virtual inverted image may be formed

(S) Object placed in front of a lens having theshape as shown (T) Object placed in front of a silvered lens as shown.

(T) Concave mirror in case of real object

EXERCISES Section 14.1 Reflection and Refraction at a Plane Surface

14.1 . The image of a tree just covers the length of a plane mirror 4.00 cm tall when the mirror is held 35.0 cm from the eye. The tree is 28.0 m from the mirror. What is its height?

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488

CHAPTER 14 Geometric Optics

14.2 . A pencil that is 9.0 cm long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead 12.0 cm from the mirror surface and the end of the eraser 21.0 cm from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?

Section 14.2 Reflection at a Spherical Surface

above the swimmer. What is the actual height of the diving board above the surface of the water? 14.12 .. A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl? 14.13 .. The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted? 14.14 .. Repeat Exercise 14.13 for the case in which the end of the rod is ground to a concave hemispherical surface with radius 4.00 cm.

14.3 . An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image. 14.4 .. The diameter of Mars is 6794 km, and its minimum distance from the earth is 5.58 * 107 km. When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of 1.75 m. 14.5 .. An object is 24.0 cm from the center of a convex silvered spherical glass Christmas tree ornament 6.00 cm in diameter. What are the position and magnification of its image? 14.6 . A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 cm. Reflection from the surface of the shell forms an image of the 1.5-cm-tall coin that is 6.00 cm behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image. 14.7 . The thin glass shell shown in Fig. E14.7 has a spherical shape with a radius of curvature of 12.0 cm, and both of its Figure E14.7 surfaces can act as mirrors. A seed 3.30 mm high is placed 15.0 cm from the center of the mirror along the optic axis, as 3.30 mm 15.0 cm shown in the figure. (a) Calculate the location and height of the image of this seed. (b) Suppose now that the shell is reversed. Find the location and height of the seed’s image. 14.8 . Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an erect image with a magnification of 2.00 when the mirror is 1.25 cm from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror? (c) Draw a principal-ray diagram to check your answer in part (b).

14.15 . An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed. 14.16 . A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram. 14.17 .. A converging lens forms an image of an 8.00-mm-tall real object. The image is 12.0 cm to the left of the lens, 3.40 cm tall, and erect. What is the focal length of the lens? Where is the object located? 14.18 .. A double-convex thin lens has surfaces with equal radii of curvature of magnitude 2.50 cm. Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 cm from the lens. What is the index of refraction of the lens? 14.19 . For each thin lens shown in Fig. E14.19, calculate the location of the image of an object that is 18.0 cm to the left of the lens. The lens material has a refractive index of 1.50, and the radii of curvature shown are only the magnitudes.

Section 14.3 Refraction at a Spherical Surface

Figure E14.19

Section 14.4 Thin Lenses

.

14.9 A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice 1n = 1.3092. What is its apparent depth when viewed at normal incidence? 14.10 .. A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence? 14.11 . A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is at a height of 5.20 m

(a)

R 5 10.0 cm

R 5 10.0 cm

R 5 10.0 cm

R 5 10.0 cm

R 5 15.0 cm

Flat

R 5 15.0 cm

R 5 15.0 cm

(b)

(c)

(d)

14.20 . A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

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Problems

14.21 . Repeat Exercise 14.20 for the case in which the lens is diverging, with a focal length of - 48.0 cm. 14.22 .. Combination of Lenses I. A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm. (b) I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses. 14.23 .. Combination of Lenses II. Two thin lenses with a focal length of magnitude 12.0 cm, the first diverging and the second converging, are located 9.00 cm apart. An object 2.50 mm tall is placed 20.0 cm to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding problem.)

Section 14.5 Cameras

14.24 . You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. (a) If the slide is placed 15.0 cm from the lens, what focal length lens is required? (b) If the dimensions of the picture on a 35-mm color slide are 24 mm * 36 mm, what is the minimum size of the projector screen required to accommodate the image? 14.25 .. A camera lens has a focal length of 200 mm. How far from the lens should the subject for the photo be if the lens is 20.4 cm from the film?

Section 14.6 The Eye

14.26 .. BIO Curvature of the Cornea. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 cm from the cornea’s vertex is focused on the retina? 14.27 .. BIO (a) Where is the near point of an eye for which a contact lens with a power of + 2.75 diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of - 1.30 diopters is prescribed for distant vision? 14.28 . BIO A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?

Section 14.8 Microscopes and Telescopes

14.29 . A certain microscope is provided with objectives that have focal lengths of 16 mm, 4 mm, and 1.9 mm and with eyepieces that have angular magnifications of 5* and 10 * . Each objective forms an image 120 mm beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable. 14.30 .. The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

489

14.31 .. A reflecting tele- Figure E14.31 scope (Fig. E14.31) is to be made by using a spherical mirror with a radius of curvature of 1.30 m and an eyepiece with a focal length of 1.10 cm. The final image is at infinity. (a) What should the distance between the eyepiece and the mirror vertex be if the object is taken to be at infinity? (b) What will the angular magnification be?

PROBLEMS 14.32 . Where

must you place an object in front of a concave mirror with radius R so that the image is erect and 2 12 times the size of the object? Where is the image? 14.33 . If you run away from a plane mirror at 3.60 m > s, at what speed does your image move away from you? 14.34 . An object is placed between two plane mirrors arranged at right angles to each other at a distance d1 from the surface of one mirror and a distance d2 from the other. (a) How many images are formed? Show the location of the images in a diagram. (b) Draw the paths of rays from the object to the eye of an observer. 14.35 .. A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.00 m from the mirror. The filament is 6.00 mm tall, and the image is to be 24.0 cm tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror? 14.36 .. Rear-View Mirror. A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 cm. (a) Another car is behind your car, 9.00 m from the mirror, and this car is viewed in the mirror by your passenger. If this car is 1.5 m tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so? 14.37 .. A layer of benzene 1n = 1.502 4.20 cm deep floats on water 1n = 1.332 that is 6.50 cm deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence? 14.38 .. CP CALC You are in your car driving on a highway at 25 m > s when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 m > s when the truck is 2.0 m from the mirror, what is the speed of the truck relative to the highway? 14.39 .. Figure P14.39 shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is 2.0 cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram for the following: (a) Using only the ray shown, decide what type of lens (converging or diverging) this is. (b) What is the focal length of the lens? (c) Locate the image by drawing the other two principal rays. (d) Calculate where the image should be, and compare this result with the graphical solution in part (c). '

'

'

'

'

Figure P14.39 ?

?

Optic axis

Plant

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? Lens

'

490

CHAPTER 14 Geometric Optics

14.40 ... A microscope is focused on the upper surface of a glass plate. A second plate is then placed over the first. To focus on the bottom surface of the second plate, the microscope must be raised 0.780 mm. To focus on the upper surface, it must be raised another 2.50 mm. Find the index of refraction of the second plate. 14.41 .. What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence? 14.42 .. A Glass Rod. Both ends of a glass rod with index of refraction 1.60 are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is 6.00 cm, and the radius of curvature at the right end is 12.0 cm. The length of the rod between vertices is 40.0 cm. The object for the surface at the left end is an arrow that lies 23.0 cm to the left of the vertex of this surface. The arrow is 1.50 mm tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image? 14.43 .. The rod in Problem 14.42 is shortened to a distance of 25.0 cm between its vertices; the curvatures of its ends remain the same. As in Problem 14.42, the object for the surface at the left end is an arrow that lies 23.0 cm to the left of the vertex of this surface. The arrow is 1.50 mm tall and at right angles to the axis. (a) What is the object distance for the surface at the right end of the rod? (b) Is the object for this surface real or virtual? (c) What is the position of the final image? (d) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (e) What is the height of the final image? 14.44 . Figure P14.44 shows an object and its image formed by a thin lens. (a) What is the focal length of the lens, and what type of lens (converging or diverging) is it? (b) What is the height of the image? Is it real or virtual? Figure P14.44 Object Image 6.50 mm 5.00 cm

3.00 cm

? ? Optic ? axis ? Lens

14.45 ... A transparent rod 30.0 cm long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 cm at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 cm from the flat end and 15.0 cm from the vertex of the curved end. When viewed from the flat end of the rod, the apparent depth of the object is 9.50 cm from the flat end. What is its apparent depth when viewed from the curved end? 14.46 ... A transparent rod 50.0 cm long and with a refractive index of 1.60 is cut flat at the right end and rounded to a hemispherical surface with a 15.0-cm radius at the left end. An object is placed on the axis of the rod 12.0 cm to the left of the vertex of the hemispherical end. (a) What is the position of the final image? (b) What is its magnification? 14.47 . An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens. 14.48 .. A convex mirror and a concave mirror are placed on the same optic axis, separated by a distance L = 0.600 m. The radius

of curvature of each mirror has Figure P14.48 a magnitude of 0.360 m. A light source is located a distance x x from the concave mirror, as shown in Fig. P14.48. (a) What S distance x will result in the rays from the source returning to the L 5 0.600 m source after reflecting first from the convex mirror and then from the concave mirror? (b) Repeat part (a), but now let the rays reflect first from the concave mirror and then from the convex one. 14.49 .. As shown in Fig. P14.49 the candle is at the center of curvature of the concave mirror, whose focal length is 10.0 cm. The converging lens has a focal length of 32.0 cm and is 85.0 cm to the right of the candle. The candle is viewed looking through the lens from the right. The lens forms two images of the candle. The first is formed by light passing directly through the lens. The second image is formed from the light that goes from the candle to the mirror, is reflected, and then passes through the lens. (a) For each of these two images, draw a principal-ray diagram that locates the image. (b) For each image, answer the following questions: (i) Where is the image? (ii) Is the image real or virtual? (iii) Is the image erect or inverted with respect to the original object? Figure P14.49

C 85.0 cm

14.50 ... One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of 1.55. When a small leaf is placed 20.0 cm in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 cm from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33) but nothing else were changed? 14.51 .. A Lens in a Liquid. A lens obeys Snell’s law, bending light rays at each surface an amount determined by the index of refraction of the lens and the index of the medium in which the lens is located. (a) Equation (14.19) assumes that the lens is surrounded by air. Consider instead a thin lens immersed in a liquid with refractive index nliq. Prove that the focal length ƒ¿ is then given by Eq. (14.19) with n replaced by n > nliq. (b) A thin lens with index n has focal length ƒ in vacuum. Use the result of part (a) to show that when this lens is immersed in a liquid of index nliq, it will have a new focal length given by '

ƒ¿ = c

nliq 1n - 12 '

n - nliq

'

dƒ

14.52 ... When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens? 14.53 ... A glass plate 3.50 cm thick, with an index of refraction of 1.55 and plane parallel faces, is held with its faces horizontal and its lower face 6.00 cm above a printed page. Find the position of the image of the page formed by rays making a small angle with the normal to the plate.

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Problems

14.54 .. A symmetric, double-convex, thin lens made of glass with index of refraction 1.52 has a focal length in air of 40.0 cm. The lens is sealed into an opening in the left-hand end of a tank filled with water. At the right-hand end of the tank, opposite the lens, is a plane mirror 90.0 cm from the lens. The index of refraction of the water is 43. (a) Find the position of the image formed by the lens–water–mirror system of a small object outside the tank on the lens axis and 70.0 cm to the left of the lens. (b) Is the image real or virtual? (c) Is it erect or inverted? (d) If the object has a height of 4.00 mm, what is the height of the image? 14.55 . Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens. 14.56 .. BIO A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead? 14.57 .. The Galilean Telescope. Figure P14.57 is a diagram of a Galilean telescope, or opera glass, with both the object and its final image at infinity. The image I serves as a virtual object for the eyepiece. The final image is virtual and erect. (a) Prove that the angular magnification is M = - ƒ1 > ƒ2. (b) Find the length of the telescope. '

'

491

14.60 An object 'O' is kept in air in front of a thin plano convex lens of radius of curvature 10 cm. It's refractive index is 3/2 and the medium towards right of plane surface is water of refractive index 4/3. What should be the distance 'x' of the object so that the rays become parallel finally. x n w= 4 3

O

ng = 3 2

14.61 A small air bubble O (as shown in fig.) is formed inside the solid glass sphere of radius 20 cm. The refractive index of glass is 3 m = . The position of air bubble is at a distance 5 cm from cen2 tre. If one half of sphere is polished. Then find the distance between the images seen by observer on the left side of sphere. Assume paraxial rays conditions are satisfied air "

µ = 1.5

"

C

" O " 5 cm "

14.62 A light beam of diameter 23R is incident symmertically on a glass hemisphere of radius R and ofrefractive index n = 43. Find the radius of the beam at the base of hemisphere.

Figure P14.57 3

R air

f1 f2

f2 F19, F2

u9

u

F29

n= 3

I R Eyepiece

Objective

A

B

14.58 Figure A shows two identical planoconvex lenses in contact as shown. The combination has focal length 24 cm. Figure B shows the same with a liquid introduced between them. If refractive index of glass of the lenses is 1.50 and that of the liquid is 1.60. Find the focal length of the system figure B. 14.59 What should be the value of distance d so that final image is formed on the object itself. (focal lengths of the lenses are written on the lenses.) 10 cm

−20 cm

O

10 cm d

14.63 A point object is placed at a distance of 0.3 m from a convex lens (focal length 0.2 m) cut into two halves each of which is displaced by 0.0005 m as shown in the figure . Find the position of the image. If more than one image is formed, find their number and the distance between them. 14.64 The rectangular box is the place of a lens. By looking at the ray diagram, answer the following questions: (2)

(1)

Object Image (2)

(1)

3 cm

(i) What is the focal length of the lens? (ii) The magnification factor is equal to 14.65 The figure illustrates an aligned system consisting of three thin lenses. The system is located in air.Determine: (a) the position (relative to right most lens) of the point of convergence of a parallel ray incoming from the left after passing through the system ;

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492

CHAPTER 14 Geometric Optics

(b) The distance between the first lens and a point lying on the axis to the left of the system, at which that point and its image are located symmetrically with respect to the lens system? L2

L1

5 cm

L3

5 cm

O

O'

+ 10.0D

− 10.0D

+ 10.0D

14.66 A concavo-convex lens made of glass of refractive index 1.5 has surfaces of radii 20 cm and 60 cm. A concave mirror of radius of curvature 20 cm is placed co-axially to the lens. A glass slab of thickness 3 cm and refractive index 1.5 is placed closed to the mirror in the space between the mirror and lens as shown in figure. The distance of the nearest surface of the slab from lens is 248 cm. An object is placed 80 cm to the left of the lens. If final position of the image formed after refraction from lens, refraction from slab, reflection from mirror, refraction from the slab and lens is at the object O, find the distance x (in cm) of the mirror from nearer surface of slab. 3 cm

O

80 cm

248 cm x cm

14.67 A thin converging lens L1 forms a real image of an object located far away from the lens as shown in the figure. The image is located at a distance 4l and has height h. A diverging lens of focal length l is placed 2l from lens L1 at A. Another converging lens of focal length 2l is placed 3l from lens L1 at B. Find the height of final image thus formed?

O

l

A B 2l 3l

h 4l

14.68 Consider a 'beam expander' which consists of two converging lenses of focal lengths 40 cm and 100 cm having a common optical axis. A laser beam of diameter 4 mm is incident on the 40 cm focal length lens. The diameter of the final beam will be (see figure) 14.69 An object AB is at a distance of 18 cm from a lens with focal length 15 cm. A flat mirror turned through 45° with respect to optical axis of the lens is placed at a distance 0.5 m behind the lens as shown in figure. A sharp image of the object is formed at the bottom of the vessel filled with water upto a height of 20 cm. Find the distance of the image from the optical axis of the system. (Use mwater = 4/3)

14.70 A thin plano-convex lens fits exactly into a plano concave lens with µ = 3/2 1 their plane surface parallel to each other as µ2 = 5/4 shown in the figure. The radius of curvature of the curved surface R = 30 cm. The lens are made of difference material 3 having refractive index m1 = and 2 5 m2 = as shown in figure. 4 (i) if plane surface of the plano-convex lens is silvered, then calculate the equivalent focal length of this system and also calculate the nature of this equivalent mirror. (ii) An object having transverse length 5 cm in placed on the axis of equivalent mirror (in part 1), at a distance 15 cm from the equivalent mirror along principal axis. Find the transverse magnification produced by equivalent mirror. 3 90˚ 14.71 An isoceles glass prism (m = 2 ) has one of its equal faces coated with silver. A ray is normally incident on the other face and is reflected twice and then emerges through the base of the prism perpendicularly as shown in q 90˚ figure. Find the angle ‘u’ of the prism 14.72 The refractive indices of the crown glass for violet and red lights are 1.51 and 1.49 respecFlint glass tively and those of the flint glass are 1.77 and 1.73 2˚ respectively. A narrow beam of white light is inci4˚ dent at a small angle of incident on the shown combination of thin prisms. Find the value of a for which the mean deviation of the incident beam is a zero. Also calculate the magnitude of dispersion produced by the combination & state order of colCrown glass ors in emerging light from top to bottom. 14.73 A ray of light passes through a prism in a principle plane the deviation being equal to angle of incidence which is equal to 2a. It (m2 - 1) is given that a is the angle of prism. Show that cos2a = , 8 where m is the refractive index of the material of prism. 14.74 A light ray is incident on a right angled triangular prism and refracted at the first surface. If the refractive index of prism is 4 mg = 1.5 & refractive index of water mw = . Calculate the small3 est angle of incidence at the first surface for which the refracted ray does not undergo internal reflection at the second surface. π2

Water

i "

" Prism "

14.75 A parallel beam of light ray parallel to the X-axis is incident on a parbolic reflecting surface x = 2by2 as shown in the figure. After reflecting, it passes through focal point F. Find the focal length of the reflecting surface. Y

B 45˚ A F 20 cm

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X

Answers

14.76 When a pin is moved along the principal axis of a small concave mirror, the image position coincides with the object at a point 0.5 m from the mirror, refer figure. If the mirror is placed at a depth of 0.2 m in a transparent liquid, the same phenomenon occurs when the pin is placed 0.4 m from the mirror. Find the refractive index of the liquid

0.2 m 0.2 m

4 14.77 Light is incident from glass (m = 1.5) to water cm = 3 d . Find the range of deviation for which there are two angle of incidence.

CHALLENGE PROBLEMS 14.78 ... CALC (a) For a lens with focal length ƒ, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)? 14.79 ... BIO People with normal vision cannot focus their eyes underwater if they aren’t wearing a face mask or goggles and there

493

is water in contact with their eyes (see Discussion Question Q14.23). (a) Why not? (b) With the simplified model of the eye described in Exercise 14.51, what corrective lens (specified by focal length as measured in air) would be needed to enable a person underwater to focus an infinitely distant object? (Be careful— the focal length of a lens underwater is not the same as in air! See Problem 14.51. Assume that the corrective lens has a refractive index of 1.62 and that the lens is used in eyeglasses, not goggles, so there is water on both sides of the lens. Assume that the eyeglasses are 2.00 cm in front of the eye.) 14.80 . The refracting angle of a prism is 90°. If g is the angle of minimum deviation and b is the deviation of ray which enters at grazing incidence, prove that sing = sin2 b and cosg = mcosb , m is refractive index of material of the prism. 14.81 . Water level in the tank is decreasing at a constant rate of 1 cm/s. A small metal sphere is moving downwards with a constant velocity 5 cm/s. Base of the tank is a concave 1 cm/s mirror of radius 40 cm. Find the velocity of the image seen [Take amw = 4/3 5 cm/s 10 cm (a) directly R = 40 cm (b) after reflection at the mirror.

Answers

Chapter Opening Question

?

A magnifying lens (simple magnifier) produces a virtual image with a large angular size that is infinitely far away, so you can see it in sharp focus with your eyes relaxed. (A surgeon doing microsurgery would not appreciate having to strain his eyes while working.) The object should be at the focal point of the lens, so the object and lens are separated by one focal length.

Test Your Understanding Questions 14.1 (iv) When you are a distance s from the mirror, your image is a distance s on the other side of the mirror and the distance from you to your image is 2s. As you move toward the mirror, the distance 2s changes at twice the rate of the distance s, so your image moves toward you at speed 2v. 14.2 (a) concave, (b) (ii) A convex mirror always produces an erect image, but that image is smaller than the object (see Fig. 14.16b). Hence a concave mirror must be used. The image will be erect and enlarged only if the distance from the object (your face) to the mirror is less than the focal length of the mirror, as in Fig. 14.20d. 14.3 no The sun is very far away, so the object distance is essentially infinite: s = q and 1>s = 0. Material a is air 1na = 1.002 and material b is water 1nb = 1.332, so the image position s¿ is given by na nb - na nb = or + s s¿ R 1.33 s¿ = R = 4.0R 0.33

0 +

1.33 1.33 - 1.00 = s¿ R

The image would be formed 4.0 drop radii from the front surface of the drop. But since each drop is only a part of a complete sphere, the distance from the front to the back of the drop is less

than 2R. Thus the rays of sunlight never reach the image point, and the drops do not form an image of the sun on the leaf. While the rays are not focused to a point, they are nonetheless concentrated and can cause damage to the leaf. 14.4 A and C When rays A and D are extended backward, they pass through focal point F2; thus, before they passed through the lens, they were parallel to the optic axis. The figures show that ray A emanated from point Q, but ray D did not. Ray B is parallel to the optic axis, so before it passed through the lens, it was directed toward focal point F1. Hence it cannot have come from point Q. Ray C passes through the center of the lens and hence is not deflected by its passage; tracing the ray backward shows that it emanates from point Q. A Q

Q 2F2

F2

F1 2F1

Q 2F2

2F2

B F2

F1 2F1

F2

F1 2F1

Q F2

F1 2F

1

2F2

C

14.5 (iii) The smaller image area of the CCD array means that the angle of view is decreased for a given focal length. Individual objects make images of the same size in either case; when a smaller light-sensitive area is used, fewer images fit into the area and the field of view is narrower. 14.6 (iii) This lens is designed to correct for a type of astigmatism. Along the vertical axis, the lens is configured as a converging lens; along the horizontal axis, the lens is configured as a diverging lens. Hence the eye is hyperopic (see Fig. 14.46) for objects that are ori-

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494

CHAPTER 14 Geometric Optics

ented vertically but myopic for objects that are oriented horizontally (see Fig. 14.47). Without correction, the eye focuses vertical objects behind the retina but horizontal objects in front of the retina. 14.7 (ii) The object must be held at the focal point, which is twice as far away if the focal length ƒ is twice as great. Equation (24.22) shows that the angular magnification M is inversely proportional to ƒ, so doubling the focal length makes M 12 as great. To improve the magnification, you should use a magnifier with a shorter focal length. 14.8 (i) The objective lens of a microscope is designed to make enlarged images of small objects, so the absolute value of its lateral magnification m is greater than 1. By contrast, the objective lens of a refracting telescope is designed to make reduced images. For example, the moon is thousands of kilometers in diameter, but its image may fit on a CCD array a few centimeters across. Thus ƒ m ƒ is much less than 1 for a refracting telescope. (In both cases m is negative because the objective makes an inverted image, which is why the question asks about the absolute value of m.)

Bridging Problem (a) 29.9 cm to the left of the goblet (b) 3.73 cm to the right of the goblet

Discussion Questions Q14.1 Yes, at the same place. Q14.2 Image distance is negative and final image is virtual. Q14.3 For better reception of signals as concave surfaces converges it will increase the signal intensity distance should be equal to focal length. Q14.4 It neither converges nor diverges. Q14.5 No, for mirror it only depends upon radius of curvature. Q14.6 For object placed beyond focus. Convex mirror does not forms real image for real objects. Q14.7 Rays will bounce back between two mirror therefore infine no. of images will be formed. For higher order reflection intensity will decrease as after each reflection distance will be increasing & power is decreasing. Q14.8 It means image is very large. Q14.9 To capture complete image of user. Size of face size and average working distance. Q14.10 This is possible as concave mirror will converge light due to increased power at a point. This is not possible with convex surface. Q14.11 If distance is greater than focus image will be inverted and if distance is less than focal length image will be erect. For convex side it will always be erect. Q14.12 Image at ∞ means that image is not formed as rays will be parallal. Q14.13 m = 1 refers to transverse magnification, where as in the case of vertical arrow the object is lying along optical axis hence longitudinal magnification would apply. Q14.14 This is true as we perceive image size according to the angle substended on eye. Due to reflection from convex glass image subsends small angle on edge therefore it appears to be at large distance.

Q14.15 Simply place it under direct sun light and measure the focal length. No for convex lens. Q14.16 Theoretically, yes. Q14.17 No, its focal length increases as its power decreases. Q14.18 It will work as diverging lens. Focal length ∝ ROC Q14.19 Yes, No, Q14.20 No, can take photograph of virtual image. Q14.21 Both converging lens. One with large aperture but small thickness & second one small in aperture more thick. (Not very thick). Q14.22 Due to water power of eye lens will decrease. To maintain the same power lenses should be converging. Q14.23 Complete image will be formed but intensity will be halved

Single Correct Q14.1 (c) Q14.2 (b) Q14.3 (c) Q14.4 (d) Q14.5 (a) Q14.6 (d) Q14.7 (a) Q14.8 (d) Q14.9 (a) Q14.10 (c) Q14.11 (b) Q14.12 (b) Q14.13 (a) Q14.14 (b) Q14.15 (a), (c) Q14.16 (d) Q14.17 (c) Q14.18 (d) Q14.19 (b) Q14.20 (c) Q14.21 (b) Q14.22 (b) Q14.23 (d)

One or More Correct Q14.24 (a), (c) Q14.25 (a), (c), (d) Q14.26 (a), (c), (d) Q14.27 (a), (c) Q14.28 (c), (d) Q14.29 (b), (c) Q14.30 (a), (b), (c), (d) Q14.31 (a), (b), (c), (d) Q14.32 (a), (d) Q14.33 (a), (b)

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Answers

Match the Column

14.38 51 m/sec

Q14.34 (a) Q (b) P, R, S, T (c) R, T (d) Q Q14.35 (a) Q (b) P, S (c) P, S (d) R, T Q14.36 (a) Q, R, T (b) Q, R, T (c) Q, S, T (d) R, S

14.39 (a) ray will bent toward opticle axies

495

(b) f = 18 cm (c) 22.5 cm to the Left 14.40 1.31

Exercises

14.41 2.00

14.1 28 m behind the mirror and 3.24 m tall 14.2 9 cm, tip of the lead 14.3 (b) 33 cm left of the mirror vertex, inverted and 1.2 cm, Real. 14.4 0.213 mm 14.5 1.6 cm from centre, + 0.0667 14.6 18 cm from vertex, 0.5 cm tall erect and virtual 14.7 (a) 10 cm from vertex and vertex and 2.2 mm (b) 4.29 cm from vertex and 0.944 mm 14.8 (a) concave (b) 5 cm 14.9 2.67 cm 14.10 (a) 5.25 cm (b) 24.8 cm 14.11 3.30 m 14.12 (a) 1.33 (b) No 14.13 14.8 cm and 0.578 mm, Inverted and Real. 14.14 8.35 cm and 0.326 mm, eracted and virtual 14.15 (a) 107 cm and 17.8 mm tall (b) f = 18.6 cm 14.16 (a) diverging lens f = 48 cm (b) 6.38 mm 14.17 2.82 cm and f = 3.69 cm 14.18 1.67 14.19 (a) 36 cm (b) 180 cm left (c) 7.20 cm left (d) 13.8 cm left 14.20 7 cm and 0.34 cm tall erect 14.21 26.3 cm and 1.24 cm in size 14.22 (a) 200 cm, 4.8 cm tall (b) 150 cm, 7.2 cm tall 14.23 (a) 44 cm to right of lens (2) and 53 cm right of (1) (b) Real (c) 2.5 mm 14.24 (a) f = 148 mm (b) area = 1.44 m * 2.16 14.25 10.2 m

14.42 (a) The image from the left end acts as the object for the right end of the rod.

14.26 R = 0.71 cm. 14.27 (a) 80 cm (b) 76.9 cm 14.28 (a) near sighted (b) Diverging lens (c) 1.33 D 14.29 (a) 640 (b) 43 14.30 (a) 8.37 mm (b) 21.4 (c) - 297 14.31 (a) 0.66 (b) 59.1 14.32 Object is 0.3 R infront image is 0.75 R behind the mirror 14.33 7.20 m/sec 14.34 (a) three images will be formed. 14.35 (a) 20 cm (b) 39 cm 14.36 (a) 1.49 cm (b) Height of Image s downward (b)

59 cm>s upwards 4

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15

INTERFERENCE

LEARNING GOALS By studying this chapter, you will learn: • What happens when two waves combine, or interfere, in space. • How to understand the interference pattern formed by the interference of two coherent light waves. • How to calculate the intensity at various points in an interference pattern.

?

Soapy water is colorless, but when blown into bubbles it shows vibrant colors. How does the thickness of the bubble walls determine the particular colors that appear?

n ugly black oil spot on the pavement can become a thing of beauty after a rain, when the oil reflects a rainbow of colors. Multicolored reflections can also be seen from the surfaces of soap bubbles and DVDs. How is it possible for colorless objects to produce these remarkable colors? In our discussion of lenses, mirrors, and optical instruments we used the model of geometric optics, in which we represent light as rays, straight lines that are bent at a reflecting or refracting surface. But many aspects of the behavior of light can’t be understood on the basis of rays. We have already learned that light is fundamentally a wave, and in some situations we have to consider its wave properties explicitly. If two or more light waves of the same frequency overlap at a point, the total effect depends on the phases of the waves as well as their amplitudes. The resulting patterns of light are a result of the wave nature of light and cannot be understood on the basis of rays. Optical effects that depend on the wave nature of light are grouped under the heading physical optics. In this chapter we’ll look at interference phenomena that occur when two waves combine. The colors seen in oil films and soap bubbles are a result of interference between light reflected from the front and back surfaces of a thin film of oil or soap solution. Effects that occur when many sources of waves are present are called diffraction phenomena; we’ll study these in Chapter 16. In that chapter we’ll see that diffraction effects occur whenever a wave passes through an aperture or around an obstacle. They are important in practical applications of physical optics such as diffraction gratings, x-ray diffraction, and holography. While our primary concern is with light, interference and diffraction can occur with waves of any kind. As we go along, we’ll point out applications to other types of waves such as sound and water waves.

A

• How interference occurs when light reflects from the two surfaces of a thin film. • How interference makes it possible to measure extremely small distances.

497

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498

CHAPTER 15 Interference

15.1

Interference and Coherent Sources

As we discussed in Chapter 15, the term interference refers to any situation in which two or more waves overlap in space. When this occurs, the total wave at any point at any instant of time is governed by the principle of superposition, which we introduced in Section 15.6 in the context of waves on a string. This principle also applies to electromagnetic waves and is the most important principle in all of physical optics. The principle of superposition states: When two or more waves overlap, the resultant displacement at any point and at any instant is found by adding the instantaneous displacements that would be produced at the point by the individual waves if each were present alone.

(In some special situations, such as electromagnetic waves propagating in a crystal, this principle may not apply. A discussion of these is beyond our scope.) We use the term “displacement” in a general sense. With waves on the surface of a liquid, we mean the actual displacement of the surface above or below its normal level. With sound waves, the term refers to the excess or deficiency of pressure. For electromagnetic waves, we usually mean a specific component of electric or magnetic field.

Interference in Two or Three Dimensions

15.1 A “snapshot” of sinusoidal waves of frequency ƒ and wavelength l spreading out from source S1 in all directions. Wave fronts: crests of the wave (frequency f ) separated by one wavelength l

S1

We have already discussed one important case of interference, in which two identical waves propagating in opposite directions combine to produce a standing wave. We saw this in Chapters 15 and 16 for transverse waves on a string and for longitudinal waves in a fluid filling a pipe; we described the same phenomenon for electromagnetic waves in Section 12.5. In all of these cases the waves propagated along only a single axis: along a string, along the length of a fluid-filled pipe, or along the propagation direction of an electromagnetic plane wave. But light waves can (and do) travel in two or three dimensions, as can any kind of wave that propagates in a two- or three-dimensional medium. In this section we’ll see what happens when we combine waves that spread out in two or three dimensions from a pair of identical wave sources. Interference effects are most easily seen when we combine sinusoidal waves with a single frequency ƒ and wavelength l. Figure 15.1 shows a “snapshot” of a single source S1 of sinusoidal waves and some of the wave fronts produced by this source. The figure shows only the wave fronts corresponding to wave crests, so the spacing between successive wave fronts is one wavelength. The material surrounding S1 is uniform, so the wave speed is the same in all directions, and there is no refraction (and hence no bending of the wave fronts). If the waves are two-dimensional, like waves on the surface of a liquid, the circles in Fig. 15.1 represent circular wave fronts; if the waves propagate in three dimensions, the circles represent spherical wave fronts spreading away from S1. In optics, sinusoidal waves are characteristic of monochromatic light (light of a single color). While it’s fairly easy to make water waves or sound waves of a single frequency, common sources of light do not emit monochromatic (single-frequency) light. For example, incandescent light bulbs and flames emit a continuous distribution of wavelengths. By far the most nearly monochromatic source that is available at present is the laser. An example is the helium–neon laser, which emits red light at 632.8 nm with a wavelength range of the order of 60.000001 nm, or about one part in 109. As we analyze interference and diffraction effects in this chapter and the next, we will assume that we are working with monochromatic waves (unless we explicitly state otherwise). '

l The wave fronts move outward from source S1 at the wave speed v 5 fl.

Constructive and Destructive Interference Two identical sources of monochromatic waves, S1 and S2, are shown in Fig. 15.2a. The two sources produce waves of the same amplitude and the same wavelength l. '

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499

15.1 Interference and Coherent Sources

In addition, the two sources are permanently in phase; they vibrate in unison. They might be two loudspeakers driven by the same amplifier, two radio anten nas powered by the same transmitter, or two small slits in an opaque screen, illuminated by the same monochromatic light source. We will see that if there were not a constant phase relationship between the two sources, the phenomena we are about to discuss would not occur. Two monochromatic sources of the same frequency and with a constant phase relationship (not necessarily in phase) are said to be coherent. We also use the term coherent waves (or, for light waves, coherent light) to refer to the waves emitted by two such sources. If the waves emitted by the two coherent sources are transverse, like electromagnetic waves, then we will also assume that the wave disturbances produced by both sources have the same polarization (that is, they lie along the same line). For example, the sources S1 and S2 in Fig. 15.2a could be two radio antennas in the form of long rods oriented parallel to the z-axis (perpendicular to the plane of the figure); at any point in the xy-plane the waves produced by both antennas S have E fields with only a z-component. Then we need only a single scalar function to describe each wave; this makes the analysis much easier. We position the two sources of equal amplitude, equal wavelength, and (if the waves are transverse) the same polarization along the y-axis in Fig. 15.2a, equidistant from the origin. Consider a point a on the x-axis. From symmetry the two distances from S1 to a and from S2 to a are equal; waves from the two sources thus require equal times to travel to a. Hence waves that leave S1 and S2 in phase arrive at a in phase. The two waves add, and the total amplitude at a is twice the amplitude of each individual wave. This is true for any point on the x-axis. Similarly, the distance from S2 to point b is exactly two wavelengths greater than the distance from S1 to b. A wave crest from S1 arrives at b exactly two cycles earlier than a crest emitted at the same time from S2, and again the two waves arrive in phase. As at point a, the total amplitude is the sum of the amplitudes of the waves from S1 and S2. In general, when waves from two or more sources arrive at a point in phase, they reinforce each other: The amplitude of the resultant wave is the sum of the amplitudes of the individual waves. This is called constructive interference (Fig. 15.2b). Let the distance from S1 to any point P be r1, and let the distance from S2 to P be r2. For constructive interference to occur at P, the path difference r2 - r1 for the two sources must be an integral multiple of the wavelength l:

15.2 (a) A “snapshot” of sinusoidal waves spreading out from two coherent sources S1 and S2. Constructive interference occurs at point a (equidistant from the two sources) and (b) at point b. (c) Destructive interference occurs at point c. '

(a) Two coherent wave sources separated by a distance 4l y

b

S1

x

a S2 c

(b) Conditions for constructive interference: Waves interfere constructively if their path lengths differ by an integral number of wavelengths: r2 2 r1 5 ml. b 7l r1 5 S1

'

r2

5

r2 2 r1 5 2l

9l

'

'

'

r2 2 r1 5 22.50l (15.1)

5 9.7

(constructive interference, sources in phase)

S1

5

1m = 0, 61, 62, 63, Á 2

(c) Conditions for destructive interference: Waves interfere destructively if their path lengths differ by a half-integral number of 1 wavelengths: r2 2 r1 5 (m 1 2 )l.

r1

r2 - r1 = ml

l

S2

S2

l

In Fig. 15.2a, points a and b satisfy Eq. (15.1) with m = 0 and m = + 2, respectively. Something different occurs at point c in Fig. 15.2a. At this point, the path difference r2 - r1 = -2.50l, which is a half-integral number of wavelengths. Waves from the two sources arrive at point c exactly a half-cycle out of phase. A crest of one wave arrives at the same time as a crest in the opposite direction (a “trough”) of the other wave (Fig. 15.2c). The resultant amplitude is the difference between the two individual amplitudes. If the individual amplitudes are equal, then the total amplitude is zero! This cancellation or partial cancellation of the individual waves is called destructive interference. The condition for destructive interference in the situation shown in Fig. 15.2a is (destructive interference, r2 - r1 = A m + 12 B l 1m = 0, 61, 62, 63, Á 2 (15.2) sources in phase) The path difference at point c in Fig. 15.2a satisfies Eq. (15.2) with m = -3.

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r2 5 7.2

5l

l c

500

CHAPTER 15 Interference

15.3 The same as Fig. 15.2a, but with red antinodal curves (curves of maximum amplitude) superimposed. All points on each curve satisfy Eq. (15.1) with the value of m shown. The nodal curves (not shown) lie between each adjacent pair of antinodal curves. Antinodal curves (red) mark positions where the waves from S1 and S2 interfere constructively. At a and b, the waves arrive in phase and interfere constructively.

y

m53 m52

b S1

m51 m50 m 5 21 m 5 22 m 5 23

a

Figure 15.3 shows the same situation as in Fig. 15.2a, but with red curves that show all positions where constructive interference occurs. On each curve, the path difference r2 - r1 is equal to an integer m times the wavelength, as in Eq. (15.1). These curves are called antinodal curves. They are directly analogous to antinodes in the standing-wave patterns described in Chapters 15 and 16 and Section 12.5. In a standing wave formed by interference between waves propagating in opposite directions, the antinodes are points at which the amplitude is maximum; likewise, the wave amplitude in the situation of Fig. 15.3 is maximum along the antinodal curves. Not shown in Fig. 15.3 are the nodal curves, which are the curves that show where destructive interference occurs in accordance with Eq. (15.2); these are analogous to the nodes in a standing-wave pattern. A nodal curve lies between each two adjacent antinodal curves in Fig. 15.3; one such curve, corresponding to r2 - r1 = - 2.50l, passes through point c. In some cases, such as two loudspeakers or two radio-transmitter antennas, the interference pattern is three-dimensional. Think of rotating the color curves of Fig. 15.3 around the y-axis; then maximum constructive interference occurs at all points on the resulting surfaces of revolution.

x S2 c

At c, the waves arrive one-half cycle out of phase and interfere destructively.

m 5 the number of wavelengths l by which the path lengths from S1 and S2 differ.

CAUTION Interference patterns are not standing waves The interference patterns in Figs. 15.2a and 15.3 are not standing waves, though they have some similarities to the standing-wave patterns described in Chapters 15 and 16 and Section 12.5. In a standing wave, the interference is between two waves propagating in opposite directions; there is no net energy flow in either direction (the energy in the wave is left “standing”). In the situations shown in Figs. 15.2a and 15.3, there is likewise a stationary pattern of antinodal and nodal curves, but there is a net flow of energy outward from the two sources. All that interference does is to “channel” the energy flow so that it is greatest along the antinodal curves and least along the nodal curves. y

For Eqs. (15.1) and (15.2) to hold, the two sources must have the same wavelength and must always be in phase. These conditions are rather easy to satisfy for sound waves. But with light waves there is no practical way to achieve a constant phase relationship (coherence) with two independent sources. This is because of the way light is emitted. In ordinary light sources, atoms gain excess energy by thermal agitation or by impact with accelerated electrons. Such an “excited” atom begins to radiate energy and continues until it has lost all the energy it can, typically in a time of the order of 10 -8 s. The many atoms in a source ordinarily radiate in an unsynchronized and random phase relationship, and the light that is emitted from two such sources has no definite phase relationship. However, the light from a single source can be split so that parts of it emerge from two or more regions of space, forming two or more secondary sources. Then any random phase change in the source affects these secondary sources equally and does not change their relative phase. The distinguishing feature of light from a laser is that the emission of light from many atoms is synchronized in frequency and phase. As a result, the random phase changes mentioned above occur much less frequently. Definite phase relationships are preserved over correspondingly much greater lengths in the beam, and laser light is much more coherent than ordinary light. Test Your Understanding of Section 15.1 Consider a point in Fig. 15.3 on the positive y-axis above S1. Does this point lie on (i) an antinodal curve; (ii) a nodal curve; or (iii) neither? (Hint: The distance between S1 and S2 is 4l.) '

15.2

y

Two-Source Interference of Light

The interference pattern produced by two coherent sources of water waves of the same wavelength can be readily seen in a ripple tank with a shallow layer of water (Fig. 15.4). This pattern is not directly visible when the interference is

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15.2 Two-Source Interference of Light

between light waves, since light traveling in a uniform medium cannot be seen. (A shaft of afternoon sunlight in a room is made visible by scattering from airborne dust particles.) One of the earliest quantitative experiments to reveal the interference of light from two sources was performed in 1800 by the English scientist Thomas Young. We will refer back to this experiment several times in this and later chapters, so it’s important to understand it in detail. Young’s apparatus is shown in perspec tive in Fig. 15.5a. A light source (not shown) emits monochromatic light; however, this light is not suitable for use in an interference experiment because emissions from different parts of an ordinary source are not synchronized. To remedy this, the light is directed at a screen with a narrow slit S0, 1 mm or so wide. The light emerging from the slit originated from only a small region of the light source; thus slit S0 behaves more nearly like the idealized source shown in Fig. 15.1. (In modern versions of the experiment, a laser is used as a source of coherent light, and the slit S0 isn’t needed.) The light from slit S0 falls on a screen with two other narrow slits S1 and S2, each 1 mm or so wide and a few tens or hundreds of micrometers apart. Cylindrical wave fronts spread out from slit S0 and reach slits S1 and S2 in phase because they travel equal distances from S0. The waves emerging from slits S1 and S2 are therefore also always in phase, so S1 and S2 are coherent sources. The interference of waves from S1 and S2 produces a pattern in space like that to the right of the sources in Figs. 15.2a and 15.3. To visualize the interference pattern, a screen is placed so that the light from S1 and S2 falls on it (Fig. 15.5b). The screen will be most brightly illuminated at points P, where the light waves from the slits interfere constructively, and will be darkest at points where the interference is destructive. To simplify the analysis of Young’s experiment, we assume that the distance R from the slits to the screen is so large in comparison to the distance d between the slits that the lines from S1 and S2 to P are very nearly parallel, as in Fig. 15.5c. This is usually the case for experiments with light; the slit separation is typically a few millimeters, while the screen may be a meter or more away. The difference in path length is then given by

501

15.4 The concepts of constructive interference and destructive interference apply to these water waves as well as to light waves and sound waves.

'

'

'

(a) Interference of light waves passing through two slits Coherent wave fronts from two slits

Cylindrical wave fronts

y

Monochromatic light

Bright bands where wave fronts arrive in phase and interfere constructively

S2 S0

Dark bands where wave fronts arrive out of phase and interfere destructively

S1

15.5 (a) Young’s experiment to show interference of light passing through two slits. A pattern of bright and dark areas appears on the screen (see Fig. 15.6). (b) Geometrical analysis of Young’s experiment. For the case shown, r2 7 r1 and both y and u are positive. If point P is on the other side of the screen’s center, r2 6 r1 and both y and u are negative. (c) Approximate geometry when the distance R to the screen is much greater than the distance d between the slits.

Screen (b) Actual geometry (seen from the side)

(c) Approximate geometry

S2

S2

d sin u u

Screen

r2

d S1

d sin u

d

u r 2

y

r1 R

P

In real situations, the distance R to the screen is usually very much greater than the distance d between the slits ...

u

S1 r1

To screen

... so we can treat the rays as parallel, in which case the path-length difference is simply r2 2 r1 5 d sin u.

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502

CHAPTER 15 Interference

r2 - r1 = d sin u PhET: Wave Interference ActivPhysics 16.1: Two-Source Interference: Introduction ActivPhysics 16.2: Two-Source Interference: Qualitative Questions ActivPhysics 16.3: Two-Source Interference: Problems

(15.3)

where u is the angle between a line from slits to screen (shown in blue in Fig. 15.5c) and the normal to the plane of the slits (shown as a thin black line).

Constructive and Destructive Two-Slit Interference We found in Section 15.1 that constructive interference (reinforcement) occurs at points where the path difference is an integral number of wavelengths, ml, where m = 0, 61, 62, 63, Á . So the bright regions on the screen in Fig. 15.5a occur at angles u for which 1m = 0, 61, 62, Á 2

d sin u = ml

(constructive interference, two slits)

(15.4)

Similarly, destructive interference (cancellation) occurs, forming dark regions on the screen, at points for which the path difference is a half-integral number of wavelengths, A m + 12 B l: d sin u = A m +

15.6 Photograph of interference fringes produced on a screen in Young’s doubleslit experiment. m (constructive interference, bright regions)

m 1 1/2 (destructive interference, dark regions) 11/2

5 9/2 4 7/2

1 2

1 1/2

ym = R tan um In experiments such as this, the distances ym are often much smaller than the distance R from the slits to the screen. Hence um is very small, tan um is very nearly equal to sin um, and '

ym = R sin um

0 21/2 21

23/2

Combining this with Eq. (15.4), we find that for small angles only,

22 25/2 23

(15.5)

'

5/2 3/2

(destructive interference, two slits)

Thus the pattern on the screen of Figs. 15.5a and 15.5b is a succession of bright and dark bands, or interference fringes, parallel to the slits S1 and S2. A photograph of such a pattern is shown in Fig. 15.6. The center of the pattern is a bright band corresponding to m = 0 in Eq. (15.4); this point on the screen is equidistant from the two slits. We can derive an expression for the positions of the centers of the bright bands on the screen. In Fig. 15.5b, y is measured from the center of the pattern, corresponding to the distance from the center of Fig. 15.6. Let ym be the distance from the center of the pattern 1u = 02 to the center of the mth bright band. Let um be the corresponding value of u; then

3 2

B l 1m = 0, 61, 62, Á 2

ym = R

ml d

(constructive interference in Young’s experiment)

(15.6)

27/2 24 25

29/2 211/2

We can measure R and d, as well as the positions ym of the bright fringes, so this experiment provides a direct measurement of the wavelength l. Young’s experiment was in fact the first direct measurement of wavelengths of light. CAUTION Equation (15.6) is for small angles only While Eqs. (15.4) and (15.5) are valid for any angle, Eq. (15.6) is valid only for small angles. It can be used only if the distance R from slits to screen is much greater than the slit separation d and if R is much greater than the distance ym from the center of the interference pattern to the mth bright fringe. y

The distance between adjacent bright bands in the pattern is inversely proportional to the distance d between the slits. The closer together the slits are, the more the pattern spreads out. When the slits are far apart, the bands in the pattern are closer together.

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15.2 Two-Source Interference of Light

503

While we have described the experiment that Young performed with visible light, the results given in Eqs. (15.4) and (15.5) are valid for any type of wave, provided that the resultant wave from two coherent sources is detected at a point that is far away in comparison to the separation d.

Example 15.1

Two-slit interference

Figure 15.7 shows a two-slit interference experiment in which the slits are 0.200 mm apart and the screen is 1.00 m from the slits. The m = 3 bright fringe in the figure is 9.49 mm from the central fringe. Find the wavelength of the light. 15.7 Using a two-slit interference experiment to measure the wavelength of light. y

Slits

m53 m52 m51

9.49 mm

x m 5 21 m 5 22 m 5 23

d 5 0.200 mm

R 5 1.00 m

SOLUTION IDENTIFY and SET UP: Our target variable in this two-slit interference problem is the wavelength l. We are given the slit separation d = 0.200 mm, the distance from slits to screen R = 1.00 m, and the distance y3 = 9.49 mm on the screen from the center of the interference pattern to the m = 3 bright fringe. We may use Eq. (15.6) to find l, since the value of R is so much greater than the values of d or y3. EXECUTE: We solve Eq. (15.6) for l for the case m = 3: l =

ymd 19.49 * 10-3 m210.200 * 10-3 m2 = mR 13211.00 m2

= 633 * 10-9 m = 633 nm

EVALUATE: This bright fringe could also correspond to m = - 3. Can you show that this gives the same result for l?

Screen

Example 15.2

Broadcast pattern of a radio station

It is often desirable to radiate most of the energy from a radio transmitter in particular directions rather than uniformly in all directions. Pairs or rows of antennas are often used to produce the desired radiation pattern. As an example, consider two identical ver tical antennas 400 m apart, operating at 1500 kHz = 1.5 * 106 Hz (near the top end of the AM broadcast band) and oscillating in phase. At distances much greater than 400 m, in what directions is the intensity from the two antennas greatest?

the ideas of two-slit interference to this problem. Since the resultant wave is detected at distances much greater than d = 400 m, we may use Eq. (15.4) to give the directions of the intensity maxima, the values of u for which the path difference is zero or a whole number of wavelengths. EXECUTE: The wavelength is l = c> ƒ = 200 m. From Eq. (15.4) with m = 0, 61, and 62, the intensity maxima are given by '

sin u =

S O L U T IO N IDENTIFY and SET UP: The antennas, shown in Fig. 15.8, correspond to sources S1 and S2 in Fig. 15.5. Hence we can apply 15.8 Two radio antennas broadcasting in phase. The purple arrows indicate the directions of maximum intensity. The waves that are emitted toward the lower half of the figure are not shown. m50 u 5 0°

m 5 21 u 5 230°

30°

m1200 m2 m ml = = d 400 m 2

EVALUATE: We can check our result by calculating the angles for minimum intensity, using Eq. (15.5). There should be one intensity minimum between each pair of intensity maxima, just as in Fig. 15.6. From Eq. (15.5), with m = - 2, - 1, 0, and 1,

m 5 11 u 5 130°

30°

m 5 12 u 5 190°

90° S1

S2 400 m

u = 0, 630°, 690°

In this example, values of m greater than 2 or less than -2 give values of sin u greater than 1 or less than -1, which is impossible. There is no direction for which the path difference is three or more wavelengths, so values of m of 63 or beyond have no meaning in this example.

sin u = m 5 22 u 5 290°

'

1m + 122l d

=

m + 2

1 2

u = 614.5°, 648.6°

These angles fall between the angles for intensity maxima, as they should. The angles are not small, so the angles for the minima are not exactly halfway between the angles for the maxima.

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504

CHAPTER 15 Interference Test Your Understanding of Section 15.2 You shine a tunable laser (whose wavelength can be adjusted by turning a knob) on a pair of closely spaced slits. The light emerging from the two slits produces an interference pattern on a screen like that shown in Fig. 15.6. If you adjust the wavelength so that the laser light changes from red to blue, how will the spacing between bright fringes change? (i) The spacing increases; (ii) the spacing decreases; (iii) the spacing is unchanged; (iv) not enough information is given to decide.

15.3

y

Intensity in Interference Patterns

In Section 15.2 we found the positions of maximum and minimum intensity in a two-source interference pattern. Let’s now see how to find the intensity at any point in the pattern. To do this, we have to combine the two sinusoidally varying fields (from the two sources) at a point P in the radiation pattern, taking proper account of the phase difference of the two waves at point P, which results from the path difference. The intensity is then proportional to the square of the resultant electric-field amplitude, as we learned in Chapter 12. To calculate the intensity, we will assume that the two sinusoidal functions (corresponding to waves from the two sources) have equal amplitude E and that S the E fields lie along the same line (have the same polarization). This assumes that the sources are identical and neglects the slight amplitude difference caused by the unequal path lengths (the amplitude decreases with increasing distance from the source). From Eq. (32.29), each source by itself would give an intensity 1 2 2 P0 cE at point P. If the two sources are in phase, then the waves that arrive at P differ in phase by an amount proportional to the difference in their path lengths, 1r2 - r12. If the phase angle between these arriving waves is f, then we can use the following expressions for the two electric fields superposed at P: '

E11t2 = E cos1vt + f2 '

E21t2 = E cos vt '

The superposition of the two fields at P is a sinusoidal function with some amplitude EP that depends on E and the phase difference f. First we’ll work on finding the amplitude EP if E and f are known. Then we’ll find the intensity I of the resultant wave, which is proportional to EP2. Finally, we’ll relate the phase difference f to the path difference, which is determined by the geometry of the situation.

Amplitude in Two-Source Interference

15.9 Phasor diagram for the superposition at a point P of two waves of equal amplitude E with a phase difference f. All phasors rotate counterclockwise with angular speed v.

To add the two sinusoidal functions with a phase difference, we use the same phasor representation that we used for simple harmonic motion (see Section 14.2) and for voltages and currents in ac circuits (see Section 11.1). We suggest that you review these sections now. Each sinusoidal function is represented by a rotating vector (phasor) whose projection on the horizontal axis at any instant represents the instantaneous value of the sinusoidal function. In Fig. 15.9, E1 is the horizontal component of the phasor representing the wave from source S1, and E2 is the horizontal component of the phasor for the wave from S2. As shown in the diagram, both phasors have the same magni tude E, but E1 is ahead of E2 in phase by an angle f. Both phasors rotate counterclockwise with constant angular speed v, and the sum of the projections on the horizontal axis at any time gives the instantaneous value of the total E field at point P. Thus the amplitude EP of the resultant sinusoidal wave at P is the magnitude of the dark red phasor in the diagram (labeled EP); this is the vector sum of the other two phasors. To find EP, we use the law of cosines and the trigonometric identity cos1p - f2 = - cos f: '

The resultant wave has amplitude f EP 5 2E cos . 2

'

EP

E p2f f

'

E vt O

E2 5 E cos vt E1 5 E cos (vt 1 f)

EP2 = E2 + E2 - 2E2 cos1p - f2 = E2 + E2 + 2E2 cos f

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15.3 Intensity in Interference Patterns

Then, using the identity 1 + cos f = 2 cos21f>22, we obtain f EP 2 = 2E211 + cos f2 = 4E2 cos2 a b 2 EP = 2 P Ecos

f P 2

(amplitude in two-source interference)

(15.7)

You can also obtain this result without using phasors (see Problem 15.50). When the two waves are in phase, f = 0 and EP = 2E. When they are exactly a half-cycle out of phase, f = p rad = 180°, cos1f>22 = cos1p>22 = 0, and EP = 0. Thus the superposition of two sinusoidal waves with the same frequency and amplitude but with a phase difference yields a sinusoidal wave with the same frequency and an amplitude between zero and twice the individual amplitudes, depending on the phase difference.

Intensity in Two-Source Interference To obtain the intensity I at point P, we recall from Section 12.4 that I is equal to the average magnitude of the Poynting vector, Sav. For a sinusoidal wave with electric-field amplitude EP, this is given by Eq. (32.29) with Emax replaced by EP. Thus we can express the intensity in several equivalent forms: '

'

'

I = Sav =

P0 2 1 EP 2 = 12 EP = 2 P0cEP 2 m 2m0c Ç 0

(15.8)

'

'

The essential content of these expressions is that I is proportional to EP2. When we substitute Eq. (15.7) into the last expression in Eq. (15.8), we get I = 12 P0cEP 2 = 2P0cE2 cos2 '

'

f 2

(15.9)

In particular, the maximum intensity I0, which occurs at points where the phase difference is zero 1f = 02, is '

I0 = 2P0cE2 '

Note that the maximum intensity I0 is four times (not twice) as great as the intensity 12 P0cE2 from each individual source. Substituting the expression for I0 into Eq. (15.9), we can express the intensity I at any point very simply in terms of the maximum intensity I0: '

'

I = I0 cos2

f 2

(intensity in two-source interference)

(15.10)

For some phase angles f the intensity is I0, four times as great as for an individual wave source, but for other phase angles the intensity is zero. If we average 2 Eq. (15.10) over all possible phase differences, the result is I0>2 = P0cE (the 1 2 average of cos 1f>22 is 2 ). This is just twice the intensity from each individual source, as we should expect. The total energy output from the two sources isn’t changed by the interference effects, but the energy is redistributed (as we mentioned in Section 15.1). '

'

Phase Difference and Path Difference Our next task is to find the phase difference f between the two fields at any point P. We know that f is proportional to the difference in path length from the two sources to point P. When the path difference is one wavelength, the phase difference is one cycle, and f = 2p rad = 360°. When the path difference is

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505

506

CHAPTER 15 Interference

l>2, f = p rad = 180°, and so on. That is, the ratio of the phase difference f to 2p is equal to the ratio of the path difference r2 - r1 to l: f r2 - r1 = 2p l Thus a path difference 1r2 - r12 causes a phase difference given by f =

2p 1r - r12 = k1r2 - r12 l 2

(phase difference related to path difference)

(15.11)

where k = 2p>l is the wave number introduced in Section 15.3. If the material in the space between the sources and P is anything other than vacuum, we must use the wavelength in the material in Eq. (15.11). If the material has index of refraction n, then l =

l0 n

and

k = nk0

(15.12)

where l0 and k0 are the wavelength and the wave number, respectively, in vacuum. Finally, if the point P is far away from the sources in comparison to their separation d, the path difference is given by Eq. (15.3): r2 - r1 = d sin u Combining this with Eq. (15.11), we find f = k1r2 - r12 = kd sin u =

2pd sin u l

(15.13)

When we substitute this into Eq. (15.10), we find I = I0 cos2 A 12 kd sin u B = I0 cos2 a

pd sin ub l

(intensity far from two sources)

(15.14)

The directions of maximum intensity occur when the cosine has the values 6 1— that is, when pd sin u = mp l

1m = 0, 61, 62, Á 2

or d sin u = ml in agreement with Eq. (15.4). You can also derive Eq. (15.5) for the zero-intensity directions from Eq. (15.14). As we noted in Section 15.2, in experiments with light we visualize the interference pattern due to two slits by using a screen placed at a distance R from the slits. We can describe positions on the screen with the coordinate y; the positions of the bright fringes are given by Eq. (15.6), where ordinarily y V R. In that case, sin u is approximately equal to y>R, and we obtain the following expressions for the intensity at any point on the screen as a function of y: I = I0 cos2 a

kdy pdy b = I0 cos2 a b (intensity in two-slit interference) 2R lR

(15.15)

Figure 15.10 shows a graph of Eq. (15.15); we can compare this with the photographically recorded pattern of Fig. 15.6. The peaks in Fig. 15.10 all have the same intensity, while those in Fig. 15.6 fade off as we go away from the center. We’ll explore the reasons for this variation in peak intensity in Chapter 16.

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15.4 Interference in Thin Films

507

15.10 Intensity distribution in the interference pattern from two identical slits.

I

Intensity maxima occur where f is an integral multiple of 2p and d sin u is an integral multiple of l.

I0 y 5 distance of a point in the pattern from the center (y 5 0) 23lR/d

22lR/d

2lR/d

0

lR/d

2lR/d

3lR/d

26p

24p

22p

0

2p

4p

6p

23l

22l

2l

0

l

2l

3l

Example 15.3

y

f 5 phase difference between the two waves at each point in the pattern

f d sin u

d sin u 5 path difference from the two slits at each point in the pattern

A directional transmitting antenna array

Suppose the two identical radio antennas of Fig. 15.8 are moved to be only 10.0 m apart and the broadcast frequency is increased to ƒ = 60.0 MHz. At a distance of 700 m from the point midway between the antennas and in the direction u = 0 (see Fig. 15.8), the intensity is I0 = 0.020 W > m2. At this same distance, find (a) the intensity in the direction u = 4.0°; (b) the direction near u = 0 for which the intensity is I0> 2; and (c) the directions in which the intensity is zero. '

'

'

'

S O L U T IO N IDENTIFY and SET UP: This problem involves the intensity distribution as a function of angle. Because the 700-m distance from the antennas to the point at which the intensity is measured is much greater than the distance d = 10.0 m between the antennas, the amplitudes of the waves from the two antennas are very nearly equal. Hence we can use Eq. (15.14) to relate intensity I and angle u. EXECUTE: The wavelength is l = c> ƒ = 5.00 m. The spacing d = 10.0 m between the antennas is just twice the wavelength (as was the case in Example 15.2), so d>l 5 2.00 and Eq. (15.14) becomes '

I = I0 cos2 a

'

pd sin u b = I0 cos2312.00p rad2sin u4 l

(a) When u = 4.0°, I = I0 cos2312.00p rad2 sin 4.0°4 = 0.82I0 = 10.82210.020 W>m22 = 0.016 W>m2

(b) The intensity I equals I0>2 when the cosine in Eq. (15.14) has the value 61> Ë2. The smallest angles at which this occurs correspond to 2.00p sin u = 6p>4 rad, so that sin u = 611>8.002 = 60.125 and u = 67.2°. (c) The intensity is zero when cos312.00p rad2sin u4 = 0. This occurs for 2.00p sin u = 6p>2, 63p>2, 65p>2, Á , or sin u = 60.250, 60.750, 61.25, Á . Values of sin u greater than 1 have no meaning, so the answers are u = 614.5°, 648.6° EVALUATE: The condition in part (b) that I = I0>2, so that 12.00p rad)sin u = 6p>4 rad, is also satisfied when sin u = 60.375, 60.625, or 60.875 so that u = 622.0°, 638.7°, or 661.0°. (Can you verify this?) It would be incorrect to include these angles in the solution, however, because the problem asked for the angle near u = 0 at which I = I0>2. These additional values of u aren’t the ones we’re looking for.

Test Your Understanding of Section 15.3 A two-slit interference experiment uses coherent light of wavelength 5.00 * 10-7 m. Rank the following points in the interference pattern according to the intensity at each point, from highest to lowest. (i) a point that is closer to one slit than the other by 4.00 * 10-7 m; (ii) a point where the light waves received from the two slits are out of phase by 4.00 rad; (iii) a point that is closer to one slit than the other by 7.50 * 10-7 m; (iv) a point where the light waves received by the two slits are out of phase by 2.00 rad. y

15.4

Interference in Thin Films

?

You often see bright bands of color when light reflects from a thin layer of oil floating on water or from a soap bubble (see the photograph that opens this chapter). These are the results of interference. Light waves are reflected from the front and back surfaces of such thin films, and constructive interference between the two reflected waves (with different path lengths) occurs in different

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508

CHAPTER 15 Interference

15.11 (a) A diagram and (b) a photograph showing interference of light reflected from a thin film. (a) Interference between rays reflected from the two surfaces of a thin film Light reflected from the upper and lower surfaces of the film comes together in the eye at P and undergoes interference. Some colors interfere constructively and others destructively, creating P the color bands we see. c a

f

Thin-Film Interference and Phase Shifts During Reflection

b

Air

e

Film

places for different wavelengths. Figure 15.11a shows the situation. Light shining on the upper surface of a thin film with thickness t is partly reflected at the upper surface (path abc). Light transmitted through the upper surface is partly reflected at the lower surface (path abdef ). The two reflected waves come together at point P on the retina of the eye. Depending on the phase relationship, they may inter fere constructively or destructively. Different colors have different wavelengths, so the interference may be constructive for some colors and destructive for oth ers. That’s why we see colored rings or fringes in Fig. 15.11b (which shows a thin film of oil floating on water) and in the photograph that opens this chapter (which shows thin films of soap solution that make up the bubble walls). The complex shapes of the colored rings in each photograph result from variations in the thick ness of the film.

Index n

t

d

(b) The rainbow fringes of an oil slick on water

15.12 Interference between light waves reflected from the two sides of an air wedge separating two glass plates. The angles and the thickness of the air wedge have been exaggerated for clarity; in the text we assume that the light strikes the upper plate at normal incidence and that the distances h and t are much less than l.

Let’s look at a simplified situation in which monochromatic light reflects from two nearly parallel surfaces at nearly normal incidence. Figure 15.12 shows two plates of glass separated by a thin wedge, or film, of air. We want to consider interference between the two light waves reflected from the surfaces adjacent to the air wedge, as shown. (Reflections also occur at the top surface of the upper plate and the bottom surface of the lower plate; to keep our discussion simple, we won’t include these.) The situation is the same as in Fig. 15.11a except that the film (wedge) thickness is not uniform. The path difference between the two waves is just twice the thickness t of the air wedge at each point. At points where 2t is an integer number of wavelengths, we expect to see constructive interfer ence and a bright area; where it is a half-integer number of wavelengths, we expect to see destructive interference and a dark area. Along the line where the plates are in contact, there is practically no path difference, and we expect a bright area. When we carry out the experiment, the bright and dark fringes appear, but they are interchanged! Along the line where the plates are in contact, we find a dark fringe, not a bright one. This suggests that one or the other of the reflected waves has undergone a half-cycle phase shift during its reflection. In that case the two waves that are reflected at the line of contact are a half-cycle out of phase even though they have the same path length. In fact, this phase shift can be predicted from Maxwell’s equations and the electromagnetic nature of light. The details of the derivation are beyond our scope, but here is the result. Suppose a light wave with electric-field amplitude Ei is traveling in an optical material with index of refraction na. It strikes, at normal incidence, an interface with another optical material with index nb. The amplitude Er of the wave reflected from the interface is proportional to the amplitude Ei of the incident wave and is given by '

'

Er =

na - nb E (normal incidence) na + nb i

This result shows that the incident and reflected amplitudes have the same sign when na is larger than nb and opposite sign when nb is larger than na. We can distinguish three cases, as shown in Fig. 15.13:

Glass

'

Air

Figure 15.13a: When na 7 nb, light travels more slowly in the first material than in the second. In this case, Er and Ei have the same sign, and the phase shift of the reflected wave relative to the incident wave is zero. This is analogous to reflection of a transverse mechanical wave on a heavy rope at a point where it is tied to a lighter rope or a ring that can move vertically without friction. Figure 15.13b: When na = nb, the amplitude Er of the reflected wave is zero. The incident light wave can’t “see” the interface, and there is no reflected wave. '

t x l

(15.16)

h

'

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15.4 Interference in Thin Films

509

15.13 Upper figures: electromagnetic waves striking an interface between optical materials at normal incidence (shown as a small angle for clarity). Lower figures: mechanical wave pulses on ropes. (a) If the transmitted wave moves faster than the incident wave … Electromagnetic waves propagating in optical materials

(b) If the incident and transmitted waves have the same speed … Material a

Material a (slow) Material b (fast) na . nb Transmitted

Incident

na 5 nb

(c) If the transmitted wave moves slower than the incident wave …

Material b (same as a) Transmitted

Material a (fast) Material b (slow) na , nb Incident

Transmitted

Incident Reflected

Reflected

… the reflected wave undergoes no phase change. Mechanical waves propagating on ropes

… there is no reflection.

… the reflected wave undergoes a half-cycle phase shift.

BEFORE Incident

Incident

Incident AFTER Reflected

Transmitted

Transmitted

Waves travel slower on thick ropes than on thin ropes.

Figure 15.13c: When na 6 nb, light travels more slowly in the second material than in the first. In this case, Er and Ei have opposite signs, and the phase shift of the reflected wave relative to the incident wave is p rad 1180° or a half-cycle2. This is analogous to reflection (with inversion) of a transverse mechanical wave on a light rope at a point where it is tied to a heavier rope or a rigid support. '

Let’s compare with the situation of Fig. 15.12. For the wave reflected from the upper surface of the air wedge, na (glass) is greater than nb, so this wave has zero phase shift. For the wave reflected from the lower surface, na (air) is less than nb (glass), so this wave has a half-cycle phase shift. Waves that are reflected from the line of contact have no path difference to give additional phase shifts, and they interfere destructively; this is what we observe. You can use the above principle to show that for normal incidence, the wave reflected at point b in Fig. 15.11a is shifted by a half-cycle, while the wave reflected at d is not (if there is air below the film). We can summarize this discussion mathematically. If the film has thickness t, the light is at normal incidence and has wavelength l in the film; if neither or both of the reflected waves from the two surfaces have a half-cycle reflection phase shift, the conditions for constructive and destructive interference are '

1m = 0, 1, 2, Á 2

2t = ml 2t = A m +

1 2

B l 1m = 0, 1, 2, Á 2

(constructive reflection from thin film, no relative phase shift) (destructive reflection from thin film, no relative phase shift)

(15.17a)

(15.17b)

If one of the two waves has a half-cycle reflection phase shift, the conditions for constructive and destructive interference are reversed: 2t = A m +

2t = ml

1 2

B l 1m = 0, 1, 2, Á 2

1m = 0, 1, 2, Á 2

(constructive reflection from thin film, half-cycle relative phase shift)

(15.18a)

(destructive reflection from thin film, half-cycle relative phase shift)

(15.18b)

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Reflected

Transmitted

510

CHAPTER 15 Interference

15.14 (a) Light reflecting from a thin film produces a steady interference pattern, but (b) light reflecting from a thick film does not.

(a) Light reflecting from a thin film

(b) Light reflecting from a thick film The waves reflected from the two surfaces are from different bursts and are not coherent.

Bursts of light a few mm long The waves reflected from the two surfaces are part of the same burst and are coherent.

Thick film

Thin film

Thin and Thick Films We have emphasized thin films in our discussion because of a principle we introduced in Section 15.1: In order for two waves to cause a steady interference pattern, the waves must be coherent, with a definite and constant phase relationship. The sun and light bulbs emit light in a stream of short bursts, each of which is only a few micrometers long 11 micrometer = 1 mm = 10-6 m2. If light reflects from the two surfaces of a thin film, the two reflected waves are part of the same burst (Fig. 15.14a). Hence these waves are coherent and interference occurs as we have described. If the film is too thick, however, the two reflected waves will belong to different bursts (Fig. 15.14b). There is no definite phase relationship between different light bursts, so the two waves are incoherent and there is no fixed interference pattern. That’s why you see interference colors in light reflected from an oil slick a few micrometers thick (see Fig. 15.11b), but you do not see such colors in the light reflected from a pane of window glass with a thickness of a few millimeters (a thousand times greater).

Problem-Solving Strategy 15.1

Interference in Thin Films

SET UP the problem using the following steps: 1. Make a drawing showing the geometry of the film. Identify the materials that adjoin the film; their properties determine whether one or both of the reflected waves have a half-cycle phase shift. 2. Identify the target variable.

2. If neither reflected wave undergoes a phase shift, or if both do, use Eqs. (15.17). If only one reflected wave undergoes a phase shift, use Eqs. (15.18). 3. Solve the resulting equation for the target variable. Use the wavelength l = l0>n of light in the film in your calculations, where n is the index of refraction of the film. (For air, n = 1.000 to four-figure precision.) 4. If you are asked about the wave that is transmitted through the film, remember that minimum intensity in the reflected wave corresponds to maximum transmitted intensity, and vice versa.

EXECUTE the solution as follows: 1. Apply the rule for phase changes to each reflected wave: There is a half-cycle phase shift when nb 7 na and none when nb 6 na.

EVALUATE your answer: Interpret your results by examining what would happen if the wavelength were changed or if the film had a different thickness.

IDENTIFY the relevant concepts: Problems with thin films involve interference of two waves, one reflected from the film’s front surface and one reflected from the back surface. Typically you will be asked to relate the wavelength, the film thickness, and the index of refraction of the film.

Example 15.4

Thin-film interference I

Suppose the two glass plates in Fig. 15.12 are two microscope slides 10.0 cm long. At one end they are in contact; at the other end they are separated by a piece of paper 0.0200 mm thick. What is the spacing of the interference fringes seen by reflection? Is the fringe at the line of contact bright or dark? Assume monochromatic light with a wavelength in air of l = l0 = 500 nm.

SOLUTION IDENTIFY and SET UP: Figure 15.15 depicts the situation. We’ll consider only interference between the light reflected from the upper and lower surfaces of the air wedge between the microscope slides. [The top slide has a relatively great thickness, about 1 mm,

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15.4 Interference in Thin Films 15.15 Our sketch for this problem.

511

From similar triangles in Fig. 15.15 the thickness t of the air wedge at each point is proportional to the distance x from the line of contact: t h = x l Combining this with Eq. (15.18b), we find 2xh = ml0 l ll0 10.100 m21500 * 10-9 m2 = m = m11.25 mm2 x = m 2h 12210.0200 * 10-3 m2

so we can ignore interference between the light reflected from its upper and lower surfaces (see Fig. 15.14b).] Light travels more slowly in the glass of the slides than it does in air. Hence the wave reflected from the upper surface of the air wedge has no phase shift (see Fig. 15.13a), while the wave reflected from the lower surface has a half-cycle phase shift (see Fig. 15.13c). EXECUTE: Since only one of the reflected waves undergoes a phase shift, the condition for destructive interference (a dark fringe) is Eq. (15.18b): 2t = ml0 1m = 0, 1, 2, Á 2

Example 15.5

EVALUATE: Our result shows that the fringe spacing is proportional to the wavelength of the light used; the fringes would be farther apart with red light (larger l0) than with blue light (smaller l0). If we use white light, the reflected light at any point is a mixture of wavelengths for which constructive interference occurs; the wavelengths that interfere destructively are weak or absent in the reflected light. (This same effect explains the colors seen when an oil film on water is illuminated by white light, as in Fig. 15.11b).

Thin-film interference II

Suppose the glass plates of Example 15.4 have n = 1.52 and the space between plates contains water 1n = 1.332 instead of air. What happens now? S O L U T IO N IDENTIFY and SET UP: The index of refraction of the water wedge is still less than that of the glass on either side of it, so the phase shifts are the same as in Example 15.4. Once again we use Eq. (15.18b) to find the positions of the dark fringes; the only difference is that the wavelength l in this equation is now the wavelength in water instead of in air.

Example 15.6

Successive dark fringes, corresponding to m = 1, 2, 3, Á , are spaced 1.25 mm apart. Substituting m = 0 into this equation gives x = 0, which is where the two slides touch (at the left-hand side of Fig. 15.15). Hence there is a dark fringe at the line of contact.

EXECUTE: In the film of water 1n = 1.332, the wavelength is l = l0>n = 1500 nm2>11 .332 = 376 nm. When we replace l0 by l in the expression from Example 15.4 for the position x of the mth dark fringe, we find that the fringe spacing is reduced by the same factor of 1.33 and is equal to 0.940 mm. There is still a dark fringe at the line of contact. EVALUATE: Can you see that to obtain the same fringe spacing as in Example 15.4, the dimension h in Fig. 15.15 would have to be reduced to 10.0200 mm2>1.33 5 0.0150 mm? This shows that what matters in thin-film interference is the ratio t>l between film thickness and wavelength. [You can see this by considering Eqs. (15.17) and (15.18).]

Thin-film interference III

Suppose the upper of the two plates of Example 15.4 is a plastic with n = 1.40, the wedge is filled with a silicone grease with n = 1.50, and the bottom plate is a dense flint glass with n = 1.60. What happens now? S O L U T IO N IDENTIFY and SET UP: The geometry is again the same as shown in Fig. 15.15, but now half-cycle phase shifts occur at both surfaces of the grease wedge (see Fig. 15.13c). Hence there is no relative phase shift and we must use Eq. (15.17b) to find the positions of the dark fringes.

EXECUTE: The value of l to use in Eq. (15.17b) is the wavelength in the silicone grease, l = l0>n = 1500 nm2>1.50 = 333 nm. You can readily show that the fringe spacing is 0.833 mm. Note that the two reflected waves from the line of contact are in phase (they both undergo the same phase shift), so the line of contact is at a bright fringe. EVALUATE: What would happen if you carefully removed the upper microscope slide so that the grease wedge retained its shape? There would still be half-cycle phase changes at the upper and lower surfaces of the wedge, so the pattern of fringes would be the same as with the upper slide present.

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512

CHAPTER 15 Interference

15.16 (a) Air film between a convex lens and a plane surface. The thickness of the film t increases from zero as we move out from the center, giving (b) a series of alternating dark and bright rings for monochromatic light.

(a) A convex lens in contact with a glass plane

(b) Newton’s rings: circular interference fringes

t

Newton’s Rings 15.17 The surface of a telescope objective lens under inspection during manufacture. Fringes map lack of fit between lens and master. Master Lens being tested

Figure 15.16a shows the convex surface of a lens in contact with a plane glass plate. A thin film of air is formed between the two surfaces. When you view the setup with monochromatic light, you see circular interference fringes (Fig. 15.16b). These were studied by Newton and are called Newton’s rings. We can use interference fringes to compare the surfaces of two optical parts by placing the two in contact and observing the interference fringes. Figure 15.17 is a photograph made during the grinding of a telescope objective lens. The lower, larger-diameter, thicker disk is the correctly shaped master, and the smaller, upper disk is the lens under test. The “contour lines” are Newton’s interference fringes; each one indicates an additional distance between the specimen and the master of one half-wavelength. At 10 lines from the center spot the distance between the two surfaces is five wavelengths, or about 0.003 mm. This isn’t very good; high-quality lenses are routinely ground with a precision of less than one wavelength. The surface of the primary mirror of the Hubble Space Telescope 1 was ground to a precision of better than 50 wavelength. Unfortunately, it was ground to incorrect specifications, creating one of the most precise errors in the history of optical technology (see Section 14.2).

Nonreflective and Reflective Coatings

15.18 A nonreflective coating has an index of refraction intermediate between those of glass and air. Destructive interference occurs when • the film is about 14 l thick and • the light undergoes a phase change at both reflecting surfaces, so that the two reflected waves emerge from the film 1 about 2 cycle out of phase.

nglass . nfilm . nair Air Film Glass

“Nonreflecting” film t5

1 l 4

Nonreflective coatings for lens surfaces make use of thin-film interference. A thin layer or film of hard transparent material with an index of refraction smaller than that of the glass is deposited on the lens surface, as in Fig. 15.18. Light is reflected from both surfaces of the layer. In both reflections the light is reflected from a medium of greater index than that in which it is traveling, so the same phase change occurs in both reflections. If the film thickness is a quarter (onefourth) of the wavelength in the film (assuming normal incidence), the total path difference is a half-wavelength. Light reflected from the first surface is then a half-cycle out of phase with light reflected from the second, and there is destructive interference. The thickness of the nonreflective coating can be a quarter-wavelength for only one particular wavelength. This is usually chosen in the central yellowgreen portion of the spectrum 1l = 550 nm2, where the eye is most sensitive. Then there is somewhat more reflection at both longer (red) and shorter (blue) wavelengths, and the reflected light has a purple hue. The overall reflection from a lens or prism surface can be reduced in this way from 4–5% to less than 1%. This also increases the net amount of light that is transmitted through the lens, since light that is not reflected will be transmitted. The same principle is used to minimize reflection from silicon photovoltaic solar cells 1n = 3.52 by use of a thin surface layer of silicon monoxide 1SiO, n = 1.452; this helps to increase the amount of light that actually reaches the solar cells.

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15.5 The Michelson Interferometer

513

If a quarter-wavelength thickness of a material with an index of refraction greater than that of glass is deposited on glass, then the reflectivity is increased, and the deposited material is called a reflective coating. In this case there is a half-cycle phase shift at the air–film interface but none at the film–glass interface, and reflections from the two sides of the film interfere constructively. For example, a coating with refractive index 2.5 causes 38% of the incident energy to be reflected, compared with 4% or so with no coating. By use of multiple-layer coatings, it is possible to achieve nearly 100% transmission or reflection for particular wavelengths. Some practical applications of these coatings are for color separation in television cameras and for infrared “heat reflectors” in motion-picture projectors, solar cells, and astronauts’ visors.

Example 15.7

A nonreflective coating

A common lens coating material is magnesium fluoride 1MgF22, with n = 1.38. What thickness should a nonreflective coating have for 550-nm light if it is applied to glass with n = 1.52? S O L U T IO N IDENTIFY and SET UP: This coating is of the sort shown in Fig. 15.18. The thickness must be one-quarter of the wavelength of this light in the coating. EXECUTE: The wavelength in air is l0 = 550 nm, so its wavelength in the MgF2 coating is l = l0>n = 1550 nm2>1.38 = 400 nm. The coating thickness should be one-quarter of this, or l>4 = 100 nm.

EVALUATE: This is a very thin film, no more than a few hundred molecules thick. Note that this coating is reflective for light whose wavelength is twice the coating thickness; light of that wavelength reflected from the coating’s lower surface travels one wavelength farther than light reflected from the upper surface, so the two waves are in phase and interfere constructively. This occurs for light with a wavelength in MgF2 of 200 nm and a wavelength in air of 1200 nm211.382 = 276 nm. This is an ultraviolet wavelength (see Section 12.1), so designers of optical lenses with nonreflective coatings need not worry about such enhanced reflection.

Test Your Understanding of Section 15.4 A thin layer of benzene 1n = 1.5012 ties on top of a sheet of fluorite 1n = 1.4342. It is illuminated from above with light whose wavelength in benzene is 400 nm. Which of the following possible thicknesses of the benzene layer will maximize the brightness of the reflected light? (i) 100 nm; (ii) 200 nm; (iii) 300 nm; (iv) 400 nm.

15.5

y

The Michelson Interferometer

An important experimental device that uses interference is the Michelson interferometer. Michelson interferometers are used to make precise measurements of wavelengths and of very small distances, such as the minute changes in thickness of an axon when a nerve impulse propagates along its length. Like the Young two-slit experiment, a Michelson interferometer takes monochromatic light from a single source and divides it into two waves that follow different paths. In Young’s experiment, this is done by sending part of the light through one slit and part through another; in a Michelson interferometer a device called a beam splitter is used. Interference occurs in both experiments when the two light waves are recombined.

How a Michelson Interferometer Works Figure 15.19 shows the principal components of a Michelson interferometer. A ray of light from a monochromatic source A strikes the beam splitter C, which is a glass plate with a thin coating of silver on its right side. Part of the light (ray 1) passes through the silvered surface and the compensator plate D and is reflected from mirror M1. It then returns through D and is reflected from the silvered surface of C to the observer. The remainder of the light (ray 2) is reflected from the silvered surface at point P to the mirror M2 and back through C to the observer’s eye. '

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514

CHAPTER 15 Interference M2

15.19 A schematic Michelson interferometer. The observer sees an interference pattern that results from the difference in path lengths for rays 1 and 2.

Movable mirror

1 Monochromatic light is sent from light source A to beam splitter C.

L2

3 Ray 1 reflects off M1, passes through the compensator plate D, and reflects off the silvered surface P; ray 2 reflects off M2 and passes through the beam splitter C.

M1 2

Monochromatic light

Fixed mirror

1

A P 2 Rays 1 and 2 emerge

C from the beam splitter Beam and travel toward mirrors splitter M1 and M2, respectively.

D Compensator plate L1 Eye

4 Finally the two rays combine and reach the observer’s eye.

The purpose of the compensator plate D is to ensure that rays 1 and 2 pass through the same thickness of glass; plate D is cut from the same piece of glass as plate C, so their thicknesses are identical to within a fraction of a wavelength. The whole apparatus in Fig. 15.19 is mounted on a very rigid frame, and the position of mirror M2 can be adjusted with a fine, very accurate micrometer screw. If the distances L1 and L2 are exactly equal and the mirrors M1 and M2 are exactly at right angles, the virtual image of M1 formed by reflection at the silvered surface of plate C coincides with mirror M2. If L1 and L2 are not exactly equal, the image of M1 is displaced slightly from M2; and if the mirrors are not exactly perpendicular, the image of M1 makes a slight angle with M2. Then the mirror M2 and the virtual image of M1 play the same roles as the two surfaces of a wedge-shaped thin film (see Section 15.4), and light reflected from these surfaces forms the same sort of interference fringes. Suppose the angle between mirror M2 and the virtual image of M1 is just large enough that five or six vertical fringes are present in the field of view. If we now move the mirror M2 slowly either backward or forward a distance l>2, the difference in path length between rays 1 and 2 changes by l, and each fringe moves to the left or right a distance equal to the fringe spacing. If we observe the fringe positions through a telescope with a crosshair eyepiece and m fringes cross the crosshairs when we move the mirror a distance y, then '

'

y = m

l 2

or

l =

2y m

(15.19)

If m is several thousand, the distance y is large enough that it can be measured with good accuracy, and we can obtain an accurate value for the wavelength l. Alternatively, if the wavelength is known, a distance y can be measured by simply counting fringes when M2 is moved by this distance. In this way, distances that are comparable to a wavelength of light can be measured with relative ease.

The Michelson-Morley Experiment The original application of the Michelson interferometer was to the historic Michelson-Morley experiment. Before the electromagnetic theory of light became established, most physicists thought that the propagation of light waves occurred in a medium called the ether, which was believed to permeate all space. In 1887 the American scientists Albert Michelson and Edward Morley used the Michelson interferometer in an attempt to detect the motion of the earth through the ether. Suppose the interferometer in Fig. 15.19 is moving from left to right relative to the ether. According to the ether theory, this would lead to changes in the speed of light in the portions of the path shown as horizontal lines in the

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15.5 The Michelson Interferometer

figure. There would be fringe shifts relative to the positions that the fringes would have if the instrument were at rest in the ether. Then when the entire instrument was rotated 90°, the other portions of the paths would be similarly affected, giving a fringe shift in the opposite direction. Michelson and Morley expected that the motion of the earth through the ether would cause a shift of about four-tenths of a fringe when the instrument was rotated. The shift that was actually observed was less than a hundredth of a fringe and, within the limits of experimental uncertainty, appeared to be exactly zero. Despite its orbital motion around the sun, the earth appeared to be at rest relative to the ether. This negative result baffled physicists until 1905, when Albert Einstein developed the special theory of relativity (which we will study in detail in Chapter 17). Einstein postulated that the speed of a light wave in vacuum has the same magnitude c relative to all inertial reference frames, no matter what their velocity may be relative to each other. The presumed ether then plays no role, and the concept of an ether has been abandoned. Test Your Understanding of Section 15.5 You are observing the pattern of fringes in a Michelson interferometer like that shown in Fig. 15.19. If you change the index of refraction (but not the thickness) of the compensator plate, will the pattern change? y

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515

CHAPTER

15

SUMMARY y

Interference and coherent sources: Monochromatic light is light with a single frequency. Coherence is a definite, unchanging phase relationship between two waves. The overlap of waves from two coherent sources of monochromatic light forms an interference pattern. The principle of superposition states that the total wave disturbance at any point is the sum of the disturbances from the separate waves.

b

S1

x

a S2 c

Two-source interference of light: When two sources are in phase, constructive interference occurs where the difference in path length from the two sources is zero or an integer number of wavelengths; destructive interference occurs where the path difference is a half-integer number of wavelengths. If two sources separated by a distance d are both very far from a point P, and the line from the sources to P makes an angle u with the line perpendicular to the line of the sources, then the condition for constructive interference at P is Eq. (15.4). The condition for destructive interference is Eq. (15.5). When u is very small, the position ym of the mth bright fringe on a screen located a distance R from the sources is given by Eq. (15.6). (See Examples 15.1 and 15.2.)

Intensity in interference patterns: When two sinusoidal waves with equal amplitude E and phase difference f are superimposed, the resultant amplitude EP and intensity I are given by Eqs. (15.7) and (15.10), respectively. If the two sources emit in phase, the phase difference f at a point P (located a distance r1 from source 1 and a distance r2 from source 2) is directly proportional to the difference in path length r2 - r1. (See Example 15.3.) '

Interference in thin films: When light is reflected from both sides of a thin film of thickness t and no phase shift occurs at either surface, constructive interference between the reflected waves occurs when 2t is equal to an integral number of wavelengths. If a half-cycle phase shift occurs at one surface, this is the condition for destructive interference. A half-cycle phase shift occurs during reflection whenever the index of refraction in the second material is greater than that in the first. (See Examples 15.4–15.7.)

d sin u = ml 1m = 0, 61, 62, Á2 (constructive interference) (15.4) d sin u = A m +

1 2

Bl

(15.6)

EP = 2E P cos

(15.7)

f =

u r1

ml d (bright fringes)

I = I0 cos2

r2 S1

(15.5)

f P 2

d sin u

d

1m = 0, 61, 62, Á 2 (destructive interference) ym = R

S2

Phasors rotate counterclockwise. EP E

f 2

(15.10)

2p 1r - r12 = k1r2 - r12 (15.11) l 2

p2f

f E vt O E2 5 E cos vt E1 5 E cos (vt 1 f)

2t = ml 1m = 0, 1, 2, Á ) (constructive reflection from thin film, no relative phase shift) (15.17a) 2t = A m +

1 2

2t = A m +

1 2

To screen

Bl

1m = 0, 1, 2, Á 2

Bl

1m = 0, 1, 2, Á 2

P c a

(destructive reflection from thin film, no relative phase shift) (15.17b)

(constructive reflection from thin film, half-cycle relative phase shift) (15.18a) 2t = ml 1m = 0, 1, 2, Á 2 (destructive reflection from thin film, half-cycle relative phase shift) (15.18b)

516

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f Air b Film

e d

Index n

t

Discussion Questions

Michelson interferometer: The Michelson interferometer uses a monochromatic light source and can be used for high-precision measurements of wavelengths. Its original purpose was to detect motion of the earth relative to a hypothetical ether, the supposed medium for electromagnetic waves. The ether has never been detected, and the concept has been abandoned; the speed of light is the same relative to all observers. This is part of the foundation of the special theory of relativity.

M2

517

Movable mirror

L2 Monochromatic 2 light 1 A P C D Beam Compensator plate splitter L

M1 Fixed mirror

1

Eye

BRIDGING PROBLEM

Modifying a Two-Slit Experiment

An oil tanker spills a large amount of oil 1n = 1.452 into the sea 1n = 1.332. (a) If you look down onto the oil spill from overhead, what predominant wavelength of light do you see at a point where the oil is 380 nm thick? What color is the light? (Hint: See Table 32.1.) (b) In the water under the slick, what visible wavelength (as measured in air) is predominant in the transmitted light at the same place in the slick as in part (a)? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. The oil layer acts as a thin film, so we must consider interference between light that reflects from the top and bottom surfaces of the oil. If a wavelength is prominent in the transmitted light, there is destructive interference for that wavelength in the reflected light.

Problems

2. Choose the appropriate interference equations that relate the thickness of the oil film and the wavelength of light. Take account of the indexes of refraction of the air, oil, and water. EXECUTE 3. For part (a), find the wavelengths for which there is constructive interference as seen from above the oil film. Which of these are in the visible spectrum? 4. For part (b), find the visible wavelength for which there is destructive interference as seen from above the film. (This will ensure that there is substantial transmitted light at the wavelength.) EVALUATE 5. If a diver below the water’s surface shines a light up at the bottom of the oil film, at what wavelengths would there be constructive interference in the light that reflects back downward?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q15.1 A two-slit interference experiment is set up, and the fringes are displayed on a screen. Then the whole apparatus is immersed in the nearest swimming pool. How does the fringe pattern change? Q15.2 Could an experiment similar to Young’s two-slit experiment be performed with sound? How might this be carried out? Does it matter that sound waves are longitudinal and electromagnetic waves are transverse? Explain. Q15.3 Monochromatic coherent light passing through two thin slits is viewed on a distant screen. Are the bright fringes equally spaced on the screen? If so, why? If not, which ones are closest to being equally spaced? Q15.4 In a two-slit interference pattern on a distant screen, are the bright fringes midway between the dark fringes? Is this ever a good approximation? Q15.5 Would the headlights of a distant car form a two-source interference pattern? If so, how might it be observed? If not, why not?

Q15.6 The two sources S1 and S2 shown in Fig. 15.3 emit waves of the same wavelength l and are in phase with each other. Suppose S1 is a weaker source, so that the waves emitted by S1 have half the amplitude of the waves emitted by S2. How would this affect the positions of the antinodal lines and nodal lines? Would there be total reinforcement at points on the antinodal curves? Would there be total cancellation at points on the nodal curves? Explain your answers. Q15.7 Could the Young two-slit interference experiment be performed with gamma rays? If not, why not? If so, discuss differences in the experimental design compared to the experiment with visible light. Q15.8 Coherent red light illuminates two narrow slits that are 25 cm apart. Will a two-slit interference pattern be observed when the light from the slits falls on a screen? Explain. Q15.9 Coherent light with wavelength l falls on two narrow slits separated by a distance d. If d is less than some minimum value,

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CHAPTER 15 Interference

no dark fringes are observed. Explain. In terms of l, what is this minimum value of d? Q15.10 A fellow student, who values memorizing equations above understanding them, combines Eqs. (15.4) and (15.13) to “prove” that f can only equal 2pm. How would you explain to this student that f can have values other than 2pm? Q15.11 If the monochromatic light shown in Fig. 15.5a were replaced by white light, would a two-slit interference pattern be seen on the screen? Explain. Q15.12 In using the superposition principle to calculate intensities in interference patterns, could you add the intensities of the waves instead of their amplitudes? Explain. Q15.13 A glass windowpane with a thin film of water on it reflects less than when it is perfectly dry. Why? Q15.14 A very thin soap film 1n = 1.332, whose thickness is much less than a wavelength of visible light, looks black; it appears to reflect no light at all. Why? By contrast, an equally thin layer of soapy water 1n = 1.332 on glass 1n = 1.502 appears quite shiny. Why is there a difference? Q15.15 Interference can occur in thin films. Why is it important that the films be thin? Why don’t you get these effects with a relatively thick film? Where should you put the dividing line between “thin” and “thick”? Explain your reasoning. Q15.16 If we shine white light on an air wedge like that shown in Fig. 15.12, the colors that are weak in the light reflected from any point along the wedge are strong in the light transmitted through the wedge. Explain why this should be so. Q15.17 Monochromatic light is directed at normal incidence on a thin film. There is destructive interference for the reflected light, so the intensity of the reflected light is very low. What happened to the energy of the incident light? Q15.18 When a thin oil film spreads out on a puddle of water, the thinnest part of the film looks dark in the resulting interference pattern. What does this tell you about the relative magnitudes of the refractive indexes of oil and water?

SINGLE CORRECT Q15.1 When light enters from vacuum to a material of refractive index n, which of the following is true? (a) Its wavelength increases and its frequency decreases (b) Its wavelength decreases, but its frequency doesn’t change (c) Its wavelength decreases and its frequency increases (d) Its wavelength increases and its frequency increases Q15.2 The phenomenon of interference demonstrates the fact that: (a) light possesses transverse wave nature (b) light possesses longitudinal wave nature (c) light possesses particle nature (d) light possesses wave nature Q15.3 Coherent light is incident on two fine parallel slits S1 and S2 as shown in the diagram below.A dark fringe occurs at P when, n being an integer, the phase difference between the wave trains from S1 and S2 is (a) np rad S1 (b) (n + 1/2)p rad (c) (2n - 1)p rad S2 P (d) (2n + 1/2)p rad Q15.4 In a Young’s double–slit experiment, the central bright fringe can be identified (a) as it has greater intensity than the other bright fringes (b) as it is wider than the other bright fringes

(c) as it is narrower than the other bright fringes (d) by using white light instead of monochromatic light Q15.5 In a Young’s double–slit experiment, if the slits are of unequal width, (a) fringes will not be formed (b) the positions of minimum intensity will not be completely dark (c) bright fringe will not be formed at the centre of the screen (d) distance between two consecutive bright fringes will not be equal to the distance between two consecutive dark fringes Q15.6 In a Young double slit experiment, green light is incident on the two slits. The interference pattern is observed on a screen. Which one of the following changes would cause the observed fringes to be more closely spaced? (a) Reducing the separation Fringes between the slits S (b) Using blue light instead of Incoming light waves green light (c) Using red light instead of Screen green light (d) Moving the light source further away from the slits Q15.7 In a biprism experiment, using glass (m = 1.5) biprism when the set up is shifted from air to the inside of a still, clear lake water (m = 4/3) (a) the fringe pattern gets enlarged (b) the fringe pattern gets shrunk (c) the fringe pattern gets shifted. (d) the fringe pattern remains unchanged Q15.8 White light is normally incident on a soap bubble in air, and the reflected light is viewed. As the wall thickness of the bubble approaches zero (just before it breaks) the reflected light: (a) Becomes dim for all wavelengths. (b) Becomes unusually bright for all wavelengths. (c) Becomes bright only for certain wavelengths, dependent on the index of refraction of the bubble. (d) None of the above occurs. Q15.9 The contrast in the fringes in any interference pattern depends on (a) fringe width (b) ratio of width of slits (c) distance between the slits (d) wavelength

ONE OR MORE CORRECT Q15.10 In a Young’s Double slit experiment, the slits are illuminated by monochromatic light. The entire set up is immersed in pure water. In order to restore the original fringe width, what can be done ? (a) The slits can be brought closed together. (b) The screen can be moved away from the slit plane. (c) The incident light can be replaced by that of a longer wavelength. (d) A thin transparent slab can be introduced in front of one of the slits. Q15.11 For the biprism experiment shown in the figure, the fringe width increases when (a) Biprism is moved towards the slit (b) Screen is moved away from the biprism

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S1

a

S S2 Screen

Exercises

Screen

(c) A biprism having smaller angle a is used (d) The slit width is reduced. Q15.12 Consider a case of thin film interference as shown. Thickness of film is equal to wavelength of light in m2. (a) Reflected light will be maxima if µ 1 m1 < m2 < m3 (b) Reflected light will be maxima if µ2 m1 < m2 > m3 µ3 (c) Transmitted light will be maxima if m1 > m2 > m3 (d) Transmitted light will be maxima if m1 > m2 < m3 Q15.13 When monochromatic light is incident normally on a wedge-shaped thin air film, refer figure, an interference pattern may be seen by reflection. Which of the following is/are correct? (a) Parallel fringes are observed (b) If water is introduced into the region between the plates, the fringe separation decreases (c) If the angle of the wedge is increased, the fringe separation decreases (d) When white light is used there will not be a completely dark fringe Q15.14 A soap film with index of refraction greater than air is formed on a circular wire frame that is held in a vertical plane. The film is viewed by reflected light from a white-light source. Bands of color are observed at the lower parts of the soap film, but the area near the top appears black. A correct explanation for this phenomenon would involve which of the following? Choose the correct option(s) (a) The top of the soap film absorbs all of the light incident on it; none is reflected (b) The thickness of the top part of the soap film has become much less than a wavelength of visible light (c) There is a phase change of 180° for all wavelengths of light reflected from the front surfaces of the soap film (d) There is no phase change for any wavelength of light reflected from the back surface of the soap film Q15.15 In a YDSE setup, we plot the phase difference (f) between both waves at point P on the screen against the angular position u of point P on the screen. The graph is as shown below. Choose the correct statement(s) (a) The distance S1S2 = 2l (b) There are a total of 4 minima P S1 4p on the screen f q (c) The first maxima above the p S2 centre is at u = q π2 4 p (d) At u = intensity is minimum 3 Q15.16 In the following Young’s double slit experiment picture, OP = y.Symbols have usual meaning (wavelength l > d) Choose the CORRECT option(s) (a) If a film of thickness t and refrac tive index 1.5 is placed before S1, S1 effective path difference in air at P yd d t is ¢x = O D 2 (b) If a film of thickness t refractive S2 index 1.5 is placed before S2, effecD tive path difference in air at P is yd t ¢x = + D 2

519

(c) If a film of thickness t and refractive index 1.5 is placed before S1 and S2 both, effective path difference in air at P is yd t ¢x = + D 4 (d) If a film of thickness t and refractive index 1.5 is placed before S1 and another film of thickness t and refractive index 2 is placed before S2, effective path difference in air at P is yd ¢x = + t D

MATCH THE COLUMNS Q15.17 In a typical Young’s double slit experiment, S1 and S2 are identical slits and equidistant from a point monochromatic source S of light having wavelength l. The distance between slits is represented by d and that between the plane of slits and screen is reprelD0 sented by D. P is a fixed point on the screen at a distance y = 2d0 from central bright fringe on the screen : where D0, d0 are initial values of D and d respectively. In each statement of Column-I some changes are made to above mentioned situation. The effect of corresponding changes is given in Column-II. Match the statements in Column-I with resulting changes in Column-II. Column I

Column II

(a) The distance d between the (P) Fringe width increases slits is doubled. (b) The distance D between slit (Q) Magnitude of optical path and screen is doubled difference between interfering waves at P will decrease (c) The width of slit S1 is (R) Magnitude of optical path difference between interdecreased (such that intenfering waves at P will sity of light due to slit S1 on increases screen decreases) and the distance D between slit and screen is doubled by shifting screen to right (d) The whole setup is sub- (S) The intensity at P will merged in water of refracincrease 4 tive index 3 (T) Fringe width decreases.

EXERCISES Section 15.1 Interference and Coherent Sources

15.1 .. Two speakers that are 15.0 m apart produce in-phase sound waves of frequency 250.0 Hz in a room where the speed of sound is 340.0 m> s. A woman starts out at the midpoint between the two speakers. The room’s walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear: constructive or destructive interference? Why? (b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity? (c) How far from the center must she walk before she first hears the sound maximally enhanced?

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CHAPTER 15 Interference

15.2 .. Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and 600 Hz. The speed of sound is 340 m> s. You find that where you are standing, you hear minimumintensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 cm toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity? 15.3 . Radio Interference. Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point Q? (b) What is the longest wavelength for which there will be constructive interference at point Q? 15.4 .. A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is 9.00 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P? 15.5 . Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 mm apart, and in line with an observer, so that one source is 2.04 mm farther from the observer than the other. (a) For what visible wavelengths (380 to 750 nm) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04 mm farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

Section 15.2 Two-Source Interference of Light

15.6 . Young’s experiment is performed with light from excited helium atoms 1l = 502 nm2. Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits? 15.7 .. Coherent light with wavelength 450 nm falls on a double slit. On a screen 1.80 m away, the distance between dark fringes is 4.20 mm. What is the separation of the slits? 15.8 .. Two thin parallel slits that are 0.0116 mm apart are illuminated by a laser beam of wavelength 585 nm. (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that sin u can be? What does this tell you is the largest value of m?) 15.9 . Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by 0.200 mm, and the interference pattern is observed on a screen 4.00 m from the slits. (a) What is the width (in mm) of the central interference maximum? (b) What is the width of the first-order bright fringe? 15.10 .. Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits separated by 0.300 mm, and the interference pattern is observed on a screen

5.00 m from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths? 15.11 .. Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fringe is at 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

Section 15.3 Intensity in Interference Patterns

15.12 .. In a two-slit interference pattern, the intensity at the peak of the central maximum is I0. (a) At a point in the pattern where the phase difference between the waves from the two slits is 60.0°, what is the intensity? (b) What is the path difference for 480-nm light from the two slits at a point where the phase angle is 60.0°? 15.13 .. Points A and B are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5-MHz signal horizontally. These transmitters are in phase with each other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 km north of the AB line and initially placed at point C, directly opposite the midpoint of AB. The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than 14 of its maximum value. How far from point C (along an east-west line) can the receiver be moved and always be able to pick up the signal?

Section 15.4 Interference in Thin Films

15.14 . What is the thinnest film of a coating with n = 1.42 on glass 1n = 1.522 for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection? 15.15 .. Nonglare Glass. When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use TiO2, which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 505 nm? (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones. 15.16 .. Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546-nm light from a mercuryvapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge. 15.17 . A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction 1.52. (a) What minimum thickness is required if light with wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference? 15.18 . The walls of a soap bubble have about the same index of refraction as that of plain water, n = 1.33. There is air both inside and outside the bubble. (a) What wavelength (in air) of visible light is most strongly reflected from a point on a soap bubble where its

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Problems

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wall is 290 nm thick? (b) Repeat part (a) for a wall thickness of 340 nm. 15.19 .. Compact Disc Player. A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength 790 nm passing through a plastic substrate of refractive index 1.8. When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so these two beams interfere with each other (Fig. E15.19). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)

occur at 6 30° instead. What is the index of refraction of this liquid?

Figure E15.19

wave is approximately 14 cycle out of phase with either of the original waves.

Pits

Reflective coating

Plastic substrate

Laser beam

15.20 . What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.

PROBLEMS 15.21 . BIO Coating Eyeglass Lenses.

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

15.22.. Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is 11.0 cm long and has a refractive index of 1.55. A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end. When you view the glass plates from above with reflected white light, you observe that, at 1.15 mm from the line where the sheets are in contact, the violet light of wavelength 400.0 nm is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength 550 nm) and orange light (of wavelength 600.0 nm) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart? 15.23 ... Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference minima at 6 45° on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima

15.24 ... CP The electric fields received at point P from two identical, coherent wave sources are E11t2 = E cos1vt + f2 and E21t2 = E cos vt. (a) Use the trigonometric identities in Appendix B to show that the resultant wave is EP1t2 = 2E cos1f>22 cos1vt + f>22. (b) Show that the amplitude A of this resultant wave is given f by A = 2E ƒ cos ƒ . (c) Use the result of part (a) to show that at an 2 interference maximum, the amplitude of the resultant wave is in phase with the original waves E11t2 and E21t2. (d) Use the result of part (a) to show that near an interference minimum, the resultant

15.25 .. Consider a two-slit interference pattern, for which the intensity distribution is given by Eq. (15.14). Let um be the angular position of the mth bright fringe, where the intensity is I0. Assume that um is small, Let um + and um - be the two angles on either side of um for which I = 12 I0. The quantity ¢um = ƒ um + - um - ƒ is the half-width of the mth fringe. Calculate ¢um. How does ¢um depend on m? 15.26 .. White light reflects at normal incidence from the top and bottom surfaces of a glass plate 1n = 1.522. There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 nm. What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 nm? 15.27 ... A source S of monochromatic light and a detector D are both located in air a distance h above a horizontal plane sheet of glass and are separated by a horizontal distance x. Waves reaching D directly from S interfere with waves that reflect off the glass. The distance x is small compared to h so that the reflection is at close to normal incidence. Show that the condition for constructive interference is 2x2 + 4h2 - x = A m + 12 B l and the condition

for destructive interference is 2x2 + 4h2 - x = ml. (Hint: Take into account the phase change on reflection.) 15.28 .. Red light with wavelength 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe 1m = 32 of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? 15.29 ... In a Young’s two-slit experiment a piece of glass with an index of refraction n and a thickness L is placed in front of the upper slit. It is the max intensity in absence of glass piece and l0 is the wavelength of light in the air. d is the separation of two slits. (a) Describe qualitatively what happens to the interference pattern. (b) Derive an expression for the intensity I of the light at points on a screen as a function given quantites. u is the usual downward angle measured from the center of the two slits. (c) From your result in part (b) derive an expression for the values of u that locate the maxima in the interference pattern.

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CHAPTER 15 Interference Figure P15.30

CHALLENGE PROBLEMS 15.30 ... CP Figure P15.30 shows

an interferometer known as Fresnel’s biprism. The magnitude of the prism angle A is extremely small. (a) If S0 is a very narrow source slit, show that the separation of the two virtual coherent sources S1 and S2 is given by d = 2aA1n - 12, where n is the index of refraction of the material of the prism.

A

S1

P

S0 d

O

S2 A a

b

Answers Chapter Opening Question

?

ber of wavelengths within the thickness of the plate. Hence this has the same effect as changing the distance L1 from the beam splitter to mirror M1 , which would change the interference pattern. '

The colors appear due to constructive interference between light waves reflected from the outer and inner surfaces of the soap bubble. The thickness of the bubble walls at each point determines the wavelength of light for which the most constructive interference occurs and hence the color that appears the brightest at that point (see Example 15.4 in Section 15.4).

Test Your Understanding Questions 15.1 (i) At any point P on the positive y-axis above S1 , the distance r2 from S2 to P is greater than the distance r1 from S1 to P by 4l. This corresponds to m = 4 in Eq. (15.1), the equation for constructive interference. Hence all such points make up an antinodal curve. 15.2 (ii) Blue light has a shorter wavelength than red light (see Section 12.1). Equation (15.6) tells us that the distance ym from the center of the pattern to the mth bright fringe is proportional to the wavelength l. Hence all of the fringes will move toward the center of the pattern as the wavelength decreases, and the spacing between fringes will decrease. 15.3 (i), (iv), (ii), (iii) In cases (i) and (iii) we are given the wavelength l and path difference d sin u. Hence we use Eq. (15.14), I = I0 cos231pd sin u)>l4. In parts (ii) and (iii) we are given the phase difference f and we use Eq. (15.10), I = I0 cos21f>22. We find: (i) I = I0 cos23p14.00 * 10-7 m2>15.00 * 10-7 m24 = I0 cos210.800p rad2 = 0.655I0; (ii) I = I0 cos2314.00 rad2>24 = I0 cos212.00 rad2 = 0.173I0; (iii) I = I0 cos23p17.50 * 10-7 m2>15.00 * 10-7 m24 = I0 cos211.50p rad2 = 0; (iv) I = I0 cos2312.00 rad2>24 = I0 cos211.00 rad2 = 0.292I0 . 15.4 (i) and (iii) Benzene has a larger index of refraction than air, so light that reflects off the upper surface of the benzene undergoes a half-cycle phase shift. Fluorite has a smaller index of refraction than benzene, so light that reflects off the benzene–fluorite interface does not undergo a phase shift. Hence the equation for constructive reflection is Eq. (15.18a), 2t = A m + 12 B l, which we can rewrite as t = A m + 12 B l>2 = A m + 12 B 1400 mm2>2 = 100 nm, 300 nm, 500 nm, Á . 15.5 yes Changing the index of refraction changes the wavelength of the light inside the compensator plate, and so changes the num'

'

Bridging Problem (a) 441 nm (b) 551 nm

Discussion Questions Q15.1 All the fringes will come closer due to decrease in wavelength. Q15.2 Yes, in same manner use two sound sources. It does not depend upon nature. Q15.3 No, only fringes which are near to central bright fringe (central maximum) will be equally spaced. Q15.4 Not, always it can be good approximation for fringes near central maxima if distance between two slits is large compared to wavelength. Q15.5 No, they are not coherent sources. Q15.6 Position of antinodal lines & nodal lines will not be affected but now there will not be total cancellation at nodal curves due to unequal amplitude. Q15.7 Not possible as for this situation slit width should very small. Only theoritically possible Q15.8 No, distance between slits should be comparable. l Q15.9 If d < . No dark fringe will be obtained. 2 Q15.10 dsinu = nl is condition for contructive interference but it can take any value therefore f can take any value Q15.11 Yes, it will be seen on screen. Central maxima will be white. Then near to central maxima very few colored fringes will be seen and then pattern will disapper due to overlapping. Q15.12 No, we can’t add intensities as intensinty q A2. Q15.13 Due to thin film reflection decreases. Due to destructive interference in reflected light. Q15.14 In first case there is destructive interference while in case two it is construtive due to extra phase difference of p. (Due to reflection from denser medium) Q15.15 If it is comparable to wavelength of light then it will be vissible otherwise not visible. If thickness much greater then l then we call it is as thick. Q15.16 Due to energy conservation principle.

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Answers

Q15.17 Maximum light will be transmitted. Q15.18 Oil is more denser than water.

Single Correct Q15.1 (b) Q15.2 (d) Q15.3 (c) Q15.4 (d) Q15.5 (b) Q15.6 (b) Q15.7 (a) Q15.8 (a) Q15.9 (b)

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Section 15.2 Two-Source Interference of Light 15.6 1.14 mm 15.7 0.193 mm 15.8 39 15.9 (a) 8 mm (b) 8 mm 15.10 3.17 mm 15.11 1200 nm

Section 15.3 Intensity in Interference Patterns 15.12 (a) 0.75 I0 (b) 80 nm 15.13 71.4 m

Section 15.4 Interference in Thin Films

One or More Correct Q15.10 (a), (b), (c) Q15.11 (a), (b), (c) Q15.12 (a), (d) Q15.13 (a), (b), (c) Q15.14 (b), (c), (d) Q15.15 (a), (b) Q15.16 (a) 6

Match the Columns

15.14 114 nm 15.15 (a) 96.4 nm (b) 192 nm , 289 nm , 386 nm 15.16 4.09 * 10 - 4 rad 15.17 (a) 74.3 nm (b) 223 nm 15.18 (a) 514 nm (b) 603 nm 15.19 110 nm 15.20 180 nm

Problems 15.21 (a) 96 nm (b) No 15.22 (a) 1.58 mm, 1.72 mm (b) 3.45 mm, 4.74 mm, 5.16 mm (c) 9.57 mm

Q15.17 (a) R, S, T (b) P, Q, S (c) P, Q, S (d) R, S, T

15.23 Í2

EXERCISES

15.25 ¢Om =

Section 15.1 Interference and Coherent Sources

fringe.

15.1 (a) constructive (b) 34 cm (c) 68 cm 15.2 (a) destructive interference, (b) 427 Hz (c) 0.796 m 15.3 (a) 240 m (b) 120 m 15.4 0.75 m , 2 m , 3.25 m , 4.50 m , 5.75 m , 7.00 m , 8.25 m 15.5 (a) 680 nm , 510 nm , 408 nm (b) ans would be same as a part (a) (c) 583 nm , 453 nm

15.26 1334 nm

l , There is no dependence on the m-value of the 2d

15.28 600 nm, 467 nm 15.29 (b) I = I0 cos2 c a

p b (d sinu + L (n - 1)) d l0 ml0 - L (n - 1) (c) sinu = ; m = 0, ; 1, ; 2, Á . d

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16

DIFFRACTION

LEARNING GOALS By studying this chapter, you will learn: • What happens when coherent light shines on an object with an edge or aperture. • How to understand the diffraction pattern formed when coherent light passes through a narrow slit. • How to calculate the intensity at various points in a single-slit diffraction pattern. • What happens when coherent light shines on an array of narrow, closely spaced slits. • How scientists use diffraction gratings for precise measurements of wavelength. • How diffraction sets limits on the smallest details that can be seen with a telescope.

?

The laser used to read a DVD has a wavelength of 650 nm, while the laser used to read a Blu-ray disc has a shorter 405-nm wavelength. How does this make it possible for a Blu-ray disc to hold more information than a DVD?

veryone is used to the idea that sound bends around corners. If sound didn’t behave this way, you couldn’t hear a police siren that’s out of sight around a corner or the speech of a person whose back is turned to you. What may surprise you (and certainly surprised many scientists of the early 19th century) is that light can bend around corners as well. When light from a point source falls on a straightedge and casts a shadow, the edge of the shadow is never perfectly sharp. Some light appears in the area that we expect to be in the shadow, and we find alternating bright and dark fringes in the illuminated area. In general, light emerging from apertures doesn’t behave precisely according to the predictions of the straight-line ray model of geometric optics. The reason for these effects is that light, like sound, has wave characteristics. In Chapter 35 we studied the interference patterns that can arise when two light waves are combined. In this chapter we’ll investigate interference effects due to combining many light waves. Such effects are referred to as diffraction. We’ll find that the behavior of waves after they pass through an aperture is an example of diffraction; each infinitesimal part of the aperture acts as a source of waves, and the resulting pattern of light and dark is a result of interference among the waves emanating from these sources. Light emerging from arrays of apertures also forms patterns whose character depends on the color of the light and the size and spacing of the apertures. Examples of this effect include the colors of iridescent butterflies and the “rainbow” you see reflected from the surface of a compact disc. We’ll explore similar effects with x rays that are used to study the atomic structure of solids and liquids. Finally, we’ll look at the physics of a hologram, a special kind of interference pattern recorded on photographic film and reproduced. When properly illuminated, it forms a three-dimensional image of the original object.

E

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16.1 Fresnel and Fraunhofer Diffraction

16.1

Fresnel and Fraunhofer Diffraction

According to geometric optics, when an opaque object is placed between a point light source and a screen, as in Fig. 16.1, the shadow of the object forms a perfectly sharp line. No light at all strikes the screen at points within the shadow, and the area outside the shadow is illuminated nearly uniformly. But as we saw in Chapter 35, the wave nature of light causes effects that can’t be understood with geometric optics. An important class of such effects occurs when light strikes a barrier that has an aperture or an edge. The interference patterns formed in such a situation are grouped under the heading diffraction. Figure 16.2 shows an example of diffraction. The photograph in Fig. 16.2a was made by placing a razor blade halfway between a pinhole, illuminated by monochromatic light, and a photographic film. The film recorded the shadow cast by the blade. Figure 16.2b is an enlargement of a region near the shadow of the right edge of the blade. The position of the geometric shadow line is indicated by arrows. The area outside the geometric shadow is bordered by alternating bright and dark bands. There is some light in the shadow region, although this is not very visible in the photograph. The first bright band in Fig. 16.2b, just to the right of the geometric shadow, is considerably brighter than in the region of uniform illumination to the extreme right. This simple experiment gives us some idea of the richness and complexity of what might seem to be a simple idea, the casting of a shadow by an opaque object. We don’t often observe diffraction patterns such as Fig. 16.2 in everyday life because most ordinary light sources are neither monochromatic nor point sources. If we use a white frosted light bulb instead of a point source to illuminate the razor blade in Fig. 16.2, each wavelength of the light from every point of the bulb forms its own diffraction pattern, but the patterns overlap so much that we can’t see any individual pattern.

16.1 A point source of light illuminates a straightedge. Geometric optics predicts that this situation should produce a sharp boundary between illumination and solid shadow. DOESN’T HAPPEN That’s NOT what really happens! Point source

Area of illumination

Geometric shadow Straightedge

Diffraction and Huygens’s Principle We can analyze diffraction patterns using Huygens’s principle (see Section 33.7). This principle states that we can consider every point of a wave front as a source of secondary wavelets. These spread out in all directions with a speed equal to the speed of propagation of the wave. The position of the wave front at any later time is the envelope of the secondary wavelets at that time. To find the resultant displacement at any point, we combine all the individual displacements produced by these secondary waves, using the superposition principle and taking into account their amplitudes and relative phases. 16.2 An example of diffraction. (a)

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(b) Photograph of a razor blade illuminated by monochromatic light from a point source (a Enlarged view of the area outside the pinhole). Notice the fringe around the geometric shadow of the blade’s edge blade outline. Position of geometric shadow

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Screen

526

CHAPTER 16 Diffraction

In Fig. 16.1, both the point source and the screen are relatively close to the obstacle forming the diffraction pattern. This situation is described as near-field diffraction or Fresnel diffraction, pronounced “Freh-nell” (after the French scientist Augustin Jean Fresnel, 1788–1827). By contrast, we use the term Fraunhofer diffraction (after the German physicist Joseph von Fraunhofer, 1787–1826) for situations in which the source, obstacle, and screen are far enough apart that we can consider all lines from the source to the obstacle to be parallel, and can likewise consider all lines from the obstacle to a given point on the screen to be parallel. We will restrict the following discussion to Fraunhofer diffraction, which is usually simpler to analyze in detail than Fresnel diffraction. Diffraction is sometimes described as “the bending of light around an obstacle.” But the process that causes diffraction is present in the propagation of every wave. When part of the wave is cut off by some obstacle, we observe diffraction effects that result from interference of the remaining parts of the wave fronts. Optical instruments typically use only a limited portion of a wave; for example, a telescope uses only the part of a wave that is admitted by its objective lens or mirror. Thus diffraction plays a role in nearly all optical phenomena. Finally, we emphasize that there is no fundamental distinction between and diffraction. In Chapter 35 we used the term interference for effects involving waves from a small number of sources, usually two. Diffraction usually involves a continuous distribution of Huygens’s wavelets across the area of an aperture, or a very large number of sources or apertures. But both interference and diffraction are consequences of superposition and Huygens’s principle. Test Your Understanding of Section 16.1 Can sound waves undergo diffrac tion around an edge?

16.2 PhET: Wave Interference ActivPhysics 16.6: Single-Slit Diffraction

Diffraction from a Single Slit

In this section we’ll discuss the diffraction pattern formed by plane-wave (parallelray) monochromatic light when it emerges from a long, narrow slit, as shown in Fig. 16.3. We call the narrow dimension the width, even though in this figure it is a vertical dimension. According to geometric optics, the transmitted beam should have the same cross section as the slit, as in Fig. 16.3a. What is actually observed is the pattern shown in Fig. 16.3b. The beam spreads out vertically after passing through the slit. The diffraction pattern consists of a central bright band, which may be much broader than the width of the slit, bordered by alternating dark and bright bands with rapidly decreasing intensity. About 85% of the power in the

16.3 (a) The “shadow” of a horizontal slit as incorrectly predicted by geometric optics. (b) A horizontal slit actually produces a diffraction pattern. The slit width has been greatly exaggerated. (a) PREDICTED OUTCOME: Geometric optics predicts that this setup will produce a single bright band the same size as the slit.

a

y

Screen

(b) WHAT REALLY HAPPENS: In reality, we see a diffraction pattern—a set of interference fringes.

a

Width

Parallel-ray monochromatic light

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527

16.2 Diffraction from a Single Slit 16.4 Diffraction by a single rectangular slit. The long sides of the slit are perpendicular to the figure. (a) A slit as a source of wavelets We divide the slit into imaginary strips parallel to the slit’s long axis. Slit width a

(b) Fresnel (near-field) diffraction

(c) Fraunhofer (far-field) diffraction

If the screen is close, the rays from the different strips to a point P on the screen are not parallel.

(d) Imaging Fraunhofer diffraction

If the screen is distant, the rays to P are approximately parallel.

A converging lens images a Fraunhofer pattern on a nearby screen. Converging cylindrical lens a

Each strip is a source of Huygens’s wavelets. Plane waves incident on the slit

P

P f Screen

transmitted beam is in the central bright band, whose width is inversely proportional to the width of the slit. In general, the smaller the width of the slit, the broader the entire diffraction pattern. (The horizontal spreading of the beam in Fig. 16.3b is negligible because the horizontal dimension of the slit is relatively large.) You can observe a similar diffraction pattern by looking at a point source, such as a distant street light, through a narrow slit formed between your two thumbs held in front of your eye; the retina of your eye corresponds to the screen.

Single-Slit Diffraction: Locating the Dark Fringes Figure 16.4 shows a side view of the same setup; the long sides of the slit are perpendicular to the figure, and plane waves are incident on the slit from the left. According to Huygens’s principle, each element of area of the slit opening can be considered as a source of secondary waves. In particular, imagine dividing the slit into several narrow strips of equal width, parallel to the long edges and perpendicular to the page. Figure 16.4a shows two such strips. Cylindrical secondary wavelets, shown in cross section, spread out from each strip. In Fig. 16.4b a screen is placed to the right of the slit. We can calculate the resultant intensity at a point P on the screen by adding the contributions from the individual wavelets, taking proper account of their various phases and amplitudes. It’s easiest to do this calculation if we assume that the screen is far enough away that all the rays from various parts of the slit to a particular point P on the screen are parallel, as in Fig. 16.4c. An equivalent situation is Fig. 16.4d, in which the rays to the lens are parallel and the lens forms a reduced image of the same pattern that would be formed on an infinitely distant screen without the lens. We might expect that the various light paths through the lens would introduce additional phase shifts, but in fact it can be shown that all the paths have equal phase shifts, so this is not a problem. The situation of Fig. 16.4b is Fresnel diffraction; those in Figs. 16.4c and 16.4d, where the outgoing rays are considered parallel, are Fraunhofer diffraction. We can derive quite simply the most important characteristics of the Fraunhofer diffraction pattern from a single slit. First consider two narrow strips, one just below the top edge of the drawing of the slit and one at its center, shown in end view in Fig. 16.5. The difference in path length to point P is 1a>22 sin u, where a is the slit width and u is the angle between the perpendicular to the slit and a line from the center of the slit to P. Suppose this path difference happens to be equal to l>2; then light from these two strips arrives at point P with a halfcycle phase difference, and cancellation occurs. Similarly, light from two strips immediately below the two in the figure also arrives at P a half-cycle out of phase. In fact, the light from every strip in the top half of the slit cancels out the light from a corresponding strip in the bottom half. [

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Screen

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CHAPTER 16 Diffraction

16.5 Side view of a horizontal slit. When the distance x to the screen is much greater than the slit width a, the rays from a distance a/2 apart may be considered parallel. (a) P y

u

x

a

O

/

For the two strips shown, the path difference to P is (a 2) sin u. When 1 a 22 sin u 5 l 2, the light cancels at P. This is true for the whole slit, so P represents a dark fringe.

/

/

(b) Enlarged view of the top half of the slit

a 2

u

u is usually very small, so we can use the approximations sin u 5 u and tan u 5 u. Then the condition for a dark band is

u

ym 5 x ml a

a sin u 2

Hence the combined light from the entire slit completely cancels at P, giving a dark fringe in the interference pattern. A dark fringe occurs whenever a l sin u = 6 2 2

or

sin u = 6

l a

(16.1)

The plus-or-minus 162 sign in Eq. (16.1) says that there are symmetric dark fringes above and below point O in Fig. 16.5a. The upper fringe 1u 7 02 occurs at a point P where light from the bottom half of the slit travels l>2 farther to P than does light from the top half; the lower fringe 1u 6 02 occurs where light from the top half travels l>2 farther than light from the bottom half. We may also divide the screen into quarters, sixths, and so on, and use the above argument to show that a dark fringe occurs whenever sin u = 62l>a, 63l>a, and so on. Thus the condition for a dark fringe is

sin u =

ml a

1m = 61, 62, 63, Á 2

(dark fringes in single-slit diffraction)

(16.2)

For example, if the slit width is equal to ten wavelengths 1a = 10l2, dark fringes 1 2 3 occur at sin u = 6 10 , 6 10 , 6 10 , Á . Between the dark fringes are bright fringes. We also note that sin u = 0 corresponds to a bright band; in this case, light from the entire slit arrives at P in phase. Thus it would be wrong to put m = 0 in Eq. (16.2). The central bright fringe is wider than the other bright fringes, as Fig. 16.3b shows. In the small-angle approximation that we will use below, it is exactly twice as wide. With light, the wavelength l is of the order of 500 nm = 5 * 10-7 m. This is often much smaller than the slit width a; a typical slit width is 10-2 cm = 10-4 m. Therefore the values of u in Eq. (16.2) are often so small that the approximation sin u L u (where u is in radians) is a very good one. In that case we can rewrite this equation as u =

ml a

1m = 61, 62, 63, Á 2

(for small angles u)

where u is in radians. Also, if the distance from slit to screen is x, as in Fig. 16.5a, and the vertical distance of the mth dark band from the center of the pattern is ym , '

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529

16.3 Intensity in the Single-Slit Pattern 16.6 Photograph of the Fraunhofer diffraction pattern of a single horizontal slit.

then tan u = ym>x. For small u we may also approximate tan u by u (in radians), and we then find ym = x

m53 m52

1for ym V x2

CAUTION Single-slit diffraction vs. two-slit interference Equation (16.3) has the same form as the equation for the two-slit pattern, Eq. (35.6), except that in Eq. (16.3) we use x rather than R for the distance to the screen. But Eq. (16.3) gives the positions of the dark fringes in a single-slit pattern rather than the bright fringes in a double-slit pattern. Also, m = 0 in Eq. (16.2) is not a dark fringe. Be careful! y

m 5 21 m 5 22 m 5 23

Single-slit diffraction

You pass 633-nm laser light through a narrow slit and observe the diffraction pattern on a screen 6.0 m away. The distance on the screen between the centers of the first minima on either side of the central bright fringe is 32 mm (Fig. 16.7). How wide is the slit? S O L U T IO N IDENTIFY and SET UP: This problem involves the relationship between the positions of dark fringes in a single-slit diffraction

pattern and the slit width a (our target variable). The distances between fringes on the screen are much smaller than the slit-toscreen distance, so the angle u shown in Fig. 16.5a is very small and we can use Eq. (16.3) to solve for a. EXECUTE: The first minimum corresponds to m = 1 in Eq. (16.3). The distance y1 from the central maximum to the first minimum on either side is half the distance between the two first minima, so y1 = (32 mm2>2 = 16 mm. Solving Eq. (16.3) for a, we find

16.7 A single-slit diffraction experiment. a =

y

Slit width 5 ? x

32 mm

16.0 m21633 * 10-9 m2 xl = = 2.4 * 10-4 m = 0.24 mm y1 16 * 10-3 m

EVALUATE: The angle u is small only if the wavelength is small compared to the slit width. Since l = 633 nm = 6.33 * 10-7 m and we have found a = 0.24 mm = 2.4 * 10-4 m, our result is consistent with this: The wavelength is 16.33 * 10-7 m2 > 12.4 * 10-4 m2 = 0.0026 as large as the slit width. Can you show that the distance between the second minima on either side is 2132 mm2 = 64 mm, and so on?

x 5 6.0 m Screen

Test Your Understanding of Section 16.2 Rank the following singleslit diffraction experiments in order of the size of the angle from the center of the diffraction pattern to the first dark fringe, from largest to smallest: (i) wavelength 400 nm, slit width 0.20 mm; (ii) wavelength 600 nm, slit width 0.20 mm; (iii) wavelength 400 nm, slit width 0.30 mm; (iv) wavelength 600 nm, slit width 0.30 mm. y

16.3

(16.3)

Figure 16.6 is a photograph of a single-slit diffraction pattern with the m = 61, 62, and 63 minima labeled.

m51

Example 16.1

ml a

Intensity in the Single-Slit Pattern

We can derive an expression for the intensity distribution for the single-slit diffraction pattern by the same phasor-addition method that we used in Section 35.3 to obtain Eqs. (35.10) and (35.14) for the two-slit interference pattern. We again imagine a plane wave front at the slit subdivided into a large number of strips. We superpose the contributions of the Huygens wavelets from all the strips at a point P on a distant screen at an angle u from the normal to the slit plane S(Fig. 16.8a). To do this, we use a phasor to represent the sinusoidally varying E field from

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'

'

530

CHAPTER 16 Diffraction

16.8 Using phasor diagrams to find the S amplitude of the E field in single-slitSdiffraction. Each phasor represents the E field from a single strip within the slit. (a)

Strips within slit

Slit width a

each individual strip. The magnitude ofS the vector sum of the phasors at each point P is the amplitude EP of the total E field at that point. The intensity at P is proportional to EP 2. At the point O shown in Fig. 16.8a, corresponding to the center of the pattern where u = 0, there are negligible path differences for x W a; the phasors are all essentially in phase (that is, have the same direction). In Fig. 16.8b we draw the phasors at time t = 0 and denote the resultant amplitude at O by E0 . In this illustration we have divided the slit into 14 strips. Now consider wavelets arriving from different strips at point P in Fig. 16.8a, at an angle u from point O. Because of the differences in path length, there are now phase differences between wavelets coming from adjacent strips; the corresponding phasor diagram is shown in Fig. 16.8c. The vector sum of the phasors is now part of the perimeter of a many-sided polygon, and EP , the amplitude of the resultant electric field at P, is the chord. The angle b is the total phase difference between the wave from the top strip of Fig. 16.8a and the wave from the bottom strip; that is, b is the phase of the wave received at P from the top strip with respect to the wave received at P from the bottom strip. We may imagine dividing the slit into narrower and narrower strips. In the limit that there is an infinite number of infinitesimally narrow strips, the curved trail of phasors becomes an arc of a circle (Fig. 16.8d), with arc length equal to the length E0 in Fig. 16.8b. The center C of this arc is found by constructing perpendiculars at A and B. From the relationship among arc length, radius, and angle, the radius of the arc is E0>b; the amplitude EP of the resultant electric field at P is equal to the chord AB, which is 21E0>b2 sin1b>22. (Note that b must be in radians!) We then have '

O P Distant screen

Plane waves incident on the slit

'

(b) At the center of the diffraction pattern (point O), the phasors from all strips within the slit are in phase. E0

(c) Phasor diagram at a point slightly off the center of the pattern; b 5 total phase difference between the first and last phasors.

EP

EP = E0

b

E0

(d) As in (c), but in the limit that the slit is subdivided into infinitely many strips

b E0 b

b 2

b 2

b n 2

b n 2

E 0 si b A

E0 b

B

E 0 si b EP E0

(amplitude in single-slit diffraction)

(16.4)

The intensity at each point on the screen is proportional to the square of the amplitude given by Eq. (16.4). If I0 is the intensity in the straight-ahead direction where u = 0 and b = 0, then the intensity I at any point is I = I0 c

C

sin1b>22 b>2

sin1b>22 2 d b>2

(intensity in single-slit diffraction)

(16.5)

We can express the phase difference b in terms of geometric quantities, as we did for the two-slit pattern. From Eq. (35.11) the phase difference is 2p>l times the path difference. Figure 16.5 shows that the path difference between the ray from the top of the slit and the ray from the middle of the slit is 1a>22 sin u. The path difference between the rays from the top of the slit and the bottom of the slit is twice this, so

b

b =

2p a sin u l

(16.6)

and Eq. (16.5) becomes I = I0 b '

sin3pa1sin u2>l4 2 r pa1sin u2>l

(intensity in single-slit diffraction)

(16.7)

This equation expresses the intensity directly in terms of the angle u. In many calculations it is easier first to calculate the phase angle b, using Eq. (16.6), and then to use Eq. (16.5). Equation (16.7) is plotted in Fig. 16.9a. Note that the central intensity peak is much larger than any of the others. This means that most of the power in the wave remains within an angle u from the perpendicular to the slit, where sin u = l>a (the first diffraction minimum). You can see this easily in Fig. 16.9b, which is a photograph of water waves undergoing single-slit diffraction. Note

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16.3 Intensity in the Single-Slit Pattern

also that the peak intensities in Fig. 16.9a decrease rapidly as we go away from the center of the pattern. (Compare Fig. 16.6, which shows a single-slit diffraction pattern for light.) The dark fringes in the pattern are the places where I = 0. These occur at points for which the numerator of Eq. (16.5) is zero so that b is a multiple of 2p. From Eq. (16.6) this corresponds to a sin u = m l

1m = 61, 62, Á 2

ml a

1m = 61, 62, Á 2

sin u =

(16.8)

16.9 (a) Intensity versus angle in singleslit diffraction. The values of m label intensity minima given by Eq. (16.8). Most of the wave power goes into the central intensity peak (between the m = 1 and m = - 1 intensity minima). (b) These water waves passing through a small aperture behave exactly like light waves in single-slit diffraction. Only the diffracted waves within the central intensity peak are visible; the waves at larger angles are too faint to see. (a)

u

I 5 0.0083I0

This agrees with our previous result, Eq. (16.2). Note again that b = 0 (corresponding to u = 0) is not a minimum. Equation (16.5) is indeterminate at b = 0, but we can evaluate the limit as b S 0 using L’Hôpital’s rule. We find that at b = 0, I = I0 , as we should expect.

m53 I 5 0.0165I0 m52 I 5 0.0472I0 m51

'

u

Intensity Maxima in the Single-Slit Pattern

1m = 0, 1, 2, Á 2

Im L

I0

A m + 12 B 2p2

m 5 21 m 5 22 m 5 23

(16.9)

This is approximately correct, but because of the factor 1b>222 in the denominator of Eq. (16.5), the maxima don’t occur precisely at these points. When we take the derivative of Eq. (16.5) with respect to b and set it equal to zero to try to find the maxima and minima, we get a transcendental equation that has to be solved numerically. In fact there is no maximum near b = 6p. The first maxima on either side of the central maximum, near b = 63p, actually occur at 62.860p. The second side maxima, near b = 65p, are actually at 64.918p, and so on. The error in Eq. (16.9) vanishes in the limit of large m—that is, for intensity maxima far from the center of the pattern. To find the intensities at the side maxima, we substitute these values of b back into Eq. (16.5). Using the approximate expression in Eq. (16.9), we get

(b)

(16.10)

where Im is the intensity of the mth side maximum and I0 is the intensity of the central maximum. Equation (16.10) gives the series of intensities 0.0450I0

0.0162I0

0.0083I0

and so on. As we have pointed out, this equation is only approximately correct. The actual intensities of the side maxima turn out to be 0.0472I0

0.0165I0

0.0083I0

O

I 5 I0

We can also use Eq. (16.5) to calculate the positions of the peaks, or intensity maxima, and the intensities at these peaks. This is not quite as simple as it may appear. We might expect the peaks to occur where the sine function reaches the value 61 —namely, where b = 6 p, 63p, 65p, or in general, b L 612m + 12p

531

Á

Note that the intensities of the side maxima decrease very rapidly, as Fig. 16.9a also shows. Even the first side maxima have less than 5% of the intensity of the central maximum.

Width of the Single-Slit Pattern For small angles the angular spread of the diffraction pattern is inversely proportional to the slit width a or, more precisely, to the ratio of a to the wavelength l. Figure 16.10 shows graphs of intensity I as a function of the angle u for three values of the ratio a>l.

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532

CHAPTER 16 Diffraction

16.10 The single-slit diffraction pattern depends on the ratio of the slit width a to the wavelength l. (a) a 5 l

(b) a 5 5l

If the slit width is equal to or narrower than the wavelength, only one broad maximum forms. I I0

220°

210°

0°

10°

20°

u

(c) a 5 8l

I

The wider the slit (or the shorter the wavelength), the narrower and sharper is the central peak.

I0

220°

210°

0°

I I0

10°

20°

u

220°

210°

0°

10°

20°

u

With light waves, the wavelength l is often much smaller than the slit width a, and the values of u in Eqs. (16.6) and (16.7) are so small that the approximation sin u = u is very good. With this approximation the position u1 of the first minimum beside the central maximum, corresponding to b>2 = p, is, from Eq. (16.7), u1 =

16.11 The sound waves used in speech have a long wavelength (about 1 m) and can easily bend around this instructor’s head. By contrast, light waves have very short wavelengths and undergo very little diffraction. Hence you can’t see around his head!

Example 16.2

l a

(16.11)

This characterizes the width (angular spread) of the central maximum, and we see that it is inversely proportional to the slit width a. When the small-angle approximation is valid, the central maximum is exactly twice as wide as each side maximum. When a is of the order of a centimeter or more, u1 is so small that we can consider practically all the light to be concentrated at the geometrical focus. But when a is less than l, the central maximum spreads over 180°, and the fringe pattern is not seen at all. It’s important to keep in mind that diffraction occurs for all kinds of waves, not just light. Sound waves undergo diffraction when they pass through a slit or aperture such as an ordinary doorway. The sound waves used in speech have wavelengths of about a meter or greater, and a typical doorway is less than 1 m wide; in this situation, a is less than l, and the central intensity maximum extends over 180°. This is why the sounds coming through an open doorway can easily be heard by an eavesdropper hiding out of sight around the corner. In the same way, sound waves can bend around the head of an instructor who faces the blackboard while lecturing (Fig. 16.11). By contrast, there is essentially no diffraction of visible light through a doorway because the width a is very much greater than the wavelength l 1of order 5 * 10-7 m2. You can hear around corners because typical sound waves have relatively long wavelengths; you cannot see around corners because the wavelength of visible light is very short.

Single-slit diffraction: Intensity I

(a) The intensity at the center of a single-slit diffraction pattern is I 0. What is the intensity at a point in the pattern where there is a 66-radian phase difference between wavelets from the two edges of the slit? (b) If this point is 7.0° away from the central maximum, how many wavelengths wide is the slit? SOLUTION

variable is a>l. We are given the angular position u of the point where b = 66 rad, so we can use Eq. (16.6) to solve for a>l. EXECUTE: (a) We have b>2 = 33 rad, so from Eq. (16.5), I = I0 c

sin133 rad2 2 d = 19.2 * 10-42I0 33 rad

(b) From Eq. (16.6),

IDENTIFY and SET UP: In our analysis of Fig. 16.8 we used the symbol b for the phase difference between wavelets from the two edges of the slit. In part (a) we use Eq. (16.5) to find the intensity I at the point in the pattern where b = 66 rad. In part (b) we need to find the slit width a as a multiple of the wavelength l so our target

b 66 rad a = = = 86 l 2p sin u 12p rad2sin 7.0°

For example, for 550-nm light the slit width is a = 18621550 nm2 = 1 4.7 * 10-5 m 5 0.047 mm, or roughly 20 mm.

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16.4 Circular Apertures and Resolving Power EVALUATE: To what point in the diffraction pattern does this value of b correspond? To find out, note that b = 66 rad is approximately equal to 21p. This is an odd multiple of p, corresponding to the form 12m + 12p found in Eq. (16.9) for the intensity

Example 16.3

maxima. Hence b = 66 rad corresponds to a point near the tenth 1m = 102 maximum. This is well beyond the range shown in Fig. 16.9a, which shows only maxima out to m = 63.

Single-slit diffraction: Intensity II

In the experiment described in Example 16.1 (Section 16.2), the intensity at the center of the pattern is I0. What is the intensity at a point on the screen 3.0 mm from the center of the pattern?

p12.4 * 10-4 m215.0 * 10-42 pa sin u = 0.60 = l 6.33 * 10-7 m I = I0 a

S O L U T IO N IDENTIFY and SET UP: This is similar to Example 16.2, except that we are not given the value of the phase difference b at the point in question. We use geometry to determine the angle u for our point and then use Eq. (16.7) to find the intensity I (the target variable). EXECUTE: Referring to Fig. 16.5a, we have y = 3.0 mm and x = 6.0 m, so tan u = y>x = 13.0 * 10-3 m2>16.0 m2 = 5.0 * 10-4. This is so small that the values of tan u, sin u, and u (in radians) are all nearly the same. Then, using Eq. (16.7),

sin 0.60 2 b = 0.89I0 0.60

EVALUATE: Figure 16.9a shows that an intensity this high can occur only within the central intensity maximum. This checks out; from Example 16.1, the first intensity minimum (m = 1 in Fig. 16.9a) is (32 mm2>2 = 16 mm from the center of the pattern, so the point in question here at y = 3 mm does, indeed, lie within the central maximum.

Test Your Understanding of Section 16.3 Coherent electromagnetic radiation is sent through a slit of width 0.0100 mm. For which of the following wavelengths will there be no points in the diffraction pattern where the intensity is zero? (i) blue light of wavelength 500 nm; (ii) infrared light of wavelength 10.6 mm; (iii) microwaves of wavelength 1.00 mm; (iv) ultraviolet light of wavelength 50.0 nm.

16.4

533

y

Circular Apertures and Resolving Power

We have studied in detail the diffraction patterns formed by long, thin slits or arrays of slits. But an aperture of any shape forms a diffraction pattern. The diffraction pattern formed by a circular aperture is of special interest because of its role in limiting how well an optical instrument can resolve fine details. In principle, we could compute the intensity at any point P in the diffraction pattern by dividing the area of the aperture into small elements, finding the resulting wave amplitude and phase at P, and then integrating over the aperture area to find the resultant amplitude and intensity at P. In practice, the integration cannot be carried out in terms of elementary functions. We will simply describe the pattern and quote a few relevant numbers. The diffraction pattern formed by a circular aperture consists of a central bright spot surrounded by a series of bright and dark rings, as Fig. 16.12 shows.

u1 is the angle between the center of the pattern and the first minimum. Airy disk

ActivPhysics 16.7: Circular Hole Diffraction ActivPhysics 16.8: Resolving Power

16.12 Diffraction pattern formed by a circular aperture of diameter D. The pattern consists of a central bright spot and alternating dark and bright rings. The angular radius u1 of the first dark ring is shown. (This diagram is not drawn to scale.)

u1 D

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534

CHAPTER 16 Diffraction

16.13 Photograph of the diffraction pattern formed by a circular aperture.

We can describe the pattern in terms of the angle u, representing the angular radius of each ring. If the aperture diameter is D and the wavelength is l, the angular radius u1 of the first dark ring is given by

Airy disk

sin u1 = 1.22

l D

(diffraction by a circular aperture)

(16.12)

The angular radii of the next two dark rings are given by sin u2 = 2.23

l D

sin u3 = 3.24

l D

(16.13)

Between these are bright rings with angular radii given by sin u = 1.63

16.14 Diffraction patterns of four very small (“point”) sources of light. The photographs were made with a circular aperture in front of the lens. (a) The aperture is so small that the patterns of sources 3 and 4 overlap and are barely resolved by Rayleigh’s criterion. Increasing the size of the aperture decreases the size of the diffraction patterns, as shown in (b) and (c). (a) Small aperture

1

2

3

4

2

3

4

2

3

4

(b) Medium aperture

1

(c) Large aperture

1

l , D

2.68

l , D

3.70

l D

(16.14)

and so on. The central bright spot is called the Airy disk, in honor of Sir George Airy (1801–1892), Astronomer Royal of England, who first derived the expression for the intensity in the pattern. The angular radius of the Airy disk is that of the first dark ring, given by Eq. (16.12). The intensities in the bright rings drop off very quickly with increasing angle. When D is much larger than the wavelength l, the usual case for optical instruments, the peak intensity in the first ring is only 1.7% of the value at the center of the Airy disk, and the peak intensity of the second ring is only 0.4%. Most (85%) of the light energy falls within the Airy disk. Figure 16.13 shows a diffraction pattern from a circular aperture.

Diffraction and Image Formation Diffraction has far-reaching implications for image formation by lenses and mirrors. In our study of optical instruments in Chapter 34 we assumed that a lens with focal length ƒ focuses a parallel beam (plane wave) to a point at a distance ƒ from the lens. This assumption ignored diffraction effects. We now see that what we get is not a point but the diffraction pattern just described. If we have two point objects, their images are not two points but two diffraction patterns. When the objects are close together, their diffraction patterns overlap; if they are close enough, their patterns overlap almost completely and cannot be distinguished. The effect is shown in Fig. 16.14, which presents the patterns for four very small “point” sources of light. In Fig. 16.14a the image of the left-hand source is well separated from the others, but the images of the middle and right-hand sources have merged. In Fig. 16.14b, with a larger aperture diameter and hence smaller Airy disks, the middle and right-hand images are better resolved. In Fig. 16.14c, with a still larger aperture, they are well resolved. A widely used criterion for resolution of two point objects, proposed by the English physicist Lord Rayleigh (1842–1919) and called Rayleigh’s criterion, is that the objects are just barely resolved (that is, distinguishable) if the center of one diffraction pattern coincides with the first minimum of the other. In that case the angular separation of the image centers is given by Eq. (16.12). The angular separation of the objects is the same as that of the images made by a telescope, microscope, or other optical device. So two point objects are barely resolved, according to Rayleigh’s criterion, when their angular separation is given by Eq. (16.12). The minimum separation of two objects that can just be resolved by an optical instrument is called the limit of resolution of the instrument. The smaller the limit of resolution, the greater the resolution, or resolving power, of the instrument. Diffraction sets the ultimate limits on resolution of lenses. Geometric optics may make it seem that we can make images as large as we like. Eventually, though, we always reach a point at which the image becomes larger but does not

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16.4 Circular Apertures and Resolving Power

535

gain in detail. The images in Fig. 16.14 would not become sharper with further enlargement. CAUTION Resolving power vs. chromatic resolving power Be careful not to confuse the resolving power of an optical instrument with the chromatic resolving power of a grating (described in Section 16.5). Resolving power refers to the ability to distinguish the images of objects that appear close to each other, when looking either through an optical instrument or at a photograph made with the instrument. Chromatic resolving power describes how well different wavelengths can be distinguished in a spectrum formed by a diffraction grating. y

?

Rayleigh’s criterion combined with Eq. (16.12) shows that resolution (resolving power) improves with larger diameter; it also improves with shorter wavelengths. Ultraviolet microscopes have higher resolution than visiblelight microscopes. In electron microscopes the resolution is limited by the wavelengths associated with the electrons, which have wavelike aspects (to be discussed further in Chapter 39). These wavelengths can be made 100,000 times smaller than wavelengths of visible light, with a corresponding gain in resolution. Resolving power also explains the difference in storage capacity between DVDs (introduced in 1995) and Blu-ray discs (introduced in 2003). Information is stored in both of these in a series of tiny pits. In order not to lose information in the scanning process, the scanning optics must be able to resolve two adjacent pits so that they do not seem to blend into a single pit (see sources 3 and 4 in Fig. 16.14). The blue scanning laser used in a Blu-ray player has a shorter wavelength (405 nm) and hence better resolving power than the 650-nm red laser in a DVD player. Hence pits can be spaced closer together in a Blu-ray disc than in a DVD, and more information can be stored on a disc of the same size (50 gigabytes on a Blu-ray disc versus 4.7 gigabytes on a DVD).

Example 16.4

Application Bigger Telescope, Better Resolution One reason for building very large telescopes is to increase the aperture diameter and thus minimize diffraction effects. The effective diameter of a telescope can be increased by using arrays of smaller telescopes. The Very Large Array (VLA) in New Mexico is a collection of 27 radio telescopes, each 25 m in diameter, that can be spread out in a Y-shaped arrangement 36 km across. Hence the effective aperture diameter is 36 km, giving the VLA a limit of resolution of 5 * 10-8 rad at a radio wavelength of 1.5 cm. If your eye had this angular resolution, you could read the “20> 20” line on an eye chart more than 30 km away!

Resolving power of a camera lens

A camera lens with focal length ƒ = 50 mm and maximum aperture ƒ>2 forms an image of an object 9.0 m away. (a) If the resolution is limited by diffraction, what is the minimum distance between two points on the object that are barely resolved? What is the corresponding distance between image points? (b) How does the situation change if the lens is “stopped down” to ƒ>16? Use l = 500 nm in both cases. S O L U T IO N IDENTIFY and SET UP: This example uses the ideas about resolving power, image formation by a lens (Section 34.4), and ƒ-number (Section 34.5). From Eq. (34.20), the ƒ-number of a lens is its focal length ƒ divided by the aperture diameter D. We use this equation to determine D and then use Eq. (16.12) (the Rayleigh criterion) to find the angular separation u between two barely resolved points on the object. We then use the geometry of image formation by a lens to determine the distance y between those points and the distance y¿ between the corresponding image points. EXECUTE: (a) The aperture diameter is D = ƒ>1ƒ-number2 = 150 mm2>2 = 25 mm = 25 * 10-3 m. From Eq. (16.12) the angular separation u of two object points that are barely resolved is u L sin u = 1.22

500 * 10-9 m l = 1.22 = 2.4 * 10-5 rad D 25 * 10-3 m

We know from our thin-lens analysis in Section 34.4 that, apart from sign, y>s = y¿>s¿ [see Eq. (34.14)]. Thus the angular separations of the object points and the corresponding image points are both equal to u. Because the object distance s is much greater than the focal length ƒ = 50 mm, the image distance s¿ is approximately equal to ƒ. Thus y = 2.4 * 10-5 9.0 m y¿ = 2.4 * 10-5 50 mm

y = 2.2 * 10-4 m = 0.22 mm y¿ = 1.2 * 10-3 mm = 0.0012 mm L

1 800

mm

(b) The aperture diameter is now 150 mm2>16, or one-eighth as large as before. The angular separation between barely resolved points is eight times as great, and the values of y and y¿ are also eight times as great as before: y = 1.8 mm

y¿ = 0.0096 mm =

1 100

mm

Only the best camera lenses can approach this resolving power. EVALUATE: Many photographers use the smallest possible aperture for maximum sharpness, since lens aberrations cause light rays that are far from the optic axis to converge to a different image point than do rays near the axis. But as this example shows, diffraction effects become more significant at small apertures. One cause of fuzzy images has to be balanced against another.

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536

CHAPTER 16 Diffraction Test Your Understanding of Section 16.4 You have been asked to compare four different proposals for telescopes to be placed in orbit, above the blurring effects of the earth’s atmosphere. Rank the proposed telescopes in order of their ability to resolve small details, from best to worst. (i) a radio telescope 100 m in diameter observing at a wavelength of 21 cm; (ii) an optical telescope 2.0 m in diameter observing at a wavelength of 500 nm; (iii) an ultraviolet telescope 1.0 m in diameter observing at a wavelength of 100 nm; (iv) an infrared telescope 2.0 m in diameter observing at a wavelength of 10 mm. y

16.5

Holography

Holography is a technique for recording and reproducing an image of an object through the use of interference effects. Unlike the two-dimensional images recorded by an ordinary photograph or television system, a holographic image is truly three-dimensional. Such an image can be viewed from different directions to reveal different sides and from various distances to reveal changing perspective. If you had never seen a hologram, you wouldn’t believe it was possible! Figure 16.15a shows the basic procedure for making a hologram. We illuminate the object to be holographed with monochromatic light, and we place a photographic film so that it is struck by scattered light from the object and also by direct light from the source. In practice, the light source must be a laser, for reasons we will discuss later. Interference between the direct and scattered light leads to the formation and recording of a complex interference pattern on the film. To form the images, we simply project light through the developed film (Fig. 16.15b). Two images are formed: a virtual image on the side of the film nearer the source and a real image on the opposite side.

Holography and Interference Patterns A complete analysis of holography is beyond our scope, but we can gain some insight into the process by looking at how a single point is holographed and imaged. Consider the interference pattern that is formed on a sheet of photographic negative film by the superposition of an incident plane wave and a

16.15 (a) A hologram is the record on film of the interference pattern formed with light from the coherent source and light scattered from the object. (b) Images are formed when light is projected through the hologram. The observer sees the virtual image formed behind the hologram. (a) Recording a hologram

Laser

(b) Viewing the hologram Beam splitter

Photographic film

Mirror

Observer

Object beam

Real image

Reference beam

Mirror

Transparent positive film

Mirror

Object

Mirror

Reconstruction beam

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Virtual image

16.5 Holography

537

16.16 (a) Constructive interference of the plane and spherical waves occurs in the plane of the film at every point Q for which the distance bm from P is greater than the distance b 0 from P to O by an integral number of wavelengths ml. For the point Q shown, m = 2. (b) When a plane wave strikes a transparent positive print of the developed film, the diffracted wave consists of a wave converging to P¿ and then diverging again and a diverging wave that appears to originate at P. These waves form the real and virtual images, respectively. ''

(b)

(a)

Plane wave Plane wave

Spherical wave Q

Q bm

l

rm b0

P

O

P

O

Transparent positive print

Film

spherical wave, as shown in Fig. 16.16a. The spherical wave originates at a point source P at a distance b 0 from the film; P may in fact be a small object that scatters part of the incident plane wave. We assume that the two waves are monochromatic and coherent and that the phase relationship is such that constructive interference occurs at point O on the diagram. Then constructive interference will also occur at any point Q on the film that is farther from P than O is by an integer number of wavelengths. That is, if bm - b 0 = ml, where m is an integer, then constructive interference occurs. The points where this condition is satisfied form circles on the film centered at O, with radii rm given by '

'

bm - b 0 = 2b02 + rm2 - b 0 = ml ''

''

1m = 1, 2, 3, Á 2

(16.15)

Solving this for rm2, we find rm2 = l12mb 0 + m2l2 '

Ordinarily, b 0 is very much larger than l, so we neglect the second term in parentheses and obtain '

rm = 22mlb 0 '

1m = 1, 2, 3, Á 2

(16.16)

The interference pattern consists of a series of concentric bright circular fringes with radii given by Eq. (16.16). Between these bright fringes are dark fringes. Now we develop the film and make a transparent positive print, so the brightfringe areas have the greatest transparency on the film. Then we illuminate it with monochromatic plane-wave light of the same wavelength l that we used initially. In Fig. 16.16b, consider a point P¿ at a distance b 0 along the axis from the film. The centers of successive bright fringes differ in their distances from P¿ by an integer number of wavelengths, and therefore a strong maximum in the diffracted wave occurs at P¿. That is, light converges to P¿ and then diverges from it on the opposite side. Therefore P¿ is a real image of point P. This is not the entire diffracted wave, however. The interference of the wavelets that spread out from all the transparent areas forms a second spherical wave that is diverging rather than converging. When this wave is traced back behind the film in Fig. 16.16b, it appears to be spreading out from point P. Thus the total diffracted wave from the hologram is a superposition of a spherical wave converging to form a real image at P¿ and a spherical wave that diverges as though it had come from the virtual image point P. '

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P9

538

CHAPTER 16 Diffraction

16.17 Two views of the same hologram seen from different angles.

Because of the principle of superposition for waves, what is true for the imaging of a single point is also true for the imaging of any number of points. The film records the superposed interference pattern from the various points, and when light is projected through the film, the various image points are reproduced simultaneously. Thus the images of an extended object can be recorded and reproduced just as for a single point object. Figure 16.17 shows photographs of a holographic image from two different angles, showing the changing perspective in this three-dimensional image. In making a hologram, we have to overcome two practical problems. First, the light used must be coherent over distances that are large in comparison to the dimensions of the object and its distance from the film. Ordinary light sources do not satisfy this requirement, for reasons that we discussed in Section 35.1. Therefore laser light is essential for making a hologram. (Ordinary white light can be used for viewing certain types of hologram, such as those used on credit cards.) Second, extreme mechanical stability is needed. If any relative motion of source, object, or film occurs during exposure, even by as much as a quarter of a wavelength, the interference pattern on the film is blurred enough to prevent satisfactory image formation. These obstacles are not insurmountable, however, and holography has become important in research, entertainment, and a wide variety of technological applications.

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CHAPTER

16

SUMMARY

Fresnel and Fraunhofer diffraction: Diffraction occurs when light passes through an aperture or around an edge. When the source and the observer are so far away from the obstructing surface that the outgoing rays can be considered parallel, it is called Fraunhofer diffraction. When the source or the observer is relatively close to the obstructing surface, it is Fresnel diffraction.

Fresnel (near-field) diffraction

Fraunhofer (farfield) diffraction

P

Single-slit diffraction: Monochromatic light sent through a narrow slit of width a produces a diffraction pattern on a distant screen. Equation (16.2) gives the condition for destructive interference (a dark fringe) at a point P in the pattern at angle u. Equation (16.7) gives the intensity in the pattern as a function of u. (See Examples 16.1–16.3.)

Circular apertures and resolving power: The diffraction pattern from a circular aperture of diameter D consists of a central bright spot, called the Airy disk, and a series of concentric dark and bright rings. Equation (16.12) gives the angular radius u1 of the first dark ring, equal to the angular size of the Airy disk. Diffraction sets the ultimate limit on resolution (image sharpness) of optical instruments. According to Rayleigh’s criterion, two point objects are just barely resolved when their angular separation u is given by Eq. (16.12). (See Example 16.4.)

sin u =

ml a

1m = 61, 62, Á 2

I 5 0.0083I0 I 5 0.0165I0 I 5 0.0472I0

(16.2)

I = I0 b

sin3pa1sin u2>l4

'

pa1sin u2>l

sin u1 = 1.22

l D

r

I 5 I0

2

(16.7)

u

u m5 m5 m5 O m5 m5 m5

3 2 1 21 22 23

(16.12)

Airy disk

539

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540

CHAPTER 16 Diffraction

BRIDGING PROBLEM

Observing the Expanding Universe

An astronomer who is studying the light from a galaxy has identified the spectrum of hydrogen but finds that the wavelengths are somewhat shifted from those found in the laboratory. In the lab, the H a line in the hydrogen spectrum has a wavelength of 656.3 nm. The astronomer is using a transmission diffraction grating having 5758 lines>cm in the first order and finds that the first bright fringe for the H a line occurs at 623.41° from the central spot. How fast is the galaxy moving? Express your answer in m> s and as a percentage of the speed of light. Is the galaxy moving toward us or away from us? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. You can use the information about the grating to find the wavelength of the Ha line in the galaxy’s spectrum. 2. In Section 16.8 we learned about the Doppler effect for electromagnetic radiation: The frequency that we receive from a mov-

Problems

ing source, such as the galaxy, is different from the frequency that is emitted. Equation (16.30) relates the emitted frequency, the received frequency, and the velocity of the source (the target variable). The equation c = ƒl relates the frequency ƒ and wavelength l through the speed of light c. EXECUTE 3. Find the wavelength of the Ha spectral line in the received light. 4. Rewrite Eq. (16.30) as a formula for the velocity v of the galaxy in terms of the received wavelength and the wavelength emitted by the source. 5. Solve for v. Express it in m> s and as a percentage of c, and decide whether the galaxy is moving toward us or moving away. EVALUATE 6. Is your answer consistent with the relative sizes of the received wavelength and the emitted wavelength?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q16.1 Why can we readily observe diffraction effects for sound waves and water waves, but not for light? Is this because light travels so much faster than these other waves? Explain. Q16.2 What is the difference between Fresnel and Fraunhofer diffraction? Are they different physical processes? Explain. Q16.3 You use a lens of diameter D and light of wavelength l and frequency ƒ to form an image of two closely spaced and distant objects. Which of the following will increase the resolving power? (a) Use a lens with a smaller diameter; (b) use light of higher frequency; (c) use light of longer wavelength. In each case justify your answer. Q16.4 Light of wavelength l and frequency ƒ passes through a single slit of width a. The diffraction pattern is observed on a screen a distance x from the slit. Which of the following will decrease the width of the central maximum? (a) Decrease the slit width; (b) decrease the frequency ƒ of the light; (c) decrease the wavelength l of the light; (d) decrease the distance x of the screen from the slit. In each case justify your answer. Q16.5 In a diffraction experiment with waves of wavelength l, there will be no intensity minima (that is, no dark fringes) if the slit width is small enough. What is the maximum slit width for which this occurs? Explain your answer. Q16.6 The predominant sound waves used in human speech have wavelengths in the range from 1.0 to 3.0 meters. Using the ideas of diffraction, explain how it is possible to hear a person’s voice even when he is facing away from you. Q16.7 In single-slit diffraction, what is sin1b>22 when u = 0? In view of your answer, why is the single-slit intensity not equal to zero at the center?

Q16.8 Some loudspeaker horns for outdoor concerts (at which the entire audience is seated on the ground) are wider vertically than horizontally. Use diffraction ideas to explain why this is more efficient at spreading the sound uniformly over the audience than either a square speaker horn or a horn that is wider horizontally than vertically. Would this still be the case if the audience were seated at different elevations, as in an amphitheater? Why or why not? Q16.9 Figure 31.12 (Section 31.2) shows a loudspeaker system. Low-frequency sounds are produced by the woofer, which is a speaker with large diameter; the tweeter, a speaker with smaller diameter, produces high-frequency sounds. Use diffraction ideas to explain why the tweeter is more effective for distributing high-frequency sounds uniformly over a room than is the woofer. Q16.10 Information is stored on an audio compact disc, CD-ROM, or DVD disc in a series of pits on the disc. These pits are scanned by a laser beam. An important limitation on the amount of information that can be stored on such a disc is the width of the laser beam. Explain why this should be, and explain how using a shorter-wavelength laser allows more information to be stored on a disc of the same size. Q16.11 With which color of light can the Hubble Space Telescope see finer detail in a distant astronomical object: red, blue, or ultraviolet? Explain your answer. Q16.12 If a hologram is made using 600-nm light and then viewed with 500-nm light, how will the images look compared to those observed when viewed with 600-nm light? Explain. Q16.13 A hologram is made using 600-nm light and then viewed by using white light from an incandescent bulb. What will be seen? Explain.

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Problems

Q16.14 Ordinary photographic film reverses black and white, in the sense that the most brightly illuminated areas become blackest upon development (hence the term negative). Suppose a hologram negative is viewed directly, without making a positive transparency. How will the resulting images differ from those obtained with the positive? Explain.

EXERCISES Section 16.2 Diffraction from a Single Slit

16.1 .. Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm. Calculate the wavelength of the light. 16.2 . Parallel rays of green mercury light with a wavelength of 545 nm pass through a slit covering a lens with a focal length of 60.0 cm. In the focal plane of the lens the distance from the central maximum to the first minimum is 10.9 mm. What is the width of the slit? 16.3 .. Light of wavelength 585 nm falls on a slit 0.0666 mm wide. On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin u can be? What does this tell you is the largest that m can be?) 16.4 . Light of wavelength 600 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. What is the distance between the two dark fringes on either side of the central bright fringe? 16.5 . Monochromatic electromagnetic radiation with wavelength l from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width a if the wavelength is (a) 500 nm (visible light); (b) 50.0 mm (infrared radiation); (c) 0.500 nm (x rays)? 16.6 .. Parallel rays of light with wavelength 620 nm pass through a slit covering a lens with a focal length of 40.0 cm. The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is 30 cm. What is the width of the slit? (Note: The angle that locates the first minimum is not small.) 16.7 .. Red light of wavelength 630 nm from a helium–neon laser passes through a slit 0.350 mm wide. The diffraction pattern is observed on a screen 3.00 m away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

Section 16.3 Intensity in the Single-Slit Pattern (Optional)

16.8 . Consider a single-slit diffraction experiment in which the amplitude of the wave at point O in Fig. 16.5a is E0. For each of the following cases, draw a phasor diagram like that in Fig. 16.8c and determine graphically the amplitude of the wave at the point in question. (Hint: Use Eq. (16.6) to determine the value of b for each case.) Compute the intensity and compare to Eq. (16.5). (a) sin u = l>2a; (b) sin u = l>a; (c) sin u = 3l>2a.

Section 16.4 Circular Apertures and Resolving Power

16.9 . Two satellites at an altitude of 1400 km are separated by 28 km. If they broadcast 3.6-cm microwaves, what minimum receiv-

541

ing-dish diameter is needed to resolve (by Rayleigh’s criterion) the two transmissions? 16.10 .. The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter of about 8000 km. When this radio telescope is focusing radio waves of wavelength 2.0 cm, what would have to be the diameter of the mirror of a visible-light telescope focusing light of wavelength 550 nm so that the visible-light telescope has the same resolution as the radio telescope? 16.11 . Photography. A wildlife photographer uses a moderate telephoto lens of focal length 135 mm and maximum aperture ƒ>4.00 to photograph a bear that is 11.5 m away. Assume the wavelength is 550 nm. (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to ƒ>22.0, what would be the width of the smallest resolvable feature on the bear? 16.12 . Observing Jupiter. You are asked to design a space telescope for earth orbit. When Jupiter is 6 * 108 km away, the telescope is to resolve, by Rayleigh’s criterion, features on Jupiter that are 240 km apart. What minimum-diameter mirror is required? Assume a wavelength of 500 nm. 16.13 . A converging lens 7.20 cm in diameter has a focal length of 300 mm. If the resolution is diffraction limited, how far away can an object be if points on it 4.00 mm apart are to be resolved (according to Rayleigh’s criterion)? Use l = 550 nm.

PROBLEMS 16.14 ... Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 mm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 W>m2. Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. 16.15 .. A slit 0.360 mm wide is illuminated by parallel rays of light that have a wavelength of 540 nm. The diffraction pattern is observed on a screen that is 1.20 m from the slit. The intensity at the center of the central maximum (u = 0°) is I0. (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) (Optional) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to I0>2? 16.16 (Optional) ... CALC The intensity of light in the Fraunhofer diffraction pattern of a single slit is I = I0 a

sin g 2 b g

where g =

pa sin u l

Show that the equation for the values of g at which I is a maximum is tan g = g. 1617 .. Underwater Photography. An underwater camera has a lens of focal length 35.0 mm and a maximum aperture of ƒ>2.80. The film it uses has an emulsion that is sensitive to light of frequency 6.00 * 1014 Hz. If the photographer takes a picture of an object 2.75 m in front of the camera with the lens wide open, what is the width of the smallest resolvable detail on the subject if the object is (a) a fish underwater with the camera in the water and (b) a person on the beach with the camera out of the water?

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542

CHAPTER 16 Diffraction

Screen

16.18 .. An astronaut in the space shuttle can just resolve two point sources on earth that are 65.0 m apart. Assume that the resolution is diffraction limited and use Rayleigh’s criterion. What is the astronaut’s altitude above the earth? Treat his eye as a circular aperture with a diameter of 4.00 mm (the diameter of his pupil), and take the wavelength of the light to be 550 nm. Ignore the effect of fluid in the eye. 16.19 .. BIO Resolution of the Eye. The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0 mm in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 mm . (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50-mm-tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells? 16.20 A hair is placed at one edge between two flat galss plates. When this arrangement is illuminated with yellow light of wavelength l = 600 nm, a total of 121 dark bands are counted starting at the contact point between the plates, and ending at the hair. How thick is the hair? 16.21 Waves from a radio station with wavelength of 600 m arrive at a home receiver a distance 50 km away from the transmitter by two paths. One is a direct line path and the second by reflection from a mountain directly behind the receiver. What is the minimum distance between the mountain and receiver such that destructive interference occurs at the location of the listener? 16.22 In a YDSE apparatus, two identical slits are separated by 1 mm and distance between slits and screen is 1 m. The wavelength of light used is 6000 Å. Find the minimum distance between two points on the screen having 75% intensity of the maximum intensity. 16.23 In a standard YDSE apparatus a thin film (µ = 1.5, t = 2.1 µm) is placed in front of upper slit. How far above & below the centre point of the screen are two nearest maxima located? Take D= 1 m, d =1 mm, l=4500 Å. (Symbols have usual meaning) 16.24 In the Young’s experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a micasheet of µ = 1.6 and thickness t = 1.964 micron is introduced in the path of one of the slits. The mica sheet is now removed and the distance between the slits and screen is doubled. It is found that the distance between successive maxima (or minima) is the same as the observed fringe shift upon the introduction of mica. Calculate l. 16.25 A narrow slit S allows monochromatic light of wavelength l =4000 Å to fall on a prism of very small angle as shown in the fig- S ure. A screen is placed at a distance l = 150 cm l = 150 cm from the source to obtain an interference pattern. To determine the distance between the virtual images formed by the prism an experiment is done. The prism and screen are kept fixed and a converging lens is moved between the prism and the screen. For two positions of the lens (between the prism and the screen) we get two sharp point images on the screen in each case. The images are separated from each other by distance

4.5 mm in one case and 2 mm in the other. Now the lens is removed and interference pattern is obtained on the screen. Calculate: (i) the fringe width of the pattern on the screen. (ii) focal length of the lens. 16.26 Two coherent point sources S1 and S2 are located at a distance of 4.6 l from each O other. Perpendicular to the straight line join- S1 S2 ing S1 and S2, there is a large screen at a disScreen tance of 1000 l from S1 as shown. Find the radius of the third maxima starting from the point O. 16.27 In Lloyd’s mirror experiment, the source slit S and its virtual image S' lie in a plane 20 cm to the left from the left edge of the mirror. The mirror is 30 cm long and a screen is placed touching the right edge of the mirror. The distance of S from plane of mirror is 2 mm and wavelength of light used = 6400 Å. Find the number of maxima formed. 16.28 Two radio transmitter each transmitting a radio of frequency 12 MHz in phase with each other are placed at a disl = 100 m tance l = 100 m on either side of the q runway on the towers as shown in the figure. An aircraft with a ground speed of V is flying towards the airport such that its velocity makes an angle u with the runway. The aircraft is heading directly for the mid-point of the towers. (a) At the aircraft’s current position, the intensity of the signal from each tower separately would be I0. Find the intensity of the combined signal from both towers received by the aircraft for heading of u = 0 and u =p/2. (b) For what heading would the aircraft receive no signal. Assuming distance of plane much greater than l. (c) If the aircraft is heading from large distance directly towards one of the transmitter in the direction perpendicular to ‘l’find the position of the aircraft from the nearest tower when the signal dropped to a minimum first time. 16.29 In Young’s double slit experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 and thickness 2 micron. While the other slit is covered by another glass plate of thickness 4 micron and refractive index 1.6. Interference pattern is observed using light of wavelength 4800 Å. The intensity of the central maximum point P is I0 without the plates in the experiment. Find the intensity of the point P after the insertion of plates in the path of the wavefronts. 16.30 A convex lens of focal length 50 cm is cut along the diameter into two identical halves A and B and in the process a layer C of the lens thickness 1 mm is lost. Then the two halves A and B are put together to form a composite lens. Now infront of this composite lens a source of light emitting wavelength l = 6000 Å is placed at a distance of 25 cm as shown in the figure. Behind the lens there is a screen at a distance 50 cm from it. Find the fringe width of the interference pattern obtained on the screen. Screen A C

A

Source 1 mm *

B

B 25 cm

50 cm

-3

16.31 A glass sheet 12 * 10 mm thick is placed in the path of one of the interfering beams in a Young’s double slit interference arrangement using monochromatic light of wavelength 6000 Å. If the central bright fringe shifts a distance equal to width of 10

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Answers

bands, find the refractive index of glass. What is the thickness of the sheet of diamond of refractive index 2.5 that has to be introduced in the path of second beam to bring the central bright fringe to original position?

CHALLENGE PROBLEMS 16.32 ... CALC It is possible to calculate the intensity in the single-slit Fraunhofer diffraction pattern without using the phasor method of Section 16.3. Let y¿ represent the position of a point within the slit of width a in Fig. 16.5a, with y¿ = 0 at the center of the slit so that the slit extends from y¿ = - a>2 to y¿ = a>2. We imagine dividing the slit up into infinitesimal strips of width dy¿, each of which acts as a source of secondary wavelets. (a) The amplitude of the total wave at the point O on the distant screen in Fig. 16.5a is E0. Explain why the amplitude of the wavelet from each infinitesimal strip within the slit is E01dy¿>a2, so that the electric field of the wavelet a distance x from the infinitesimal strip is dE = E01dy¿>a2 sin1kx - vt2. (b) Explain why the wavelet from each strip as detected at point P in Fig. 16.5a can be expressed as

dE = E0

543

dy¿ sin3k1D - y¿ sin u2 - vt4 a

where D is the distance from the center of the slit to point P and k = 2p>l. (c) By integrating the contributions dE from all parts of the slit, show that the total wave detected at point P is E = E0 sin1kD - vt2

sin3ka1sin u2>24

= E0 sin1kD - vt2

ka1sin u2>2 sin3pa1sin u2>l4

'

pa1sin u2>l

(The trigonometric identities in Appendix B will be useful.) Show that at u = 0, corresponding to point O in Fig. 16.5a, the wave is E = E0 sin1kD - vt2 and has amplitude E0, as stated in part (a). (d) Use the result of part (c) to show that if the intensity at point O is I0, then the intensity at a point P is given by Eq. (16.7). '

Answers Chapter Opening Question

?

The shorter wavelength of a Blu-ray scanning laser gives it superior resolving power, so information can be more tightly packed onto a Blu-ray disc than a DVD. See Section 16.7 for details.

Test Your Understanding Questions 16.1 yes When you hear the voice of someone standing around a corner, you are hearing sound waves that underwent diffraction. If there were no diffraction of sound, you could hear sounds only from objects that were in plain view. 16.2 (ii), (i) and (iv) (tie), (iii) The angle u of the first dark fringe is given by Eq. (16.2) with m = 1, or sin u = l>a. The larger the value of the ratio l>a, the larger the value of sin u and hence the value of u. The ratio l>a in each case is (i) 1400 nm2> 10.20 mm2 = 14.0 * 10 -7 m2>12.0 * 10-4 m2 = 2.0 * 10-3; (ii) -4 (600 nm)/(0.20 mm) = (6.0 * 10-7)/(2.0 * 10 m) = 3.0 * 10-3; (iii) 1400 nm2>10.30 mm2 = 14.0 * 10-7 m2>13.0 *

10-4 m2 = 1.3 * 10-3; (iv) 1600 nm2>10.30 mm2 = 16.0 *

10-7 m2>13.0 * 10-4 m2 = 2.0 * 10-3. 16.3 (ii) and (iii) If the slit width a is less than the wavelength l, there are no points in the diffraction pattern at which the intensity is zero (see Fig. 16.10a). The slit width is 0.0100 mm = 1.00 * 10-5 m, so this condition is satisfied for (ii) 1l = 10.6 mm = 1.06 * 10-5 m2 and (iii) 1l = 1.00 mm = 1.00 * 10-3 m2 but not for (i) 1l = 500 nm = 5.00 * 10-7 m2 or (iv) 1l = 50.0 nm = 5.00 * 10-8 m2. 16.4 yes; mi 5 65, 610, Á A “missing maximum” satisfies both d sin u = mil (the condition for an interference maximum) and a sin u = mdl (the condition for a diffraction minimum). Substituting d = 2.5a, we can combine these two conditions into the relationship mi = 2.5md . This is satisfied for mi = 65 and md = 62 (the fifth interference maximum is missing because it coincides with the second diffraction minimum), mi = 610 and md = 64 (the tenth interference maximum is '

missing because it coincides with the fourth diffraction minimum), and so on. 16.5 (i) As described in the text, the resolving power needed is R = Nm = 1000. In the first order 1m = 12 we need N = 1000 slits, but in the fourth order 1m = 42 we need only N = R>m = 1000>4 = 250 slits. (These numbers are only approximate because of the arbitrary nature of our criterion for resolution and because real gratings always have slight imperfections in the shapes and spacings of the slits.) 16.6 no The angular position of the mth maximum is given by Eq. (16.16), 2d sin u = ml. With d = 0.200 nm, l = 0.100 nm, and m = 5, this gives sin u = ml>2d = 15210.100 nm2>12210.200 nm2 = 1.25. Since the sine function can never be greater than 1, this means that there is no solution to this equation and the m = 5 maximum does not appear. 16.7 (iii), (ii), (iv), (i) Rayleigh’s criterion combined with Eq. (16.17) shows that the smaller the value of the ratio l>D, the better the resolving power of a telescope of diameter D. For the four telescopes, this ratio is equal to (i) 121 cm2>1100 m2 = 10.21 m2>1100 m2 = 2.1 * 10-3; (ii) 1500 nm2>12.0 m2 = 15.0 * 10-7 m2>12.0 m2 = 2.5 * 10-7; (iii) 1100 nm2>11.0 m2 = 11.0 * 10-7 m2>11.0 m2 = 1.0 * 10-7; (iv) 110 mm2>12.0 m2 = 11.0 * 10-5 m2>12.0 m2 = 5.0 * 10-6.

Bridging Problem

Answers: 1.501 * 107 m> s or 5.00% of c; away from us

Discussion Questions 16.1 Wavelength for light is very small compared to sound & waterwaves. 16.2 One is for finite distance while second is for practically infinite separation. Physically both are same 16.3 Using high frequency will increase resolving power as 1 Resolving power r l 2lx 16.4 (a) and (c) will decrease, W = d

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544

CHAPTER 16 Diffraction

16.5 Slit width should be lesss than l. 16.6 Size of our mouth is much less than the wavelength of sound so the wave spreads in all directions. 16.7 It is zero but at this angle both term approaches zero which will gives maximum value. 16.8 If the droplets are small enough to cause significant differaction. The colours of rainbow overlap resulting in a white rainbow. Size of the drop should be comparable to the wavelength of light. 16.9 Diffraction is more effective in the plane where dimension is small. Therefore vertical loudspeaker will ensure more spread in horizontal plane. 16.10 Diffraction is effective when l used is comparable to size of instrumetns. 16.11 Shorter wavelength means shorter width, which means more turns for a given area hence more data. 16.12 Small l ensure more resolving power hence one should use UV. 16.13 (a) In first case all vectors will form a close polygon which means result will be zero. (b) There will be total N possible case. Corresponding to integral 2p multiple of when integer is multiple of N. Then it will be maxN imum. Therefore (N - 1) minimum in between. 16.14 Wavelength should be of order of bond length of crystal parameters. Therefore X-ray is better option.

Exercises Section 16.2 Diffraction from a Single Slit 16.1 506 nm 16.2 3 * 10 - 5 m 16.3 226 16.4 S 5.6 mm 16.5 (a) 4.17 * 10 - 4m (b) 4.2 cm (c) 4.17 * 10 - 7m 16.6 1.033 µm 16.7 (a) 10.8 mm (b) 5.4 mm

Section 16.3 Intensity in the Single-Slit Pattern (optional) 16.8 (a) 0.405 I0, agrees with Eq. (16.5) (b) zero, agrees with Eq. (16.5)

4I0 , agrees with Eq. (16.5) gp2 I0 is the intensity in straight ahead direction where u = 0 (c)

Section 16.4 Circular Abertures and Resolving Power. 16.9 2.2 m 16.10 220 m 16.11 (a) 0.22 mm (b) 1.2 mm 16.12 1.53 m 16.13 429 m

Problems 16.14 78 16.15 (a) 1.8 mm (b) 0.796 mm 16.17 (a) 0.101 mm (b) 0.134 mm 16.18 387 km 16.19 (a) No (b) 85 µm (c) 3.4 * 10 - 4 rad (d) diffraction 16.20 3.6 * 10 - 5 m 16.21 300 m 16.22 0.20 mm 16.23 0.15 mm, 0.3 mm 16.24 5892 Å 16.25 (i) 0.2 mm, (ii) f = 36 cm 16.26 Ë4.29 * 1000 l 16.27 38 16.28 (a) 4I0, 2n + 1 b n = 0, 1, 2, 3 (b) u = sin - 1 a 8 (c) 393.75 m 16.29 I0/4 16.30 0.6 mm 3 16.31 , 4 * 10 - 6 m 2

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PHOTONS: LIGHT WAVES BEHAVING AS PARTICLES

17

LEARNING GOALS By studying this chapter, you will learn: • How experiments involving the photoelectric effect and x rays pointed the way to a radical reinterpretation of the nature of light. • How Einstein’s photon picture of light explains the photoelectric effect.

?

This plastic surgeon is using two light sources: a headlamp that emits a beam of visible light and a handheld laser that emits infrared light. The light from both sources is emitted in the form of packets of energy called photons. For which source are the photons more energetic: the headlamp or the laser?

n Chapter 12 we saw how Maxwell, Hertz, and others established firmly that light is an electromagnetic wave. Interference, diffraction, and polarization, discussed in Chapters 15 and 16, further demonstrate this wave nature of light. When we look more closely at the emission, absorption, and scattering of electromagnetic radiation, however, we discover a completely different aspect of light. We find that the energy of an electromagnetic wave is quantized; it is emitted and absorbed in particle-like packages of definite energy, called photons. The energy of a single photon is proportional to the frequency of the radiation. We’ll find that light and other electromagnetic radiation exhibits wave–particle duality: Light acts sometimes like waves and sometimes like particles. Interference and diffraction demonstrate wave behavior, while emission and absorption of photons demonstrate the particle behavior. This radical reinterpretation of light will lead us in the next chapter to no less radical changes in our views of the nature of matter.

I

17.1

• How experiments with x rays and gamma rays helped confirm the photon picture of light. • How the wave and particle pictures of light complement each other. • How the Heisenberg uncertainty principle imposes fundamental limits on what can be measured.

17.1 The photoelectric effect. Light

Light Absorbed as Photons: The Photoelectric Effect

A phenomenon that gives insight into the nature of light is the photoelectric effect, in which a material emits electrons from its surface when illuminated (Fig. 17.1). To escape from the surface, an electron must absorb enough energy from the incident light to overcome the attraction of positive ions in the material. These attractions constitute a potential-energy barrier; the light supplies the “kick” that enables the electron to escape. The photoelectric effect has a number of applications. Digital cameras and night-vision scopes use it to convert light energy into an electric signal that is

Photoelectric effect: Light absorbed by a surface causes electrons to be ejected.

Electrons

To eject an electron the light must supply enough energy to overcome the forces holding the electron in the material.

545

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546

CHAPTER 17 Photons: Light Waves Behaving as Particles

17.2 (a) A night-vision scope makes use of the photoelectric effect. Photons entering the scope strike a plate, ejecting electrons that pass through a thin disk in which there are millions of tiny channels. The current through each channel is amplified electronically and then directed toward a screen that glows when hit by electrons. (b) The image formed on the screen, which is a combination of these millions of glowing spots, is thousands of times brighter than the naked-eye view. (a)

(b)

reconstructed into an image (Fig. 17.2). On the moon, sunlight striking the surface causes surface dust to eject electrons, leaving the dust particles with a positive charge. The mutual electric repulsion of these charged dust particles causes them to rise above the moon’s surface, a phenomenon that was observed from lunar orbit by the Apollo astronauts.

Threshold Frequency and Stopping Potential In Section 12.1 we explored the wave model of light, which Maxwell formulated two decades before the photoelectric effect was observed. Is the photoelectric effect consistent with this model? Figure 17.3a shows a modern version of one of the experiments that explored this question. Two conducting electrodes are enclosed in an evacuated glass tube and connected by a battery, and the cathode is illuminated. Depending on the potential difference VAC between the two electrodes, electrons emitted by the illuminated cathode (called photoelectrons) may travel across to the anode, producing a photocurrent in the external circuit. (The tube is evacuated to a pressure of 0.01 Pa or less to minimize collisions between the electrons and gas molecules.) The illuminated cathode emits photoelectrons with various kinetic energies. If the electric field points toward the cathode, as in Fig. 17.3a, all the electrons are accelerated toward the anode and contribute to the photocurrent. But by reversing the field and adjusting its strength as in Fig. 17.3b, we can prevent the less energetic electrons from reaching the anode. In fact, we can determine the maximum kinetic energy Kmax of the emitted electrons by making the potential of the anode relative to the cathode, VAC, just negative enough so that the current stops. This occurs for VAC = - V0, where V0 is called the stopping potential. As an electron moves from the cathode to the anode, the potential decreases by V0 and negative work -eV0 is done on the (negatively charged) electron. The most energetic electron leaves the cathode with kinetic energy Kmax = 12 mvmax 2 and has zero kinetic energy at the anode. Using the work–energy theorem, we have Wtot = - eV0 = ¢K = 0 - Kmax Kmax = 12 mv max 2 = eV0

(maximum kinetic energy of photoelectrons)

(17.1)

Hence by measuring the stopping potential V0, we can determine the maximum kinetic energy with which electrons leave the cathode. (We are ignoring any effects due to differences in the materials of the cathode and anode.) In this experiment, how do we expect the photocurrent to depend on the voltage across the electrodes and on the frequency and intensity of the light? Based on Maxwell’s picture of light as an electromagnetic wave, here is what we would expect: PhET: Photoelectric Effect ActivPhysics 17.3: Photoelectric Effect

Wave-Model Prediction 1: We saw in Section 12.4 that the intensity of an electromagnetic wave depends on its amplitude but not on its frequency. So the photoelectric effect should occur for light of any frequency, and the magnitude of the photocurrent should not depend on the frequency of the light. Wave-Model Prediction 2: It takes a certain minimum amount of energy, called the work function, to eject a single electron from a particular surface (see Fig. 17.1). If the light falling on the surface is very faint, some time may elapse before the total energy absorbed by the surface equals the work function. Hence, for faint illumination, we expect a time delay between when we switch on the light and when photoelectrons appear. Wave-Model Prediction 3: Because the energy delivered to the cathode surface depends on the intensity of illumination, we expect the stopping potential to increase with increasing light intensity. Since intensity does not depend on frequency, we further expect that the stopping potential should not depend on the frequency of the light.

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17.1 Light Absorbed as Photons: The Photoelectric Effect

The experimental results proved to be very different from these predictions. Here is what was found in the years between 1877 and 1905: Experimental Result 1: The photocurrent depends on the light frequency. For a given material, monochromatic light with a frequency below a minimum threshold frequency produces no photocurrent, regardless of intensity. For most metals the threshold frequency is in the ultraviolet (corresponding to wavelengths l between 200 and 300 nm), but for other materials like potassium oxide and cesium oxide it is in the visible spectrum (l between 380 and 750 nm).

17.3 An experiment testing whether the photoelectric effect is consistent with the wave model of light. V

(a)

Vacuum tube

Light causes cathode to emit electrons. S E field pushes electrons to anode.

Experimental Result 2: There is no measurable time delay between when the light is turned on and when the cathode emits photoelectrons (assuming the frequency of the light exceeds the threshold frequency). This is true no matter how faint the light is.

–

Cathode

Einstein’s Photon Explanation

v

E

–

i VAC 2e

G

i

E Electrons return to cathode via circuit; galvanometer measures current. (b)

We now reverse the electric field so that it tends to repel electrons from the anode. Above a certain field strength, electrons no longer reach the anode.

S

E

–

(energy of a photon)

Electron trajectory v

v –

'

hc l

Potential of anode relative to cathode

+

?

E = hƒ =

Anode

v

Einstein made the radical postulate that a beam of light consists of small packages of energy called photons or quanta. This postulate was an extension of an idea developed five years earlier by Max Planck to explain the properties of blackbody radiation, which we discussed in Section 17.7. (We’ll explore Planck’s ideas in Section 18.5.) In Einstein’s picture, the energy E of an individual photon is equal to a constant h times the photon frequency ƒ. From the relationship ƒ = c > l for electromagnetic waves in vacuum, we have '

Monochromatic light

S

Experimental Result 3: The stopping potential does not depend on intensity, but does depend on frequency. Figure 17.4 shows graphs of photocurrent as a function of potential difference VAC for light of a given frequency and two different intensities. The reverse potential difference - V0 needed to reduce the current to zero is the same for both intensities. The only effect of increasing the intensity is to increase the number of electrons per second and hence the photocurrent i. (The curves level off when VAC is large and positive because at that point all the emitted electrons are being collected by the anode.) If the intensity is held constant but the frequency is increased, the stopping potential also increases. In other words, the greater the light frequency, the higher the energy of the ejected photoelectrons.

These results directly contradict Maxwell’s description of light as an electromagnetic wave. A solution to this dilemma was provided by Albert Einstein in 1905. His proposal involved nothing less than a new picture of the nature of light.

(17.2) G

i50

where h is a universal constant called Planck’s constant. The numerical value of this constant, to the accuracy known at present, is

+

h = 6.626068961332 * 10-34 J # s CAUTION Photons are not “particles” in the usual sense It’s common to envision photons as miniature billiard balls or pellets. While that’s a convenient mental picture, it’s not very accurate. For one thing, billiard balls and bullets have a rest mass and travel slower than the speed of light c, while photons travel at the speed of light and have zero rest mass. For another thing, photons have wave aspects (frequency and wavelength) that are easy to observe. The fact is that the photon concept is a very strange one, and the true nature of photons is difficult to visualize in a simple way. We’ll discuss the dual personality of photons in more detail in Section 17.4. y

547

E The stopping potential at which the current ceases has absolute value V0.

In Einstein’s picture, an individual photon arriving at the surface in Fig. 17.1a or 17.2 is absorbed by a single electron. This energy transfer is an all-or-nothing process, in contrast to the continuous transfer of energy in the wave theory of

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548

CHAPTER 17 Photons: Light Waves Behaving as Particles

17.4 Photocurrent i for a constant light frequency ƒ as a function of the potential VAC of the anode with respect to the cathode. The stopping potential V0 is independent of the light intensity ... ... but the photocurrent i for large positive VAC is directly i proportional to the intensity. f is constant.

Constant intensity 2I

Constant intensity I

2V0

VAC

0

17.5 Stopping potential as a function of frequency for a particular cathode material.

light; the electron gets all of the photon’s energy or none at all. The electron can escape from the surface only if the energy it acquires is greater than the work function f. Thus photoelectrons will be ejected only if hƒ 7 f, or ƒ 7 f>h. Einstein’s postulate therefore explains why the photoelectric effect occurs only for frequencies greater than a minimum threshold frequency. This postulate is also consistent with the observation that greater intensity causes a greater photocurrent (Fig. 17.4). Greater intensity at a particular frequency means a greater number of photons per second absorbed, and thus a greater number of electrons emitted per second and a greater photocurrent. Einstein’s postulate also explains why there is no delay between illumination and the emission of photoelectrons. As soon as photons of sufficient energy strike the surface, electrons can absorb them and be ejected. Finally, Einstein’s postulate explains why the stopping potential for a given surface depends only on the light frequency. Recall that f is the minimum energy needed to remove an electron from the surface. Einstein applied conservation of energy to find that the maximum kinetic energy Kmax = 12 mvmax 2 for an emitted electron is the energy hƒ gained from a photon minus the work function f: Kmax = 12 mvmax2 = hƒ - f

V0 (V)

(17.3)

Substituting Kmax = eV0 from Eq. (17.1), we find

3

eV0 = hƒ - f

2 1 15

f (10 Hz)

0 0.25

0.50 0.75 1.0

21

Table 17.1 Work Functions of Several Elements Element

Work Function (eV)

Aluminum Carbon Copper Gold Nickel Silicon Silver Sodium

4.3 5.0 4.7 5.1 5.1 4.8 4.3 2.7

2f2/e

Equation (17.4) shows that the stopping potential V0 increases with increasing frequency ƒ. The intensity doesn’t appear in Eq. (17.4), so V0 is independent of intensity. As a check of Eq. (17.4), we can measure the stopping potential V0 for each of several values of frequency ƒ for a given cathode material (Fig. 17.5). A graph of V0 as a function of ƒ turns out to be a straight line, verifying Eq. (17.4), and from such a graph we can determine both the work function f for the material and the value of the quantity h > e. After the electron charge - e was measured by Robert Millikan in 1909, Planck’s constant h could also be determined from these measurements. Electron energies and work functions are usually expressed in electron volts 1eV2, defined in Section 3.2. To four significant figures, '

'

To this accuracy, Planck’s constant is

Stopping potential Material 2 V0 Material 1 f2 . f1

Frequency f

2f1/e

(17.4)

1 eV = 1.602 * 10-19 J

17.6 Stopping potential as a function of frequency for two cathode materials having different work functions f.

0

(photoelectric effect)

Threshold frequency Stopping potential is zero at threshold frequency (electrons emerge with zero kinetic energy).

h = 6.626 * 10-34 J # s = 4.136 * 10-15 eV # s Table 17.1 lists the work functions of several elements. These values are approximate because they are very sensitive to surface impurities. The greater the work function, the higher the minimum frequency needed to emit photoelectrons (Fig. 17.6). The photon picture explains a number of other phenomena in which light is absorbed. One example is a suntan, which is caused when the energy in sunlight triggers a chemical reaction in skin cells that leads to increased production of the pigment melanin. This reaction can occur only if a specific molecule in the cell absorbs a certain minimum amount of energy. A short-wavelength ultraviolet photon has enough energy to trigger the reaction, but a longer-wavelength visible-light photon does not. Hence ultraviolet light causes tanning, while visible light cannot.

For each material,

Photon Momentum

f h eV 5 hf 2 f or V0 5 2 e e so the plots have same slope h/e but different intercepts 2f/e on the vertical axis.

Einstein’s photon concept applies to all regions of the electromagnetic spectrum, including radio waves, x rays, and so on. A photon of any electromagnetic radiation with frequency ƒ and wavelength l has energy E given by Eq. (17.2).

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549

17.1 Light Absorbed as Photons: The Photoelectric Effect

Furthermore, according to the special theory of relativity, every particle that has energy must also have momentum, even if it has no rest mass. Photons have zero rest mass. As we saw in Eq. (37.40), a particle with zero rest mass and energy E has momentum with magnitude p given by E = pc. Thus the wavelength l of a photon and the magnitude of its momentum p are related simply by p =

hƒ h E = = c c l

(momentum of a photon)

(17.5)

The direction of the photon’s momentum is simply the direction in which the electromagnetic wave is moving.

Problem-Solving Strategy 17.1

Photons

IDENTIFY the relevant concepts: The energy and momentum of an individual photon are proportional to the frequency and inversely proportional to the wavelength. Einstein’s interpretation of the photoelectric effect is that energy is conserved as a photon ejects an electron from a material surface. SET UP the problem: Identify the target variable. It could be the photon’s wavelength l, frequency ƒ, energy E, or momentum p. If the problem involves the photoelectric effect, the target variable could be the maximum kinetic energy of photoelectrons Kmax, the stopping potential V0, or the work function f. EXECUTE the solution as follows: 1. Use Eqs. (17.2) and (17.5) to relate the energy and momentum of a photon to its wavelength and frequency. If the problem involves the photoelectric effect, use Eqs. (17.1), (17.3), and

Example 17.1

(17.4) to relate the photon frequency, stopping potential, work function, and maximum photoelectron kinetic energy. 2. The electron volt (eV), which we introduced in Section 3.2, is a convenient unit. It is the kinetic energy gained by an electron when it moves freely through an increase of potential of one volt: 1 eV = 1.602 * 10-19 J. If the photon energy E is given in electron volts, use h = 4.136 * 10-15 eV # s; if E is in joules, use h = 6.626 * 10-34 J # s. EVALUATE your answer: In problems involving photons, at first the numbers will be unfamiliar to you and errors will not be obvious. It helps to remember that a visible-light photon with l = 600 nm and ƒ = 5 * 1014 Hz has an energy E of about 2 eV, or about 3 * 10-19 J.

Laser-pointer photons

A laser pointer with a power output of 5.00 mW emits red light 1l = 650 nm2. (a) What is the magnitude of the momentum of each photon? (b) How many photons does the laser pointer emit each second? S O L U T IO N IDENTIFY and SET UP: This problem involves the ideas of (a) photon momentum and (b) photon energy. In part (a) we’ll use Eq. (17.5) and the given wavelength to find the magnitude of each photon’s momentum. In part (b), Eq. (17.2) gives the energy per photon, and the power output tells us the energy emitted per second. We can combine these quantities to calculate the number of photons emitted per second. EXECUTE: (a) We have l = 650 nm = 6.50 * 10-7 m, so from Eq. (17.5) the photon momentum is p =

6.626 * 10-34 J # s h = l 6.50 * 10-7 m '

= 1.02 * 10 (Recall that 1 J = 1 kg # m2 > s2.) '

'

-27

kg # m > s '

'

(b) From Eq. (17.2), the energy of a single photon is E = pc = 11.02 * 10-27 kg # m > s213.00 * 108 m > s2 '

= 3.06 * 10

-19

'

'

'

J = 1.91 eV

The laser pointer emits energy at the rate of 5.00 * 10-3 J> s, so it emits photons at the rate of 5.00 * 10-3 J > s '

'

3.06 * 10-19 J > photon '

'

= 1.63 * 1016 photons > s '

'

EVALUATE: The result in part (a) is very small; a typical oxygen molecule in room-temperature air has 2500 times more momentum. As a check on part (b), we can calculate the photon energy using Eq. (17.2): 16.626 * 10-34 J # s213.00 * 108 m > s2 hc = l 6.50 * 10-7 m '

E = hf =

'

= 3.06 * 10-19 J = 1.91 eV Our result in part (b) shows that a huge number of photons leave the laser pointer each second, each of which has an infinitesimal amount of energy. Hence the discreteness of the photons isn’t noticed, and the radiated energy appears to be a continuous flow.

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550

CHAPTER 17 Photons: Light Waves Behaving as Particles

Example 17.2

A photoelectric-effect experiment

While conducting a photoelectric-effect experiment with light of a certain frequency, you find that a reverse potential difference of 1.25 V is required to reduce the current to zero. Find (a) the maximum kinetic energy and (b) the maximum speed of the emitted photoelectrons.

Kmax = eV0 = e11.25 V2 = 1.25 eV since the electron volt (eV) is the magnitude of the electron charge e times one volt (1 V). (b) From Kmax = 12 mvmax 2 we get vmax =

SOLUTION IDENTIFY and SET UP: The value of 1.25 V is the stopping potential V0 for this experiment. We’ll use this in Eq. (17.1) to find the maximum photoelectron kinetic energy Kmax, and from this we’ll find the maximum photoelectron speed. EXECUTE: (a) From Eq. (17.1), Kmax = eV0 = 11.60 * 10-19 C211.25 V2 = 2.00 * 10-19 J

(Recall that 1 V = 1 J > C.) In terms of electron volts, '

Example 17.3

2Kmax 212.00 * 10-19 J2 = B m B 9.11 * 10-31 kg

= 6.63 * 105 m > s '

'

EVALUATE The value of vmax is about 0.2% of the speed of light, so we are justified in using the nonrelativistic expression for kinetic energy. (An equivalent justification is that the electron’s 1.25-eV kinetic energy is much less than its rest energy mc2 = 0.511 MeV = 5.11 * 105 eV.)

'

Determining F and h experimentally

For a particular cathode material in a photoelectric-effect experiment, you measure stopping potentials V0 = 1.0 V for light of wavelength l = 600 nm, 2.0 V for 400 nm, and 3.0 V for 300 nm. Determine the work function f for this material and the implied value of Planck’s constant h.

obtained from ƒ = c > l and c = 3.00 * 108 m > s, are 0.50 * 1015 Hz, 0.75 * 1015 Hz, and 1.0 * 1015 Hz, respectively. From a graph of these data (see Fig. 17.6), we find '

-

SOLUTION

'

'

'

f = vertical intercept = - 1.0 V e f = 1.0 eV = 1.6 * 10-19 J

IDENTIFY and SET UP: This example uses the relationship among stopping potential V0, frequency ƒ, and work function f in the photoelectric effect. According to Eq. (17.4), a graph of V0 versus ƒ should be a straight line as in Fig. 17.5 or 17.6. Such a graph is completely determined by its slope and the value at which it intercepts the vertical axis; we will use these to determine the values of the target variables f and h.

and

EXECUTE: We rewrite Eq. (17.4) as

EVALUATE: The value of Planck’s constant h determined from your experiment differs from the accepted value by only about 3%. The small value f = 1.0 eV tells us that the cathode surface is not composed solely of one of the elements in Table 17.1.

f h V0 = ƒ e e In this form we see that the slope of the line is h > e and the verticalaxis intercept (corresponding to ƒ = 0) is - f > e. The frequencies, '

'

Application Sterilizing with High-Energy Photons One technique for killing harmful microorganisms is to illuminate them with ultraviolet light with a wavelength shorter than 254 nm. If a photon of such short wavelength strikes a DNA molecule within a microorganism, the energy of the photon is great enough to break the bonds within the molecule. This renders the microorganism unable to grow or reproduce. Such ultraviolet germicidal irradiation is used for medical sanitation, to keep laboratories sterile (as shown here), and to treat both drinking water and wastewater.

¢V0 3.0 V - 1- 1.0 V2 = 4.0 * 10-15 J # s > C ¢ƒ 1.00 * 1015 s-1 - 0 h = slope * e = 14.0 * 10-15 J # s > C211.60 * 10-19 C2

Slope =

'

'

= 6.4 * 10

-34

'

'

J#s

'

'

Test Your Understanding of Section 17.1 Silicon films become better electrical conductors when illuminated by photons with energies of 1.14 eV or greater, an effect called photoconductivity. Which of the following wavelengths of electromagnetic radiation can cause photoconductivity in silicon films? (i) ultraviolet light with l = 300 nm; (ii) red light with l = 600 nm; (iii) infrared light with l = 1200 nm.

17.2

y

Light Emitted as Photons: X-Ray Production

The photoelectric effect provides convincing evidence that light is absorbed in the form of photons. For physicists to accept Einstein’s radical photon concept, however, it was also necessary to show that light is emitted as photons. An experiment

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17.2 Light Emitted as Photons: X-Ray Production

551

that demonstrates this convincingly is the inverse of the photoelectric effect: Instead of releasing electrons from a surface by shining electromagnetic radiation on it, we cause a surface to emit radiation—specifically, x rays—by bombarding it with fast-moving electrons.

X-Ray Photons 17.7 An apparatus used to produce x rays, similar to Röntgen’s 1895 apparatus. Electrons are emitted thermionically from the heated cathode and are accelerated toward the anode; when they strike it, x rays are produced. Heated cathode Power supply for heater

Anode

+

X rays were first produced in 1895 by the German physicist Wilhelm Röntgen, using an apparatus similar in principle to the setup shown in Fig. 17.7. Electrons are released from the cathode by thermionic emission, in which the escape energy is supplied by heating the cathode to a very high temperature. (As in the photoelectric effect, the minimum energy that an individual electron must be given to escape from the cathode’s surface is equal to the work function for the surface. In this case the energy is provided to the electrons by heat rather than by light.) The electrons are then accelerated toward the anode by a potential difference VAC. The bulb is evacuated (residual pressure 10-7 atm or less), so the electrons can travel from the cathode to the anode without colliding with air molecules. When VAC is a few thousand volts or more, x rays are emitted from the anode surface. The anode produces x rays in part simply by slowing the electrons abruptly. (Recall from Section 12.1 that accelerated charges emit electromagnetic waves.) This process is called bremsstrahlung (German for “braking radiation”). Because the electrons undergo accelerations of very great magnitude, they emit much of their radiation at short wavelengths in the x-ray range, about 1029 to 10212 m (1 nm to 1 pm). (X-ray wavelengths can be measured quite precisely by crystal diffraction techniques, which we discussed in Section 16.6.) Most electrons are braked by a series of collisions and interactions with anode atoms, so bremsstrahlung produces a continuous spectrum of electromagnetic radiation. Just as we did for the photoelectric effect in Section 17.1, let’s compare what Maxwell’s wave theory of electromagnetic radiation would predict about this radiation to what is observed experimentally.

X-ray beam + Accelerating voltage V

Wave-Model Prediction: The electromagnetic waves produced when an electron slams into the anode should be analogous to the sound waves produced by crashing cymbals together. These waves include sounds of all frequencies. By analogy, the x rays produced by bremsstrahlung should have a spectrum that includes all frequencies and hence all wavelengths. Experimental Result: Figure 17.8 shows bremsstrahlung spectra using the same cathode and anode with four different accelerating voltages. We see that not all x-ray frequencies and wavelengths are emitted: Each spectrum has a maximum frequency ƒmax and a corresponding minimum wavelength lmin. The greater the potential difference VAC, the higher the maximum frequency and the shorter the minimum wavelength.

The wave model of electromagnetic radiation cannot explain these experimental results. But we can readily understand them using the photon model. An electron has charge -e and gains kinetic energy eVAC when accelerated through a potential increase VAC. The most energetic photon (highest frequency and shortest wavelength) is produced if the electron is braked to a stop all at once when it hits the anode, so that all of its kinetic energy goes to produce one photon; that is, eVAC = hƒmax

hc = lmin

(bremsstrahlung)

(17.6)

(In this equation we neglect the work function of the target anode and the initial kinetic energy of the electrons “boiled off” from the cathode. These energies are very small compared to the kinetic energy eVAC gained due to the potential

17.8 The continuous spectrum of x rays produced when a tungsten target is struck by electrons accelerated through a voltage VAC. The curves represent different values of VAC; points a, b, c, and d show the minimum wavelength for each voltage. Vertical axis: x-ray intensity per unit wavelength I (l) 10

50 kV

8 40 kV 6 4

30 kV

2

0

a b c 20 40

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20 kV d 60 80

100

l (pm)

Horizontal axis: x-ray wavelength in picometers (1 pm 5 10212 m)

552

CHAPTER 17 Photons: Light Waves Behaving as Particles

difference.) If only a portion of an electron’s kinetic energy goes into producing a photon, the photon energy will be less than eVAC and the wavelength will be greater than lmin. As further support for the photon model, the measured values for lmin for different values of eVAC (see Fig. 17.8) agree with Eq. (17.6). Note that according to Eq. (17.6), the maximum frequency and minimum wavelength in the bremsstrahlung process do not depend on the target material; this also agrees with experiment. So we can conclude that the photon picture of electromagnetic radiation is valid for the emission as well as the absorption of radiation. The apparatus shown in Fig. 17.7 can also produce x rays by a second process in which electrons transfer their kinetic energy partly or completely to individual atoms within the target. It turns out that this process not only is consistent with the photon model of electromagnetic radiation, but also provides insight into the structure of atoms. We’ll return to this process in Section 41.5.

Example 17.4

Producing x rays

Electrons in an x-ray tube accelerate through a potential difference of 10.0 kV before striking a target. If an electron produces one photon on impact with the target, what is the minimum wavelength of the resulting x rays? Find the answer by expressing energies in both SI units and electron volts. SOLUTION IDENTIFY and SET UP: To produce an x-ray photon with minimum wavelength and hence maximum energy, all of the electron’s kinetic energy must go into producing a single x-ray photon. We’ll use Eq. (17.6) to determine the wavelength. EXECUTE: From Eq. (17.6), using SI units we have

16.626 * 10-34 J # s213.00 * 108 m > s2 hc = eVAC 11.602 * 10-19 C2110.0 * 103 V2 '

lmin =

'

Using electron volts, we have 14.136 * 10-15 eV # s213.00 * 108 m > s2 hc = eVAC e110.0 * 103 V2 '

lmin =

'

= 1.24 * 10-10 m = 0.124 nm In the second calculation, the “e” for the magnitude of the electron charge cancels the “e” in the unit “eV,” because the electron volt (eV) is the magnitude of the electron charge e times one volt (1 V). EVALUATE: To check our result, recall from Example 17.1 that a 1.91-eV photon has a wavelength of 650 nm. Here the electron energy, and therefore the x-ray photon energy, is 10.0 * 103 eV = 10.0 keV, about 5000 times greater than in Example 17.1, and the 1 wavelength is about 5000 as great as in Example 17.1. This makes sense, since wavelength and photon energy are inversely proportional.

= 1.24 * 10-10 m = 0.124 nm

17.9 This radiologist is operating a CT scanner (seen through the window) from a separate room to avoid repeated exposure to x rays.

Applications of X Rays X rays have many practical applications in medicine and industry. Because x-ray photons are of such high energy, they can penetrate several centimeters of solid matter. Hence they can be used to visualize the interiors of materials that are opaque to ordinary light, such as broken bones or defects in structural steel. The object to be visualized is placed between an x-ray source and an electronic detector (like that used in a digital camera) or a piece of photographic film. The darker an area in the image recorded by such a detector, the greater the radiation exposure. Bones are much more effective x-ray absorbers than soft tissue, so bones appear as light areas. A crack or air bubble allows greater transmission and shows as a dark area. A widely used and vastly improved x-ray technique is computed tomography; the corresponding instrument is called a CT scanner. The x-ray source produces a thin, fan-shaped beam that is detected on the opposite side of the subject by an array of several hundred detectors in a line. Each detector measures absorption along a thin line through the subject. The entire apparatus is rotated around the subject in the plane of the beam, and the changing photon-counting rates of the detectors are recorded digitally. A computer processes this information and

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17.3 Light Scattered as Photons: Compton Scattering and Pair Production

reconstructs a picture of absorption over an entire cross section of the subject (see Fig. 17.9). Differences in absorption as small as 1% or less can be detected with CT scans, and tumors and other anomalies that are much too small to be seen with older x-ray techniques can be detected. X rays cause damage to living tissues. As x-ray photons are absorbed in tissues, their energy breaks molecular bonds and creates highly reactive free radicals (such as neutral H and OH), which in turn can disturb the molecular structure of proteins and especially genetic material. Young and rapidly growing cells are particularly susceptible, which is why x rays are useful for selective destruction of cancer cells. Conversely, however, a cell may be damaged by radiation but survive, continue dividing, and produce generations of defective cells; thus x rays can cause cancer. Even when the organism itself shows no apparent damage, excessive exposure to x rays can cause changes in the organism’s reproductive system that will affect its offspring. A careful assessment of the balance between risks and benefits of radiation exposure is essential in each individual case.

553

X-Ray Absorption and Medical Imaging

Application

Atomic electrons can absorb x rays. Hence materials with many electrons per atom tend to be better x-ray absorbers than materials with few electrons. In this x-ray image the lighter areas show where x rays are absorbed as they pass through the body, while the darker areas indicate regions that are relatively transparent to x rays. Bones contain large amounts of elements such as phosphorus and calcium, with 15 and 20 electrons per atom, respectively. In soft tissue, the predominant elements are hydrogen, carbon, and oxygen, with only 1, 6, and 8 electrons per atom, respectively. Hence x rays are absorbed by bone but can pass relatively easily through soft tissue.

Test Your Understanding of Section 17.2 In the apparatus shown in Fig. 17.7, suppose you increase the number of electrons that are emitted from the cathode per second while keeping the potential difference VAC the same. How will this affect the intensity I and minimum wavelength lmin of the emitted x rays? (i) I and lmin will both increase; (ii) I will increase but lmin will be unchanged; (iii) I will increase but lmin will decrease; (iv) I will remain the same but lmin will decrease; (v) none of these. y

17.3

Light Scattered as Photons: Compton Scattering and Pair Production

The final aspect of light that we must test against Einstein’s photon model is its behavior after the light is produced and before it is eventually absorbed. We can do this by considering the scattering of light. As we discussed in Section 13.6, scattering is what happens when light bounces off particles such as molecules in the air.

Compton Scattering Let’s see what Maxwell’s wave model and Einstein’s photon model predict for how light behaves when it undergoes scattering by a single electron, such as an individual electron within an atom. Wave-Model Prediction: In the wave description, scattering would be a process of absorption and re-radiation. Part of the energy of the light wave would be absorbed by the electron, which would oscillate in response to the oscillating electric field of the wave. The oscillating electron would act like a miniature antenna (see Section 12.1), re-radiating its acquired energy as scattered waves in a variety of directions. The frequency at which the electron oscillates would be the same as the frequency of the incident light, and the re-radiated light would have the same frequency as the oscillations of the electron. So, in the wave model, the scattered light and incident light have the same frequency and same wavelength.

17.10 The photon model of light scattering by an electron. (a) Before collision: The target electron is at rest. Incident photon: wavelength l, S momentum p

Target electron (at rest)

(b) After collision: The angle between the directions of the scattered photon and the incident photon is f.

Photon-Model Prediction: In the photon model we imagine the scattering process as a collision of two particles, the incident photon and an electron that is initially at rest (Fig. 17.10a). The incident photon would give up part of its energy and momentum to the electron, which recoils as a result of this impact. The scattered photon that remains can fly off at a variety of angles f with respect to the incident direction, but it has less energy and less momentum than the incident photon (Fig. 17.10b). The energy and momentum of a photon are given by E = hƒ = hc>l (Eq. 17.2) and p = hƒ>c = h>l (Eq. 17.5). Therefore, in the photon model, the scattered light has a lower frequency ƒ and longer wavelength l than the incident light.

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Scattered photon: wavelength lr, S momentum pr f S

Pe Recoiling electron: S momentum Pe

554

CHAPTER 17 Photons: Light Waves Behaving as Particles 17.11 A Compton-effect experiment.

ActivPhysics 17.4: Compton Scattering

Scattered photons X-ray source

lr l

The change in wavelength depends on the angle at which the photons are scattered.

f

Incident photons

Detector

Target

The definitive experiment that tested these predictions of the wave and photon models was carried out in 1922 by the American physicist Arthur H. Compton. In his experiment Compton aimed a beam of x rays at a solid target and measured the wavelength of the radiation scattered from the target (Fig. 17.11). He discovered that some of the scattered radiation has smaller frequency (longer wavelength) than the incident radiation and that the change in wavelength depends on the angle through which the radiation is scattered. This is precisely what the photon model predicts for light scattered from electrons in the target, a process that is now called Compton scattering. Specifically, if the scattered radiation emerges at an angle f with respect to the incident direction, as shown in Fig. 17.11, and if l and l¿ are the wavelengths of the incident and scattered radiation, respectively, Compton found that l¿ - l =

h 11 - cos f2 mc

(Compton scattering)

(17.7)

where m is the electron rest mass. In other words, l¿ is greater than l. The quantity h > mc that appears in Eq. (17.7) has units of length. Its numerical value is '

'

6.626 * 10-34 J # s h = 2.426 * 10-12 m = mc 19.109 * 10-31 kg212.998 * 108 m > s2 '

17.12 Vector diagram showing conservation of momentum in Compton scattering. S

pr

'

Compton showed that Einstein’s photon theory, combined with the principles of conservation of energy and conservation of momentum, provides a beautifully clear explanation of his experimental results. We outline the derivation below. The electron recoil energy may be in the relativistic range, so we have to use the relativistic energy–momentum relationships, Eqs. (37.39) and (37.40). The incident S photon has momentum p , with magnitude p and energy pc. The scattered photon has S momentum p ¿, with magnitude p¿ and energy p¿c. The electron is initially at rest, so its initial momentumSis zero and its initial energy is its rest energy mc2. The final electron momentum Pe has magnitude Pe, and the final electron energy is given by Ee2 = 1mc222 + 1Pec22. Then energy conservation gives us the relationship pc + mc2 = p¿c + Ee

S

p

f

Rearranging, we find S

Pe

1pc - p¿c + mc222 = Ee 2 = 1mc222 + 1Pec22

(17.8)

S

Conservation of S pr momentum during Compton scattering

S

f S

p

We can eliminate the electron momentum Pe from Eq. (17.8) by using momenS S S tum conservation. From Fig. 17.12 we see that p 5 p ¿ 1 Pe, or

Pe S

S

S

Pe 5 p 2 p¿

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(17.9)

17.3 Light Scattered as Photons: Compton Scattering and Pair Production

By taking the scalar product of each side of Eq. (17.9) with itself, we find Pe2

2

2

= p + p¿ - 2pp¿ cos f

17.13 Intensity as a function of wavelength for photons scattered at an angle of 135° in a Compton-scattering experiment.

(17.10)

We now substitute this expression for Pe2 into Eq. (17.8) and multiply out the left side. We divide out a common factor c2; several terms cancel, and when the resulting equation is divided through by 1 pp¿2, the result is mc mc = 1 - cos f p p¿

(17.11)

Example 17.5

'

'

'

'

Intensity per unit wavelength

Finally, we substitute p¿ = h > l¿ and p = h > l, then multiply by h > mc to obtain Eq. (17.7). When the wavelengths of x rays scattered at a certain angle are measured, the curve of intensity per unit wavelength as a function of wavelength has two peaks (Fig. 17.13). The longer-wavelength peak represents Compton scattering. The shorter-wavelength peak, labeled l0, is at the wavelength of the incident x rays and corresponds to x-ray scattering from tightly bound electrons. In such scattering processes the entire atom must recoil, so the m in Eq. (17.7) is the mass of the entire atom rather than of a single electron. The resulting wavelength shifts are negligible. '

555

'

Photons scattered from tightly bound electrons undergo a negligible wavelength shift.

Photons scattered from loosely bound electrons undergo a wavelength shift given by Eq. (17.7).

λ l0

l9

Compton scattering

You use 0.124-nm x-ray photons in a Compton-scattering experiment. (a) At what angle is the wavelength of the scattered x rays 1.0% longer than that of the incident x rays? (b) At what angle is it 0.050% longer?

¢l 1.24 * 10-12 m = 1 = 0.4889 h > mc 2.426 * 10-12 m f = 60.7°

cos f = 1 -

'

'

(b) For ¢l to be 0.050% of 0.124 nm, or 6.2 * 10-14 m, S O L U T IO N IDENTIFY and SET UP: We’ll use the relationship between scattering angle and wavelength shift in the Compton effect. In each case our target variable is the angle f (see Fig. 17.10b). We solve for f using Eq. (17.7). EXECUTE: (a) In Eq. (17.7) we want ¢l = l¿ - l to be 1.0% of 0.124 nm, so ¢l = 0.00124 nm = 1.24 * 10-12 m. Using the value h > mc = 2.426 * 10-12 m, we find '

'

¢l =

h 11 - cos f2 mc

cos f = 1 -

6.2 * 10-14 m

2.426 * 10-12 m f = 13.0°

= 0.9744

EVALUATE: Our results show that smaller scattering angles give smaller wavelength shifts. Thus in a grazing collision the photon energy loss and the electron recoil energy are smaller than when the scattering angle is larger. This is just what we would expect for an elastic collision, whether between a photon and an electron or between two billiard balls.

Pair Production Another effect that can be explained only with the photon picture involves gamma rays, the shortest-wavelength and highest-frequency variety of electromagnetic radiation. If a gamma-ray photon of sufficiently short wavelength is fired at a target, it may not scatter. Instead, as depicted in Fig. 17.14, it may disappear completely and be replaced by two new particles: an electron and a positron (a particle that has the same rest mass m as an electron but has a positive charge +e rather than the negative charge - e of the electron). This process, called pair production, was first observed by the physicists Patrick Blackett and Giuseppe Occhialini in 1933. The electron and positron have to be produced in pairs in order to conserve electric charge: The incident photon has zero charge, and the electron–positron pair has net charge 1- e2 + 1 +e2 = 0. Enough energy must be available to account for the rest energy 2mc2 of the two particles. To four significant figures, this minimum energy is

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556

CHAPTER 17 Photons: Light Waves Behaving as Particles

17.14 (a) Photograph of bubblechamber tracks of electron–positron pairs that are produced when 300-MeV photons strike a lead sheet. A magnetic field directed out of the photograph made the electrons (e- ) and positrons (e + ) curve in opposite directions. (b) Diagram showing the pair-production process for two of the gamma-ray photons (g).

E min = 2mc2 = 219.109 * 10-31 kg212.998 * 108 m>s22 = 1.637 * 10-13 J = 1.022 MeV Thus the photon must have at least this much energy to produce an electron– positron pair. From Eq. (17.2), E = hc>l, the photon wavelength has to be shorter than 16.626 * 10-34 J # s212.998 * 108 m > s2 hc = E min 1.637 * 10-13 J '

(a)

Electron–positron pair

l max =

'

= 1.213 * 10-12 m = 1.213 * 10-3 nm = 1.213 pm

(b) g

e2

e1 S

1 This is a very short wavelength, about 1000 as large as the x-ray wavelengths that Compton used in his scattering experiments. (The requisite minimum photon energy is actually a bit higher than 1.022 MeV, so the photon wavelength must be a bit shorter than 1.213 pm. The reason is that when the incident photon encounters an atomic nucleus in the target, some of the photon energy goes into the kinetic energy of the recoiling nucleus.) Just as for the photoelectric effect, the wave model of electromagnetic radiation cannot explain why pair production occurs only when very short wavelengths are used. The inverse process, electron–positron pair annihilation, occurs when a positron and an electron collide. Both particles disappear, and two (or occasionally three) photons can appear, with total energy of at least 2mec2 = 1.022 MeV. Decay into a single photon is impossible because such a process could not conserve both energy and momentum. It’s easiest to analyze this annihilation process in the frame of reference called the center-of-momentum system, in which the total momentum is zero. It is the relativistic generalization of the center-of-mass system that we discussed in Section 8.5.

e2

B g

e1

Example 17.6

Pair annihilation

An electron and a positron, initially far apart, move toward each other with the same speed. They collide head-on, annihilating each other and producing two photons. Find the energies, wavelengths, and frequencies of the photons if the initial kinetic energies of the electron and positron are (a) both negligible and (b) both 5.000 MeV. The electron rest energy is 0.511 MeV. SOLUTION IDENTIFY and SET UP: Just as in the elastic collisions we studied in Chapter 8, both momentum and energy are conserved in pair annihilation. The electron and positron are initially far apart, so the initial electric potential energy is zero and the initial energy is the sum of the particle kinetic and rest energies. The final energy is the sum of the photon energies. The total initial momentum is zero; the total momentum of the two photons must likewise be zero. We find the photon energy E using conservation of energy, conservation of momentum, and the relationship E = pc (see Section 17.1). We then calculate the wavelengths and frequencies using E = hc > l = hƒ. '

'

EXECUTE: If the total momentum of the two photons is to be zero, their momenta must have equal magnitudes p and opposite directions. From E = pc = hc > l = hƒ, the two photons must also have the same energy E, wavelength l, and frequency ƒ. '

Before the collision the energy of each electron is K + mc2, where K is its kinetic energy and mc2 = 0.511 MeV. Conservation of energy then gives 1K + mc22 + 1K + mc22 = E + E

Hence the energy of each photon is E = K + mc2. (a) In this case the electron kinetic energy K is negligible compared to its rest energy mc2, so each photon has energy E = mc2 = 0.511 MeV. The corresponding photon wavelength and frequency are 14.136 * 10-15 eV # s213.00 * 108 m > s2 hc = E 0.511 * 106 eV -12 = 2.43 * 10 m = 2.43 pm '

'

l =

ƒ =

0.511 * 106 eV E = = 1.24 * 1020 Hz h 4.136 * 10-15 eV # s

(b) In this case K = 5.000 MeV, so each photon has energy E = 5.000 MeV + 0.511 MeV = 5.511 MeV. Proceeding as in part (a), you can show that the photon wavelength is 0.2250 pm and the frequency is 1.333 * 1021 Hz .

'

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17.4 Wave–Particle Duality, Probability, and Uncertainty EVALUATE: As a check, recall from Example 17.1 that a 650-nm visible-light photon has energy 1.91 eV and frequency 4.62 * 1014 Hz. The photon energy in part (a) is about 2.5 * 105 times

557

greater. As expected, the photon’s wavelength is shorter and its frequency higher than those for a visible-light photon by the same factor. You can check the results for part (b) in the same way.

Test Your Understanding of Section 17.3 If you used visible-light photons in the experiment shown in Fig. 17.11, would the photons undergo a wavelength shift due to the scattering? If so, is it possible to detect the shift with the human eye? y

17.4

Wave–Particle Duality, Probability, and Uncertainty

We have studied many examples of the behavior of light and other electromagnetic radiation. Some, including the interference and diffraction effects described in Chapters 15 and 16, demonstrate conclusively the wave nature of light. Others, the subject of the present chapter, point with equal force to the particle nature of light. At first glance these two aspects seem to be in direct conflict. How can light be a wave and a particle at the same time? We can find the answer to this apparent wave–particle conflict in the principle of complementarity, first stated by the Danish physicist Niels Bohr in 1928. The wave descriptions and the particle descriptions are complementary. That is, we need both to complete our model of nature, but we will never need to use both at the same time to describe a single part of an occurrence.

17.15 Single-slit diffraction pattern of light observed with a movable photomultiplier. The curve shows the intensity distribution predicted by the wave picture. The photon distribution is shown by the numbers of photons counted at various positions. Movable photomultiplier detector Counter

Diffraction and Interference in the Photon Picture

009

Slit

Let’s start by considering again the diffraction pattern for a single slit, which we analyzed in Sections 16.2 and 16.3. Instead of recording the pattern on a digital camera chip or photographic film, we use a detector called a photomultiplier that can actually detect individual photons. Using the setup shown in Fig. 17.15, we place the photomultiplier at various positions for equal time intervals, count the photons at each position, and plot out the intensity distribution. We find that, on average, the distribution of photons agrees with our predictions from Section 16.3. At points corresponding to the maxima of the pattern, we count many photons; at minimum points, we count almost none; and so on. The graph of the counts at various points gives the same diffraction pattern that we predicted with Eq. (16.7). But suppose we now reduce the intensity to such a low level that only a few photons per second pass through the slit. We now record a series of discrete strikes, each representing a single photon. While we cannot predict where any given photon will strike, over time the accumulating strikes build up the familiar diffraction pattern we expect for a wave. To reconcile the wave and particle aspects of this pattern, we have to regard the pattern as a statistical distribution that tells us how many photons, on average, go to each spot. Equivalently, the pattern tells us the probability that any individual photon will land at a given spot. If we shine our faint light beam on a two-slit apparatus, we get an analogous result (Fig. 17.16). Again we can’t predict exactly where an individual photon will go; the interference pattern is a statistical distribution. How does the principle of complementarity apply to these diffraction and interference experiments? The wave description, not the particle description, explains the single- and double-slit patterns. But the particle description, not the wave description, explains why the photomultiplier records discrete packages of energy. The two descriptions complete our understanding of the results. For instance, suppose we consider an individual photon and ask how it knows “which way to go” when passing through the slit. This question seems like a conundrum, but that is because it is framed in terms of a particle description—whereas it is the wave nature of light that determines the distribution of photons. Conversely,

026 243 576

Intensity

Monochromatic light Screen

17.16 These images record the positions where individual photons in a two-slit interference experiment strike the screen. As more photons reach the screen, a recognizable interference pattern appears. After 21 photons reach the screen

After 1000 photons reach the screen

After 10,000 photons reach the screen

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558

CHAPTER 17 Photons: Light Waves Behaving as Particles

the fact that the photomultiplier detects faint light as a sequence of individual “spots” can’t be explained in wave terms.

Probability and Uncertainty PhET: Fourier: Making Waves PhET: Quantum Wave Interference ActivPhysics 17.6: Uncertainty Principle

Although photons have energy and momentum, they are nonetheless very different from the particle model we used for Newtonian mechanics in Chapters 4 through 8. The Newtonian particle model treats an object as a point mass. We can describe the location and state of motion of such a particle at any instant with three spatial coordinates and three components of momentum, and we can then predict the particle’s future motion. This model doesn’t work at all for photons, however: We cannot treat a photon as a point object. This is because there are fundamental limitations on the precision with which we can simultaneously determine the position and momentum of a photon. Many aspects of a photon’s behavior can be stated only in terms of probabilities. (In Chapter 18 we will find that the non-Newtonian ideas we develop for photons in this section also apply to particles such as electrons.) To get more insight into the problem of measuring a photon’s position and momentum simultaneously, let’s look again at the single-slit diffraction of light (Fig. 17.17). Suppose the wavelength l is much less than the slit width a. Then most (85%) of the photons go into the central maximum of the diffraction pattern, and the remainder go into other parts of the pattern. We use u1 to denote the angle between the central maximum and the first minimum. Using Eq. (16.2) with m = 1, we find that u1 is given by sin u1 = l > a. Since we assume l = a, it follows that u1 is very small, sin u1 is very nearly equal to u1 (in radians), and '

u1 =

'

l a

(17.12)

Even though the photons all have the same initial state of motion, they don’t all follow the same path. We can’t predict the exact trajectory of any individual photon from knowledge of its initial state; we can only describe the probability that an individual photon will strike a given spot on the screen. This fundamental indeterminacy has no counterpart in Newtonian mechanics. Furthermore, there are fundamental uncertainties in both the position and the momentum of an individual particle, and these uncertainties are related inseparably. To clarify this point, let’s go back to Fig. 17.17. A photon that strikes the screen at the outer edge of the central maximum, at angle u1, must have a component of momentum py in the y-direction, as well as a component px in the x-direction, despite the fact that initially the beam was directed along the x-axis. From the geometry of the situation the two components are related by py > px = tan u1. Since u1 is small, we may use the approximation tan u1 = u1, and '

17.17 Interpreting single-slit diffraction in terms of photon momentum.

px and py are the momentum components for a photon striking the outer edge of the central maximum, at angle u1.

Diffraction pattern

a S

p

py u1 px Photons of monochromatic light

'

Screen Slit

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17.4 Wave–Particle Duality, Probability, and Uncertainty

py = px u1

(17.13)

Substituting Eq. (17.12), u1 = l > a, into Eq. (17.13) gives '

'

py = px

l a

(17.14)

Equation (17.14) says that for the 85% of the photons that strike the detector within the central maximum (that is, at angles between -l > a and +l > a), the y-component of momentum is spread out over a range from -pxl > a to +pxl > a. Now let’s consider all the photons that pass through the slit and strike the screen. Again, they may hit above or below the center of the pattern, so their component py may be positive or negative. However the symmetry of the diffraction pattern shows us the average value 1py2av = 0. There will be an uncertainty ¢py in the y-component of momentum at least as great as pxl > a. That is, '

'

'

'

'

¢py Ú px

'

'

'

'

'

l a

(17.15)

The narrower the slit width a, the broader is the diffraction pattern and the greater is the uncertainty in the y-component of momentum py. The photon wavelength l is related to the momentum px by Eq. (17.5), which we can rewrite as l = h > px. Using this relationship in Eq. (17.15) and simplifying, we find '

'

¢py Ú px

h h = px a a

¢py a Ú h '

(17.16)

What does Eq. (17.16) mean? The slit width a represents an uncertainty in the y-component of the position of a photon as it passes through the slit. We don’t know exactly where in the slit each photon passes through. So both the y-position and the y-component of momentum have uncertainties, and the two uncertainties are related by Eq. (17.16). We can reduce the momentum uncertainty ¢py only by reducing the width of the diffraction pattern. To do this, we have to increase the slit width a, which increases the position uncertainty. Conversely, when we decrease the position uncertainty by narrowing the slit, the diffraction pattern broadens and the corresponding momentum uncertainty increases. You may protest that it doesn’t seem to be consistent with common sense for a photon not to have a definite position and momentum. We reply that what we call common sense is based on familiarity gained through experience. Our usual experience includes very little contact with the microscopic behavior of particles like photons. Sometimes we have to accept conclusions that violate our intuition when we are dealing with areas that are far removed from everyday experience.

The Uncertainty Principle In more general discussions of uncertainty relationships, the uncertainty of a quantity is usually described in terms of the statistical concept of standard deviation, which is a measure of the spread or dispersion of a set of numbers around their average value. Suppose we now begin to describe uncertainties in this way [neither ¢py nor a in Eq. (17.16) is a standard deviation]. If a coordinate x has an uncertainty ¢x and if the corresponding momentum component px has an uncertainty ¢px, then those standard-deviation uncertainties are found to be related in general by the inequality ¢x ¢px Ú U>2

(Heisenberg uncertainty principle for position and momentum)

(17.17)

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559

560

CHAPTER 17 Photons: Light Waves Behaving as Particles

In this expression the quantity U (pronounced “h-bar”) is Planck’s constant divided by 2p : '

U =

h = 1.0545716281532 * 10-34 J # s 2p

We will use this quantity frequently to avoid writing a lot of factors of 2p in later equations. CAUTION h versus h-bar It’s common for students to plug in the value of h when what they really wanted was U = h>2p, or vice versa. Be careful not to make the same mistake, or you’ll find yourself wondering why your answer is off by a factor of 2p! y 17.18 The Heisenberg uncertainty principle for position and momentum components. It is impossible for the product ¢x¢px to be less than U> 2 = h > 4p. '

'

'

Small position uncertainty; large momentum uncertainty

Dpx

Allowed: Dx Dpx $ " 2

/

/

Dx Dpx 5 " 2 Impossible: Dx Dpx , " 2

/

O

Dx Large position uncertainty; small momentum uncertainty

Equation (17.17) is one form of the Heisenberg uncertainty principle, first discovered by the German physicist Werner Heisenberg (1901–1976). It states that, in general, it is impossible to simultaneously determine both the position and the momentum of a particle with arbitrarily great precision, as classical physics would predict. Instead, the uncertainties in the two quantities play complementary roles, as we have described. Figure 17.18 shows the relationship between the two uncertainties. Our derivation of Eq. (17.16), a less refined form of the uncertainty principle given by Eq. (17.17), shows that this principle has its roots in the wave aspect of photons. We will see in Chapter 18 that electrons and other subatomic particles also have a wave aspect, and the same uncertainty principle applies to them as well. It is tempting to suppose that we could get greater precision by using more sophisticated detectors of position and momentum. This turns out not to be possible. To detect a particle, the detector must interact with it, and this interaction unavoidably changes the state of motion of the particle, introducing uncertainty about its original state. For example, we could imagine placing an electron at a certain point in the middle of the slit in Fig. 17.17. If the photon passes through the middle, we would see the electron recoil. We would then know that the photon passed through that point in the slit, and we would be much more certain about the x-coordinate of the photon. However, the collision between the photon and the electron would change the photon momentum, giving us greater uncertainty in the value of that momentum. A more detailed analysis of such hypothetical experiments shows that the uncertainties we have described are fundamental and intrinsic. They cannot be circumvented even in principle by any experimental technique, no matter how sophisticated. There is nothing special about the x-axis. In a three-dimensional situation with coordinates 1x, y, z2 there is an uncertainty relationship for each coordinate and its corresponding momentum component: ¢x¢px Ú U>2, ¢y¢py Ú U>2, and ¢z¢pz Ú U>2. However, the uncertainty in one coordinate is not related to the uncertainty in a different component of momentum. For example, ¢x is not related directly to ¢py. '

'

'

Waves and Uncertainty Here’s an alternative way to understand the Heisenberg uncertainty principle in terms of the properties of waves. Consider a sinusoidal electromagnetic wave propagating in the positive x-direction with its electric field polarized in the y-direction. If the wave has wavelength l , frequency ƒ, and amplitude A, we can write the wave function as '

Ey1x, t2 = A sin1kx - vt2

(17.18)

In this expression the wave number is k = 2p>l and the angular frequency is v = 2pƒ. We can think of the wave function in Eq. (17.18) as a description of a photon with a definite wavelength and a definite frequency. In terms of k and v we can express the momentum and energy of the photon as (photon momentum in terms of h h 2p px = = = Uk (17.19a) wave number) l 2p l

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17.4 Wave–Particle Duality, Probability, and Uncertainty

E = hƒ =

h 2pƒ = Uv 2p

(photon energy in terms of angular frequency)

(17.19b)

Using Eqs. (17.19) in Eq. (17.18), we can rewrite our photon wave equation as (wave function for a photon with x-momentum px and energy E)

Ey1x, t2 = A sin31px x - Et2>U4

(17.20)

Since this wave function has a definite value of x-momentum px, there is no uncertainty in the value of this quantity: ¢px = 0. The Heisenberg uncertainty principle, Eq. (17.17), says that ¢x ¢px Ú U>2. If ¢px is zero, then ¢x must be infinite. Indeed, the wave described by Eq. (17.20) extends along the entire x-axis and has the same amplitude everywhere. The price we pay for knowing the photon’s momentum precisely is that we have no idea where the photon is! In practical situations we always have some idea where a photon is. To describe this situation, we need a wave function that is more localized in space. We can create one by superimposing two or more sinusoidal functions. To keep things simple, we’ll consider only waves propagating in the positive x-direction. For example, let’s add together two sinusoidal wave functions like those in Eqs. (17.18) and (17.20), but with slightly different wavelengths and frequencies and hence slightly different values px1 and px2 of x-momentum and slightly different values E1 and E2 of energy. The total wave function is '

Ey1x, t2 = A1 sin31p1x x - E1t2>U4 + A2 sin31p2x x - E2 t2>U4

(17.21)

Consider what this wave function looks like at a particular instant of time, say t = 0. At this instant Eq. (17.21) becomes Ey1x, t = 02 = A1 sin1p1x x>U2 + A2 sin1p2x x>U2

(17.22)

Figure 17.19a is a graph of the individual wave functions at t = 0 for the case A2 = -A1, and Fig. 17.19b graphs the combined wave function Ey1x, t = 02 given by Eq. (17.22). We saw something very similar to Fig. 17.19b in our discussion of beats in Section 16.7: When we superimposed two sinusoidal waves with slightly different frequencies (see Fig. 16.24), the resulting wave exhibited amplitude variations not present in the original waves. In the same way, a photon represented by the wave function in Eq. (17.21) is most likely to be found in the regions where the wave function’s amplitude is greatest. That is, the photon is localized. However, the photon’s momentum no longer has a definite value because we began with two different x-momentum values, px1 and px2. This agrees with the Heisenberg uncertainty principle: By decreasing the uncertainty in the photon’s position, we have increased the uncertainty in its momentum. 17.19 (a) Two sinusoidal waves with slightly different wave numbers k and hence slightly different values of momentum px = Uk shown at one instant of time. (b) The superposition of these waves has a momentum equal to the average of the two individual values of momentum. The amplitude varies, giving the total wave a lumpy character not possessed by either individual wave. Ey(x) (a) 0

x

Ey(x)

(b) 0

x

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561

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CHAPTER 17 Photons: Light Waves Behaving as Particles

Uncertainty in Energy Our discussion of combining waves also shows that there is an uncertainty principle that involves energy and time. To see why this is so, imagine measuring the combined wave function described by Eq. (17.21) at a certain position, say x = 0, over a period of time. At x = 0, the wave function from Eq. (17.21) becomes Ey1x, t2 = A1 sin1- E1 t>U2 + A2 sin1- E2 t>U2 = -A1 sin1E1 t>U2 - A2 sin1E2 t>U2

(17.23)

What we measure at x = 0 is a combination of two oscillating electric fields with slightly different angular frequencies v1 = E1>U and v2 = E2>U. This is exactly the phenomenon of beats that we discussed in Section 16.7 (compare Fig. 16.24). The amplitude of the combined field rises and falls, so the photon described by this field is localized in time as well as in position. The photon is most likely to be found at the times when the amplitude is large. The price we pay for localizing the photon in time is that the wave does not have a definite energy. By contrast, if the photon is described by a sinousoidal wave like that in Eq. (17.20) that does have a definite energy E but that has the same amplitude at all times, we have no idea when the photon will appear at x = 0. So the better we know the photon’s energy, the less certain we are of when we will observe the photon. Just as for the momentum–position uncertainty principle, we can write a mathematical expression for the uncertainty principle that relates energy and time. In fact, except for an overall minus sign, Eq. (17.23) is identical to Eq. (17.22) if we replace the x-momentum px by energy E and the position x by time t. This tells us that in the momentum–position uncertainty relation, Eq. (17.17), we can replace the momentum uncertainty ¢px with the energy uncertainty ¢E and replace the position uncertainty ¢x with the time uncertainty ¢t. The result is ¢t ¢E Ú U>2

(Heisenberg uncertainty principle for energy and time)

(17.24)

In practice, any real photon has a limited spatial extent and hence passes any point in a limited amount of time. The following example illustrates how this affects the momentum and energy of the photon. Example 17.7

Ultrashort laser pulses and the uncertainty principle

Many varieties of lasers emit light in the form of pulses rather than a steady beam. A tellurium–sapphire laser can produce light at a wavelength of 800 nm in ultrashort pulses that last only 4.00 * 10-15 s (4.00 femtoseconds, or 4.00 fs). The energy in a single pulse produced by one such laser is 2.00 mJ = 2.00 * 10-6 J, and the pulses propagate in the positive x-direction. Find (a) the frequency of the light; (b) the energy and minimum energy uncertainty of a single photon in the pulse; (c) the minimum frequency uncertainty of the light in the pulse; (d) the spatial length of the pulse, in meters and as a multiple of the wavelength; (e) the momentum and minimum momentum uncertainty of a single photon in the pulse; and (f ) the approximate number of photons in the pulse.

the pulse, since we don’t know when during the pulse that photon emerges. Similarly, the position uncertainty of a photon is the spatial length of the pulse, since a given photon could be found anywhere within the pulse. To find our target variables, we’ll use the relationships for photon energy and momentum from Section 17.1 and the two Heisenberg uncertainty principles, Eqs. (17.17) and (17.24). EXECUTE: (a) From the relationship c = lƒ, the frequency of 800-nm light is 3.00 * 108 m > s c = 3.75 * 1014 Hz = l 8.00 * 10-7 m '

ƒ =

'

'

(b) From Eq. (17.2) the energy of a single 800-nm photon is SOLUTION

E = hƒ = 16.626 * 10-34 J # s213.75 * 1014 Hz2

'

IDENTIFY and SET UP: It’s important to distinguish between the light pulse as a whole (which contains a very large number of photons) and an individual photon within the pulse. The 5.00-fs pulse duration represents the time it takes the pulse to emerge from the laser; it is also the time uncertainty for an individual photon within

= 2.48 * 10

-19

J

The time uncertainty equals the pulse duration, ¢t = 4.00 * 10-15 s . From Eq. (17.24) the minimum uncertainty in energy corresponds to the case ¢t ¢E = U>2, so

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17.4 Wave–Particle Duality, Probability, and Uncertainty

¢E =

U 1.055 * 10-34 J # s = 1.32 * 10-20 J = 2¢t 214.00 * 10-15 s2 '

This is 5.3% of the photon energy E = 2.48 * 10-19 J, so the energy of a given photon is uncertain by at least 5.3%. The uncertainty could be greater, depending on the shape of the pulse. (c) From the relationship ƒ = E>h, the minimum frequency uncertainty is ¢ƒ =

1.32 * 10-20 J ¢E = = 1.99 * 1013 Hz h 6.626 * 10-34 J # s '

This is 5.3% of the frequency ƒ = 3.75 * 1014 Hz we found in part (a). Hence these ultrashort pulses do not have a definite frequency; the average frequency of many such pulses will be 3.75 * 1014 Hz , but the frequency of any individual pulse can be anywhere from 5.3% higher to 5.3% lower. (d) The spatial length ¢x of the pulse is the distance that the front of the pulse travels during the time ¢t = 4.00 * 10-15 s it takes the pulse to emerge from the laser: '

'

¢x = c¢t = 13.00 * 108 m > s214.00 * 10-15 s2 '

= 1.20 * 10

¢x =

-6

'

''

m

1.20 * 10-6 m 8.00 * 10-7 m>wavelength

= 1.50 wavelengths '

This justifies the term ultrashort. The pulse is less than two wavelengths long! (e) From Eq. (17.5), the momentum of an average photon in the pulse is px =

E 2.48 * 10-19 J = 8.28 * 10-28 kg # m > s = c 3.00 * 108 m > s ''

'

'

'

'

563

The spatial uncertainty is ¢x = 1.20 * 10-6 m . From Eq. (17.17) minimum momentum uncertainty corresponds to ¢x ¢px = U>2, so ¢px =

1.055 * 10-34 J # s U = = 4.40 * 10-29 kg # m > s 2¢x 211.20 * 10-6 m2 ''

'

'

This is 5.3% of the average photon momentum px. An individual photon within the pulse can have a momentum that is 5.3% greater or less than the average. (f ) To estimate the number of photons in the pulse, we divide the total pulse energy by the average photon energy: 2.00 * 10-6 J>pulse 2.48 * 10-19 J>photon

= 8.06 * 1012 photons>pulse '

The energy of an individual photon is uncertain, so this is the average number of photons per pulse. EVALUATE: The percentage uncertainties in energy and momentum are large because this laser pulse is so short. If the pulse were longer, both ¢t and ¢x would be greater and the corresponding uncertainties in photon energy and photon momentum would be smaller. Our calculation in part (f ) shows an important distinction between photons and other kinds of particles. In principle it is possible to make an exact count of the number of electrons, protons, and neutrons in an object such as this book. If you repeated the count, you would get the same answer as the first time. By contrast, if you counted the number of photons in a laser pulse you would not necessarily get the same answer every time! The uncertainty in photon energy means that on each count there could be a different number of photons whose individual energies sum to 2.00 * 10-6 J. That’s yet another of the many strange properties of photons.

Test Your Understanding of Section 17.4 Through which of the following angles is a photon of wavelength l most likely to be deflected after passing through a slit of width a? Assume that l is much less than a. (i) u = l>a; (ii) u = 3l>2a; (iii) u = 2l>a; (iv) u = 3l>a; (v) not enough information given to decide. y

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CHAPTER

17

SUMMARY hc l

Photons: Electromagnetic radiation behaves as both waves and particles. The energy in an electromagnetic wave is carried in units called photons. The energy E of one photon is proportional to the wave frequency ƒ and inversely proportional to the wavelength l, and is proportional to a universal quantity h called Planck’s constant. The momentum of a photon has magnitude E > c. (See Example 17.1.)

E = hf =

The photoelectric effect: In the photoelectric effect, a surface can eject an electron by absorbing a photon whose energy hƒ is greater than or equal to the work function f of the material. The stopping potential V0 is the voltage required to stop a current of ejected electrons from reaching an anode. (See Examples 17.2 and 17.3.)

eV0 = hƒ - f

'

p =

(17.2)

hf E h = = c c l

(17.5)

'

(17.4)

Monochromatic light – S

Cathode

v

E

Anode i

v –

G 2e

i + E

Photon production, photon scattering, and pair production: X rays can be produced when electrons accelerated to high kinetic energy across a potential increase VAC strike a target. The photon model explains why the maximum frequency and minimum wavelength produced are given by Eq. (17.6). (See Example 17.4.) In Compton scattering a photon transfers some of its energy and momentum to an electron with which it collides. For free electrons (mass m2, the wavelengths of incident and scattered photons are related to the photon scattering angle f by Eq. (17.7). (See Example 17.5.) In pair production a photon of sufficient energy can disappear and be replaced by an electron–positron pair. In the inverse process, an electron and a positron can annihilate and be replaced by a pair of photons. (See Example 17.6.)

The Heisenberg uncertainty principle: It is impossible to determine both a photon’s position and its momentum at the same time to arbitrarily high precision. The precision of such measurements for the x-components is limited by the Heisenberg uncertainty principle, Eq. (17.17); there are corresponding relationships for the y- and z-components. The uncertainty ¢E in the energy of a state that is occupied for a time ¢t is given by Eq. (17.24). In these expressions, U = h > 2p. (See Example 17.7.) '

eVAC = hƒmax =

hc lmin

(17.6) Incident photon: wavelength l, S momentum p

(bremsstrahlung) h 11 - cos f2 mc (Compton scattering) l¿ - l =

Scattered photon: wavelength lr, S momentum pr f S

Pe

(17.7) Recoiling electron: S momentum Pe

¢x ¢px Ú U>2 (Heisenberg uncertainty principle for position and momentum)

(17.17)

¢t ¢E Ú U>2 (Heisenberg uncertainty principle for energy and time)

(17.24)

Dpx

Allowed: Dx Dpx $ "/2 Dx Dpx 5 "/2 Impossible: Dx Dpx , "/2 O

'

564

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Dx

Discussion Questions

BRIDGING PROBLEM

565

Compton Scattering and Electron Recoil

An incident x-ray photon is scattered from a free electron that is initially at rest. The photon is scattered straight back at an angle of 180° from its initial direction. The wavelength of the scattered photon is 0.0830 nm. (a) What is the wavelength of the incident photon? (b) What are the magnitude of the momentum and the speed of the electron after the collision? (c) What is the kinetic energy of the electron after the collision? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

IDENTIFY and SET UP 1. In this problem a photon is scattered by an electron initially at rest. In Section 17.3 you learned how to relate the wavelengths of the incident and scattered photons; in this problem you must also find the momentum, speed, and kinetic energy of the recoiling electron. You can find these because momentum and energy are conserved in the collision. 2. Which key equation can be used to find the incident photon wavelength? What is the photon scattering angle f in this problem?

Problems

EXECUTE 3. Use the equation you selected in step 2 to find the wavelength of the incident photon. 4. Use momentum conservation and your result from step 3 to find the momentum of the recoiling electron. (Hint: All of the momentum vectors are along the same line, but not all point in the same direction. Be careful with signs.) 5. Find the speed of the recoiling electron from your result in step 4. (Hint: Assume that the electron is nonrelativistic, so you can use the relationship between momentum and speed from Chapter 8. This is acceptable if the speed of the electron is less than about 0.1c. Is it?) 6. Use your result from step 4 or step 5 to find the electron kinetic energy. EVALUATE 7. You can check your answer in step 6 by finding the difference between the energies of the incident and scattered photons. Is your result consistent with conservation of energy?

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q17.1 In what ways do photons resemble other particles such as electrons? In what ways do they differ? Do photons have mass? Do they have electric charge? Can they be accelerated? What mechanical properties do they have? Q17.2 There is a certain probability that a single electron may simultaneously absorb two identical photons from a high-intensity laser. How would such an occurrence affect the threshold frequency and the equations of Section 17.1? Explain. Q17.3 According to the photon model, light carries its energy in packets called quanta or photons. Why then don’t we see a series of flashes when we look at things? Q17.4 Would you expect effects due to the photon nature of light to be generally more important at the low-frequency end of the electromagnetic spectrum (radio waves) or at the high-frequency end (x rays and gamma rays)? Why? Q17.5 During the photoelectric effect, light knocks electrons out of metals. So why don’t the metals in your home lose their electrons when you turn on the lights? Q17.6 Most black-and-white photographic film (with the exception of some special-purpose films) is less sensitive to red light than blue light and has almost no sensitivity to infrared. How can these properties be understood on the basis of photons? Q17.7 Human skin is relatively insensitive to visible light, but ultraviolet radiation can cause severe burns. Does this have anything to do with photon energies? Explain.

Q17.8 Explain why Fig. 17.4 shows that most photoelectrons have kinetic energies less than hƒ - f, and also explain how these smaller kinetic energies occur. Q17.9 In a photoelectric-effect experiment, the photocurrent i for large positive values of VAC has the same value no matter what the light frequency ƒ (provided that ƒ is higher than the threshold frequency ƒ0). Explain why. Q17.10 In an experiment involving the photoelectric effect, if the intensity of the incident light (having frequency higher than the threshold frequency) is reduced by a factor of 10 without changing anything else, which (if any) of the following statements about this process will be true? (a) The number of photoelectrons will most likely be reduced by a factor of 10. (b) The maximum kinetic energy of the ejected photoelectrons will most likely be reduced by a factor of 10. (c) The maximum speed of the ejected photoelectrons will most likely be reduced by a factor of 10. (d) The maximum speed of the ejected photoelectrons will most likely be reduced by a factor of 110. (e) The time for the first photoelectron to be ejected will be increased by a factor of 10. Q17.11 The materials called phosphors that coat the inside of a fluorescent lamp convert ultraviolet radiation (from the mercuryvapor discharge inside the tube) into visible light. Could one also make a phosphor that converts visible light to ultraviolet? Explain. Q17.12 In a photoelectric-effect experiment, which of the following will increase the maximum kinetic energy of the photoelectrons? (a) Use light of greater intensity; (b) use light of higher

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CHAPTER 17 Photons: Light Waves Behaving as Particles

frequency; (c) use light of longer wavelength; (d) use a metal surface with a larger work function. In each case justify your answer. Q17.13 Why must engineers and scientists shield against x-ray production in high-voltage equipment?

SINGLE CORRECT Q17.1 In a photoelectric effect, electrons are emitted (a) at a rate that is proportional to the square of the amplitude of the incident radiation (b) with a maximum velocity proportional to the frequency of the incident radiation (c) at a rate that is independent of the emitter (d) only if the frequency of the incident radiation is above a certain threshold value Q17.2 A nonmonochromatic light is used in an experiment on photoelectric effect. The stopping potential. (a) is related to the mean wavelength (b) is related to the shortest wavelength (c) is related to the longest wavelength (d) is not related to the wavelength. Q17.3 By increasing the intensity of incident light keeping frequency (v > v0) fixed on the surface of metal (a) kinetic energy of the photoelectrons increases (b) number of emitted electrons increases (c) kinetic energy and number of electrons increases (d) no effect Q17.4 The maximum energies of photoelectrons emitted from various metals are measured at a number of frequencies of incident light. If the graphs of maximum energy against frequency are plotted or each metal on the same set of axes, all of the graphs (a) cut the frequency axis at the same point (b) pass through the origin (c) are parallel to each other (d) flatten out to approach the same maximum energy value Q17.5 The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation : (a) The intensity increases (b) The minimum wavelength increases (c) The intensity remains unchanged (d) The minimum wavelength decreases.

ONE OR MORE CORRECT Q17.6 Photoelectric effect supports quantum nature of light because (a) there is minimum frequency of light below which no photoelectrons are emitted (b) the maximum kinetic energy of photo-electrons depends only on the frequency of light and not on its intensity (c) even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediately (d) electric charge of photo-electrons is quantized Q17.7 Which of the following are not dependent on the intensity of the incident radiation in a photoelectric experiment? (a) Amount of photoelectric current (b) Stopping potential to reduce the photoelectric current to zero (c) Work function of the surface (d) Maximum kinetic energy of photoelectrons

MATCH THE COLUMN Q17.8 Consider a typical experimental setup of photelectric effect. Match entries of column-I with those of column-II on which it may depend. Column-I (a) Threshold frequency

Column-II (P) Wavelength of incident light (b) Maximum kinetic energy of photo (Q) Intensity of light electron (just after their ejection) (c) Saturation photo current (d) Stopping potential

(R) Material of cathode (S) Potential difference applied accross the tube (T)Work function

EXERCISES Section 17.1 Light Absorbed as Photons: The Photoelectric Effect 17.1 (a) A proton is moving at a speed much slower than the speed of light. It has kinetic energy K1 and momentum p1. If the momentum of the proton is doubled, so p2 = 2p1, how is its new kinetic energy K2 related to K1? (b) A photon with energy E1 has momentum p1. If another photon has momentum p2 that is twice p1, how is the energy E2 of the second photon related to E1? 17.2 A photon of green light has a wavelength of 600 nm. Find the photon’s frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts. 17.3 The graph in Fig. E17.3 shows the stopping potential as a Figure E17.3 Stopping potential (V) 7.0 6.0 5.0 4.0 3.0 2.0 1.0 O

0.50 1.0

1.5

2.0

2.5

3.0

f 1 1015 Hz2

function of the frequency of the incident light falling on a metal surface. (a) Find the photoelectric work function for this metal. (b) What value of Planck’s constant does the graph yield? (c) Why does the graph not extend below the x-axis? (d) If a different metal were used, which characteristics of the graph would you expect to be the same and which ones would be different? 17.4 .. What would the minimum work function for a metal have to be for visible light (380–700 nm) to eject photoelectrons? 17.5 .. When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0 nm falls on the same surface? 17.6 .. The photoelectric work function of potassium is 2.3 eV. If light having a wavelength of 250 nm falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy in electron volts of the most energetic electrons ejected; (c) the speed of these electrons.

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Answers

Section 17.2 Light Emitted as Photons: X-Ray Production

17.7 . The cathode-ray tubes that generated the picture in early color televisions were sources of x rays. If the acceleration voltage in a television tube is 15.0 kV, what are the shortest-wavelength x rays produced by the television? (Modern televisions contain shielding to stop these x rays.) 17.8 .. (a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x rays with a wavelength of 0.150 nm? (b) What is the shortest wavelength produced in an x-ray tube operated at 30.0 kV?

Section 17.3 Light Scattered as Photons: Compton Scattering and Pair Production

17.9 . X rays are produced in a tube operating at 18.0 kV. After emerging from the tube, x rays with the minimum wavelength produced strike a target and are Compton-scattered through an angle of 45.0°. (a) What is the original x-ray wavelength? (b) What is the wavelength of the scattered x rays? (c) What is the energy of the scattered x rays (in electron volts)? 17.10 .. A photon scatters in the backward direction 1f = 180°2 from a free proton that is initially at rest. What must the wavelength of the incident photon be if it is to undergo a 10.0% change in wavelength as a result of the scattering?

PROBLEMS 17.11 .. (a) If

the average frequency emitted by a 200-W light bulb is 5.00 * 1014 Hz, and 10.0% of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second? (b) At what distance would this correspond to 1.00 * 1011 visible-light photons per square centimeter per second if the light is emitted uniformly in all directions? 17.12 . A 2.50-W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 eV. Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal? (b) How many photoelectrons are ejected each second from this metal? (c) If the power of the light beam, but not its wavelength, were reduced by half, what would be the answer to part (b)? (d) If the wavelength of the beam, but not its power, were reduced by half, what would be the answer to part (b)? 17.13 . The photoelectric work functions for particular samples of certain metals are as follows: cesium, 2.1 eV; copper, 4.7 eV; potassium, 2.3 eV; and zinc, 4.3 eV. (a) What is the threshold wavelength for each metal surface? (b) Which of these metals could not emit photoelectrons when irradiated with visible light (380–750 nm)? 17.14 .. CP An x-ray tube is operating at voltage V and current I. (a) If only a fraction p of the electric power supplied is converted into x rays, at what rate is energy being delivered to the target? (b) If the target has mass m and specific heat c (in J > kg # K), at what '

567

average rate would its temperature rise if there were no thermal losses? (c) Evaluate your results from parts (a) and (b) for an x-ray tube operating at 26 kV and 60.0 mA that converts 1.0% of the electric power into x rays. Assume that the 0.250-kg target is made of lead 1c = 130 J > kg # K2. (d) What must the physical properties of a practical target material be? What would be some suitable target elements? 17.15 .. A photon with wavelength 0.1100 nm collides with a free electron that is initially at rest. After the collision the wavelength is 0.1132 nm. (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon? 17.16 Polychromatic light described at a place by the equation E = 100[sin(0.5p×1015t) + cos(p×1015t) + sin(2p×1015t)] where E is in V/m and t in sec, falls on a metal surface having work function 2.0 eV. Find the maximum kinetic energy of the photoelectron [Take h = planck’s constant = 6.4 * 10 - 34J - s] 17.17 A potassium surface is placed 75 cm away from a 100 W bulb. It is found that the energy radiated by the bulb is 5% of the input power. Consider each K-atom as a circular disc of diameter 1Å. Determine the time required for each atom to absorb an amount of energy equal to its work function of 2.0 eV considering wave nature of light. 17.18 An isolated copper ball of radius 9 mm having work function 4.47 eV is illuminated by electromagnetic radiation of wavelength 1243 Å. Find the maximum charge on the ball. 17.19 A parallel beam of uniform monochromatic light of wavelength 546 nm has an intensity of 200 w/m2. Find the number of photons in 1 mm3 of this radiation. 17.20 A point source of P = 12 W is located on the axis of a circular mirror plate of radius R = 3 cm. If distance of source from the plate is a = 39 cm and reflection coefficient of mirror plate is a = 0.7 , calculate force exerted by light rays on the plate. '

'

CHALLENGE PROBLEM 17.21 ... Consider Compton

scattering of a photon by a moving electron. Before the collision the photon has wavelength l and is moving in the +x-direction, and the electron is moving in the -x-direction with total energy E (including its rest energy mc2). The photon and electron collide headon. After the collision, both are moving in the -x-direction (that is, the photon has been scattered by 180°). (a) Derive an expression for the wavelength l¿ of the scattered photon. Show that if E W mc2, where m is the rest mass of the electron, your result reduces to l¿ =

m2c4l hc a1 + b E 4hcE

'

Answers Chapter Opening Question

?

The energy of a photon E is inversely proportional to its wavelength l: The shorter the wavelength, the more energetic is the pho-

ton. Since visible light has shorter wavelengths than infrared light, the headlamp emits photons of greater energy. However, the light from the infrared laser is far more intense (delivers much more energy per second per unit area to the patient’s skin) because it

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568

CHAPTER 17 Photons: Light Waves Behaving as Particles

emits many more photons per second than does the headlamp and concentrates them onto a very small spot.

Test Your Understanding Questions 17.1 (i) and (ii) From Eq. (17.2), a photon of energy E = 1.14 eV has wavelength l = hc > E = 14.136 * 10-15 eV2 (3.00 * 108) m/s·>11.14 eV2 = 1.09 * 10-6 m = 1090 nm.This is in the infrared part of the spectrum. Since wavelength is inversely proportional to photon energy, the minimum photon energy of 1.14 eV corresponds to the maximum wavelength that causes photoconductivity in silicon. Thus the wavelength must be 1090 nm or less. 17.2 (ii) Equation (17.6) shows that the minimum wavelength of x rays produced by bremsstrahlung depends on the potential difference VAC but does not depend on the rate at which electrons strike the anode. Each electron produces at most one photon, so increasing the number of electrons per second causes an increase in the number of x-ray photons emitted per second (that is, the x-ray intensity). 17.3 yes, no Equation (17.7) shows that the wavelength shift ¢l = l¿ - l depends only on the photon scattering angle f, not on the wavelength of the incident photon. So a visible-light photon scattered through an angle f undergoes the same wavelength shift as an x-ray photon. Equation (17.7) also shows that this shift is of the order of h > mc = 2.426 * 10-12 m = 0.002426 nm. This is a few percent of the wavelength of x rays (see Example 17.5), so the effect is noticeable in x-ray scattering. However, h > mc is a tiny fraction of the wavelength of visible light (between 380 and 750 nm). The human eye cannot distinguish such minuscule differences in wavelength (that is, differences in color). 17.4 (ii) There is zero probability that a photon will be deflected by one of the angles where the diffraction pattern has zero intensity. These angles are given by a sin u = ml with m = 61, 62, 63, Á . Since l is much less than a, we can write these angles as u = ml>a = 6l>a, 62l>a, 63l>a, Á . These values include answers (i), (iii), and (iv), so it is impossible for a photon to be deflected through any of these angles. The intensity is not zero at u = 3l>2a (located between two zeros in the diffraction pattern), so there is some probability that a photon will be deflected through this angle. '

'

'

'

'

'

Bridging Problem Answers: (a) 0.0781 nm (b) 1.65 * 10-23 kg # m>s, 1.81 * 107 m>s (c) 1.49 * 10-16 J

Discussion Questions Q17.1 Electron radius is considered to be about 10 - 15 meter but high energy photons have wavelength 10 - 11 meter. These photons can be treated as particles of this size. Photons have no charge electrons are charged. photons have no mass but me = 9 * 10 - 31kg. Photons can't be accelerated. Electrons can be accelerated. Photons have momentum but not in the sense of product of mass and velocity. Q17.2 There are a few categories of laser which can show the phenomenon of two photon absorption. According to Einestein photons can ionize an atom if they have energy greater than particular threshold energy corresponding to ionization energy of an atom.

An atom absorbing multiple photons simultaneously might be ionized by photons with energy less than the threshold energy but with intense beam of radiation like special laser named Nd-YAG laser. Q17.3 One of the reason behind this is the perceptibility of vision. A household light source will emit more than a million-million photons every second. So with these many photons our vision gets smoothed out with continuity suppressing the effect of quantized nature of light. Q17.4 Infact high frequency photons are more important for photon nature of light. At high frequencies energy is very high and photoelectric effect is caused by individual photons. The energy of the photons must be higher than the energy which keeps the electrons bounded with metal plate. Q17.5 For metals to lose the electrons due to incident light, the frequency of falling light must be greater than the threshold frequencies of metals. Q17.6 Blue light photons have less wavelength hence high frequency so they can affect the photographic plates abruptly. In some cases even red light-photons are harmful for photographic plates. In most of the black and white films we can not use red light of wavelength approximately 620 nm. Most of the black and white photos are "Panchromatic" means sensitive from 250 nm (UV) till 680 (dark red). Q17.7 UV photons (light) are more energetic than visible light. Our skin only evolved to handle certain amount of UV light but it is for visible light. The exposure to high UV radiations can cause severe burns while very little UV light can help the skin to make vitamin D. Q17.8 The photoelectrons on the surface which are having almost no-binding energy with metal plates can have maximum KE(hf - f), but other inner electrons lose some energy due to neighbouring electrons and other factors. Q17.9 This is because photo current can be changed by changing the intensity not only by frequency. Intensity does not appear in eVAC = hfmax Q17.10 Intensity is I = hf . A¢t Number of photoelectrons emitted is directly proportional to the intensity of incident radiation so (a) is true. The energy of photoelectrons does not change with intensity so (b), (c), (d), (e) are false. Q17.11 No, as photons of ultra violet light have more energy than that of visible light. Q17.12 (a) No, KE is independent of intensity of incident radiation. (b) hv = hv0 + (KE)max So, increasing v, KE can be increased. (c) No, (longer wavelength means lesser frequency (KE)max can not be increased) (d) Yes (KE)max = hv - (work function) So lesser work function greater KE. Q17.13 Any vacuum tube (or equipment operating on high voltage) operating on several thousand volts or more can produce x-rays and unwanted byproduct. They can effect human organs dangerously which may not be immediately detectable. So scientists and engineers must take care of these equipments with proper safely measures.

Single Correct Q17.1 (d) Q17.2 (b) Q17.3 (b)

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Answers

Q17.4 (c) Q17.5 (a)

Section 17.3 Light Scattered as Photons: Compton Scattering and Pair Production

One or More Correct

17.9 (a) 0.069nm (b) 0.0698nm (c) 17.8 keV

Q17.6 (a), (b), (c) Q17.7 (b), (c), (d)

Match the Column Q17.8 (a) R, T (b) P, R, T (c) Q (d) P, R, T

EXERCISES Section 17.1 Light Absorbed as Photons: The Photoelectric Effect 17.1 (a) K2 = 4K1 (b) E2 = 2E1 17.2 3.32 * 10 - 19J, 2.07eV 17.3 (a) 4.8eV (b) 6.1 * 10 - 34J.s (c) Slope same, but different intercepts with horizontal axis. 17.4 1.77eV 17.5 2.14eV 17.6 (a) 2.7 volts (b) 2.7eV (c) 9.7 * 105m/s

Section 17.2 Light Emitted as Photons: X-Ray Production 17.7 0.0829nm 17.8 (a) 8.29 kv (b) 0.0414 nm.

569

17.10 (a) 2.65 * 10 - 14m.

PROBLEMS 17.11 (a) 6.03 * 1019photons/sec (b) 69.3m 17.12 (a) 5.85eV (b) 1.56 * 1018 (c) 7.8 * 1017 (d) 7.8 * 1017 17.13 (a) 590nm (Cesium), 264nm (Copper), 539nm (potassium) and 288nm (zinc). (b) copper and zinc. 17.14 (a) 15.6W (b) 0.48 k/sec 17.15 (a) 320eV, 1.05 * 107m/s (b) 3.98nm 17.16 2 eV 17.17 57.6 sec 17.18 5.5 * 10 - 12C 17.19 1830 17.20 10 - 10N

CHALLENGE PROBLEM 17.21 (a)

l(E - Pc) + 2hc E + Pc

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18

PARTICLES BEHAVING AS WAVES T bacteriophage virus Viral DNA

LEARNING GOALS By studying this chapter, you will learn: • De Broglie’s proposal that electrons, protons, and other particles can behave like waves. • How electron diffraction experiments provided evidence for de Broglie’s ideas. • How electron microscopes can provide much higher magnification than visible-light microscopes.

100 nm 5 0.1 mm

?

• How physicists discovered the atomic nucleus. • How Bohr’s model of electron orbits explained the spectra of hydrogen and hydrogenlike atoms.

Viruses (shown in blue) have landed on an E. coli bacterium and injected their DNA, converting the bacterium into a virus factory. This false-color image was made using a beam of electrons rather than a light beam. What properties of electrons make them useful for imaging such fine details?

n Chapter 38 we discovered one aspect of nature’s wave–particle duality: Light and other electromagnetic radiation act sometimes like waves and sometimes like particles. Interference and diffraction demonstrate wave behavior, while emission and absorption of photons demonstrate particle behavior. If light waves can behave like particles, can the particles of matter behave like waves? As we will discover, the answer is a resounding yes. Electrons can be made to interfere and diffract just like other kinds of waves. We will see that the wave nature of electrons is not merely a laboratory curiosity: It is the fundamental reason why atoms, which according to classical physics should be profoundly unstable, are able to exist. In this chapter we’ll use the wave nature of matter to help us understand the structure of atoms, the operating principles of a laser, and the curious properties of the light emitted by a heated, glowing object. Without the wave picture of matter, there would be no way to explain these phenomena. In Chapter 40 we’ll introduce an even more complete wave picture of matter called quantum mechanics. Through the remainder of this book we’ll use the ideas of quantum mechanics to understand the nature of molecules, solids, atomic nuclei, and the fundamental particles that are the building blocks of our universe.

I

18.1

Electron Waves

In 1924 a French physicist and nobleman, Prince Louis de Broglie (pronounced “de broy”; Fig. 18.1), made a remarkable proposal about the nature of matter. His reasoning, freely paraphrased, went like this: Nature loves symmetry. Light is dualistic in nature, behaving in some situations like waves and in others like particles. If nature is symmetric, this duality should also hold for matter. Electrons and protons, which we usually think of as particles, may in some situations behave like waves. 570

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18.1 Electron Waves

If a particle acts like a wave, it should have a wavelength and a frequency. De Broglie postulated that a free particle with rest mass m, moving with nonrelativistic speed v, should have a wavelength l related to its momentum p = mv in exactly the same way as for a photon, as expressed by Eq. (38.5): l = h > p. The de Broglie wavelength of a particle is then '

l =

h h = p mv

(de Broglie wavelength of a particle)

'

(18.1)

571

18.1 Louis-Victor de Broglie, the seventh Duke de Broglie (1892–1987), broke with family tradition by choosing a career in physics rather than as a diplomat. His revolutionary proposal that particles have wave characteristics—for which de Broglie won the 1929 Nobel Prize in physics—was published in his doctoral thesis.

where h is Planck’s constant. If the particle’s speed is an appreciable fraction of the speed of light c, we use Eq. (37.27) to replace mv in Eq. (18.1) with gmv = mv > "1 - v2> c2. The frequency ƒ, according to de Broglie, is also related to the particle’s energy E in the same way as for a photon—namely, '

'

'

E = hƒ

(18.2)

Thus in de Broglie’s hypothesis, the relationships of wavelength to momentum and of frequency to energy are exactly the same for free particles as for photons. CAUTION Not all photon equations apply to particles with mass Be careful when applying the relationship E = hƒ to particles with nonzero rest mass, such as electrons and protons. Unlike a photon, they do not travel at speed c, so the equations ƒ = c > l and E = pc do not apply to them! y '

'

Observing the Wave Nature of Electrons De Broglie’s proposal was a bold one, made at a time when there was no direct experimental evidence that particles have wave characteristics. But within a few years of de Broglie’s publication of his ideas, they were resoundingly verified by a diffraction experiment with electrons. This experiment was analogous to those we described in Section 18.6, in which atoms in a crystal act as a three-dimensional diffraction grating for x rays. An x-ray beam is strongly reflected when it strikes a crystal at an angle that gives constructive interference among the waves scattered from the various atoms in the crystal. These interference effects demonstrate the wave nature of x rays. In 1927 the American physicists Clinton Davisson and Lester Germer, working at the Bell Telephone Laboratories, were studying the surface of a piece of nickel by directing a beam of electrons at the surface and observing how many electrons bounced off at various angles. Figure 18.2 shows an experimental setup like theirs. Like many ordinary metals, the sample was polycrystalline: It consisted of many randomly oriented microscopic crystals bonded together. As a result, the electron beam reflected diffusely, like light bouncing off a rough surface (see Fig. 33.6b), with a smooth distribution of intensity as a function of the angle u. 1 A heated filament emits electrons.

2 The electrons are accelerated by electrodes and directed at a crystal.

a

4 The detector can be moved to detect scattered electrons at any angle u.

PhET: Davisson-Germer: Electron Diffraction ActivPhysics 17.5: Electron Interference

18.2 An apparatus similar to that used by Davisson and Germer to discover electron diffraction.

b u Voltage source

Electron beam (in vacuum) 3 Electrons strike a nickel crystal.

Vba 5

Vb 2 V

a

Vba . 0, so electrons speed up in moving from a to b.

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572

CHAPTER 18 Particles Behaving as Waves

18.3 (a) Intensity of the scattered electron beam in Fig. 18.2 as a function of the scattering angle u. (b) Electron waves scattered from two adjacent atoms interfere constructively when d sin u = ml. In the case shown here, u = 50° and m = 1. (a)

This peak in the intensity of scattered electrons is due to constructive interference between electron waves scattered by different surface atoms.

I Vba 5 54 V

u 5 50°

u O

15°

(b)

If the scattered waves are in phase, there is a peak in the intensity of scattered electrons. Incident waves in phase

30°

45°

60°

75°

90°

During the experiment an accident occurred that permitted air to enter the vacuum chamber, and an oxide film formed on the metal surface. To remove this film, Davisson and Germer baked the sample in a high-temperature oven, almost hot enough to melt it. Unknown to them, this had the effect of creating large regions within the nickel with crystal planes that were continuous over the width of the electron beam. From the perspective of the electrons, the sample looked like a single crystal of nickel. When the observations were repeated with this sample, the results were quite different. Now strong maxima in the intensity of the reflected electron beam occurred at specific angles (Fig. 18.3a), in contrast to the smooth variation of intensity with angle that Davisson and Germer had observed before the accident. The angular positions of the maxima depended on the accelerating voltage Vba used to produce the electron beam. Davisson and Germer were familiar with de Broglie’s hypothesis, and they noticed the similarity of this behavior to x-ray diffraction. This was not the effect they had been looking for, but they immediately recognized that the electron beam was being diffracted. They had discovered a very direct experimental confirmation of the wave hypothesis. Davisson and Germer could determine the speeds of the electrons from the accelerating voltage, so they could compute the de Broglie wavelength from Eq. (18.1). If an electron is accelerated from rest at point a to point b through a potential increase Vba = Vb - Va as shown in Fig. 18.2, the work done on the electron eVba equals its kinetic energy K. Using K = 1122mv2 = p2 > 2m for a nonrelativistic particle, we have '

eVba =

p2 2m

'

p = !2meVba

We substitute this into Eq. (18.1), the expression for the de Broglie wavelength of the electron:

l u 5 50°

d Atoms on surface of crystal

18.4 X-ray and electron diffraction. The upper half of the photo shows the diffraction pattern for 71-pm x rays passing through aluminum foil. The lower half, with a different scale, shows the diffraction pattern for 600-eV electrons from aluminum. The similarity shows that electrons undergo the same kind of diffraction as x rays. Top: x-ray diffraction

Bottom: electron diffraction

l =

h h = p !2meVba

(de Broglie wavelength of an electron)

(18.3)

The greater the accelerating voltage Vba, the shorter the wavelength of the electron. To predict the angles at which strong reflection occurs, note that the electrons were scattered primarily by the planes of atoms near the surface of the crystal. Atoms in a surface plane are arranged in rows, with a distance d that can be measured by x-ray diffraction techniques. These rows act like a reflecting diffraction grating; the angles at which strong reflection occurs are the same as for a grating with center-to-center distance d between its slits (Fig. 18.3b). From Eq. (18.13) the angles of maximum reflection are given by d sin u = ml

1m = 1, 2, 3, Á 2

(18.4)

where u is the angle shown in Fig. 18.2. (Note that the geometry in Fig. 18.3b is different from that for Fig. 18.22, so Eq. (18.4) is different from Eq. (18.16).) Davisson and Germer found that the angles predicted by this equation, using the de Broglie wavelength given by Eq. (18.3), agreed with the observed values (Fig. 18.3a). Thus the accidental discovery of electron diffraction was the first direct evidence confirming de Broglie’s hypothesis. In 1928, just a year after the Davisson–Germer discovery, the English physicist G. P. Thomson carried out electron-diffraction experiments using a thin, polycrystalline, metallic foil as a target. Debye and Sherrer had used a similar technique several years earlier to study x-ray diffraction from polycrystalline specimens. In these experiments the beam passes through the target rather than being reflected from it. Because of the random orientations of the individual microscopic crystals in the foil, the diffraction pattern consists of intensity maxima forming rings around the direction of the incident beam. Thomson’s results again confirmed the de Broglie relationship. Figure 18.4 shows both x-ray and electron diffraction

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18.1 Electron Waves

573

patterns for a polycrystalline aluminum foil. (G. P. Thomson was the son of J. J. Thomson, who 31 years earlier discovered the electron. Davisson and the younger Thomson shared the 1937 Nobel Prize in physics for their discoveries.) Additional diffraction experiments were soon carried out in many laboratories using not only electrons but also various ions and low-energy neutrons. All of these are in agreement with de Broglie’s bold predictions. Thus the wave nature of particles, so strange in 1924, became firmly established in the years that followed.

Problem-Solving Strategy 18.1

Wavelike Properties of Particles

IDENTIFY the relevant concepts: Particles have wavelike properties. A particle’s (de Broglie) wavelength is inversely proportional to its momentum, and its frequency is proportional to its energy. SET UP the problem: Identify the target variables and decide which equations you will use to calculate them. EXECUTE the solution as follows: 1. Use Eq. (18.1) to relate a particle’s momentum p to its wavelength l; use Eq. (18.2) to relate its energy E to its frequency ƒ. 2. Nonrelativistic kinetic energy may be expressed as either K = 12 mv2 or (because p = mv) K = p2 > 2m. The latter form is useful in calculations involving the de Broglie wavelength. 3. You may express energies in either joules or electron volts, using h = 6.626 * 10-34 J # s or h = 4.136 * 10-15 eV # s as appropriate. '

Example 18.1

'

'

'

An electron-diffraction experiment

In an electron-diffraction experiment using an accelerating voltage of 54 V, an intensity maximum occurs for u = 50° (see Fig. 18.3a). X-ray diffraction indicates that the atomic spacing in the target is d = 2.18 * 10-10 m = 0.218 nm. The electrons have negligible kinetic energy before being accelerated. Find the electron wavelength. SOLUTION IDENTIFY, SET UP, and EXECUTE: We’ll determine l from both de Broglie’s equation, Eq. (18.3), and the diffraction equation, Eq. (18.4). From Eq. (18.3),

Example 18.2

EVALUATE your answer: To check numerical results, it helps to remember some approximate orders of magnitude. Here’s a partial list: Size of an atom: 10-10 m = 0.1 nm Mass of an atom: 10-26 kg Mass of an electron: m = 10-30 kg; mc2 = 0.511 MeV Electron charge magnitude: 10-19 C 1 kT at room temperature: 40 eV Difference between energy levels of an atom (to be discussed in Section 18.3): 1 to 10 eV Speed of an electron in the Bohr model of a hydrogen atom (to be discussed in Section 18.3): 106 m > s

l =

6.626 * 10-34 J # s

!219.109 * 10-31 kg211.602 * 10-19 C2154 V2

= 1.7 * 10-10 m = 0.17 nm

Alternatively, using Eq. (18.4) and assuming m = 1, l = d sin u = 12.18 * 10-10 m2 sin 50° = 1.7 * 10-10 m EVALUATE: The two numbers agree within the accuracy of the experimental results, which gives us an excellent check on our calculations. Note that this electron wavelength is less than the spacing between the atoms.

Energy of a thermal neutron

Find the speed and kinetic energy of a neutron 1m = 1.675 * 10-27 kg2 with de Broglie wavelength l = 0.200 nm, a typical interatomic spacing in crystals. Compare this energy with the average translational kinetic energy of an ideal-gas molecule at room temperature 1T = 20°C = 293 K2. SOLUTION

K = 12 mv2. We’ll use Eq. (18.16) to find the average kinetic energy of a gas molecule. EXECUTE: From Eq. (18.1), the neutron speed is v =

6.626 * 10-34 J # s h = lm 10.200 * 10-9 m211.675 * 10-27 kg2

= 1.98 * 103 m > s '

IDENTIFY and SET UP: This problem uses the relationships between particle speed and wavelength, between particle speed and kinetic energy, and between gas temperature and the average kinetic energy of a gas molecule. We’ll find the neutron speed v using Eq. (18.1) and from that calculate the neutron kinetic energy

'

The neutron kinetic energy is K = 12 mv2 = 1211.675 * 10-27 kg211.98 * 103 m > s22 '

= 3.28 * 10

-21

J = 0.0205 eV

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'

Continued

574

CHAPTER 18 Particles Behaving as Waves

From Eq. (18.16), the average translational kinetic energy of an ideal-gas molecule at T = 293 K is 1 2 2 m1v 2av

= 32 kT = 3211.38 * 10-23 J > K21293 K2 '

= 6.07 * 10

-21

'

J = 0.0379 eV

The two energies are comparable in magnitude, which is why a neutron with kinetic energy in this range is called a thermal neutron.

Diffraction of thermal neutrons is used to study crystal and molecular structure in the same way as x-ray diffraction. Neutron diffraction has proved to be especially useful in the study of large organic molecules. EVALUATE: Note that the calculated neutron speed is much less than the speed of light. This justifies our use of the nonrelativistic form of Eq. (18.1).

De Broglie Waves and the Macroscopic World If the de Broglie picture is correct and matter has wave aspects, you might wonder why we don’t see these aspects in everyday life. As an example, we know that waves diffract when sent through a single slit. Yet when we walk through a doorway (a kind of single slit), we don’t worry about our body diffracting! The principal reason we don’t see these effects on human scales is that Planck’s constant h has such a minuscule value. As a result, the de Broglie wavelengths of even the smallest ordinary objects that you can see are extremely small, and the wave effects are unimportant. For instance, what is the wavelength of a falling grain of sand? If the grain’s mass is 5 * 10 - 10 kg and its diameter is 0.07 mm = 7 * 10 - 5 m, it will fall in air with a terminal speed of about 0.4 m> s. The magnitude of its momentum is then p = mv = 15 * 10 - 10 kg2 * 10.4 m>s2 = 2 * 10 - 10 kg # m> s. The de Broglie wavelength of this falling sand grain is then l =

6.626 * 10-34 J # s h = 3 * 10-24 m = p 2 * 10-10 kg # m/s

Not only is this wavelength far smaller than the diameter of the sand grain, but it’s also far smaller than the size of a typical atom (about 10–10 m). A more massive, faster-moving object would have an even larger momentum and an even smaller de Broglie wavelength. The effects of such tiny wavelengths are so small that they are never noticed in daily life.

The Electron Microscope The electron microscope offers an important and interesting example of the interplay of wave and particle properties of electrons. An electron beam can be used to form an image of an object in much the same way as a light beam. A ray of light can be bent by reflection or refraction, and an electron trajectory can be bent by an electric or magnetic field. Rays of light diverging from a point on an object can be brought to convergence by a converging lens or concave mirror, and electrons diverging from a small region can be brought to convergence by electric and> or magnetic fields. The analogy between light rays and electrons goes deeper. The ray model of geometric optics is an approximate representation of the more general wave model. Geometric optics (ray optics) is valid whenever interference and diffraction effects can be neglected. Similarly, the model of an electron as a point particle following a line trajectory is an approximate description of the actual behavior of the electron; this model is useful when we can neglect effects associated with the wave nature of electrons. How is an electron microscope superior to an optical microscope? The resolution of an optical microscope is limited by diffraction effects, as we discussed in Section 18.7. Since an optical microscope uses wavelengths around 500 nm, it can’t resolve objects smaller than a few hundred nanometers, no matter how carefully its lenses are made. The resolution of an electron microscope is similarly limited by the wavelengths of the electrons, but these wavelengths may be many thousands of times smaller than wavelengths of visible light. As a result, the useful magnification of an electron microscope can be thousands of times greater than that of an optical microscope.

?

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18.2 The Nuclear Atom and Atomic Spectra

575

Note that the ability of the electron microscope to form a magnified image does not depend on the wave properties of electrons. Within the limitations of the Heisenberg uncertainty principle (which we’ll discuss in Section 18.6), we can compute the electron trajectories by treating them as classical charged particles under the action of electric and magnetic forces. Only when we talk about resolution do the wave properties become important.

Example 18.3

An electron microscope

In an electron microscope, the nonrelativistic electron beam is formed by a setup similar to the electron gun used in the Davisson– Germer experiment (see Fig. 18.2). The electrons have negligible kinetic energy before they are accelerated. What accelerating voltage is needed to produce electrons with wavelength 10 pm = 0.010 nm (roughly 50,000 times smaller than typical visible-light wavelengths)? SOLUTION IDENTIFY, SET UP, and EXECUTE: We can use the same concepts we used to understand the Davisson–Germer experiment. The accelerating voltage is the quantity Vba in Eq. (18.3). Rewrite this equation to solve for Vba:

Vba = =

h2 2mel2

16.626 * 10-34 J # s22

219.109 * 10-31 kg211.602 * 10-19 C2110 * 10-12 m22

= 1.5 * 104 V = 15,000 V EVALUATE: It is easy to attain 15-kV accelerating voltages from 120-V or 240-V line voltage using a step-up transformer (Section 31.6) and a rectifier (Section 31.1). The accelerated electrons have kinetic energy 15 keV; since the electron rest energy is 0.511 MeV = 511 keV, these electrons are indeed nonrelativistic.

Test Your Understanding of Section 18.1 (a) A proton has a slightly smaller mass than a neutron. Compared to the neutron described in Example 18.2, would a proton of the same wavelength have (i) more kinetic energy; (ii) less kinetic energy; or (iii) the same kinetic energy? (b) Example 18.1 shows that to give electrons a wavelength of 1.7 * 10-10 m, they must be accelerated from rest through a voltage of 54 V and so acquire a kinetic energy of 54 eV. Does a photon of this same energy also have a wavelength of 1.7 * 10-10 m ? y '

18.2

The Nuclear Atom and Atomic Spectra

Every neutral atom contains at least one electron. How does the wave aspect of electrons affect atomic structure? As we will see, it is crucial for understanding not only the structure of atoms but also how they interact with light. Historically, the quest to understand the nature of the atom was intimately linked with both the idea that electrons have wave characteristics and the notion that light has particle characteristics. Before we explore how these ideas shaped atomic theory, it’s useful to look at what was known about atoms—as well as what remained mysterious— by the first decade of the twentieth century.

Line Spectra Everyone knows that heated materials emit light, and that different materials emit different kinds of light. The coils of a toaster glow red when in operation, the flame of a match has a characteristic yellow color, and the flame from a gas range is a distinct blue. To analyze these different types of light, we can use a prism or a diffraction grating to separate the various wavelengths in a beam of light into a spectrum. If the light source is a hot solid (such as the filament of an incandescent light bulb) or liquid, the spectrum is continuous; light of all wavelengths is present (Fig. 18.5a). But if the source is a heated gas, such as the neon in an advertising sign or the sodium vapor formed when table salt is thrown into a campfire, the spectrum includes only a few colors in the form of isolated sharp parallel lines (Fig. 18.5b). (Each “line” is an image of the spectrograph slit, deviated through an angle that depends on the wavelength of the light forming that image; see

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576

CHAPTER 18 Particles Behaving as Waves (a) Continuous spectrum: light of all wavelengths is present.

18.5 (a) Continuous spectrum produced by a glowing light bulb filament. (b) Emission line spectrum emitted by a lamp containing a heated gas.

(b) Line spectrum: only certain discrete wavelengths are present.

Screen

Screen

Diffraction grating Slit

Diffraction grating

Lens

Slit Lens Light bulb with heated filament

Lamp with heated gas

Section 18.5.) A spectrum of this sort is called an emission line spectrum, and the lines are called spectral lines. Each spectral line corresponds to a definite wavelength and frequency. It was discovered early in the 19th century that each element in its gaseous state has a unique set of wavelengths in its line spectrum. The spectrum of hydrogen always contains a certain set of wavelengths; mercury produces a different set, neon still another, and so on (Fig. 18.6). Scientists find the use of spectra to identify elements and compounds to be an invaluable tool. For instance, astronomers have detected the spectra from more than 100 different molecules in interstellar space, including some that are not found naturally on earth. While a heated gas selectively emits only certain wavelengths, a cool gas selectively absorbs certain wavelengths. If we pass white (continuous-spectrum) light through a gas and look at the transmitted light with a spectrometer, we find a series of dark lines corresponding to the wavelengths that have been absorbed (Fig. 18.7). This is called an absorption line spectrum. What’s more, a given kind of atom or molecule absorbs the same characteristic set of wavelengths when it’s cool as it emits when heated. Hence scientists can use absorption line spectra to identify substances in the same manner that they use emission line spectra. As useful as emission line spectra and absorption line spectra are, they presented a quandary to scientists: Why does a given kind of atom emit and absorb only certain very specific wavelengths? To answer this question, we need to have a better idea of what the inside of an atom is like. We know that atoms are much smaller than the wavelengths of visible light, so there is no hope of actually seeing an atom using that light. But we can still describe how the mass and electric charge are distributed throughout the volume of the atom. 18.6 The emission line spectra of several kinds of atoms and molecules. No two are alike. Note that the spectrum of water vapor (H2O) is similar to that of hydrogen (H2), but there are important differences that make it straightforward to distinguish these two spectra. Helium (He) Hydrogen (H2) Krypton (Kr) Mercury (Hg) Neon (Ne) Water vapor (H2O) Xenon

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18.2 The Nuclear Atom and Atomic Spectra

Here’s where things stood in 1910. In 1897 the English physicist J. J. Thomson (Nobel Prize 1906) had discovered the electron and measured its charge-to-mass ratio e > m. By 1909, the American physicist Robert Millikan (Nobel Prize 1923) had made the first measurements of the electron charge -e. These and other experiments showed that almost all the mass of an atom had to be associated with the positive charge, not with the electrons. It was also known that the overall size of atoms is of the order of 10-10 m and that all atoms except hydrogen contain more than one electron. In 1910 the best available model of atomic structure was one developed by Thomson. He envisioned the atom as a sphere of some as yet unidentified positively charged substance, within which the electrons were embedded like raisins in cake. This model offered an explanation for line spectra. If the atom collided with another atom, as in a heated gas, each electron would oscillate around its equilibrium position with a characteristic frequency and emit electromagnetic radiation with that frequency. If the atom were illuminated with light of many frequencies, each electron would selectively absorb only light whose frequency matched the electron’s natural oscillation frequency. (This is the phenomenon of resonance that we discussed in Section 14.8.) '

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18.7 The absorption line spectrum of the sun. (The spectrum “lines” read from left to right and from top to bottom, like text on a page.) The spectrum is produced by the sun’s relatively cool atmosphere, which absorbs photons from deeper, hotter layers. The absorption lines thus indicate what kinds of atoms are present in the solar atmosphere.

Rutherford’s Exploration of the Atom The first experiments designed to test Thomson’s model by probing the interior structure of the atom were carried out in 1910–1911 by Ernest Rutherford (Fig. 18.8) and two of his students, Hans Geiger and Ernest Marsden, at the University of Manchester in England. These experiments consisted of shooting a beam of charged particles at thin foils of various elements and observing how the foil deflected the particles. The particle accelerators now in common use in laboratories had not yet been invented, and Rutherford’s projectiles were alpha particles emitted from naturally radioactive elements. The nature of these alpha particles was not completely understood, but it was known that they are ejected from unstable nuclei with speeds of the order of 107 m > s, are positively charged, and can travel several centimeters through air or 0.1 mm or so through solid matter before they are brought to rest by collisions. Figure 18.9 is a schematic view of Rutherford’s experimental setup. A radioactive substance at the left emits alpha particles. Thick lead screens stop all particles except those in a narrow beam. The beam passes through the foil target (consisting of gold, silver, or copper) and strikes screens coated with zinc sulfide, creating a momentary flash, or scintillation. Rutherford and his students counted the numbers of particles deflected through various angles. The atoms in a metal foil are packed together like marbles in a box (not spaced apart). Because the particle beam passes through the foil, the alpha particles must pass through the interior of atoms. Within an atom, the charged alpha particle will interact with the electrons and the positive charge. (Because the total charge of the atom is zero, alpha particles feel little electrical force outside an atom.) An electron has about 7300 times less mass than an alpha particle, so momentum considerations indicate that the atom’s electrons cannot appreciably deflect the alpha particle—any more than a swarm of gnats deflects a tossed pebble. Any deflection will be due to the positively charged material that makes up almost all of the atom’s mass. In the Thomson model, the positive charge and the negative electrons are distributed through the whole atom. Hence the electric field inside the atom should be quite small, and the electric force on an alpha particle that enters the atom should be quite weak. The maximum deflection to be expected is then only a few degrees (Fig. 18.10a). The results of the Rutherford experiments were very different from the Thomson prediction. Some alpha particles were scattered by nearly 180°—that is, almost straight backward (Fig. 18.10b). Rutherford later wrote: '

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Application Using Spectra to Analyze an Interstellar Gas Cloud The light from this glowing gas cloud—located in the Small Magellanic Cloud, a small satellite galaxy of the Milky Way some 200,000 lightyears (1.9 ∞ 1018 km) from earth—has an emission line spectrum. Despite its immense distance, astronomers can tell that this cloud is composed mostly of hydrogen because its spectrum is dominated by red light at a wavelength of 656.3 nm, a wavelength emitted by hydrogen and no other element.

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578

CHAPTER 18 Particles Behaving as Waves

18.8 Born in New Zealand, Ernest Rutherford (1871–1937) spent his professional life in England and Canada. Before carrying out the experiments that established the existence of atomic nuclei, he shared (with Frederick Soddy) the 1908 Nobel Prize in chemistry for showing that radioactivity results from the disintegration of atoms.

18.9 The Rutherford scattering experiments investigated what happens to alpha particles fired at a thin gold foil. The results of this experiment helped reveal the structure of atoms. 3 Alpha particles strike foil and are scattered by gold atoms.

Zinc sulfide scintillation screen

2 Small holes in a

pair of lead screens create a narrow beam of alpha particles. 1 Alpha particles are emitted by a radioactive element such as radium.

Gold-foil target

4 A scattered alpha particle produces a flash of light when it hits a scintillation screen, showing the direction in which it was scattered.

It was quite the most incredible event that ever happened to me in my life. It was almost as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you. PhET: Rutherford Scattering ActivPhysics 19.1: Particle Scattering

18.10 A comparison of Thomson’s and Rutherford’s models of the atom. (a) Thomson’s model of the atom: An alpha particle is scattered through only a small angle. a

Clearly the Thomson model was wrong and a new model was needed. Suppose the positive charge, instead of being distributed through a sphere with atomic dimensions (of the order of 10-10 m), is all concentrated in a much smaller volume. Then it would act like a point charge down to much smaller distances. The maximum electric field repelling the alpha particle would be much larger, and the amazing large-angle scattering that Rutherford observed could occur. Rutherford developed this model and called the concentration of positive charge the nucleus. He again computed the numbers of particles expected to be scattered through various angles. Within the accuracy of his experiments, the computed and measured results agreed, down to distances of the order of 10-14 m. His experiments therefore established that the atom does have a nucleus—a very small, very dense structure, no larger than 10-14 m in diameter. The nucleus occupies only about 10-12 of the total volume of the atom or less, but it contains all the positive charge and at least 99.95% of the total mass of the atom. Figure 18.11 shows a computer simulation of alpha particles with a kinetic energy of 5.0 MeV being scattered from a gold nucleus of radius 7.0 * 10-15 m 18.11 Computer simulation of scattering of 5.0-MeV alpha particles from a gold nucleus. Each curve shows a possible alpha-particle trajectory. (a) The scattering curves match Rutherford’s experimental data if a radius of 7.0 * 10-15 m is assumed for a gold nucleus. (b) A model with a much larger radius for the gold nucleus does not match the data. (a) A gold nucleus with radius 7.0 3 10215 m gives large-angle scattering.

(b) A nucleus with 10 times the radius of the nucleus in (a) shows no large-scale scattering.

(b) Rutherford’s model of the atom: An alpha particle can be scattered through a large angle by the compact, positively charged nucleus (not drawn to scale). a

Nucleus

Motion of incident 5.0-MeV alpha particles

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18.2 The Nuclear Atom and Atomic Spectra

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(the actual value) and from a nucleus with a hypothetical radius ten times larger. In the second case there is no large-angle scattering. The presence of large-angle scattering in Rutherford’s experiments thus attested to the small size of the nucleus. Later experiments showed that all nuclei are composed of positively charged protons (discovered in 1918) and electrically neutral neutrons (discovered in 1930). For example, the gold atoms in Rutherford’s experiments have 79 protons and 118 neutrons. In fact, an alpha particle is itself the nucleus of a helium atom, with two protons and two neutrons. It is much more massive than an electron but only about 2% as massive as a gold nucleus, which helps explain why alpha particles are scattered by gold nuclei but not by electrons.

Example 18.4

A Rutherford experiment

An alpha particle (charge 2e) is aimed directly at a gold nucleus (charge 79e). What minimum initial kinetic energy must the alpha particle have to approach within 5.0 * 10-14 m of the center of the gold nucleus before reversing direction? Assume that the gold nucleus, which has about 50 times the mass of an alpha particle, remains at rest. SOLUTION IDENTIFY: The repulsive electric force exerted by the gold nucleus makes the the alpha particle slow to a halt as it approaches, then reverse direction. This force is conservative, so the total mechanical energy (kinetic energy of the alpha particle plus electric potential energy of the system) is conserved. SET UP: Let point 1 be the initial position of the alpha particle, very far from the gold nucleus, and let point 2 be 5.0 * 10-14 m from the center of the gold nucleus. Our target variable is the kinetic energy K1 of the alpha particle at point 1 that allows it to reach point 2 with K2 = 0. To find this we’ll use the law of conservation of energy and Eq. (23.9) for electric potential energy, U = qq0>4pP0r. EXECUTE: At point 1 the separation r of the alpha particle and gold nucleus is effectively infinite, so from Eq. (23.9) U1 = 0. At point 2 the potential energy is U2 =

1 qq0 4pP0 r

122179211.60 * 10-19 C22

= 19.0 * 109 N # m2 > C22

5.0 * 10-14 m J = 4.6 * 10 eV = 4.6 MeV '

= 7.3 * 10

-13

'

6

By energy conservation K1 + U1 = K2 + U2, so K1 = K2 + U2 - U1 = 0 + 4.6 MeV - 0 = 4.6 MeV. Thus, to approach within 5.0 * 10-14 m, the alpha particle must have initial kinetic energy K1 = 4.6 MeV. EVALUATE: Alpha particles emitted from naturally occurring radioactive elements typically have energies in the range 4 to 6 MeV. For example, the common isotope of radium, 226Ra, emits an alpha particle with energy 4.78 MeV. Was it valid to assume that the gold nucleus remains at rest? To find out, note that when the alpha particle stops momentarily, all of its initial momentum has been transferred to the gold nucleus. An alpha particle has a mass ma = 6.64 * 10-27 kg; if its initial kinetic energy K1 = 12 mv12 is 7.3 * 10-13 J, you can show that its initial speed is v1 = 1.5 * 107 m > s and its initial momentum is p1 = mav1 = 9.8 * 10-20 kg # m > s. A gold nucleus (mass mAu = 3.27 * 10-25 kg) with this much momentum has a much 5 slower speed vAu = 3.0 * 10 m > s and kinetic energy KAu = 1 2 -14 1.5 * 10 J = 0.092 MeV. This recoil kinetic 2 mvAu = energy of the gold nucleus is only 2% of the total energy in this situation, so we are justified in neglecting it.

The Failure of Classical Physics Rutherford’s discovery of the atomic nucleus raised a serious question: What prevented the negatively charged electrons from falling into the positively charged nucleus due to the strong electrostatic attraction? Rutherford suggested that perhaps the electrons revolve in orbits about the nucleus, just as the planets revolve around the sun. But according to classical electromagnetic theory, any accelerating electric charge (either oscillating or revolving) radiates electromagnetic waves. An example is the radiation from an oscillating point charge that we depicted in Fig. 32.3 (Section 32.1). An electron orbiting inside an atom would always have a centripetal acceleration toward the nucleus, and so should be emitting radiation at all times. The energy of an orbiting electron should therefore decrease continuously, its orbit should become smaller and smaller, and it should spiral into the nucleus within a fraction of a second (Fig. 18.12). Even worse, according to classical theory the frequency of the electromagnetic waves emitted should equal the frequency of revolution. As the electrons radiated energy, their angular speeds

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580

CHAPTER 18 Particles Behaving as Waves

18.12 Classical physics makes predictions about the behavior of atoms that do not match reality. ACCORDING TO CLASSICAL PHYSICS: • An orbiting electron is accelerating, so it should radiate electromagnetic waves. • The waves would carry away energy, so the electron should lose energy and spiral inward. • The electron’s angular speed would increase as its orbit shrank, so the frequency of the radiated waves should increase. Thus, classical physics says that atoms should collapse within a fraction of a second and should emit light with a continuous spectrum as they do so. PREDICTED BEHAVIOR

Longer wavelength

would change continuously, and they would emit a continuous spectrum (a mixture of all frequencies), not the line spectrum actually observed. Thus Rutherford’s model of electrons orbiting the nucleus, which is based on Newtonian mechanics and classical electromagnetic theory, makes three entirely wrong predictions about atoms: They should emit light continuously, they should be unstable, and the light they emit should have a continuous spectrum. Clearly a radical reappraisal of physics on the scale of the atom was needed. In the next section we will see the bold idea that led to an new understanding of the atom, and see how this idea meshes with de Broglie’s no less bold notion that electrons have wave attributes. Test Your Understanding of Section 18.2 Suppose you repeated Rutherford’s scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14.0 K.) The nucleus of a hydrogen atom is a single proton, with about one-fourth the mass of an alpha particle. Compared to the original experiment with gold foil, would you expect the alpha particles in this experiment to undergo (i) more large-angle scattering; (ii) the same amount of large-angle scattering; or (iii) less large-angle scattering? y

18.3 Shorter wavelength IN FACT: • Atoms are stable. • They emit light only when excited, and only at specific frequencies (as a line spectrum).

18.13 Niels Bohr (1885–1962) was a young postdoctoral researcher when he proposed the novel idea that the energy of an atom could have only certain discrete values. He won the 1922 Nobel Prize in physics for these ideas. Bohr went on to make seminal contributions to nuclear physics and to become a passionate advocate for the free exchange of scientific ideas among all nations.

Energy Levels and the Bohr Model of the Atom

In 1913 a young Danish physicist working with Ernest Rutherford at the University of Manchester made a revolutionary proposal to explain both the stability of atoms and their emission and absorption line spectra. The physicist was Niels Bohr (Fig. 18.13), and his innovation was to combine the photon concept that we introduced in Chapter 38 with a fundamentally new idea: The energy of an atom can have only certain particular values. His hypothesis represented a clean break from 19th-century ideas.

Photon Emission and Absorption by Atoms Bohr’s reasoning went like this. The emission line spectrum of an element tells us that atoms of that element emit photons with only certain specific frequencies ƒ and hence certain specific energies E 5 hƒ. During the emission of a photon, the internal energy of the atom changes by an amount equal to the energy of the photon. Therefore, said Bohr, each atom must be able to exist with only certain specific values of internal energy. Each atom has a set of possible energy levels. An atom can have an amount of internal energy equal to any one of these levels, but it cannot have an energy intermediate between two levels. All isolated atoms of a given element have the same set of energy levels, but atoms of different elements have different sets. Suppose an atom is raised, or excited, to a high energy level. (In a hot gas this happens when fast-moving atoms undergo inelastic collisions with each other or with the walls of the gas container. In an electric discharge tube, such as those used in a neon light fixture, atoms are excited by collisions with fast-moving electrons.) According to Bohr, an excited atom can make a transition from one energy level to a lower level by emitting a photon with energy equal to the energy difference between the initial and final levels (Fig. 18.14). If Ei is the initial energy of the atom before such a transition, Ef is its final energy after the transition, and the photon’s energy is hƒ = hc > l, then conservation of energy gives '

hƒ =

hc = Ei - Ef l

'

(energy of emitted photon)

(18.5)

For example, an excited lithium atom emits red light with wavelength l = 671 nm. The corresponding photon energy is

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18.3 Energy Levels and the Bohr Model of the Atom

16.63 * 10-34 J # s213.00 * 108 m > s2 hc = l 671 * 10-9 m '

E =

= 2.96 * 10

-19

'

J = 1.85 eV

This photon is emitted during a transition like that shown in Fig. 18.14 between two levels of the atom that differ in energy by Ei - Ef = 1.85 eV. The emission line spectra shown in Fig. 18.6 show that many different wavelengths are emitted by each atom. Hence each kind of atom must have a number of energy levels, with different spacings in energy between them. Each wavelength in the spectrum corresponds to a transition between two specific energy levels of the atom. CAUTION Producing a line spectrum The lines of an emission line spectrum, such as the helium spectrum shown at the top of Fig. 18.6, are not all produced by a single atom. The sample of helium gas that produced the spectrum in Fig. 18.6 contained a large number of helium atoms; these were excited in an electric discharge tube to various energy levels. The spectrum of the gas shows the light emitted from all the different transitions that occurred in different atoms of the sample. y

The observation that atoms are stable means that each atom has a lowest energy level, called the ground level. Levels with energies greater than the ground level are called excited levels. An atom in an excited level, called an excited atom, can make a transition into the ground level by emitting a photon as in Fig. 18.14. But since there are no levels below the ground level, an atom in the ground level cannot lose energy and so cannot emit a photon. Collisions are not the only way that an atom’s energy can be raised from one level to a higher level. If an atom initially in the lower energy level in Fig. 18.14 is struck by a photon with just the right amount of energy, the photon can be absorbed and the atom will end up in the higher level (Fig. 18.15). As an example, we previously mentioned two levels in the lithium atom with an energy difference of 1.85 eV. For a photon to be absorbed and excite the atom from the lower level to the higher one, the photon must have an energy of 1.85 eV and a wavelength of 671 nm. In other words, an atom absorbs the same wavelengths that it emits. This explains the correspondence between an element’s emission line spectrum and its absorption line spectrum that we described in Section 18.2. Note that a lithium atom cannot absorb a photon with a slightly longer wavelength (say, 672 nm) or one with a slightly shorter wavelength (say, 670 nm). That’s because these photons have, respectively, slightly too little or slightly too much energy to raise the atom’s energy from one level to the next, and an atom cannot have an energy that’s intermediate between levels. This explains why absorption line spectra have distinct dark lines (see Fig. 18.7): Atoms can absorb only photons with specific wavelengths. An atom that’s been excited into a high energy level, either by photon absorption or by collisions, does not stay there for long. After a short time, called the lifetime of the level (typically around 10–8 s), the excited atom will emit a photon and make a transition into a lower excited level or the ground level. A cool gas that’s illuminated by white light to make an absorption line spectrum thus also produces an emission line spectrum when viewed from the side, since when the atoms de-excite they emit photons in all directions (Fig. 18.16). To keep a gas of atoms glowing, you have to continually provide energy to the gas in order to re-excite atoms so that they can emit more photons. If you turn off the energy supply (for example, by turning off the electric current through a neon light fixture, or by shutting off the light source in Fig. 18.16), the atoms drop back into their ground levels and cease to emit light. By working backward from the observed emission line spectrum of an element, physicists can deduce the arrangement of energy levels in an atom of that element. As an example, Fig. 18.17a shows some of the energy levels for a sodium atom. You may have noticed the yellow-orange light emitted by sodium vapor street lights. Sodium atoms emit this characteristic yellow-orange light

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18.14 An excited atom emitting a photon. i An atom drops from an initial level i to a lower-energy final level f by emitting a photon with energy equal to Ei 2 Ef. f

Ei

hf 5 Ei 2 Ef Ef

PhET: Models of the Hydrogen Atom ActivPhysics 18.1: The Bohr Model

18.15 An atom absorbing a photon. (Compare with Fig. 18.14.) f

hf 5 Ef 2 Ei i

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Ef An atom is raised from an initial level i to a higherenergy final level f by absorbing a photon with energy equal to Ef 2 Ei. Ei

M39-(5)_7133_Ch39_pp1286-1327 10-08-2015 PM 05:06 Page 582

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CHAPTER 18 Particles Behaving as Waves

Diffraction grating

Vacuum White light Mirrors redirect light

Continuous spectrum

Wavelength

Transmitted light

Intensity

Absorption spectrum Container of cool gas

Source of white light

Intensity

18.16 When a beam of white light with a continuous spectrum passes through a cool gas, the transmitted light has an absorption spectrum. The absorbed light energy excites the gas and causes it to emit light of its own, which has an emission spectrum.

Wavelength

Intensity

White light

Light emitted by gas

Emission spectrum

Wavelength

Diffraction grating for light emitted by gas

with wavelengths 589.0 and 589.6 nm when they make transitions from the two closely spaced levels labeled lowest excited levels to the ground level. A standard test for the presence of sodium compounds is to look for this yellow-orange light from a sample placed in a flame (Fig. 18.17b). 18.17 (a) Energy levels of the sodium atom relative to the ground level. Numbers on the lines between levels are wavelengths of the light emitted or absorbed during transitions between those levels. The column labels, such as 2S1 > 2, refer to some quantum states of the atom. (b) When a sodium compound is placed in a flame, sodium atoms are excited into the lowest excited levels. As they drop back to the ground level, the atoms emit photons of yellow-orange light with wavelengths 589.0 and 589.6 nm. '

'

(a) Energy (eV) 2S

2P

1/2

2P

3/2

2D 5/2, 3/2

1/2

2F

7/2, 5/2

5.12

5

0

nm

1 m

Excited levels

nm

81 8

81

nm

.3 n

8

8.2

n 58 m 9. 6

nm

Lowest excited levels

2.7

330.2 2n85.3 nm m 589 .0 n m

nm

1

9.5

113

34

.4

40

11

56

8.

3

2

9 nm

nm nm

2208.4

.0

6 84

497.

nm 9 4. m .1 n 616

31

4

(b)

These wavelengths give excited sodium a yellow-orange color.

Ground level

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18.3 Energy Levels and the Bohr Model of the Atom

Example 18.5

Emission and absorption spectra

A hypothetical atom (Fig. 18.18a) has energy levels at 0.00 eV (the ground level), 1.00 eV, and 3.00 eV. (a) What are the frequencies and wavelengths of the spectral lines this atom can emit when excited? (b) What wavelengths can this atom absorb if it is in its ground level?

18.18 (a) Energy-level diagram for the hypothetical atom, showing the possible transitions for emission from excited levels and for absorption from the ground level. (b) Emission spectrum of this hypothetical atom. (a)

SOLUTION

3.00 eV

IDENTIFY and SET UP: Energy is conserved when a photon is emitted or absorbed. In each transition the photon energy is equal to the difference between the energies of the levels involved in the transition.

2.00 eV 3.00 eV 1.00 eV Ground level

E 1.00 eV = = 2.42 * 1014 Hz h 4.136 * 10-15 eV # s

Emission transitions

1.00 eV Absorption transitions

(b)

For 2.00 eV and 3.00 eV, ƒ = 4.84 * 1014 Hz and 7.25 * 1014 Hz, respectively. For 1.00-eV photons,

414 nm

620 nm

1240 nm

3.00 * 108 m > s c = = 1.24 * 10-6 m = 1240 nm ƒ 2.42 * 1014 Hz '

l =

3.00 eV

1.00 eV

EXECUTE: (a) The possible energies of emitted photons are 1.00 eV, 2.00 eV, and 3.00 eV. For 1.00 eV, Eq. (18.2) gives ƒ =

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This is in the infrared region of the spectrum (Fig. 18.18b). For 2.00 eV and 3.00 eV, the wavelengths are 620 nm (red) and 414 nm (violet), respectively. (b) From the ground level, only a 1.00-eV or a 3.00-eV photon can be absorbed (Fig. 18.18a); a 2.00-eV photon cannot be absorbed because the atom has no energy level 2.00 eV above the ground level. Passing light from a hot solid through a gas of these hypothetical atoms (almost all of which would be in the ground

state if the gas were cool) would yield a continuous spectrum with dark absorption lines at 1240 nm and 414 nm. EVALUATE: Note that if a gas of these atoms were at a sufficiently high temperature, collisions would excite a number of atoms into the 1.00-eV energy level. Such excited atoms can absorb 2.00-eV photons, as Fig. 18.18a shows, and an absorption line at 620 nm would appear in the spectrum. Thus the observed spectrum of a given substance depends on its energy levels and its temperature.

Suppose we take a gas of the hypothetical atoms in Example 18.5 and illuminate it with violet light of wavelength 414 nm. Atoms in the ground level can absorb this photon and make a transition to the 3.00-eV level. Some of these atoms will make a transition back to the ground level by emitting a 414-nm photon. But other atoms will return to the ground level in two steps, first emitting a 620-nm photon to transition to the 1.00-eV level, then a 1240-nm photon to transition back to the ground level. Thus this gas will emit longer-wavelength radiation than it absorbs, a phenomenon called fluorescence. For example, the electric discharge in a fluorescent lamp causes the mercury vapor in the tube to emit ultraviolet radiation. This radiation is absorbed by the atoms of the coating on the inside of the tube. The coating atoms then re-emit light in the longer-wavelength, visible portion of the spectrum. Fluorescent lamps are more efficient than incandescent lamps in converting electrical energy to visible light because they do not waste as much energy producing (invisible) infrared photons. Our discussion of energy levels and spectra has concentrated on atoms, but the same ideas apply to molecules. Figure 18.6 shows the emission line spectra of two molecules, hydrogen (H2) and water (H2O). Just as for sodium or other atoms, physicists can work backward from these molecular spectra and deduce the arrangement of energy levels for each kind of molecule. We’ll return to molecules and molecular structure in Chapter 42.

Application Fish Fluorescence When illuminated by blue light, this tropical lizardfish (family Synodont idae) fluoresces and emits longer-wavelength green light. The fluorescence may be a sexual signal or a way for the fish to camouflage itself among coral (which also have a green fluorescence).

The Franck–Hertz Experiment: Are Energy Levels Real? Are atomic energy levels real, or just a convenient fiction that helps us to explain spectra? In 1914, the German physicists James Franck and Gustav Hertz answered this question when they found direct experimental evidence for the existence of atomic energy levels.

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584

CHAPTER 18 Particles Behaving as Waves

Franck and Hertz studied the motion of electrons through mercury vapor under the action of an electric field. They found that when the electron kinetic energy was 4.9 eV or greater, the vapor emitted ultraviolet light of wavelength 250 nm. Suppose mercury atoms have an excited energy level 4.9 eV above the ground level. An atom can be raised to this level by collision with an electron; it later decays back to the ground level by emitting a photon. From the photon formula E = hc>l, the wavelength of the photon should be 14.136 * 10-15 eV # s213.00 * 108 m > s2 hc = E 4.9 eV -7 = 2.5 * 10 m = 250 nm '

'

l =

This is equal to the wavelength that Franck and Hertz measured, which demonstrates that this energy level actually exists in the mercury atom. Similar experiments with other atoms yield the same kind of evidence for atomic energy levels. Franck and Hertz shared the 1925 Nobel Prize in physics for their research.

Electron Waves and the Bohr Model of Hydrogen Bohr’s hypothesis established the relationship between atomic spectra and energy levels. By itself, however, it provided no general principles for predicting the energy levels of a particular atom. Bohr addressed this problem for the case of the simplest atom, hydrogen, which has just one electron. Let’s look at the ideas behind the Bohr model of the hydrogen atom. Bohr postulated that each energy level of a hydrogen atom corresponds to a specific stable circular orbit of the electron around the nucleus. In a break with classical physics, Bohr further postulated that an electron in such an orbit does not radiate. Instead, an atom radiates energy only when an electron makes a transition from an orbit of energy Ei to a different orbit with lower energy Ef, emitting a photon of energy hƒ = Ei - Ef in the process. As a result of a rather complicated argument that related the angular frequency of the light emitted to the angular speed of the electron in highly excited energy levels, Bohr found that the magnitude of the electron’s angular momentum is quantized; that is, this magnitude must be an integral multiple of h > 2p. (Because 1 J = 1 kg # m2 > s2, the SI units of Planck’s constant h, J # s, are the same as the SI units of angular momentum, usually written as kg # m2 > s.) Let’s number the orbits by an integer n, where n = 1, 2, 3, Á , and call the radius of orbit n rn and the speed of the electron in that orbit vn. The value of n for each orbit is called the principal quantum number for the orbit. From Section 10.5, Eq. (10.28), the magnitude of the angular momentum of an electron of mass m in such an orbit is Ln = mvnrn (Fig. 18.19). So Bohr’s argument led to '

'

'

18.19 Calculating the angular momentum of an electron in a circular orbit around an atomic nucleus.

Ln = mvnrn = n

r

Angular momentum Ln of orbiting electron is perpendicular to plane of orbit (since we take origin to be at nucleus) and has magnitude L 5 mvnrn sin f 5 mvnrn sin 90° 5 mvnrn. z nth allowed electron orbit

r

Ln

y

x

Nucleus rrn Electron

pr 5 mvr n f 5 90°

'

'

h 2p

'

(quantization of angular momentum)

(18.6)

Instead of going through Bohr’s argument to justify Eq. (18.6), we can use de Broglie’s picture of electron waves. Rather then visualizing the orbiting electron as a particle moving around the nucleus in a circular path, think of it as a sinusoidal standing wave with wavelength l that extends around the circle. A standing wave on a string transmits no energy (see Section 15.7), and electrons in Bohr’s orbits radiate no energy. For the wave to “come out even” and join onto itself smoothly, the circumference of this circle must include some whole number of wavelengths, as Fig. 18.20 suggests. Hence for an orbit with radius rn and circumference 2prn, we must have 2prn = nln, where ln is the wavelength and n = 1, 2, 3, Á . According to the de Broglie relationship, Eq. (18.1), the wavelength of a particle with rest mass m moving with nonrelativistic speed vn is ln = h > mvn. Combining 2prn = nln and ln = h > mvn, we find 2prn = nh > mvn or '

'

'

'

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'

'

585

18.3 Energy Levels and the Bohr Model of the Atom

h 2p This is the same as Bohr’s result, Eq. (18.6). Thus a wave picture of the electron leads naturally to the quantization of the electron’s angular momentum. Now let’s consider a model of the hydrogen atom that is Newtonian in spirit but incorporates this quantization assumption (Fig. 18.21). This atom consists of a single electron with mass m and charge -e in a circular orbit around a single proton with charge +e. The proton is nearly 2000 times as massive as the electron, so we can assume that the proton does not move. We learned in Section 5.4 that when a particle with mass m moves with speed vn in a circular orbit with radius rn, its centripetal (inward) acceleration is vn 2>rn. According to Newton’s second law, a radially inward net force with magnitude F = mvn 2>rn is needed to cause this acceleration. We discussed in Section 12.4 how the gravitational attraction provides that inward force for satellite orbits. In hydrogen the force F is provided by the electrical attraction between the positive proton and the negative electron. From Coulomb’s law, Eq. (21.2), mvnrn = n

F =

18.20 These diagrams show the idea of fitting a standing electron wave around a circular orbit. For the wave to join onto itself smoothly, the circumference of the orbit must be an integral number n of wavelengths.

l n52

1 e2 4pP0 rn2

Hence Newton’s second law states that

l

mvn2 1 e2 = rn 4pP0 rn2

(18.7) n53

When we solve Eqs. (18.6) and (18.7) simultaneously for rn and vn, we get rn = P0

n2h2 pme2

1 e2 vn = P0 2nh

(orbital radii in the Bohr model)

(18.8) l

(orbital speeds in the Bohr model)

(18.9)

Equation (18.8) shows that the orbit radius rn is proportional to n2, so the smallest orbit radius corresponds to n = 1. We’ll denote this minimum radius, called the Bohr radius, as a0: a0 = P0

h2 pme2

(Bohr radius)

n54

(18.10)

Then we can rewrite Eq. (18.8) as rn = n2a0

(18.11)

The permitted orbits have radii a0, 4a0, 9a0, and so on. You can find the numerical values of the quantities on the right-hand side of Eq. (18.10) in Appendix F. Using these values, we find that the radius a0 of the smallest Bohr orbit is

Proton is assumed to be stationary. Electron revolves in a circle of radius rn with speed vn.

18.854 * 10-12 C2 > N # m2216.626 * 10-34 J # s22 '

a0 =

18.21 The Bohr model of the hydrogen atom.

p19.109 * 10

'

-31

kg211.602 * 10-19 C22

vn

= 5.29 * 10-11 m This gives an atomic diameter of about 10-10 m = 0.1 nm, which is consistent with atomic dimensions estimated by other methods. Equation (18.9) shows that the orbital speed vn is proportional to 1>n. Hence the greater the value of n, the larger the orbital radius of the electron and the slower its orbital speed. (We saw the same relationship between orbital radius and speed for satellite orbits in Section 13.4.) We leave it to you to calculate the speed

Proton M, 1e

rn

Electrostatic attraction provides centripetal acceleration.

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F

Electron m, 2e

586

CHAPTER 18 Particles Behaving as Waves

in the n = 1 orbit, which is the greatest possible speed of the electron in the hydrogen atom (see Exercise 18.29); the result is v1 = 2.19 * 106 m > s. This is less than 1% of the speed of light, so relativistic considerations aren’t significant. '

'

Hydrogen Energy Levels in the Bohr Model We can now use Eqs. (18.8) and (18.9) to find the kinetic and potential energies Kn and Un when the electron is in the orbit with quantum number n: Kn = 12 mvn2 = Un = -

1 me4 P02 8n2h2

1 e2 1 me4 = - 2 2 2 4pP0 rn P0 4n h

(kinetic energies in the Bohr model)

(18.12)

(potential energies in the Bohr model)

(18.13)

The potential energy has a negative sign because we have taken the electric potential energy to be zero when the electron is infinitely far from the nucleus. We are interested only in the differences in energy between orbits, so the reference position doesn’t matter. The total energy En is the sum of the kinetic and potential energies: (total energies in the 1 me4 En = Kn + Un = - 2 2 2 (18.14) Bohr model) P0 8n h Since En in Eq. (18.14) has a different value for each n, you can see that this equation gives the energy levels of the hydrogen atom in the Bohr model. Each distinct orbit corresponds to a distinct energy level. Figure 18.22 depicts the orbits and energy levels. We label the possible energy levels of the atom by values of the quantum number n. For each value of n there are corresponding values of orbit radius rn, speed vn, angular momentum Ln = nh>2p, and total energy En. The energy of the atom is least when n = 1 and En has its most negative value. This is the ground level of the hydrogen atom; it is the level with the smallest orbit, of radius a0. For n = 2, 3, Á , the absolute value of En is smaller and the energy is progressively larger (less negative). [

18.22 Two ways to represent the energy levels of the hydrogen atom and the transitions between them. Note that the radius of the nth permitted orbit is actually n2 times the radius of the n 5 1 orbit. (a) Permitted orbits of an electron in the Bohr model of a hydrogen atom (not to scale). Arrows indicate the transitions responsible for some of the lines of various series. Balmer series (visible light and ultraviolet)

n5 n5 n5 n5 n5

Paschen series (infrared) Brackett series (infrared)

Lyman series (ultraviolet)

(b) Energy-level diagram for hydrogen, showing some transitions corresponding to the various series

7 6 5 4 3

n52

Lyman series

Paschen series

Pfund series

Brackett series Balmer series

20.28 eV 20.38 eV 20.54 eV 20.85 eV 21.51 eV 23.40 eV

Pfund series (infrared)

n51 n52 n53 n54 n55

n51

n56

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213.60 eV

18.3 Energy Levels and the Bohr Model of the Atom

587

Figure 18.22 also shows some of the possible transitions from one electron orbit to an orbit of lower energy. Consider a transition from orbit nU (for “upper”) to a smaller orbit nL (for “lower”), with nL 6 nU—or, equivalently, from level nU to a lower level nL. Then the energy hc>l of the emitted photon of wavelength l is equal to EnU - EnL. Before we use this relationship to solve for l, it’s convenient to rewrite Eq. (18.14) for the energies as En = -

hcR , n2

R =

where

me4 8P02 h3c

(total energies in the Bohr model)

(18.15)

The quantity R in Eq. (18.15) is called the Rydberg constant (named for the Swedish physicist Johannes Rydberg, who did pioneering work on the hydrogen spectrum). When we substitute the numerical values of the fundamental physical constants m, c, e, h, and P0, all of which can be determined quite independently of the Bohr theory, we find that R = 1.097 * 107 m–1. Now we solve for the wavelength of the photon emitted in a transition from level nU to level nL: hc hcR hcR 1 1 = EnU - EnL = a - 2 b - a - 2 b = hcRa 2 - 2 b l nU nL nL nU 1 1 1 = Ra 2 - 2 b l nL nU

(hydrogen wavelengths in the Bohr model, nL 6 nU)

(18.16)

Equation (18.16) is a theoretical prediction of the wavelengths found in the emission line spectrum of hydrogen atoms. When a hydrogen atom absorbs a photon, an electron makes a transition from a level nL to a higher level nU. This can happen only if the photon energy hc>l is equal to EnU - EnL, which is the same condition expressed by Eq. (18.16). So this equation also predicts the wavelengths found in the absorption line spectrum of hydrogen. How does this prediction compare with experiment? If nL = 2, corresponding to transitions to the second energy level in Fig. 18.22, the wavelengths predicted by Eq. (18.16) are all in the visible and ultraviolet parts of the electromagnetic spectrum. These wavelengths are collectively called the Balmer series (Fig. 18.23). If we let nL 5 2 and nU 5 3 in Eq. (18.16) we obtain the wavelength of the Ha line: 1 = 11.097 * 107 m-12 A 14 l

1 9

PhET: Neon Lights and Other Discharge Lamps ActivPhysics 18.2: Spectroscopy

B or l = 656.3 nm

With nL = 2 and nU = 4 we obtain the wavelength of the Hb line, and so on. With nL = 2 and nU = q we obtain the shortest wavelength in the series, l = 364.6 nm. These theoretical predictions are within 0.1% of the observed

18.23 The Balmer series of spectral lines for atomic hydrogen. You can see these same lines in the spectrum of molecular hydrogen (H2) shown in Fig. 18.6, as well as additional lines that are present only when two hydrogen atoms are combined to make a molecule. 364.6 nm

H`

410.2 nm

Hd

All Balmer lines beyond Hd are in the ultraviolet spectrum.

434.1 nm

Hg

486.1 nm

Hb Ha, Hb, Hg, and Hd are in the visible region of the spectrum.

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656.3 nm

Ha

588

CHAPTER 18 Particles Behaving as Waves

hydrogen wavelengths! This close agreement provides very strong and direct confirmation of Bohr’s theory. The Bohr model also predicts many other wavelengths in the hydrogen spectrum, as Fig. 18.22 shows. The observed wavelengths of all of these series, each of which is named for its discoverer, match the predicted values with the same percent accuracy as for the Balmer series. The Lyman series of spectral lines is caused by transitions between the ground level and the excited levels, corresponding to nL = 1 and nU = 2, 3, 4, Á in Eq. (18.16). The energy difference between the ground level and any of the excited levels is large, so the emitted photons have wavelengths in the ultraviolet part of the electromagnetic spectrum. Transitions among the higher energy levels involve a much smaller energy difference, so the photons emitted in these transitions have little energy and long, infrared wavelengths. That’s the case for both the Brackett series (nL = 3 and nU = 4, 5, 6, Á , corresponding to transitions between the third and higher energy levels) and the Pfund series (nL = 4 and nU = 5, 6, 7, Á , with transitions between the fourth and higher energy levels). Figure 18.22 shows only transitions in which a hydrogen atom loses energy and a photon is emitted. But as we discussed previously, the wavelengths of those photons that an atom can absorb are the same as those that it can emit. For example, a hydrogen atom in the n 5 2 level can absorb a 656.3-nm photon and end up in the n 5 3 level. One additional test of the Bohr model is its predicted value of the ionization energy of the hydrogen atom. This is the energy required to remove the electron completely from the atom. Ionization corresponds to a transition from the ground level 1n = 12 to an infinitely large orbit radius 1n = q 2, so the energy that must be added to the atom is E q - E1 = 0 - E1 = - E1 (recall that E1 is negative). Substituting the constants from Appendix F into Eq. (18.15) gives an ionization energy of 13.606 eV. The ionization energy can also be measured directly; the result is 13.60 eV. These two values agree within 0.1%.

Example 18.6

Exploring the Bohr model

Find the kinetic, potential, and total energies of the hydrogen atom in the first excited level, and find the wavelength of the photon emitted in a transition from that level to the ground level. SOLUTION IDENTIFY and SET UP: This problem uses the ideas of the Bohr model. We use simplified versions of Eqs. (18.12), (18.13), and (18.14) to find the energies of the atom, and Eq. (18.16), hc>l = EnU - EnL, to find the photon wavelength l in a transition from nU = 2 (the first excited level) to nL = 1 (the ground level). EXECUTE: We could evaluate Eqs. (18.12), (18.13), and (18.14) for the nth level by substituting the values of m, e, P0, and h. But we can simplify the calculation by comparing with Eq. (18.15), which shows that the constant me4/8P02h2 that appears in Eqs. (18.12), (18.13), and (18.14) is equal to hcR: me4 8P02h2

= hcR = 16.626 * 10-34 J # s212.998 * 108 m>s2 * 11.097 * 107 m-12

= 2.179 * 10

-18

This allows us to rewrite Eqs. (18.12), (18.13), and (18.14) as Kn =

13.60 eV n2

Un =

- 27.20 eV n2

En =

-13.60 eV n2

For the first excited level 1n = 22, we have K2 = 3.40 eV, U2 = -6.80 eV, and E2 = -3.40 eV. For the ground level 1n = 12, E1 = - 13.60 eV. The energy of the emitted photon is then E2 - E1 = -3.40 eV - 1- 13.60 eV2 = 10.20 eV, and 14.136 * 10-15 eV # s213.00 * 108 m > s2 hc = E2 - E1 10.20 eV '

l =

'

= 1.22 * 10-7 m = 122 nm This is the wavelength of the Lyman-alpha 1La2 line, the longestwavelength line in the Lyman series of ultraviolet lines in the hydrogen spectrum (see Fig. 18.22). EVALUATE: The total mechanical energy for any level is negative and is equal to one-half the potential energy. We found the same energy relationship for Newtonian satellite orbits in Section 12.4. The situations are similar because both the electrostatic and gravitational forces are inversely proportional to 1 > r 2.

J = 13.60 eV

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'

'

'

18.3 Energy Levels and the Bohr Model of the Atom

Nuclear Motion and the Reduced Mass of an Atom The Bohr model is so successful that we can justifiably ask why its predictions for the wavelengths and ionization energy of hydrogen differ from the measured values by about 0.1%. The explanation is that we assumed that the nucleus (a proton) remains at rest. However, as Fig. 18.24 shows, the proton and electron both revolve in circular orbits about their common center of mass (see Section 8.5). It turns out that we can take this motion into account very simply by using in Bohr’s equations not the electron rest mass m but a quantity called the reduced mass mr of the system. For a system composed of two bodies of masses m1 and m2, the reduced mass is mr =

m1m2 m1 + m2

18.24 The nucleus and the electron both orbit around their common center of mass. The distance rN has been exaggerated for clarity; for ordinary hydrogen it actually equals re > 1836.2. '

'

Nucleus mN

(18.17)

For ordinary hydrogen we let m1 equal m and m2 equal the proton mass, mp = 1836.2m. Thus the proton–electron system of ordinary hydrogen has a reduced mass of mr =

589

m11836.2m2 = 0.99946m m + 1836.2m

When this value is used instead of the electron mass m in the Bohr equations, the predicted values agree very well with the measured values. In an atom of deuterium, also called heavy hydrogen, the nucleus is not a single proton but a proton and a neutron bound together to form a composite body called the deuteron. The reduced mass of the deuterium atom turns out to be 0.99973m. Equations (18.15) and (18.16) (with m replaced by mr2 show that all wavelengths are inversely proportional to mr. Thus the wavelengths of the deuterium spectrum should be those of hydrogen divided by 10.99973m2> 10.99946m2 = 1.00027. This is a small effect but well within the precision of modern spectrometers. This small wavelength shift led the American scientist Harold Urey to the discovery of deuterium in 1932, an achievement that earned him the 1934 Nobel Prize in chemistry.

Hydrogenlike Atoms We can extend the Bohr model to other one-electron atoms, such as singly ionized helium 1He+2, doubly ionized lithium 1Li2 + 2, and so on. Such atoms are called hydrogenlike atoms. In such atoms, the nuclear charge is not e but Ze, where Z is the atomic number, equal to the number of protons in the nucleus. The effect in the previous analysis is to replace e2 everywhere by Ze2. In particular, the orbital radii rn given by Eq. (18.8) become smaller by a factor of Z, and the energy levels En given by Eq. (18.14) are multiplied by Z 2. We invite you to verify these statements. The reduced-mass correction in these cases is even less than 0.1% because the nuclei are more massive than the single proton of ordinary hydrogen. Figure 18.25 compares the energy levels for H and for He+, which has Z = 2. Atoms of the alkali metals (at the far left-hand side of the periodic table; see Appendix D) have one electron outside a core consisting of the nucleus and the inner electrons, with net core charge + e. These atoms are approximately hydrogenlike, especially in excited levels. Physicists have studied alkali atoms in which the outer electron has been excited into a very large orbit with n = 1000. In accordance with Eq. (18.8), the radius of such a Rydberg atom with n = 1000 is n2 = 106 times the Bohr radius, or about 0.05 mm—about the same size as a small grain of sand. Although the Bohr model predicted the energy levels of the hydrogen atom correctly, it raised as many questions as it answered. It combined elements of classical physics with new postulates that were inconsistent with classical ideas.

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Electron m

cm rN

re r

590

CHAPTER 18 Particles Behaving as Waves

18.25 Energy levels of H and He +. The energy expression, Eq. (18.14), is multi+ plied by Z2 = 4 for He , so the energy of an He + ion with a given n is almost exactly four times that of an H atom with the same n. (There are small differences of the order of 0.05% because the reduced masses are slightly different.)

Hydrogen (H)

Helium ion (He1)

E n53

E3 5 21.5 eV

n52

E2 5 23.4 eV

n51

E1 5 213.6 eV

n56 n55 n54

E6 5 21.5 eV E5 5 22.2 eV E4 5 23.4 eV

n53

E3 5 26.0 eV

n52

E2 5 213.6 eV

n51

E1 5 254.4 eV

The model provided no insight into what happens during a transition from one orbit to another; the angular speeds of the electron motion were not in general the angular frequencies of the emitted radiation, a result that is contrary to classical electrodynamics. Attempts to extend the model to atoms with two or more electrons were not successful. An electron moving in one of Bohr’s circular orbits forms a current loop and should produce a magnetic dipole moment (see Section 27.7). However, a hydrogen atom in its ground level has no magnetic moment due to orbital motion. In Chapters 40 and 41 we will find that an even more radical departure from classical concepts was needed before the understanding of atomic structure could progress further.

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CHAPTER

18

SUMMARY

De Broglie waves and electron diffraction: Electrons and other particles have wave properties. A particle’s wavelength depends on its momentum in the same way as for photons. A nonrelativistic electron accelerated from rest through a potential difference Vba has a wavelength given by Eq. (18.3). Electron microscopes use the very small wavelengths of fast-moving electrons to make images with resolution thousands of times finer than is possible with visible light. (See Examples 18.1–18.3.)

l =

h h = p mv

Incident electron waves in phase Scattered electron waves in phase

(18.1)

E = hƒ

l

(18.2)

(18.3) d

The nuclear atom: The Rutherford scattering experiments show that most of an atom’s mass and all of its positive charge are concentrated in a tiny, dense nucleus at the center of the atom. (See Example 18.4.)

hc = Ei - Ef l

Atomic line spectra and energy levels: The energies of atoms are quantized: They can have only certain definite values, called energy levels. When an atom makes a transition from an energy level Ei to a lower level Ef, it emits a photon of energy Ei - Ef. The same photon can be absorbed by an atom in the lower energy level, which excites the atom to the upper level. (See Example 18.5.)

hƒ =

The Bohr model: In the Bohr model of the hydrogen atom, the permitted values of angular momentum are integral multiples of h > 2p. The integer multiplier n is called the principal quantum number for the level. The orbital radii are proportional to n2 and the orbital speeds are proportional to 1 > n. The energy levels of the hydrogen atom are given by Eq. (18.15), where R is the Rydberg constant. (See Example 18.6.)

h 2p 1n = 1, 2, 3, Á 2

'

'

'

'

Atom on crystal surface

50°

h h = l = p !2meVba

(18.5)

Nucleus

a

Ei

i

hf 5 Ei 2 Ef Ef

f

Ln = mvnrn = n

rn = P0

n2h2

pme2

(18.6)

= n2a0

= n 15.29 * 10 2

-11

(18.8)

F rn

m2

Electron m, 2e vn

(18.10)

2.19 * 10 m > s 1 e = (18.9) n P0 2nh 6

2

'

vn =

Proton M, 1e

En = -

hcR n2

= -

1n = 1, 2, 3, Á 2

13.60 eV n2

'

(18.15)

591

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592

CHAPTER 18 Particles Behaving as Waves

BRIDGING PROBLEM

Hot Stars and Hydrogen Clouds

Figure 18.36 shows a cloud, or nebula, of glowing hydrogen in interstellar space. The atoms in this cloud are excited by shortwavelength radiation emitted by the bright blue stars at the center of the nebula. (a) The blue stars act as blackbodies and emit light with a continuous spectrum. What is the wavelength at which a star with a surface temperature of 15,100 K (about 212 times the surface temperature of the sun) has the maximum spectral emittance? In what region of the electromagnetic spectrum is this? (b) Figure 18.32 shows that most of the energy radiated by a blackbody is at wavelengths between about one half and three times the wavelength of maximum emittance. If a hydrogen atom near the star in part (a) is initially in the ground level, what is the principal quantum number of the highest energy level to which it could be excited by a photon in this wavelength range? (c) The red color of the nebula is primarily due to hydrogen atoms making a transition from n 5 3 to n 5 2 and emitting photons of wavelength 656.3 nm. In the Bohr model as interpreted by de Broglie, what are the electron wavelengths in the n 5 2 and n 5 3 levels? SOLUTION GUIDE See MasteringPhysics® study area for a Video Tutor solution.

18.36 The Rosette Nebula.

IDENTIFY and SET UP 1. To solve this problem you need to use your knowledge of both blackbody radiation (Section 18.5) and the Bohr model of the hydrogen atom (Section 18.3). 2. In part (a) the target variable is the wavelength at which the star emits most strongly; in part (b) the target variable is a principal quantum number, and in part (c) it is the de Broglie wavelength of an electron in the n 5 2 and n 53 Bohr orbits (see Fig. 18.22). Select the equations you will need to find the target variables. (Hint: In Section 18.5 you learned how to find the energy change involved in a transition between two given levels of a hydrogen atom. Part (b) is a variation on this: You are to find the final level in a transition that starts in the n 5 1 level and involves the absorption of a photon of a given wavelength and hence a given energy.) EXECUTE 3. Use the Wien displacement law to find the wavelength at which the star has maximum spectral emittance. In what part of the electromagnetic spectrum is this wavelength? 4. Use your result from step 3 to find the range of wavelengths in which the star radiates most of its energy. Which end of this range corresponds to a photon with the greatest energy? 5. Write an expression for the wavelength of a photon that must be absorbed to cause an electron transition from the ground level 1n = 12 to a higher level n. Solve for the value of n that corresponds to the highest-energy photon in the range you calculated in step 4. (Hint: Remember than n must be an integer.) 6. Find the electron wavelengths that correspond to the n 5 2 and n 5 3 orbits shown in Fig. 18.20. EVALUATE 7. Check your result in step 5 by calculating the wavelength needed to excite a hydrogen atom from the ground level into the level above the highest-energy level that you found in step 5. Is it possible for light in the range of wavelengths you found in step 4 to excite hydrogen atoms from the ground level into this level? 8. How do the electron wavelengths you found in step 6 compare to the wavelength of a photon emitted in a transition from the n 5 3 level to the n 5 2 level?

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Single Correct

Problems

593

For instructor-assigned homework, go to www.masteringphysics.com

. , .. , ... : Problems of increasing difficulty. CP: Cumulative problems incorporating material from earlier chapters. CALC: Problems requiring calculus. BIO: Biosciences problems.

DISCUSSION QUESTIONS Q18.1 If a proton and an electron have the same speed, which has the longer de Broglie wavelength? Explain. Q18.2 If a proton and an electron have the same kinetic energy, which has the longer de Broglie wavelength? Explain. Q18.3 Does a photon have a de Broglie wavelength? If so, how is it related to the wavelength of the associated electromagnetic wave? Explain. Q18.4 When an electron beam goes through a very small hole, it produces a diffraction pattern on a screen, just like that of light. Does this mean that an electron spreads out as it goes through the hole? What does this pattern mean? Q18.5 Galaxies tend to be strong emitters of Lyman-a photons (from the n = 2 to n = 1 transition in atomic hydrogen). But the intergalactic medium—the very thin gas between the galaxies— tends to absorb Lyman-a photons. What can you infer from these observations about the temperature in these two environments? Explain. Q18.6 A doubly ionized lithium atom 1Li + + 2 is one that has had two of its three electrons removed. The energy levels of the remaining single-electron ion are closely related to those of the hydrogen atom. The nuclear charge for lithium is 63e instead of just +e. How are the energy levels related to those of hydrogen? How is the radius of the ion in the ground level related to that of the hydrogen atom? Explain. Q18.7 The emission of a photon by an isolated atom is a recoil process in which momentum is conserved. Thus Eq. (18.5) should include a recoil kinetic energy Kr for the atom. Why is this energy negligible in that equation? Q18.8 How might the energy levels of an atom be measured directly—that is, without recourse to analysis of spectra? Q18.9 Elements in the gaseous state emit line spectra with welldefined wavelengths. But hot solid bodies always emit a continuous spectrum—that is, a continuous smear of wavelengths. Can you account for this difference? Q18.10 As a body is heated to a very high temperature and becomes self-luminous, the apparent color of the emitted radiation shifts from red to yellow and finally to blue as the temperature increases. Why does the color shift? What other changes in the character of the radiation occur? Q18.11 The peak-intensity wavelength of red dwarf stars, which have surface temperatures around 3000 K, is about 1000 nm, which is beyond the visible spectrum. So why are we able to see these stars, and why do they appear red? Q18.12 You have been asked to design a magnet system to steer a beam of 54-eV electrons like those described in Example 18.1 (Section 18.1). The goal is to be able to direct the electron beam to a specific target location with an accuracy of 61.0 mm. In your design, do you need to take the wave nature of electrons into account? Explain. Q18.13 Why go through the expense of building an electron microscope for studying very small objects such as organic molecules? Why not just use extremely short electromagnetic waves, which are much cheaper to generate?

Q18.14 Which has more total energy: a hydrogen atom with an electron in a high shell (large n) or in a low shell (small n)? Which is moving faster: the high-shell electron or the low-shell electron? Is there a contradiction here? Explain. Q18.15 Does the uncertainty principle have anything to do with marksmanship? That is, is the accuracy with which a bullet can be aimed at a target limited by the uncertainty principle? Explain. Q18.16 Suppose a two-slit interference experiment is carried out using an electron beam. Would the same interference pattern result if one slit at a time is uncovered instead of both at once? If not, why not? Doesn’t each electron go through one slit or the other? Or does every electron go through both slits? Discuss the latter possibility in light of the principle of complementarity. Q18.17 Equation (18.30) states that the energy of a system can have uncertainty. Does this mean that the principle of conservation of energy is no longer valid? Explain. Q18.18 Laser light results from transitions from long-lived metastable states. Why is it more monochromatic than ordinary light? Q18.19 Why can an electron microscope have greater magnification than an ordinary microscope? Q18.20 When you check the air pressure in a tire, a little air always escapes; the process of making the measurement changes the quantity being measured. Think of other examples of measurements that change or disturb the quantity being measured.

SINGLE CORRECT Q18.1 The wavelength of first line of Lyman series will be maximum for (a) hydrogen atom (b) deuterium atom (c) helium ion (He+) (d) lithium ion (Li2+) n=5 Q18.2 The energy level diagram of a hypothetn=4 ical atom is shown in the figure. A transition n=3 n=2 from n = 3 to n = 2 produces visible radiation. The possible transition to obtain infrared n=1 radiation is (a) 4 S 3 (b) 2 S 1 (c) 4 S 1 (d) none of these Q18.3 Of the following transitions in hydrogen atom, the one which gives an absorption line of highest frequency is: (a) n = 1 to n = 2 (b) n = 3 to n = 8 (c) n = 2 to n = 1 (d) n = 8 to n = 3 Q18.4 A hydrogen atom in an excited state emits a photon which has the longest wavelength of the Paschen series. Further emissions from the atom cannot include the (a) longest wavelength of the Lyman series (b) second longest wavelength of the Lyman series (c) longest wavelength of the Balmer series (d) second longest wavelength of the Balmer series

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CHAPTER 18 Particles Behaving as Waves

Q18.5 Mosley law relates : (a) Frequency of emitted X-ray with applied voltage (b) Wavelength and intensity of X-ray. (c) Frequency of emitted X-ray with atomic number (d) Wavelength and angle of scattering. Q18.6 If the frequencies of Ka, Kb and La X-rays for a material vKa, vKb, vLa respectively, then (a) vKa = vKb + vLa (b) vLa = vKa + vKb (c) vKb = vKa + vLa (d) none of these Q18.7 If the potential difference applied across a Coolidge tube is increased : (a) the wavelength of the Ka line will increase (b) the wavelength of the Kbline will decrease (c) the difference in wavelength between the Ka and Kb will decrease (d) none of these Q18.8 In X-ray tube, when the accelerating voltage V is doubled, the difference between the wavelength of Ka line and the minimum cut off of continuous X-ray spectrum : (a) remains constant (b) becomes more than two times (c) becomes half (d) becomes less than 2 times. Q18.9 X-rays from a given X-ray tube operating under specified condition have a sharply defined minimum wavelength. The value of this minimum wavelength could be reduced by (a) increasing the temperature of the filament. (b) increasing the p.d. between the cathode and the target. (c) reducing the pressure in the tube. (d) using a target material of higher relative atomic mass. Q18.10 When a high speed electron beam hits a metal target, which of the following must happen? (i) Heat is produced (ii) Continuous X-rays are produced (iii) Characteristic X-rays are produced (a) (ii) & (iii) only (b) (ii) only (c) (i) & (ii) only (d) (i), (ii) & (ii) Q18.11 A beam of electrons striking a copper target produces X rays with the spectrum shown. Characteristic lines

Relative intensity

Wavelength Cutoff wavelength

If the copper target is replaced with a different metal, how will wavelength and characteristic lines of the new spectrum compare with the old? Cutoff Wavelength Characteristic lines (a) Unchanged Different (b) Unchanged Unchanged (c) Different Different (d) Different Unchanged

Q18.12 Characteristic X-rays (a) Have only discrete wavelength which are characteristic of the target (b) Have all the possible wavelength (c) Are characteristic of speed of projectile electrons (d) None of these

ONE OR MORE CORRECT Q18.13 Which of the following transitions of He+ ion will give rise to spectral line which has same wavelength as some spectral line in hydrogen atom? (a) n = 4 to n = 2 (b) n = 6 to n = 5 (c) n = 6 to n = 3 (d) n = 8 to n = 4 Q18.14 A beam of ultraviolet light of a mixture of several wavelengths passes through hydrogen gas at room temperature, in the x direction. Assume that all the photons emitted due to electron transitions inside the gas emerge in the y direction. Let A and B denote the lights emerging from the gas in the x and y directions (a) some of the incident wavelengths will be absent in A (b) Only those wavelengths will be present in B which are absent in A (c) B will contain some visible light (d) B will contain some infrared light Q18.15 If electron of the hydrogen atom is replaced by another particle of same charge but of double the mass, then: (a) Bohr radius will increase (b) ionisation energy of the atom will be doubled (c) speed of the new particle in a given state will be lesser than the electron's speed in same orbit (d) gap between energy levels will now be doubled. Q18.16 An electron of the kinetic energy 10 eV collidies with a hydrogen atom in 1st excited state. Assuming loss of kinetic energy in the collision to be quantized, the collision : (a) may be perfectly inelastic (b) may be inelastic (c) may be elastic (d) must be inelastic Q18.17 When in a hydrogen atom an electron makes a transition from a higher state to a lower state (a) angular momentum of the electron remains constant. (b) its kinetic energy increases (c) its potential energy decreases (d) its deBroglie wavelength decreases Q18.18 A hydrogen like gas atoms absorb radiations of wavelength l0 and consequently emit radiations of 6 difference wavelengths of which, two wavelengths are shorter than l0. Choose the correct alternative(s). (a) The final excited state of the atoms is n = 3. (b) The final excited state of the atoms is n = 4. (c) The initial state of the atoms is n = 2 (d) The initial state of the atoms is n = 3. Q18.19 If the potential energy is taken zero at ground state of hydrogen atom in place of infinity then (a) Total energy of atom will get changed. (b) Binding energy of hydrogen atom in ground state will remain unchanged. (c) Kinetic energy of electron in ground state will get changed. (d) Energy of photon generated becuase of de-excitation of atom from first excited state will remain unchanged.

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Problems

Q18.20 Using Bohr's theory in a one-electron atom. Choose the correct option(s) (n S principal quantum number, Z is number of protons in the nucleus) (a) Total energy of electron increases with n. (b) Total energy and potential energy of electrons is negative in all orbits. (c) Magnitude of angular momentum of electron increases with n. (d) Linear momentum of electron decreases with z for same n.

EXERCISES (Take h =

20 * 10 - 34J # s, me = 9 * 10 - 31Kg) 3

Section 18.1 Electron Waves

18.1 . (a) An electron moves with a speed of 5 * 106 m > s. What is its de Broglie wavelength? (b) A proton moves with the same speed. Determine its de Broglie wavelength. 18.2 . An electron has a de Broglie wavelength of 2 * 10-10 m. Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts). 18.3 . In the Bohr model of the hydrogen atom, what is the de Broglie wavelength for the electron when it is in (a) the n = 1 level and (b) the n = 4 level? In each case, compare the de Broglie wavelength to the circumference 2prn of the orbit. 18.4 . (a) A nonrelativistic free particle with mass m has kinetic energy K. Derive an expression for the de Broglie wavelength of the particle in terms of m and K. (b) What is the de Broglie wavelength of an 800-eV electron? 18.5 . Why Don’t We Diffract? (a) Calculate the de Broglie wavelength of a typical person walking through a doorway. Make reasonable approximations for the necessary quantities. (b) Will the person in part (a) exhibit wavelike behavior when walking through the “single slit” of a doorway? Why? 18.6 .. What is the de Broglie wavelength for an electron with speed (a) v = 0.480c and (b) v = 0.960c ? (Hint: Use the correct relativistic expression for linear momentum if necessary.) 18.7 .. (a) What accelerating potential is needed to produce electrons of wavelength 5.00 nm? (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)? '

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Section 18.2 The Nuclear Atom and Atomic Spectra

18.8 .. CP A 5 MeV alpha particle from a 226Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

Section 18.3 Energy Levels and the Bohr Model of the Atom

18.9 .. A hydrogen atom is in a state with energy - 1.51 eV. In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus? 18.10 . A triply ionized beryllium ion, Be3 + (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the

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ground-level energy of Be3 + ? How does this compare to the groundlevel energy of the hydrogen atom? (b) What is the ionization energy of Be3 + ? How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the n = 2 to n = 1 transition is 122 nm (see Example 18.6). What is the wavelength of the photon emitted when a Be3 + ion undergoes this transition? (d) For a given value of n, how does the radius of an orbit in Be3 + compare to that for hydrogen? 18.11 .. (a) Show that, as n gets very large, the energy levels of the hydrogen atom get closer and closer together in energy. (b) Do the radii of these energy levels also get closer together? 18.12 . CP The energy-level scheme for the hypothetical one- Figure E18.12 electron element Searsium is n 5 4 22 eV shown in Fig. E18.12. The n 5 3 25 eV potential energy is taken to be zero for an electron at an infi- n 5 2 210 eV nite distance from the nucleus. (a) How much energy (in electron volts) does it take to ionize an electron from the ground n 5 1 220 eV level? (b) An 18-eV photon is absorbed by a Searsium atom in its ground level. As the atom returns to its ground level, what possible energies can the emitted photons have? Assume that there can be transitions between all pairs of levels. (c) What will happen if a photon with an energy of 8 eV strikes a Searsium atom in its ground level? Why? (d) Photons emitted in the Searsium transitions n = 3 S n = 2 and n = 3 S n = 1 will eject photoelectrons from an unknown metal, but the photon emitted from the transition n = 4 S n = 3 will not. What are the limits (maximum and minimum possible values) of the work function of the metal? 18.13 . Find the longest and shortest wavelengths in the Lyman and Paschen series for hydrogen. In what region of the electromagnetic spectrum does each series lie? 18.14 .. Use Balmer’s formula to calculate (a) the wavelength, (b) the frequency, and (c) the photon energy for the Hg line of the Balmer series for hydrogen.

PROBLEMS 18.15 .. The negative muon has a charge equal to that of an electron but a mass that is 200 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the n = 2 level to the n = 1 level? 18.16 . An atom with mass m emits a photon of wavelength l. (a) What is the recoil speed of the atom? (b) What is the kinetic energy K of the recoiling atom? (c) Find the ratio K > E, where E is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate K (in electron volts) and K > E for a hydrogen atom (mass 1.67 * 10-27 kg) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process? 18.17 ... A sample of hydrogen atoms is irradiated with light with wavelength 80 nm, and electrons are observed leaving the gas. (a) If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of '

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596

CHAPTER 18 Particles Behaving as Waves

these photoelectrons? (b) A few electrons are detected with energies as much as 10.2 eV greater than the maximum kinetic energy calculated in part (a). How can this be? 18.18 .. The Red Supergiant Betelgeuse. The star Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that large, we would be inside it!) Assume that it radiates like an ideal blackbody. (a) If Betelgeuse were to radiate all of its energy at the peak-intensity wavelength, how many photons per second would it radiate? (b) Find the ratio of the power radiated by Betelgeuse to the power radiated by our sun (at 5800 K). 18.19 ... When a photon is emitted by an atom, the atom must recoil to conserve momentum. This means that the photon and the recoiling atom share the transition energy. (a) For an atom with mass m, calculate the correction ¢l due to recoil to the wavelength of an emitted photon. Let l be the wavelength of the photon if recoil is not taken into consideration. (b) Evaluate the correction for a hydrogen atom in which an electron in the nth level returns to the ground level. How does the answer depend on n? 18.20 . (a) What is the energy of a photon that has wavelength 0.10 mm ? (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10 mm in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons? 18.21 .. CP An electron beam and a photon beam pass through identical slits. On a distant screen, the first dark fringe occurs at the same angle for both of the beams. The electron speeds are much slower than that of light. (a) Express the energy of a photon in terms of the kinetic energy K of one of the electrons. (b) Which is greater, the energy of a photon or the kinetic energy of an electron? 18.22 .. CALC Zero-Point Energy. Consider a particle with mass m moving in a potential U = 12 kx2, as in a mass–spring system. The total energy of the particle is E = p2 > 2m + 12 kx2. Assume that p and x are approximately related by the Heisenberg uncertainty principle, so px L h. (a) Calculate the minimum possible value of the energy E, and the value of x that gives this minimum E. This lowest possible energy, which is not zero, is called the zero-point energy. (b) For the x calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle? 18.23 A monochromatic light source of frequency v illuminates a metallic surface and ejects photoelectrons. The photoelectrons emitted are able to excite the hydrogen atom which then emits a photon of wavelength 1237Å . When the whole experiment is repeated with light source of frequency 13v/10 , the photoelectrons emitted are just able to ionize the hydrogen atom in ground state. Find the work function of the metal and frequency v of the source. 18.24 A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV. (i) What is the initial state of hydrogen atom? (ii) What is the final state of hydrogen atom? (iii) What is the wavelength of the photon emitted? '

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18.25 A Bohr hydrogen atom undergoes a transition n = 5 S n = 4 and emits a photon of frequency v. Frequency of circular motion of electron in n = 4 orbit is v4. Find the ratio v/v4. 18.26 The K, L and M energy levels of platinum lie roughly at 78, 12, and 3 KeV respectively. The ratio of wavelength of Ka line to that of Kb line in X-ray spectrum is 22 3 22 (c) 25 (a)

3 22 25 (d) 22 (b)

18.27 The X-ray energy levels of rubidium are 1.76 × 104 eV for the K-shell 4.68 × 103 eV for the L-shell and 2.64 × 103 eV for the M-shell. Minimum energy of the incident electrons to produce Ka and Kblines is __________. 18.28 The energy of a K-electron is tungsten is –20keV and of an L-electron is –2keV. The wavelength of ka X-ray photon emitted when a K-electron is knocked out is ___________Å. 18.29 In an X-ray tube the accelerating voltage is 20 KV. Two targets A and B are used one by one. For ‘A’ the wavelength of the Ka line is 62 pm. For ‘B’ the wavelength of the La line is 124 pm. The energy of the ‘B’ ion with vacancy in ‘L’ shell is 15.5 KeV higher than the atom of B. [Take hc = 12400eVÅ] (i) Find lmin in Å. (ii) Can Ka – photon be emitted by ‘A’? Explain with reason. (iii) Can L – photons be emitted by ‘B’? What is the minimum wavelength (in Å) of the characteristic L series X-ray that will be emitted by ‘B’. 18.30 Figure shows Ka & Kb X-rays along with continuous X-ray. Find the energy of La X-ray. (Use hc = 12420evÅ).

Ka 1Å

Kb

2Å

CHALLENGE PROBLEMS 18.31 ... (a) Show that in the Bohr model, the frequency of revolution of an electron in its circular orbit around a stationary hydrogen nucleus is ƒ = me4 > 4P02n3h3. (b) In classical physics, the frequency of revolution of the electron is equal to the frequency of the radiation that it emits. Show that when n is very large, the frequency of revolution does indeed equal the radiated frequency calculated from Eq. (18.5) for a transition from n1 = n + 1 to n2 = n. (This illustrates Bohr’s correspondence principle, which is often used as a check on quantum calculations. When n is small, quantum physics gives results that are very different from those of classical physics. When n is large, the differences are not significant, and the two methods then “correspond.” In fact, when Bohr first tackled the hydrogen atom problem, he sought to determine ƒ as a function of n such that it would correspond to classical results for large n) '

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Answers

597

Answers Chapter Opening Question

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The smallest detail visible in an image is comparable to the wavelength used to make the image. Electrons can easily be given a large momentum p and hence a short wavelength l = h > p, and so can be used to resolve extremely fine details. (See Section 18.1.) '

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Test Your Understanding Questions 18.1 (a) (i), (b) no From Example 18.2, the speed of a particle is v = h > lm and the kinetic energy is K = 12 mv2 = 1m > 221h > lm22 = h2 > 2l2m. This shows that for a given wavelength, the kinetic energy is inversely proportional to the mass. Hence the proton, with a smaller mass, has more kinetic energy than the neutron. is photon a of energy the (b), part For E = hf, and the frequency of a photon is ƒ = c > l. Hence E = hc > l and l = hc > E = 14.136 * 10-15 eV # s212.998 * This is more than 100 times greater than the wavelength of an electron of the same energy. While both photons and electrons have wavelike properties, they have different relationships between their energy and momentum and hence between their frequency and wavelength. 18.2 (iii) Because the alpha particle is more massive, it won’t bounce back in even a head-on collision with a proton that’s initially at rest, any more than a bowling ball would when colliding with a Ping-Pong ball at rest (see Fig. 8.22b). Thus there would be no large-angle scattering in this case. Rutherford saw large-angle scattering in his experiment because gold nuclei are more massive than alpha particles (see Fig. 8.22a). 18.3 (iv) Figure 18.25 shows that many (though not all) of the energy levels of are the same as those of H. Hence photons emitted during transitions between corresponding pairs of levels in and H have the same energy E and the same wavelength An H atom that drops from the level to the level emits a photon of energy 10.20 eV and wavelength 122 nm (see Example 18.6); a ion emits a photon of the same energy and wavelength when it drops from the level to the level. Inspecting Fig. 18.25 will show you that every even-numbered level in matches a level in H, while none of the odd-numbered levels do. The first three transitions given in the question (n 5 2 to n 5 1, n 5 3 to n 5 2, and n 5 4 to n 5 3) all involve an odd-numbered level, so none of their wavelengths match a wavelength emitted by H atoms. 18.4 (i) In a neon light fixture, a large potential difference is applied between the ends of a neon-filled glass tube. This ionizes some of the neon atoms, allowing a current of electrons to flow through the gas. Some of the neon atoms are struck by fast-moving electrons, making them transition to an excited level. From this level the atoms undergo spontaneous emission, as depicted in Fig. 18.28b, and emit 632.8-nm photons in the process. No population inversion occurs and the photons are not trapped by mirrors as shown in Fig. 18.29d, so there is no stimulated emission. Hence there is no laser action. 18.5 (a) yes, (b) yes The Planck radiation law, Eq. (18.24), shows that an ideal blackbody emits radiation at all wavelengths: The spectral emittance is equal to zero only for and in the limit So a blackbody at 2000 K does indeed emit both x rays and radio waves. However, Fig. 18.32 shows that the spectral emittance for this temperature is very low for wavelengths much shorter than '

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Bridging Problem Answers: (a) 192 nm; ultraviolet (b) n 5 4

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(including x rays) and for wavelengths much longer than a few (including radio waves). Hence such a blackbody emits very little in the way of x rays or radio waves. 18.6 (i) and (iii) (tie), (ii) and (iv) (tie) According to the Heisenberg uncertainty principle, the smaller the uncertainty in the xcoordinate, the greater the uncertainty in the x-momentum. The relationship between and does not depend on the mass of the particle, and so is the same for a proton as for an electron.

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Discussion Questions h So electron's de Broglie wavelength will be greater. mv h as mp>me, so le>lp Q18.2 l = 2(KE)m È Q18.3 Photons are massless entities. In that sense photons do not have de Broglie wavelength. h As l = mv On the other hand if we use c hv then l = Momentum = v c But this wavelength is generally not categorised as de' Broglie wavelength. Q18.4 Interaction with slit scatters the electrons creating a spread ¢x in px of width . This is similar to diffraction of light. h

Q18.1 l =

x

Px = 0

Px = 0 ±

∆x h

Q18.5 Temperature of galaxies must be much higher than temperature in inter galactic space. Q18.6 Doubly ionized Lithium (Li2+) is a single electron element, like hydrogen but it has 122.4 eV as ionization energy for hydrogen ionization energy is -13.6eV. Q18.7 In equation(18.5) hc = Ei - Ef hf = l the recoil KE is almost negligible p2 E2 or KE = as KE = 2m 2mc2 but E ≈ 1.85 eV So, KE is negligible. Q18.8 It takes a minimum amount of energy(ionization energy) to remove electron from the hydrogen atom. The different energy levels of E0 a hydrogen atoms are given by the equation En = 2 . So by calculating n ionization energy. Energy level of atom can be ascertained. Q18.9 Each gas emits a particular set of spectral lines or has line spectrum; this provides a method of element identification. This is produced due to specific transitions of electrons in energy levels.

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598

CHAPTER 18 Particles Behaving as Waves

In solids (hot) electrons undergo thermal agitation and emit emradiations of many different wavelength, hence producing continuous spectrum. Q18.10 With the increase in temperature the wavelength of radiations decreases, this makes the colour change from red to yellow to blue. Q18.11 Red dwarf stars burn a little bit of fuel at a time the coolest part of the fire is at the top of the flame where it grows red; the hotter part in the middle glows yellow. We see the red part of it. Q18.12 No, the wave nature will not be considere. The magnetic force on electrons in calculated considering paticle nature (F = Bqvsinu) Q18.13 Resolving power increases due to smaller wavelength of matter waves of e - as compared to short electromagnetic waves. 13.6 Q18.14 Energy in nth orbit is En = - 2 eV . So, greater n, n mv2n (Ze)e will give greater energy state.We know = K 2 rn rn ‹

vn =

k ä

ze2 mrn

So, greater rn will give smaller vn. Energy in an orbit infact sum of KE and electrostatic potential energy. This sum comes out to - ve indicating binding energy. Q18.15 Uncertainty principle is related to quantum mechanical concepts. The bullets speed and size are not of that order over which quantum mechanical concepts can be suitably considered. Q18.16 Electrons like bullets strike the target one at a time. Yet like waves, they create an interference pattern. Although each electron arrives at the target at a single place and time, it seems that each has passed through or somehow felt the presence of both the slits at once. Thus the electrons show wave-particle nature. Q18.17 Principle of energy conservation is not violated. In the case under consideration mechanical form of energy and electromagnetic form of energy both are to be considered. Q18.18 Because the wavelength variation width (≤l) is very very small. Q18.19 Wavelength of an electron can be upto 1000000 times shorter than that of visible photons the electron microscope has a higher resolving power than an ordinary microscope. Its resolution in also very high. Q18.20 We can think of measuring of current by ammeter or potential difference by voltmeter. Ammeter is not ideal (R Z 0) and so as the voltmeter.

Single correct Q18.1 (a) Q18.2 (a) Q18.3 (a) Q18.4 (d) Q18.5 (c) Q18.6 (c) Q18.7 (d) Q18.8 (d) Q18.9 (b) Q18.10 (c) Q18.11 (a) Q18.12 (a)

One or More Correct Q18.13 (a), (d) Q18.14 (a), (c), (d) Q18.15 (b), (d) Q18.16 (a), (b), (c) Q18.17 (b), (c), (d) Q18.18 (b), (c) Q18.19 (a), (b), (d) Q18.20 (a), (b), (c)

Exercises 18.1 (a) 0.15 nm (b) 8.1×10 - 14 m. 18.2 (a) 3.3×10 - 24kg - m/s, (b) 6.2×10 - 18J, 38.75 eV. 18.3 (a) 3.32×10 - 10 m, equal to circumfrence of orbit 1 (b) 1.33×10 - 9 m, equal to times circumference. 4 h 18.4 (a) , (b) 4.3×10 - 11m È2Km 18.5 (a) 8.8×10 - 36m, (b) No

18.6 (a) 4.43×10 - 12m (b) 7.07×10 - 13m 18.7 (a) 0.0607 V (b) 248 eV (c) 20.5mm 18.8 (a) 5.3×10 - 14m (b) 15 N 18.9 3.16×10 - 34 kg - m2/s. 18.10 (a) -218 eV, 16 times, (b) 218 eV, 16 times (c) 7.6 nm, (d) one fourth. 18.11 (b) No 18.12 (a) 20 eV, (b) 3 eV, 5 eV, 8 eV, 10 eV, 15 eV, 18 eV. (c) No absorption of photon (d) 3 eV to 5 eV 18.13 Lyman: lmax = 121.5 nm, lmin = 91.16 nm Paschen: lmax = 1875 nm, lmin = 820 nm 18.14 (a) 434 nm (b) 6.9×1014HZ (c) 2.856 eV. 18.15 (a) 1.21×10 - 28 Kg (b) 1.83 KeV (c) 0.91 nm h h h2 18.16 (a) (b) (c) , important for small m and l 2 ml 2mcl 2ml (d) 5.42×10 - 9, No 18.17 (a) 2 eV 18.18 (a) 5×1049 (b) 3×104 h 18.19 (a) (b) 6.6×10 - 16m 2mc 18.20 (a) 12 eV (b) 1.5×10 - 4V, 7.4×103m/s (c) 8.2×10 - 8V, 4m/s 18.21 (a) CÈ2mK (b) photon

18.22 (a) Emin = hÈK/m,x = a 18.23 2 eV; 3 * 1015 Hz

h 1/2 b (b) 1 ÈmK

18.24 (i) n1 = 4, (ii) n2 = 2, (iii) l = 4860Å 18.25 0.72 18.26 D 18.27 17.6 keV 18.28 0.69Å 18.29 (i) 0.62Å, (ii) No, (iii) Yes 0.8Å 18.30 6210 eV

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19

NUCLEAR PHYSICS

LEARNING GOALS By studying this chapter, you will learn: • Some key properties of atomic nuclei, including radii, densities, spins, and magnetic moments. • How the binding energy of a nucleus depends on the numbers of protons and neutrons that it contains.

?

This sculpture of a wooly mammoth, just 3.7 cm (1.5 in.) in length, was carved from a mammoth’s ivory tusk by an artist who lived in southwestern Germany 35,000 years ago. What physical principles make it possible to date biological specimens such as these?

uring the past century, applications of nuclear physics have had enormous effects on humankind, some beneficial, some catastrophic. Many people have strong opinions about applications such as bombs and reactors. Ideally, those opinions should be based on understanding, not on prejudice or emotion, and we hope this chapter will help you to reach that ideal. Every atom contains at its center an extremely dense, positively charged nucleus, which is much smaller than the overall size of the atom but contains most of its total mass. We will look at several important general properties of nuclei and of the nuclear force that holds them together. The stability or instability of a particular nucleus is determined by the competition between the attractive nuclear force among the protons and neutrons and the repulsive electrical interactions among the protons. Unstable nuclei decay, transforming themselves spontaneously into other nuclei by a variety of processes. Nuclear reactions can also be induced by impact on a nucleus of a particle or another nucleus. Two classes of reactions of special interest are fission and fusion. We could not survive without the energy released by one nearby fusion reactor, our sun.

D

19.1

• The most important ways in which unstable nuclei undergo radioactive decay. • How the decay rate of a radioactive substance depends on time. • Some of the biological hazards and medical uses of radiation. • How to analyze some important types of nuclear reactions. • What happens in a nuclear fission chain reaction, and how it can be controlled. • The sequence of nuclear reactions that allow the sun and stars to shine.

Properties of Nuclei

As we described in Section 18.2, Rutherford found that the nucleus is tens of thousands of times smaller in radius than the atom itself. Since Rutherford’s initial experiments, many additional scattering experiments have been performed, using high-energy protons, electrons, and neutrons as well as alpha particles (helium-4 nuclei). These experiments show that we can model a nucleus as a sphere with a radius R that depends on the total number of nucleons (neutrons 599

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600

CHAPTER 19 Nuclear Physics

and protons) in the nucleus. This number is called the nucleon number A. The radii of most nuclei are represented quite well by the equation R = R0 A1>3 '

1radius of a nucleus2

(19.1)

where R0 is an experimentally determined constant: R0 = 1.2 * 10-15 m = 1.2 fm The nucleon number A in Eq. (19.1) is also called the mass number because it is the nearest whole number to the mass of the nucleus measured in unified atomic mass units (u). (The proton mass and the neutron mass are both approximately 1 u.) The best current conversion factor is 1 u = 1.6605387821832 * 10-27 kg In Section 19.2 we’ll discuss the masses of nuclei in more detail. Note that when we speak of the masses of nuclei and particles, we mean their rest masses.

Nuclear Density

The volume V of a sphere is equal to 4pR3>3, so Eq. (19.1) shows that the volume of a nucleus is proportional to A. Dividing A (the approximate mass in u) by the volume gives us the approximate density and cancels out A. Thus all nuclei have approximately the same density. This fact is of crucial importance in understanding nuclear structure.

Example 19.1

Calculating nuclear properties

The most common kind of iron nucleus has mass number A = 56. Find the radius, approximate mass, and approximate density of the nucleus.

The volume V of the nucleus (which we treat as a sphere of radius R) and its density r are V = 43 pR3 = 43 pR03A = 43 p14.6 * 10-15 m23 = 4.1 * 10-43 m3

SOLUTION IDENTIFY and SET UP: Equation (19.1) tells us how the nuclear radius R depends on the mass number A. The mass of the nucleus in atomic mass units is approximately equal to the value of A, and the density r is mass divided by volume. EXECUTE: The radius and approximate mass are R = R0 A1>3 = 11.2 * 10-15 m215621>3 = 4.6 * 10-15 m = 4.6 fm

m L 156 u211.66 * 10-27 kg>u2 = 9.3 * 10-26 kg

r =

9.3 * 10-26 kg m L = 2.3 * 1017 kg>m3 V 4.1 * 10-43 m3

EVALUATE: As we mentioned above, all nuclei have approximately this same density. The density of solid iron is about 7000 kg>m3; the iron nucleus is more than 1013 times as dense as iron in bulk. Such densities are also found in neutron stars, which are similar to gigantic nuclei made almost entirely of neutrons. A 1-cm cube of material with this density would have a mass of 2.3 * 1011 kg, or 230 million metric tons!

Nuclides and Isotopes The building blocks of the nucleus are the proton and the neutron. In a neutral atom, the nucleus is surrounded by one electron for every proton in the nucleus. We introduced these particles in Section 1.1; we’ll recount the discovery of the neutron and proton. The masses of these particles are Proton:

mp = 1.007276 u = 1.672622 * 10-27 kg

Neutron:

mn = 1.008665 u = 1.674927 * 10-27 kg

Electron:

me = 0.000548580 u = 9.10938 * 10-31 kg

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19.1 Properties of Nuclei

The number of protons in a nucleus is the atomic number Z. The number of neutrons is the neutron number N. The nucleon number or mass number A is the sum of the number of protons Z and the number of neutrons N: A = Z + N

(19.2)

A single nuclear species having specific values of both Z and N is called a nuclide. Table 19.1 lists values of A, Z, and N for some nuclides. The electron structure of an atom, which is responsible for its chemical properties, is determined by the charge Ze of the nucleus. The table shows some nuclides that have the same Z but different N. These nuclides are called isotopes of that element; they have different masses because they have different numbers of neutrons in their nuclei. A familiar example is chlorine (Cl, Z = 17). About 76% of chlorine nuclei have N = 18; the other 24% have N = 20. Different isotopes of an element usually have slightly different physical properties such as melting and boiling temperatures and diffusion rates. The two common isotopes of uranium with A = 235 and 238 are usually separated industrially by taking advantage of the different diffusion rates of gaseous uranium hexafluoride 1UF62 containing the two isotopes. Table 19.1 also shows the usual notation for individual nuclides: the symbol of the element, with a pre-subscript equal to Z and a pre-superscript equal to the mass number A. The general format for an element El is AZ El. The isotopes of chlorine 37 mentioned above, with A = 35 and 37, are written 35 17Cl and 17Cl and pronounced “chlorine-35” and “chlorine-37,” respectively. This name of the element determines the atomic number Z, so the pre-subscript Z is sometimes omitted, as in 35Cl. Table 19.2 gives the masses of some common atoms, including their electrons. Note that this table gives masses of neutral atoms (with Z electrons) rather than masses of bare nuclei, because it is much more difficult to measure masses of bare nuclei with high precision. The mass of a neutral carbon-12 atom is exactly 12 u; that’s how the unified atomic mass unit is defined. The masses of other atoms are approximately equal to A atomic mass units, as we stated earlier. In fact, the atomic masses are less than the sum of the masses of their parts (the Z protons, the Z electrons, and the N neutrons). We’ll explain this very important mass difference in the next section. Table 19.1 Compositions of Some Common Nuclides Z = atomic number (number of protons) N = neutron number A = Z + N = mass number (total number of nucleons) Z

N

A5Z1N

1 1H 2 1H 4 2He 6 3Li 7 3Li 9 4Be 10 5B 11 5B 12 6C 13 6C 14 7N 16 8O 23 11Na 65 29Cu 200 80 Hg

1 1 2 3 3 4 5 5 6 6 7 8 11 29 80

0 1 2 3 4 5 5 6 6 7 7 8 12 36 120

1 2 4 6 7 9 10 11 12 13 14 16 23 65 200

235 92 U

92

143

235

238 92 U

92

146

238

Nucleus

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601

602

CHAPTER 19 Nuclear Physics Table 19.2 Neutral Atomic Masses for Some Light Nuclides Element and Isotope

Atomic Number, Z

Hydrogen 111H2 Deuterium 121H2 Tritium 131H2 Helium 132He2 Helium 142He2 Lithium 163Li2 Lithium 173Li2 Beryllium 194Be2 Boron 1105B2 Boron 1115B2 Carbon 1126C2 Carbon 1136C2 Nitrogen 1147N2 Nitrogen 1157N2 Oxygen 1168O2 Oxygen 1178O2

1 1 1 2 2 3 3 4 5 5 6 6 7 7 8 8

Neutron Number, N

Atomic Mass (u)

0 1 2 1 2 3 4 5 5 6 6 7 7 8 8 9

Mass Number, A

1.007825 2.014102 3.016049 3.016029 4.002603 6.015122 7.016004 9.012182 10.012937 11.009305 12.000000 13.003355 14.003074 15.000109 15.994915 16.999132

1 2 3 3 4 6 7 9 10 11 12 13 14 15 16 17

Oxygen 1188O2 8 10 17.999160 Source: A. H. Wapstra and G. Audi, Nuclear Physics A595, 4 (1995).

19.2

18

Nuclear Binding and Nuclear Structure

Because energy must be added to a nucleus to separate it into its individual protons and neutrons, the total rest energy E0 of the separated nucleons is greater than the rest energy of the nucleus. The energy that must be added to separate the nucleons is called the binding energy EB; it is the magnitude of the energy by which the nucleons are bound together. Thus the rest energy of the nucleus is E0 - EB . Using the equivalence of rest mass and energy, we see that the total mass of the nucleons is always greater than the mass of the nucleus by an amount EB>c2 called the mass defect. The binding energy for a nucleus containing Z protons and N neutrons is defined as '

Deuterium and Heavy Water Toxicity

Application

A crucial step in plant and animal cell division is the formation of a spindle, which separates the two sets of daughter chromosomes. If a plant is given only heavy water—in which one or both of the hydrogen atoms in an H2O molecule are replaced with a deuterium atom—cell division stops and the plant stops growing. The reason is that deuterium is more massive than ordinary hydrogen, so the O–H bond in heavy water has a slightly different binding energy and heavy water has slightly different properties as a solvent. The biochemical reactions that occur during cell division are very sensitive to these solvent properties, so a spindle never forms and the cell cannot reproduce.

Spindle

EB = 1ZMH + Nmn - AZM2c2

(nuclear binding energy)

(19.3)

where AZ M is the mass of the neutral atom containing the nucleus, the quantity in the parentheses is the mass defect, and c2 = 931.5 MeV>u. Note that Eq. (19.3) does not include Zmp , the mass of Z protons. Rather, it contains ZMH , the mass of Z protons and Z electrons combined as Z neutral 11H atoms, to balance the Z electrons included in AZ M, the mass of the neutral atom. The simplest nucleus is that of hydrogen, a single proton. Next comes the nucleus of 21H, the isotope of hydrogen with mass number 2, usually called deuterium. Its nucleus consists of a proton and a neutron bound together to form a particle called the deuteron. By using values from Table 19.2 in Eq. (19.3), we find that the binding energy of the deuteron is '

'

EB = 11.007825 u + 1.008665 u - 2.014102 u21931.5 MeV>u2 = 2.224 MeV

Chromosomes

This much energy would be required to pull the deuteron apart into a proton and a neutron. An important measure of how tightly a nucleus is bound is the binding energy per nucleon, EB>A. At 12.224 MeV2>12 nucleons2 = 1.112 MeV per nucleon, 21H has the lowest binding energy per nucleon of all nuclides.

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19.2 Nuclear Binding and Nuclear Structure

Problem-Solving Strategy 19.1

Nuclear Properties

IDENTIFY the relevant concepts: The key properties of a nucleus are its mass, radius, binding energy, mass defect, binding energy per nucleon, and angular momentum. SET UP the problem: Once you have identified the target variables, assemble the equations needed to solve the problem. A relatively small number of equations from this section and Section 19.1 are all you need. EXECUTE the solution: Solve for the target variables. Bindingenergy calculations using Eq. (19.3) often involve subtracting two nearly equal quantities. To get enough precision in the difference, you may need to carry as many as nine significant figures, if that many are available.

Example 19.2

603

EVALUATE your answer: It’s useful to be familiar with the following benchmark magnitudes. Protons and neutrons are about 1840 times as massive as electrons. Nuclear radii are of the order of 10 -15 m. The electric potential energy of two protons in a nucleus is roughly 10 -13 J or 1 MeV, so nuclear interaction energies are typically a few MeV rather than a few eV as with atoms. The binding energy per nucleon is about 1% of the nucleon rest energy. (The ionization energy of the hydrogen atom is only 0.003% of the electron’s rest energy.) Angular momenta are determined only by the value of U, so they are of the same order of magnitude in both nuclei and atoms. Nuclear magnetic moments, however, are about a factor of 1000 smaller than those of electrons in atoms because nuclei are so much more massive than electrons.

The most strongly bound nuclide

Find the mass defect, the total binding energy, and the binding energy per nucleon of 62 28Ni, which has the highest binding energy per nucleon of all nuclides (Fig. 19.1). The neutral atomic mass of 62 28Ni is 61.928349 u. S O L U T IO N IDENTIFY and SET UP: The mass defect ¢M is the difference between the mass of the nucleus and the combined mass of its constituent nucleons. The binding energy EB is this quantity multiplied by c2, and the binding energy per nucleon is EB divided by the mass number A. We use Eq. (19.3), ¢M = ZMH + Nmn - AZM, to determine both the mass defect and the binding energy.

Eq. (19.3) gives ¢M 5 0.585361 u. The binding energy is then EB = 10.585361 u21931.5 MeV>u2 = 545.3 MeV

The binding energy per nucleon is EB>A = 1545.3 MeV2>62, or 8.795 MeV per nucleon. EVALUATE: Our result means that it would take a minimum of 545.3 MeV to pull a 62 28Ni completely apart into 28 protons and 34 neutrons. The mass defect of 62 28Ni is about 1% of the atomic (or the nuclear) mass. The binding energy is therefore about 1% of the rest energy of the nucleus, and the binding energy per nucleon is about 1% of the rest energy of a nucleon. Note that the mass defect is more than half the mass of a nucleon, which suggests how tightly bound nuclei are.

EXECUTE: With Z = 28, MH = 1.007825 u, N = A - Z = 62 - 28 = 34, mn = 1.008665 u, and AZM = 61.928349 u,

Nearly all stable nuclides, from the lightest to the most massive, have binding energies in the range of 7–9 MeV per nucleon. Figure 19.1 is a graph of binding energy per nucleon as a function of the mass number A. Note the spike at A = 4, showing the unusually large binding energy per nucleon of the 42He nucleus (alpha particle) relative to its neighbors. To explain this curve, we must consider the interactions among the nucleons.

The Nuclear Force The force that binds protons and neutrons together in the nucleus, despite the electrical repulsion of the protons, is an example of the strong interaction. In the context of nuclear structure, this interaction is called the nuclear force. Here are some of its characteristics. First, it does not depend on charge; neutrons as well as protons are bound, and the binding is the same for both. Second, it has short range, of the order of nuclear dimensions—that is, 10-15 m. (Otherwise, the nucleus would grow by pulling in additional protons and neutrons.) But within its range, the nuclear force is much stronger than electrical forces; otherwise, the nucleus could never be stable. It would be nice if we could write a simple equation like Newton’s law of

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604

CHAPTER 19 Nuclear Physics

19.1 Approximate binding energy per nucleon as a function of mass number A (the total number of nucleons) for stable nuclides.

/

/

EB A (MeV nucleon) 9 62 Ni 28

8

12 C 6

238 U 92

7 6

4 He 2

5

/

The curve reaches a peak of about 8.8 MeV nucleon at A 5 62, corresponding to the element nickel. The spike at A 5 4 shows the unusual stability of the 42 He structure.

4 3 2 1

2H 1

50

O

100

150

200

250

A

gravitation or Coulomb’s law for this force, but physicists have yet to fully determine its dependence on the separation r. Third, the nearly constant density of nuclear matter and the nearly constant binding energy per nucleon of larger nuclides show that a particular nucleon cannot interact simultaneously with all the other nucleons in a nucleus, but only with those few in its immediate vicinity. This is different from electrical forces; every proton in the nucleus repels every other one. This limited number of interactions is called saturation; it is analogous to covalent bonding in molecules and solids. Finally, the nuclear force favors binding of pairs of protons or neutrons with opposite spins and of pairs of pairs—that is, a pair of protons and a pair of neutrons, each pair having opposite spins. Hence the alpha particle (two protons and two neutrons) is an exceptionally stable nucleus for its mass number. We’ll see other evidence for pairing effects in nuclei in the next subsection. The analysis of nuclear structure is more complex than the analysis of manyelectron atoms. Two different kinds of interactions are involved (electrical and nuclear), and the nuclear force is not yet completely understood. Even so, we can gain some insight into nuclear structure by the use of simple models. We’ll discuss briefly two rather different but successful models, the liquid-drop model and the shell model.

The Liquid-Drop Model The liquid-drop model, first proposed in 1928 by the Russian physicist George Gamow and later expanded on by Niels Bohr, is suggested by the observation that all nuclei have nearly the same density. The individual nucleons are analogous to molecules of a liquid, held together by short-range interactions and surface-tension effects. We can use this simple picture to derive a formula for the estimated total binding energy of a nucleus. We’ll include five contributions: 1. We’ve remarked that nuclear forces show saturation; an individual nucleon interacts only with a few of its nearest neighbors. This effect gives a bindingenergy term that is proportional to the number of nucleons. We write this term as C1 A, where C1 is an experimentally determined constant. 2. The nucleons on the surface of the nucleus are less tightly bound than those in the interior because they have no neighbors outside the surface. This decrease in the binding energy gives a negative energy term proportional to the surface area 4pR2. Because R is proportional to A1>3, this term is proportional to A2>3; we write it as -C2 A2>3, where C2 is another constant. '

'

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19.2 Nuclear Binding and Nuclear Structure

605

3. Every one of the Z protons repels every one of the 1Z - 12 other protons. The total repulsive electric potential energy is proportional to Z1Z - 12 and inversely proportional to the radius R and thus to A1/3. This energy term is negative because the nucleons are less tightly bound than they would be without the electrical repulsion. We write this correction as -C3 Z(Z - 1)>A1/3. 4. To be in a stable, low-energy state, the nucleus must have a balance between the energies associated with the neutrons and with the protons. This means that N is close to Z for small A and N is greater than Z (but not too much greater) for larger A. We need a negative energy term corresponding to the difference ƒ N - Z ƒ . The best agreement with observed binding energies is obtained if this term is proportional to 1N - Z22>A. If we use N = A - Z to express this energy in terms of A and Z, this correction is -C41A - 2Z22>A. 5. Finally, the nuclear force favors pairing of protons and of neutrons. This energy term is positive (more binding) if both Z and N are even, negative (less binding) if both Z and N are odd, and zero otherwise. The best fit to the data occurs with the form 6C5 A-4>3 for this term. '

'

The total estimated binding energy EB is the sum of these five terms: EB = C1 A - C2 A2>3 - C3 '

'

Z1Z - 12

- C4

1A - 2Z22 6 C5 A-4>3 A

A1>3 (nuclear binding energy)

'

(19.4)

The constants C1 , C2 , C3 , C4 , and C5 , chosen to make this formula best fit the observed binding energies of nuclides, are '

'

'

'

'

C1 = 15.75 MeV C2 = 17.80 MeV C3 = 0.7100 MeV C4 = 23.69 MeV C5 = 39 MeV The constant C1 is the binding energy per nucleon due to the saturated nuclear force. This energy is almost 16 MeV per nucleon, about double the total binding energy per nucleon in most nuclides. If we estimate the binding energy EB using Eq. (19.4), we can solve Eq. (19.3) to use it to estimate the mass of any neutral atom: A ZM

= ZMH + Nmn -

EB c2

(semiempirical mass formula)

(19.5)

Equation (19.5) is called the semiempirical mass formula. The name is apt; it is empirical in the sense that the C’s have to be determined empirically (experimentally), yet it does have a sound theoretical basis.

Example 19.3

Estimating binding energy and mass

For the nuclide 62 28Ni of Example 19.2, (a) calculate the five terms in the binding energy and the total estimated binding energy, and (b) find the neutral atomic mass using the semiempirical mass formula. S O L U T IO N IDENTIFY and SET UP: We use the liquid-drop model of the nucleus and its five contributions to the binding energy, as given by

Eq. (19.4), to calculate the total binding energy EB. We then use Eq. (19.5) to find the neutral atomic mass 62 28M. EXECUTE: (a) With Z = 28, A = 62, and N = 34, the five terms in Eq. (19.4) are 1. C1A = 115.75 MeV21622 = 976.5 MeV

2. - C2 A2>3 = -117.80 MeV216222>3 = - 278.8 MeV

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Continued

606

CHAPTER 19 Nuclear Physics

3. - C3

Z1Z - 12 A

1>3

= - 10.7100 MeV2

12821272

= - 135.6 MeV

1622

1A - 2Z22 162 - 5622 = - 123.69 MeV2 4. - C4 A 62 = - 13.8 MeV '

5. + C5 A-4>3 = 139 MeV21622-4>3 = 0.2 MeV

The pairing correction (term 5) is by far the smallest of all the terms; it is positive because both Z and N are even. The sum of all five terms is the total estimated binding energy, EB 5 548.5 MeV. (b) We use EB 5 548.5 MeV in Eq. (19.5): 62 28M

-

1>3

= 2811.007825 u2 + 3411.008665 u2

548.5 MeV = 61.925 u 931.5 MeV>u

EVALUATE: The binding energy of 62 28Ni calculated in part (a) is only about 0.6% larger than the true value of 545.3 MeV found in Example 19.2, and the mass calculated in part (b) is only about 0.005% smaller than the measured value of 61.928349 u. The semiempirical mass formula can be quite accurate!

The liquid-drop model and the mass formula derived from it are quite successful in correlating nuclear masses, and we will see later that they are a great help in understanding decay processes of unstable nuclides. Some other aspects of nuclei, such as angular momentum and excited states, are better approached with different models.

The Shell Model

19.2 Approximate potential-energy functions for a nucleon in a nucleus. The approximate nuclear radius is R. (a) The potential energy Unuc due to the nuclear force is the same for protons and neutrons. For neutrons, it is the total potential energy. Unuc (MeV) 20 R r

0 220 240

(b) For protons, the total potential energy Utot is the sum of the nuclear (Unuc) and electric (Uel) potential energies. U (MeV) 20 0 220 240

Uel r

R

Utot 5 Uel 1 Unuc Unuc

The shell model of nuclear structure is analogous to the central-field approximation in atomic physics. We picture each nucleon as moving in a potential that represents the averaged-out effect of all the other nucleons. This may not seem to be a very promising approach; the nuclear force is very strong, very short range, and therefore strongly distance dependent. However, in some respects, this model turns out to work fairly well. The potential-energy function for the nuclear force is the same for protons as for neutrons. Figure 19.2a shows a reasonable assumption for the shape of this function: a spherical version of the square-well potential. The corners are somewhat rounded because the nucleus doesn’t have a sharply defined surface. For protons there is an additional potential energy associated with electrical repulsion. We consider each proton to interact with a sphere of uniform charge density, with radius R and total charge 1Z - 12e. Figure 19.2b shows the nuclear, electric, and total potential energies for a proton as functions of the distance r from the center of the nucleus. In principle, we could solve the Schrödinger equation for a proton or neutron moving in such a potential. For any spherically symmetric potential energy, the angular-momentum states are the same as for the electrons in the central-field approximation in atomic physics. In particular, we can use the concept of filled shells and subshells and their relationship to stability. In atomic structure we found that the values Z = 2, 10, 18, 36, 54, and 86 (the atomic numbers of the noble gases) correspond to particularly stable electron arrangements. A comparable effect occurs in nuclear structure. The numbers are different because the potential-energy function is different and the nuclear spin-orbit interaction is much stronger and of opposite sign than in atoms, so the subshells fill up in a different order from those for electrons in an atom. It is found that when the number of neutrons or the number of protons is 2, 8, 20, 28, 50, 82, or 126, the resulting structure is unusually stable—that is, has an unusually high binding energy. (Nuclides with Z = 126 have not been observed in nature.) These numbers are called magic numbers. Nuclides in which Z is a magic number tend to have an above-average number of stable isotopes. There are several doubly magic nuclides for which both Z and N are magic, including 4 2He

16 8O

40 20Ca

48 20Ca

208 82Pb

All these nuclides have substantially higher binding energy per nucleon than do nuclides with neighboring values of N or Z. They also all have zero nuclear spin. The magic numbers correspond to filled-shell or -subshell configurations of nucleon energy levels with a relatively large jump in energy to the next allowed level.

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19.3 Nuclear Stability and Radioactivity

607

Test Your Understanding of Section 19.2 Rank the following nuclei in order from largest to smallest value of the binding energy per nucleon. (i) 42He; (ii) 52 24Cr; 200 252 (iii) 152 (iv) (v) y Sm; Hg; Cf. 62 80 92

19.3

Nuclear Stability and Radioactivity

Among about 2500 known nuclides, fewer than 300 are stable. The others are unstable structures that decay to form other nuclides by emitting particles and electromagnetic radiation, a process called radioactivity. The time scale of these decay processes ranges from a small fraction of a microsecond to billions of years. The stable nuclides are shown by dots on the graph in Fig. 19.3, where the neutron number N and proton number (or atomic number) Z for each nuclide are plotted. Such a chart is called a Segrè chart, after its inventor, the ItalianAmerican physicist Emilio Segrè (1905–1989). Each blue line perpendicular to the line N = Z represents a specific value of the mass number A = Z + N. Most lines of constant A pass through only one or two stable nuclides; that is, there is usually a very narrow range of stability for a given mass number. The lines at A = 20, A = 40, A = 60, and A = 80 are examples. In four cases these lines pass through three stable nuclides—namely, at A = 96, 124, 130, and 136. Four stable nuclides have both odd Z and odd N: 2 1H

6 3Li

10 5B

ActivPhysics 19.2: Nuclear Binding Energy ActivPhysics 19.4: Radioactivity

14 7N

These are called odd-odd nuclides. The absence of other odd-odd nuclides shows the influence of pairing. Also, there is no stable nuclide with A = 5 or A = 8. The doubly magic 42He nucleus, with a pair of protons and a pair of neutrons, has no interest in accepting a fifth particle into its structure. Collections of eight nucleons decay to smaller nuclides, with a 84Be nucleus immediately splitting into two 42He nuclei. The points on the Segrè chart representing stable nuclides define a rather narrow stability region. For low mass numbers, the numbers of protons and neutrons are approximately equal, N L Z. The ratio N>Z increases gradually with A, up to about 1.6 at large mass numbers, because of the increasing influence of the electrical repulsion of the protons. Points to the right of the stability region represent nuclides that have too many protons relative to neutrons to be stable. In these cases, repulsion wins, and the nucleus comes apart. To the left are nuclides with too many neutrons relative to protons. In these cases the energy associated with the neutrons is out of balance with that associated with the protons, and the nuclides decay in a process that converts neutrons to protons. The graph also shows that no nuclide with A 7 209 or Z 7 83 is stable. A nucleus is unstable if it is too big. Note that there is no stable nuclide with Z = 43 (technetium) or 61 (promethium).

Alpha Decay Nearly 90% of the 2500 known nuclides are radioactive; they are not stable but decay into other nuclides. When unstable nuclides decay into different nuclides, they usually emit alpha 1a2 or beta 1b2 particles. An alpha particle is a 4He nucleus, two protons and two neutrons bound together, with total spin zero. Alpha emission occurs principally with nuclei that are too large to be stable. When a nucleus emits an alpha particle, its N and Z values each decrease by 2 and A decreases by 4, moving it closer to stable territory on the Segrè chart.

PhET: Alpha Decay

A familiar example of an alpha emitter is radium, 226 88Ra (Fig. 19.4a). The speed of the emitted alpha particle, determined from the curvature of its path in a transverse magnetic field, is about 1.52 * 107 m>s. This speed, although large, is

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CHAPTER 19 Nuclear Physics

19.3 Segrè chart showing neutron number and proton number for stable nuclides.

N 144

A

136

A

5

128

22 0

20

0

120

A

112

5

18

0

A

104

5

16

0

96

A

A A 5 5 5 13 14 6 A 1 0 3 5 0 A 12 5 4 12 0

88 80 72

5

5

A

56

Z

A 5 A 1 5 00 96

64

N

Neutron number

5

80

48

A 40

A

32

5

5

60

40

As the mass number A increases, stable nuclides have an increasing ratio of neutrons to protons.

24

A 16

5

20

8 0

8

16

24

32

40 48 56 Proton number

64

72

80

88

96

Z

19.4 Alpha decay of the unstable radium nuclide 226 88Ra. (a)

(b) Potential-energy curve for an a particle and 222 Rn nucleus 86

226 Ra 88

U (MeV)

The a particle tunnels through the potential-energy barrier.

88p 138n The nuclide 226 88 Ra decays by alpha emission to 222 86 Rn. 86p 136n 222 86 Rn

(c ) Energy-level diagram for the system 226 88 Ra

can a-decay directly to the 222 86 Rn ground level ...

4.685 4.871 MeV MeV

4.871

a (2p, 2n) 4 He 2

a

a g

0

R

r 0.186 MeV

... or it can a-decay to an excited level 222 Rn*, 86 which can then decay to 226 Ra the 222 Rn ground level by 86 88 emitting a 0.186-MeV photon (g). 222 Rn* 86 222 Rn 86

only 5% of the speed of light, so we can use the nonrelativistic kinetic-energy expression K = 12 mv2: K = 12 16.64 * 10-27 kg211.52 * 107 m>s22 = 7.67 * 10-13 J = 4.79 MeV Alpha particles are always emitted with definite kinetic energies, determined by conservation of momentum and energy. Because of their charge and mass, alpha

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19.3 Nuclear Stability and Radioactivity

609

particles can travel only several centimeters in air, or a few tenths or hundredths of a millimeter through solids, before they are brought to rest by collisions. Some nuclei can spontaneously decay by emission of a particles because energy is released in their alpha decay. You can use conservation of mass-energy to show that alpha decay is possible whenever the mass of the original neutral atom is greater than the sum of the masses of the final neutral atom and the neutral helium-4 atom.

In alpha decay, the a particle tunnels through a potential-energy barrier, as Fig. 19.4b shows.

Example 19.4

Alpha decay of radium

222 4 Show that the a-emission process 226 88Ra S 86Rn + 2He is energetically possible, and calculate the kinetic energy of the emitted a particle. The neutral atomic masses are 226.025403 u for 226 88Ra and 222.017571 u for 222 . Rn 86 '

'

S O L U T IO N IDENTIFY and SET UP: Alpha emission is possible if the mass of the 226 88Ra atom is greater than the sum of the atomic masses of 222 4 86Rn and 2He. The mass difference between the initial radium atom and the final radon and helium atoms corresponds (through E = mc2) to the energy E released in the decay. Because momentum is conserved as well as energy, both the alpha particle and the 222 86Rn atom are in motion after the decay; we will have to account for this in determining the kinetic energy of the alpha particle. EXECUTE: From Table 19.2, the mass of the 42He atom is 4.002603 u. The difference in mass between the original nucleus and the decay products is 226.025403 u - 1222.017571 u + 4.002603 u2 = +0.005229 u

Since this is positive, a decay is energetically possible. The energy equivalent of this mass difference is E = 10.005229 u21931.5 MeV>u2 = 4.871 MeV '

Thus we expect the decay products to emerge with total kinetic energy 4.871 MeV. Momentum is also conserved; if the parent 226 222 88Ra nucleus is at rest, the daughter 86Rn nucleus and the a particle have momenta of equal magnitude p but opposite direction. Kinetic energy is K = 12 mv2 = p2>2m : Since p is the same for the two particles, the kinetic energy divides inversely as their masses. Hence the a particle gets 222>1222 + 42 of the total, or 4.78 MeV. EVALUATE: Experiment shows that 226 88Ra does undergo alpha decay, and the observed a-particle energy is 4.78 MeV. You can check your results by verifying that the alpha particle and the 222 86Rn nucleus produced in the decay have the same magnitude of momentum p = mv. You can calculate the speed v of each of the decay products from its respective kinetic energy. You’ll find that the alpha particle moves at a sprightly 0.0506c = 1.52 * 107 m>s; if momentum is conserved, you should find that the 222 86Rn nucleus 4 moves 222 as fast. Does it?

Beta Decay There are three different simple types of beta decay: beta-minus, beta-plus, and electron capture. A beta-minus particle (b -) is an electron. It’s not obvious how a nucleus can emit an electron if there aren’t any electrons in the nucleus. Emission of a involves transformation of a neutron into a proton, an electron, and a third particle called an antineutrino. In fact, if you freed a neutron from a nucleus, it would decay into a proton, an electron, and an antineutrino in an average time of about 15 minutes. Beta particles can be identified and their speeds can be measured with techniques that are similar to the Thomson experiments we described in Section 7.5. The speeds of beta particles range up to 0.9995 of the speed of light, so their motion is highly relativistic. They are emitted with a continuous spectrum of energies. This would not be possible if the only two particles were the b - and the recoiling nucleus, since energy and momentum conservation would then require a definite speed for the b -. Thus there must be a third particle involved. From conservation of charge, it must be neutral, and from conservation of angular momentum, it must be a spin-12 particle. This third particle is an antineutrino, the antiparticle of a neutrino. The symbol for a neutrino is ne (the Greek letter nu). Both the neutrino and the antineutrino

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CHAPTER 19 Nuclear Physics

have zero charge and zero (or very small) mass and therefore produce very little observable effect when passing through matter. Both evaded detection until 1953, when Frederick Reines and Clyde Cowan succeeded in observing the antineutrino directly. We now know that there are at least three varieties of neutrinos, each with its corresponding antineutrino; one is associated with beta decay and the other two are associated with the decay of two unstable particles, the muon and the tau particle. The antineutrino that is emitted in b - decay is denoted as ne. The basic process of b - decay is n S p + b - + ne

(19.6)

Beta-minus decay usually occurs with nuclides for which the neutron-toproton ratio N>Z is too large for stability. In b - decay, N decreases by 1, Z increases by 1, and A doesn’t change. You can use conservation of mass-energy to show that beta-minus decay can occur whenever the mass of the original neutral atom is larger than that of the final atom.

Example 19.5

Why cobalt-60 is a beta-minus emitter

The nuclide 60 27Co, an odd-odd unstable nucleus, is used in medical and industrial applications of radiation. Show that it is unstable relative to b - decay. The atomic masses you need are 59.933822 u 60 for 60 27Co and 59.930791 u for 28Ni . SOLUTION IDENTIFY and SET UP: Beta-minus decay is possible if the mass of the original neutral atom is greater than that of the final atom. We must first identify the nuclide that will result if 60 27Co undergoes b - decay and then compare its neutral atomic mass to that of 60 27Co.

EXECUTE: In the presumed b - decay of 60 27Co, Z increases by 1 from 27 to 28 and A remains at 60, so the final nuclide is 60 28Ni. 60 The neutral atomic mass of 60 is greater than that of by Co Ni 27 28 0.003031 u, so b - decay can occur. EVALUATE: With three decay products in b - decay—the 60 28Ni nucleus, the electron, and the antineutrino—the energy can be shared in many different ways that are consistent with conservation of energy and momentum. It’s impossible to predict precisely how the energy will be shared for the decay of a particular 60 27Co nucleus. By contrast, in alpha decay there are just two decay products, and their energies and momenta are determined uniquely (see Example 19.4).

We have noted that b - decay occurs with nuclides that have too large a neutron-to-proton ratio N>Z. Nuclides for which N>Z is too small for stability can emit a positron, the electron’s antiparticle, which is identical to the electron but with positive charge. The basic process, called beta-plus decay 1b +2, is p S n + b + + ne

(19.7)

+

where b is a positron and ne is the electron neutrino. Beta-plus decay can occur whenever the mass of the original neutral atom is at least two electron masses larger than that of the final atom.

You can show this using conservation of mass-energy. The third type of beta decay is electron capture. There are a few nuclides for which b + emission is not energetically possible but in which an orbital electron (usually in the K shell) can combine with a proton in the nucleus to form a neutron and a neutrino. The neutron remains in the nucleus and the neutrino is emitted. The basic process is p + b - S n + ne

(19.8)

You can use conservation of mass-energy to show that electron capture can occur whenever the mass of the original neutral atom is larger than that of the final atom.

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19.3 Nuclear Stability and Radioactivity

611

In all types of beta decay, A remains constant. However, in beta-plus decay and electron capture, N increases by 1 and Z decreases by 1 as the neutron–proton ratio increases toward a more stable value. The reaction of Eq. (19.8) also helps to explain the formation of a neutron star, mentioned in Example 19.1. CAUTION Beta decay inside and outside nuclei The beta-decay reactions given by Eqs. (19.6), (19.7), and (19.8) occur within a nucleus. Although the decay of a neutron outside the nucleus proceeds through the reaction of Eq. (19.6), the reaction of Eq. (19.7) is forbidden by conservation of mass-energy for a proton outside the nucleus. The reaction of Eq. (19.8) can occur outside the nucleus only with the addition of some extra energy, as in a collision. y

Example 19.6

Why cobalt-57 is not a beta-plus emitter

The nuclide 57 27Co is an odd-even unstable nucleus. Show that it cannot undergo b + decay, but that it can decay by electron capture. The atomic masses you need are 56.936296 u for 57 27Co and . 56.935399 u for 57 Fe 26 S O L U T IO N IDENTIFY and SET UP: Beta-plus decay is possible if the mass of the original neutral atom is greater than that of the final atom plus two electron masses (0.001097 u). Electron capture is possible if the mass of the original atom is greater than that of the final atom. We must first identify the nuclide that will result if 57 27Co undergoes b + decay or electron capture and then find the corresponding mass difference.

+ EXECUTE: The original nuclide is 57 27Co. In both the presumed b decay and electron capture, Z decreases by 1 from 27 to 26, and A remains at 57, so the final nuclide is 57 26Fe. Its mass is less than that of 57 by a value smaller than 0.001097 u (two Co 0.000897 u, 27 electron masses), so b + decay cannot occur. However, the mass of the original atom is greater than the mass of the final atom, so electron capture can occur.

EVALUATE: In electron capture there are just two decay products, the final nucleus and the emitted neutrino. As in alpha decay (Example 19.4) but unlike in b - decay (Example 19.5), the decay products of electron capture have unique energies and momenta. In Section 19.4 we’ll see how to relate the probability that electron capture will occur to the half-life of this nuclide.

Gamma Decay The energy of internal motion of a nucleus is quantized. A typical nucleus has a set of allowed energy levels, including a ground state (state of lowest energy) and several excited states. Because of the great strength of nuclear interactions, excitation energies of nuclei are typically of the order of 1 MeV, compared with a few eV for atomic energy levels. In ordinary physical and chemical transformations the nucleus always remains in its ground state. When a nucleus is placed in an excited state, either by bombardment with high-energy particles or by a radioactive transformation, it can decay to the ground state by emission of one or more photons called gamma rays or gamma-ray photons, with typical energies of 10 keV to 5 MeV. This process is called gamma 1g2 decay. For example, alpha particles emitted from 226Ra have two possible kinetic energies, either 4.784 MeV or 4.602 MeV. Including the recoil energy of the resulting 222Rn nucleus, these correspond to a total released energy of 4.871 MeV or 4.685 MeV, respectively. When an alpha particle with the smaller energy is emitted, the 222Rn nucleus is left in an excited state. It then decays to its ground state by emitting a gamma-ray photon with energy 14.871 - 4.6852 MeV = 0.186 MeV

A photon with this energy is observed during this decay (Fig. 19.4c). CAUTION G decay vs. A and B decay In both a and b decay, the Z value of a nucleus changes and the nucleus of one element becomes the nucleus of a different element. In g decay, the element does not change; the nucleus merely goes from an excited state to a less excited state. y

Natural Radioactivity Many radioactive elements occur in nature. For example, you are very slightly radioactive because of unstable nuclides such as carbon-14 and potassium-40

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CHAPTER 19 Nuclear Physics

19.5 Earthquakes are caused in part by the radioactive decay of 238U in the earth’s interior. These decays release energy that helps to produce convection currents in the earth’s interior. Such currents drive the motions of the earth’s crust, including the sudden sharp motions that we call earthquakes (like the one that caused this damage).

that are present throughout your body. The study of natural radioactivity began in 1896, one year after Röntgen discovered x rays. Henri Becquerel discovered a radiation from uranium salts that seemed similar to x rays. Intensive investigation in the following two decades by Marie and Pierre Curie, Ernest Rutherford, and many others revealed that the emissions consist of positively and negatively charged particles and neutral rays; they were given the names alpha, beta, and gamma because of their differing penetration characteristics. The decaying nucleus is usually called the parent nucleus; the resulting nucleus is the daughter nucleus. When a radioactive nucleus decays, the daughter nucleus may also be unstable. In this case a series of successive decays occurs until a stable configuration is reached. Several such series are found in nature. The most abundant radioactive nuclide found on earth is the uranium isotope 238U, which undergoes a series of 14 decays, including eight a emissions and six b - emissions, terminating at a stable isotope of lead, 206Pb (Fig. 19.5). Radioactive decay series can be represented on a Segrè chart, as in Fig. 19.6. The neutron number N is plotted vertically, and the atomic number Z is plotted horizontally. In alpha emission, both N and Z decrease by 2. In b - emission, N decreases by 1 and Z increases by 1. The decays can also be represented in equation form; the first two decays in the series are written as

N 146

238U

4.47 3 109 y

145 234Th

144

24.10 d

143

234Pa

70 s

142

234U

2.46 3 105 y

141 140

7.54 3 104 y

139

230Th

138 226Ra

1600 y

137 136

134

27 m

131

214Bi

19.9 m

19.9 m

130

b2 decay

214Po

210 Tl

1.637 3 1024 s

1.30 m 210Pb

22.3 y

127

19.6 Segrè chart showing the uranium 238 U decay series, terminating with the stable nuclide 206Pb. The times are halflives (discussed in the next section), given in years (y), days (d), hours (h), minutes (m), or seconds (s).

a decay

214Pb

132

128

218Po

3.10 m

133

129

222Rn

3.82 d

135

210Bi

5.01 d

126 125

206Pb

210Po

138.38 d

124 80

81

82

83

84

85

86

87

88

89

90

91

92

93

Hg

Tl

Pb

Bi

Po

At

Rn

Fr

Ra

Ac

Th

Pa

U

Np

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Z

19.4 Activities and Half-Lives 238

U S 234Th + a 234 Th S 234Pa + b - + ne or more briefly as a

238

U S 234Th b-

234

Th S 234Pa

In the second process, the beta decay leaves the daughter nucleus 234Pa in an excited state, from which it decays to the ground state by emitting a gamma-ray photon. An excited state is denoted by an asterisk, so we can represent the g emission as Pa* S 234Pa + g

234

or g

Pa* S 234Pa

234

An interesting feature of the 238U decay series is the branching that occurs at Bi. This nuclide decays to 210Pb by emission of an a and a b -, which can occur in either order. We also note that the series includes unstable isotopes of several elements that also have stable isotopes, including thallium (Tl), lead (Pb), and bismuth (Bi). The unstable isotopes of these elements that occur in the 238U series all have too many neutrons to be stable. Many other decay series are known. Two of these occur in nature, one starting with the uncommon isotope 235U and ending with 207Pb, the other starting with thorium 1232Th2 and ending with 208Pb. 214

Test Your Understanding of Section 19.3 A nucleus with atomic number Z and neutron number N undergoes two decay processes. The result is a nucleus with atomic number Z - 3 and neutron number N - 1. Which decay processes may have taken place? (i) two b - decays; (ii) two b + decays; (iii) two a decays; (iv) an a decay and a b - decay; (v) an a decay and a b + decay.

19.4

y

Activities and Half-Lives

Suppose you need to dispose of some radioactive waste that contains a certain number of nuclei of a particular radioactive nuclide. If no more are produced, that number decreases in a simple manner as the nuclei decay. This decrease is a statistical process; there is no way to predict when any individual nucleus will decay. No change in physical or chemical environment, such as chemical reactions or heating or cooling, greatly affects most decay rates. The rate varies over an extremely wide range for different nuclides.

Radioactive Decay Rates Let N1t2 be the (very large) number of radioactive nuclei in a sample at time t, and let dN1t2 be the (negative) change in that number during a short time interval dt. (We’ll use N1t2 to minimize confusion with the neutron number N.) The number of decays during the interval dt is - dN1t2. The rate of change of N1t2 is the negative quantity dN1t2>dt; thus -dN1t2>dt is called the decay rate or the activity of the specimen. The larger the number of nuclei in the specimen, the more nuclei decay during any time interval. That is, the activity is directly proportional to N1t2; it equals a constant l multiplied by N1t2: -

dN1t2 = lN1t2 dt

(19.9)

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613

614

CHAPTER 19 Nuclear Physics

The constant l is called the decay constant, and it has different values for different nuclides. A large value of l corresponds to rapid decay; a small value corresponds to slower decay. Solving Eq. (19.9) for l shows us that l is the ratio of the number of decays per time to the number of remaining radioactive nuclei; l can then be interpreted as the probability per unit time that any individual nucleus will decay. The situation is reminiscent of a discharging capacitor, which we studied in Section 6.4. Equation (19.9) has the same form as the negative of Eq. (6.15), with q and 1>RC replaced by N1t2 and l. Then we can make the same substitutions in Eq. (6.16), with the initial number of nuclei N102 = N0 , to find the exponential function: '

N1t2 = N0 e-lt '

(number of remaining nuclei)

(19.10)

Figure 19.7 is a graph of this function, showing the number of remaining nuclei N1t2 as a function of time. The half-life T1>2 is the time required for the number of radioactive nuclei to decrease to one-half the original number N0 . Then half of the remaining radioactive nuclei decay during a second interval T1>2 , and so on. The numbers remaining after successive half-lives are N0>2, N0>4, N0>8, Á . To get the relationship between the half-life T1>2 and the decay constant l, we set N1t2>N0 = 12 and t = T1>2 in Eq. (19.10), obtaining '

'

1 2

= e-lT1>2

We take logarithms of both sides and solve for T1>2 : '

T1>2 =

ln 2 0.693 = l l

(19.11)

The mean lifetime Tmean , generally called the lifetime, of a nucleus or unstable particle is proportional to the half-life T1>2 : '

'

Tmean =

T1>2 T1>2 1 = = l ln 2 0.693

(lifetime Tmean , decay constant l, and half-life T1>2) '

(19.12)

In particle physics the life of an unstable particle is usually described by the lifetime, not the half-life. Because the activity - dN1t2>dt at any time equals lN1t2, Eq. (19.10) tells us that the activity also depends on time as e-lt. Thus the graph of activity versus time has the same shape as Fig. 19.7. Also, after successive half-lives, the activity is one-half, one-fourth, one-eighth, and so on of the original activity. 19.7 The number of nuclei in a sample of a radioactive element as a function of time. The sample’s activity has an exponential decay curve with the same shape. N(t) N0

A common unit of activity is the curie, abbreviated Ci, which is defined to be 3.70 * 1010 decays per second. This is approximately equal to the activity of one gram of radium. The SI unit of activity is the becquerel, abbreviated Bq. One becquerel is one decay per second, so

/

N0 2

/ /

N0 4 N0 8 N0 16 0

/

CAUTION A half-life may not be enough It is sometimes implied that any radioactive sample will be safe after a half-life has passed. That’s wrong. If your radioactive waste initially has ten times too much activity for safety, it is not safe after one half-life, when it still has five times too much. Even after three half-lives it still has 25% more activity than is safe. The number of radioactive nuclei and the activity approach zero only as t approaches infinity. y

1 Ci = 3.70 * 1010 Bq = 3.70 * 1010 decays>s T1/2 2T1/2 3T1/2 4T1/2

t

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19.4 Activities and Half-Lives

Example 19.7

Activity of

57

615

Co

The isotope 57Co decays by electron capture to 57Fe with a halflife of 272 d. The 57Fe nucleus is produced in an excited state, and it almost instantaneously emits gamma rays that we can detect. (a) Find the mean lifetime and decay constant for 57Co. (b) If the activity of a 57Co radiation source is now 2.00 mCi, how many 57 Co nuclei does the source contain? (c) What will be the activity after one year? S O L U T IO N IDENTIFY and SET UP: This problem uses the relationships among decay constant l, lifetime Tmean, and activity - dN1t2>dt. In part (a) we use Eq. (19.12) to find l and Tmean from T1>2. In part (b), we use Eq. (19.9) to calculate the number of nuclei N1t2 from the activity. Finally, in part (c) we use Eqs. (19.9) and (19.10) to find the activity after one year. EXECUTE: (a) It’s convenient to convert the half-life to seconds: T1>2 = 1272 d2186,400 s>d2 = 2.35 * 107 s From Eq. (19.12), the mean lifetime and decay constant are T1>2 2.35 * 10 s = = 3.39 * 107 s = 392 days ln 2 0.693 1 = 2.95 * 10-8 s-1 l = Tmean 7

Tmean =

(b) The activity -dN1t2>dt is given as 2.00 mCi, so

-

dN1t2 = 2.00 mCi = 12.00 * 10-6213.70 * 1010 s-12 dt = 7.40 * 104 decays>s

From Eq. (19.9) this is equal to lN(t), so we find N1t2 = -

dN1t2>dt l

=

7.40 * 104 s-1 2.95 * 10-8 s-1

= 2.51 * 1012 nuclei

If you feel we’re being too cavalier about the “units” decays and nuclei, you can use decays>1nucleus # s2 as the unit for l. (c) From Eq. (19.10) the number N1t2 of nuclei remaining after one year 13.156 * 107 s2 is N1t2 = N0 e-lt = N0 e- 12.95 * 10

s 213.156 * 107 s2

-8 -1

= 0.394N0

The number of nuclei has decreased to 0.394 of the original number. Equation (19.9) says that the activity is proportional to the number of nuclei, so the activity has decreased by this same factor to 10.394212.00 mCi2 = 0.788 mCi. EVALUATE: The number of nuclei found in part (b) is equivalent to 4.17 * 10-12 mol, with a mass of 2.38 * 10-10 g. This is a far smaller mass than even the most sensitive balance can measure. After one 272-day half-life, the number of 57Co nuclei has decreased to N0>2; after 21272 d2 = 544 d, it has decreased to N0>22 = N0>4. This result agrees with our answer to part (c), which says that after 365 d the number of nuclei is between N0>2 and N0>4.

Radioactive Dating

?

An interesting application of radioactivity is the dating of archaeological and geological specimens by measuring the concentration of radioactive isotopes. The most familiar example is carbon dating. The unstable isotope 14C, produced during nuclear reactions in the atmosphere that result from cosmic-ray bombardment, gives a small proportion of 14C in the CO2 in the atmosphere. Plants that obtain their carbon from this source contain the same proportion of 14C as the atmosphere. When a plant dies, it stops taking in carbon, and its 14C b - decays to 14 N with a half-life of 5730 years. By measuring the proportion of 14C in the remains, we can determine how long ago the organism died. One difficulty with radiocarbon dating is that the 14C concentration in the atmosphere changes over long time intervals. Corrections can be made on the basis of other data such as measurements of tree rings that show annual growth cycles. Similar radioactive techniques are used with other isotopes for dating geological specimens. Some rocks, for example, contain the unstable potassium isotope 40K, a beta emitter that decays to the stable nuclide 40Ar with a half-life of 2.4 * 108 y. The age of the rock can be determined by comparing the concentrations of 40K and 40Ar. Example 19.8

Radiocarbon dating

Before 1900 the activity per unit mass of atmospheric carbon due to the presence of 14C averaged about 0.255 Bq per gram of carbon. (a) What fraction of carbon atoms were 14C? (b) In analyzing an archaeological specimen containing 500 mg of carbon, you

observe 174 decays in one hour. What is the age of the specimen, assuming that its activity per unit mass of carbon when it died was that average value of the air? Continued

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CHAPTER 19 Nuclear Physics

SOLUTION IDENTIFY and SET UP: The key idea is that the present-day activity of a biological sample containing 14C is related to both the elapsed time since it stopped taking in atmospheric carbon and its activity at that time. We use Eqs. (19.9) and (19.10) to solve for the age t of the specimen. In part (a) we determine the number of 14C atoms N1t2 from the activity -dN1t2>dt using Eq. (19.9). We find the total number of carbon atoms in 500 mg by using the molar mass of carbon (12.011 g>mol, given in Appendix D), and we use the result to calculate the fraction of carbon atoms that are 14C. The activity decays at the same rate as the number of 14C nuclei; we use this and Eq. (19.10) to solve for the age t of the specimen. EXECUTE: (a) To use Eq. (19.9), we must first find the decay constant l from Eq. (19.11): T1>2 = 5730 y = 15730 y213.156 * 107 s>y2 = 1.808 * 1011 s l =

ln 2 0.693 = = 3.83 * 10-12 s-1 T1>2 1.808 * 1011 s

Then, from Eq. (19.9), -dN>dt N1t2 =

= l

0.255 s-1 3.83 * 10-12 s-1

= 6.65 * 1010 atoms

The total number of C atoms in 1 gram 11>12.011 mol2 is 11>12.011216.022 * 10232 = 5.01 * 1022. The ratio of 14C atoms to all C atoms is 6.65 * 1010 5.01 * 1022

= 1.33 * 10-12

Only four carbon atoms in every 3 * 1012 are 14C. (b) Assuming that the activity per gram of carbon in the specimen when it died 1t = 02 was 0.255 Bq>g = 10.255 s-1 # g-1213600 s>h2 = 918 h-1 # g-1, the activity of 500 mg of carbon then was 10.500 g21918 h-1 # g-12 = 459 h-1. The observed activity now, at time t, is 174 h-1. Since the activity is proportional to the number of radioactive nuclei, the activity ratio 174>459 = 0.379 equals the number ratio N1t2>N0 . Now we solve Eq. (19.10) for t and insert values for N1t2>N0 and l : '

t =

ln1N1t2>N02 -l

=

ln 0.379 -3.83 * 10-12 s-1

= 2.53 * 1011 s = 8020 y

EVALUATE: After 8020 y the 14C activity has decreased from 459 to 174 decays per hour. The specimen died and stopped taking CO2 out of the air about 8000 years ago.

Radiation in the Home A serious health hazard in some areas is the accumulation in houses of 222Rn, an inert, colorless, odorless radioactive gas. Looking at the 238U decay chain in Fig. 19.6, we see that the half-life of 222Rn is 3.82 days. If so, why not just move out of the house for a while and let it decay away? The answer is that 222Rn is continuously being produced by the decay of 226Ra, which is found in minute quantities in the rocks and soil on which some houses are built. It’s a dynamic equilibrium situation, in which the rate of production equals the rate of decay. The reason 222Rn is a bigger hazard than the other elements in the 238U decay series is that it’s a gas. During its short half-life of 3.82 days it can migrate from the soil into your house. If a 222Rn nucleus decays in your lungs, it emits a damaging a particle and its daughter nucleus 218Po, which is not chemically inert and is likely to stay in your lungs until it decays, emits another damaging a particle and so on down the 238U decay series. How much of a hazard is radon? Although reports indicate values as high as 3500 pCi>L, the average activity per volume in the air inside American homes due to 222Rn is about 1.5 pCi>L (over a thousand decays each second in an average-sized room). If your environment has this level of activity, it has been estimated that a lifetime exposure would reduce your life expectancy by about 40 days. For comparison, smoking one pack of cigarettes per day reduces life expectancy by 6 years, and it is estimated that the average emission from all the nuclear power plants in the world reduces life expectancy by anywhere from 0.01 day to 5 days. These figures include catastrophes such as the 1986 nuclear reactor disaster at Chernobyl, for which the local effect on life expectancy is much greater.

Test Your Understanding of Section 19.4 Which sample contains a greater number of nuclei: a 5.00-mCi sample of 240Pu (half-life 6560 y) or a 4.45-mCi sample of 243Am (half-life 7370 y)? (i) the 240Pu sample; (ii) the 243Am sample; (iii) both have the same number of nuclei.

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y

19.5 Biological Effects of Radiation

19.5

Biological Effects of Radiation

The above discussion of radon introduced the interaction of radiation with living organisms, a topic of vital interest and importance. Under radiation we include radioactivity (alpha, beta, gamma, and neutrons) and electromagnetic radiation such as x rays. As these particles pass through matter, they lose energy, breaking molecular bonds and creating ions—hence the term ionizing radiation. Charged particles interact directly with the electrons in the material. X rays and g rays interact by the photoelectric effect, in which an electron absorbs a photon and breaks loose from its site, or by Compton scattering (see Section 17.3). Neutrons cause ionization indirectly through collisions with nuclei or absorption by nuclei with subsequent radioactive decay of the resulting nuclei. These interactions are extremely complex. It is well known that excessive exposure to radiation, including sunlight, x rays, and all the nuclear radiations, can destroy tissues. In mild cases it results in a burn, as with common sunburn. Greater exposure can cause very severe illness or death by a variety of mechanisms, including massive destruction of tissue cells, alterations of genetic material, and destruction of the components in bone marrow that produce red blood cells.

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Table 19.3 Relative Biological Effectiveness (RBE) for Several Types of Radiation Radiation rem/rad) X rays and g rays Electrons Slow neutrons Protons a particles Heavy ions

RBE (Sv/Gy or 1 1.0–1.5 3–5 10 20 20

Calculating Radiation Doses Radiation dosimetry is the quantitative description of the effect of radiation on living tissue. The absorbed dose of radiation is defined as the energy delivered to the tissue per unit mass. The SI unit of absorbed dose, the joule per kilogram, is called the gray (Gy); 1 Gy = 1 J>kg. Another unit is the rad, defined as 0.01 J>kg: 1 rad = 0.01 J>kg = 0.01 Gy Absorbed dose by itself is not an adequate measure of biological effect because equal energies of different kinds of radiation cause different extents of biological effect. This variation is described by a numerical factor called the relative biological effectiveness (RBE), also called the quality factor (QF), of each specific radiation. X rays with 200 keV of energy are defined to have an RBE of unity, and the effects of other radiations can be compared experimentally. Table 19.3 shows approximate values of RBE for several radiations. All these values depend somewhat on the kind of tissue in which the radiation is absorbed and on the energy of the radiation. The biological effect is described by the product of the absorbed dose and the RBE of the radiation; this quantity is called the biologically equivalent dose, or simply the equivalent dose. The SI unit of equivalent dose for humans is the sievert (Sv): Equivalent dose 1Sv2 = RBE * Absorbed dose 1Gy2

(19.13)

Equivalent dose 1rem2 = RBE * Absorbed dose 1rad2

(19.14)

A more common unit, corresponding to the rad, is the rem (an abbreviation of röntgen equivalent for man):

Thus the unit of the RBE is 1 Sv>Gy or 1 rem/rad, and 1 rem = 0.01 Sv. Example 19.9

Dose from a medical x ray

During a diagnostic x-ray examination a 1.2-kg portion of a broken leg receives an equivalent dose of 0.40 mSv. (a) What is the equivalent dose in mrem? (b) What is the absorbed dose in mrad and in mGy? (c) If the x-ray energy is 50 keV, how many x-ray photons are absorbed?

SOLUTION IDENTIFY and SET UP: We are asked to relate the equivalent dose (the biological effect of the radiation, measured in sieverts or rems) to the absorbed dose (the energy absorbed per mass, meas-

Continued

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CHAPTER 19 Nuclear Physics

ured in grays or rads). In part (a) we use the conversion factor 1 rem = 0.01 Sv for equivalent dose. Table 19.3 gives the RBE for x rays; we use this value in part (b) to determine the absorbed dose using Eqs. (19.13) and (19.14). Finally, in part (c) we use the mass and the definition of absorbed dose to find the total energy absorbed and the total number of photons absorbed.

0.40 mSv = 0.40 mGy = 4.0 * 10-4 J>kg 1 Sv>Gy (c) The total energy absorbed is 14.0 * 10-4 J>kg211.2 kg2 = 4.8 * 10-4 J = 3.0 * 1015 eV The number of x-ray photons is

EXECUTE: (a) The equivalent dose in mrem is

3.0 * 1015 eV

0.40 mSv = 40 mrem 0.01 Sv>rem

5.0 * 104 eV>photon

(b) For x rays, RBE = 1 rem>rad or 1 Sv>Gy, so the absorbed dose is 40 mrem = 40 mrad 1 rem>rad

= 6.0 * 1010 photons

EVALUATE: The absorbed dose is relatively large because x rays have a low RBE. If the ionizing radiation had been a beam of a particles, for which RBE = 20, the absorbed dose needed for an equivalent dose of 0.40 mSv would be only 0.020 mGy, corresponding to a smaller total absorbed energy of 2.4 * 10-5 J.

Radiation Hazards

19.8 Contribution of various sources to the total average radiation exposure in the U.S. population, expressed as percentages of the total. From human activity 18% Other Occupational Fallout Nuclear fuel cycle Miscellaneous

323 = A. Therefore, doubling the volume requires increasing the mass number by a factor of 2; doubling the radius implies increasing both the volume and the mass number by a factor of 23 = 8. 19.2 (ii), (iii), (iv), (v), (i) You can find the answers by inspecting Fig. 19.1. The binding energy per nucleon is lowest for very light nuclei such as 42He, is greatest around A = 60, and then decreases with increasing A. 19.3 (v) Two protons and two neutrons are lost in an a decay, so Z and N each decrease by 2. A b + decay changes a proton to a neutron, so Z decreases by 1 and N increases by 1. The net result is that Z decreases by 3 and N decreases by 1. 19.4 The activity - dN1t2>dt of a sample is the product of the number of nuclei in the sample N1t2 and the decay constant l = 1ln 22>T1>2 . Hence N(t) = 1- dN1t2>dt2T1>2> 1ln 22. Taking the ratio of this expression for 240Pu to this same expression for 243 Am, the factors of ln 2 cancel and we get '

1 - dNPu>dt2T1>2–Pu 15.00 mCi216560 y2 NPu = 1.00 = = NAm 1- dNAm>dt2T1>2–Am 14.45 mCi217370 y2

The two samples contain equal numbers of nuclei. The 243Am sample has a longer half-life and hence a slower decay rate, so it has a lower activity than the 240Pu sample. 19.5 (ii) We saw in Section 19.3 that alpha particles can travel only a very short distance before they are stopped. By contrast, x-ray photons are very penetrating, so they can easily pass into the body. 19.6 no The reaction 11H + 73Li S 42He + 42He is a nuclear reaction,

which can take place only if a proton (a hydrogen nucleus) comes into contact with a lithium nucleus. If the hydrogen is in atomic form, the interaction between its electron cloud and the electron cloud of a lithium atom keeps the two nuclei from getting close to each other. Even if isolated protons are used, they must be fired at the lithium atoms with enough kinetic energy to overcome the electric repulsion between the protons and the lithium nuclei. The statement that the reaction is exoergic means that more energy is released by the reaction than had to be put in to make the reaction occur. 19.7 no Because the neutron has no electric charge, it experiences no electric repulsion from a 235U nucleus. Hence a slow-moving neutron can approach and enter a 235U nucleus, thereby providing the excitation needed to trigger fission. By contrast, a slow-moving proton (charge +e) feels a strong electric repulsion from a 235U nucleus (charge +92e). It never gets close to the nucleus, so it cannot trigger fission. 19.8 no Fusion reactions between sufficiently light nuclei are exoergic because the binding energy per nucleon EB>A increases. If the nuclei are too massive, however, EB>A decreases and fusion is endoergic (i.e., it takes in energy rather than releasing it). As an example, imagine fusing together two nuclei of A = 100 to make a single nucleus with A = 200. From Fig. 19.1, EB>A is more than 8.5 MeV for the A = 100 nuclei but is less than 8 MeV for the A = 200 nucleus. Such a fusion reaction is possible, but requires a substantial input of energy.

Bridging Problem (a) 128Xe (b) no; b + emission would be endoergic (c) 3.25 * 109 atoms, 1.50 * 106 Bq (d) N1t2 = 13.25 * 109 atoms211 - e- 14.62 * 10

-4

s - 12t

2

Discussion Questions Q19.2 Number of nucleons on the surface produce 2nd term. As total number of nucleons increase the area increases slower than

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636

CHAPTER 19 Nuclear Physics

volume. Hence importance of 1st term which depends on volume is greater. Q19.3 Due to mass defect the total mass of nucleus is not equal to sum of masses of its constituents. Q19.4 No, as it depends on Z and N being odd or even. Q19.5 He, O, Ca, N, Sn, Pb have nuclei having magic number. Their nucleons are arranged in to complete shells with in the atomic nucleus. They have higher binding energy per nucleon. Q19.6 Binding energy per electron increases with increase in Z. It is due to long range nature of coulomb force. Q19.7 Q value of a or b emission is positive hence energetically favorable. Q19.8 Coulomb repulsion makes proton rich nucleus unstable. Q19.9 214Pb is not stable isotope, hence it decays. Q19.10 Intermediate elements do not have same half life. Q19.11 More charge of a particle allows more interaction hence they penetrate less. Q19.12 (a) b minus decay (b) a-decay (c) b -plus decay. Q19.13 Positron S 0 + 1e + Neutrino 0 0 v0 Antineutrino 00 vq 0 Q19.14 State of matter is due to bonding between outer electrons. Q19.15 Released energy may knock out an inner shell electron which leads to emission of x-ray photon. Q19.16 Activity of atmospheric carbon depends on content of carbon in atmosphere which have changed since 1900 due to carbon emissions by industries and vehicles Q19.17 Older material would have much smaller activity hence effect of contamination would be higher. Q19.18 Infinite time is required for whole radium to decay. Q19.19 Graph of binding energy per nucleon may be used to explain the phenomena of fission and fusion. Q19.20 The released energy appears in the form of kinetic energy of fragments. Q19.21 Interior of older stars is hotter than younger stars as carbon cycle occurs at higher temperature.

Single Correct Q19.1 (c) Q19.2 (c) Q19.3 (c) Q19.4 (d) Q19.5 (b) Q19.6 (c) Q19.7 (b) Q19.8 (c) Q19.9 (a) Q19.10 (a) Q19.11 (b) Q19.12 (a) Q19.13 (c) Q19.14 (c)

Q19.20 (a), (d) Q19.21 (a), (c), (d)

Match the Column Q19.22 (a) P (b) Q (c) P (d) S Q19.23 (a) P (b) P (c) R (d) T

EXERCISES Section 19.1 Properties of Nuclei 19.1 (a) 14, 14 (b) 37, 48 (c) 81, 124

Section 19.2 Nuclear Binding and Nuclear Structure 19.2 (a) 1.93m (b) 1.8 * 103 MeV (c) 7.56 MeV per nucleon 19.3 5.58 * 10 - 13 m 19.4 (a) 92.16 MeV (b) 7.68 MeV/nucleon (c) 0.8245% 19.5 (a) 5.68 * 10 - 13J (b) 1.13 * 107m/s

Section 19.3 Nuclear Stability and Radioactivity 19.6 (a) 783 keV (b) Not possible 19.7 (a) 92U235 (b) 12Mg24 (c) 7N15 19.8 (a) 4.27 MeV (b) 2.43 * 105 m/s 19.9 (a) positron (b) alpha-particle (c) electron 19.10 (a) 0.836 MeV (b) 0.7 MeV

Section 19.4 Activities and Half-Lives

One or More Correct Q19.15 (a), (c) Q19.16 (c), (d) Q19.17 (a), (b), (c) Q19.18 (c) Q19.19 (a), (b), (c)

19.11 5.01 * 104 yrs 19.12 The source is bavely usable. 19.13 (a) 4.92 * 1018 s - 1 (b) 2.99 * 103 kg (c) 1023 19.14 0.176 decays/min. 19.15 (a) 0.429 decays/sec (b) 1.16 * 10 - 11Ci 19.16 (a) 2.02 * 1015 (b) 1.01 * 1015, 3.78 * 1011 decays/sec (c) 2.53 * 1014, 9.45 * 1010 decays/sec 19.17 5727.3 years

Section 19.5 Biological Effects of Radiation 19.18 (a) 20 rad

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Answers

(b) 5 mJ (c) 20 rem 19.19 (a) 0.497 mJ (b) 0.0828 rem

Section 19.6 Nuclear Reactions, Section 19.7 Nuclear Fission, and Section 19.8 Nuclear Fusion 19.20 (a) X is 3Li6 (b) - 10.14 MeV (c) 11.59 MeV 19.21 18.4 MeV 19.22 (a) X is 3Li7 (b) 7.152 MeV (c) 1.4 MeV 19.23 (a) X is an a particle (b) 9.984 MeV 19.24 1.586 MeV 19.25 (a) 5,10 (b) 2.79 MeV absorbed

PROBLEMS 19.28 (a) 16 MeV (b) The proton removal energy is about twice the binding energy per nucleon 19.29 (a) 4.41 MeV (b) The neutron removal energy is about half the binding energy per nucleon. 19.30 (a) 39Y90 (b) 25% (c) 112 years.

19.31 4.69 MeV 19.32 (a) 5.30 MeV (b), (c), (d), (e) S Not possible 19.33 0.96 MeV 19.34 possible 19.35 0.0012864, possible 19.36 104 yrs. 19.38 185 MeV 19.39 1.57 * 106J 19.40 (a) A = 232, Z = 90 (b) 5.3 MeV, 1823 MeV 19.41 80 19.42 18m sec Ë2 - 1 19.43 2 19.44 8 * 109years 19.45 (c) 19.46 0.23 19.47 125mCi 19.48 (a) NB = lN0t e - lt lN0 1 (b) Amax = ;t = e l 19.49 (a) m - 2Bn1 - 4 m + 1Bn2 m - 1Cn - 4; NA = 2; (b) NB NA 1 = ; (c) NB2 2 (d) NA = N0e - 3lt; lNA & 2lNA

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PHOTO CREDITS About the Author Hugh D. Young; Roger Freedman Chapter 1 Opener: iStockphoto; 1.5: Journal-Courier/The Image Works; 1.8a: Richard Megna/Fundamental Photographs; 1.29a: Reproduced from PSSC Physics, 2nd ed. (1965), D.C. Heath & Company with Education Development Center, Inc., Newton, Massachusetts; 1.30b: Tony Craddock/Photo Researchers, Inc.; App. 1.1 iStockphoto; App. 1.2: Albert Kok; P1.94: David Parker/Photo Researchers, Inc. Chapter 2 Opener: U.S. Department of Energy; 2.10: AKG London Ltd.; 2.27b: Peter Terren/Tesladownunder.com; 2.28b: Russ Kinne/Comstock; App. 2.1a: Mark Webster/Photolibrary; App. 2.2: Dr. David Furness, Keele University/SPL/Photo Researchers, Inc. Chapter 3 Opener: Ted Kurihara/Getty Images; 3.9: NASA/JPL/Caltech; 3.11: Lester V. Berman/Corbis; 3.17: Hulton Archive/Keystone/Getty Images; 3.22: iStockphoto; App. 3.1: iStockphoto; App. 3.2: 3D4Medical.com/Getty Images Chapter 4 Opener: Shutterstock; 4.3: Shutterstock; 4.4: Andrew Lambert/Photo Researchers, Inc.; 4.7: Eric Schrader - Pearson Science; 4.11: Sandia National Laboratories; App. 4.1: iStockphoto; App. 4.2: H Heller, M Schaefer, & K Schulten, “Molecular dynamics simulation of a bilayer of 200 lipids in the gel and in the liquid-crystal phases,” J. Phys. Chem. 97:8343-60, 1993. Chapter 5 Opener: iStockphoto; 5.4: Eric Schrader - Pearson Science; 5.5: PhotoDisc/Getty Images; 5.8: iStockphoto; 5.12: iStockphoto; App. 5.1: Dr. David Furness, Keele University/Photo Researchers, Inc.; App. 5.2: Andrew J. Martinez/Photo Researchers, Inc.; 5.15 Richard Megna/Fundamental Photographs Chapter 6 Opener: Maximilian Stock/Photo Researchers, Inc.; 6.2 Photos.com; 6.5: John P. Surey; 6.13: John P. Surey; App. 6.1: Dorling Kindersley Media Library; 6.18: Richard Megna/Fundamental Photographs; App. 6.2: Mehau Kulyk/Photo Researchers, Inc.; 6.25a: Banana Stock/Alamy; 6.25b: Wikipedia Chapter 7 Opener: Simon Fraser/Photo Researchers, Inc.; 7.2b: Bill Aron/ PhotoEdit Inc.; 7.14: Richard Megna/Fundamental Photographs; 7.17b: Sargent-Welch/VWR International; 7.20b: Shutterstock; 7.21: Lawrence Berkeley National Laboratory; 7.40: Jeremy Burgess/Photo Researchers, Inc.; App. 7.1: Robert Smith/Photolibrary; App. 7.2: iStockphoto Chapter 8 Opener: CERN/European Organization for Nuclear Research; 8.8: Definitive Stock; 8.11: Jeremy Walker/Photo Researchers, Inc.; App. 8.1: iStockphoto; Test Your Understanding, page 267: Mark Antman/The Image Works; App. 8.2: Scarberry, K. E., Dickerson, E. B., Zhang, Z. J., Benigno, B. B. & McDonald, J. F., “Selective removal of ovarian cancer cells from human ascites fluid using magnetic nanoparticles.” Nanomedicine: Nanotechnology, Biology and Medicine. 6, 399-408 (2010) Chapter 9 Opener: iStockphoto; App. 9.1: Simon Fraser/University of Durham/Photo Researchers, Inc.; 9.9: Maximilian Stock Ltd./Photo Researchers, Inc.; 9.18a: Photodisc Red/Getty Images; 9.18b: Toyota Motor Sales USA; 9.18c: CIRRUS Design Corporation; App. 9.2: NASA/JPL/ Caltech; 9.25: Ken Gatherum, Boeing Computer Services Chapter 10 Opener: EschCollection/Getty Images; 10.2: Shutterstock; App. 10.1: Ron Chapple/Getty Images; 10.7: Photodisc/Getty Images; App. 10.2: SOHO/NASA; 10.10: John Walsh/Photo Researchers, Inc. Chapter 11 Opener: John P. Surey & Caroline J. Robillard; 11.5: Rubberball/ Getty Images; 11.14: Lawrence Migdale/Photo Researchers, Inc.; App. 11.1: Mauro Fermariello/Photo Researchers, Inc.; App. 11.2: iStockphoto; 11.22: Shutterstock; 11.23: Roger A. Freedman

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