JEE Advanced Physics-Electrostatics and Current Electricity [3 ed.] 9789353940317, 9789353942465

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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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JEE Advanced Physics: Electrostatics and Current Electricity

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JEE Advanced Physics: Electrostatics and Current Electricity Rahul Sardana

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Copyright © 2020 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 978-93-539-4031-7 eISBN: 978-93-539-4246-5 Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh, India. Registered Office: The HIVE, 3rd Floor, Pillaiyar Koil Street, Jawaharlal Nehru Road, Anna Nagar, Chennai 600 040, Tamil Nadu, India. Phone: 044-66540100 website: in.pearson.com, Email: [email protected]

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Contents Chapter Insight  xiv Preface xix About the Author  xx

CHAPTER

1

Electrostatics������������������������������������������������������������������������� 1.1 Coulomb’s Law, Electric Field, Dipole and Applications . Introduction .

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1.1

Electric Charge .

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1.1

Conductors and Insulators .

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1.2

Methods of Charging . . . . . . . . . . . . . . . . . . . . .

1.2

Properties of Electric Charge . . . . . . . . . . . . . . . . . . . 1.3 Concept of Charge Distribution(s) . . . . . . . . . . . . . . . . . 1.3 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . Important Points Regarding Coulomb’s Law .

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1.4 . 1.5

Coulomb’s Law in Vector Form . . . . . . . . . . . . . . . . . . 1.5 Coulomb’s Law in Position Vector Form . . . . . . . . . . . . . . . 1.6 Principle of Superposition . . . . . . . . . . . . . . . . . . . 1.14 Equilibrium of Three Charges . . . . . . . . . . . . . . . . . . 1.16  Electrostatic Field ( E ) . . . . . . . . . . . . . . . . . . . . 1.20 Electrostatic Lines of Force: Properties .

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Electric Field at the Axis of a Circular Uniformly Charged Ring . . . . . . 1.24 Electric Field Due to a Large Thick Charged Sheet . . . . . . . . . . . 1.33 Motion of a Charged Particle in an Electric Field . . . . . . . . . . . 1.43 Electric Dipole and Dipole Moment . . . . . . . . . . . . . . . . 1.45 Electric Field Due to a Dipole at a Point Lying on the Axial Line (End on Position) ��������������������������������������������������������������������������������������������������������� 1.45 Electric Field Due to a Dipole at a Point Lying on the Equitorial Line (Broad Side on Position) ���������������������������������������������������������������������������������������������1.46 Electric Field Due to a Dipole at Any Point P(r, q )�������������������������������������������������������1.46 Torque on a Dipole Placed in a Uniform Electric Field������������������������������������������������1.47 Potential Energy of a Dipole Placed in a Uniform Electric Field��������������������������������1.48 Small Oscillations of a Dipole Placed in a Uniform Electric Field . . . . . . 1.49

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viii  Contents

Force on an Electric Dipole in Non-uniform Electric Field . . . . . . . . 1.50 Force on a Dipole in the Surrounding of a Long Charged Wire .

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Electric Force Between Two Dipoles . . . . . . . . . . . . . . . . 1.51 Concept of Distributed Dipole . Gauss’s Law and Applications .

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Area as a Vector . . . . . . . . . . . . . . . . . . . . . . . 1.57 Electric Flux .

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Concept of Solid Angle . . . . . . . . . . . . . . . . . . . . 1.61 Solid Angle Subtended Due to a Random Surface at a Given Point . . . . . 1.61 Solid Angle Subtended by a Surface Not Normal to Axis of Cone or Not Lying on Surface of Sphere . . . . . . . . . . . . . . . . 1.62 Relation Between Half Angle of Cone and Solid Angle at Vertex .

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Electric Flux Produced by a Point Charge . . . . . . . . . . . . . . 1.63 Electric Flux Calculation Due to a Point Charge Using Solid Angle .

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Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . 1.67 Electric Field Due to a Point Charge q at a Point P at Distance x . . . . . . 1.69 Electric Field Strength of a Charged Conducting Sphere or Conducting Shell .

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Electric Field Due to a Non-conducting Uniformly Charged Sphere . . . . . 1.71 Electric Field Due to an Infinitely Long Thin Charged Wire .

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Electric Field Due to a Long Uniformly Charged Conducting Cylinder .

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Electric Field Due to a Long Uniformly Charged Non-conducting Cylinder .

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Electric Field Due to an Infinite Non-conducting Thin Uniformly Charged Sheet . . . . . . . . . . . . . . . . .

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Electric Field Due to an Infinite Charged Conducting Sheet .

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Electrostatic Potential, Potential Energy, Conductors and Applications . . . . 1.82 Electrostatic Potential Energy and Potential .

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Electrostatic Interaction Energy of a System of Two Charged Particles .

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Electrostatic Interaction Energy for a System of Particles .

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Electrostatic Potential (V ) . . . . . . . . . . . . . . . . . . . 1.84 Electrostatic Potential Difference (ΔV ) .

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Potential Due to an Assembly of Charges . . . . . . . . . . . . . . 1.85 Points with Zero Potential Due to Two Point Charges .

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Graphical Representation of Potential of a System of Two Point Charges .

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Electrostatic Potential at the Axis of a Uniformly Charge Rod .

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Electrostatic Potential at Point P Lying on Perpendicular Bisector of Rod . . . 1.88 Electrostatic Potential Due to a Charged Ring at Its Centre .

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Contents ix

Electrostatic Potential at a Point P on the Axis of the Ring at Distance x from Its Centre . . . . . . . . . . . .

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Variation of Electrostatic Potential on the Axis of a Charged Ring .

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Electrostatic Potential Due to a Uniformly Charged Disc at a Point P on Its Axis . . . . . . . . . . . . . . . . . . . . . . . 1.91 Electric Potential of an Annulus . . . . . . . . . . . . . . . . . 1.93 Electrostatic Potential on the Edge of a Uniformly Charged Disc . . . . . . 1.94 Electrostatic Potential Due to a Charged Conducting Sphere/Charged Shell . . 1.95 Electrostatic Potential Due to a Non-conducting Uniformly Charged Sphere . . 1.96 Potential Due to a Dipole . . . . . . . . . . . . . . . . . . . 1.97 Potential at Point P(r, q ) Due to a Small Dipole . . . . . . . . . . . . 1.98 Binding Energy of a Dipole . . . . . . . . . . . . . . . . . . . 1.98 Equipotential Surfaces: Introduction .

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Equipotential Surface: Properties . . . . . . . . . . . . . . . .

1.101

Relation Between Electrostatic Field and Potential . . . . . . . . . .

1.104

Motion of Charged Particles and Conservation Laws . . . . . . . . .

1.107

Conductors and Their Properties . . . . . . . . . . . . . . . .

1.113

Charge Distribution on a System of Parallel Plates . . . . . . . . . .

1.118

Method of Images .

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Electric Pressure on a Charged Surface Due to External Electric Field . . . . 1.126 Mechanical Force on a Charged Conductor . . . . . . . . . . . . . 1.126 Electrostatic Energy Density (ue) .

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Electrostatic Field and Potential Due to Induced Charges . . . . . . . . 1.130 Earthing of Charged or Uncharged Metal Bodies . . . . . . . . . . . 1.131 Current Due to Movement of Charges Through Earth .

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1.131

Field Energy Density of Electric Field . . . . . . . . . . . . . . . 1.135 Electrostatic Self Energy of a Charged Conducting Sphere .

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Total Electrostatic Energy of a System of Charged Bodies . . . . . . . . 1.138 Electrostatic Energy of a System of Concentric Shells . . . . . . . . . Solved Problems . Practice Exercises .

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Reasoning Based Questions . . . . . . . . . . . . . . . . . . . . 1.192 Linked Comprehension Type Questions .

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x  Contents Matrix Match/Column Match Type Questions . . . . . . . . . . . . . . 1.201 Integer/Numerical Answer Type Questions . . . . . . . . . . . . . . . 1.208 Archive: JEE Main .

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Answer Keys–Test Your Concepts and Practice Exercises . . . . . . . . . . 1.232

CHAPTER

2

Capacitance and Applications������������������������������������������� 2.1 Capacitors: Introduction .

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Calculation of Capacitance . . . . . . . . . . . . . . . . . . . . .

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Definition of C .

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Energy Stored in a Charged Capacitor . . . . . . . . . . . . . . . . .

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Capacitors in Series . . . . . . . . . . . . . . . . . . . . . . . 2.10 Capacitors in Parallel . . . . . . . . . . . . . . . . . . . . . . . 2.11 Electrostatic Energy Density (uE) . . . . . . . . . . . . . . . . . . . 2.14 Energy for Series and Parallel Combinations . Calculating the Net Capacitance of Circuits .

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Simple Circuits . . . . . . . . . . . . . . . . . . . . . . . . . 2.15 Circuits With Extra Wires . . . . . . . . . . . . . . . . . . . . . 2.17 Concept of Line of Symmetry . . . . . . . . . . . . . . . . . . . . 2.18 Balanced Wheatstone Bridge . . . . . . . . . . . . . . . . . . . . 2.18 Extended Wheatstone Bridge . Infinite Chain of Capacitors .

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Network With More Than One Cell . . . . . . . . . . . . . . . . . . 2.20 Case of Compound Dielectrics .

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Dielectric Slab Inserted in a Parallel Plate Capacitor . . . . . . . . . . . . 2.26 Conducting Slab Inserted in a Parallel Plate Capacitor . . . . . . . . . . . 2.27 Charge Induced on a Dielectric and Gauss’s Law for Dielectrics . Effect of Insertion of Dielectric in a Parallel Plate Capacitor .

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Electric Energy Density of Dry Air . . . . . . . . . . . . . . . . . . 2.35 Kirchhoff’s Laws for Capacitor Circuits . . . . . . . . . . . . . . . . 2.38

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Contents xi

Flow of Charge . . . . . . . . . . . . . . . . . . . . . . . . . 2.40 Energy Supplied/Consumed by the Battery . . . . . . . . . . . . . . . 2.40 Generation of Heat .

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Spherical Capacitor . . . . . . . . . . . . . . . . . . . . . . . 2.45 Cylindrical Capacitor .

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Energy Stored in a Spherical Shell . . . . . . . . . . . . . . . . . . 2.51 Solved Problems . Practice Exercises .

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Multiple Correct Choice Type Questions . . . . . . . . . . . . . . . . . . .

2.91

Reasoning Based Questions . . . . . . . . . . . . . . . . . . . . . . .

2.98

Linked Comprehension Type Questions . . . . . . . . . . . . . . . . . . . . 2.99 Matrix Match/Column Match Type Questions . . . . . . . . . . . . . . . . . 2.105 Integer/Numerical Answer Type Questions . . . . . . . . . . . . . . . . . . 2.110 Archive: JEE Main .

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Archive: JEE Advanced . . . . . . . . . . . . . . . . . . . . . . . . 2.117

Answer Keys–Test Your Concepts and Practice Exercises .

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Electric Current and Circuits����������������������������������������� 3.1 Electric Current .

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Different Situations Producing Current . . . . . . . . . . . . . . . . . 3.2  Current Density ( J ) . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Ohm’s Law and Electrical Resistance . . . . . . . . . . . . . . . . . . 3.5 Resistivity (r), Conductivity (s ) and Conductance (G) . . . . . . . . . . . . 3.5 Recasting a Wire of Given Mass . . . . . . . . . . . . . . . . . . . . 3.7 Fractional Change in Resistance .

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Current Density and Drift Velocity . . . . . . . . . . . . . . . . . . . 3.8 Ohm’s Law: Revisited . . . . . . . . . . . . . . . . . . . . . . . 3.9 Mobility . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10 Variation of Resistance with Temperature . . . . . . . . . . . . . . . . 3.12 Colour Code for Carbon Resistances . . . . . . . . . . . . . . . . . 3.14 Thermistor . . . . . . . . . . . . . . . . . . . . . . . . . . 3.14 Electrical Energy and Power . . . . . . . . . . . . . . . . . . . . 3.15 Electromotive Force (EMF) . . . . . . . . . . . . . . . . . . . . . 3.17 Resistors in Series and in Parallel .

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Network Analysis . . . . . . . . . . . . . . . . . . . . . . . . 3.21 Earthing or Grounding in an Electrical Circuit . Delta to Star or Delta-star Transformation .

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Star to Delta or Star-delta Transformation .

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Short and Open Circuits .

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Principle of Symmetry . . . . . . . . . . . . . . . . . . . . . . 3.30 Symmetrical Circuits . . . . . . . . . . . . . . . . . . . . . . . 3.30 Identical Potential Points . . . . . . . . . . . . . . . . . . . . . 3.32 Infinite Ladder and Grid .

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Balanced Wheatstone Bridge Like Situations .

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Kirchhoff’s Circuit Rules .

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Nodal Analysis in Circuits .

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Nodal Analysis Method . . . . . . . . . . . . . . . . . . . . . . 3.48 Primary and Secondary Cells/Batteries . . . . . . . . . . . . . . . . 3.52 Ideal Voltage Source .

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Practical Voltage Source . . . . . . . . . . . . . . . . . . . . . . 3.52 Discharging of Cell . . . . . . . . . . . . . . . . . . . . . . . 3.53 Charging of Cell . . . . . . . . . . . . . . . . . . . . . . . . 3.53 Short-circuiting of a Cell .

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F01_Physics for JEE Mains and Advanced - 2 xxxx 01_Prelims.indd 12

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Contents xiii

Wheatstone Bridge: Condition of Balance . . . . . . . . . . . . . . . . 3.79 The Metre Bridge . . . . . . . . . . . . . . . . . . . . . . . . 3.81 End Corrections in Metre Bridge . . . . . . . . . . . . . . . . . . . 3.82 Post Office Box .

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Hints and Explanations Chapter 1: Electrostatics . . . . . . . . . . . . . . . . . . . H.3 Chapter 2: Capacitance and Applications . . . . . . . . . . . . . H.126 Chapter 3: Electric Current and Circuits . . . . . . . . . . . . . . H.198

F01_Physics for JEE Mains and Advanced - 2 xxxx 01_Prelims.indd 13

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1

Electrostatics Capacitance and Applications

Learning Objectives

1.60

CHAPTER

Help the students set an aim to achieve the major take-aways from a particular chapter .

2

CHAPTER

CHAPTER

ChaPter InsIGht

3

Learning Objectives After reading this chapter, you will be able to:

Electric Current and Circuits

After reading this chapter, you will be able to understand concepts and problems based on:

(a) Coulomb’s Law, Electric Field and Dipole (b) Gauss Law Learning Objectives AfterConductors reading this chapter, you will be able to: (c) Electrostatic Potential, Potential Energy and reading this chapter, you will questions be able to understand All this is followed by a variety of Exercise After Sets (fully solved) which contain as per the concepts and problems based on: latest JEE pattern. At the end of Exercise Sets,(a) a collection of problems (c) asked Energypreviously Conceptsin JEE (Main(e) Capacitor Circuits Capacitors and Advanced) are also given. (d) Effect of Dielectric (b) Capacitance (f) Spherical Capacitor and Cylindrical Insertion Capacitor

Electricity & Magnetism-1

All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main

Learning Objectives After readingythis chapter, you willand be able to: Advanced) are also given.

1.60 Electricity IllustRatIon 46& Magnetism-1

COULOMB’S LAW, ELECTRIC FIELD, DIPOLE AND APPLICATIONS

After reading this chapter, you will be able to understand concepts and problems based on: A closed surface with dimensions a = b = 0.4 m and 2l 46 S cIllustRatIon = 0.6 m is located as in figure. The left edge of the (a) Current y (f) Nodal Analysis (k) Heating Effects of Current closed surface is located at positiona =x b= =a 0. .The INTRODUCTION (b) Ohm’s Law (g) Cells ELECTRIC CHARGE(l) Wheatstone Bridge A closed surface with dimensions 4 m elecand 2.4 Electricity & Magnetism-1 2l tric field throughout the region is non-uniform and (c) Resistance (h) Cell Combinations (m) Post Office Box to the other one, giving one conductor a c S c = 0.6 m is located as in figure. The left edge of the CAPACITORS: INTRODUCTION Life without electricity and its ects would have been Charge is a scalar quantity which is categorised into 1 Q eff (d) Resistance Combinations x (i) Ammeter (n) Potentiometer and the other one a charge −Q as a resul closedby surface at −position . The elecgiven E = ( 3is+ located 2x 2 ) iˆ NC , where xx =isain meter. hard to imagine. The electromagnetic force between is a two types. l A capacitor device which stores electric charge. (e)the Kirchhoff ’sofLaws ( j) Voltmeter (o) RC Circuit and Applications tric fieldnthroughout the region isSince non-uniform and If the process is being repeated times, the potential the charge is distributed in ratio their a potential difference ΔV is created, with charged particles is one of the fundamental forces ofvary (a) (a) Calculate the net electric1 flux leaving the closed Positive charge (anciently called Capacitors shape size,which but the basic con- Vitreous) Q xof Exercise in ( 3V.+ 2Calculate All thisbegin isconductors), followed by2la variety Sets (fullyand solved) contain questions as percharged the of the large capacitor has nowby fallen capacitance (or radii in case spherical given E =to x 2 ) iˆ NC −C. , where x is in meter. tively conductor at a higher pote nature. Hereofwe shall by ldescribing some ofinvolves surface. (b) Negative charge (anciently called Resinous) fi guration two conductors carrying equal zJEE latestof pattern. At the end of Exercise a collection of problems asked previously in JEEnegatively (Main i.e., the charged conductor. Note th the basic properties electrostatic forces and then Sets, (b) net charge enclosed byleaving the surface? (a) What Calculate the netiselectric flux the closed but opposite charges (shown in fi gure). Capacitors SOLUTION A body having no charge, is said to be neutral in and are also charged or uncharged, the net charge on th 2l given. smoothly on Advanced) to the basic electrical phenomeq1′ R1 6 1 (b) moving have many important applications in electronics. surface. Using the result obtained in (a), if the charge +Q nature i.e., on a neutral body the sumasofa positive = = non = and applications y whole is zero. Let V1 be the potential of C0 after first charging, then z along with aofrigorous mathemat2 q2′ R2 12 Some examples include storing electric potential (b) What net charge isx enclosed by the surface? is now at the centre a cube of side 2l, what is charges is equal to the sum of negative charges. =a The simplest example of a capacitor ical treatment through ILLUSTRATIVE EXAMPLES. energy, delaying voltage changes when coupled with E the total flux emerging from all the six faces of The positive charge means defi ciency elecC + C V = C V (b) Using the result obtained in (a), if the charge +Q ( 0) 1 0 0 two of conducting plates of area A, which a y ⎛ 1 ⎞ ( ) The branch, known as electromagnetism is conc resistors, fi ltering out unwanted frequency signals, ⇒ q1′ = ⎜ 9 = 3 μ Cthe closed surface? trons, whereas the negative charge on a body implies ⎟ isthe now at theofcentre ofand a cube of sidephe2l, what is to each other, and separated by a distance ⎝ ⎠ x = a cerned with nature electric magnetic 1+ 2 C0V0 forming resonant circuits and making frequency excess of electrons. The SI unit of Theory current is the amperein (A), where E the the totalconnection flux emerging from all the six faces of ELECTRIC CURRENT ⇒ V1 = shown figure (called as a Parallel Plate C with nomenon and between them. I shall a 1.60 Electricity & Magnetism-1 solutIon dependent and The independent dividers C + C0 c S.I. unit is coulomb (C). z 1 A of 1charge coulomb sec −1 .when =voltage the closed surface? ⎛ 2 ⎞be( presenting together with some applications of 6 Whenever μ CThethis ⎟ 9 ) = (a) +Q combined with resistors. a charge moves (i.e. a charge moving with b and x q2′ = ⎝⎜ Illustrations electric field due to the charge +Q is given ⎠ 1 + 2 interest to you from IIT-JEE view point exclusively. Common currents range from mega-amperes (in Let V2 be the potential of C0 after second charging then a Units of Charge respect we get current. So, moving solutIon IllustRatIon 46 y by to an observer) z lightning) to nano-amperes (in your nerves). When a glass rod is rubbed with silk, it acquires The common potential is charges given by(positive and negative) constitute electric Elaborative and x C0V1 = ( C + C0 ) V2A b The electric field due toasthe +Q is given (a) S.I. unit of limit chargeΔtis→coulomb (C). One coulomb solutIon closed surface with dimensions a = b = 0.4 m and In the 0, the instantaneous current I may power (a) tocurrent. attract such pieces ˆcharge  light –Q xi zkˆ ⎞of measure of is yj aˆ +direct 1 bodies Q electric 1 charge Q ⎛small A −6 ) Flow 2l + by of charge is that charge which when placed at d ( ˆ be defi ned as S cThe = 0.electric 6 m is located as in figure. The left edge of the q + q E = r = 9 10 × simple theory paper. The bodies which acquire this power are said ⎜ ⎟ 2 1 2 field throughout the region 2 2 ⎝a collection ⎠of charges +Q of electric4πε current. Suppose –Q V =is directed = C0 ⎞ surface C0located πε 4 r ⎛ C0 ⎞ ⎛ closed ⎛ is ⎞ r r 0 charges do0 not move they are rest in vacuum at a distance of one metre from an at position x = a . The electo be charged. If these solutIon ⇒ V2 = ⎜ V V = and 4 πε R + R C + C along⎟ x,0 therefore, ˆ + yjˆ + of 1 to dA 2 0 (is 1moving ΔQ helps dQ surface A , as ⎜⎝ C + C ⎟⎠ E 0will be perpendicular  2 ) perpendicular ⎞ xi zkˆ area the students 1andQthe branch 1 to Q a⎛Physics Experimentally it has been verified that th + C ⎠ throughout ⎝ C + C0 ⎟⎠ ⎜⎝ Ctric equal and Isimilar charge repels it and = lim stationary = …(2) 0 is non-uniform and called static which shown charges E= ⎜ ⎟ over0field the fourfield faces ofthe theregion surface perpenin 4fiπε gure.2 rˆ = y4πε of The electric throughout the which region are is directed 9 charge Q stored in a capacitor is linear 0 Δt dt of 9 × 10of −6 ) ( 9) r x ⎠ 2 )ˆ −1 is repelled by tit→with a force N. r2 ⎝ ( so on 0 r is Q 0electrostatics. deals with static charges called ( 9 10 9 × 10 × given by E = 3 + 2 x i NC , where x is in meter. to understand dicular the yz plane, andbe E perpendicular will to dA along x,totherefore, E will to Il ⇒be parallel V= to ΔV, the electric potential differenc Since flow has a direction, so we tional have implicitly intron ( 18 × 10 −2 ) the two which parallel to the yzperpenplane. y fourfaces faces ofelectric theare surface which are ⎛ C0 (a) ⎞overCalculate A In Ethe uncharged state, the charge on either one illustrations of the the net flux leaving the closed the the plates. Thus, we may write duced a convention that the direction of current corredA ⇒ Vn = V = ⎜ V 0 2l conductors in the capacitor is zero. During the chargdicular to the yz plane, and E will be parallel to dA n ⎝ C + C0 ⎠⎟Therefore surface. 5 sponds to the direction in which positive ⇒ to V 4.5 plane. × 10 V Q9/24/2019 = charges C Δ9:58:07 V are z 1 the for JEE Mains and Advanced - 2_Part 1.indd AM overWhat the two which are parallel the=01_Physics yz ingE process, a charge Q is moved from onesupporting conductor y enclosed r (b) net faces charge is by the surface? flowing. This current is called the Conventional Current. dA 1 ⎤ Therefore ⎡ E Q x Charges through cross nif section theory . Please note (b) Using the resultmoving obtained ina(a), the charge +Q The flowing charges inside wires are negatively ⎛ V0 ⎞ n b ⎢ ⎥ ILLUSTRATION 4 1.60 Electricity & Magnetism-1 y l ⇒ C= ⎜ ⎟ − 1 ⎥ C0 y charged electrons that move in the opposite direcis now at the centre a ned cube siderate 2l, at what is S The electric current is rof defi asofthe which ⎣⎢ ⎝ V ⎠ ⎦ x=a that theory and AEcapacitor of 20 μF charged 500flux V, isemerging connected a E Q x six tion of the current and is called ‘Electronic Current’. the to total from all the faces charges fl ow across any cross-sectional area. If of an 2l b l02_Physics xparallel to another capacitor of 10zμ F charged in to IllustRatIon 46 y c for JEE Mains and Advanced - 2_Part 1.indd 1 Electric currentsproblem flow in conductors : solids (metals, the closed surface? a c solving amount of charge ΔQ passes through a surface in a time S ILLUSTRATION 3 200 V, potential. semiconductors), liquids (electrolytes, ionized) a A closed surface zwith dimensions a = b =find 0.4 the m common and interval Δt, then the average 1current Iavg is given by 2 ax 2 2 2 2l techniques are Two isolated sphericalc conductors have radii 6 cm solutIon 2l and gases (ionized), but the flow is impeded in where r =ΔzQx + y + z S in Cartesian coordi= 0.6 m is located as ain figure. The left edge of the z c I avg = …(1) and 12 cm respectively.closed They surface have charges of 12 μposition Cb SOLUTION x non-conductors or insulators. (a) The electric due toS 1the is given = a . The elecbased on simple tfield nates. On Δthe surface , 2x =charge l and +Q the area eleϕafter Eis A + at Ex x = a + c The A x common zlocated E = − x are 2 2 2 x = a connected and −3 μ C . Find the charges they potential,  field throughout the region is non-uniform and where r = x + y + z in Cartesian coordiby tric ment is dA = dAiˆ = ( dydz ) iˆ . So, by a conducting wire. Also find the common potenlearning program 2 )ˆ −1 −1 Q x given by x = is 2inabc meter. solutIon the surface S , ⎛x = l and the area ele)2A) ab ( 2a=+Cc1)V1 + C2V2 nates. ϕ=E−=E( 3−=+(E23xa+2x2)= xab Ai(+NC E (, where  On tial after redistribution. 1 Q 1 l Q xiˆ + yjˆ + zkˆ ⎞ Vcommon a + 3 + x2 xa= a++ cc ˆ ˆ ˆ ˆ IF → THEN → E = is dA =2QdAi r =ˆ ⎛= xi ⎞ ⎜⎝)+iˆ zk ⎟ C1 + C2 ment + yj The electric field throughout the leaving region is directed . So, ( dydz (a) Calculate thegiven net electric for flux closed 4Aπε=0 r- 2_Part 1.indd 4πε r 2 2l E ⋅ ddA Substituting the a2, b andthe c , we find ⎟⎠ ⋅r( dydz⎠) iˆ = 03_Physics for JEE Mains and Advanced 10 9/24/2019 11:16:36 AM 2 ) E (values SOLUTION 2 ⎜ ⎝ ( along x,−therefore, will be perpendicular to dA ) ( ) ( ) r = 3 + 2 a ab + 3 + 2 a + c ab = 2 abc 2 a + c surface. 4 r πε ELSE . I would 0 ϕE = 0.269 Nm 2 C −1 − 6 F , V = 500 z V Care μF = 20 × 10 1 = 20 1 over faces ofisthe surfacebyHere which perpenQ y⎛ xiˆ + yjˆ + zkˆ ⎞ Net charge = ( 12 − 3 ) μ(b) C = What 9the μ Cfour net charge enclosed the surface? Ql ˆ E ⋅ dA = ⋅ dydz i = Substituting the given values for a , b and c , we find ) 3 dydz ⎜ suggest you not 2 ⎝ −12 E will be parallel to dA dicular to= the and ⎠⎟ ( −6 4πε result charge +Q −.138 × 10 ε 0ϕEyz2=C2plane, C = 2.38 pCC = 10 μF = 10 × (b) 10 = 200 V 0 robtainedr in (a), if the4πε y Using F , V2 the 0Q – Nm ϕE =the 0r –.269 to the2 yz plane. over two faces which are parallel E side 2l, what is – is now at the centre of a cube of to attempt the Ql dA x=a – Therefore dydz 6 emerging Thus, electric placed at + + + nS the E pC the from all six rfaces 3 x =ofl 20 × 10 −6 × 500 10the 10 +total × flux × 200flux through IllustRatIon Q = ε 0ϕE 47 =–2.38 × 10 −12 C = 2.38 + 4πε illustrations – 0 ⇒ V = + c is common −6 closed surface? −6 y r the + R1 ⇒ (a) RCompute the electric flux through a square sur- 20 × 10 + 10 × 10 + 2 E Thus, the electric S placed at x = l Q flux through x – l l + without going IllustRatIon 47 – face of edges 2l due located= at a V solutIon   b to a charge a +Q + + + dy ⇒ Vcommon 400 is ϕ = E ⋅ dA = Qll Sdz z– E 12 μ C distance l from the center ofsurthe 32 (a) –perpendicular Compute the electric flux through a square through the theory x 2 2 2 4πε 0tol thel charge – b (a) The electric field due +Q is given a − l 2l − l l + y + z S  square, as shown in figure. –3 μof C edges face 2l due to a charge  dy x +Q located at a by ϕ = zE ⋅ dA = Ql dz of that section . a c E + perpendicular distance l from the center of the 32 + 4πε 0 + z solutIon l 2 +ˆ y 2 +ˆ ⎞z 2 1 2 ˆ − − S l l square, as shown in figure.  ⎛ Remark(s) + 2 1 2 Q xi + yj + zk 1 Q2 + + + where E = r = x rˆ+=y + z 2 ⎜ in Cartesian ⎟⎠ coordiThe electric field+ throughout the region is directed + 4πε 0 r ⎝ r CAPACITOR OR CONDENSER 4πε 0 r 2 + + nates. On the surface S , x = l and the area eleV + V ϕ = − E A + E A along x, therefore, E will be perpendicular to dA E x x= a x x = a+c +  of collecting An arrangement which has capability over the four faces + + of the surface which are perpenment is dA = dAiˆ y=can + ) iˆ . So, ( dydz + + + (and storing) charge and whose capacitance be 2 be parallel to dA +- 22_Part JEE Mains 60+ 2 ( E 8/27/2019 6:36:57 PM to ( the ( 2a +accapacitor ) q′1 01_Physics fordicular − Advanced 3 + 2yz a )plane, ab2.indd + ( 3and a +will c ) )varied ab = 2is abc +=and called (or condenser). +   over the two Q ⎛ xiˆ + yjˆ + zkˆ ⎞ E q′2 faces which are parallel to the yz plane. ˆ dA ⎟ ⋅ ( dydz ) i = E ⋅ dA = Substituting the given values for a , b and c , we find ⎜ 01_Physics forTherefore JEE Mains Advanced - 2_Part 2.indd 01_Prelims.indd 60 8/27/2019 6:36:57 PM ⎠n F01_Physics for JEE Mains andandAdvanced - 2 xxxx 14 10/18/2019 12:43:27 PM r 4πε 0 r 2 ⎝ ϕE = 0.269 Nm 2 C −1

(

)

(

)

(

)

(

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∫ ∫

2

= = .

=

+

+

=

=

⋅ A=

=

⎜

(

+

+

)

2

2

(

)

(

)

(

)

(

)

(

∫ ∫( ∫ ∫( )

) )

2.24

R

2R

V

Sw

Electricity & Magnetism-1 1

Finally

(q1)f = (Q – q)

2 (q2)f =

Q+q 2

20 V

⇒

⎛Q ⎞ ⎜⎝ + q ⎟⎠ (Q − q ) = V 2 − 4πε 0 ( 2R ) 4πε 0 ( R ) a

⇒

b

Q + q − 2Q + 2q = 4πε 0 ( 2R ) V 2

c

⇒

3Q 2

3 q = 8πε 0 RV +

8πε 0 RV Q ⇒ q= + The charge on3plate b2is negative on both faces. Thus, the faces aonand c is So charge the finaloncharges each of the spheres are

and

Q a8πε b 0 RV − 2 C1 3

( q1 ) f

=Q−q=

( q2 ) f

b RV c 8πε Q 0 = +q = Q+ C2 3 2

SOLUTION

Chapter Insight

xv

Test Your Concepts

20 V

As the potentials of a and c are equal, the capacitors Test C Yourare Concepts-I These topic based Cab and in parallel. Therefore, equivalent bc ε A qa = qc = 0 V capacitance is Based on General Capacitance d exercise sets are 3.24 2ε A C = Cab + Cbc = 0 ⇒ qa =Ele qcct=r 100ε 0 × 20 (Solutions on page H.126) based on simple, ic ity & Further compare this energy with d μF charged to 500 V, is contwo capacitors. 1. A capacitor of 20 Mag netisin the capacitors. ⇒ the qa initial = qc =energy 2000ε 0stored − 2 nected in parallel to another capacitor of 10 μ F single concept m ε A ε × 2 × 10 -1 Cab = Cto = 0 V, fi=nd0the common = 100 ε0 where,charged potential. 5. Consider two conducting spheres with radii 2.28 R1 Electricity & Magnetism-1 bc 200 ⇒ Q = 2 × 2000ε 0 d 0.2 × 10 −3 classification 2. Calculate the heat generated when a condenser and R2 separated by a distance much greater than of 100 of either. Q is shared between ⇒ radius Q = 4000 ε 0 A total charge 2 × εμ0F×capacity 2 × 10 −2 and charged to 200 V is disR1 technique . These Compare (1) and (2), we get ⇒ Ccharged = = 200ε 0 1 Regarding dielectrics, it i the spheres, subject R1 to the condition that the elec0.2through × 10 −3 a 2 Ω resistance. 3 (a) These are non-cond 1 σ are meant for R1 system has the smallest 3. The capacitance of a variable radio capacitor can tric potential energy of the 2 3 σ σ − = ( ) p 2.36 Electricity Magnetism-1 field depending on i be changed from 50 pF to 950 pF by turning the possible value. &The spheres being very far apart, ε 0 K ε 0 students practice R2 R3 4 A 2 3 Ω this limiting value dial from 0° to 180°. With the dial set at 180°, the charge of each is uniR1 = you Rcan assume that the R 1 ⎞ C after they study ⎛ dielectric loses its in 12 R capacitor is connected to a 400 V battery. After formly distributed over the2 surface of each. Test Your Concepts-II ⇒ σp = σ ⎜ 1− ⎟ 13 R12 +Test ⎝ K ⎠16 R23 +Your Concepts-III to conduct. charging, the capacitor is disconnected from the (a) Determine the values of charges on the spheres /9 Ωa particular topic R 1 Based & Parallel Combination of13Capacitors A 2/9 Ω 4 (b) These have either R2 = battery and the dial is turnedon at Series 0°. Based on Dielectrics Breakdown R23inR terms of Q, R1 and R2. qp qand 1 ⎛ ⎞ want to 12 (polar-dielectrics, e.g C (a) What is the potential difference across the difference between sur⇒ =D ⎜ 1 − S ⎟ 24/9and on page the 2.124) R12(b) Ω + R Find the potential (Solutions (Solutions on page 2.129) A A⎝ K⎠ 2faces andspheres. 3 + Rcapacitance dipole moment (no when the dial capacitors, reads 0°? each with of the 1. Two capacitor identical parallel-plate the total of the system change in both 8Ω 16/9on 1 practice more 3 1. A capacitor is fi lled with two dielectrics of same E R RA13 Rcapacitor placed in an electric (b) How much work is required to turndifference the dial, if 3 = 6. cases of capacitance C0 ofisplates? charged to a B capacitance C, are charged to potential by the above displacement 1⎞ ⎛ dimensions but of dielectric constant 2 and 3 2 2.30 Electricity & Magnetism-1 3 R12 + potential 4 Ωthat topic learnt . (c) The dielectric consta friction is neglected? V0 and then isolated. A small capacitor⇒C qp = q ⎝⎜ 1 − K ⎠⎟ DV and connected in parallel. Then the plate sepad/2 3. AR210 μRF capacitor is charged 15 V. It isinnext Find the ratio to of capacities the two 3 +respectively. on its temperature a 4. ration A battery of of 10the V capacitors is connected to a capacitor of is then13charged from C0, discharged and charged in one is doubled. 35/9 connected in series with an uncharged 5SμSince F Finally, in case arrangements. TThe with rise in temperat 22 to Gauss’s Law, for dielectric AR T according capacitance F. The battery now removed and 2 again, the process being repeated 10 times.ILLUSTRATION (a) Find the 0.1 total energy of is the system of two B q capacitor. The series combination is fi nally conOD d/2any difficulty A/2fallen A/2 IllU TAn  ElT of1capacityofC RAV.air Since thiscapacitors capacitor isbefore connected a second uncharged potential of the capacitor C0 has now to 1= 10 μF is connected the toplate separation is U = STRA capacitor N A nected across a 50 V battery, as shown in Figure. SFOER⋅ dA = OR( Σqenc ) = ( q − qp ) TION 2C Calculate C. capacitor. If the charge distributes equally Finon STARthey Mvoltage doubled. 13 Find the newCpotential ε 12 V. Now to aF constant the space d tthese (d/2) K1 theS Ku ATIεO0 battery differences across 1 K1 refer 25pμ hethe -of pose D 0 can of inpu q02 poin in twoFind capacitors, find the total energy across stored d d U10 (b) the potential difference each between plates isNfilled withElaTliquid dielectric wethe c 0 A ILLUSTRATION 20 ts⇒ o t and μ F capacitors. nne3.48 C2 {K r (d/2) a ∵ q q and C KC = = = = e A U C C  Electricity & Magnetism-1  } r 2 s e 3 4 q 0 0 i 1 1 ⎤ ⎡ c s and 2KC tanKc ⎛tocharge ⎞ hints t the and capacitor after the plate separation is doubled. before 5.givCalculate fromthe plate was 3 asconstant =that flows ⇒ed in ksEt=⋅ dA =e(a) q − qthe n thCompute B of0 e of ⎜ 1 −the⎠⎟capacitance ⎥ ⎢ Two parallel conducting s a h r μ μ 5 F 10 F ⎝ e the figu own to ther capacitor. ε ⎣ e res K ⎦ K ε 0 (c) Find the total energy of the system after the m battery ci(A) re. in fi fashio0ninserted. istan rcuit (B) inals to these and −q are as shown. A gure (b) What betEw1 issolutions are c ceEcapacitance b plate separation is doubled. e s b f the after the plate was 2 t e R w e onc   (aa).q 4 Ωpotential is kept constant CASE-2:AWhen ORV battery repla en te , In this step, for each nod ΔVie=en 15 constant K, thickness 2 SOLUTION o=faq exercise rm ∵1 eεRsets heplateco 2= aεnε ed,= Sinserted? eddbis eErn (d) Reconcile difference in the answers to attached PRINCIPLE OF Athe PARALLEL ⋅ dA = Kε d given 2. CA Q parallel-plate capacitor tof separation nnec c⇒ remains currents are leavin esεe0 t εr as the i{nals 1, r 0 R3 0 } thickness d iseither inserted the t thK ted i ycharge C= 2 h parts (a) and (c) with the Law of Conservation r Initially on the capacitor Q = 10 × 12 = 120 μ C r h e a n PLATE CAPACITOR e charged to a potential difference DV . A dielecs n r (c) What does the value of capacitance becomes e esR3the i pec ee the d V remains constant, i.e., V = V , 0 node),plate and then atthe positive and we x =d tive ofdthe Ω istan end elta resista rat (a) Potential4 difference of Energy. 0 tethe tric slab of thickness d and dielectric constant whenn the thin cplate are n tmedium r upper plate When κdielectric iss filled, capacitance e so and w= vsin xterm andofVnode vs xvoltages graph K ε 0 E ⋅ edA orqk a cebook . Letbattery us 6place another identical conducting plate⇒1B R1 , s can no Consider a conducting plate A which is givenbecause a 8source is a of constant potential Ω Ω D d is introduced between the plates while the battery k times, bi.e.,shortened w by a conducting 2 R connected (i.e., s shcapacitance becomes new is 2. Two identical parallel plate capacitors are first conkeep in ma and V Always is the potential cw R 23 and parallel to itEsuch that charge is induced on plate B charge Q (as shown in Figure 1(a)) such that difference. its R1 o R is A. 8Ω remains connected Areanofin2each 50 Vto the plates. R1 figu plate nected in series and then in parallel. In each case,  =5 × 10wire)? called a node but to a node the positive plate be V re (b 13 (as shown in Figure 1(b)). C ′ = kC 50 μ C = potential rises to V. Then (b) Capacitance increases and becomes C = KC0 , ⇒ D ⋅ dA = q ). a,three R3 after (a) 4Show that the ratio of energy stored the above 6. figure A capacitor composed of parallel conthe plates of one capacitor are brought closer byBa STEP-4: Solve the system In the we can issay that points b, c, d, e, Ω of because by presence of a dielectric capacitance 4. Each plate a parallel plate capacitor has an S Nstored 1Q f All dielectric is introduced to the energy ducting plates. three lUTother distance Dd and the plates of O the Final charge onand the capacitor 50 ×called 12 plates = 600 μare C of the same RNodes to find the unknown no f are c and f= are Junctions.  point ION capacitor 2  becomes K times. area A. What amount of work has toU3be performed RThe where D = K ε area A. E is called Electric Displacement fithe rst pair of plates are kept a distancecurrent d1 For are moved away by same distance Dd. How does 1 through differen 3 Nodes) can as find = κ . Give a(or physin the empty capacitor isq 2also be0 called Hence additional charge supplied by the battery is q in 0 figu apart, and the space Vector. R1 between them is filled with a + + + + + + +U0+ + + + + + + + + re in g RAB , 2 3.40 Electricity & Magnetism-1 to its w ofRa dielectric K1. The corresponding data ΔQ3:30:07 = QPMf − Qmedium ical explanation for Tthis in)9/19/2019 stored i =3 480 μ C23 heseincrease (a 02_Physics for JEE Mains and Advanced - 2_Part 1.indd 7 equi e will c onve vale for the second pair2 are d2 and K2, respectively.NODAl ANAlYSIS M energy. enUt b twKo arr n V0, rEt0,tU C, V, lE, 0, 0 h a V0 t staC V e C o n c e p t u a l N od t e (ds ) n e R r as caus gem dWhat I2 elta happens to the CS = 8 ofcharge 2d show d (b) ethe ents ILLUSTRATION 23 term on 1 2 r capacitor? × 4 Problem C DE Solving Technique(s) To solve problems on th 16 n in (b) willSolving Technique(s) inals esistanProblem = o fi c f I1 beofe a variable radio capacitor can be 18 eare gure capacitors i s a 3. Two parallel each of capacitance C W s t use the following steps. The capacitance h l A current through a resistor can be calculated by using m ; R Applying Kirchhoff’s ec ri . e sam 9 easu⎛ Rules a cally ES = 8 RD redpF1 ⎞⎟tto e 7:21:31 K1turning K2 the dial 02_Physics for JEE Mains and Advanced - 2_Part 1.indd 24 8/27/2019 in series to a (B) battery emf V. Now, one (A) SOLUTION infrom × 6connected changed 50 Since qbpoPM = R1 of node-voltages S =6 epF 24 quiby e⎠twe 950 × 4 1In2 this chapter we thqt⎝⎜h1−asbKfollows: 2 =R STEP-1: Assign potentia =the va en a set 1 have seen how Kirchhoff ’s rules can of capacitors is fi lled completely with a dielec+ = e 8 1 from 0° to 180°. With the R R a W T ny p at -180°, the capaciI3 rranI dial 18 (c) WCharge on capacitor he tw 9 2 + 1 R2 The electric field in air R doing this, we assign ze q = Kq increases, i.e., , air V (< V ) gem 0 9be used to analyzetric o pa multiloop circuits. The steps are of dielectric constant K. V tor is connected to a 400 V battery. After charging, the e 2 1 a conductor K → ∞. Hence, 1 nts. ralle RFor Kirchhoff’s junction rule R2 = node of the circuit. This because below:(a) Calculate the ratio of the 3 l res summarized redu electric fi eld strength 3 Node 1 Node 2 capacitor is disconnected from the battery and the R istan V ced In dielectric, field is 2 + potential) is called as E Re= R2 R qp = q , σ p = σ and E = 0 ces b q = CV 3 +dial in the capacitor to the electric fiR eld when a sin According to the junction rule, A the threetocurrents et wae circuit diagram, turned at 0°. 3 the is Draw and labelanall quantigle r (a) Junction. s see en S V1 - V2 d the R esist dielectric is introduced. Does the fi eld increase n fro R is zero. Hence, the E x g shown here are related by and and unknown. The number I= V1 > V2 ) the above discussion as we (when may conclude a ties, both known (a) 1 Hence, What m fig For example, in the cir ∵ 1q30 ==RC1 0of V} 0 n ce⇒of 3q5 = ( KC0 ) VB=or Ris the potential difference across the capacicKq adecrease after {insertion of+dielectric? ure R3 + R1 R under n be to the unknown quantities is equal number of lintor when the dial reads 0°? potential to the point c. W 3 E I1 = I 2 + I 3 9 field remains (b) Find the amountsoof Echarge that flows through RA Problem Solving =E (d) Electric unchanged, R(b) R I required early independent equations we must look for. 0 How muchV workσis to turn the dial, if B = 4 ⎛ 16 2 +⎜ V2(> V1) a 10 Ω to the positive ter1 The middleqplate is connected ⎞ ⎛ the battery. Pro (a) E = E = = +20 V 3 (b) Assign a direction to the current in each branch ⎝ friction is neglected? vacuum 0 b 5 ⎟ + lem V0with air as dielectric ⎞ 8 Node 1ε of aA 2 LoopTechniques Rule V V0 parallel plate capacitor 9 ⎠ ⎝⎜ minal constant voltage source V, and the exterI1 ε 0 Node 0 Sol=v Eto0 7 the Ecircuit. E0 = 4.=(IfAthe = actual E0 ∵direction V = V0 is and opposite 9 ⎠⎟ = of ingplate W nal plates are connected the other terminal of V. d has d a plate area of 200 cmd3 and Techseparation 9 what SOLUTION The sum of the techniques voltage drops DV , across any circuit Hothe you have assumed, your result at end will E E n V V These w 0 1 = 0 to R andchangeiqin u(b) The the {K = dielectric ofin4  mm. Calculate percentage Ie=the (when V >the V ) capacitance (sE)2dielectric (a) Find of theconstant} system. (e) Energy stored the capacitor elements that form a closed circuit is zero. k eme equincreases be a negative number). the dial is atK 0°,2the 1capacitance of the capaciivof alevarnish ( km )When =b 3e of( thick- R capacitance ifmin aOften, ensure that nt found +20 V (b) What is the surface charge density on the = KU becomes U to alayer r ls 03_P d (c) Apply KCL each junction. it is 0 elta is giv torSis(c) tagiven h b {as K → ∞ } thisogiven r E by = 0 middle plate? 203d V rof esisboth en binside ness mmKCL the O d DV = 0 ysics for JEE M sewhileon Delt t nc plates. toconductor more convenient to0.1 y ththe students becomeains and Adv mmarking 1 A 2 thin 1apply ehalfway a)=? 50 × 10 −12 F e sum ain resis t2erplate inalsis inserted betC 5. conducting = 50 pF closed loop t w U = CV = KC V Since, ance a 1 ( )( ) e n currents in the branches. This reduces the number o (c) Compute the energy density in the medium K . p 0 0 en a f star d-2 lus t 1 _Par 2 2 the platesceofs d ny tw he capacitor. t 1.in resis aivparallel plate capable enough to Then, Vc = 0 V , Vb = +2 dd 2 ed have of unknowns tobetween be solved. For example, ifidwe o tethe tan is atSolving Problem Technique(s) When dial 180°, capacitance is given by 4 by th prod This rule is simply saying that there cannot be a e thir uct of t ces betw r1OB as 1 de are perfectly conduc assumed currents in the branches AO and 2 een is a −systematic d sta The hnodal solve a variety 12 analysis way of applying U = KU0 ∵C = KC0 and U0 = C0V0 e pF potential difference between of a point and itself and is ⇒ r reCs 2 =es950 F o s=ta950 × 10 I1 and I2, we 2 need not mark the current in branch 2 istan ’stwCircuit So, Va = Vb = +20 V , Ve Kirchhoff Laws (KCL) at each essential node r also inproblems accordance with the Law of Conservation of c e. in an easy OC as I3. Instead we apply KCL and say that the The potential difference across capacitor C2 , isingiven a circuit and- 2_Part represents terms 02_Physics for JEEof Mains and Advanced 1.indd 28 the branch current Energy. Let the potential of node current in branch OC is ( I1 - I2 ) . by of the node voltages. This will give us a set of equaand quick manner . The rules for determining DV across a resistor Problem Solving Technique(s) STEP-2: Now at junctio V = 400 V tions that we solve together to find the node voltages. and a battery with a designated travel direction are 2 A I1 DV While solving problems of this type always keep in where I = . Assumi Once we find the node voltages, we can use them to shown below: C O Charge on capacitor C2 is R mind that, whenever a battery is disconnected then, calculate any other currents or voltages of interest. (I1 – I2) to enter the junction, we q = constant −12essential nodes and select one Travel direction Travel direction 8/27/2019 7:22:52 PM 02_Physics for JEE Mains 8/27= 950all STEP-1: the B andI2Advanced - 2_Part 1.indd 36 × 400 q = C2VIdentify 2 /2019 7 × 10 :09:5 I1 + I 2 + I 3 = 0 4 PMnode i.e. the node at zero potenand if battery remains attached, then V = constant of them as a reference Lower V Lower V Higher V I I Higher V (d) Mark the polarity of the voltage across each 10 −9reference C ⇒ tialq i.e. = 380 0 V.×This node (can generally be taken 20 - x -5 - x 0 a a b b ⇒ + + element. In a resistor, the polarity depends upon as the ground) has usually most elements tied to it. ΔV = Vb – Va = +IR ΔV = Vb – Va = –IR 10 30 2 the assumed direction of current. The end into STEP-2: Now assign voltages (with respect to the refTravel direction Travel direction ( ) ( ⇒ 6 20 x + 2 5 which the current enters is marked positive. erence node) to all the nodes except the reference ⇒ 120 - 6 x - 10 - 2x node. E E Lower V Lower V Higher V Higher V – + Assumed direction of current + – STEP-3: Apply the KCL at each node except the refer⇒ 11x = 110 8/27/2019 7:22:12 PM a b a b 02_Physics for JEE Mains and Advanced - 2_Part 1.indd 30 ence node and write the equations. ΔV = Vb – Va = –E ΔV = Vb – Va = +E I ⇒ x = 10 V R

∫ ∫ ∫

Dielectric

∫ ∫

{

}

∑

{

Convention for determining ΔV

Note that the choice of travel direction is arbitrary. The same equation is obtained whether the closed

F01_Physics for JEE Mains and Advanced - 2 xxxx 01_Prelims.indd 15

+

V(= IR)

}

–

(e) Apply the loop rule to the loops until the number of independent equations obtained is the same as

10/18/2019 12:43:30 PM

2

1

+

Qq

a)

2

2

⎞ + 2Qq ⎟ ⎠

2

2

⎛ a ⎞ ⎜⎝ ⎟ 2⎠

2

1 ⎞ ⎟ 2⎟ ⎟⎠



ElECtRIC FIElD DuE to a laRGE   tHICK sHEEt FCA CHaRGED = FCB CASE-1: OUTSIDE THE SHEET q1 q2 Cd having uniform Consider a largelsheet of thickness chargeA density ρ . On bothBsides FCAof sheet FCBelectric field strength is directed, away fromx the sheet. q1 q

ρ2

⇒

ult even if we would have

=0

)

and q2 are fixed at a sepa-

in nature the net force due to q1 and lie between the charges q1 g them, towards the charge

B = l (given)

C

l–x FCA

Net electric field at point P will be EP ( = E ) given by

EP = E = E2 − E1

q1 ⎛ r1 ⎞ ⎛ l+x⎞ = =⎜ ⎝ x ⎟⎠ q2 ⎜⎝ r2 ⎟⎠

P

⇒

E = EP =

2

d

⇒

E=

The electric field strength at a point P in front of the sheetC is o n c e p t u a l N o t e ( s )

⎛d ⎞ ⎛d ⎞ ρ⎜ + x⎟ ρ⎜ − x⎟ ⎝2 ⎠ ⎝2 ⎠ − 2ε 0 2ε 0

ρx ε0

Conceptual Note(s)

Conceptual Notes σ Electrostatic Field in some Cases …(1) 2ε 0 (a) The electric field due to a uniformly Chapter charged2:rod The point where the resultant force on a charged Capacitance & ApplicationsThe 2.19Conceptual where becomes σ is the zero charge per equilibrium unit surfaceposition. area of the AB having charge density λ at a point P at perparticle is called Notes, Remarks, sheet. Further pendicular distance a from the rod, as shown in (a) stable equilibrium: A charge is initially in equiequiC C C E figure is G Words of Advice, σ = ρposition d and is displacedCby a small dis on c e pdist -u a l N o t e ( s ) librium tance. If the charge tries to return back to the A B A Misconception ρd C C ⇒ same E = equilibrium position then this equilibrium …(2) C1 C3 0 Removals provide is called2εposition of stable If equilibrium. = , then no charge exists in the branch conF H C C2 C C C (b) unstable equilibrium: quilibrium: If charge is4displaced by warnings to the CASE-2: INSIDE THE SHEET taining C (also called fifth branch) and hence it can a small distance from its equilibrium5 position and a α To find electric field strengthbe at an interior point P of ( just This circuit is known as Extended Wheatstonestudents Bridge N P the charge has no tendency toomitted return to theneglect same it as if it was never present). about β the sheet at a distance x from its centre as shown, we and it has two branches EF and GH to the left and equilibrium position. Instead it goes away from divide the sheet in two sheet parts. One of thickness common errors right of which symmetry in the ratio of capacitances the equilibrium position. As a consequence of this we can say that C and 1 ⎛ d neutral ⎞ ⎛ dis displaced ⎞ can be seen. B (c) If charge by a other of thickness and help them ⎜⎝ − x ⎟⎠ andequilibrium: ⎜⎝ + x ⎟⎠ as shown. C1C3 2 small distance and it is still 2in series condi, C2 and C4 are in C3inare to give equilibrium C C C C1 + C3 = = =1 avoid falling λ tion then it is called neutral Eequilibrium. Ex = ( sinα + sinβC) C C 2 C2 C 4 4πε 0net a and both in parallel. So, series to give for conceptual It can be seen that ratio of capacitances in branches C + C 2 4 P d λ PART 1 and However if, 2q1– xand q2 are notcapacitance fixed, then equilibrium of and E y = cos βAE − cos α ) EG is same as that between the capacitances ( pitfalls . x 4πε 0 a of the branches AF and FH . Thus, in the bridge d have to q will exactly be the same as done before,PART but if2Cwe CC d 1C 3 AEGHFA; the branch2.58 EF can be removed. Similarly Electricity & Magnetism-1 E1 then Cnetwe = have + 2 4 keep q1 and 2q+2 xalso in equilibrium, to make C1 + C3 C2 + C4 in the bridge EGBHFE branch GH can be removed. q2 (at B ) and q (at C) the net force on q1 (at A ) due to Other Balanced Wheatstone Bridge are to get the to be zero and then repeat the same forshapes q2 also of C E C CG= CC 1 + C2 indicated desired results. See Illustration(s) below below. to learn how to ⇒ A K ε 0 ( π R2B) K ε 0 R2ϕ ε 0 R2ϕ perform in these kind of problems. C1 C2 ⇒ C= − + electrostatic equilibrium E=

ree Charges

q

⎛d ⎞ ρ⎜ + x⎟ ⎝2 ⎠ 2ε 0

E

q1 q

= xvi ⇒ Chapter 4πε 0 r12 Insight 4πε 0 r22

0

E2 =

q2 B

2

te in nature and q2 < q1 he net force due to q1 and t lie on the line joining the

D

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 33

8/27/2019 6:34:00 PM

C5

A C3

E

B

C

CAB

2C 2d

Energy of capacitor = CV INFINITE CHAIN OF CAPACITORS 2

2d

…(4)

[ Kπ + 1 − K ϕ ] V itance between A and⇒B isUto= be calculated. 4Chapter d 2: Capacitance & Applications 2.53 C1 C3 A B Suppose the effective capacitance between A For a conservative force, C2 C4 C1 and B is X . Since the network is infinite, so even ∂U C3 if weSolved remove one repetitive τ =unit − of capacitors from ProblemS the chain, then too the remaining∂ϕnetwork would still Problem 1 have infinite pair of capacitors, i.e., capaci⎤ ε R22 ∂ ⎡ 1effective ⇒ τ ⎛=still − a⎢ 2Xg0 ⎞ { Kπ + ( 1 − K ) ϕ } ⎥ V 2 C5 between P and ⇒ Q ywill = ⎜ l ∂−ϕbe AC1parallelC3plate capacitor is placed intance a cylindrical ⎣ 4 dt ⎟ ⎦ …(1) ⎝ 2A ⎠ A B withC1a liquid C3 tank filled of dielectric constant K. The P ε 0the R2 capacitor 2 C C C Capacitance of at time t is, 2 4 2 A height of ( K − 1 )V area of cross-section of the tank is A and ⇒ τ=− C1 4d C1 theC4liquid is equal to the length of the square plate C1 C2 C 2 C ε 0 l ( l −∞y ) K ε 0 yl TheCnegative sign+shows that the&torque acts opposite = 2 Chapter 2: Capacitance Applications 2.55 2 of plate area l . The separation between the plates d dSeries to angular displacement. B at the botis d . A small orifice of area a is opened C1 is (C2a+square X) Q {∵ the plate plate of side l } tom ofx the tank at t = 0. If the capacitor in the proExTENDED WHEATSTONE BRIDGE capacitance of C1 + C2 + X Problem 8 strip Parallel 2 2 X) andCby η) ⇒ remains = ( 1 −connected 1 definition of current, cess with a battery of emf CE1 and P Since, q(C=2 +EC ε Athree conducting spherical shells A, B Figure shows ηd A dC = A 0 S {∵ ASQ= area of strip = bdx } assuming the level of liquidBridges in the capacitor remains The given figure consists of two Wheatstone dqAB BC C I = Q2, ++X) and + Q respectively. and C with outside, find the current in the circuit as a connected together.same Onexas bridge is connected between X X 2 X+charges −(C 2 2 ⇒ = η(1 − η ) K K 2 dt 1 2 function of time points AEGHFA and thedother is connected between Calculate Bthe system B decreasing withofthethe passage of between time, so εcapacitance Q Since here, q is 0 bdx points EGBHFE. So, fractional change is = ⇒ dC A and C . points Solution dq x tan θ d − x tan θ I = − + x2 to Equation 2 2 According dtK1 K 2 +Q = η ( 1 − η ) = 0.of 1 ( 1Continuity, − 0.1 ) = 0.Rate 081 of Change d is av, where, v is the velocity of efflux of volume ⎛ C dq ⎞ l ⎛ dC ⎞ +Q/2 ⎛ K ε 0 l ε 0 l ⎞ dy ⇒ I = ⎜ − ⎟ = − E⎜ =bdx − E⎜ − …(2) ⎟ given by ⎝ dt ⎠⎟ε 0–Q ⎝ d ⇒ Percentage Fractional Change = 8.1% dt ⎠= d ⎠ dt ⇒ C =⎝ dC ⎛ x tan θA da− x tan θ ⎞ E From equation (1), + 0 0 ⎜ a+b K ⎝ K ⎠⎟

∫

02_Physics for JEE Mains and Advanced - 2_PartProblem 1.indd 19

The capacitance of a parallel plate capacitor with plate area A and separation d is C. The space between the plates is filled with two wedges of dielectric constants K1 and K 2 respectively. Find capacitance of y the resultant capacitor. K1



v = 2 gy

∫

B 1

4

K2

d

⇒

VA =

Q ⎛1 1 ⎜ − + 4πε 0 ⎝ b a ( a

2 7:21:00 PM 8/27/2019 2

a 2 gCl⎛ a 2g ⎞ dy =− t⎟ ⎜⎝ l − ⎠ dt A 2 A dx b ⇒ C = ε 0 bK1 K 2 K1 d +(2), K 2we − Kget ( 1 ) x tan θ Substituting in equation 0

∫

Solution

2 ⎛ ⎞ εε0 El bK Ka2 2 gl − a gt ⎟ ( K − 1 ) ⇒ C Potential VA ⎡islog given by =I = 0d at⎜⎝1A, e (2 K⎠1 d + ( K 2 − K1 ) x tan θ ) ⎤ A A ⎣ ⎦ ( K 2 − K1 ) tan θ

l



⇒

⎛ Potential ⎞ ⎛ ⎜ at C due ⎟ ⎜

( VC ) = ⎜ to charge ⎟ + ⎜ ⎜ ⎝

VC = −

∫

F01_Physics for JEE Mains and Advanced - 2 xxxx 01_Prelims.indd 16

t

∫

⇒

ΔV =

Q ⎛1 1 ⎜ − + 4πε 0 ⎝ b a ( a

⇒

ΔV =

Q ⎛ a 2 − 2b 2 + ⎜ 8πε 0 ⎝ ab ( a + b

Capacitance of the arrange

CAC =

⇒

CAC =

Q Q = ΔV VA − VC

Q ⎪⎧ Q ⎛ a 2 − 2b ⎨ ⎜ ⎩⎪ 8πε 0 ⎝ ab ( a

Problem 9

Five identical conducting fixed parallel to and equid shown in Figure. Plates 2 conductor while 1 and 3 a ductor. The junction of 1 a connected to a source of co distance between any two s the area of either face of ea 5

0

⎛ Potential ⎞ ⎛ Potential ⎞ ⎛ Potential ⎞ at AKdue ⎟ ⎜ at t A due ⎟ ⎜ at A due ⎟ ε⎜0 bK 2 V = + ⎜ ⎡ K1 d + ( K⎟2+−⎜ K1 ) l tan θ ⎤⎟ ( ) A ⎜ to 1charge ⎟log ⇒ parallel C = plate charge horizontally to chargeso ⎢located ⎥ e to A θ ⎟ is Kd ( K 2 ⎜⎝− Kcapacitor ) tan ⎜ 1on ⎣ A is ⎠just on B ⎟⎠ 1⎜⎝ into onliquid C ⎦⎟⎠ ⎝ submerged that one of its plate Problem 2

Let us consider ε 0 AK1the K 2 equilibrium ⎛ K 2 ⎞ of liquid inside the logFor ⇒ C =plate e ⎜ that parallel − K1 ) d ⎝ K1 ⎟⎠ ( K 2 capacitor. +σ

Problem 5

⎟ ⎜ ⎠ ⎝

Potential difference, ΔV = V

d while the other is over the surface. The permitivity But tan θ = and A = lb of the liquidl is equal to K and its density is ρ . To ⎝ dt ⎠ Let the length of capacitor is l and breadth be b and what height will the level of liquid in the capacitor ⎡ K d + ( K 2 − K1 ) d ⎤ ε bK1 K 2get a charge 1 d is distance between plates and area of capaci⎛ a 2 g two ⎞ − σ. rise density ⇒ after C = its 0plates log e ⎢ 1 of surface ⎥ 2 ⇒ is A. − y dy = ⎜ dt d K1 d tor ⎣ ⎦ ⎝ A ⎠⎟ K 2 − K-12_Part ( ) 02_Physics for JEE Mains and Advanced 2.indd 58 Solution l So A = lb y

on A

1 Q 1 + 4πε 0 b 4πε 0

⎛ dy ⎞ Solution ⇒ A ⎜ − ⎟ = av = a 2 gy

1 ⎛ a 2 gfirst ⎞ we will find the capaciIn this − problem, ⇒ of − small y 2 dy area = ⎜ considering dt a strip of width dx tance ⎝ A ⎟⎠ l 0 x = l to get total capacithen integrate over. x = 0 to tance of capacitor where l is length of each capacitor

4πε 0

Potential at C, VC is given b

2

2 In the following infinite circuits,1the capac- 2 ε 0 Requivalent ( )

8/27/2019 6:34:59 PM C5 C4

These are based on multiple concept usage in a single problem approach so as to expose a student’s brain to the ultimate throttle required to take the JEE examination .

2d

3 ε R2 ⇒C C = 0C [ Kπ + ( 1 − K ) ϕ ] 2d H

1

C4

C2

Chapter End Solved Problems

F

1 Q + 4πε 0 a

VA = −

10/18/2019 12:43:35 PM

4 3 2 1

(C) ε 0 mgq tan θ

a

(A)

Q 8ε 0

(B)

Q 4ε 0

(C)

Q ε0

(D) zero

185. An isolated and charged spherical soap bubble has a radius r and the pressure inside is atmospheric. If T 1.208 JEE Advanced Physics: Electrostatics and Current Electricity 1.192 Electricity & Magnetism-1

is the surface tension of soap solution, then charge on drop is

(D)

ε 0 mg tan θ 3q

187. Two similar point charges q1 and q2 are placed at a distance r, apart in air. The force between them is F1 . A dielectric slab of thickness t ( < r ) and dielectric constant K is placed between the charges. Then the F force between the same charges is F2. The ratio 1 is F2 (A) 1 (B) K

Chapter Insight 2

xvii

2 ⎛ r–t+t K ⎞ r ⎛ ⎞ (D) ⎜ (C) ⎜ ⎟⎠ ⎝ ⎝ r – t + t K ⎠⎟ r string makes an angle θ with vertical. Its time period of oscillation is T in this position. Then

2qE COLUMN-I COLUMN-II (A) Maximum stretch in the spring is k E (B) (q) stretch (B) In equilibrium position, in the spring is (A) Ebig = ....Esmall (p) n σq σq (A) tan θ = (B) tan θ = qE Multiple CorreCt ChoiCe type QueStionS ε 0 mg 2ε 0 mg praCtiCe exerCiSeS k +Q (B) Vbig = ....Vsmall (q) n3 Chapter 3: Electric Current & Circuits 3.163 qE r This(C) section containsofMultiple Correct Choice Amplitude oscillation of block is Type Questions. Each question has lfour choices (A), (B), (C) andl (D), out of k (C) T < 2 π (D) T > 2 π which ONE OR MORE is/are correct. (r) n2 3 (C) C big = ....Csmall g g Single CorreCt ChoiCe type QueStionS 33. Compute value+Q ofofbattery current series with an ammeter and two cells identical with (C) the (r) isE 2qEI, in ampere, Amplitude oscillation 1.194 JEE Advanced Physics: Electrostatics andcork Current Electricity 1. (D) A charged ball of mass 1 g is suspended on 2. Consider a point charge Q placed at the origin O of k 3 A when the cells and batshown in figure. All resistances are in ohm. the others. The current    -1 3 (D) σcontains 61. the Twocoordinate point charges Q1 and EQ2, lie along linethe at eleca disa light stringchoices in the(A), presence a (D), uniform elec (s) n Type Questions. Each question big = ....σSingle small This section Correct Choice has four (B), (C) and out ofelecsystem. EB cells and and ECa be  of 2 ALet whenAshow the battery tery aid each other and 5 -1 6 magnitude tance from 10 field as E = 3i + 5Ej0 ×exists Ishown. which ONLY ONE is correct. 59. Atric uniform electric fieldWhen of in a, m, potential B ( 1, 1, - varia1) m tric fields ateach threeother. pointsFigure A ( 1, 2, 3 ) the r NC oppose each other. How many cells in the battery are Linked Comprehensionregion Type QuesTions tion along of charges θ = 37° . There T is are the two tenthe ball equilibrium at x-axis. 45°4 with at is aninangle m due to charge Q. Then and C ( 2, 2the , 2 )line Entries in COLUMN-I hollow metal spheres 1.22. An insulating long light show rod offour length the vertical side walls L pivoted at its Electric field E exists between wrongly connected?     qb )Paragraph ishaving the charge on the ball Each set consists sion in and A (the a, 0)string and B (0, or potential VA and point V and section external radius Linked b. each O with radius This Comprehension Type Questions based andinternal balanced with aaweight of the elevator. The time taken by the-2block to Questions. come centre W atcontains a distance (A) EA ⊥ofEaB Paragraph (B) EA  EC 4 12 V Take sin 37 ° = 0 . 60 0 and d g = 10 1 ms m 36. one A parallel plate capacitor has plate of area 10 cm 2 respectively, VBhas +2Qthen Eand (D), out of which only (D) (s) each in COLUMN-I with by questions. Each question fourpoint choices (B), (C) is correct. (For the sake of q and x Match from the leftcharge end as distribution shown infollowed figure. Charges to the lowest of (A), inclined plane is (assuming the     8 separated by a distance of 1 mm. the EBfilled (C) E (D)It is = 8 Ewith corresponding graph of electric field versus radial competitiveness may be a surface few(A) questions may have2more B = 4 EC C 2qthe are fixed to the ends of the rod. Exactly belowthere each toVAbe>smooth) Vthat (B) than VA one = VBcorrect if a = boptions) B if a > b 2 mica is 1 × 10 3 Ωm , dielectric mica theQresistivityQof r from as shown in COLUMN-II. +Q h a positive charge Q is ofdistance these charges at centre a distance 1 3. Equipotential surfaces representing a uniform electric r of electrons (C) VA > VB if a < b a0θ (D) (C) VA < N VB( =if4a) ,>excess b the leakage current through capacitorr is found to Comprehension 1 fixed. Then x is 1 the 2 3 field must be 34. 1:Calculate theLaw, potential, volt,Dipole of points A , B(, C )and1.201 COLUMN-I COLUMN-II Chapter Coulomb’s Electricin Field, &(D) Applications 13 E N = 5 , excess of electrons 10 -plane ampere, where ∗ is not readable. Find the be ∗ ×(A) surfaces. In the classical model of a hydrogen atom, the electron L D60.shown in Figure 1. What would be the new potenA simple pendulum of length  has a bob of mass x (A) Q. 1 is positive and Q2 of is negative. q value(B) of ∗normal (p) (A) to the direction the field. E –Q r =polarity 0.53 Å. revolves around the proton in a circle radius tial values, in volt, if V battery is reversed m,of with a charge q Å onofit.6 A vertical sheet of charge, q 2q Comprehension 3 Q2 ishaving positive. (B) spaced Q1 is negative (C) such thatand surfaces equal differences O Matrix Match/cOLUMN TheMatch magnitudetype of the QUestiONs chargeasofshown thewith electron andσ2? proton is area, passes Ω. incharge Figure Allunit resistances are inthrough per the point37. The gap between the plates of a parallel plate capacitor −19 is more than magnitude of Q2 . (C) Magnitude of Q in potential are separated by equal distances. 1 ( at t =is0 filled ) horizontally e = 1.6 × 10 C. Based on the above facts, answer the folAn electron is injectedthe into a uni+Q of suspension of the pendulum. At equilibrium, up with an inhomogeneous poorly conducting m (A) q = 11 nC (B) q = 12 nC +q rgiven in two columns, A B D Electric field is zero at 3. Each question statements which have to be matched. The statements in C (D) having potentials decreasing in the direction of h in this section contains lowing questions. form field produced by two oppositely charged plates hav- varies linearly medium whose conductivity in a direc3 3 COLUMN-I are labelled A, B, C W and D, while the statements in COLUMN-II are labelled q, sT(and t). given field. The par(C) T × 10 (D) = 4.l55 ×Any 10separation N 1 = 5.55 2 N p, ing 3r,length and d, as shownthe in figure.  plates from σ = 10 -12 Ω -1m -1 h 1. correct The magnitude of theONE electric between the proton  tion perpendicular to the statement in COLUMN-I can have matching with ORforce MORE statement(s) in COLUMN-II. The appropriate ticle has an initial velocity v0 = v0 iˆ perpendicular to E = Ejˆ . Q Q (Continued) -12 the electron is have to be darkened as illustrated in the following examples: bubbles corresponding to the answersand to these questions to 2 × 10 Ω -1m -1 . Each plate has an area 230 cm 2 30° −6 −7 y of 8.22t;× B 10→ qNand r; C → (B) 2 ×and 10 DN→ s and t; then the correct darkening (A) IfQLq the+correct matches are A → p, p and8.q; ε 02 h 2 LW QLq +s εand and the 0 h LW L separation between the plates is d = 2 mm . −8 (A) (B) BaSed QueStionS iNteger/NUMericaL aNswer ONreaSoning 2 bubbles willhlook l 8.2 × 10QUestiONs N (D) None of these Find the current, in nanoampere, flowing through the W like the following: (C) ε 0 h 2type W 122 hV 6V P G (A) t = = out 300 of V which . capacitor a voltage p value q electric r for s field t after 01_Physics JEE Mains and Advanced - 2_Part 5.indd 184 8/27/2019 2. QLq The of the due to the proton 2 2 This section contains Reasoning type questions, having four choicesdue (A),to (C) and V (D) ONLY ONE is 6:52:03 PM In this section, a numerical obtained series ofat calculations based oneach the data y(B), gdoing 4QLq +the ε 0 hanswer LW to each question + 4ismagnitude πε 2 0 h LW p q r correct. (C) in the question(s). (D) s t Each question contains distance θ have 0V STATEMENT 1 andv0STATEMENT 2.electric You to rated mark your answer as given 2 2 r is A 38. An bulb for 500 W at 100 V is used in a E 8πε 0 h W 8π h W – 12 p −q 1 11 − 1 s x e t r B (A) 5.76 × 10 NC (B) 5=.76 × 10 2 hNC Figure 1 having a 200 V supply. Calculate the O 1 the correct (B) (A) ofIffriction both statements are TRUE and STATEMENT 2 yiscircuit explanation of STATEMENT 1. resistance 1. A ring of radius 0.1 m is made out of a thin metallic s thet tforce C p −q1 r Bubble qEbetween the drop and the air to be -6(C) 2 5.76such 2. Three charges q1 , q2 and q310are taken that R that one must be put in series with the bulb, ato +statements 10D10 NC these (B) ( gof If- both arethe TRUE but STATEMENT explanation of STATEMENT 1. so that wire point of area of cross-section m . The×ring has 0 ) the p q r Bubble proportional velocity of drop, find the time 2 is not the correct s(D) t None D A B C Screen m q1 andcharge q2 areofplaced close together to form a when the bulb delivers 500 W. Give your answer in ohm. Bubble If STATEMENT 1 is TRUE π coulomb. The change in the a uniform t,the in (C) millisecond during which the and dropSTATEMENT travels the 2 is FALSE. 3. The ratio of the magnitudes of electrical and gravi1 2 3 L from is single point charge, force q3 atofdistance Bubble (D) If STATEMENT 1 isBased FALSE 2 is A TRUE. radius of the ring,the when a on charge 10 -8 coulomb 2h onbut theSTATEMENT above facts,39. and thefuse figure same of provided, lead wireanswer has an area of cross-sectional tational force between and 2proton isafter the field is switched off. (C) t =distance combination is a given repulsion of 2 unit in magnitude. 1.thisMatch the forces in COLUMN-I with the prop- electron 2 the following questions. COLUMN-I If there COLUMN-II 3 qE exists placed at the centre of the ring is x(A) × 10 -25 .2m.× Find x, if 0.2 mm . On short circuiting, current in theconductfuse coulombic attraction 4. Statement-1: The potentialthe of an uncharged rStatement-1: 10 39 and qin3 COLUMN-II are so combined the force on q1, dependent at When q2 given erties gcharges + a0 ) - each of magnitude 4.1.onFour point 1 μC (each pos( Chapter 1: E 11 2 m reaches 30ofA. HowR,long short-circufor aafter pointthe charge q located at two bodies both of 6. them may not be ing sphere radius the young’s of the metal is Nm . The path followed electron is a/an L is modulus an attractive force of magnitude 4 36unit. distance (B)2 ×210 .2 × 10 , dependent onitive) rbetween Theby the wire at the vertices of aForce square (p) onABCD. Q charges. before (A) are Q =fixed +q, displaced q COLUMN-I COLUMN-II iting, will fuse beginitstocentre melt?( rGive answer in 39 Statement-2: In2square coulombic attraction two bodies are (A) straight line (B)distance ellipse an attractive Also q3 and q1 when combined exert ) your r from > R is . O. Length of each diagonals of the intersect at h being displaced is along x-axis V 6 V 12 (C) 2 . 2 × 10 , independent of r 2. Three identical small balls, each of mass 0.1 g, are G Electricity & Magnetism-1 (D) always t = 3.174 4πε 0 r[IIT-JEE 2012] 2. Lista charge I with List IImillisecond. and the correct answer using oppositely charged. (C) cycloid (D) select parabola L. force on q2 of magnitude 18 unit at same distance diagonal is 2 m. A particle 2 g having qE ⎞ 2of masszero (A) Electrostatic force (D) 2.2 ×each 10 36 , havindependent of r suspended from one point by(p) silkConservative threads ( g + a0 )2 - ⎛⎜⎝from ⎟⎠ h Statement-2: Electric inside An conducinfinitely long solid cylinder of the code given below the lists. The algebraic ratio of charges q1 , q2 and q3 is )-1 , the ..032intensity calg -1 ( °C melting For lead, specific heatis 0field +2 μC is released a point 1The cmtime from O and 0V m of a nucleus oncm. an What charge, in nC, must be ing a length of l = 20 7. t t taken by the electron to leave the plate q1 and q2 are 2. Statement-1: Two charges (q) 1 placed at 1 Potential energy (B) Q = -q, displaced tor is zero therefore potential at each point form on conducvolume charge density ρ . It h (A) 100motion W2 -3 (B)is 25 W with 76. [2010] (A) 1 : 2electron : 3 to each ball so as(B) 2 : –3 each : 4 thread form 99 cm from A. The of the particle SHM 2 imparted toComprehension make Figure Then +q point is . gcm 327 ° C , density is 11 34 and resistivr. Then ofofforce on each the system ischarge along x-axis tor is zero. 5. An separation electron of(C) mass m,magnitude initially at rest, moves through neither (D) Both Two conductors have the same resistance (C) 4 : –3 : 1 (D) 4 : –3 : 2 at 0 °RC with but its centre on the (0, b) 6 q q of radius − 6 1 2 π x maximum ity is Statement-1: 22 × 10ltheir Ω cm . Theainitial temperature of wire negais an (B) angle = 30°attracting with the vertical? 2lm t. distance uniform field in gravity). time An drop of radius r35. = 1.64 ×a10certain m mass iscells F =having . in a denTheα force (q) oil Action-reaction 5. (B) When neutral body of is resistance charged s. of x.(A) (Neglect period 2 1 and temperature coefficients Twelve each having the value sameelectric e.m.f. are 2 Find the are α tconnected t1 = 75. [2011] 1 = 2 −3 4 πε r 3 3. In the process, when two bodies are charged by rub0 M, also initially at rest, takeseEtime A protonto of fall mass v0αmass ρoilto = 8measure .51 × 10 kgm is series allowed from rest. sity force 20°C . Neglect heat losses. a manexperiments towards the performed as shown in the figure. The magn tively, its increases slightly. . The respective temperature coefficients  of in and are kept in a closed box. Some of the cells their 3. In classical the 2 (r) toPotential (C) Q = q,If displaced a Now wire is stretched make it energy 0.1% longer, bing against the other, one becomes positively charged isthis placed near q1 its resis- Statement-2: Statement-2: third charge to move through ana equal distance uniform E battery It, then into a regionare external field the earth 3in When a body is chargedare negatively, parallel field at theitpoint P, which is at a d combinations 5.of Tconstant An electron is will projected withofisqthe an initial nearly dseries and 2md wrongly connected. This connected in chargecentre of an of electron, a charged dropenters of oil is placed system isspeed along y-axis tance Q Q Q Q while the other becomes negatively charged. Then t = t = (D) (C) Then force on q due to q remains F. and q2. oil electric field. Neglecting the effect of gravity, the ratio 1 2 3 4 applied in the downward direction. The drop has an 1 some electrons and electron has finite mass : 1 2 gains between thebetween horizontal plates ofDepends a planeoncapacitor. u = 4 × 106 ms into the field1 between (A)-1increase eE the v00) α byuniform 0.05% minimum α1 + α 2 α 1 axis + α 2 of the cylinder, is given by th Force (r) the (–2a, 0) (–a, 0) (+a, (+2a, 1 + α0) 2 Tmagnitude (A) (C) mass of each body earth remainsunknown unchanged of thetoelectric electric (B) , α1 + α 2 Chapter 1: Electrostatics 1.229 though (A) quite small , is nearly equal Under and the sun action of an electrostatic field, thecharge drop q. The plates (B) in increase 0.2%  figure. The direction the field is 3. t parallel Statement-1: While going away fromof a point charge 2 2 nature of medium 2 ˆjby (B) mass of each body changesfield marginally (s) The charge Q is in F = mg = − mg is adjusted till the gravitational force, arChiVe: Jee Main (D) Q = -q, displaced g The value of k is it leaves 8. The velocity of the electron at Statement-1: time t1 when moves uniformly upward, covering a certain distance (C) decrease bythe 0.2% vertically downward, and field is zero except in the 6. Electric field at a point is always inversely or a small electric dipole, electric field decreases at the between the (C) mass of each body changeson slightly and hence α 1 +List α2 α 1α 2 equilibrium the oil drop is the exactly balanced 2M by along the electric force,by 0.05%stable y-axis mthe plates List Ifield decrease (C) αto or downward, when the during space the plates. The electron entersis z 1 +(α samebetween rate(D) in both the cases. 2, (A) (B) at [IIT-JEE 2006] )2 II (D) α 1 + α 2 , objects distance total the masstime t1 = 6 ms, 13. the 0 , z0 ) where z0 > 0 proportional . Then the motion ( 0, the 2 this balancing occurs when the value of Fe =interacting qE . Suppose α +α m M point 1.  [Online 2019] eE l aboutto atˆStatement-2: a April point midway between the plates. the ⎛Ifproportional ⎞ electron For spherical symmetrical charge distribution, R1 a field R1 due to a line charge 1at a 2 (t) Executes SHM sign of the on the plates changes, covering the 5varia−1 )field Electric is inversely Statement-2: Electric ˆ (D) mass of charge each body changes slightly but the total P is of ( (B) v (A) 1.92 × 10has = −200 Ey j Ω = − resistor NCa certain j, electric field of is given by EA ⎜ ⎠⎟If one 0 +x 1. (D) Force between two (s)tion Principle colour code. ⎝ just misses upper platethe as it emerges from electric potential from centre isfrom mfor v0all , Q2 , Q all positive P.the Qfield, 5 −1 with distance of the distance point charge. point is inversely proportional to distance. =of 3 ms. Assuming same distance the time 1values M MOperiodic mass remainsduring the same (A) of3 , zQ 2E onsquare the above facts, answer withtSuperposition 0 4satisfying 0 < z0 < ∞ y = 1.92 × 10is NC . Based protons (C) (D)in the 1 replaces the green code, the-new E3 q red colour by . the electric in 1NC m magnitude (B)mfield simple harmonic thegiven following questions. Vfind = the for r ≤ Rof in kept diagram. Given that R 2 4 for 2 2all2 values of z0 satisfying R2 0 applicable +q is at the 4. A small block of mass m, charge resistance will be m v + e l E (D) None of these (C) 2. -x πe 4 R 0 Jee adVanCed 0arChiVe: 0 Q. Q1, Q2 positive; Q3 , QE41 negative z0mv ≤a R 0 0 i.e., if both q1 and q2 are positive or both q1 and q2 are negative then the charges repel each other.  F12 is the force exerted on charge q1 by charge q2 and q1 q2 > 0 r12 F12



 F12 =

q1

r21

q2

F21

1 q1 q2 rˆ21 4pe 0 r 2

9/20/2019 10:07:16 AM

1.6  JEE Advanced Physics: Electrostatics and Current Electricity

 If F21 is the force exerted on q2 due to q1  then   F21 =

1 q1 q2 rˆ12 4pe 0 r 2

Both rˆ21 and rˆ12 have same magnitude i.e., unity and are oppositely directed   1 q1 q2 ˆ21 ) = - F12 F21 = r ( 4pe 0 r 2   F21 = - F12 So the forces exerted by charges on each other are equal in magnitude opposite in direction

COULOMB’S LAW IN POSITION VECTOR FORM Consider two point charges q1 and q2 located at points A and Brespectively. According to Coulomb’s Law the force F12 exerted on charge q1 by q2 is given by  1 q1 q2 F12 = rˆ21 2 4pe 0 r21  r21 ˆ Since, r21 =  r21   1 q1 q2 r21 1 q1 q2  r21 ⇒ F12 = = 2 3 4pe 0 r21 r21 4pe 0 r21 F21

q2 r21

B

q1

r2

A

F12

r1 O

Applying Triangle Law of vectors to vector triangle OBA    r2 + r21 = r1    r21 = r1 - r2  F12 =



q1 q2 1  4pe 0 r1 - r2



3

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 6



( r1 - r2 )

 Similarly F21 =

q1 q2 1  4pe 0 r2 - r1



3



( r2 - r1 )

If the charges are attracting each other i.e., are of opposite nature, then  F21 =

q1 q2 1  4pe 0 r1 - r2



3



( r1 - r2 ) = -

 and similarly for F12

q1 q2 1  4pe 0 r2 - r1



3



( r2 - r1 )

So, if charge q1 , is lying at position vector  r1 = x1iˆ + y1 j + z1 k and a charge q2 is lying at posi tion vector r2 = x2 iˆ + y 2 j + z2 k , then (a) For like charges, we have  q1 q2   1 F12 = r - r2 ) and   3 ( 1 4pe 0 r1 - r2  q1 q2   1 F21 = ( r2 - r1 ) 4pe 0 r2 - r1 3 (b) For unlike charges, we have  q1 q2   1 F12 = r - r2 ) and   3 ( 1 4pe 0 r1 - r2  q1 q2   1 F21 = r - r1 )   3 ( 2 4pe 0 r2 - r1

Problem Solving Technique(s) (a) The electrostatic force (F) is central (it is directed along the line joining the centres of two charges) and spherically symmetric (it is a function only of r). (b) F is a force obeying the inverse Square Law. (c) While applying Coulomb’s Law, I must remind you to remember the following two points. Firstly, the charges are assumed to be at rest. (As we shall see later that charges in motion also produce and experience magnetic forces). Secondly, the charges are assumed to be point particles. However, when the charge is distributed uniformly over a spherical surface, then force on a point charge outside the surface may be computed from Coulomb’s Law by treating the charge on the sphere as if it were concentrated at its centre and so

9/20/2019 10:07:27 AM

Chapter 1: Electrostatics 1.7

q Q

Centre r

r



Furthermore, if the dimensions of two charged bodies are small in comparison to their ­separation, then too Coulomb’s Law (as applied for point charges) will provide an approximate value for the force between them. In all other cases, integral calculus finds its applications by suitably applying the concepts of charge distributions (linear charge distributions, surface charge distributions or volume charge distributions, discussed later). (d) While calculating the force between the two charges from Coulomb’s Law, never take into account, the sign of the two charges. The sign just indicates the nature of the force. However it is advisable to take the sign when writing Coulomb’s Law in vector form. (e) When we are to calculate the force on charge 1 due to charge 2, then we assume charge 2 to be fixed and vice versa, unless and otherwise stated.  F 12 = m1a1  F 21 = m2 a2 i.e., m1a1 = m2 a2 (by Newton’s Third Law) where m1 and m2 are the masses of particles having charges q1 and q2 respectively. (f) When two identical bodies having charges q1 and q2 respectively are brought in contact and sepaq +q rated, then the charge on each body is 1 2 . 2 (g) Some problems can also be solved by using the concept of Lami’s Theorem, according to which “if a body is in equilibrium under   the influence  of three concurrent forces F1 , F2 and F3 i.e.,     F1 + F2 + F3 = 0, then F3

α

F2 γ

β

F1

OR

F F1 F = 2 = 3 sinα sin b sinγ

q

Q

F3

γ α

with q1 and q2 substituted in the above expressions along with the nature of the charge signified by the signs on them. (i) If we are to find the electrostatic force on a charge q0 at a point P, due to a charge distribution (may be uniform or non uniform), then we calculate the force on q0 due to an infinitesimal element of the distribution and then integrate it within appropriate limits. Illustration 2

Two point charges Q1 and Q2 are 3 m apart and their combined charge is 20 m C. (a) If one repels other with force of 0.075 N. C ­ alculate the two charges. (b) If one attracts the other with force of 0.525 N. Find the magnitude of the charges. Solution

(a) Since combined charge of Q1 and Q2 is Q1 + Q2 = 20 m C …(1) The force is repulsive, so according to Coulomb’s Law 1 Q1Q2 ⎡Q Q ⎤ 0.075 = = 9 × 109 ⎢ 1 22 ⎥ 2 4pe 0 r ⎣ (3) ⎦ ⇒ Q1Q2 = 75 × 10 -12 C 2 = 75 m C 2   Substituting value of Q from (1) 2

F1 β

(h) For two charges q1 and q2 having position vectors    r1 and r2 the electrostatic force F12 is given by the expression    q1q2 1 F12 = ( r1 - r2 ) 4pe 0 r1 - r2 3       q1q2 1 r - r1 ) and F21 =   3( 2 4pe 0 r2 - r1

F2

Q1 ( 20 - Q1 ) = 75 ⇒  Q12 - 20Q1 + 75 = 0 ⇒  Q1 = 15 , 5 So the Q ′s are 5 m C and 15 m C

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 7

9/20/2019 10:07:37 AM

1.8  JEE Advanced Physics: Electrostatics and Current Electricity

(b) Force is attractive, so one charge is negative so force equation is     

-0.525 = 9 × 109

For P N1

Q1Q2 9

F

Q1Q2 = -525 m C 2

⇒

  ⇒   Q1 ( 20 - Q1 ) = -525

T mg

30°

2 ⇒   Q1 - 20Q 1 - 525 = 0

α

α

For bead P to be in equilibrium,

From here Q values are 35 m C and -15 m C or -35 m C and 15 m C



mg cos 60° = ( T - F ) cos α …(1)

and  N1 = mg cos 30° + ( T - F ) sin α …(2) For bead Q to be in equilibrium,

Illustration 3

A rigid insulated wire frame in the form of a rightangled triangle ABC is set in a vertical plane as shown in the figure. Two beads of equal masses m each and carrying the charges q1 and q2 are connected by a cord of length l and can slide without friction on the wires. Considering the case when the beads are stationary determine q1 P

A α 90° l

Q

q2

30° B

C

(a) the angle α (b) the tension in the cord and (c) the normal reaction on the beads. If the cord is now cut, what are the values of the charges for which the beads continue to remain stationary?



mg sin 60° = ( T - F ) sin α …(3)

and  N 2 = mg cos 60° + ( T - F ) cos α …(4) Dividing (3) by (1)

tan 60° = tan α

Hence, α = 60° (b) Hence, From (3), T - F = mg ⇒   T = mg + F = mg +

(i) Weight mg acting vertically downward (ii) Tension T in the string along the length PQ (iii) The electric force F =

1 q1 q2 between the 4pe 0 l 2

beads along the length PQ (iv)  Normal reaction N1 of wire on bead.

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 8

4pe 0 l 2

From (2) and (4), we have N = mg cos 30° + ( T - F ) sin 60°   1 ⇒   N1 = mg cos 30° + mg sin 60° = 3 mg and     N 2 = mg sin 30° + mg cos 60° = mg (c) If string is cut, Tension T = 0 Hence we have,

Solution

(a) The forces acting on bead at P having charge q1 are

q1 q2

- q1 q2 4pe 0 l 2

= mg

q1 q2 = -4pe 0 mgl 2 Hence if the beads are to remain in equilibrium after the string is cut, q1 and q2 should have opposite charges satisfying the above condition. Illustration 4

A particle of mass m carrying charge q1 is revolving around a fixed charge -q2 in a circular path of radius r. Calculate the period of revolution and its speed also.

9/20/2019 10:07:50 AM

Chapter 1: Electrostatics 1.9 Solution 2

1 q1 q2 4p mr = mrω 2 = 4pe 0 r 2 T2

⇒ T2 =

( 4pe 0 ) r 2 ( 4p 2 mr ) q1 q2

Tcos ⎛⎛ dθ ⎛ 2 ⎛



Let us consider an element of arc length dl having a ⎛ q ⎞ charge dq. Then dq = ⎜ dl . ⎝ 2p r ⎟⎠

pe 0 mr ⇒ T = 4p r q1 q2 4pe 0 r 2

⇒ v=

=

dθ 2

Tcos ⎛⎛dθ ⎛ 2 T

2Tsin ⎛⎛dθ ⎛ 2 ⎛

q1 q2

Since,

dθ 2

⎛

Solution

mv 2 r r r dθ dθ 2 q 2 0

q1 q2 4pe 0 mr

Illustration 5

What would be the interaction force between two copper spheres, each of mass 1 g, separated by the distance 1 m, if the total electronic charge in them differed from the total charge of the nuclei by one percent? Solution

Total number of atoms in the sphere of mass 1 1g= × 6.023 × 10 23 63.54 So the total nuclear charge

Q=

6.023 × 10 23 × 1.6 × 10 -19 × 29 63.54

Now the charge on the sphere ( q ) = Total nuclear charge - Total electronic charge

1 q0 dq  4pe 0 r 2 For equilibrium to be there dF =

⇒



1 ( 4.398 × 10 2 ) N F= 4pe 0 12



F = 9 × 109 × 19.348 N = 1.74 × 1015 N

2

Illustration 6

A thin wire ring of radius r has an electric charge q. What will be the increment of the force stretching the wire if a point charge q0 is placed at the ring’s centre?

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 9

(radially outwards)

⎛ dq ⎞ dF = 2T sin ⎜ ⎝ 2 ⎟⎠ 1 q0 dq ⎛ dq ⎞  2T ⎜ 2 ⎝ 2 ⎟⎠ 4pe 0 r ⎧ ⎛ dq ⎞ dq ⎫ ⎨∵ for small angle, sin ⎜⎝ ⎬ ⎟ 2 ⎠ 2 ⎭ ⎩

 ⇒

1 4pe 0

⇒

T=

23

6.023 × 10 29 × 1 × 1.6 × 10 -19 × = 4.298 × 10 2 C 63.54 100 Hence force of interaction between these two spheres, ⇒   q=

If dF is the force of repulsion between the element and the charge q0 at the centre, then

⎛ q ⎞ q0 ⎜ dl ⎝ 2p r ⎟⎠ ⎛ dl ⎞ = T⎜ ⎟ 2 ⎝ r⎠ r qq0

8p 2 e 0 r 2

Illustration 7

Two small equally charged spheres, each of mass m, are suspended from the same point by light silk threads of length l. The separation between the dq spheres is x  l. Calculate the rate with which dt the charge leaks off each sphere if their velocity of α , where α is a positive approach varies as v = x constant.

9/20/2019 10:08:00 AM

1.10  JEE Advanced Physics: Electrostatics and Current Electricity Solution

In this problem, let us first calculate the value of x in terms of other known parameters. Let us consider the particles to be in equilibrium, then

T cos q = mg …(1) T sin q =

⇒ tan q =

q2 4pe 0 x 2 q

…(2)

⇒

x3 q

2

=

l 2pe 0 mg

4pe 0 mgx 2

θ θ

T

Illustration 8

Solution l

l T Fe

Before starting the problem, let us draw a diagram that gives a visualisation of the problem.

T Fe

x mg

P

mg

tan q is very small and hence tan q  q =

⇒

q2 x = 2l 4pe 0 mg x 2

x 2l

q2 l …(3) 2pe 0 mg 1

⎛ q2 l ⎞ 3 ⇒ x=⎜ ⎟ ⎝ 2pe 0 mg ⎠

Take derivative of (3) w.r.t. time, we get dx ⎛ l ⎞ ⎛ dq ⎞ 3x 2 =⎜ 2q ⎜ ⎟ dt ⎝ 2pe 0 mg ⎟⎠ ⎝ dt ⎠

But according to the problem, v =

α ⎛ x 3 ⎞ ⎛ dq ⎞ ⇒ 3x =⎜ ⎟ 2q ⎜ ⎟ x ⎝ q2 ⎠ ⎝ dt ⎠ dq 3α ⎛ q ⎞ = ⎜ ⎟ dt 2 ⎝ x3 2 ⎠

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 10

l T

α

B θ

qQ 4πε0 (AB)2

mg



2α + q = p

p q - …(1) 2 2 Further from the diagram, we observe three forces to be acting on B, ⇒ α=

(a) Weight mg (acting vertically downwards) (b) Tension T (acting along QP ) (c) Coulombic force F =

α dx = dt x

F=

Q

Let the angle of suspension be q . If ∠PAB = α , then ∠PBA = α (because angles opposite to equal sides are equal). Also, in ΔPAB

Qq 4pe 0 ( AB )

2

along AB

Free body diagram of B is shown here T

2

⇒

T

α A q

∴



θ

l

Since x  l

⇒ x3 =

2pe 0 mg l

dq 3α = dt 2

A particle A having a charge q is fixed on a vertical (insulated) wall. A second particle B of mass m, charge Q is suspended by a silk thread of length l from a point P on the wall that is at a distance l above the A. Calculate the angle made by the thread with the vertical, when B stays in equilibrium.

2

T

But

θ

(π

π – θ 2 2

B ) –θ θ mg

T

F ≡

π +θ F 2 2 B π –θ π + θ 2 2

mg

9/20/2019 10:08:14 AM

Chapter 1: Electrostatics 1.11

Now, we can make the use of Lami’s Theorem and get mg F = ⎛ p q ⎞ sin ( p - q ) sin ⎜ + ⎟ ⎝ 2 2⎠ ⇒

mg F = …(2) ⎛ q ⎞ sin q cos ⎜ ⎟ ⎝ 2⎠

where F =

Qq 4pe 0 ( AB )

⇒

After drawing the diagram and applying Coulomb’s Law, we get

Also,

F = FAB = FBC = FAC =

3q2 4pe 0 a 2

P

l AB =  ⎛q⎞ ⎛q⎞ ⎛q⎞ cos ⎜ ⎟ 2 sin ⎜ ⎟ cos ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠

p q⎫ ⎧ ⎨∵ α = - ⎬ 2 2⎭ ⎩

θ

⎛q⎞ ⇒ AB = 2l sin ⎜ ⎟ …(3) ⎝ 2⎠

F 30° A Fnet

mg Qq ⇒ = q ⎞ ⎛q⎞ ⎛ cos ⎜ ⎟ 4pe 0 ⎜ 4l 2 sin 2 ⎛⎜ ⎞⎟ ⎟ sin q ⎝ 2⎠ ⎝ 2⎠⎠ ⎝ Qq ⎛q⎞ 32pe 0 l sin ⎜ ⎟ ⎝ 2⎠



4pe 0 a 2

Let us now draw the free body diagram for A. The following forces are acting on A (see figure).

l AB = sin α sin q

⇒ mg =

q2

Fnet = F 2 + F 2 + 2 F 2 cos 60

⇒ Fnet = 3 F =

2

In triangle PAB

Solution

2

3

30° 60°

⎛q⎞ ⇒ sin 3 ⎜ ⎟ = ⎝ 2 ⎠ 32pe mgl 2 0 1

Qq ⎞3 ⎛q⎞ ⎛ ⇒ sin ⎜ ⎟ = ⎜ ⎝ 2 ⎠ ⎝ 32pe mgl 2 ⎟⎠ 0

1 ⎡ Qq ⎞3 -1 ⎢ ⎛ ⇒ q = 2 sin ⎢ ⎜⎝ 32pe mgl 2 ⎟⎠ 0 ⎣

90°

G

a = 3 cm

60° 3 cm B

T

θ

3F

Qq

C

N

Tcos θ



⎧ ⎛q⎞ ⎛q⎞⎫ ⎨∵ sin q = 2 sin ⎜⎝ ⎟⎠ cos ⎜⎝ ⎟⎠ ⎬ 2 2 ⎭ ⎩

A

Tsin θ

mg

(a) Tension T in the thread AP acting upwards (as shown) q2 (b) Force 3F, where F = 4pe 0 a 2 (c) Weight mg (vertically down) For A to be in equilibrium, we get

⎤ ⎥ ⎥ ⎦

Illustration 9

Three particles, each of mass 1 g and carrying a charge q are suspended from a common point by three insulated massless strings, each 100 cm long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side length 3 cm, calculate the charge q on each particle. (Take g = 10 ms -2 ).

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 11

F

l=1m

T sin q = 3 F

T cos q = mg

⇒ tan q =

3F …(1) mg

But since, we observe from the data provided in the problem, that AP  AB . So q must be very small and hence AG tan q  q = AP

9/20/2019 10:08:27 AM

1.12 JEE Advanced Physics: Electrostatics and Current Electricity

So, now we need to find AG (where G denotes the centroid of the equilateral triangle ABC). In ΔAGN,

cos 30 =

AN AG

⇒

a⎞ ⎟⎠ a ⇒ AG = 2 = ⎛ 3⎞ 3 ⎜⎝ ⎟⎠ 2 So, from equation (1), we get ⎛ ⎜⎝



a 3 = l

3

q2 =

4pe 0 mga 3 3l

3 ⎡ 1 ⎛ 3 ⎞ ⎤ × 10 × ⎜ ⎢ ⎟ ⎝ 100 ⎠ ⎥⎦ 1 ⎣ 1000 q2 = × 3 (1) 9 × 109

a⎫ ⎧ ⎨∵ AN = ⎬ 2⎭ ⎩

q2 4pe 0 a mg

⇒

2

1 ⎧ ⎫ ⎪⎪∵ m = 1000 kg ⎪⎪ ⎨ ⎬ ⎪ a= 3 m ⎪ ⎪⎩ ⎪⎭ 100 ⇒

q2 = 10 -17 = 10 × 10 -18

⇒

q = 3.17 × 10 -9 C

Test Your Concepts-I

Based on Coulomb’s law 1. Two similar point charges q1 and q2 are placed at a distance r apart in air. A dielectric slab of thickness t (  r ) having dielectric constant K is placed between the charges. Calculate the Coulomb force of repulsion between the charges. Now assume that a slab of thickness half the separation between the charges is placed between the charges and the Coulomb’s repulsive force is reduced in the ratio 9 : 4 . Calculate K for such a slab. 2. A copper atom consists of copper nucleus that has 29 electrons surrounding it. The atomic weight of copper is 63.5 gmol-1. Now take two pieces of copper each weighing 10 g and transfer an electron from one piece to other for every 1000 atoms in that piece. Calculate the Coulomb force between the two pieces after the transfer of electrons, if the pieces are 1 cm apart. Given NA = 6.0 × 1023 mol-1, e = 1.6 × 10 -19 C. 3. A wire of length L is placed along x-axis with one end at the origin. The linear charge density of the wire varies with distance x from the origin as ⎛ x2 ⎞ λ = λ0 ⎜ ⎟ , where λ0 is a positive constant. ⎝ L ⎠ Determine the total charge Q on the rod. 4. Two identical balls each having a density ρ are suspended from a common point by two insulating

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 12

(Solutions on page H.3) strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle q with vertical. Now, both the balls are immersed in a liquid. As a result of immersion in the liquid the angle q does not change. The density of the liquid is σ . Find the dielectric constant of the liquid. 5. Calculate the ratio of the electrostatic force to the gravitation force between two electrons, between two protons. At what value of the specific charge (ratio of charge to mass) of a particle would these forces become equal (in their absolute values) in the case of interaction of identical particles? 6. A positive point charge 50 mC is located in the  plane xy at the point with radius vector r0 = 2iˆ + 3 ˆj , where iˆ and ˆj are the unit vectors of the x and  y axes. Find the electrostatic force F and its magnitude on a charge of 2 mC placed at the point    with radius vector r = 8iˆ - 5 ˆj . Here r0 and r are expressed in metre. 7. A charge q = 1 mC is placed at a point P ( 1, 2, - 4 )m . Find the electric force on a -17 mC charge at point Q ( 4, 6, - 16 )m to the nearest approximation. 8. A ring of radius 0.1 m is made out of a thin metallic wire of area of cross-section 10-6 m2. The ring has

9/20/2019 10:08:38 AM

Chapter 1: Electrostatics 1.13

a uniform charge of p coulomb. Find the change in the radius of the ring when a charge of 10-8 coulomb is placed at the centre of the ring. Young’s modulus of the metal is 2 × 1011 Nm-2 . 9. Three identical small balls, each of mass 0.1 g, are suspended at one point on silk threads having a length of  = 20 cm. What charge should be imparted to each ball so that every thread makes an angle of α = 30° with the vertical? 10. A small point mass m has a charge q, which is constrained to move inside a narrow frictionless cylinder. At the base of the cylinder is a point mass of charge Q having the same sign as q. Show that if the mass m is displaced by a small amount from its equilibrium position and released, it will exhibit simple harmonic motion with angular frequency 2g where  is the equilibrium position of ω=  charge q. 11. A small bead of mass m, charge - q is constrained to move along a frictionless wire. A positive charge Q lies at a distance L from the wire. Initially the bead constrained to move along the wire is just above + Q. Now if the bead is displaced a distance x, where x  L , and released, it will exhibit simple harmonic motion. Obtain an expression for the time period of simple harmonic motion of the bead. 12. Two identical metallic blocks resting on a frictionless horizontal surface are connected by a light metallic spring having a spring constant k and an unstretched length L0. A total charge Q is slowly placed on the system, causing the spring to stretch to an equilibrium length L, as shown. Determine the value of Q, assuming that all the charge resides on the blocks and assuming the blocks as point charges. L0 k

m

L k

R

R

m

m R

18. Two particles each of charge 10 -7 C and mass 5 g, stay in limiting equilibrium on a horizontal surface. The particles have a separation of 10 cm between them. Assume the coefficient of friction between each particle and the table to be m. Calculate m. 19. A charge Q is divided into two parts. Calculate the ratio of the charges on the divided parts such that force between the divided parts is the maximum.

m

INITIALLY

m

13. Two identical oppositely charged metallic spheres placed 0.5 m apart, attract each other with a force of 0.108 N. When connected to each other by a copper wire for a short while, they begin to repel each other with a force of 0.036 N. Find the initial charge on each one of them. 14. A charge of 50 mC has been distributed between two spheres, such that they repel each other with a force of 1 N with their centres 2 m apart. Find the distribution. 15. In an H atom, an electron revolves round a proton in a circular orbit of 0.53 Å. Calculate the radial acceleration, the angular velocity and the period of revolution of the electron. 16. Show that the gravitational force is negligible in comparison to the electrostatic force in an Hydrogen atom in which an electron is revolving around the proton in a circle of radius 0.53 Å. 17. Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, non-conducting walls, the beads move, and at equilibrium they are a distance R apart (shown in figure). Determine the charge on each bead.

m

FINALLY

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 13

9/20/2019 10:08:41 AM

1.14  JEE Advanced Physics: Electrostatics and Current Electricity

PRINCIPLE OF SUPERPOSITION This principle is used to calculate the force on a charge due to an assembly of charges placed near it. According to this principle, “the net force on a charge due to the others in the assembly is calculated by taking the vector sum of all the forces acting on that charge, one at a time”. Figure below shows an arrangement of four interacting charged particles. +

q4 –

q2 –

q3

F12 F14

+ q1

F13

So, net force on q1 , according to the principle is     F1 = F12 + F13 + F14 In general, the net force on q1 due to assembly of N charges is      F1 = F12 + F13 + F14 + ......... + F1N Here, we must note one thing that force between any pair of charges is simply not affected by the presence of remaining others. Also, this principle will also be used to find the ­electric field due to an assembly of charges at a point.

Problem Solving Technique(s) Suppose we wish to calculate the net force on q1 due to q2, q3 and q4 (in the arrangement already shown), then we follow the following series of steps. Step-1: Find out whether the force due to a given charge on q1 is attractive or repulsive. Draw the force vector with its tail at q1, either towards q1 (in case of repulsion) or away from q1 (in case of attraction). Step-2: Find out the magnitude of each force, using Coulomb’s Law. Do not take into account the sign of the charges (the signs account for the direction of force on q1, already that has been taken into account in Step 1). For this step to be completed, you will also require the separation between q1 and others i.e., r12, r13, r14 etc., then

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 14

    F12 =

q1q2 2 4pe 0 r12

, F13 =

q1q3

, F14 =

2 4pe 0 r13

q1q4 2 4pe 0 r14

Step-3:  Now select a convenient coordinate axis, so as to calculate the components of these forces (say x and y components). Then in 2D space (say xy plane), F = ( F12 ) x + ( F13 ) x + ( F14 ) x       1x and  F1y = ( F12 ) y + ( F13 ) y + ( F14 ) y Here, please note that all vectors must be expressed in terms of unit vectors iˆ, ˆj or kˆ notations, unless otherwise indicated. Step-4: As a final step, if we wish to calculate the magnitude of the net force on q1, then

(

( )

F1 = ( F1x ) + F1y        where  F1 = ( F1x ) iˆ + ( F1y ) ˆj 2

)

2

y

F1

F1y β

F1x

x

This vector can be diagramatically expressed as  So, if we say that, let F1 makes an angle b with x-axis, then the direction is given by F1y tan b = F1x    Step-5: If need be, then we can extend the same to 3D space, where we modify the equations as



 





 













F1x = ( F12 ) x + ( F13 ) x + ( F14 ) x F1y = ( F12 ) y + ( F13 ) y + ( F14 ) y

F = ( F12 )z + ( F13 )z + ( F14 )z   1z   and then F1 = F12x + F12y + F12z     where  F1 = ( F1x ) iˆ + ( F1y ) ˆj + ( F1z ) kˆ     To have a basic understanding of all the steps, see the following Illustrative Examples.

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Chapter 1: Electrostatics 1.15 Illustration 10

Figure below shows three charges q1, q2 and q3 placed at A, B and C respectively. If q1 = -1 m C , q2 = +3 m C , q3 = -2 m C , AB = 15 cm and AC = 10 cm. Calculate the net force that acts on q1 placed at A. C q3

⇒ F1x = ( F12 )x + ( F13 )x = F12 + F13 sin ( 30° ) and F1y = ( F12 )y + ( F13 )y = 0 - F13 cos ( 30° ) ⇒ F1x = 1.2 + 1.8 sin ( 30° ) ⇒ F1x = 2.1 N

30°

and F1y = -1.8

q1

q2

A

B

(Negative sign shows that F1y acts in the downward direction)  ⇒ F1 = 2.1iˆ - 1.6 ˆj N

(

iˆ

Solution

Here we shall use the concept of Principle of Superposition along with the problem solving strategy we have learnt already. Before we start, let us consider the origin to be at A and the reconstruction of the diagram is shown. So, we have drawn/shown all the forces that are acting on the charge q1. F12 =

q1 q2 2 4pe 0 r12

=

( 9 × 109 )( 10 -6 ) ( 3 × 10 -6 ) = 1.2 N ( 15 × 10 -2 )2

⇒   F12 = 1.2 N (attractive in nature) y

A –

q1

F12

30°

q2 + B

x

F13

Similarly ⇒

F13 =

1 q1 q3 2 4pe 0 r13

( 9 × 109 )( 10 -6 ) ( 2 × 10 -6 ) F13 = = 1.8 N ( 10 × 10 -2 )2

Illustration 11

Four identical free charges each of value q are located at the corners of a square of side a. What must be the charge Q that has to be placed at the centre of the system so that the system stays in equilibrium? Solution

Let us here consider the equilibrium of any charge q at 1 (say). For equilibrium of charge at 1, we have   F1 = 0

⇒

∑ ∑F ∑F

1x

= 0 …(1)

1y

= 0 …(2)

Out of these two equations any one can be selected as both of these will give identical results. So, lets take

∑F

1x

=0 F14

q 2

⇒ F13 = 1.8 N (repulsive in nature)

a

Since both the forces have different directions, so we shall apply principle of superposition in component form along x and y -axis.

3

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 15

)

 2 2 and F1 = ( 2.1 ) + ( 1.6 ) = 4.41 + 2.56 = 2.64 N

⇒

30°

q3 C –

3 = -1.6 N 2

F15 F13

a q

Q

F12

y x

a

5

q

a

q

4

9/20/2019 10:08:59 AM

1.16  JEE Advanced Physics: Electrostatics and Current Electricity

Now, F1x =

∑F

1x

= ( F12 )x + ( F13 )x + ( F14 )x + ( F15 )x

⇒ F1x = F12 + F13 cos ( 45° ) + 0 + F15 cos ( 45° ) ⇒ F1x =

⇒ F1x =

q2 Qq 1 ⎛ q2 1 1 ⎞ + + ⎜ ⎟ 2 2 2 4pe 0 ⎜ a ( 2a ) 2 ⎛ a ⎞ 2 ⎟ ⎜⎝ ⎟ ⎜⎝ ⎟⎠ 2⎠



q1

( ∑F

1y

=0

⇒

)

CASE-1: q1 and q2 similar in nature Here, the point C, where the net force due to q1 and q2 on q will be zero must lie between the charges q1 and q2 on the line joining them, towards the charge of smaller magnitude. If AC = r1, CB = r2 and AB = l (given)   then FCA = FCB

A

q1 q 4pe 0 r12

=

FCB

q C

l–x FCA

q2 B

q2 q 4pe 0 r22 2

q1 ⎛ r1 ⎞ ⎛ x ⎞ =⎜ ⎟ =⎜ ⎝ l - x ⎟⎠ q2 ⎝ r2 ⎠

2

CASE-2: q1 and q2 opposite in nature and q2 < q1 Here the point C, where the net force due to q1 and q2 on q will be zero must lie on the line joining the

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 16

4pe 0 r12

=

⇒

C FCA x

FCB

q1 q 4pe 0 r22

q1 ⎛ r1 ⎞ ⎛ l+x⎞ =⎜ ⎟ =⎜ ⎝ x ⎟⎠ q2 ⎝ r2 ⎠

2

Conceptual Note(s)

Equilibrium of q when q1 and q2 are fixed at a separation l from each other.

x

q1 q

B

2

EQUILIBRIUM OF THREE CHARGES

q1

q2

l

A

1 ⎞ ⎛ q⎜ 1+ ⎟ + 2Q = 0 ⎝ 2 2⎠

taken F1y = 0 or

⇒

  FCA = FCB

⎛ 2 ⎞ q2 + + 2Qq ⎟ q ⎜ 2 ⎝ ⎠ 2 2 4pe 0 a

q ⇒ Q = - (1 + 2 2 ) 4 We shall get the same result even if we would have

⇒

If again AC = r1 , BC = r2 and AB = l (given), then

1

So, for F1x = 0 , we get

two charges q1 and q2 but outside them towards the charge of smaller magnitude i.e. q2.

Electrostatic Equilibrium The point where the resultant force on a charged ­particle becomes zero is called equilibrium position. (a) Stable Equilibrium: A charge is initially in equilibrium position and is displaced by a small distance. If the charge tries to return back to the same equilibrium position then this equilibrium is called position of stable equilibrium. (b) Unstable Equilibrium: If charge is displaced by a small distance from its equilibrium position and the charge has no tendency to return to the same equilibrium position. Instead it goes away from the equilibrium position. (c) Neutral Equilibrium: If charge is displaced by a small distance and it is still in equilibrium condition then it is called neutral equilibrium. However if, q1 and q2 are not fixed, then equilibrium of q will exactly be the same as done before, but if we have to keep q1 and q2 also in equilibrium, then we have to make the net force on q1 (at A ) due to q2 (at B ) and q (at C) to be zero and then repeat the same for q2 also to get the desired results. See Illustration(s) below to learn how to perform in these kind of problems.

9/20/2019 10:09:17 AM

Chapter 1: Electrostatics 1.17 Illustration 12

⇒ 9q + 4Q = 0

Two free charges +Q and +4Q are placed at a separation L. Find the magnitude, sign and the location of the third charge that makes the system to stay in equilibrium.

4Q 9 • Please note that we could also have taken ⇒ q=-

∑F

B

Solution

Since both the charges have similar nature, so irrespective of a nature of third charge, the location of the third charge must be somewhere between +Q and +4Q on the line joining them. Let the net force be zero at P at a distance x from +Q. x

Q

(L – x)

P

A

⇒ ⇒ ⇒

4Q B

  FPA = FPB Qq 4pe 0 x 2

4pe 0 ( L - x )



y

1 r1

⇒ x = -L But this coordinate lies outside A and B so, this value of x is REJECTED.

L 2L from 4Q. from Q or 3 3 To find the charge q , such that the system is in equilibrium, we proceed further by taking either ⇒ q must be placed at

∑

FA = 0  OR  So let us take

∑F

B

2

+

( 4Q ) Q 2

4pe 0 L

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 17

=0

–q3 3

r23 = r2 – r3 2 q2

r3

O

x

For equilibrium of -q3 , we have    F31 + F32 = 0  q1 q3 q2 q3     ⇒   3 ( r1 - r3 ) +   3 ( r2 - r3 ) = 0 4pe 0 r1 - r3 4pe 0 r2 - r3 …(1) Also, from figure, we see that rˆ13 = - rˆ23   r -r ⇒ 1 3 = r1 - r3

∑

Qq

=0

q1 r13 = r1 – r3

r2



FA = 0 ⇒ FAP + FAB = 0 ⎛ L⎞ 4pe 0 ⎜ ⎟ ⎝ 3⎠

Two positive charges q1 and q2 are located at points   1 and 2 with radius vectors r1 and r2 . Find a nega tive charge q3 and the radius vector r3 of a point 3 at which it has to be placed for the force acting on each of the three charges to be equal to zero.

L-x = -2 x ⇒ L - x = -2x



⇒

Illustration 13

Before we start the problem, we must keep in mind that the radius vector of any point P is actually the position vector ( PV ) of the point P which is the vector drawn from the origin to the point P. So, let the origin be at O (see figure)

2

L-x = ±2 x

L-x =2 x ⇒ L = 3x L ⇒ x= 3

4Q , then still it will 9 be in equilibrium, but then the system would not be in equilibrium.

• Also, note that even if q = +

Solution

( 4Q ) q

=

= 0 , to get the same result.

  r2 - r3   …(2) r2 - r3

Put (2) in (1), we get

q1   r1 - r3

2

q =  2 r2 - r3

3

9/20/2019 10:09:36 AM

1.18  JEE Advanced Physics: Electrostatics and Current Electricity

    ⇒ ± q1 ( r2 - r3 ) = q2 ( r1 - r3 ) Taking positive sign, we get    q1 r2 - q2 r1 = q1 - q2 r3   q1 r2 - q2 r1  ⇒ r3 = q1 - q2

(

)

This must be REJECTED as it will lie outside q1 and q2. Taking negative sign, we get    q1 r2 + q2 r1 = q1 + q2 r3   q1 r2 + q2 r1  ⇒ r3 = q1 + q2

(

)

Clearly, this position vector or radius vector will lie on the line joining q1 and q2, between them.   q1 r2 + q2 r1  So, r3 = q1 + q2 For q1 to be in equilibrium, let charge placed on 3 be q3 , then    F12 + F13 = 0  q1 q2 q1q3     ⇒   3 ( r1 - r2 ) +   3 ( r1 - r3 ) = 0 4pe 0 r1 - r2 4pe 0 r1 - r3 While writing the above expression, we have not taken any nature of q3 . The answer will have its sign attached to q3 . So, as of now q3 is assumed to be positive and then lets see what happens. Also, as per the assumption made rˆ12 = rˆ13     q2 q3 r1 - r3 ⎫ r1 - r2 ⎧ ⇒   2 = -   2  ⎨∵   =   ⎬ r1 - r2 r1 - r3 ⎭ r1 - r2 r1 - r3 ⎩   2 r1 - r3 ⇒ q3 = - q2   …(3) r1 - r2      ⎛ q1 r2 + q2 r1 ⎞ Now, r1 - r3 = r1 - ⎜ ⎟ q1 + q2 ⎠ ⎝     q1 r1 + q2 r1 - q1 r2 - q2 r1   ⇒ r1 - r3 = q1 + q2   q1 ( r1 - r2 )   ⇒ r1 - r3 = q1 + q2

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 18

⇒

  r1 - r3   r1 - r2

2

=

(

q1 q1 + q2

)

2

Using (3), we get ⇒ q3 = -

q1 q2

(

)

q1 + q2

2

Illustration 14

Two identical positive point charges, each having a charge Q are fixed at a separation 2a. A point charge q lies midway between the fixed charges. Show that for small (a) displacement (compared to a ) along the line joining the fixed charges, the charge q executes SHM, if it is positive in nature. (b) lateral displacement, the charge q executes SHM, if it is negative in nature.  Compare the periods of oscillations in the above two cases. Solution

(a) When q is displaced along the line joining the fixed charges. Let the displacement of q be x (  a ). Q

q

a

A

a

x

O

O′

Q B

The restoring force F acting on q is F=

Qq ⎡ 1 1 ⎤ 4pe 0 ⎢⎣ ( a - x )2 ( a + x )2 ⎥⎦

(Negative sign in the above expression indicates that F is directed towards the mean position) F=-

Qq 4 ax 4pe 0 ( a 2 - x 2 )2

Since x  a ∴  F = -

Qq 4 x 4pe 0 a 3

 ⎛ Qq ⇒   mx + ⎜ ⎝ pe 0 a 3

{∵ for x  a, ( a

2

- x2 ) ≅ a4 2

}

⎞ ⎟⎠ x = 0

9/20/2019 10:09:52 AM

Chapter 1: Electrostatics





⇒

⎛ Qq x + ⎜ ⎝ p me 0 a 3

⎞ ⎟⎠ x = 0

2  x + ω1 x



=0

⇒

ω1 =

⇒

pe 0 ma 3 T1 = 2p Qq

pe 0 ma 3 …(1)

(b) When q is replaced by -q and is given a lateral displacement OO ′ = x (  a ). Then the forces acting on -q are shown in figure. Qq

and r = a 2 + x 2 4pe 0 r 2 The charge -q is attracted towards Q with a force F (given above) on resolution, the component F cos q cancels, so that the net force 2F sin q restores -q back to mean position O. F cos θ r Q

⇒ ⇒

F

–q θ

O′ θ

F cos θ F

x 2F sin θ O a

a

r θ

Q

a + x2

Fnet = -2

Qqx 4pe 0 ( a 2 + x 2 )

3/2

Qq

4pe 0 ( a 2 + x 2 )

⇒

( a2 + x 2 )3/2 ≅ a3

⇒

⎛ Qq Fnet = -2 ⎜ ⎝ 4pe 0 a 3

⇒

⎛ Qq mx + ⎜ ⎝ 2pe 0 a 3

⎞ ⎟⎠ x = 0

⇒

Qq ⎛ x + ⎜ 3 2 pe ⎝ 0 ma

⎞ ⎟⎠ x = 0

⎞ ⎟⎠ x

Again compare with standard equation of SHM, we get

x + ω 22 x = 0

⇒

ω2 =

⇒

T2 = 2p

Qq 2pe 0 ma 3 2pe 0 ma 3 Qq

…(2)

Finally, from (1) and (2), we get

Fnet = -2 F sin q mx = -2

⇒

2

Since x  a

Qq

where F =

x

But sin q =

Compare with the standard equation of SHM, that is

1.19

sin q



T1 1 = T2 2

Test Your Concepts-II

Based on Principle of superposition (Solutions on page H.7) 1. Eight charges each of magnitude Q are located at the corners of a cube of side L (see figure). One corner is at the origin and the edges lie along the rectangular axes. Calculate the net electrostatic force  on the charge at r = Liˆ + Ljˆ + Lkˆ .

y

L L O

L x

z

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 19

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1.20  JEE Advanced Physics: Electrostatics and Current Electricity

2. Four equal positive charges, each of charge Q are arranged at the corners of a square of side . A unit positive charge of mass m is placed at point P at height h above the centre of the square. Calculate Q, so that the unit charge is in equilibrium. 3. Two equal positive point charges Q are separated by a distance 2a. A point test charge q0 is located in a plane normal to the line joining the two charges and midway between them. Find the radius r of a circle of symmetry in this plane for which the force on the test charge has a maximum value. Find the direction of the force on this charge if it is assumed to be positive. Also find Fmax . 4. Three identical spheres each having a charge 2q and radius R are kept such that each touches the other two. Find the magnitude of the electric force on any sphere due to the other two. 5. Five point charges each of value +q are placed on five vertices of a regular hexagon of side a. Calculate the magnitude of the force on a point charge -q placed at the centre of the hexagon. 6. (a) Twelve equal charges, q, are situated at the corners of a regular 12 sided polygon (for instance, one on each numeral of a clock face). What is the net force on a test charge Q at the center? (b) Suppose one of the 12 q’s is removed (say the one at “6 o’clock”). What is the force on Q? Explain your reasoning carefully.



(Assume the distance between q and Q pair(s) to be a) 7. Four identical charges, each having a charge +q are fixed at the corners of a square of side L. A fifth point charge -Q lies a distance x along the line perpendicular to the plane of the square. Find the force exerted by the charges on -Q. Show that for z  a the motion of -Q is simple harmonic. Also calculate the time period of oscillations of -Q if it has a mass m? (Neglect gravitational force.) 8. Consider a regular polygon with 24 sides. The distance from the center to each vertex is a. Twenty three identical charges each having a charge q are placed at the vertices of the polygon. A single charge Q is placed at the centre of the polygon experiences a force F. What is the magnitude and direction of the force experienced by the charge Q? 9. Three charges, each of magnitude 100 mC are located in vacuum at the corners A, B and C of an equilateral triangle of side length 3 m. If charges placed at A and C are positive and the one placed at B is negative, then find the magnitude and direction of force F experienced by the charge placed at C.

ELECTROSTATIC FIELD ( E )

tive charge. SI unit of electric field is newton/

The region of space around a source charge ( q ) in which it can exert a force on a test charge ( q0 ). Mathematically, electric field is the force experience per unit test charge q0 placed in the electrostatic influence of source charge q.   F E= q0 Electric field strength is a vector quantity directed away from a positive charge and towards the nega-

The dimensional formula for E is MLT -3 A -1 i.e., [ E ] = MLT -3 A -1

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 20

coulomb ( NC -1 ) or volt metre ( Vm -1 ).

 Since F =

1 qq0 rˆ 4pe 0 r 2   F q ⇒ E= = rˆ q0 4pe 0 r 2

is the electric field due to a source point charge q at a distance r from it.

9/20/2019 10:10:13 AM

Chapter 1: Electrostatics 1.21

Problem Solving Technique(s)



(a) For calculating the electric field due to a charge at a point P at distance r we proceed as follows. Step-1: Place a charge of +1 C at point P. Step-2:  Calculate the force between q and +1 C. Step-3:  This force is the value of E due to q at the point P. Step-4: The direction of force experienced by a charge of +1 C is the direction of E due to q at the point P. (b) If we are to find the electrostatic field due to assembly of charges q1, q2, …., qn at a point P, then we apply Superposition Principle and then     E p = E 1 + E 2 + ..... + E n    where we find E 1 , E 2 , …. E n at point P by the technique mentioned above. (c) If we are to find the electric field at a point P due to a uniform charge distribution, then we calculate the field due to an infinitesimal element of the distribution at the point P and then integrate it within appropriate limits.

r

P

(c) Surface charge, σ



∫



∫r

dq 2

For charge distributed on a wire with linear charge density λ , E=



(d) Volume charge, ρ

1 4pe 0

2

 For charge distributed on a surface with surface charge density σ ,



1 E= 4pe 0

Calculate the electric field due to a uniformly charged rod of length L at a point P at distance r from one end of the rod. Assume total charge on the rod to be Q. Q

∫ A

r

σ dA r2

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 21

P

L Axis of rod

Solution

For this we consider an infinitesimal element of length dx on rod as shown in figure. Charge on the infinitesimal elemental length dx is dq =

Q dx L x

dx

r

L

P dE

This dq (an infinitesimal element) can be regarded as a point charge, hence the electric field dE due to this element at point P is dE given by dE = ⇒ dE =

λd

∫r

ρ dV r2

Illustration 15

P

dV

1 E = dE = 4pe 0

∫

dl (b) Line charge, λ

(a) Continuous distribution dA

1 4pe 0

V If the charge distributions are uniform, then charge densities can be taken out of the integral to get desired results.



r

dq

r

E=

P

P

r

For charge distributed on a volume with volume charge density ρ ,

dq 4pe 0 x 2 ⎛ Q⎞ ⎜⎝ ⎟⎠ dx L 4pe 0 x 2

The net electric field strength at point P can be given by integrating this expression over the whole length of rod as r+L

EP =

Qdx

∫ dE = ∫ 4pe Lx r

0

2

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1.22  JEE Advanced Physics: Electrostatics and Current Electricity

Q ⇒ EP = 4pe 0 L

of length dx at a distance x from the point P. If dE be the electric field due to the rod at P, then

r+L

∫x

-2

dx

r

⎛ λ =⎜ ⎝ 4pe 0 4pe 0 x dq

dE =

r+L

2

⎞ -2 ⎟⎠ x dx

Q ⎡ 1⎤ 4pe 0 L ⎢⎣ x ⎥⎦ r



⇒ EP =

1 ⎤ Q ⎡1 ⎢ 4pe 0 L ⎣ r r + L ⎥⎦

⇒ E=

⎞ ⎛ d+ 2a λ ( 2a ) λ ⎜ x -2 dx ⎟ = pe 4pe 0 ⎜ 4 ⎟⎠ 0 ( d + 2a ) ⎝ d

⇒ EP =

Q 4pe 0 r ( r + L )

⇒ E=

Q 4pe 0 d ( d + 2 a )

⇒ EP =

SPECIAL CASE: This special case can be taken as a check point too. For x  L i.e., for distances far away from the rod, the rod with charge Q must be have as a point charge. In this expression, when we put r  L, Q then E ≅ ( = field due to a point charge Q at 4pe 0 r 2 a distance r from it). Illustration 16

Two identical thin rods of length 2a carry equal charges +Q that is distributed uniformly along their lengths. The rods lie along the x-axis with their centers separated by a distance b > 2 a . Show that the magnitude of the force exerted by the left rod on the right one is given by ⎛ ⎞ b2 F= log ⎜ e 2 2 ⎟ 2 ⎝ 16pe 0 a b - 4a ⎠

PART II Now after we have calculated the field due to one rod, we actually observe that the second rod will be lying in the field of the first rod. For finding the force on this rod let us again consider an infinitesimal element of length dx at a distance x from the nearest end of the left rod. If dF be the force due to this rod then

dF = Edq =

b x –a

b a

b–a

b+a

x

Solution

This problem is a very special one where we shall learn to calculate the force between two extended bodies using the concept of electric field. For this let us divide the problem in two parts. PART I Here we shall imagine the rod to the right to be absent and then let us calculate the field due to the rod on the left at a hypothetical point P (say) at a distance d from one end of the rod. Let us consider an element

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 22

a

dx b–a

b+a

x

Q 2a

⎛ Q ⎞⎛ Q⎞⎡ dx ⎤ ⇒ dF = ⎜ ⎜⎝ ⎟⎠ ⎢ ( ⎟ ) ⎝ 4pe 0 ⎠ 2 a ⎣ x x + 2 a ⎥⎦

y

–a

Q ( λ dx ) 4pe 0 x ( x + 2 a ) y

Since λ =

Q2



∫

⇒ F=

⇒ F=

∫

Q2 dF = 8pe 0 a

b

∫

b -2a

dx

x ( x + 2a )

Q2 ⎡ 1 ⎛ 2a + x ⎞ ⎤ - log e ⎜ ⎢ ⎝ x ⎟⎠ ⎥⎦ 8pe 0 a ⎣ 2 a

b b -2a

2

⇒ F=

⎡ ⎛ 2a + b ⎞ ⎛ b ⎞⎤ ⎢ - log e ⎜⎝ b ⎟⎠ + log e ⎜⎝ b - 2 a ⎟⎠ ⎥ 16pe 0 a ⎣ ⎦

⇒ F=

⎤ ⎡ b2 log ⎥ ⎢ e 2 16pe 0 a ⎣ ( b - 2a ) ( b + 2a ) ⎦

⇒ F=

⎛ ⎞ b2 log ⎜ e 2 2 2 ⎝ b - 4 a ⎟⎠ 16pe 0 a

Q

2

Q2 Q2

9/20/2019 10:10:33 AM

Chapter 1: Electrostatics 1.23 Illustration 17

Properties of Field Lines

Calculate the electric field intensity which would be just sufficient to balance the weight of a particle of charge -10 m C and mass 10 mg . (Take g = 10 ms -2 )

(a) Field lines always come (emanate) out of positive charge and enter the negative charge. (b) The number of lines per unit area through a surface held normally to the line is observed to be proportional to the magnitude of the electric field existing in a given region. (c) Field lines never cross each other; otherwise the field would be pointing in two different direction at the same point. (d) Field lines never form closed loops. (e) Field lines are always directed from higher potential to lower potential. (f) Field lines never exist inside a conductor. (g) If N1 is the number of field lines coming out of a charge q1 and N 2 is the number of field lines entering q2 , then

Solution

Force on a charge q in an electric field  E is   F = qE According to given problem, we have    F = W ⇒ q E = mg ⇒ E=

F = qE

–q, m

E

W = mg

mg = 10 NC -1, in downward direction. q

ELECTROSTATIC LINES OF FORCE: PROPERTIES A line of force is an imaginary path straight or curved such that the tangent to it at any point gives the direction of electrostatic field at that point. A field line is an imaginary line along which a unit positive charge would move when set free. The lines of force are drawn such that the number of lines per unit area of cross-section, (area held normally  to the field lines) is proportional to magnitude of E .

q1

 

q2

=

N1 N2

+2q

–q

Pattern of Field Lines The pattern of electric field lines can be obtained by considering the following points. (a) Symmetry: For every point above the line joining the two charges there is an equivalent point below it. Therefore, the pattern must be symmetrical about the line joining the two charges. (b) Near Field: Very close to a charge, the field due to that charge predominates. Therefore, the lines are radial and spherically symmetric. (c) Far Field: Far from the system of charges, the pattern should look like that of a single point charge of value Q = ΣQi . Thus, the lines should be radially outward, unless Q = 0 .   (d) Null Point: This is a point at which E = 0 , and no field lines should pass through it.

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 23

The electric field lines for a point charge +2q and a second point charge -q. Note that two lines leave the charge +2q for every one that terminates on -q.

(h) If N1 is the number of field lines coming out of a charge q1 and N 2 is the number of field lines coming out of charge q2 , then q1 N1 = q2 N 2

  This relation also exists if field lines are entering both the charges (i) Tangent to field line at a point gives the direction of field at that point.

9/20/2019 10:10:42 AM

1.24  JEE Advanced Physics: Electrostatics and Current Electricity

(j) Field lines exhibit longitudinal (length wise) contraction, thus indicating that unlike charges attract each other. (See Figure 1)

dq

Q

a

a2 + x 2

dE sin θ dE

θ θ

x

y

θ dE cos θ θ dE cos θ dE

dE sin θ

x

Consider two elements of length dl placed symmetrically on the two diametrically opposite ends. If dE is the field due to each such element, then dE sin q components will cancel out such that the net field is just due to dE cos q components summed up over the entire ring. So, Enet = Ex = Figure 1  T  he electric field lines for two equal and opposite point charges an electric dipole. Note that the number of lines that leave the positive charge equals the number that terminate at the negative charge.

(k) Field lines exhibit lateral (sideways) expansion, thus indicating that like charges repel each other. (See Figure 2)



∫ dE cosq = ∫ 4pe ( 0

⇒ Ex = ⇒ Ex =

λ cos q

4pe 0 ( a 2 + x 2 )

+

)

2

cos q

∫ dl 0

32

(along +x axis) ⎧ ⎨∵ cos q = ⎩



+

a2 + x 2

2p a

Qx 4pe 0 ( a 2 + x 2 )

dq

x

⎫ ⎬ a +x ⎭ 2

2

Remark(s) (a) For the point P to lie at far off distance from the centre of the ring i.e., for x >> a we have

( x 2 + a2 )3 2 ≅ x 3 .

Figure 2  T  he electric field lines for two equal positive point charge

(l) Field lines always enter or leave a surface at right angles.

ELECTRIC FIELD AT THE AXIS OF A CIRCULAR UNIFORMLY CHARGED RING Consider a ring of uniform charge density λ . If a is the radius of the ring and Q is the total charge on the ring, then

Q λ= ...(1) 2p a

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 24



⇒   Ex 

Q 4pe 0 x 2



This result under the specified condition just matches with the field due to a point charge at a distance x from it. (b) For this field to be a MAXIMUM dE =0 dx

⇒ 

d ⎡ ( 2 2 -3 2 ⎤ ⎣x x +a ) ⎦=0 dx



⇒ 

( x 2 + a2 )- 3 2 + x ⎛⎜ - 3 ⎞⎟ ( x 2 + a2 )-5 2 ( 2 x ) = 0 ⎝ 2⎠

9/20/2019 10:10:49 AM

Chapter 1: Electrostatics 1.25

3x 2

Enet =



⇒ 1=



⇒ x 2 + a2 = 3x 2



⇒ x=±



i.e., the field will attain a maximum value at a x=± along the axis of the ring and the maxi2 mum value equals Emax given by

x 2 + a2 a 2

Q Emax =



 



⇒

Emax =



⇒

Emax =

a 2

⎛ a2 ⎞ 4pe 0 ⎜ a2 + ⎟ ⎝ 2 ⎠

32

2Q 4pe 0 3 3a2 Q 6 3pe 0 a2

Problem Solving Technique(s) For symmetrical charge distributions, if we are asked to calculate the electric field at a point which lies symmetrically with respect to the charge distribution, then we proceed as follows : Step-1: Consider an infinitesimal element having a charge dq (say) that lies at a distance r from the point P. If dE is the electric field due to this element then dq dE = 4pe 0 r 2   Step-2: Now consider another identical mirror infinitesimal element having same charge and located at the same distance from P. Then field due to this element will also be dE. Step-3: On resolving dE due to both the elements along suitable axis we observe that one set of components cancels and the net electric field will then be calculated by taking the integral of the component of the electric field due to the contribution of the single element.

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 25

∫ ( contribution due to a single element )

However, in the case of unsymmetrical charge distributions, if we have to calculate the electric field at any point P, then we proceed as follows : Step-1: Consider an infinitesimal element having a charge dq (say) that lies at a distance r from the point P. If dE is the electric field due to this element then dq dE = 4pe 0 r 2 Step-2: Resolve dE along the selected axes (say x and y) so as to get the components dEx and dEy. Step-3: Integrate dEx and dEy separately to get Ex and Ey.   Step-4: Then E = E x iˆ + E y ˆj such that E = E = E 2x + E 2y  and if E makes an angle b with x-axis, then Ey tan b = . Ex

Illustration 18

Calculate the electric field due to a uniformly charged rod of length L at a point P that lies at a distance r on its perpendicular bisector. Assume charge on the rod to be Q. Discuss the result when r  L and r L  r . In the case when L  r , also draw a plot of L E Q (along x-axis) vs (on y-axis) where E0 = . E0 4pe 0 L2 Solution

For this we consider an element of length dx at a distance x from centre of rod as shown in Figure. dE dEcos θ dEsin θ

P

P θ r

r

L

x Equator of rod

dx

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1.26  JEE Advanced Physics: Electrostatics and Current Electricity

The charge on this element is Q dq = dx L

If the strength of electric field at point P due to this infinitesimal charge dq is dE , then dE =

dq 1 2 4pe 0 ( r + x 2 )

Let us again take a symmetrical element such that it is just the mirror image of this element. Then we can see that the component dE sin q will get cancelled and net electric field at point P will be due to integration of dE cos q only. Thus net electric field strength at point P can be given as

∫ dE cosq

E = EP =

1 ⇒ E= 4pe 0

+L 2

∫

-L 2

1 Qr ⇒ E= 4pe 0 L

Qdx

L ( r 2 + x2 ) +L 2

∫

-L 2

Let the integral be I =

r

×

2

r + x2

dx

(r

2

+ x2 )

∫ (r

⇒ E=

E

Q

4pe 0 r 2 On the other hand, when L  r , we have E

λ Q = 2pe 0 rL 2pe 0 r

In this infinite length limit, the system has cylindrical symmetry. In this case, an alternative approach based on Gauss’s Law will also be derived at a later stage. E Q The characteristic behaviour of (with E0 = ) E0 4pe 0 l 2 r as a function of is shown in figure. L

32

E/E0 10

+ x2 )

32

–3 –2 –1

–5

⇒ I=

1

2

3

r/L

Electric field of a non-conducting rod as a function of r/L

r sec 2 q dq 3

5

–10

⇒ dx = r sec 2 q dq

∫r

+

In the limit where r  L the above expression reduces to the “point-charge” limit and we get

x = r tan q

⇒ I=

⎞ ⎟ ⎟ 2 L + r 2 ⎟⎟ ⎠ 4

4pe 0 r L2 + 4 r 2

To evaluate this let us substitute

L2 + r2 4

L 2

2Q

dx 2

L 2

⎛ 1 Q⎜ ⇒ E = EP = 4pe 0 Lr ⎜ ⎜ ⎜⎝

Illustration 19

sec 3 q

Calculate the electric field due to a uniformly charged rod AB having charge density λ at a point P at perpendicular distance a from the rod, as shown in figure.

∫ cosq dq

⇒ I = sin q

A

Since x = r tan q ⇒ sin q =

x x2 + r 2 L + ⎞ 2

N

1 Q⎛ x So, E = EP = ⎜ ⎟ 4pe 0 Lr ⎝ x 2 + r 2 ⎠ - L 2

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 26

a

α β

P

B

9/20/2019 10:11:13 AM

Chapter 1: Electrostatics 1.27 Solution

Similarly, let us calculate the value of Ey, given by

Here it is very important to note the unsymmetrical placement of the point P and hence we must calculate the fields Ex and Ey separately. For this, let us



consider an infinitesimal element of length dx at a distance x as shown. A dx

y

r r

x

dθ

P dEcos θ

θ

a

dE sin θ

The net electric field at point P due to this infinitesimal element is dE . This dE is resolved into components. (b)  dEy = dE sin q The net field E can be calculated by finding Ex and Ey from the above expressions. So

∫

dx cos q r

2

=

λ 4pe 0

Also, we observe that x tan q = a

⇒ x = a tan q ⇒ dx = a sec 2 q dq ⇒ Ex =

λ dx

1

∫ dE cosq , where dE = 4pe λ 4pe 0

λ 4pe 0

2

∫

λ 4pe 0

⇒ Ey =

λ 4pe 0 a

a sec 2 q sin q dq

∫

a 2 sec 2 q α

∫ sin q dq

-b

(

)

⇒ Ey =

λ - cos q 4pe 0 a

⇒ Ey =

λ ( cos b - cos α ) 4pe 0 a

α -b

a sec 2 q cos q dq a 2 sec 2 q α

0

r2

λ ( sin α + sin b ) 4pe 0 a λ and Ey = ( cos b - cos α ) 4pe 0 a Ex =

dx cos q

∫ (a

2

+ x2 )

Misconception Removal While attempting this problem, we may think that dx = rdq, but that would be a wrong step. From the magnified diagram of the element shown PC = r and PN = r. So its CN that equals rdq and not CD. Since CD is extremely small, so ∠MCP = ∠CDP = 90 - q . Now, we observe the triangle NCD, then D dx C

∫ cosq dq

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 27

CN = rd θ –θ

-b

λ ⇒ Ex = ( sin b + sin α ) 4pe 0 a

90 – θ N 90

λ ⇒ Ex = 4pe 0 a

⇒ Ey =

So, the electric field due to a rod of length having uniform charge density λ at a point P , that subtends an angle α at one end and b at the other is given by

(a)  dEx = dE cos q

⇒ Ex =

2

0

Since x = a tan q

dE

B



⇒ Ey

y

⇒ dx = a sec 2 q dq

D C

x

Ex =

∫ dE = ∫ dE sin q λ dx sin q = ∫ 4pe (a + x )

Ey =

r dθ

M

a

θ

P

9/20/2019 10:11:29 AM

1.28  JEE Advanced Physics: Electrostatics and Current Electricity

P to wire normally and passing through the known end of wire) is calculated by taking.

CN r dq   sin( 90 - q ) = = CD dx r dq ⇒     dx = cosq a dq ⇒     dx = cos2 q ⇒    



α ⎯⎯ →

⇒ Ex =

p and b ⎯⎯ →0 2

λ λ and Ey = 4pe 0 a 4pe 0 a

dx = a sec 2 q dq

Otherwise we would have got dx = a sec q dq which definitely would not fetch us correct result for E.

So, we can also extend the results obtained to more look alike situations as the special cases of the above discussion. Situation 1: For an infinite rod having uniform charge density λ , electric field at point p P is calculated by taking α ⎯⎯ → 2 p and b ⎯⎯ → . 2

a

a

Situation 2: For a semi-infinite rod having uniform charge density λ , electric field at point P is calculated by taking p α ⎯⎯ → and b = b 2

λ λ cos b ( 1 + sin b ) and Ey = 4pe 0 a 4pe 0

45° Ey

P

λ ⇒ Ex = and Ey = 0 2pe 0 a

⇒ Ex =

P

a

Ex E

Illustration 20

A segment of a charged wire of length l, charge density λ 2 , and an infinitely long charged wire, charge density λ1 , lie in a plane at right angles to each other. The separation between the wires is r0 . Determine the force of interaction between the wires. Solution

Electric field near a long wire is given by E=

λ 2pe 0 r

The second wire lies in the non uniform field of first wire so that each element of second wire experiences a different magnitude of field due to the first wire. Wire 1

a

β

P

λ1

x r0

Situation 3: For a semi-infinite rod having uniform density λ , the electric field at point P (which lies on the line joining

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 28

dx Wire 2 l

Let us consider an infinitesimal element of length dx having a charge dq = λ 2 dx , at a distance x from the

9/20/2019 10:11:38 AM

Chapter 1: Electrostatics 1.29

long wire. The infinitesimal force acting on this element dF is given by ⎛ λ1 ⎞ λ 2 dx dF = Edq = ⎜ ⎝ 2pe 0 x ⎟⎠

F=

λ1 λ 2 dx λ1λ 2 l ⎞ ⎛ = log e ⎜ 1 + ⎟ pe 2 x r ⎝ 0 0 ⎠ 0

∫ 2pe r0

So,

F=

λ Qx dx

∫ 4pe ( R 0

0

The force acting on each element depends on x, the separation between wire 1 and wire 2. Integrating the expression for dF within the ­limits x = r0 to x = r0 + l , we obtain r0 + l

∞

Illustration 21

A system consists of a thin charged wire ring of radius R and a very long uniformly charged thread oriented along the axis of the ring, with one of its ends coinciding with the centre of the ring. The total charge of the ring is equal to Q. The charge of the thread (per unit length) is equal to λ . Find the interaction force between the ring and the thread.

+ x2 )

2

Illustration 22

Calculate the electric field at the centre of a uniformly charged wire in the form of an arc of radius r. The wire subtends an angle f at the centre. Solution

Consider an element dq ( = λ dx ) of the wire that makes an angle q with y-axis and subtends an angle dq at the centre. Then

dq = λ ( r dq )

Since, dE = ⇒ dE =

dq 4pe 0 r

2

=

λ ( r dq ) 4pe 0 r

λ dq 4pe 0 r y dq

4pe 0 ( R + x )  And from symmetry E at every point on the axis is directed along the x-axis Let us consider an element dx on thread which carries the charge dq = λ dx . The electric force experienced by the element in the field of ring.



O

E x

dF = ( λ dx ) E ( x ) =

4pe 0 ( R2 + x

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 29

)

r dθ

ϕ

dE sin θ

x

Again take another element which is the mirror image of the element already taken. On resolution, we observe that the x components cancel, while the net field is equal to integral of contribution due to a single element. Hence

E = Ey =

⇒ E= 2 32

dx

dE dE dE cos θ

∫ dE cosq

∫ dE cosq +

λ Qx dx

θ θ

=λ

θθ

⇒ E=

dx

r

dE sin θ

2 32

R

d =λ

dθ

Qx 2

dq

x

The electric field strength due to ring at a point on its axis (say x-axis) at distance x from the centre of the ring is given by E( x ) =

λQ 4pe 0 R

On integrating we get, F =

Solution



32

λ 4pe 0 r

f 2

∫ cosq dq

-

f 2

9/20/2019 10:11:48 AM

1.30  JEE Advanced Physics: Electrostatics and Current Electricity

λ ⇒ E= sin q 4pe 0 r

Now we know that electric field strength due to a ring of radius R, charge Q, at a distance x from its centre on its axis can be given as

f 2 -

f 2

1 Qx  (Derived earlier) 4pe 0 ( x 2 + R2 )3 2 So, due to the infinitesimal elemental ring the electric field strength dE at point P is Eaxis =

λ ⎡ ⎛ f⎞⎤ ⎛f⎞ ⇒ E= sin ⎜ ⎟ - sin ⎜ - ⎟ ⎥ ⎢ ⎝ 2⎠ ⎦ ⎝ ⎠ 4pe 0 r ⎣ 2 ⇒ E= ⇒ E=

λ ⎛f⎞ 2 sin ⎜ ⎟ ⎝ 2⎠ 4pe 0 r λ ⎛f⎞ sin ⎜ ⎟ ⎝ 2⎠ 2pe 0 r

dE =

( dq ) x 1 4pe 0 x 2 + y 2 3 2

⇒ dE =

1 σ 2p y dy x 4pe 0 x 2 + y 2 3 2



Problem Solving Technique(s)

( (

R

σ

p (b) For Quarter-circle f = 2 λ ⇒ E= 2 2pe 0 r

P

α

x

E

R

(c) For circle, f = 2p



⇒ E=0

E = Eaxis =

∫ dE 0

⇒ E=

Illustration 23

Calculate the electric field at point P that lies on the axis of a uniformly charged disc of radius R having charge density σ at a distance x from the centre.

⇒ E=

Solution

To find electric field at point P due to this disc let us consider an elemental ring of radius y and width dy. If dq be the charge on this infinitesimal element of area dA = 2pydy, then

)

Please note that, here once point P is taken, then value of x remains fixed. So,

(a) For semi-circle, f = p λ ⇒ E = 2pe 0 r



)

dq = σ dA = σ ( 2p ydy ) dy y

x P

σ

01_Physics for JEE Mains and Advanced - 2_Part 1.indd 30

dE

⇒ E=

∫

1 dE = 4pe 0

σ px 4pe 0

R

∫ (x 0

R

σ 2p xy dy

∫ (x 0

2

+ y2

)

32

2 y dy 2

+ y2

)

32

2 σx ⎡ ⎢2 4pe 0 ⎢ x + y2 ⎣

y=R

y =0

⎤ ⎥ ⎥ ⎦

σ 2e 0

x ⎤ ⎡ ⎢1 2 2 ⎥ x +R ⎦ ⎣ Also, we observe here that, if we assume the complete disc to subtend an apex angle α at P , then, the field can also be expressed as σ ⎡ σ x ⎤ ( 1 - cos α )  1E= = 2 2 ⎥ 2e 0 ⎢⎣ 2 e0 R + x ⎦ x  (Because cos α = ) 2 R + x2 ⇒ E=

9/20/2019 10:12:03 AM

Chapter 1: Electrostatics 1.31

So, finally, we get E=

σ 2ε 0

E/E0 1

σ x ⎤ ⎡ ( 1 − cos α ) = ⎢1 − 2 2 ⎥ 2 ε0 x +R ⎦ ⎣

0.5 –3 –2 –1 1 2 –0.5

(a) Please note that here we have used dE =

( dq ) x 32 4πε 0 ( a2 + x 2 )



σx 2ε 0



1 ⎤ ⎡1 …(2) ⎢x− 2 2 ⎥ R +x ⎦ ⎣

where it would not be possible to calculate E at x = 0. So, to be precise with the boundaries here, we just calculate E ( at x = 0 ) and E ( as x → 0 ) and both will yield different results.



…(1)

But actually the result obtained from this e­ xpression is

 E ( at x = 0 ) = 0 

{from expression (2)}

Notice the discontinuity in electric field as we cross the plane. The discontinuity is given by ΔE x = E x + − E x − =

E σ (d) The electric field x , where E0 = as a func2ε 0 E0 x tion of is shown in figure R

σ  (i.e., very close to the 2ε 0 disc)  {from expression (1)} So, we can say that E has two different values i.e., at the centre and just outside the disc and hence E shows discontinuity at x = 0. x ⎡ σ ⎛ ⎞ , x>0 ⎢ 2ε ⎜ 1− 2 ⎟ 2 0 ⎝ R +x ⎠ ⎢ (b) So, E x = ⎢ x σ ⎛ ⎞ ⎢ −1− , x 0  ⎪⎪ 2ε 0 E=⎨ ⎪ − σ iˆ, x < 0 ⎪⎩ 2ε 0

The plot of the electric field in this limit is shown in figure

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 31

σ ⎛ σ ⎞ σ − − = 2ε 0 ⎜⎝ 2ε 0 ⎟⎠ ε 0

As we shall see during the study of Gauss’s Law that if a given surface has a charge density σ , then the normal component of the electric field across that surface always exhibits a discontinuity with σ ΔEn = . ε0

Ex σ 2ε 0

and E ( as x → 0 ) =



x/R

–1

Remark(s)

E=

3

x – σ 2ε 0 Electric field of an infinitely large non-conducting plane

(e) To show that the “point-charge” limit is retained for x  R, we make use of the expansion series, in which ⎛ R2 ⎞ 1− = 1 − ⎜ 1+ 2 ⎟ ⎝ x ⎠ x 2 + R2 x



This gives Ex =

−1 2

=

⎛ ⎞ R2 1 R2 .... 1− ⎜ 1− + ⎟⎠ ≈ 2 ⎝ 2 x2 2x

σ R2 1 σπ R2 1 Q = = 2 2 2ε 0 2 x 4πε 0 x 4πε 0 x 2

which is indeed the expected “point-charge” result.

9/24/2019 10:30:38 AM

1.32  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 24

Calculate the electric field strength due to a uniformly charged infinite sheet having charge density s at a point P at perpendicular distance x from the sheet.

⇒ E=-

dq = s ( 2p y dy )



1⎞ ⎛ ⎜⎝ 0 - ⎟⎠ x

s  2ε 0

⇒ E=

Solution

Let us consider an infinitesimal elemental ring of radius y having width dy with centre at O as shown. If dq be the charge on this ring, then

sx 2ε 0

{independent of value of x}

Illustration 25

Calculate the electric field due to a thin uniformly charged hemispherical shell (of radius R having uniform charge density s ) at its centre. Solution

Let us consider an infinitesimal elemental ring on its surface having angular width dq at an angle q from its axis as shown. The surface area dA of this ring is

σ y O

x

P



dA = ( 2p R sin q )( Rdq ) r

dy R

The electric field strength due to this elemental ring at point P at distance x is dE = ⇒

E=

⇒ E=

dqx 1 4pε 0 x 2 + y 2

(

1 4pε 0

ps x 4pε 0

sx ⇒ E= 4ε 0 Since,

∫

∞

0

2

∞

∫ (x

)

dq = s dA = s ( 2p R2 sin q dq ) Now due to this ring, electric field strength at centre C is

0

)

+ y2

32

dE =

2 y dy 2

+y



)

∫ ( x2 + y2 )

-3 2

⎧⎪ ( dq ) x ⎨∵ dE = 32 4pε 0 ( a 2 + x 2 ) ⎪⎩



2 y dy

0

⎡⎣ f ( x ) ⎤⎦ ⎡⎣ f ( x ) ⎤⎦ f ′ ( x ) dx = n+1 n

⎛ 3 2 2 - 2 +1 ⎜ x + y sx ⇒ E= ⎜ 3 4ε 0 ⎜ - +1 ⎜⎝ 2 ⇒ E=-

dq ( R cos q ) ⎤ 1 ⎡ ⎢ ⎥ 3 2 4pε 0 ⎢ ( R2 sin 2 q + R2 cos 2 q ) ⎥ ⎣ ⎦

2 32

∞

(

dθ O r = R sin θ x = R cos θ

Charge on this elemental ring is 32

s ( 2p y dy ) x

∫ (x

dr

xθ

)

2s x ⎛ 1 ⎜ 4ε 0 ⎜ x 2 + y 2 ⎝

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 32

y →∞

y =0

y →∞

y =0

⎞ ⎟ ⎟⎠

⎞ ⎟ ⎟ ⎟ ⎟⎠

)

⇒ dE =

2 1 ⎡ s 2p R sin q dq R cos q ⎤ ⎢ ⎥ 4pε 0 ⎣ R3 ⎦

⇒ dE =

ps sin ( 2q ) dq 4pε 0

n+1

, n ≠ -1

(

⎫⎪ ⎬ ⎪⎭

Net electric field at center is obtained by integrating the above expression for dE between the limits zero p to . So, 2 E=

∫

s dE = 4ε 0

p 2

∫ sin 2q dq 0

9/20/2019 10:07:32 AM

Chapter 1: Electrostatics 1.33

⎛ cos 2q ⎜⎝ 2

⇒ E=

s 4ε 0

⇒ E=

s ⎛ 1 1⎞ ⎜ + ⎟ 4ε 0 ⎝ 2 2 ⎠

⇒ E=

s 4ε 0

Due to the thinner sheet (PART 1) electric field at point P is in downward direction, say it is E1 which is given by using equation (2).

p 2⎞ 0

⎟⎠

⎛d ⎞ r⎜ - x⎟ ⎝2 ⎠ E1 = 2ε 0

Similarly due to thicker sheet (PART 2) electric field at P is in upward direction, say it is E2 which is given again by using equation (2).

ELECTRIC FIELD DUE TO A LARGE THICK CHARGED SHEET CASE-1: OUTSIDE THE SHEET Consider a large sheet of thickness d having uniform charge density r . On both sides of sheet electric field strength is directed, away from the sheet.

⎛d ⎞ r⎜ + x⎟ ⎝2 ⎠ E2 = 2ε 0

Net electric field at point P will be EP ( = E ) given by EP = E = E2 - E1



⎛d ⎞ ⎛d ⎞ r⎜ + x⎟ r⎜ - x⎟ ⎝2 ⎠ ⎝2 ⎠ ⇒ E = EP = 2ε 0 2ε 0

E P ρ

⇒ E=

d

The electric field strength at a point P in front of the sheet is E=

s …(1) 2ε 0

where s is the charge per unit surface area of the sheet. Further

s = rd

⇒ E=

rx ε0

Conceptual Note(s) Electrostatic Field in Some Cases (a) The electric field due to a uniformly charged rod AB having charge density l at a point P at perpendicular distance a from the rod, as shown in figure is A

rd …(2) 2ε 0

CASE-2: INSIDE THE SHEET To find electric field strength at an interior point P of the sheet at a distance x from its centre as shown, we divide the sheet in two sheet parts. One of thickness ⎛d ⎞ ⎛d ⎞ ⎜⎝ - x ⎟⎠ and other of thickness ⎜⎝ + x ⎟⎠ as shown. 2 2

N

d +x 2

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 33

P

PART 2 E1

P

l ( sina + sinb ) 4pε 0 a l and E y = ( cos b - cos a ) 4pε 0 a Ex =

PART 1

x

α β

B

E2 d –x 2

a

d



9/20/2019 10:07:42 AM

1.34  JEE Advanced Physics: Electrostatics and Current Electricity



So, we can also extend the result obtained to more look alike situations. (b) For an infinite rod having uniform charge density l , electric field at point P is calculated by taking p p a ⎯⎯ → and b ⎯⎯ → . 2 2

⇒ Ex =

l and E y = 0 2pε 0 a

(c) For a semi-infinite rod having uniform charge density l , electric field at point P is calculated by p taking a ⎯⎯ → and b = b 2

l l and E y = ⇒   Ex = 4pε 0 a 4pε 0 a

Ex E

(e) The electric field at the centre of a uniformly charged wire in the form of a circular arc of radius r subtending an angle f at the centre is l ⎛f⎞ E= sin ⎜ ⎟ , along the angle bisector. 2pε 0 r ⎝ 2 ⎠





(iii) For circle, f = 2p

⇒  E=0 (f) The electric field at point P that lies on the axis of a uniformly charged disc of radius R having charge density s at a distance x from the centre is

x

 E =

s 2ε 0

α

P

E

x ⎤ s ( ⎡ 1- cos a ) ⎢ 1- 2 ⎥= R + x 2 ⎦ 2ε 0 ⎣

If Q is the total charge on the disc, then E=

x 2Q ⎡ ⎤ 12 2 ⎥ 4pε 0R2 ⎢⎣ x +R ⎦

(i) Due to a charged spherical conductor (spherical shell) of charge Q, radius R at a distance r from its centre.



(i)  For semi-circle, f = p

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 34

R

σ

  (g) The electric field strength due to a uniformly charged infinite sheet having charge density s at a point P at perpendicular distance x from the s sheet is E = (independent of value of x). 2ε 0

45° Ey

l So, E = 2pε 0 r

l 2 2pε 0 r



p 2

(h) The electric field due to a thin uniformly charged shell (of radius R having uniform charge density s s ) at its centre is E = . 4ε 0

P



⇒  E=



l l cos b ⇒   Ex = ( 1+ sinb ) and E y = 4pε 0 a 4pε 0

(d) For a semi-infinite rod having uniform density l , the electric field at point P (which lies on the line joining P to wire normally and passing through the known end of wire) is calculated by taking p a ⎯⎯ → and b ⎯⎯ →0 2

(ii) For Quarter-circle, f =



⎧ Q for r ≥ R ⎪ E = ⎨ 4pε 0 r 2 ⎪0 for r < R ⎩   If the conducting sphere has uniform surface charge density s , then ⎧ s R2 ⎪ E = ⎨ ε0 r 2 ⎪0 ⎩  

for r ≥ R for r < R

9/20/2019 10:07:52 AM

Chapter 1: Electrostatics 1.35

Suppose Q charge is given to

E Q = σ 4πε0R2 ε 0

E ∝ 1r

(j) Due to a uniformly charged non-conducting sphere of charge Q, radius R at a distance r from the centre of the sphere.





⎧ Q ⎪ 4pε r 2 ⎪ 0 E=⎨ ⎪ Qr ⎪⎩ 4pε 0R3  

Electric field for both the cases E=

for r ≥ R for r < R

E=

 

for r ≥ R

E∝

E ∝ 12 r

θ

O

ρ 3ε 0

Q A

Because Q is distributed in area A.

  (o) Due to two parallel plane thin sheets with surface charge density sA and sB in the regions specified in figure

r=R

2l 4pε 0 r

s ε0

s 2ε 0

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 35

B

A

  (m) Due to an infinite non conducting plane sheet of charge with surface charge density s .

 

Where σnon-conducting =

r

  (l) Due to an infinite thin conducting plane sheet of charge with surface charge density s .



2ε 0

s ε0

I

(k) Due to an infinite thread having uniform charge distribution l at distance r from it.

E=

σnon-conducting

r

Slope = tanθ =

E=

E=

Because Q is distributed in area 2A.

E=

E

E=

σconducting ε0

(n) Near a conductor of any shape,

for r < R

ρ Q = R 4πε0 R2 3 ε 0

Q 2Aε 0

Q Where σconducting = 2A

If the sphere has uniform volume charge density r, then ⎧ rR3 ⎪ ⎪ 3ε r 2 E=⎨ 0 ⎪ rr ⎪⎩ 3ε 0

Non-conducting plate

r

r=R

O

Conducting plate

II σA



   

EI = -

III

+

σB

1 (s A + sB ) 2ε 0

EII =

1 (s A - sB ) 2ε 0

EIII =

1 (s A - sB ) 2ε 0

  If s A = +s and s B = -s , then

s   EI = EIII = 0 and EII = ε 0

9/20/2019 10:07:58 AM

1.36  JEE Advanced Physics: Electrostatics and Current Electricity

(p) At the axis of a uniformly charged ring of radius R at a distance x from the centre of the ring. E=  

Qx

and



4pε 0 ( R2 + x 2 )

32

( E2 )x = 0

 So, EC  ⇒ EC

Illustration 26

In the given arrangement find the electric field at C in the figure. Here the U-shaped wire uniformly charged with linear charge density l .

l (along +y axis) 2pε 0 a    = E1 + E2 + E3  =0

( E2 )y =

Illustration 27

Calculate the electric field due to a rod of length L at a point P at a distance l from its end, when the charge density of the rod varies with distance x from the other end of the rod as l = l0 x where l0 is a positive constant. Solution

a

The rod AB shown here is of length L having linear charge density that depends on distance x from end A of rod as

C



Solution

l = l0 x

Here we shall divide the arrangement into three parts as shown and then apply the principle of superposition.

1

(E2)y (E1)x

(E3)x

C (E3)y (E1)y

L A

3 y x



( E1 )x =

l (along +x axis) 4pε 0 a

( E1 )y =

l (along -y axis) 4pε 0 a



( E3 )x



dE =

l = (along -x axis) 4pε 0 a l (along -y axis) 4pε 0 a

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 36

l

P

dq 1 4pε 0 ( L + l - x )2

Net electric field at P due to complete rod (from A to B ) is calculated by integrating the above expression within limits taken from zero to L . Thus we have E=

( E3 )y =

B

dq = l dx = l0 x dx Now electric field strength E at point P due to infinitesimal charge dq is dE (say), given by

Here we have used the results obtained previously

Similarly

dx

and we have to calculate electric field strength at point P. For this let us consider an infinitesimal element of length dx at a distance x from the end A as shown. Here charge on this element is

2



x

⇒ E=

∫

1 dE = 4pε 0

l0 4pε 0

L

L

∫ 0

l0 x dx

l = 0 2 ( L + l - x ) 4pε 0

L

x dx

∫ (L+ l - x)

2

0

x dx

∫ (L+ l - x)

2

0

9/20/2019 10:08:11 AM

Chapter 1: Electrostatics 1.37

To calculate the integral, let us denote the integral by I=

x dx

∫ (L+ l - x)

2

(a) the centre of the ring (b) on the axis of the ring as a function of the distance x from its centre. Investigate the obtained function at x  R.

Substituting L + l - x = y ⇒ - dx = dy ⇒ I=

∫

- ⎡⎣ ( L + l ) - y ⎤⎦ dy y

∫

⇒ I=

(L+ l) y

Solution

2

⇒ I = - ( L + l ) y -2 dy + ⇒ I = -(L + l)

angle. Calculate the magnitude of the electric field strength at

y

∫y

2

(a) Here by observing the function l = l0 cos q , we can state that the first and fourth quadrant are positively charged and second and third quadrant are negatively charged.

dy

y -2 +1 + log e y -2 + 1

y θ Rd

+ log e y

dθ

dE cos θ θ

L+l + log e ( L + l - x ) ⇒ I= L+l-x L

l ⇒ E= 0 4pε 0

∫ (L+ l - x)

l ⇒ E= 0 4pε 0

⎡ ⎢ ⎛⎜ L + l ⎞⎟ ⎢⎝ L + l - x ⎠ ⎣

⇒ E=

x

x

dE sin θ ρ0

x dx

2ε 0

2

0

L 0

⎤ + log e ( L + l - x ) ⎥ ⎥ 0 ⎦ L

l0 ( L + l ) ⎡ 1 1 ⎤ + ⎢ 4pε 0 ⎣ L + l - L L + l ⎥⎦ l ⎛ L+l-L⎞ log e ⎜ ⎝ L + l ⎟⎠ 4pε 0

⇒ E=

l0 ( L + l ) ⎡ 1 l 1 ⎤ ⎛ l ⎞ + 0 log e ⎜ ⎢ ⎥ ⎝ L + l ⎟⎠ 4pε 0 ⎣ l L + l ⎦ 4pε 0

⇒ E=

l0 ( L + l ) L l ⎛ l ⎞ + 0 log e ⎜ ⎝ L + l ⎟⎠ 4pε 0 l ( L + l ) 4pε 0

⇒ E=

l0 ⎡ L ⎛ l ⎞⎤ + log e ⎜ ⎢ ⎝ L + l ⎟⎠ ⎥⎦ 4pε 0 ⎣ l

⇒ E=

l0 ⎡ L L⎞ ⎤ ⎛ - log e ⎜ 1 + ⎟ ⎥ ⎝ 4pε 0 ⎢⎣ l l⎠⎦

Illustration 28

A thin non conducting ring of radius R has linear charge density l varying as l = l0 cos f , where l0 is a positive constant and f is the azimuthal

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 37

y

θ

To find electric field strength at the centre of the ring, we consider an infinitesimal element on ring having polar width dq and making an angle q with x-axis. The charge on this infinitesimal element is     dq = l dx ⇒   dq = l ( Rdq ) ⇒   dq = ( l0 cos q ) ( Rdq ) …(1) The electric field strength at the centre of the ring due to this element is

dE =

1 ⎛ dq ⎞ ⎜ ⎟ …(2) 4pε 0 ⎝ R2 ⎠

To find net electric field at centre of ring due to this semicircular distribution, we integrate the components of this electric field taken for the semi-circumference of ring. Here components dE sin q will cancel each other and the component dE cos q will be contributing to the net field. Thus net electric field strength at centre due to semicircular distribution will be given as

9/20/2019 10:08:25 AM

1.38  JEE Advanced Physics: Electrostatics and Current Electricity

  E = Edue to semi-circular + Edue to semi-circular positive charge distribution

negative charge distribution

dl ϕ

p 2

∫ dE cosq

⇒  E=2

-

x-ax

is

p 2

2l 0 ⇒   E = 4pε R 0

l0 ⇒   E = 2pε R 0

dEsin θ

∫ cos q dq p 2

p 2

+

1 ⇒   E1 = 4pε 0

⎛ 1 + cos 2q ⎞ ⎟⎠ dq 2

∫ ⎜⎝

-

p 2

p 2 -

p 2

+

sin ( 2q ) 2

p 2 -

p 2

⎞ ⎟ ⎟ ⎠

1 ⇒   E1 = 4pε 0

l0 ⇒   E = 4pε R ( p + 0 ) 0

1 ⇒   E1 = 4pε

l0 ⇒   E = 4ε R 0

⇒   E1 =

(b) Take an infinitesimal element dl at an azimuthal angle f . This element subtends an angle df at the centre O . The charge on this element dq = l dl = l ( Rdf )

    ⇒   dq = ( l0 cos f ) ( Rdf )

{∵ l = l0 cos f and dl = Rdf }

   

dE =

dq 1 2 4pε 0 ( R + x 2 )

Electric field due to positive half ring at P on the axis is +

E1 =  

p 2

∫ dE sin q cos f

p 2

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 38

θ

P

dE

ϕ θ

dE

co

sθ

dEsin θ cos ϕ

2

-

x21 ) /2

dEsin θ sin ϕ

p 2

⎛ l0 ⎜ ⇒   E = 4pε R ⎜ q 0 ⎝



(R 2 +

dϕ

⇒

p 2

∫ (R

-

p 2

+

p 2

∫

-

dq 2

cos f

( R 2 + x 2 )3 2

p 2

2

( R2 + x 2 ) 2l 0 R 2

4pε 0 ( R + x 2

l0 R

E1 =

R

×

) ( R2 + x 2 )1 2

l0 R cos f df × R cos f

l0 R

0

+x

2

32

)

2 32

⎛ p ⎞ ⎜ 2 ⎡ 1 + cos 2f ⎤ ⎟ ⎜2 ⎢ ⎥⎦ df ⎟⎟ ⎣ 2 ⎜⎝ ⎠ 0

∫

⎛p⎞ ⎜⎝ ⎟⎠ 4

2

8ε 0 ( R 2 + x 2 )

32

Similarly, for negative part, we have E2 =

l0 R2

8ε 0 ( R 2 + x 2 )

32

   So, net electric field at point P is given by Enet = E1 + E2 =

l0 R2

4ε 0 ( R 2 + x 2 )    The direction will be perpendicular to the axis of ring For x  R , Enet =   

32

l0 R2

4ε 0 ( x 3 )

9/20/2019 10:08:36 AM

1.39

Chapter 1: Electrostatics IllustRatIon 29

(a) Consider a uniformly charged thin walled right circular cylindrical shell having total charge Q, radius R, and length l . Determine the electric field at a point a distance d from the right side of the cylinder as shown in figure. l R

d

(b) Now consider a solid cylinder with the same dimensions and carrying the same charge, uniformly distributed through its volume. Find the field it creates at the same point. solutIon

(a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has charge Qdx and produces, at the chosen point, a field l  1 x Qdx ˆ dE = i 3 4pε 0 2 l ( x + R2 ) 2 l R

d

Q E= 8pε 0 l

⇒

Since

∫

E=

d+

Q

∫ dE = 4pε l ∫ ( x 0

d

+ R2 )

d

3

1 ⎞ Q ⎛ ⇒ E = - 4pε l ⎜ ⎟ 2 2 0 ⎝ x +R ⎠

2 2

x + R2 d+

d

1 1 ⎞ Q ⎛ ⇒ E = 4pε l ⎜ 2 2 2 2 ⎟ ( d + l) + R ⎠ 0 ⎝ d +R (b) Think of the cylinder as a stack of disks, each Qdx , and charge per with thickness dx, charge l Qdx area s = . One disk produces a field p R2 l  s ⎛ x 1dE = ⎜ 2ε 0 ⎝ x 2 + R2

⎞ˆ i ⎟⎠

So, the field due to the entire cylinder is given by E=

d+

Q

x

⎛

∫ dE = 2pε R l ∫ ⎜⎝ 1 0



2

2

x +R

d

⇒

d+l ⎛ d+l 1 ⎜ dx x 2 + R2 E= 2 2pε 0 R2 l ⎜⎝ d d

⇒

E=

⇒

E=

Q

∫

(x 2pε R l Q 0

xdx 2

∫

3

( x 2 + R2 )- 2 2xdx

( x 2 + R2 )- 2 2xdx = -

dx

The total field is

d+

2

Q 2

2pε 0 R l

(l +

∫(

x 2 + R2

)

)

2

-

1 2

⎞ dx ⎟⎠ ⎞ 2xdx ⎟ ⎟⎠

d+l d

2 d 2 + R2 - ( d + l ) + R2

)

3 2

Test Your Concepts-III

Based on Electric Field 1. Four particles each having a charge q, are placed on the four vertices of a regular pentagon. The distance of each corner from the centre is a. Find the electric field at the centre of the pentagon.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 39

(Solutions on page H.10) 2. A circular wire loop of radius R carries a total charge Q distributed uniformly over its length. A small length x (  R ) of the wire is cut off. Find the electric field at the centre due to the remaining wire.

9/20/2019 10:08:44 AM

1.40  JEE Advanced Physics: Electrostatics and Current Electricity

3. An electric field of 105 NC -1 points due North at a certain spot. Find the magnitude and direction of the force that acts on a charge of +2 mC and -5 mC at this spot? 4. Three identical positive charges q are arranged at the vertices of an equilateral triangle of side . Calculate the intensity of the field at the vertex of a regular tetrahedron of which the triangle is the base. 5. A square frame ABCD is made of four thin rods, each of length L and mass m and charged with a charge q. The frame is hanging from one of its sides as shown in figure along a horizontal axis XY. At t = 0 a horizontal electric field is switched on in horizontal direction, perpendicular to the plane of the frame. Find the minimum value of E so that the square frame rotates upto horizontal level. Y

D E

A X

C B

6. A particle of mass m and charge -Q is constrained to move along the axis of a ring of radius R. The ring carries a uniform charge density + l along its length. Initially the particle is in the plane of the ring where the force on it is zero. Show that the period of oscillation of the particle when it is displaced slightly from its equilibrium position is given by





T = 2p

2ε 0 mR2 lQ

7. Two point charges are positioned at points 1 and 2. The field intensity to the right of the charge Q2 on the line that passes through the two charges varies according to a law that is represented schematically

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 40

in the figure. The field intensity is assumed to be positive if its direction coincides with the positive direction on the x-axis. The distance between the charges is . E

a

1

x

2 l b



(a)  Find the sign of each charge. (b) Find the ratio of the absolute values of the Q charges 1 Q2



(c) Find the value of b where the field intensity is maximum. 8. A non-conducting ring of mass m and radius R is charged as shown. The charge density, i.e., charge per unit length is l . It is then placed on a rough non-conducting horizontal plane. At time t = 0 , a  uniform electric field E = E0 iˆ is switched on and the ring starts rolling without sliding. Determine the frictional force (magnitude and direction) acting on the ring when it starts moving. E0 + +

++

y –– –– – ––

x

9. A thin insulating wire is stretched along the diameter of an insulated circular hoop of radius R. A small bead of mass m and charge -q is threaded onto the wire. Two small identical charges are tied to the hoop at points opposite to each other, so that the line connecting them is perpendicular to the thread (shown in figure). The bead is released at a point which is a distance x0 from the centre of the hoop.

9/20/2019 10:08:47 AM

Chapter 1: Electrostatics 1.41

+Q Y R

X

(–q, m) x0 +Q



(a)  Calculate the resultant force acting on the charged bead. (b) Describe (qualitatively) the motion of the bead after it is released. (c)  Write the exact equation that governs the motion of the bead along the thread. x (d) After the assumption that 0  1 is used to R obtain an approximate equation of motion, find the displacement and velocity of the bead as functions of time. (e) When will the velocity of the bead vanish for the first time? 10. A non-conducting ring of mass m and radius R is lying at rest in the vertical XY plane on a smooth non-conducting horizontal XZ plane. Charge +q and -q are distributed uniformly on the ring, on the two sides of the vertical diameter of  the ring. A constant and uniform electric field E is set up along the x direction. The ring is given a small angular rotation about an axis perpendicular to its plane and released. Find the period of oscillation of the ring. 11. In the given arrangement find electric field at C. Complete wire is uniformly charged at linear charge density l .

12. A pitch ball covered with tin foil having a mass m hangs by a fine silk thread of length  in a horizontal electric field E. When the ball is given an electric charge q, it stands out a distance d from the vertical line. Show that the electric field is given by mgd E= q 2 - d 2 13. A bob of mass m carrying a positive charge q is suspended from a light inextensible string of length  inside a parallel plate condenser with its plates making an angle b with the horizontal. The upper plate of the condenser is negatively charged and the intensity of electric field inside the condenser is E. Find the period of vibration of the pendulum and the angle between the thread in equilibrium position and the vertical. 14. A block of mass m containing a net positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure. The distance of the block from the wall is d. A horizontal electric field E towards right is switched on. Assuming collisions (if any) to be only of elastic nature, find the time period of resulting oscillatory motion. Is it a simple harmonic motion. d E m q

15. In the given arrangement of a charged square frame find electric field at centre. The linear charged density is as shown in figure. –3λ 3 l λ

1

1

2

2λ

4 4λ C R 2 3

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 41

16. In the figure shown, three charged rods each of length L, having charge densities l , - l and l form a triangle. Calculate the electric field due to the arrangement at its centre.

9/20/2019 10:08:50 AM

1.42  JEE Advanced Physics: Electrostatics and Current Electricity



A

λ

C

B

λ

C

–λ

17. A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field as shown in figure. When an electric field -1 given by E = ( Aiˆ + Bjˆ )NC is switched on (where



A and B are positive numbers) the ball attains equilibrium making an angle q with the vertical. Find (a)  the charge on the ball and (b)  the tension in the string. θ

(a) Suppose the electric field is produced by two concentric cylindrical electrodes (not shown in the diagram) and hence is radial. Calculate the magnitude of the electric field. (b) If the field is produced by two flat plates and is uniform in direction, what value should the field have in this case? 19. A small, 2 g plastic ball is suspended by a 20 cm long string in a uniform electric field as shown in Figure. If the ball is in equilibrium when the string makes a 15° angle with the vertical, what is the net charge on the ball?

E = 1 × 103 i NC–1

y x

20 cm 15° m = 2g

E

y x

q

18. A physicist studying the properties of ions in the upper atmosphere wishes to construct an apparatus having the following characteristics. In the setup used, an electric field is applied such that a beam of ions, each having charge q, mass m, and initial velocity viˆ , is turned through an angle of 90° such that each ion undergoes displacement Riˆ + Rjˆ. The ions enter a chamber as shown in figure, and leave through the exit port with the same speed they had when they entered the chamber. The electric field acting on the ions is to have constant magnitude. y v

20. Consider n equal positive point charges each of Q magnitude placed symmetrically around a n circle of radius R. Calculate the magnitude of the electric field at a point a distance x on the line passing through the centre of the circle and perpendicular to the plane of the circle. 21. A line of charge starts at x = + x 0 and extends to positive infinity. The linear charge density is l x l = 0 0 . Determine the electric field at the x origin. 22. A rod lies along x-axis with one end at origin and other at x → ∞ . The rod has a uniform charge density of l Cm-1 . Find the electric field at the point x = - a on x-axis. Find E x and E y for a point on the y-axis where y = b .

R +

x

v R

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 42

9/20/2019 10:08:54 AM

Chapter 1: Electrostatics 1.43

MOTION OF A CHARGED PARTICLE IN AN ELECTRIC FIELD CASE-1: In uniform electric field when a positively charge particle is released from rest, it starts moving in the direction of electric field with uniform acceleration a given by

 

a=

 

θ

q, m

A particle of mass m and charge q is thrown from ground at an angle q with initial speed u. A uniform  electric field E exists in downward direction as shown. Here during motion of particle we can consider the effective acceleration to be

1 2 1 ⎛ qE ⎞ 2 at = ⎜ ⎟t 2 2⎝ m ⎠

KE =

1 1 ⎛ q 2 E2 ⎞ mv 2 = ⎜ ⎟ t = y qE 2 2⎝ m ⎠

CASE-2: In uniform electric field when a positively charge particle is released with non-zero initial velocity u perpendicular to the electric field, then Velocity at any instant    Also, v = u + a t  ⎛ qEt ⎞  ⇒ v = ui + ⎜ j ⎝ m ⎟⎠



{∵ E = E ˆj ( say ) , then u = uiˆ }

Comparing with  v = vx i + vy j we have vx = u

v vy β

vx

qEt vy = m  Since, v = v = vx2 + vy2 ⇒ v = u2 +

q 2 E2 t 2 m

tan b =

vy vx

=

a= g+

qE m

a= g-

qE m

Here in above example we can use all the concepts of projectile motion by replacing g by a. If electric field in space is in upward direction, effective acceleration due to gravity will be

Illustration 30

A particle of mass 9 × 10 -31 kg and a negative charge of 1.6 × 10 -19 C is projected horizontally with a velocity of 10 5 ms -1 into a region between two infinite horizontal parallel plates of metal. The distance between the plates is 0.3 cm and the particle enters 0.1  cm below the top plate. The top and bottom plates are connected respectively to the positive and ­negative terminals of a 30 V battery. Calculate the component of the velocity of the particle just before it hits one of the plates. Solution

2

If b is the angle made by v with x-axis, then



g u

  and kinetic energy,

E

qE m

⎛ qE ⎞ Then v = at = ⎜ t ⎝ m ⎟⎠ y=

CASE-3: If the particle is thrown in a uniform electric field in the direction different from the direction of electric field, it will follow a parabolic trajectory like projectile motion. Consider the following case.

qEt mu

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 43



E=

⇒ E=

V d 30 3 × 10 -3

= 10 4 NC -1 

{∵ d = 3 × 10

-3

m

}

9/20/2019 10:09:04 AM

1.44  JEE Advanced Physics: Electrostatics and Current Electricity

Force on the particle of negative charge moving between the plates is

and ax =

F = eE = 1.6 × 10 -19 × 10 4 = 1.6 × 10 -15 N

( 1.6 × 10 -15 ) ( 9 × 10 -31 )

E

y

The direction of force will be towards the positive plate i.e., upwards Acceleration of the particle is eE a= m ⇒ a=

qE sin b m β

u α

q

Time of flight T =

⇒ a = 1.77 × 1015 ms -2 The electric intensity E is acting in the vertical direction so the horizontal velocity v of the particle remains same. If y is the displacement of the particle, in upward direction, then 1 y = at 2 2

where y = 0.1 cm = 10 -3 m , a = 1.77 × 1015 ms -2 ⇒ 10 -3 =

1 ( × 1.77 × 1015 )( t 2 ) 2

x

2vy ay

=

In x direction range of projectile is given by

x = ux t +

1 2 ax t 2

⇒ R = ux T +

1 ax T 2 2

⇒ R = u cos a

⎛ 2u sin a ⎞ + ⎜ qE ⎟ cos b + g ⎟ ⎜⎝ ⎠ m 1 ⎛ qE sin b ⎞ ⎛ 2u sin a ⎞ ⎜ ⎟ ⎟ 2⎝ m ⎠ ⎜ qE cos b + g ⎟ ⎜⎝ ⎠ m

⇒ t = 1.063 × 10 -9 s Component of velocity in the direction of field is given by

vy = at

⇒ vy = ( 1.77 × 1015 ) ( 1.063 × 10 -9 ) ⇒ vy = 1.881 × 106 ms -1 and vx = 10 5 ms -1 Illustration 31

A projectile of mass m having charge q is thrown with initial velocity u at an angle a with the horizontal. An electric field of strength E exists, making an angle b with the downward vertical away from the point of projection. Find the time of flight and range of projectile. Solution

Initial velocity is given by (in components)

uy = u sin a , ux = u cos a

⎛ qE ⎞ cos b + g ⎟ and acceleration ay = ⎜ ⎝ m ⎠

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 44

2u sin a ⎛ qE ⎞ cos b + g ⎟ ⎜⎝ ⎠ m

2

Illustration 32

Two balls of charge q1 and q2 initially have a velocities of equal magnitude and same direction. After a uniform electric field has been applied for a certain time interval, the direction of first ball changes by 60° and the velocity magnitude is reduced by half. The direction of velocity of the second ball changes thereby 90°. E

q1 m1

E

v

q2 m2 Initial condition

v

(a) Determine the magnitude of the charge to mass ratio of the second ball if it is equal to a 1 for the first ball.

9/20/2019 10:09:17 AM

Chapter 1: Electrostatics 1.45

(b) In what ratio will the velocity of the second ball change? Ignore the electrostatic interaction between the balls Solution

(a) Let the electric field acting on each ball be given by

E = Ex iˆ + Ey ˆj v

v = v1 2

⇒ 

Final condition

From Impulse Momentum equation, we have Impulse = Change in momentum Let the final velocities of the balls be v1 and v2. v Noting that v1 = , we have 2

)

q2 4 q1 4 = = a1 m2 3 m1 3

q1 m1 3v = (b) Also q2 4 v2 m2 ⇒ 

60°

(

    

⎛ q1 ⎞ ⎜⎝ m ⎟⎠ 3 1 = …(7) q ⎛ 2 ⎞ 4 ⎜⎝ m ⎟⎠ 2

3v 3 = 4 v2 4

⇒   v2 =

Two equal and opposite charges separated by a distance together constitute a dipole. Equitorial line

v ⎛v ⎞ m1 ⎜ cos 60°iˆ + sin 60° ˆj ⎟ - m1viˆ …(1) ⎝2 ⎠ 2

(

)

q2 Ex iˆ + Ey ˆj Dt =

(

)

m2 v2 cos 90°iˆ + v2 sin 90° ˆj - m2 viˆ …(2)

  On comparing the x and y-components on both sides of equation (1), we get

 

q1 3 Ex Dt = - v and 4 m1

…(3)

q1 3 Ey Dt = v …(4) 4 m1

  Similarly, from equation (2), we get  

q2 Ex Dt = -v and…(5) m2 q2 Ey Dt = v2 …(6) m2

  From equations (3) and (5), by dividing the ­equations expressing x-components, we get

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 45

3

ELECTRIC DIPOLE AND DIPOLE MOMENT

q1 Ex iˆ + Ey ˆj Dt =  

v

+q

–q

O a

Axial line

a

 Dipole moment p is defined as the simple product of magnitude of either charge and the distance of separation between the two charges.   p = q ( 2a )  Dipole moment p always points from -q to +q . Its SI unit is coulombmetre ( Cm ).

( )

ELECTRIC FIELD DUE TO A DIPOLE AT A POINT LYING ON THE AXIAL LINE (END ON POSITION) The electric field due to a dipole at point P at distance r from the centre of the dipole is

Eaxial = EB - EA

⇒ Eaxial =

q 2 4pε 0 ( r - a )

-

q 2 4pε 0 ( r + a )

9/20/2019 10:09:27 AM

1.46  JEE Advanced Physics: Electrostatics and Current Electricity +q

–q EB

O

A

a

B

a

r

(r + a)

⇒   Eaxial

1 = 4pε 0

P EA

 Vectorially Eequitorial =

(r – a)

For r >> a

4 raq

Eaxial

For r >> a



Eaxial 

Eequitorial 



( r 2 - a 2 )2

( r 2 - a 2 )2

  

1 2p 4pε 0 r 3

Eequitorial = 2EA cos q where EA =

Solution

q

⇒

(Because the components EA sin q and EA sin q cancel out)

⇒ ⇒

θ 2EA cos θ θ

P EA sinθ

EB θ

O a

2 r

3

r3 r ′3

=

1 r ′3

=2

a

r = 23 r′

ELECTRIC FIELD DUE TO A DIPOLE AT ANY POINT P(r, q )

|EA| = |EB|

r 2 + a2

r

θ

1 2p 1 p = 3 4pε 0 r 4pε 0 r ′ 3

1

EA sinθ EA

A

Eaxial  2Eequitorial

The electric field due to a short dipole at a distance r, on the axial line, from its mid-point is the same as that of electric field at a distance r ′, on the equatorial line, r from its mid-point. Determine the ratio . r′

4pε 0 ( r 2 + a 2 )

+q

1 p 4pε 0 r 3

Illustration 33

The electric field due to the dipole at the point P at distance r from O is

r 2 + a2

( r 2 + a 2 )3 2

So we conclude that at points lying far away from the centre of the dipole { for r >> a }

2 pr

ELECTRIC FIELD DUE TO A DIPOLE AT A POINT LYING ON THE EQUITORIAL LINE (BROAD SIDE ON POSITION)





( -p )

Remark(s)

Since, p = q ( 2 a ). So, 1 = 4pε 0

1 4pε 0

Consider a point P ( r, q ) at a large distance r from  the centre O of the dipole at an angle q with p (as shown).

–q B

E

⇒ Eequitorial = 2

q

4pε 0 ( r 2 + a 2 )

2

r +a

1 = 4pε 0

a

⎫ ⎬ a +r ⎭ 2

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 46

r

β

E1 = Eaxial

P (r, θ )

2

⎧ from positive to ⎫ ⎨ ⎬ 3 2 ( r 2 + a2 ) ⎩ negative charge ⎭ p

β

E2 = Eequitorial

2

⎧ ⎨∵ cos q = ⎩



⇒ Eequitorial

a

Dipole 2 Dipole 1

θ

O

p cos θ p

α =θ +β

p sin θ

9/20/2019 10:09:39 AM

Chapter 1: Electrostatics 1.47

 On resolving p in two components p1 = p cos q and p2 = p sin q , we observe that the point P lies at the axial line of Dipole ➀ (having dipole moment p1 = p cos q ) and on the equatorial line of Dipole ➁ (having dipole moment p2 = p sin q ). So,

E1 =

1 2 p1 1 2 p cos q = and 3 4pε 0 r 4pε 0 r3

E2 =

1 p2 1 p sin q = 3 4pε 0 r 4pε 0 r 3

⇒ E = E12 + E22 ⇒ E=

1 p 1 + 3 cos 2 q 4pε 0 r 3

If b is the angle made by the net field E with E1 then, E 1 tan b = 2 = tan q E1 2 1 ⇒ tan b = ( tan q ) 2 So, we observe that the net field E makes an angle  a = q + b with the dipole moment p.

⇒ tan 2 q + 3 tan q - 2 = 0 Solving this Quadratic in tan q , we get tan q =

⇒ q = 29°18 ′

TORQUE ON A DIPOLE PLACED IN A UNIFORM ELECTRIC FIELD Suppose an electric dipole is placed in a uniform  external electric field E where the dipole moment makes an angle q with the field. The forces on the two charges are equal and opposite as shown, each having a magnitude F = qE Thus, we see that the net force on the dipole is zero. However, the two forces produce a net torque on the dipole, and the dipole tends to rotate such that its axis gets aligned with the field. +q O

Illustration 34

 Find the locus of all the points where the resultant E field will always have a bearing of 45° with the axis of a short dipole. Solution

p (Given) …(1) 4 1 Since, tan f = tan q (from the theory of dipole) …(2) 2 q +f =

E P O

θ

ϕ

θ + ϕ = 45° p



01_Physics for JEE Mains and Advanced - 2_Part 2.indd 47

–q

–F

F

θ E

The torque due to the force on the positive charge about an axis through O is given by Fa sin q where, a sin q is the moment arm of F about O. This force tends to produce a clockwise rotation. Likewise, the torque on the negative charge about O is also Fa sin q , and so the net torque τ about O is given by

τ = 2Fa sin q Since F = qE and p = 2 aq ⇒

τ = 2aqE sin q

⇒

τ = pE sin q

It is convenient to express the torque in vector form  as the cross product of the vectors p and E , so vectorially,

So, from (1) and (2), we get ⎛p ⎞ 1 tan ⎜ - q ⎟ = tan q ⎝4 ⎠ 2

-3 ± 17 = 0.56 or -3.56 2



iˆ    τ = p × E = px

ˆj py

kˆ pz

Ex

Ey

Ez

9/20/2019 10:09:54 AM

1.48  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 35

A charge e is placed at (1, 2 , 1) and another charge -e is placed at (0, 1, 0) such that they form an electric dipole. There exists a uniform electric field E = 2iˆ + 3 ˆj . Calculate the dipole moment vector

(

)

and torque experienced by the dipole. Solution

  (i) p = qr –e    (r1) ⇒   p = e ( r2 - r1 )  Now, r1 = ˆj  ˆ ˆ ˆ    r2 = i + 2 j + k   ˆ ˆ ˆ ⇒   r2 - r1 = i + j + k  ˆ ˆ ˆ ⇒  p= e i + j+k  and p = 3 e    (ii) τ = p × E  ˆ ˆ ˆ ˆ ˆ ⇒   τ = e i + j + k × 2i + 3 j

(

r = r2 – r1

) (

q

U = pE ( - cos q ) q = pE ( cos q0 - cos q ) 0

The term involving cos q0 is a constant that depends on the initial orientation of the dipole. It is convenient to choose q0 = 90° , so that cos q0 = cos 90° = 0 . In this case, we can express U as U = - pE cos q  This is  equivalent to the dot product of the vectors p and E . So,   U = - p ⋅ E = - ( px Ex + py Ey + pz Ez ) Illustration 36

When an electric dipole is placed in a uniform electric field making angle q with electric field, it experiences a torque τ . Calculate the minimum work done in changing the orientation to 2q .

)

(

Solution



)

τ = pE sin q

⇒ pE =

iˆ ˆj kˆ  τ = e 1 1 1 = e -3iˆ + 2 ˆj + kˆ Nm 2 3 0

(

 

+e (r2)

⇒

)



τ sin q

W = DU = - pE cos q 2 + pE cos q1

τ ( cos q - cos 2q ) sin q

POTENTIAL ENERGY OF A DIPOLE PLACED IN A UNIFORM ELECTRIC FIELD

⇒ W=

Work must be done by an external agent to rotate the dipole through a given angle in the field. This work done is then stored as potential energy in the system, that is, the dipole and the external field. The work dW required to rotate the dipole through an angle dq is given by

Illustration 37

dW = τ dq Since,

q

U=

∫

q0

q

τ dq =

In the arrangement shown, if the dipole is released from q = 53° , then calculate +q, m l θ = 53° O –q, m

τ = pE sin q This work is transformed into potential energy U . We find this for a rotation from q0 to q . So,

∫

q

∫

pE sin qdq = pE sin qdq

q0

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 48

q0

{∵ q1 = q , q 2 = 2q }

⇒ W = pE ( cos q - cos 2q ) 

E

(i) Its angular acceleration just after releasing. (ii) Its angular velocity when it passes through s­ table equilibrium. (iii) The work required to rotate it by 180° Solution

(i) τ net = pE sin ( 53° ) = Ia

9/20/2019 10:10:10 AM

Chapter 1: Electrostatics 1.49

4

( ql ) E ⎛⎜⎝ 3 ⎞⎟⎠

⇒  a=

2

2

ml 2 ⎛ l⎞ ⎛ l⎞ + m⎜ ⎟ + m⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ 12 4 qlE 16 ⎛ qE ⎞ = ⇒ a= 3 ⎜ ⎟ 7 7 ⎝ ml ⎠ ml 2 12 ( ii) By Law of Conservation of Energy, we get

The electric field is given as  E = Eiˆ    Since τ = p × E iˆ

)

y-axis

+q

θ

( K + U )initial = ( K + U )final

1 ⇒ 0 + ( - pE ( cos 53° ) ) = Iω 2 + ( - pE cos 0° ) 2 2

⇒ ω =

4 pE = 5I

4 qlE = ⎛ 7 ⎞ 5 ⎜ ml 2 ⎟ ⎝ 12 ⎠

48qE 35ml

(iii) W = U f - Ui ⇒   W = ( - pE cos ( 180° + 53° ) ) - ( - pE cos 53° ) ⎛ 4⎞ ⎛ 4⎞ W = ( ql ) E ⎜ ⎟ + ( ql ) E ⎜ ⎟ ⎝ 3⎠ ⎝ 3⎠

To calculate angular speed, applying Law of Conservation of Mechanical Energy, i.e. Ui + Ki = U f + K f   where, Ui = - p ⋅ E = - pE cos q   U f = - p ⋅ E = - pE  {∵ rod becomes parallel to x-axis}



⎛ 8⎞ ⇒   W = ⎜ ⎟ qlE ⎝ 3⎠



Illustration 38

A light massless rod of length l lies in x -y plane with its centre at origin and it makes an angle q with x-axis. A particle of mass m and charge -q is attached at its left end and another particle of mass m and charge q at the other end. Write the dipole moment vector. Now an electric field of constant magnitude E and directed along x-axis is switched on. Write an expression for the torque on the rod at the initial positon. What is the angular speed of the rod at the instant when it becomes parallel to x-axis ? Solution

The dipole moment has a magnitude p = q ( 2 a ) = ql And it is directed along a unit vector cos q iˆ + sin q ˆj  ⇒ p = ql ⎡⎣ cos q iˆ + sin q ˆj ⎤⎦

x-axis

–q

2

7 ml 2 ⎛ l⎞ ⎛ l⎞ Where I = + m⎜ ⎟ + m⎜ ⎟ = ml 2 ⎝ 2⎠ ⎝ 2⎠ 12 12

 

(

 ⇒ τ = ql cos q iˆ + sin q ˆj × Eiˆ = - ( qlE sin q ) kˆ

Ki = 0 and Kf =

2 2 1 2 1⎡ ⎛ l ⎞ ⎛ l⎞ ⎤ Iω = ⎢ m ⎜ ⎟ + m ⎜ ⎟ ⎥ ω 2 ⎝ 2⎠ ⎦ 2 2 ⎣ ⎝ 2⎠

⇒ - pE cos q = - pE + ⇒ ω=

ml 2 2 ω 4

4 qE ( 1 - cos q ) ml

SMALL OSCILLATIONS OF A DIPOLE PLACED IN A UNIFORM ELECTRIC FIELD Let the dipole possess a moment of inertia I about O . Since

τ = -pE sin q where negative sign indicates that this is the restoring torque which tries to restore dipole to its mean position. For small q , sin q ≅ q

iˆ

⇒

τ = -pEq

iˆ

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 49

9/20/2019 10:10:26 AM

1.50  JEE Advanced Physics: Electrostatics and Current Electricity Solution

ii

Further τ = Ia = I q

(a) We know that electric field at a distance x from the rod

ii

⇒

I q = - pEq

⇒

⎛ pE ⎞ q+⎜ q=0 ⎝ I ⎟⎠



ii

This equation shows that the dipole will execute oscillations when given a small angular displacement from the mean position. The time period of oscillations is T given by T = 2p

q ii

= 2p

q

I I = 2p CM pE pE

where ICM is the moment of inertia of the dipole about the centre of mass of the dipole.

FORCE ON AN ELECTRIC DIPOLE IN NON-UNIFORM ELECTRIC FIELD In a non-uniform electric field, if a dipole is placed at  a point where electric field is E, then the interaction energy of dipole at this point is given by   U = -p ⋅ E The force on dipole due to electric field is   F = -∇U For unidirectional variation in electric field, we get  d   F =p⋅E dx If the dipole is placed in the direction of electric field, then

(



)

dE F = -p dx

Illustration 39

E=

l  2pε 0 x

{linear charge distribution}

directed along the axis of the dipole. The direction can be determined by the fact that electric field is always perpendicular to equipotential surface. Net force on dipole due to electric field of rod dE F=q dx As we know potential difference = dV = -Edr Change in electrostatic Potential Energy

= dU =

dV = -Edr q

dU = - qEdr

dU pdE =  {∵ U = - pE as p  E } dr dr As dipole is lying along x-axis F=-



F=p F=

dE d ⎡ l ⎤ = 2 aq ⎢ dx dx ⎣ 2pε 0 x ⎥⎦

l aq

pε 0 x 2 (b) Workdone = Difference in P.E. at two positions ⇒ W = - pE cos 180° - ( - pE cos 0 ) ⎛ l ⎞ 2l aq ⇒ W = 2 ( 2 aq ) ⎜ 2pε x ⎟ = pε x ⎝ 0 ⎠ 0 (c) When the dipole is slightly disturbed from its equilibrium position by an angle q . Restoring torque = -pE sin q

{ sin q  q for small displacement }



An electric dipole is placed at distance x from an infinitely long rod of linear charge density l .

- l aq q lq ⇒ τ = - pEq = -2 aq × 2pε x = pε x 0 0

(a) What is net force acting on dipole? (b) What is work done in rotating dipole through 180° ? (c) If dipole is slightly rotated about its equilibrium position. Find the time period of oscillation.

Now torque is also given as

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 50

τ = Ia = I

d 2q

dt 2 where I is the moment of inertia and a is angular acceleration

9/20/2019 10:10:36 AM

Chapter 1: Electrostatics 1.51



I



d 2q dt 2

2 ⇒ 2ma

⇒

d 2q 2

=d 2q dt

+

2

l aq q  pε 0 x =

{for dipole I = 2ma 2 }

l aq q pε 0 x

ELECTRIC FORCE BETWEEN TWO DIPOLES Electrostatic force between two dipoles of dipole moments p1 and p2 lying at a separation r ( r  a ) is

l qq =0 2pε 0 max

dt Comparing it with standard equation of S.H.M. d 2q

dt 2

⇒ T =

lq 2pε 0 max

+ ω 2q = 0 , where ω = 2pε 0 max 2p = 2p ω lq

FORCE ON A DIPOLE IN THE SURROUNDING OF A LONG CHARGED WIRE In the situation shown in figure, the electric field strength due to the wire, at the position of dipole is E=

2l l = 4pε 0 r 2pε 0 r

Thus force on dipole is



l ⎤ ⎛ dE ⎞ ⎡ F = -p ⎜ = -p ⎢ 2 ⎥ ⎝ dr ⎟⎠ ⎣ 2pε 0 r ⎦

⇒ F=

lp

4

r 3 p1 p2 r4

,

when dipoles are placed coaxially to each oth her

,

when dipoles are placed perpendicular to each other

The above results have been derived here to give you a view of the fundamentals related to dipoles. Please keep in mind that the above results are obtained on the basis of assumption that the centres of the dipoles lie far apart from each other i.e., r  2 a for each one of them. CASE-1: When the dipoles are placed coaxially p1

A

1

p2

B

2

r

E

 Consider two dipoles with dipole moments p1 and  p2 (as shown) with their centres at distance r (  2 a ). The electric field of the dipole 1 on the left side exerts a net force on the dipole 2 placed to its right. The electric field due to 1 at the centre of 2 is

2pε 0 r 2

6 p1 p2

⎧ 1 ⎪ 4pε ⎪ 0 F=⎨ ⎪ 1 ⎪⎩ 4pε 0

E=

1 2 p1 {from A to B} 4pε 0 r 3

⇒ dE =

2 p1 d ( r -3 ) 4pε 0

⇒ dE = p

6 p1 4pε 0 r 4 –q

r

A

dr …(1) p2

+q C

B

C is midpoint of dipole 2 with AB = dr λ

Here negative charge of dipole is placed close to wire and hence net force on the dipole due to wire will be attractive.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 51

Let the -q charge of dipole be placed at A and the +q charge be placed at B . So, the electric field at A (where -q lies) is

EA = E + dE

9/20/2019 10:10:48 AM

1.52  JEE Advanced Physics: Electrostatics and Current Electricity

{obviously field due to dipole 1 is stronger at A in comparison to B }

as the force between the dipoles placed perpendicular to each other.

and EB = E - dE

Illustration 40

As a result of this, force on -q at A is F1 = qEA 

{towards left}

and that on +q at B is F2 = qEB  ⇒ Net force on the dipole is

Find the magnitude of the electric field at the point P in the configuration shown in figure for d  a. Take 2qa = p.

{towards right}

F = F1 - F2  ⇒ F = qEA - qEB

{towards left}

⇒ F = q [ ( E + dE ) - ( E - dE

P

P

P

d

d

d

–q

q

)]

a

(a)

a

+q

(b)

–q

+q +q (c)

⇒ F = 2q dE ⎛ 6 p1 ⇒ F = 2q ⎜ ⎝ 4pε 0 r 4 ⇒ F= ⇒ F=

Solution

⎞ ⎟⎠ dr

(a) The electric field due to a point charge at P is given by

6 p1 ( 2qdr ) 4pε 0 r 4 6 p1 p2 4pε 0 r 4



{∵ 2q ( dr ) = p2 }



CASE-2: When the dipoles are placed perpendicular to each other. 1

2 r

Proceeding exactly the same way as in case-1, we get F=

3 p1 p2

4pε 0 r 4 All assumptions here are exactly the same as before, other than this that the field due to dipole 1 at the centre of 2 will be E=

p1 3

4pε 0 r and hence we get F=

3 p1 p2 4pε 0 r 4

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 52

EP =

q 4pε 0 d 2

(b) As d  a a combination of -q and +q can be treated as a dipole. Hence, the electric field at P is

EP =

q ( 2a ) 4pε 0 d 3



(towards left)

Direction will be horizontal towards left (c) This case is the super position of above two cases  Thus we have E =  ⇒ E=

1 ⎡ q ˆ ⎛ p j+⎜ - 3 ⎝ d 4pε 0 ⎢⎣ d 2

⎞ ˆ⎤ ⎟⎠ i ⎥ ⎦

⎛ ˆ p ˆ⎞ ⎜ qi - j ⎟⎠ d 4pε 0 d ⎝ 1

2

Illustration 41

A water molecule is placed at a distance l from the line carrying linear charge density l . Find the maximum force exerted on the water molecule. The shape of water molecule and the partial charges on H and O atoms are shown in figure. Assume l to be much larger than the dimensions of the molecule.

9/20/2019 10:11:01 AM

Chapter 1: Electrostatics 1.53

l ⎛ 1 ⎞ ⎛q⎞ ⇒ Fmax = 2qd cos ⎜ ⎟ × ⎝ 2 ⎠ 2pε 0 ⎜⎝ x 2 ⎟⎠

+q

θ

λ

–2q

⇒ Fmax =

+q

⇒

Now

⎛q⎞ pnet = 2qd cos ⎜ ⎟ ⎝ 2⎠



pε 0 x 2

⎛q⎞ l qd cos ⎜ ⎟  ⎝ 2⎠ Fmax = pε 0 x 2

Solution

The figure can be resolved as combination of 2 dipoles Dipole moments of each p = qd Here total dipole moment of system is

⎛q⎞ l qd cos ⎜ ⎟ ⎝ 2⎠

Fmax =

⎛q⎞ l qd cos ⎜ ⎟ ⎝ 2⎠ l2

CONCEPT OF DISTRIBUTED DIPOLE Till now we have discussed about an electric dipole that has two equal and opposite charges separated by  a small distance 2 a as shown. Its dipole moment is   defined, as p = q ( 2 a ) from -q to +q.

   ⎛ dE ⎞ F = pnet ⎜ ⎝ dx ⎟⎠

d –q

d

+q p = q(2a)

θ

d

  dE For maximum force, the angle between pnet and dx is 0° λ

Sometimes the two charges of dipole are not concentrated at its ends. The system, instead of having two charges has more than two charges. In such a case we select a convenient origin. Then we locate the coordinates of the charges. Consider a system having charges q1 , q2 and q3 at

points ( x1 , y1 , z1 ) , ( x2 , y 2 , z2 ) and ( x3 , y 3 , z3 ) then

pnet = 2qd cos

θ 2

l

⎛q⎞⎡ d ⎛ l ⎞⎤ ⇒ Fmax = 2qd cos ⎜ ⎟ ⎢ ⎜ ⎝ 2 ⎠ dx ⎝ 2pε 0 x ⎟⎠ ⎥ ⎣ ⎦

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 53



px = q1 x1 + q2 x2 + q3 x3 , py = q1 y1 + q2 y 2 + q3 y 3

and pz = q1 z1 + q2 z2 + q3 z3 and the net dipole moment is then given by  p = px iˆ + py ˆj + pz kˆ  where p = px2 + py2 + pz2

9/20/2019 10:11:13 AM

1.54  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 42

In figure shown, an electric dipole lies at a distance x from the centre of the axis of a charged ring of radius R with charge Q uniformly distributed over it. R

m–q

C

x

+q m 2a

(a) Find the net force acting on the dipole. (b) What is the work done in rotating the dipole through 180° ? (c) The dipole is slightly rotated about its equilibrium position. Find the time period of oscillation. Assume that the dipole is linearly restrained. Solution

(a) Field at a distance x on the axis of a ring is given by E=

F=p

F=

d 2q dt

2

=-

pE 2ma 2

q

I 2ma 2 ⇒   T = 2 p = 2 p pE pE 2ma 2 ( R2 + x 2 ) ⇒   T = 2p 2 aq Qx

32

⇒  T=

dE d ⎡ x 1 ⎤ Q = 2 aq dx 4pε 0 dx ⎢ ( R2 + x 2 )3 2 ⎥ ⎣ ⎦

aQq R2 - 2x 2 2pε 0 ( R2 + x 2 )5 2

(b) Work done in rotating a dipole is equal to change in its potential energy.

Illustration 43

On a half ring a charge +q is uniformly distributed and another equal negative charge -q is placed at the centre of ring. +q

In this system negative charge is a point charge and positive charge is distributed on the ring. This system is called, distributed dipole. Calculate the dipole moment of this system. Solution

For this we consider a polar element of angular width dq at an angle q from the vertical axis as shown.

dp

–q

aQqx 2 ( 2a ) Qx = ⇒   W = 4pε 32 32 2 2 0 (R + x ) pε 0 ( R2 + x 2 ) (c) Restoring torque τ = - pE sin q = - pEq (for small q)



dt

2

dq

θ dθ

⇒   W = 2 pE

d 2q

R

–q

    W = DU   ⇒   Ui = - p ⋅ E = - pE cos ( 0° ) = - pE   and U f = - p ⋅ E = - pE cos ( 180° ) = + pE

     τ=I

4pε 0

32 16p 2 ε ma ( 2 R + x2 ) qQx

1 Qx 4pε 0 ( R2 + x 2 )3 2

The net force on the dipole is



⇒ 

= 2ma 2

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 54

d 2q dt 2

Charge on this element is dq =

q dq p

Now in the point charge -q , we consider an element charge -dq which forms a dipole with the polar element of charge +dq . This element dipole moment can be given as

9/20/2019 10:11:27 AM

Chapter 1: Electrostatics



dp = dqR =

q Rdq p

+

⇒

Total dipole moment of system can be given by integrating all such elemental dipole moments as

ptotal =

∫

ptotal =

p 2

q

∫ p R cosq dq

-

p 2

+

dp cos q

⇒

{The components dp sin q cancel out}

ptotal

1.55

p

2 qR [ sin q ] = p p 2

⇒

ptotal =

2qR p

Test Your Concepts-IV

Based on Dipoles and Dipole Moment 1. Three charges -2q, q and q are arranged in the configuration shown. Calculate resultant dipole moment of the system. +q

θ

–2q

(Solutions on page H.15) 4. Figure shows an assembly of two charged rods each of mass m, length L connected by non-conducting massless rods of length d. The system is free to rotate about the axis shown. The system is given an anticlockwise small angular displacement. Show that the motion is simple harmonic and determine its time period.

+q

2. Three point charges 2q, q and -q are located respectively at ( 0, a, a ) , ( 0, - a, a ) and ( 0, 0, - a ) as shown. Find the dipole moment of this distribution. z q

2q

O x

y

–q

3. A small freely oriented dipole with dipole moment p lies at the centre of a uniformly charged hemisphere, charge density s and radius R. Determine the potential energy of the dipole and the period of small oscillations about an axis perpendicular to the axis of the hemisphere. The moment of inertia of the dipole about the rotation axis is Ι .

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 55

E

+λ

Free to rotate

–λ

What is the work done required to turn the system p through an angle (a) (b) p ? 2 5. Figure shows a charged rod of mass m bent in the form of an arc of a circle of radius R. The charge distribution on the rod is shown in the figure. The assembly is kept in a uniform electric field.

E

+λ + + + + – – – –λ –

Non conducting massless support θ0 θ0

Pivot

9/20/2019 10:11:32 AM

1.56  JEE Advanced Physics: Electrostatics and Current Electricity



(a) Show that for small angular displacement the system will perform SHM. Determine its time period. (b) Considering the size of the system to be very small, determine the work done required to turn it through 180°. Neglect any displacement of centre of charge. 6. A dipole with an electric moment p is located at a distance r from a long thread charged uniformly with a linear density l . Find the force F acting on  the dipole if the vector p is oriented (a)  along the thread  (b)  along the radius vector r (c) at right angles to the thread and the radius vec tor r . 7. A thin non-conducting ring of radius R has a linear charge density l = l0 cosq , where l0 is the value of

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 56

l at q = 0. Find the net electric dipole moment for this charge distribution. λ θ R

λ0

8. A positive point charge q is fixed at origin. A small  dipole with a dipole moment p is placed along  the x-axis far away from the origin with p pointing along positive x-axis. Find (a) the kinetic energy of the dipole when it reaches a distance d from the origin, and (b) the force experienced by the charge q at this moment.

9/20/2019 10:11:34 AM

Chapter 1: Electrostatics 1.57

Gauss’s Law and Applications AREA AS A VECTOR The physical quantity, area is a pseudo vector and we treat area of a surface as a vector quantity whose direction is along the outward normal to the surface.  The area vector A of a surface which has surface area A can be written as  A = Anˆ A



fE = EA Area = A

E

n

Where nˆ is the unit vector in the direction along outward normal to the surface. When a surface is three dimensional we consider a small infinitesimal area element dA on this surface and direction of this elemental area vector is along the local outward normal to the surface at the point where elemental area is chosen as shown. n dA

So,

E and surface area A perpendicular to the field is called the electric flux fE (Greek phi)

 A = ( dA ) nˆ

Here nˆ is the unit vector in the direction along the outward normal at elemental area dA.

Field lines representing a uniform electric field penetrating a plane of area A perpendicular to the field. The electric flux f E through this area is equal to EA.

From the SI units of E and A , we see that fE has SI units Nm 2 C -1. Electric flux is proportional to the number of electric field lines penetrating some surface. If the surface under consideration is not perpendicular to the field, the flux through it must be less than EA. We can understand this by considering figure, where the normal to the surface of area A is at an angle q to the uniform electric field. Note that the number of lines that cross this area A is equal to the number that cross the area A ′ , which is a projection of area A onto a plane oriented perpendicular to the field. From figure we see that the two areas are related by A ′ = A cos q . Normal

Electric Flux Consider an electric field that is uniform in both magnitude and direction, as shown in figure. The field lines penetrate a rectangular surface of area A, whose plane is oriented perpendicular to the field. Since the number of lines per unit area (in other words, the line density) is proportional to the magnitude of the electric field. Therefore, the total number of lines penetrating the surface is proportional to the product EA. This product of the magnitude of the electric field

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 57

A θ

θ

E

A′ = A cos θ Field lines representing a uniform electric field penetrating an area A that is at an angle q to the field. Because the number of lines that go through A′ is the same as the number that go through A, the flux through A′ is equal to the flux through A and is given Φ E = EA cos q.

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1.58  JEE Advanced Physics: Electrostatics and Current Electricity

Because the flux through A equals the flux through A ′ , we conclude that the flux through A is

fE = EA ′ = EA cos q …(1) From this result, we see that the flux through a surface of fixed area A has a maximum value EA when the surface is perpendicular to the field (when the normal to the surface is parallel to the field, that is, q = 0° in figure); the flux is zero when the surface is parallel to the field (when the normal to the surface is perpendicular to the field, that is q = 90° ). dA θ

Ei

region, so that one cannot move from one region to the other without crossing the surface. The surface of a sphere, for example, is a closed surface. Consider the closed surface in figure. The vectors  D Ai point in different directions for the various surface elements, but at each point they are normal to the surface and, by convention, always point outward. For element ➀, where the field lines are crossing the surface from outside to inside, 180° > q > 90° and the flux is negative because cosq is negative. For element ➁, the field lines graze the surface  (perpendicular to the vector D A2 ); thus, q = 90° and the flux is zero. E



A small element of surface area  dA. The electric field makes an angle q with the vector dA, defined as being normal to the surface element, and the flux through the element is equal to EdA cos q.

fE =

∫

2

  E ⋅ dA …(2)

Δ A1

surface Equation (2) is a surface integral, which means it must be evaluated over the surface in question. In general, the value of Φ E depends both on the field pattern

and on the surface. Using the symbol

∫

Δ A3

to represent



∫

  E ⋅ dA =

∫ E dA …(3) n

where En represents the component of the electric field normal to the surface. If the field is normal to the surface at each point and constant in magnitude, the calculation is straightforward.

Remark(s) (a) We are often interested in evaluating the flux through a closed surface, which is defined as one that divides space into an inside and an outside

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 58

3

2

1

θ

E

E En

En θ

E

Δ A2

A closed surface in an electric field. The area vectors DAi are, by convention, normal to the surface and point outward. The flux through an area element can be negative (element ➀), zero (element ➁), or positive (element ➂).

an integral over a closed surface, we can write the net flux fE through a closed surface as

fE =

3

1



For element ➂, the field lines are crossing the surface from the inside to  the  outside and q < 90°; hence, the flux DfE = E ⋅ DA1 through this element is positive. (b) The net flux through the surface is proportional to the net number of lines leaving the surface, where the net number means the number leaving the surface minus the number entering the surface. If more lines are leaving than entering, the net flux is positive. If more lines are entering than leaving, the net flux is negative.

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Chapter 1: Electrostatics 1.59 Illustration 44

Calculate the electric flux through the surface of the disc of radius R due to a point charge q placed at a point P on its axis at a distance l from its centre.

ql ⇒ f= 2ε 0

3 ⎤ ⎡ R ⎢ 1 ( l 2 + x 2 ) 2 2x dx ⎥ ⎢2 ⎥ ⎣ 0 ⎦

∫

Solution

R

Since a point charge q has electric field originating in radially outward direction.

α

l

q cos α =

Since,

q l

To calculate this flux, we consider an infinitesimal ring on disc surface of radius x and width dx as shown. Area of this strip is

⎡⎣ f ( x ) ⎤⎦ ⎡⎣ f ( x ) ⎤⎦ f ′ ( x ) dx = n+1

E=

q 1 2 4pε 0 ( x + l 2 ) R

⇒ f=-

dA

l

If df is the flux due to this elemental ring, then   d f = E ⋅ dA = EdA cos q q l 1 ⎛ ⎞ ⇒ df = 2p x dx ) ( 2 2 ⎜ 4pε 0 ( l + x ) ⎝ l 2 + x 2 ⎟⎠

⇒ f=

∫

ql df = 2ε 0

R

∫ (l 0

x dx 2

+ x2 )

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 59

32

R 0

ql ⎡ 1 1 ⎤ 2 2 ⎥ 2ε 0 ⎢⎣ l R +l ⎦

⇒ f=

q 2ε 0

⎛ 1⎜⎝

0

, n ≠ -1

⎤ ⎥ ⎥ ⎥ ⎥⎦

⎤ ⎥ ⎥⎦

⇒ f=

q ⎞ ( 1 - cos a ) = ⎟⎠ 2ε 0 R +l l

2

2

Illustration 45

A point charge Q is placed at the corner of a square of side a , then find the flux through the square.

dr

ql x dx ⇒ df = 2ε 0 ( l 2 + x 2 ) 3 2

ql ⎡ 1 ⎢ 2ε 0 ⎢ l 2 + x 2 ⎣

R

n+1

E

x

θ

q

θ

n

⎡ 3 - +1 ql ⎢ 1 ( l 2 + x 2 ) 2 So, f = ⎢ 3 2ε 0 ⎢ 2 - +1 ⎢⎣ 2

dA = ( 2p x ) dx

The electric field due to q at this elemental ring is



∫

l l2 + R2

Q

a

a

Solution

The electric field due to Q at any point of the square will be along the plane of square and the electric field line are perpendicular to square, so f = 0. In other words we can say that no line is crossing the square so flux = 0.

9/20/2019 10:12:01 AM

1.60  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 46

y

A closed surface with dimensions a = b = 0.4 m and c = 0.6 m is located as in figure. The left edge of the closed surface is located at position x = a . The electric field throughout the region is non-uniform and given by E = ( 3 + 2x 2 ) iˆ NC -1 , where x is in meter.

Q

E

a

z

2l z

(b) Using the result obtained in (a), if the charge +Q is now at the centre of a cube of side 2l, what is the total flux emerging from all the six faces of the closed surface?

y

c

x

b

Solution

The electric field throughout the region is directed along x, therefore, E will be perpendicular to dA over the four faces of the surface which are perpendicular to the yz plane, and E will be parallel to dA over the two faces which are parallel to the yz plane. Therefore

Solution

(a) The electric field due to the charge +Q is given by  E=

   

1 Q 1 Q ⎛ xiˆ + yjˆ + zkˆ ⎞ rˆ = ⎜ ⎟⎠ 2 4pε 0 r 4pε 0 r 2 ⎝ r y

c

Q



(

x= a

(

2

) ab

fE = 2 abc ( 2 a + c )

Q = ε 0fE = 2.38 × 10 -12 C = 2.38 pC

Illustration 47

(a) Compute the electric flux through a square surface of edges 2l due to a charge +Q located at a perpendicular distance l from the center of the square, as shown in figure.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 60

2l

where r = x 2 + y 2 + z 2

Substituting the given values for a , b and c , we find fE = 0.269 Nm 2 C -1

S

z

) A + ( Ex x = a+ c ) A

⇒ fE = - ( 3 + 2 a 2 ) ab + ( 3 + 2 ( a + c ) ⇒

x

l

x

z

fE = - Ex

n

r

E

a a

E

dA

y b

x l

(a) Calculate the net electric flux leaving the closed surface. (b) What net charge is enclosed by the surface? x=a

2l

S

)

12

in Cartesian coordi-

nates. On the surface S , x = l and the area ele ment is dA = dAiˆ = ( dydz ) iˆ . So, ⎛ xiˆ + yjˆ + zkˆ ⎞ ⎟⎠ ⋅ ( dydz ) iˆ 2 ⎜ ⎝ r r pε 4 0      Ql ⇒ E ⋅ dA = dydz 4pε 0 r 3   E ⋅ dA =

Q

Thus, the electric flux through S placed at x = l is

fE =   

∫ S

  Ql E ⋅ dA = 4pε 0

l

l

-l

-l

∫ dz∫ ( l

dy 2

+ y 2 + z2

)

32

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Chapter 1: Electrostatics 1.61

Ql ⇒ fE = 4pε 0 Ql ⇒ fE = 2pε 0

l

l

∫ dz ( l -l

2

(

+ z2 ) l2 + y 2 + z2

l

∫ (l -l

)

12

CONCEPT OF SOLID ANGLE

+ z 2 ) ( 2l 2 + z

2

the charge enclosed. In addition, the result further strengthens the notion that Φ E is independent of the shape of the closed surface.

-l

l dz

Q -1 ⎛ ⇒ fE = 2pε tan ⎜ ⎝ 0 Q ⇒ fE = 2pε 0

y

z z 2 + 2l 2

⎞ ⎟⎠

)

2 12

The three dimensional angle enclosed by the lateral surface of a cone at its vertex (as shown in figure) is the Solid Angle.

l

-l

S

1 ⎞⎤ ⎡ -1 ⎛ 1 ⎞ -1 ⎛ ⎟⎥ ⎟⎠ - tan ⎜⎝ ⎢ tan ⎜⎝ 3 3⎠⎦ ⎣

R

Q ⎛p p⎞ Q ⇒ fE = 2pε ⎜⎝ 6 + 6 ⎟⎠ = 6ε 0 0

Ω (Solid angle)

In evaluating the above integrals, the following results have been used.



∫ (x



∫ (x

dx 2

+a

)

2 32

=

dx 2

+a

2

) ( x 2 + b 2 )1 2 1

12 a ( b 2 - a2 )



Solid angle is also defined as the three dimensional angle subtended by a spherical section at its centre of curvature. As in the figure shown point C is the centre of curvature of a spherical section S of radius R which subtends a solid angle Ω (Greek symbol omega) at point C .

x a2 ( x 2 + a

)

2 12

=

tan -1

b 2 - a2

a2 ( x 2 + b 2 )

, b 2 > a2

y E dA n

r Q

2l

x S1

z

2l

2l

(b) From symmetry arguments, the flux through each face must be the same. Thus, the total flux through the cube is just six times that through one face. So,

⎛ Q ⎞ Q ΦE = 6 ⎜ = ⎝ 6ε 0 ⎟⎠ ε 0

The result shows that the electric flux Φ E passing through a closed surface is proportional to

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 61

C

SOLID ANGLE SUBTENDED DUE TO A RANDOM SURFACE AT A GIVEN POINT Consider a random surface S as shown in figure. To find the solid angle subtended by this surface at a point P , let us join all the points of the periphery or boundary of the surface S to the point P by straight lines. This gives a cone with vertex at P . Now by taking centre at P, we construct several spherical sections on this cone of different radii as shown. Let the area of spherical section which is of radius r1 be A1 and the area of section of radius r2 be A2 . A1

S

r1 P

A2

Ω r2

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1.62  JEE Advanced Physics: Electrostatics and Current Electricity

Mathematically we observe that the ratio of area of any sphere enclosed by the cone to the square of radius of that sphere is a constant and this constant is called solid angle Ω . In the figure shown the solid angle Ω subtended at P is given by Ω=

A1 r12

=

A2

r22 Solid angle is a dimensionless physical quantity and its S.I. unit is steradian. One steradian is the solid angle subtended at the centre of sphere by the surface of the sphere having area equal to square of the radius of sphere. Also, Ωmax =



Area of the sphere

( Radius of sphere )2

=

Consider a spherical section S of radius R , which subtends a half angle f (in radian) at the centre of curvature. To find the area of this section, we consider an infinitesimal circular strip on this section having angular width dq and radius R sin q . The surface area dA of this strip is

dA = ( 2p R sin q )( Rdq ) Rsin θ Rdθ

r2

dθ

R θ

SOLID ANGLE SUBTENDED BY A SURFACE NOT NORMAL TO AXIS OF CONE OR NOT LYING ON SURFACE OF SPHERE  Consider a small surface AB of area dA as shown. Let PQ be the axis of cone formed by this surface. Here PQ is not normal to the surface 1 (in light grey shade). The solid angle Ω subtended at point P is given by dΩ =



dA cos q r2

…(1)

1

dA cos θ dA

dΩ

 where q is the angle between surface area vector dA and the axis of cone PQ as shown in the Figure. Since, the total solid angle that any closed surface can subtend is 4p steradian. So, r2

O

The total area of spherical section is calculated by integrating the area of this elemental strip within limits from 0 to f . Total area of spherical section is f

A=

∫ dA = ∫ 2p R sin q dq 2

0

⇒ A = 2p R2 ( 1 - cos f ) …(1)

dA

θ

dA cos q

Ω

f

2

∫

ϕ

⇒ A = 2p R2 [ - cos q ]0

θ

P

S

4p r 2

Ωmax = 4p steradian

⇒



RELATION BETWEEN HALF ANGLE OF CONE AND SOLID ANGLE AT VERTEX

= 4p

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 62

If this section subtends a solid angle Ω at its centre O is Ω then, by definition Ω=

A

…(2) From (1) and (2), we get R2

Ω = 2p ( 1 - cos f ) …(3) This equation gives the relation in half angle of a cone f and the solid angle enclosed by the lateral surface of cone at its vertex.

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Chapter 1: Electrostatics 1.63

ELECTRIC FLUX PRODUCED BY A POINT CHARGE The figure shows a point charge placed at the centre of a spherical surface of radius R . For convenience and picture clarity only lower half of sphere is drawn in the figure.

At every point on spherical surface the magnitude of electric field remains same and hence we have

f= ⇒ f= ⇒ f=

dA dA E

Since the charge q lies inside the sphere, so the total flux that originates comes out from the spherical surface. To find the total flux, let us consider an elemental area dA on surface. The electric field on the points on surface of sphere is 1 q E= 4pε 0 R2

The electric flux coming out from the surface dA is   df = E ⋅ dA = EdA cos 0    {because E and dA are parallel} ⇒ df =

q 4pε 0 R2

q 4pε 0 R2 q 4pε 0 R2

∫ dA ( 4p R2 )

q ε0

q Thus total flux, originating from charge q is . ε0 q electric lines (flux) Similarly a charge -q absorbs ε0 into it. Figure shows a charge q enclosed in a closed surface S of random shape. Here we can say that the total electric flux emerging out from the surface S is the flux that originates from charge q and hence flux emerging from surface is

fS =

q ε0

Please note that, the above result is independent of the shape of surface it only depends on the amount of charge enclosed by the surface. Also field lines leaving a surface will give positive flux and the field lines entering a surface give negative flux. Illustration 48

A point charge q is fixed at the origin. Calculate the electric flux through the infinite plane y = a.

dA

Solution

Consider a ring of radius r on the plane y = a. The thickness of ring is dr. Electric field on the ring due to charge q is, q

S

E=

q 1 2 4pε 0 ( r + a 2 ) θ E

dr

r

a

So, total flux coming out from the spherical surface is

f=

∫

df =

y

q

∫ 4pε R 0

plane y=a

2

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 63

dA z

q

x

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1.64 JEE Advanced Physics: Electrostatics and Current Electricity

Area of the ring dA = ( 2p r ) dr cos q =

Total flux passing through the plane is given by

a 2

∞

fe =

2

r +a Flux passing through the infinitesimal ring is



q 1 a dfe = EdA cos q = 2p r dr ) 2 2 ( 2 4pε 0 r + a r + a2

⇒

q ar dr dfe = 2ε 0 ( r 2 + a 2 )3 2

⇒

∫ 0

qa dfe = 2ε 0

∞

∫ (r 0

r dr 2

+ a2 )

32

⎧ ⎪ ⎨∵ ⎪⎩

q fe = 2ε 0

∞

∫ (r 0

r dr 2

+ a2 )

32

=

⎫ 1⎪ ⎬ a⎪ ⎭

Test Your Concepts-V

Based on Flux 1. Two point charges q and -q separated by a distance 2. Find the flux of electric field strength vector across the circle of radius R placed with its centre coinciding with the midpoint of line joining the two charges in the perpendicular plane. R

+q

E

R

H

–q

l

l

2. A point charge q is kept at the centre of the cylinder of length L and radius R. Calculate the flux through the curved surface of the cylinder. 3. A very long uniformly charged thread oriented along the axis of a circle of radius R rests on its centre with one of the ends. The charge of the thread per unit length is equal to l . Find the flux of the vector E across the circle area. 4. Consider the situation shown in figure. A point charge q is placed at a depth h = 3 R exactly below the centre of mouth of a vessel whose open end is circular having a radius R. Calculate the electric flux through the lateral surface of this vessel. R h q

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 64

(Solutions on page H.18) 5. A cylinder of height H and radius R is placed in a uniform electric field as shown in figure. Find flux crossing the cylinder.

6. A cylinder of height H and radius R is placed in a uniform electric field (as shown) such that the axis of the cylinder makes an angle q with the vertical. Calculate the flux through the cylinder. E

θ

7. A non-uniform electric field is given by the expression E = ayiˆ + bzjˆ + cxkˆ , where a, b, and c are constants. Determine the electric flux through a rectangular surface in the xy plane, extending from x = 0 to x = . 8. An electric field having a magnitude of 3.5 kNC -1 is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.35 m wide and 0.7 m long, assuming that the plane

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Chapter 1: Electrostatics 1.65



(a)  is parallel to the yz plane. (b)  is parallel to the xy plane. (c) contains the y-axis, and its normal makes an angle of 60° with the x-axis. 9. On a day, above the Earth’s surface, when a thunderstorm is brewing a vertical electric field of magnitude 2 × 104 NC -1 exists. A car having a rectangular size of 6 m by 3 m is travelling along the road sloping downward at 60°. Determine the electric flux through the bottom of the car. 10. A loop of radius 40 cm is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 5.04 × 105 Nm2 C -1 . Calculate the magnitude of the electric field. 11. Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.80 × 104 NC -1 as shown.

13. A cone with base radius R and height h is located on a horizontal table. A horizontal uniform field E penetrates the cone, as shown in Figure. Determine the electric flux that enters the left-hand side of the cone.

h E R

14. A point charge Q is located just above the center of the flat face of a hemisphere of radius R as shown in figure . What is the electric flux through

30 cm

10 cm

δ

60°

Calculate the electric flux through the (a)  vertical rectangular surface. (b)  slanted surface. (c)  entire surface of the box. 12. A pyramid with horizontal square base, 6 m on each side, and a height of 4 m is placed in a vertical electric field of 52 NC -1. Calculate the total electric flux through the pyramid’s four slanted surfaces.



↓

E

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 65

R

(a)  the curved surface and (b)  the flat face?

ELECTRIC FLUX CALCULATION DUE TO A POINT CHARGE USING SOLID ANGLE Consider a point charge q, placed at a distance l from the centre of a circular disc of radius R. Let us find the electric flux passing through the surface of disc due to the charge q.

Q

0

2

R l2 +

R

ϕ

q

l Ω

9/20/2019 10:12:34 AM

1.66  JEE Advanced Physics: Electrostatics and Current Electricity

q originates from a point charge q ε0

Since a total flux in all directions

R

q We can say that from a point charge q , flux origiε0 nates in a solid angle 4p .

b

Q

Solution

Q 2ε 0

Since fE , disc =

According to the problem, we have

q

fE , disc = The solid angle enclosed by cone subtended by the disc at the point charge is given by

Ω = 2p ( 1 - cos f )

l ⎛ ⎞ ⇒ Ω = 2p 1  ⎜⎝ 2 2 ⎟ l +R ⎠

⎧ ⎨∵ cos f = ⎩

⇒

Q Q = 4 ε 0 2ε 0

⇒

1 b = 12 2 R + b2

l

⎫ ⎬ l +R ⎭ 2

2

1⎛ Q ⎞ 4 ⎜⎝ ε 0 ⎟⎠

⇒

b2 2

R +b

2

=

Total flux associated with ⎛ ⎞ ⎜ closed surface due to the charrge ⎟ ⎛ Solid angle ⎞ f=⎜ ⎟⎠ ⎜ subtended ⎟ ⎝ ⎝ ⎠ 4p

⇒ R= 3b



⎞ ⎟⎠ Ω

⇒ fdisc =

q l ⎛ ⎞ × 2p 1 ⎜⎝ 2 2 ⎟ 4pε 0 l +R ⎠

⇒ fdisc =

q 2ε 0

⎛ 1⎜⎝

2

R +b

⎞ ⎟⎠

Illustration 50

A point charge q is placed on the top of a cone of semi vertex angle q. Show that the electric flux associated with the base of the cone due to the charge is q ( 1 - cos q ) . 2ε 0

C q θ θ R

q ⎞ = 1 - cos f ) ⎟⎠ 2ε 0 ( l +R l

2

2

1 4

⇒ 4b 2 = R 2 + b 2

⎛ q ⎞ ⎜⎝ ε ⎟⎠ ⎛ q 0 ×Ω = ⎜ = 4p ⎝ 4pε 0

b

⎛ 1⎜⎝

The flux of q associated with the surface of the disc

fdisc

b ⎤ ⎡ ⎢1 2 2 ⎥ R +b ⎦ ⎣

2

A

B

Illustration 49

A point charge Q is located on the axis of a disk of radius R at a distance b from the plane of the disk. Show that if one fourth of the electric flux from the charge passes through the disk, then R = 3b .

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 66

Solution

Consider a Gaussian surface, a sphere with its centre at the top and radius the slant length of the cone. The q flux through the whole sphere is . Therefore, the ε0

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Chapter 1: Electrostatics 1.67

flux through the base of the cone can be calculated by using the following formula,

f=

q Ω 4pε 0 q ( 1 - cos q ) 2ε 0

GAUSS’S LAW

Illustration 51

Two charges +q1 and -q2 are placed at A and B respectively. A line of force emanates from q1 at angle a with the line AB. At what angle will it terminate at -q2 ?

A

α

q1 ⎛a⎞ sin ⎜ ⎟ ⎝ 2⎠ q2

⎡ q ⎛a⎞⎤ ⇒ b = 2 sin -1 ⎢ 1 sin ⎜ ⎟ ⎥ ⎝ 2⎠⎥ ⎢⎣ q2 ⎦

Since, Ω = 2p ( 1 - cos q ) ⇒ f=

⎛ b⎞ ⇒ sin ⎜ ⎟ = ⎝ 2⎠

β

q1

This law gives the mathematical relation between the electric flux associated with a closed surface (number of field lines entering or leaving a closed surface) and the charge enclosed by the surface.

q1 q4

B

q2

–q3

A line can leave +q1 in a cone of apex angle a and then enter -q2 in a cone of apex angle b. q So, flux due to the charge +q1 is f1 = 1 ( 1 - cos a ) 2ε 0 and that due to the charge -q2 is f2 =

α α

q2 ( 1 - cos b ) . 2ε 0

β β

B –q2

Since, we know that only one line is leaving q1 to enter -q2. So, we can say

⇒

–q8

S

Solution

+q1 A

q5

q2

f1 N1 = =1 f2 N 2 q1 q ( 1 - cos a ) = 2 ( 1 - cos b ) 2ε 0 2ε 0

⎡ ⎡ ⎛a⎞⎤ ⎛ b⎞⎤ ⇒ q1 ⎢ 2 sin 2 ⎜ ⎟ ⎥ = q2 ⎢ 2 sin 2 ⎜ ⎟ ⎥ ⎝ ⎠ ⎝ 2⎠⎦ 2 ⎦ ⎣ ⎣

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 67

–q6

q7

Gauss’s Law states that the total flux due to a closed 1 surface is equal to the times the total charge ε0 enclosed by the surface. Mathematically Gauss’s Law is written as



  

∫ S

  Σq E ⋅ dA = enc ε0

Here the sign

∫

represents the integration over a

closed surface S which encloses a total charge Σqenc . Let us consider a surface S shown which encloses four charges q1 , -q3 , q4 and -q8 . For the surface S , if we find surface integral of electric field   E ⋅ dA , it gives the total electric flux coming out

∫ S

from the surface, which can be given as



∫ S

  1 E ⋅ dA = ( q1 + q4 - q3 - q8 )  ε0

{Gauss’s Law}

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1.68  JEE Advanced Physics: Electrostatics and Current Electricity

Remark(s)

 (a) Here electric field E is the net electric field at the points on the surface of S. Remember that the electric field we use to find the flux must be the net electric field of the system due to all the charges inside and outside the Gaussian surface, but the total flux associated with the surface is due to the charges that are enclosed inside the closed surface i.e. the Gaussian surface. (b) Using Gauss’s Law we can find electric field strength due to some symmetrical distribution of charges. (c) For application of Gauss’s Law, we choose a closed imaginary surface over which we apply Gauss’s Law, called Gaussian surface. (d) Gauss’s Law can be used to calculate electric field strength, for this we first choose a proper Gaussian surface on which the electric field strength is to be calculated. (e) Sometimes, when a random Gaussian surface   is chosen then the integral E ⋅ dA involves

∫

complex calculations. To make these calculations easier, we choose a Gaussian surface keeping following points in mind. (i)  The Gaussian surface should be chosen in such a way that at every point of surface the magnitude of electric field is either uniform or zero. (ii) The surface should be chosen in such a way that at every point of surface electric field strength is either parallel or perpendicular to the surface. Keeping the above two points e) (i) and e) (ii) in mind, let us conclude what type of Gaussian surface should be selected in different situations A. SPHERICAL SYMMETRY Make the Gaussian surface a concentric sphere.

R r

Gaussian surface

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 68

B. CYLINDRICAL SYMMETRY Make the Gaussian surface a coaxial cylinder.

Gaussian surface

C. PLANE SYMMETRY Use a Gaussian Pillbox which passes through the surface. Gaussian pillbox

Illustration 52

Four closed surfaces, S1 , S2 , S3 and S4 , together with the charges -2Q , Q , and -Q are sketched in figure. Find the electric flux through each surface. S1 –2Q

S4

S3

+Q –Q S2

Solution

fE =

qenc ε0

fE =

-2Q + Q Q =ε0 ε0

fE =

+Q - Q =0 ε0

Through S1

Through S2

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Chapter 1: Electrostatics 1.69

Through S3

fE =

fill solid angle equal to one-eighth of a sphere as seen 1⎛ q ⎞ from q , and together pass flux ⎜ ⎟ . 8 ⎝ ε0 ⎠

-2Q + Q - Q 2Q =ε0 ε0

Through S4

Each face containing a intercepts equal flux going into the cube so,

fE = 0

Φ E, net = 0



Conceptual Note(s) (a) Electric field intensity at the Gaussian surface is due to all the charges present inside as well as outside the Gaussian surface. (b) Flux through Gaussian surface is independent of its shape. (c) Flux through Gaussian surface depends only on total charge present inside Gaussian surface. (d) Flux through Gaussian surface is independent of position of charges inside Gaussian surface. (e) Flux due to field lines entering a closed surface is taken as negative and flux due to field lines leaving a surface is taken as positive. This is because nˆ is taken positive in outward direction. (f) In a Gaussian surface f = 0 does not imply E = 0 at every point of the surface but E = 0 at every point implies f = 0.

Illustration 53

The line ag in figure is a diagonal of a cube. A point charge q is located at the vertex a of the cube. Determine the electric flux through each of the sides of the cube which meet at the point a. d

⇒ 3Φ E , abcd +

ELECTRIC FIELD DUE TO A POINT CHARGE q AT A POINT P AT DISTANCE x To find electric field strength at P , we first consider a Gaussian surface so that point P lies on its surface. dA E P q

E=

e

g

Solution

No charge is inside the cube. The net flux through the cube is zero. Positive flux comes out through the three faces meeting at g . These three faces together

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 69

1 q  4pε 0 x 2

∫

  q E ⋅ dA = ε0 q

∫ EdA cos ( 0° ) = ε ∫

⇒ E dA = f

x

{directed radially outwards} According to Gauss’s Law

⇒ h

q

x

Here, the Gaussian surface taken has to be spherical such that at every point on this surface electric field due to the charge q is



a

q 24 ε 0

⇒ Φ E , abcd = -

c b

q =0 8ε 0

q ε0

⇒ E ( 4p x 2 ) = ⇒ E=

0

q ε0

1 q 4pε 0 x 2

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1.70  JEE Advanced Physics: Electrostatics and Current Electricity

Also, we observe that for the Gaussian surface  selected the electric field vector is parallel to dA and also the magnitude of E is uniform at every point,   thus the integral E ⋅ dA can be easily evaluated.

∫

Problem Solving Technique(s)

ELECTRIC FIELD STRENGTH OF A CHARGED CONDUCTING SPHERE OR CONDUCTING SHELL CASE-1: OUTSIDE ( x > R ) To find electric field at an outer point at a distance x ( > R ) from the centre of sphere, we consider a spherical Gaussian surface of radius x. E dA

Gauss’s Law provides a convenient tool for evaluating electric field. However, its application is limited only to systems that possess certain symmetry, namely, systems with cylindrical, planar and spherical symmetry. In the table below, we give some examples of systems in which Gauss’s Law is applicable for determining electric field, with the corresponding Gaussian surfaces: System

Symmetry

Gaussian Surface

Sphere, Spherical shell

Spherical

Concentric Sphere

Infinite rod

Cylindrical

Coaxial Cylinder

Infinite plane

Planar

Gaussian “Pillbox”

P Q

R

Due to symmetry the electric field strength at every point of this surface is E . Using Gauss’s Law we have   q E ⋅ dA = enc ε0 1 ⇒ EdA cos 0 = ( qenc ) ε0

∫ ∫

The following steps may be useful when applying Gauss’s Law.

Here we have

(a) Identify the symmetry associated with the charge distribution. (b) Determine the direction of the electric field, and a “Gaussian surface” on which the magnitude of the electric field is constant over portions of the surface. (c) Divide the space into different regions associated with the charge distribution. For each region, calculate qenc , the charge enclosed by the Gaussian surface. (d) Calculate the electric flux fE (or f ) through the Gaussian surface for each region. q (e) Now fE = enc (due to Gauss’s Law) and deduce ε0 the magnitude of the electric field. Let us apply this technique to the situations discussed hereafter.



01_Physics for JEE Mains and Advanced - 2_Part 2.indd 70

E

∫

dA =

Q ε0

⇒ E ( 4p x 2 ) =

Q Q   ⇒  E= ε0 4pε 0 x 2

CASE-2: AT THE SURFACE Similarly for points on the surface, we can consider a spherical Gaussian surface of radius R which gives electric field strength on the sphere surface as E=

Q

4pε 0 R2 In both these cases, it seems as if the net field is as if the charge Q is concentrated at the centre of the sphere. CASE-3: INSIDE THE SPHERE ( x < R ) To find electric field strength at an interior point of the sphere, we consider an inner spherical Gaussian surface of radius x ( x < R ) .

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Chapter 1: Electrostatics 1.71 Q

C

P



∫

{as all the charge resides on surface so qenc = 0}

One important thing that we observe regarding the electric field due to conducting sphere is that it exhibits discontinuity at x = R where as the field due to the non conducting uniformly charged sphere is continuous in nature and the field due to a uniformly charged sphere is given by

ELECTRIC FIELD DUE TO A NON-CONDUCTING UNIFORMLY CHARGED SPHERE CASE-1 and 2: For points outside and at the surface of sphere, the electric field strength is calculated by using Gauss’s Law similar to the previous case of conducting sphere. Here too we get the net field at these points (outside and at the surface) as if the net charge is concentrated at the centre.



Q 4πε 0 R2

If we apply Gauss’s Law for this surface, then we get

∫

  q E ⋅ dA = enc ε0

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 71

x≥R

( outside and at the surface )

x R} ε0 r

⎡ sR ⎢ So, E = ⎢ ε 0 r ⎢ 0 ⎣

for for

r≥R

( outside and at the surface)

r R

Solution l

P

III

To find electric field inside the cylinder at a distance r from the axis, we consider a small cylindrical Gaussian surface of radius r and length l . Applying Gauss’s Law for this surface, we get



∫

  q E ⋅ dA = enc ε0

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 75

In this case the charge density is not uniform, and   1 rdV . We E ⋅ dA = Gauss’s Law is written as ε0 use a Gaussian surface which is a cylinder of radius r, length l and is coaxial with the charge distribution.

∫

r (a) When r < R, this becomes E(2p rl) = 0 ε0

∫

r

⎛

r⎞

∫ ⎜⎝ a - b ⎟⎠ dV. 0

The element of volume is a cylindrical shell of radius r , length l and thickness dr so that dV = 2p rldr .

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1.76  JEE Advanced Physics: Electrostatics and Current Electricity

⎛ 2p r 2 lr0 ⎞ ⎛ a r ⎞ E ( 2p rl ) = ⎜ ⎜ - ⎟ ⎝ ε 0 ⎟⎠ ⎝ 2 3b ⎠  So, inside the ­cylinder, we get

r r⎛ 2r ⎞ E= 0 ⎜ a- ⎟ 2ε 0 ⎝ 3b ⎠



(b) When r > R , Gauss’s Law becomes R

r ⎛ r⎞ E ( 2p rl ) = 0 ⎜ a - ⎟ ( 2p rldr ) ε0 ⎝ b⎠ 0  So, outside the cylinder, we get

∫

E=



r0 R2 2ε 0 r

2R ⎞ ⎛ ⎜⎝ a ⎟ 3b ⎠

ELECTRIC FIELD DUE TO AN INFINITE NON-CONDUCTING THIN UNIFORMLY CHARGED SHEET To find the electric field strength at a point P at distance r infront of the charged sheet we consider a cylindrical Gaussian surface (also called as Gaussian Pill box) as shown in figure of face area A . Applying Gauss’s Law for this surface, we have   q E ⋅ dA = enc ε0       sA ⇒ E ⋅ dA + E ⋅ dA + E ⋅ dA =  {∵ qenc = s A } ε0

∫ ∫

∫

I

II



∫

I

⇒ 2EA = 

II

sA 2ε 0

sA  ε0

{As electric field is uniform on both sides}

⇒ E=

s 2ε 0

{This field is independent of the distance of point P from the sheet}

ELECTRIC FIELD DUE TO AN INFINITE CHARGED CONDUCTING SHEET Figure shows a large charged conducting sheet, charged on both the surfaces with surface charge density s . There is no charge within the volume of the sheet and also the electric field inside the metal sheet is zero. To find electric field strength at a point P in front of the sheet we consider a cylindrical Gaussian surface (or Gaussian Pill Box) as shown. E1 Gaussian pillbox

∫

dA1

E3

S1 S3

dA3

III

E1 Gaussian pillbox

∫

ftotal = EdA + EdA =

dA1

S2

E3 dA2

S1 S3

dA3

E2

Applying Gauss’s Law to this surface, we get S2 dA2

In this case

∫

E2

  E ⋅ dA = 0 , as the lateral surface of cylin-

II

der is parallel to the direction of electric  field strength. So, E becomes perpendicular to A for curved surface and hence no flux is associated with the curved surface.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 76

⇒

∫ ∫ I

  q E ⋅ dA = enc ε0

    E ⋅ dA + E ⋅ dA +

∫ II

∫

III

  2s A E ⋅ dA =  ε0

{Here qenc = 2s A} Since the field inside the sheet is zero, so the Gaussian Pill box will enclose a charge s A at the left and s A at the right too. So that the total charge enclosed is

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Chapter 1: Electrostatics 1.77



The charge dq in this infinitesimal shell is

qenc = 2s A

⇒ ftotal =

∫

    E ⋅ dA + E ⋅ dA +

∫

I

⇒ 2EA = ⇒ E=

II

∫

III

  2s A E ⋅ dA = ε0



dq = r dV

(

2 ⇒   dq = r 4p x dx

)

r0 2 ⇒   dq = x 4p x dx

2s A ε0

(

)

⇒   dq = 4pr0 x dx

s ε0

{Again here E is independent of the distance of point P from the sheet}

⇒  Q=

∫

R

dq =

∫ 4pr x dx 0

0

R

Illustration 57

A sphere of radius R is charged with a non-uniform charge density which varies with the distance x from the centre as

r r= 0 x

where r0 is a positive constant. Find the electric field strength at a point P situated at a distance r from centre of sphere (a) outside it (b) inside it (c) also plot the variation of E with r. Solution

(a) EP =

Q 4pε 0 r 2



⎡ x2 ⎤ 4 ⇒   Q = pr ⎥ 0⎢ ⎣ 2 ⎦0 2 ⇒   Q = 2pr0 R

Thus electric field strength at point outside the sphere at distance r from the centre is

   

Ep =

Q 4pε 0 r

Eoutside = ⇒   

2

=

(

)

2 r0 R2 1 2pr0 R = 4pε 0 r2 2ε 0 r 2

r0 R2 2ε 0 r 2

(b) To find electric field strength at an interior point at distance r from the centre of sphere, we first find the charge enclosed within the inner sphere of radius r such that point P is on the surface. Thus enclosed charge is

{where Q is the total charge of sphere} P

For outer points we see that the entire charge of sphere to be concentrated at its centre. We can also prove this again by taking a spherical Gaussian surface as done in earlier cases. Let us now calculate Q . This is done by integrating the charge of an elemental shell of radius x and thickness dx as shown in figure.

r

Q enc

r

P r

R

x

Qenc =     

r0

∫ R ( 4p x dx ) 2

0

dx x

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 77

R

2 ⇒   Q = 2pr0 r Here electric field strength at point will be due to the inner charged core having charge Qenc as if it were again concentrated at the centre. So

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1.78  JEE Advanced Physics: Electrostatics and Current Electricity

Einside = EP =





(

2 1 Qenc 1 2pr0 r = 4pε 0 r 2 4pε 0 r2

)

 

for r ≥ R

d

for r < R

EP =

( inside the sphere )



r = 0  2ε 0

{by putting r = R } Esurface   So, the field has a continuity as shown

0

Inside

⇒



EP =

∫

d

dE =

∫ 0

r0 x 2 dx 2ε 0

r0 d 3 6ε 0

Illustration 59

Inside a ball charged uniformly with volume density r , there is a spherical cavity. The centre of the cavity  is displaced with respect to the centre of the ball by a. Calculate the field strength inside the cavity, assuming the permittivity equal to unity.

E ρ

2ε 0

x

(l, 0)

Net electric field at P is given by

( outside and at the surface )

Also, we observe that

E∝ 1 r2 Outside

r=R

O

P dE

O

 {Independent of distance r from the centre} (c) Since ⎡ r0 R2 ⎢ 2ε r 2 E=⎢ 0 ⎢ r ⎢ 0 ⎢⎣ 2ε 0

x

z

dx

x

r0 2ε 0

⇒   EP =

y

y

Solution

Illustration 58

A region between yz plane and another plane parallel to yz plane at x = d has charge density r which varies with distance x from origin O as

r = r0 x 2

where r0 is a positive constant. Calculate the electric field strength due to this slab at point P ( l , O ) .

Consider a point P in the cavity at a position vector  r from the centre of sphere and at a position vector  r1 from the centre of cavity as shown. E2 r C a

P

E1 r1 C1

Solution

Let us consider an elemental sheet of thickness dx at a distance x from the yz plane. Since we know that due rd to a sheet of thickness d the electric field is E = . 2ε 0 Electric field strength at P due to this sheet is



dE =

⇒ dE =

rdx 2ε 0



2

r0 x dx 2ε 0

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 78

{∵ r = r x } 0

2

 If E be the electric field strength at P due to the complete charge of the sphere (inside cavity also) then we know electric field strength inside a uniformly charged sphere is given as  rr E= 3ε 0

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Chapter 1: Electrostatics

Similarly if we assume that the charge is only there in the region of cavity and hence this  will also be a uniformly charged small sphere. If E1 be the electric field only due to the cavity charge then,   rr1 E1 = 3ε 0 Now the electric field due to the charged sphere in the cavity at point P can be given as    E = E - E1 net {As now charge of cavity is removed}  r      ∵ a + r1 = r } ⇒ Enet = ( r - r1 ) { 3ε 0   ra    ⇒ Enet = { As r - r1 = a } 3ε 0

Problem solving technique(s) “Cavity can be considered to be negative charge embedded in positive charge so that the common region having size of cavity has no charge on it”. This shows that the net electric field inside the  cavity is uniform and in the direction of a i.e., along the line joining the centre of spheres and cavity.

1.79

a

ρ

Similarly we can find the electric field strength inside a cylindrical cavity of a long uniformly charged cylinder. If cavity axis is displaced from axis of cylinder  by a displacement vector a, then by the analysis we have already done for a sphere, we can say that the electric field strength inside the cavity is also uniform and hence will be given as  ra E= 2ε 0

Test Your Concepts-VI

Based on Gauss’s law 1. An electric dipole is placed at the centre of a sphere. Find the electric flux passing through the sphere. 2. A point charge q is placed at the centre of a cube of edge a. Calculate the flux linked (a) with all the faces of the cube. (b) with each face of the cube. If charge is not at the centre, then what will be the answers for parts (a) and (b). 3. The intensity of an electric field depends only on  a ( xiˆ + yjˆ ) the coordinates x and y as E = 2 , where, x + y2 a is a constant and iˆ and ˆj are the unit vectors of the x and y axes. Calculate the charge within a sphere of radius R having its centre at the origin.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 79

(Solutions on page H.20) 4. Figure shows an imaginary cube of side a. A uniformly charged rod of length a moves towards right at a constant speed v. At t = 0, the right end of the rod just touches the left face of the cube. Plot a graph between electric flux passing through the cube versus time. λ

V a a

5. A long cylinder (shown in figure) carries a charge density that is proportional to the distance x from the axis as r = kx, for some constant k. Find the electric field inside this cylinder.

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1.80  JEE Advanced Physics: Electrostatics and Current Electricity

E l

Gaussian surface

E

9. An infinitely long line charge having a uniform charge per unit length l lies a distance d from point O as shown in figure. Determine the total electric flux through the surface of a sphere of radius R having centre at O resulting from this line charge. Consider both cases, where R < d and R > d .

6. (a) A point charge q is located a distance d from an infinite plane. Determine the electric flux through the plane due to the point charge.



(b) If a point charge q is located a very small distance from the centre of a very large square on the line perpendicular to the square and going through its centre. Determine the approximate electric flux through the square due to the point charge. (c) Explain why the answers to parts (a) and (b) are identical.

7. A positive point charge Q is located at the centre of a cube of edge L. In addition, six other identical negative point charges q are positioned symmetrically around Q as shown in figure. Determine the electric flux through one face of the cube. Also find the condition when no flux is associated with the cube. L q q L

q

q

Q q q

d

λ

R O

10. A solid insulating sphere of radius R has a non-uniform charge density that varies with r according to the expression r = r0 r 2 , where r0 is a constant and r < R is measured from the centre of the sphere. Show that the magnitude of the electric field

(a) outside ( r > R ) the sphere is E =



(b) inside ( r < R ) the sphere is E =

r0R5 . 5ε 0 r 2

r0 r 3 . 5ε 0

11. A slab of insulating material has a non-uniform positive charge density r = r0 x 2 , where x is measured from the centre of the slab as shown in Figure, and r0 is a constant. The slab is infinite in the y and z directions. Derive expressions for the electric field in the y

L

8. An insulating solid sphere of radius a has a uniform volume charge density and carries a total positive charge Q. A spherical Gaussian surface of radius r, which shares a common center with the insulating sphere, is inflated starting from r = 0. (a) Find an expression for the electric flux passing through the surface of the Gaussian sphere as a function of r for r < a. (b)  Find an expression for the electric flux for r > a. (c)  Plot the flux versus r.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 80

O

x

d





(a)  exterior regions of d d ⎛ ⎞ ⎜⎝ i.e., x > and x < - ⎟⎠ 2 2

the

slab

d d⎞ ⎛ (b)  interior region of the slab ⎜ i.e., - < x < ⎟ . ⎝ 2 2⎠

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Chapter 1: Electrostatics 1.81

12. A spherically symmetric charge distribution has a r charge density given by r = 0 , where r0 is conr stant. Find the electric field as a function of r. 13. (a)  Using the mathematical similarity between Coulomb’s Law and Newton’s Law of Universal Gravitation, show that Gauss’s Law for gravitation expressed as

O



(a) Show that the magnitude of the electric field a distance x from its centre and inside the slab is rx E= . ε0



(b) Suppose an electron of charge -e and mass m can move freely within the slab. It is released from rest at a distance x from the centre. Show that the electron exhibits simple harmonic motion with a period

enc

enc

 Fg the Gaussian surface and g = represents m the gravitational field at any point on the Gaussian surface. (b) Determine the gravitational field at a distance r from the centre of the Earth where r < RE , assuming that the Earth’s mass density is uniform. 14. A slab of insulating material (infinite in two of its three dimensions) has a uniform positive charge density r. An edge view of the slab is shown in figure.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 81

x

d



∫ g⋅ dA = -4pG∑ m    where ∑ m is the net mass enclosed by  

y

T = 2p

mε 0 re

   15. Inside an infinitely long circular cylinder charged uniformly with volume density r there is a circular cylindrical cavity. The distance between the axes of  the cylinder and the cavity is equal to a . Find the  electric field strength E inside the cavity. The permittivity is assumed to be equal to unity.

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1.82  JEE Advanced Physics: Electrostatics and Current Electricity

Electrostatic Potential, Potential Energy, Conductors and Applications ELECTROSTATIC POTENTIAL ENERGY AND POTENTIAL Electrostatic Potential Energy Potential energy of a system of particles is defined as the work done in assembling the system of charges in a given configuration against the interaction forces of particles. Potential energy of a system of particles is a concept associated with conservative fields only. Since, electric field is also conservative in nature so, we associate the concept of potential energy with it. When all particles of a system are separated far apart by infinite distance there exists no interaction between them. This state, we take as reference of Zero Potential Energy (ZPE). Electrostatic potential energy is categorised in two ways. (a) Electrostatic interaction energy of charged particles of a system. (b) Electrostatic self energy of a charged object.

bringing test charge q0 from infinity to the given point P at a distance r from the source charge q. Let, at any instant the test charge q0 be at a distance x from the source charge q. Let q0 be displaced through dx towards q. The electrostatic force of repulsion F acts on q0 away from q. r

q

ELECTROSTATIC INTERACTION ENERGY OF A SYSTEM OF TWO CHARGED PARTICLES Figure shows two positive charges q and q0 separated by a distance r . The electrostatic interaction energy of this system can be given as work done in

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 82

q0

dx q0 F

∞

If dW be the work done, then   dW = F ⋅ dx qq0 ⇒ dW = dx cos ( 180° ) 4pε 0 x 2 ⇒ dW = -

qq0 -2 x dx 4pε 0

qq ⇒ W=- 0 4pε 0

Electrostatic Interaction Energy Electrostatic interaction energy of a system of charged particles is defined as the external work required to assemble the particles from infinity to a given configuration (as that required in system). When charged particles are lying at infinity, their potential energy is taken to be zero, because no interaction exists between them. When these charges are brought close to a given configuration, external work is required if the force between these particles is repulsive and hence energy is supplied to the system so that final potential energy of system will be positive. If the force between the particles is attractive, work will be done by the system and final potential energy of system will be negative. In giving these arguments please keep this thing in mind that work done by a conservative force equals the decrease in potential energy of the system.

P

x

⇒ W=⇒ W=

r

∫x

-2

dx

0

qq0 ⎛ x -2 +1 ⎜ 4pε 0 ⎝ -2 + 1

⎞ ⎟ ∞⎠ r

qq0 ⎛ 1 1 ⎞ ⎜ - ⎟ 4pε 0 ⎝ r ∞ ⎠

⇒ W∞→ P =

qq0 =W 4pε 0 r

So, work done by the external force in bringing the test charge q0 from ∞ to the point P under the influence of source charge q is W = W∞→ P =

qq0 4pε 0 r

This work done is stored in the form of electrostatic interaction energy ( U ) of the system. So, U=

qq0 4pε 0 r

Please note that if the two charges are of opposite nature, then

U=-

qq0 4pε 0 r

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Chapter 1: Electrostatics 1.83 Illustration 60

q1

Two charges, +q and -q , each of mass m , are revolving in a circle of radius R , under mutual electrostatic force. Find (a) speed of each charge (b) kinetic energy of system (c) potential energy of the system (d) total energy of the system

r12

r23

q2

Solution

In this situation, the centripetal force is provided by the electrostatic force of attraction. (a) By Newton’s second law we get,

r13

F=

⇒  v=

q2 1 mv 2 = 4pε 0 ( 2R )2 R

For example if a system of three particles having charges q1 , q2 and q3 is given as shown in figure. The total interaction energy of this system can be given as U=

1 ⎡ q1 q2 q2 q3 q1 q3 + + r23 r13 4pε 0 ⎢⎣ r12

and q4 (having total number of six interaction = 4 C2), we have

v

r12

q1 +q

F

R O

F

–q

r14 q4

1 1 (b) Kinectic energy of the system is K = mv 2 + mv 2 2 2 q2 q2 ⇒   K = 4pε ( 4 R ) = 16pε R 0 0 (c) Potential energy of the system is U = q2 -1 q2 ⇒   U = 4pε ( 2R ) = - 8pε R 0 0

1 q ( -q ) 4pε 0 2R

U= ⇒

-q q ⇒   E = 4pε ( 4 R ) = - 16pε R 0 0 So, we observe that T .E. = - ( K .E. ) =

1 ( P.E. ) 2

1 ⎡ q1 q2 q2 q3 q3 q4 q4 q1 q1 q3 q2 q4 ⎤ + + + + + 4pε 0 ⎢⎣ r12 r23 r34 r14 r13 r24 ⎥⎦

U = U12 + U 23 + U 34 + U 41 + U13 + U 24

U=

1 4pε 0

N

N

qi q j

∑∑ r i =1 j =1 ( j>i )

ij

where j > i assures that the pairs are not repeated. Other way round, we can count every pair twice and 1 divide the result by . So, 2

ELECTROSTATIC INTERACTION ENERGY FOR A SYSTEM OF PARTICLES

1 U= 8pε 0

When more than two charged particles are there in a system, the interaction energy can be given by sum of interaction energy of all the pairs of particles with no pair repeated.

U=

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 83

q3

r34

From above we see that the total potential energy is simply the sum of contributions due to distinct pairs. Generalising to N charges, we get

(d) Total energy of the system is E = K .E. + P.E. 2

q2 r13

r24

v

2

⎤ ⎥ = U12 + U13 + U 23 ⎦

Similarly, for an assembly of four charges q1 , q2 , q3

q2 1 4pε 0 ( 4 R ) m

R

q3

⇒

1 2

N

N

qi q j

∑∑ r i =1 j =1

ij

1 = 2

N

∑ i =1

⎛ 1 qi ⎜ ⎜ 4pε 0 ⎜⎝

N

∑ j =1 j≠i

qi ⎞ ⎟ rij ⎟ ⎟⎠

N

∑q V(r ) i

i

i =1

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1.84  JEE Advanced Physics: Electrostatics and Current Electricity

where V(ri) is the potential due to all other charges at the location of charge qi ( i.e., ri )

+q

Illustration 61

+q

l

l

Calculate the work required to be done to make an arrangement of three particles each having a charge +q such that the particles lie at the vertices of an equilateral triangle of side a. What work will be done by electric field when the particles are shifted away so that the side of triangle becomes 2a?

l

+q

4pε 0 ( 2 a )

3q2 3q2 3q2 = 4pε 0 a 8pε 0 a 8pε 0 a





+q

l

+q

l

(a)

+q

–q

l

l

+q

–q

(b)

l

l

l

l +q

l

+q

–q

–q

(c)

Ub = 4

- q2 q2 q2 ( -4 + 2 ) +2 = 4pε 0 l 4pε 0 ( 2l ) 4pε 0 l q2 2q 2 2q 2 2q 2 =4pε 0 l 4pε 0 l 4pε 0 ( 2l ) 4pε 0 l

VP = Potential at point P = Q

Determine the interaction energy of the point charges located at the corners of a square of side l in the ­figures shown. l

l

q2 q2 q2 (4+ 2 ) +2 = 4pε 0 l 4pε 0 ( 2l ) 4pε 0 l

Illustration 62

+q

(b)

l

The electrostatic potential at a point P due to a source charge is the work done per unit test charge in bringing the test charge ( q0 ) from infinity to the point P under the electrostatic influence of source charge Q. So,

×3

⇒ WE⋅Field = Ui - U f =

l

ELECTROSTATIC POTENTIAL (V )

q2 Since Ui = × 3 and 4pε 0 ( a )



–q

+q

Ua = 4

and Uc = 2

q2

l

l

Let interaction energy of the system in Figure (a), (b) and (c) be U a , Ub and Uc respectively, then

Work done by electric field equals the decrease in potential energy. So, we have

Uf =

+q

(a)



WE⋅Field = -DU = Ui - U f

+q

–q

l

3q2 q2 ⇒ W= ×3 = 4pε 0 a 4pε 0 a



l

l

l

Solution

The work required to make an arrangement of charges is equal to potential energy of the system

+q

+q

l l

+q

–q

(c)

–q

Since W∞→ P = ⇒ VP =

W∞→ P q0

r

P

Qq0 4pε 0 r

W∞→ P Q = 4pε 0 r q0

Electrostatic potential is a scalar physical quantity with SI unit joule per coulomb ( JC -1 ) and 1 JC -1 = 1 volt = 1 V .

Solution

ELECTROSTATIC POTENTIAL DIFFERENCE (∆V )

Interaction energy of any two point charges q1 and qq q2 at separation r is given by 1 2 4pε 0 r

Consider a charge +q to be located at the origin  O. Let us arbitrarily choose two points A and B. Let E be the electric field strength at point P on this curve. The

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 84

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Chapter 1: Electrostatics 1.85

 test charge q0 will experience an electric force q0 E directed radially away from charge +q. In order to prevent the  test charge from accelerating in the direction of q0 E , the external agent is required to apply a force F in the opposite direction. So,   F = - q0 E Y

q

Before moving further with this discussion, let us know this and keep in mind that electrostatic potential is a scalar quantity and is positive due to positive charges and negative due to negative charges. Let us now calculate the electrostatic potential due to an assembly of charges at point P shown.

q0E

A rA

POTENTIAL DUE TO AN ASSEMBLY OF CHARGES

q1

P F

B rB





∫ F ⋅ dl

q4

B



B







∫ -q E ⋅ dl = -q ∫ E ⋅ dl 0

A

0

A

B

  WAB = - E ⋅ dl ⇒ q0

∫

A

But VB - VA =

The electrostatic potential V at point P due to the assembly of charges shown is the algebraic sum of the potential due to each of the charge in the assembly. So, V=

A

⇒ WAB =

r4 –q5

If dW is the infinitesimally small amount of work done  by external agent in giving small displacement dl , then   dW = F ⋅ dl ⇒ WAB =

–q3

r3 r5

Z

B

r2

P

X

O

q2

r1

dl

θ

1 ⎛ q1 q2 q3 q4 q5 ⎞ + - + 4pε 0 ⎜⎝ r1 r2 r3 r4 r5 ⎟⎠

POINTS WITH ZERO POTENTIAL DUE TO TWO POINT CHARGES Case-1: If both charges are like, then there is no point at a finite distance where net potential zero. Case-2: If the charges are equal and unlike then the net potential is zero at all points on the equatorial plane as shown below.

WAB q0

B

  ⇒ - E ⋅ dl = VB - VA

∫

q

–q

A B

  W ⇒ VB - VA = - E ⋅ dl = A→B q0

∫

A

So, electrostatic potential difference between two points is work done by external agent in moving unit positive charge from one point to another.

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 85

V=0

Case-3: If the charges are unequal and unlike then all such points where resultant potential is zero lie on a closed surface.

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1.86  JEE Advanced Physics: Electrostatics and Current Electricity

However, on the line joining the two charges, only two such points exist, one between the charges and the other outside the charges. Both the above points lie nearer to charge of smaller magnitude.

Illustration 63

Two charges 6 m C and -4 m C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Solution

GRAPHICAL REPRESENTATION OF POTENTIAL OF A SYSTEM OF TWO POINT CHARGES

Case-1: Between the charges, let V = 0 at distance x from 6 × 10 -6 C charge. So, we have

As we move on the line joining two charges from one charge towards the other, the variation of potential V with distance x is shown below.





x

–4 μ C

3 2 = x 15 - x





⇒ 





⇒   x = 9 cm

Case-2: Outside the charges, let V = 0 at distance

V –q

–q

x from 6 × 10 -6 C charge. So, we have

x

Case-2: For unlike charges of equal magnitude.

1 ⎛ 6 × 10 -6 4 × 10 -6 ⎞ ⎜ ⎟ =0 4pε 0 ⎝ x x - 15 ⎠





   





⇒   x = 45 cm

V

15 cm 6 μC +q

–q

x

Case-3: For unlike charges of unequal magnitude. V

E = – dV = 0 dx

(15 – x)

v=0

6 μC

V

+q

P

x

Case-1: For like charges of equal magnitude.

+q

  

1 ⎛ 6 × 10 -6 4 × 10 -6 ⎞ ⎜ ⎟ =0 4pε 0 ⎝ x 15 - x ⎠

q1

01_Physics for JEE Mains and Advanced - 2_Part 2.indd 86

x –4 μ C

(x – 15)

P v=0

So, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Conceptual Note(s) –q2

x q1 < q2

The electric field for such an arrangement of charges is zero only at one point where as the potential is zero for two different points (other than ∞ ).

9/20/2019 10:14:38 AM

Chapter 1: Electrostatics 1.87 Illustration 64

Consider a system of two charges shown in figure. The arrangement is also called an Electric Dipole. Find the electric potential at an arbitrary point on V(x) x the x-axis and make a plot of Vs where V0 a q V0 = . 4pe 0 a y

a

a

–q

x

q

As can be seen from the graph, V(x) diverges at x = ±1 , where the charges are located. a

ELECTROSTATIC POTENTIAL AT THE AXIS OF A UNIFORMLY CHARGE ROD Consider a non-conducting charged rod of length L, having a uniform charge Q . Let us find the electrostatic potential at a point P due to the rod at a distance r from one end of the rod. x dx

P

Q

Electric dipole

r

L

Solution

The electric potential can be found by the superposition principle. At a point P ( x, 0 ) on the x-axis, we have

⇒

V(x) =

q 1 1 ( -q ) + 4pe 0 x - a 4pe 0 x + a

V(x) =

q ⎡ 1 1 ⎤ 4pe 0 ⎢⎣ x - a x + a ⎥⎦

The above expression is rewritten as



V(x) 1 1 = x x V0 -1 +1 a a

–q

dq =

+q

dV =

⇒ V=

x (x–a)

q . The plot of the dimensionless elec4pe 0 a x tric potential as a function of is shown in figure. a

where V0 =

V(x)/V0 40

1 2 3 –20

1 dq 4pe 0 x

1 Q dx 4pe 0 Lx

Net electric potential at point P is obtained by integrating this expression

P(x, 0)

20 –3 –2 –1

Q dx L

The potential dV due to this element at point P is

⇒ V=

(x–a)

a





y

a

For this we consider an infinitesimal element of length dx at a distance x from the point P . Charge on this element is

⇒ V= ⇒ V=

∫

1 dV = 4pe 0

r+L

Q

∫ Lx dx

1 Q log e x 4pe 0 L

r

r+L r

1 Q ⎛ r+L⎞ log e ⎜ ⎝ r ⎟⎠ 4pe 0 L

If charge density of the rod is l , then l = x/a



V=

Q . So, L

l ⎛ r+L⎞ log e ⎜ ⎝ r ⎟⎠ 4pe 0

–40

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1.88  JEE Advanced Physics: Electrostatics and Current Electricity

ELECTROSTATIC POTENTIAL AT POINT P LYING ON PERPENDICULAR BISECTOR OF ROD

V(y)

A plot of

V0 is shown in figure.

Consider a non-conducting rod of length l having a uniform charge density l . The electric potential at P, a perpendicular distance y above the midpoint of the rod is calculated by considering a differential element of length dx which carries a charge dq ( = l dx ) as shown in figure. The source element is located at ( x, 0 ) , while the observation point P is located on the y-axis at ( 0, y ) . The distance of P from element is r = x 2 + y 2 . Its contribution to the potential dV is given by dV =

l dx 1 dq 1 = 4pe 0 r 4pe 0 x 2 + y 2

(

4 3 2

1 –4

)



r dx

Taking V to be zero at infinity, the total potential due to the entire rod is

l 4pe 0

  Using

∫

l 2

∫

l 2

dx x2 + y2

x +y

l log e ⎡⎣ x + x 2 + y 2 ⎤⎦ 4pe 0

2

(

= log e x + x 2 + y 2

⎡ l ⎢ ⎛⎜ ⎞⎟ + ⎝ 2⎠ l log e ⎢⎢ ⇒ V= 4pe 0 ⎢ -⎛ l ⎞ + ⎜ ⎟ ⎣⎢ ⎝ 2 ⎠

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 88

⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦

)

2 ⎤ ⎛ l⎞ 2 ⎜⎝ ⎟⎠ + y ⎥ 2 ⎥ ⎥ 2 ⎛ l⎞ 2 ⎥ ⎜⎝ ⎟⎠ + y ⎥ 2 ⎦

l 2

2y 2 ⎛ 2y ⎞ 1+ ⎜  1 + ⎝ l ⎟⎠ l2

⎡ 2y 2 2+ 2 ⎢ l l ⇒ V= log e ⎢ 4pe 0 ⎢ 2y 2 ⎢ ⎣ l2 Since,

-

dx 2

=

y/l

2 ⎡ l l ⎛ 2y ⎞ ⎢ + 1+ ⎜ ⎟ ⎝ l ⎠ l ⎢ 2 2 V= log e ⎢ 2 4pe 0 ⎢ - ⎛ l ⎞ + l 1 + ⎛ 2y ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ l ⎠ ⎣⎢ ⎝ 2 ⎠ 2

Since

  A  non-conducting rod of length l and uniform charge density l.

V=

4

2

x

l

l 2

2

2 ⎡ 2y ⎞ ⎤ ⎥ ⎢ 1 + 1 + ⎛⎜ ⎟ ⎝ l ⎠ ⎥ l ⎢ ⇒ V= log e ⎢ 2 ⎥ 4pe 0 ⎢ -1 + 1 + ⎛ 2 y ⎞ ⎥ ⎜ ⎟ ⎢⎣ ⎝ l ⎠ ⎥⎦

θ

O

–2

In the limit l  y , the potential becomes

12

P

y l , as a function of 4pe 0 l

V( y)/V0

y

y

, where V0 =

y2

1

l2

⇒ 2+

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

2y 2

⇒ V≈

⇒ V=

l2

2

⎛ l2 ⎞ l l ⎛ 2 ⎞ log e = log e⎜ 2 ⎟ ⎜ 2 y 2 ⎟ 4pe 0 4pe 0 ⎝y ⎠ ⎜ 2 ⎟ ⎝ l ⎠

l ⎛ l⎞ log e ⎜ ⎟ 2pe 0 ⎝ y⎠

9/20/2019 10:23:28 AM

Chapter 1: Electrostatics 1.89

The corresponding electric field can be obtained by using



l l ∂V 2 Ey = = ∂y 2pe 0 y ⎛ l ⎞ 2 2 ⎜⎝ ⎟⎠ + y 2

⇒ VB - V0 =

which is in complete agreement with the result.

On a thin rod of length l = 1 m, lying along the x-axis with one end at the origin x = 0, there is uniformly distributed charge per unit length l = Kx, where K = 10 -9 Cm -2 . Find the work done in displacing a

(

charge q = 1000 m C from a point A 0 , B ( 0, l ) m.

)

Let us calculate the potential due to the rod at A and then at B.

(

Potential at A 0 ,

0 .44

)m

Potential at A due to rod is calculated by taking an infinitesimal element of length dx at a distance x from O ( 0 , 0 ) on the rod. Then

⇒ dV = VA

⇒

∫

V0

dq l dx = …(1) 4pe 0 r 4pe OA 2 + x 2 0 kx dx l =1 m

0

⎤ ⎥ ⎥⎦

⇒ WA→B =

qk [ 1.414 - 1 - 1.2 + 0.66 ] 4pe 0

⇒ WA→B = 9 × 109 × 1000 × 10 -6 × 10 -9 [ -0.126 ] ⇒ WA→B = -1.1 × 10 -3 J

ELECTROSTATIC POTENTIAL DUE TO A CHARGED RING AT ITS CENTRE Let us first find potential dV at centre C due to an infinitesimal charge dq on ring which is



dV =

1 dq 4pe 0 R

⇒ V=

1 4pe 0

dq

∫ dV

Q

∫ R = 4pe R 0

Q

dq OA = 0.44 m = constant OB = l = 1 m = constant

A

R C

r

0.44 m

x x

dx

Potential at B ( 0, 1 ) m



(Please observe, that V0 will cancel in the process)

k [ 1.44 - 0.44 ] …(2) 4pe 0

B

dV =

qk [ 2 - 1 - 1.44 + 0.44 ] 4pe 0

2

⎡ k ⎢ 0.44 + x 2 dV = 4pe 0 ⎢⎣

O

2 - 1 ) …(3)

Total potential at C is V =

4pe 0 0.44 + x

⇒ VA - V0 =

(

⎤ ⎥ ⎥ 0⎦

1

Now, by definition, WA→B = q ( VB - VA )

0.44 m to

Solution



k 4pe 0

⇒ WA→B = W =

Illustration 65

dV =

⎡ k ⎢ 1 + x2 ⇒ VB - V0 = 4pe 0 ⎢⎣

dq kx dx kx dx = = 2 2 4pe 0 r ′ 4pe OB + x 4pe 0 1 + x 2 0

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 89

Since all the infinitesimal dq’s of the ring are situated at same distance R from the ring centre C so, we can directly say that the total electric potential at centre of Q ring is VC = . 4pe 0 R

9/20/2019 10:23:39 AM

1.90  JEE Advanced Physics: Electrostatics and Current Electricity

Here we must see that even if charge Q is nonuniformly distributed on ring, the electric potential C will remain same (we shall discuss this with an example below) because in that case too



∫

dq 1 dq = 4pe 0 R 4pe 0 R

∫

V=

⇒ Vcentre =

Qtotal on ring 4pe 0 R

Illustration 66

Consider a charged ring of radius R . Let the charge on this ring be varying according to the relation l = l0 cos 2 ϕ , where l0 is a positive constant and ϕ is the azimuthal angle. Find the electrostatic potential due to this ring at the centre. Solution

Let us first calculate the total charge possessed by the ring. Since, ⇒

∫ l dx dq = l Rdϕ ∫ dq =

⇒ q = l0 Rp + 0 ⇒ q = l0p R Since, dV = ⇒ V=

1 dq 4pe 0 R

⇒ V=

l0p R 4pe 0 R

⇒ V=

l0 4e 0

∫

ELECTROSTATIC POTENTIAL AT A POINT P ON THE AXIS OF THE RING AT DISTANCE x FROM ITS CENTRE If we wish to find the electrostatic potential at a point P lying on the axis of ring, we can directly calculate the result because here too, all points of ring are at same distance R2 + x 2 from the point P. So, potential at P is

2p

dq 4pe 0 R

VP =

1 4pe 0

Q 2

R + x2 Q

∫

⇒ dq = l0 R cos 2 ϕ dϕ 0

⇒ q=

∫

l R dq = 0 2

P

2p

∫ ⎡⎣ 1 + cos ( 2ϕ ) ⎤⎦ dϕ

R

0

2p ⎤ ⎡ 2p l0 R ⎢ ⇒ q= dϕ + cos 2ϕ dϕ ⎥ ⎥ 2 ⎢ 0 ⎣0 ⎦

∫

x r = R2 + x2

∫

2p

sin ( 2ϕ ) l R⎡ ⇒ q = 0 ⎢ 2p + 2 ⎢⎣ 2

0

dq dϕ ϕ

VARIATION OF ELECTROSTATIC POTENTIAL ON THE AXIS OF A CHARGED RING

⎤ ⎥ ⎥⎦ dx

Since  Vcentre = and 

Vaxis =

Q 4pe 0 R Q

4pe 0 R2 + x 2

From the above two expressions, we conclude that electrostatic potential is maximum at the centre and

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 90

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Chapter 1: Electrostatics 1.91

goes on decreasing when we move away from the centre on its axis.

⇒ ΔV =

q 4pe 0

⇒ ΔV =

2q ⎡ 1 1⎤ 4pe 0 ⎢⎣ R2 + a 2 R ⎥⎦

⇒ ΔV =

q ⎡ 1 1⎤ 2pe 0 ⎢⎣ R2 + a 2 R ⎥⎦

V V0 =

Q 4πε0R x

x=0

Since, WA→B = q0 ( VB - VA )

Illustration 67

There are two thin wire rings, each of radius R, whose axes coincide. The charges of the rings are q and -q. Find the potential difference between the centres of the rings separated by a distance a. Also calculate the work done to move a charge q0 from centre of first ring to the centre of other ring. Solution

Net potential at centre of ring A is



⎛ Potential at A ⎞ ⎛ Potential at A ⎞ VA = ⎜ + ⎝ due to itself ⎟⎠ ⎜⎝ due to B ⎟⎠ +q

1 1 1 ⎤ ⎡ 1 - + ⎢- R + 2 2 ⎥ 2 2 R R +a R +a ⎦ ⎣

R 2+

2

a 2 R +

a2

R

R

–q

⇒ WA→B =

qq0 ⎡ 1 1⎤ 2pe 0 ⎢⎣ R2 + a 2 R ⎥⎦

ELECTROSTATIC POTENTIAL DUE TO A UNIFORMLY CHARGED DISC AT A POINT P ON ITS AXIS Consider a uniformly charged disc of radius R with surface charge density s . We wish to find electric potential at point P lying on its axis at distance x from the centre. For this we consider an infinitesimal elemental ring of radius y and thickness dy concentric with the disc. Then charge on this infinitesimal element is dq = s (Area of element) ⇒ dq = s ( 2p y dy )

a

dy A

B

( -q ) ⎞ 1 ⎛ q + ⇒ VA = ⎜ ⎟ 4pe 0 ⎝ R R2 + a2 ⎠ ⇒ VA =

q 4pe 0



⎛ Potential at B ⎞ ⎛ Potential at B ⎞ VB = ⎜ + ⎝ due to itself ⎟⎠ ⎜⎝ due to A ⎟⎠

q ⇒ VB = 4pe 0

1 ⎤ ⎡ 1 ⎢- R + 2 2 ⎥ R +a ⎦ ⎣

Thus potential difference,

x

1 ⎤ ⎡1 ⎢R2 2 ⎥ R +a ⎦ ⎣

Similarly net potential at centre of ring B is

ΔV = VB - VA

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 91

x2 + y2

y

P

σ

The electric potential at point P due to this infinitesimal element is dV =

1 4pe 0

dq

x + y2 (Please note here that once the point P is taken, then x becomes a fixed value) ⇒ dV =

2

1 s ( 2p y dy ) 4pe 0 x 2 + y 2

9/20/2019 10:24:02 AM

1.92  JEE Advanced Physics: Electrostatics and Current Electricity

Net electric potential at point P due to entire disc is calculated by integrating this expression. R

⇒ V=

0

s ⇒ V= 2e 0 s ⇒ V= 4e 0 Using

∫

s

∫ dV = ∫ 2 e R

∫ 0

y dy x2 + y2

0

x2 + y2

R

∫ ( x2 + y2 )

-

1 2

2 y dy

0

⇒ V=

(

n+1

, n ≠ -1

R 1 2 - 2 +1 y

)

2 s x + 1 4e 0 - +1 2

s ⎡ 2 x + y2 2e 0 ⎣⎢

⇒ VP =

s 2e 0

⇒ V=

Q = Vdue to a point charge 4pe 0 x at distance x from it

Illustration 68

y dy

⎡⎣ f ( x ) ⎤⎦ n ⎡⎣ f ( x ) ⎤⎦ f ′ ( x ) dx = n+1

⇒ V=

⎡ ⎤ s R2 s ( p R2 ) R2 x + x = ⎢ ⎥= 2x 4pe 0 x ⎣ ⎦ 4e 0 x

⇒ V=

0 R 0

⎤ ⎦⎥

A non-conducting disc of radius a and uniform positive surface charge density s is placed on the ground with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc from a height H with zero initial velocity. The partiq 4e 0 g cle has = . m s (a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position. Solution

Since, Vp =

s ⎡ 2 2 ⎣ R + x - x ⎤⎦ 2e 0

Potential at centre, ( O ) will be

V/V0 6 5 4 3 2 1 –3 –2.5 –2 –1.5 –1 –0.5

s ⎡ 2 2 ⎣ a + H - H ⎤⎦ 2e 0

point charge disk

0.5 1 1.5 2 2.5 3

2/R

 Comparison of the electric potentials of a non-conducting disk and a poin charge. The electric potential is measured in terms of V0 = Q/4p e0 R

 

VO =

sa  2e 0

{H = 0}

(a) Particle is released from P and it just reaches point O . Therefore, from conservation of mechanical energy decrease in gravitational potential energy = increase in electrostatic potential energy  

( ΔKE = 0 because Ki = K f P

= 0)

q, m

Also, at the centre of the disc, we have x = 0 , so Vcentre = VC =

H

sR 2e 0

In the limit x  R , we have 1 R2 ⎞ 2



⎛ ⎛ ⎞ R2 R2 + x 2 = x ⎜ 1 + 2 ⎟  x ⎜ 1 + 2 + ..... ⎟ ⎝ ⎝ ⎠ 2x x ⎠

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 92

a

O

⇒   mgH = q [ VO - VP ]

9/20/2019 10:24:14 AM

Chapter 1: Electrostatics 1.93

⎛ q ⎞⎛ s ⎞⎡ 2 2 ⎤ ⇒   gH = ⎜⎝ m ⎟⎠ ⎜ 2e ⎟ ⎣ a - a + H + H ⎦  ⎝ 0⎠ …(1) q 4e 0 g Since, = s m ⇒ 

qs = 2g 2e 0 m

2 2 ⎤ ⎡   gH = 2 g ⎣ a + H - a + H ⎦ H 2 2 ⇒   2 = (a+ H)- a + H

H H2 a 2 + H 2 = a + or a 2 + H 2 = a 2 + + aH 2 4 3 2 H = aH 4

2H

⇒   1+ ⇒ 

Substituting in equation (1), we get

⇒ 

H ⎤ ⎡ - 1⎥ = 0 ⇒   mg + 2mg ⎢ 2 2 ⎣ a +H ⎦

⇒ 

-2= 0

a2 + H 2 2H 2

a + H2 H2 2

a +H

2

=1

=

1 4

2 2 ⇒   3H = a

⇒  H=

a 3

From equation (2), we can see that,

U = 2mga at H = 0 and

4 ⇒   H = 3 a and H = 0



U = Umin = 3 mga at H =

⎛ 4⎞ ⇒   H = ⎜⎝ 3 ⎟⎠ a

Therefore, U -H graph will be as shown.

⇒ 

2mga

U = Ue + Ug

where U e = qV =

3mga

sq ⎡ 2 2 ⎣ a + H - H ⎤⎦ 2e 0

O

sq ⎡ 2 2 ⎤ ⇒   U = 2e ⎣ a + H - H ⎦ + mgH …(2) 0 At equilibrium position

position.

- dU =0 dH

  Differentiating equation (2) w.r.t. H , we get mg +  

sq 2e 0

1 ⎤ ⎡⎛ 1 ⎞ - 1⎥ = 0 ⎢ ⎜⎝ 2 ⎟⎠ ( 2 H ) 2 2 a +H ⎣ ⎦



01_Physics for JEE Mains and Advanced - 2_Part 3.indd 93

Also we note that at H = Therefore, H =

sq ⎧ ⎫ = 2mg ⎬ ⎨∵ ⎩ 2e 0 ⎭

H

a/ 3

and  Ug = mgH

F=

3

U

(b) If U be the potential energy, Ue be the electrostatic energy and Ug be the gravitational potential energy of the particle at height h, then

a

a 3

a 3

, U is minimum.

is the stable equilibrium

ELECTRIC POTENTIAL OF AN ANNULUS Consider an annulus of uniform charge density s , as shown in figure. The electric potential at a point P along the symmetric axis is calculated by taking an infinitesimal elemental ring of radius y and thickness dy concentric with the annulus. If dq be the charge on this infinitesimal element, then

9/20/2019 10:24:29 AM

1.94  JEE Advanced Physics: Electrostatics and Current Electricity

P

z

R

P

θ

C

r

b

σ

a

dr An annulus of uniform charge density



dq = s dA = s ( 2p y dy )

Its contribution to the electric potential at P is dV =

1 dq 1 s ( 2p y dy ) = 4pe 0 r 4pe 0 z2 + y 2

Integrating over the entire annulus and using

∫

y dy

= y 2 + x2

y 2 + x2 We obtain



s V= 4pe 0

2ps ⇒ V= 4pe 0 ⇒ V=

s 2e 0

(

∫ a

∫ a



dq 4pe 0 r

where dq = s ( Area of infinitesimal element )

Diameter 2R

x2 + y2

θ

=

= 2rθ }

2s rθ dr s = θ dr 4pe 0 r 2pe 0

x2 + y2 y dy

{∵ length of element

⇒ dq = s ( 2rθ ) dr 

Also, we observe that (see figure)

s ( 2p y dy )

b 2 + z 2 - a2 + z 2

)

s ⎡ 2 2 ⎣ R + z - z ⎤⎦ 2e 0

which coincides with the result of a non-conducting disk of radius R

ELECTROSTATIC POTENTIAL ON THE EDGE OF A UNIFORMLY CHARGED DISC Consider a uniformly charged disc of radius R , having surface charge density s . Let us take a point P on the edge of the disc.

C r

s ⎡ 2 2 2 2 ⎣ b + z - a + z ⎤⎦ 2e 0

In the limit a → 0 and b → R , the potential becomes V=

dV =

⇒ dV =

b

b

For this, assume the disc to be made up of a large number of concentric rings with P as the centre. Let us consider one such infinitesimal ring of radius r , thickness dr and charge dq . Then potential due to this infinitesimal ring is

Semicircle

r = 2R cos θ ⇒ dr = -2R sin θ dθ ⇒ dV = 0

⇒ V=

∫ p 2

s 2R θ sin θ dθ 2pe 0 0

sR dV = θ sin θ dθ pe 0

∫ p 2

a

b

∫

Since - f ( x ) dx = b

∫ f ( x ) dx a

p 2

⇒ V=

sR θ sin θ dθ pe 0

∫ 0

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 94

9/20/2019 10:24:41 AM

Chapter 1: Electrostatics 1.95

Using ILATE property to calculate the integral, we get p 2



p 2

p 2

⎤ ⎡ ⎢ ⎥ sR ⎢ θ sin θ dθ - ( - cos θ ) dθ ⎥ V= pe 0 ⎢ ⎥ 0 ⎣ 0 ⎦

∫

∫

⎡ ⎢ sR ⎢ ⇒ V= -θ cos θ pe 0 ⎢ ⎣

p 2

p 2

⎤ ⎥ ⎥ ⎥ ⎦

+ sin θ

0

0

sR ⎡ ⎤ ⎛p⎞ ⇒ V= 0 + sin ⎜ ⎟ - sin 0 ⎥ ⎝ 2⎠ pe 0 ⎢⎣ ⎦ ⇒ V=

Potential due to the hemisphere is

sR pe 0



⎡ sR sR ⎢ V= sin θ dθ = - cos θ 2e 0 2e 0 ⎢⎣

∫ 0

⇒ V=

p 2 0

⎤ ⎥ ⎥⎦

sR 2e 0

Illustration 70

The electrostatic potential at surface of thin nonconducting sheet with charge density s is V0. Find the potential at a distance x from infinite sheet. Solution

Potential at point P at a distance x is

Also, we know that at the centre of the disc VC =

sR 2e 0

V0

Hence the potential at the centre is more than that at the edge

P x

Illustration 69

Find the electric field potential at the centre of a hemisphere of radius R having uniform surface charge density s . Solution

Consider an infinitesimal ring element as shown. The charge on this element is given by 

dq = ( 2p R sin θ ) ( R dθ ) s {Because radius of ring is r = R sin θ } r R

x θ

dθ

O

r = R sin θ x = R cos θ



⇒ dV =

1 dq 1 = × ( 2p Rs sin θ ) dθ 4pe 0 R 4pe 0

sR sin θ dθ 2e 0

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 95

∫

VP = V0 - E dx

0

x

⇒ VP = V0 -

s

∫ 2e 0

⇒ VP = V0 -

⇒ dr

The potential due to this element at the centre O of the hemisphere dV =

x

dx 0

s x 2e 0

ELECTROSTATIC POTENTIAL DUE TO A CHARGED CONDUCTING SPHERE/ CHARGED SHELL For points located outside the charged sphere we can assume the entire charge to be concentrated at its centre. Thus electrostatic potential at a distance r from the centre of sphere, outside it will be

Voutside = V =

Q 4pe 0 r

9/20/2019 10:24:52 AM

1.96  JEE Advanced Physics: Electrostatics and Current Electricity Q

(b) Electric potential at any point outside the sphere is

R

x P

At the points on surface of sphere, the potential is given by Vsurface



 



For example, in the figure shown, potential at A is

 

{ r = distance of the point from the centre } VA =

1 ⎡ q A qB qC ⎤ + + 4pe 0 ⎢⎣ rA rB rC ⎥⎦ C

At the interior points of sphere the electric field is zero at all points inside, so, we can say that the inner portion of the conducting sphere / charged shell is an equipotential region and hence at every interior point potential is same as that on its surface. Thus we have Vinside = Vsurface =

1 q ⋅ 4pe 0 r





Q = VS = 4pe 0 R

V=

qB

qC

qA

B A



Q 4pe 0 R

The variation of V with distance r from the centre is shown in figure Also, the same results are valid for a uniformly charged hollow sphere

Similarly, potential at B is VB =

1 ⎡ q A qB qC ⎤ + + 4pe 0 ⎢⎣ rB rB rC ⎥⎦



 



and potential at C is ,



 

VC =

1 ⎡ q A qB qC ⎤ + + 4pe 0 ⎢⎣ rC rC rC ⎥⎦

V Q 4πε0R

O

ELECTROSTATIC POTENTIAL DUE TO A NON-CONDUCTING UNIFORMLY CHARGED SPHERE

V ∝1 r r=R

r

Problem Solving Technique(s) To find the electric potential due to a conducting sphere (or shell) we should keep in mind the following two points: (a) Electric potential on the surface and at any point inside the sphere is V=

1 q ⋅ 4pe 0 R

  (R = radius of sphere)

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 96

Here too, for outer and surface points the potential remains same as that of a conducting sphere (as if the entire charge is concentrated at the centre). So, Q Voutside = Vo =  { for r > R } 4pe 0 r

Vsurface = VS =

Q  4pe 0 R

{ for r = R }

For an interior point, unlike to a conducting sphere, potential will not remain uniform as electric field exists inside this region. Inside a uniformly charged sphere electric field is in radially outward direction, so as we move away from centre, in the direction of electric field the potential decreases.

9/20/2019 10:24:57 AM

Chapter 1: Electrostatics 1.97 S r

⎡ rR3 ⎢ 3e 0 r V=⎢ ⎢ r ( 3R2 - r 2 ) ⎢ 6 e ⎣ 0

P Q

R

r≥R

( outside and at surface )

r a . The outer shell has charge Q but the inner shell is grounded. Find the charge on the inner shell. 12. At a certain distance from a point charge, the magnitude of the electric field is 500 Vm-1 and the electric potential is -3 kV. (a)  What is the distance to the charge? (b)  What is the magnitude of the charge?

9/20/2019 10:25:26 AM

1.100  JEE Advanced Physics: Electrostatics and Current Electricity

13. Find out the points on the line joining two charges +q and -3q (kept at a distance of 1 m) where electric potential is zero. 14. A light unstressed spring has length d. Two identical particles, each with charge q, are connected to the opposite ends of the spring. The particles are held stationary a distance d apart and then released at the same time. The system then oscillates on a horizontal frictionless table. The spring has a bit of internal kinetic friction, so the oscillation is damped. The particles eventually stop vibrating when the distance between them is 3d. Find the increase in internal energy that appears in the spring during the oscillations. Assume that the system of the spring and two charges is isolated. 15. Consider a ring of radius R with the total charge Q spread uniformly over its perimeter. What is the potential difference between the point at the center of the ring and a point on its axis a distance 2R from the center? 16. A uniform electric field E0 is directed along positive y-direction. Find the change in electric potential energy of a positive test charge q0 when it is displaced in this field from yi = a to y f = 2a along the y-axis. 17. A wire having a uniform linear charge density l is bent into the shape shown in figure. Find the electric potential at point O. R

2R

2R

19. An infinite number of charges each equal to q are placed along the x axis at x = 1, x = 4, x = 8, …. and so on. Find the potential and electric field at the point x = 0 due to this set of charges. What will be potential and electric field if in the above set up the consecutive charge have opposite sign? 20. Two concentric spheres of radii R and 2R are charged. The inner sphere has a charge of 1 mC and the outer sphere has a charge of 2 mC of the same sign. The potential is 9000 V at a distance 3R from the common center. What is the value of R? 1μC

2μC

R

P 3R

2R

21. A point charge Q is held fixed at origin. A second point charge -q having mass m is placed on the x-axis, at ( 2, 0 ) from the origin. The second point charge is released from rest. What is its speed when ⎛ ⎞ it is at ⎜ , 0 ⎟ from the origin? ⎝4 ⎠ 22. Eight point charges are placed at the corners of a cube of edge a as shown in figure. Find the work done in disassembling this system of charges. +q

–q

O

18. A disk of radius R has a non-uniform surface charge density s = s 0 r , where s0 is a positive constant and r is measured from the center of the disk. Find the potential at P.

R

P x

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 100

+q

–q +q

–q +q

–q

23. A conducting bubble of radius a, thickness t ( t  a ) has potential V. Now the bubble collapses into a droplet. Find the potential of the droplet. 24. Two point charges 2 mC and -4 mC are kept 6  m apart. At what points on the line joining the charges, the electric potential will be zero.

9/20/2019 10:25:30 AM

Chapter 1: Electrostatics 1.101

EQUIPOTENTIAL SURFACES: INTRODUCTION Consider a situation in which a charge is displaced from a point A to B on a surface S which is perpendicular to the direction of electric field. Then work done in displacing the charge will be zero, because electric force is normal to the direction of displacement.

Planar equipotential surface

S

EQUIPOTENTIAL SURFACE: PROPERTIES

B

A E

Since no work is done in moving the charge from A to B , hence we can say that A and B are at same potential or we can say that all the points lying on surface S are at same potential. So, we call the surface S to be an Equipotential Surface. Following figures show equipotential surfaces in the surrounding of point charge, a long charged wire and at far off distance from both. Mathematically equipotential surface is the locus of all points which have the same potential due to a charge distribution. Any surface in an electric field at every point of which the direction of electric field is normal to the surface can be regarded as equipotential surface.

Since, we have already read that an equipotential surface is the locus of all points which have the same electrostatic potential due to a charge distribution. Equipotential surfaces possess the following properties. (a) The work done in moving a charge along an equipotential surface is always zero. (b) An electric field line must cut the equipotential surface at right angles. (c) The tangential component of the electric field along the equipotential surface is zero, otherwise non-zero work would be done to move a charge from one point on the surface to another point on it. (d) No two equipotential surfaces cross each other. (e) An equipotential surface for a point charge must be a family of concentric spheres, for a line charge must be a family of coaxial cylinders. (f) The equipotential surface for a point charge or a line charge at far off distance is a family of planes perpendicular to the field lines. Illustration 71

A charge q0 is transported from point A to B along the arc AB with centre at C as shown in figure near a long charged wire with linear density l lying in the same plane. Find the work done in doing so. q 2r Spherical equipotential Cylindrical equipotential surfaces for point charge    surfaces for line charge (Continued)

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 101

C

r

A

r B

9/20/2019 10:25:35 AM

1.102  JEE Advanced Physics: Electrostatics and Current Electricity Solution

Since, electric field is conservative in nature, so work done does not depend upon the path followed between the two points.

y



∫

  Vx - Vy = Edx x

⇒ WA→B = WA→C + WC →B But WC →B = 0 (because B and C are at same potentials or B and C lie on the same equipotential line)

x

y

⇒ WA→B = WA→C 2r

  and WA→C = - q0 E ⋅ dr

∫

3r

⇒ WA→C = ⇒ WA→C =

q0 2pe 0

3r

l

l ⎫ ⎧ ⎨∵ E = ⎬ 2 pe 0r ⎭ ⎩

∫ r dr 

2r

q0 l ⎛ 3⎞ log e ⎜ ⎟ ⎝ 2⎠ 2pe 0

(e) Figure shows two equipotential surfaces in a uniform electric field E. If we have to calculate the potential difference between two points A and B as shown in figure, we simply find the potential difference between the two equipotential surfaces on which the points lie. So,

  VA - VB = Ed d

B

Remark(s) (a) A region in which at every point electric field is zero, can be regarded as an equipotential region. (b) As we have already discussed whenever charge is given to a metal body, it is distributed on its outer surface in such a way that net electric field at every interior point of body is zero. Thus if inside a metal body, a charge is displaced, no work is done in the process because electric field at every point is zero. Hence we can say that the whole metal body is equipotential at the inside. (c) For uniform electric field as shown in figure, we have

  VA - VB = Ed {As E = constant}

A

B d

(d) For a non-uniform electric field like shown in ­figure, we have (between two points),

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 102

A E S1

S2

(f) Similarly, the figure shows a line charge with linear charge density l . To find potential difference between points A and B which lie on equipotential surfaces S1 and S2, we find the potential difference between these surfaces and for that we consider a point P at a distance x from wire as shown. The electric field at point P is



 

E=

l 2pe 0 x λ + + + + + + + + +

S1 r2

B P

x r1

S2

E

A

9/20/2019 10:25:42 AM

Chapter 1: Electrostatics

Now the potential difference between surface S1 and S2 is given by

⇒ VA - VB =

∫

VB - VA = - E dx

l

∫ 2pe x dx 0

r1

r2



r2

⇒ VA - VB =

r1

1.103

l ⎛r ⎞ log e ⎜ 2 ⎟ 2pe 0 ⎝ r1 ⎠

Test Your Concepts-VIII

Based on Equipotential surfaces (Solutions on page H.30) A

-1

1. A uniform electric field of magnitude 325 Vm is directed in the negative y direction in figure. The coordinates of point A are ( -20, - 30 ) cm, and those of point B are ( 40, 50 ) cm . Calculate the potential difference VB - VA , along the path shown.

0

2

4

6

8

y B

B

x

A

E

2. Figure shows several equipotential lines each labelled by its potential in volt. The distance between the lines of the square grid represents 2 cm. (a) Where the magnitude of the field is larger, at A or at B? Explain. (b) What is the electric field at B? (c) Represent what the field looks like by drawing at least eight field lines in all directions of the page.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 103

3. A point charge q is located at x = -R, and a point charge -2q is located at the origin. Prove that the equipotential surface that has zero potential is ⎛ 4R ⎞ a sphere centered at ⎜ - , 0, 0 ⎟ and having a ⎝ 3 ⎠ 2R radius r = . 3 4. Three infinitely long linear charges of charge density l , l and -2l are placed in space. A point in space is specified by its perpendicular distance r1, r2 and r3 respectively from the linear charges. Prove that for the points which are equipotential, we have



r1r2 r32

= constant

9/20/2019 10:25:47 AM

1.104  JEE Advanced Physics: Electrostatics and Current Electricity

RELATION BETWEEN ELECTROSTATIC FIELD AND POTENTIAL   Since dV = -E ⋅ dl In cartesian coordinates, we have   E = Ex iˆ + Ey ˆj + Ez kˆ and dl = dx iˆ + dy ˆj + dz kˆ , so we get

(

)(

dV = - Ex iˆ + Ey ˆj + Ez kˆ ⋅ dxiˆ + dyjˆ + dzkˆ ⇒ dV = - ( Ex dx + Ey dy + Ez dz ) ⇒ Ex = -

)

∂V ∂V ∂V , Ey = and Ez = ∂x ∂z ∂y

Now, by introducing a new differential quantity  called the “del (gradient) operator ( ∇ ) ”, we get electrostatic field as the negative gradient of potential. So,   E = -∇ V …(1)  ∂ ∂ ∂ where ∇ = i + j + k ∂x ∂y ∂z  ⎛ ∂ V  ∂V  ∂V ⎞ ⇒ E = - ⎜ i +j +k ∂y ∂z ⎟⎠ ⎝ ∂x where







Partial Derivative of V w.r.t. x OR ⎞ ∂V ⎛ = ⎜ Derivative of V w.r.t. x keeping y and ⎟ ∂x ⎜ ⎟⎠ z constant ⎝ Partial Derivative of V w.r.t. y OR ⎞ ∂V ⎛ = ⎜ Derivative of V w.r.t. y keeping x and ⎟ ∂y ⎜ ⎟⎠ z constant ⎝ Partial Derivative of V w.r.t. z OR ⎞ ∂V ⎛ = ⎜ Derivative of V w.r.t. z keeping x and ⎟ ∂z ⎜ ⎟⎠ y constant ⎝

∂V ∂V ⎛ ∂y ⎞ ⎛ ∂x ⎞ = kx ⎜ ⎟ = kx = ky ⎜ ⎟ = ky; ⎝ ⎠ ∂y ∂x ∂x ⎝ ∂y ⎠ ∂V and =0 ∂z  ⇒   E = -k yi+ xj

where



(

)

 Notice that ∇ operates on a scalar quantity (electric potential) and results in a vector quantity (elec tric field). Mathematically, we can think of E as the negative of the gradient of the electric potential V. Physically, the negative sign implies that if V increases as a positive charge moves along some ∂V direction, say x , with > 0, then there is a non ∂x  vanishing component of E in the opposite direction

( -Ex ≠ 0 ) , because

E always goes from higher V to

lower V. If the charge distribution possesses spherical symmetry, then the resulting electric field is a function of the radial distance r , i.e., E = Er rˆ . In this  case, dV = -Er dr . If V ( r ) is known, then E may be obtained as  ⎛ dV ⎞ ˆ E = Er rˆ = - ⎜ r ⎝ dr ⎟⎠ For example, the electric potential due to a point q charge q is V ( r ) = . Using the above formula, 4pe 0 r the electric field is simply  ⎛ q ⎞ E =⎜ rˆ 4pe 0 r 2 ⎟⎠ ⎝ Illustration 72

Just to make you understand the concept discussed here, I have a small problem discussed here for your comfort.  PROBLEM: Find the electrostatic field E for the potential V = kxy . SOLUTION: Since  ⎛ ∂V  ∂V  ∂V ⎞ +j +k ,    E = - ⎜ i ∂y ∂z ⎟⎠ ⎝ ∂x

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 104

Suppose that the electric potential in some region of space is given by V ( x , y , z ) = V0 exp ( - az ) cos ax , where a is a positive constant. Find the electric field everywhere. Solution



  ⎛ ∂V ˆ ∂V ˆ ∂V ˆ ⎞ E = -∇V = - ⎜ i+ j+ k ∂y ∂z ⎟⎠ ⎝ ∂x

9/20/2019 10:26:01 AM

Chapter 1: Electrostatics 1.105



∂V ∂ = -V0 exp ( - az ) [ cos ( ax ) ] ∂x ∂x

Remark(s) (a) For an attractive system U is always NEGATIVE. (b) For a repulsive system U is always POSITIVE. (c) For a stable system U must be MINIMUM

∂V ⇒ Ex = = + aV0 sin ( ax ) exp ( - az ) ∂x Ey =

∂V =0 ∂y

{∵ the given expression has no dependence only} Ez = -



⇒ Ez = -

∂V ∂ = -V0 cos ( ax ) ⎡⎣ exp ( - az ) ⎤⎦ ∂z ∂z

dU =0 dx

⇒  F=-

y

=0

}

 E = aV0 exp ( - az ) ⎡⎣ sin ( ax ) iˆ + cos ( ax ) kˆ ⎤⎦

  ∂V ⎞ ⎛ ∂V ˆ ∂V +j +k E = -∇V = - ⎜ iˆ ∂y ∂z ⎟⎠ ⎝ ∂x    (e) If we are given E ( x , y, z ) = E x iˆ + E y ˆj + E z kˆ , then iˆ

iˆ

Illustration 73

Find the potential function V ( x , y ) of an electrostatic  field E = 2 axy iˆ + a x 2 - y 2 ˆj where a is a constant.

)

Solution

  V = - E ⋅d

∫



 



⇒  V = - ⎢⎡ E x dx + E y dy + E z dx ⎤⎥ ⎣ ⎦ The integrals are to be calculated within specified limits.

 where, d  = dxiˆ + dyjˆ + dzkˆ



Let V0 be the potential at origin.   Since, dV = -E ⋅ d 

∫

∫

⇒ V ( x , y ) - V ( 0, 0 ) = -

( 2axy dx + ax ∫ ( )

2

)

ak

2

dy - ay dy

0, 0

∫ { d ( ax y ) - ay dy } 2

2

( 0, 0 )

( x, y ) ⎤ ⎡⎛ ay 3 ⎞ 2 ⎢ ⎥ ⇒   V ( x , y ) - V0 = ⎜ - ax y + ⎢⎝ ⎥ 3 ⎟⎠ ( 0 , 0 ) ⎥⎦ ⎢⎣

⇒   V ( x , y ) = V0 - ax 2 y +

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 105

ay 3

eg

ra d

ie n t of V to

ge

t

E=–∇V

( x, y )

⇒   V ( x , y ) - V0 = -

V (a scalar)

E (a vector)

T

( x, y )

)

Ta

) (

(

∫

integral of E to ge line e k

tV

( x,y ) ⇒ dV = - ⎡⎣ 2 axy iˆ + a x 2 - y 2 ˆj ⎤⎦ ⋅ dx iˆ + dy ˆj + dz kˆ ( 0, 0 ) ( 0, 0 )

∫

∫

V = – ∫if E ⋅ dl

iˆ

( x, y )

(FOR A STABLE SYSTEM)

(d) If we are given V ( x , y, z ), then

{∵ E

(

dU dx

Since F =

∂V = aV0 cos ( ax ) exp ( - az ) ∂z

⇒ E = Ex iˆ + Ez kˆ  ⇒

dU d 2U i.e., = 0 and 2 > 0 dx dx

E

Ex = -

PROBLEM: An electrostatic field is given by  E = -k yi+ xj . Find the electrostatic potential generating such a field.   SOLUTION:  Since V = - E ⋅ dr

(

)

∫



 Let dr = idx + j dy +  k dz



⇒ V = k yi+ xj ⋅ idx + j dy +  k dz

3

∫

(

)(

)

9/20/2019 10:26:13 AM

1.106 JEE Advanced Physics: Electrostatics and Current Electricity

∫ V = k d ( xy ) ∫



⇒ V = k y dx + x dy



⇒

{ ∵ d ( xy ) = x dy + y dx }

⇒ V = k ( xy ) + constant (f) Electrostatic potential at a point due to a positive charge is positive and due to a negative charge is negative. (g) Electrostatic potential due an assembly of charges q1, q2, …., qn at a point P at distance r1, r2, …., rn respectively from the charges is given by V=

1 4pe 0

n

∑r

qi

charge always moves from lower potential to higher potential. (j) Commonly used unit of electrostatic energy is electronvolt (eV) and 1 eV = 1.6 × 10 -19 J. (k) If a charge q of mass m accelerates through a potential V, then velocity of q is calculated by using the Work-Energy Theorem, according to which work done equals the change in kinetic energy i.e.,

i

i =1 (h) If two points A and B are at potentials VA and VB, then work done in taking a test charge q0 from A to B is WA→B = q0 ( VB - VA )

(i) A positive charge always moves from higher potential to lower potential whereas a negative

1 qV = mv 2 2 ⇒

v=

2qV m

(l) Consider two points A and B situated in a uniform electric field at a distance d such that the line joining A and B is parallel to the field. If V be the potential difference between them, then V = Ed ( in magnitude )

Test Your Concepts-Ix

Based on relation Between Electrostatic Field and Potential (Solutions on page H.31) 1. Over a certain region of space, the electric potential is V = 5 x - 3 x 2 y + 2 yz 2 . Find the expressions for the x, y and z components of the electric field over this region. Also calculate the magnitude of the field at the point P that has coordinates ( 1, 0, - 2 ) m ? 2. When an uncharged conducting sphere of radius a is placed at the origin of an xyz cartesian coordinate system that  lies in an initially uniform electric field E = E0 k , the resulting electric potential is V ( x , y, z ) = V0 for points inside the sphere and V ( x , y, z ) = V0 - Eo z + E0 a 3

z

(

3 2 2 z

)

x 2 + y2 + for points outside the sphere, where V0 is the (constant) electric potential on the conductor. Use this equation to determine the x, y and z components of the resulting electric field.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 106

3. In a region of space, the electric potential is represented by V = 2 x + 3 y - z . Obtain an expression for the electric field strength. 4. In a region of space the electric field is given by  E = xiˆ - 2 yjˆ + zkˆ Vm-1 . Calculate the potential

(

)

difference VAB between A(2, 1, 0) m and B(0, 2, 4) m.

5. Find potential difference VAB between A ( 0, 0, 0 ) m and B ( 1, 1, 1) m in an electric field given by  (a) E = yiˆ + xjˆ  (b) E = 3 x 2 yiˆ + x 3 ˆj. What can you say about the nature of the fields? Explain. 6. Determine the electric field strength vector if the potential of this field depends on x, y coordinates as

(

(a) φ = a x 2 - y 2 (b) φ = axy

)

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Chapter 1: Electrostatics 1.107

where a is a positive constant. Draw the approximate shape of these fields using lines of force (in the x, y plane). 7. The potential of a certain electrostatic field has the form V = a x 2 + y 2 + bz 2 , where a and b are

(

)

constants. Find the magnitude and direction of the

MOTION OF CHARGED PARTICLES AND CONSERVATION LAWS For a variety of problems, we have to understand the Conservation Laws that can be applied to the electrostatic system of particles. CASE-1: Law of Conservation of Linear Momentum For an electrostatic system of charges, when no external force acts on the system, then we observe that no net internal forces will be acting on the system because electrostatic forces always form an action reaction pair. So, by a suitable selection of a system or subsystem we can apply Law of Conservation of Linear Momentum.

electric field strength vector. What shape will the equipotential surface have for a > 0, b > 0?

8. Determine the potential V ( x , y, z ) of an electrostatic field E = ayiˆ + ( ax + bz ) ˆj + bykˆ , where a and b are constants, iˆ, ˆj, kˆ are the unit vectors of the axes x, y, z.

CASE-3: Law of Conservation of Angular Momentum Since electrostatic forces are central in nature, so torque due to electrostatic force will be zero. So, we can easily use Law of Conservation of Angular Momentum too. Illustration 74

A proton is fixed at origin. Another proton is released from rest, from a point at a distance r from origin. Taking charge of proton as e and mass as m , find the speed of the proton (a) at a distance 2r from origin (b) at a large distance from origin. e

Total Initial Momentum = Total Final Momentum CASE-2: Law of Conservation of Energy Since electrostatic forces are conservative in nature, so, in the absence of an external force and a dissipative force (such as forces of friction) we can easily use this law as ⎛ Total Initial Energy ⎞ ⎛ Total Final Energy ⎞ = ⎜⎝ of the system ⎟⎠ ⎜⎝ of the system ⎟⎠

Here we need to read the problem carefully for the energies possessed by the system (as selected) initially and finally. Another point which should be kept in mind is that electric field is a conservative field. Work done by electric field on a moving charge only depends on the positions of the charge. If a charge is freely moving in electric field, workdone by electric field on charge is equal to the change in kinetic energy of the charge i.e., W = ΔK In the presence of an external force and the dissipative non conservative force, the law will be suitably modified as below i.e.,

Wext + Wnc = ΔU + ΔK

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 107

Proton

O Proton

r

e

Solution

(a) The proton moves away under electrostatic repulsion. So, by Conservation of Energy, we have     Ui + K i = U f + K f ⇒ 

e2 e2 1 +0 = + mv 2 4pe 0 r 4pe 0 ( 2r ) 2

⇒ 

1 e2 mv 2 = 2 4pe 0 ( 2r )

⇒  v=

e2 4pe 0 rm

(b) Similarly, we have     Ui + K i = U∞ + K∞ ⇒ 

1 e2 mv 2 = 2 4pe 0 r

⇒  v=

2e 2 4pe 0 rm

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1.108  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 75

Illustration 76

A charge q of mass m initially lying at infinity is projected head on with speed v toward another stationary particle of same mass and charge, initially at rest. Find the distance of closest approach of the two particles.

Two particles, each of charge q , and masses m1 and m2 , are released from rest, from points A and B , r distance apart. Calculate the final relative velocity of separation of the particles at large separation from each other. (Neglect gravitational interaction of the particles)

Solution

As the moving charge comes closer to the charge initially at rest, due to electric repulsion it also starts moving and the first one starts retarding. q m +

q m +

v Initially

q m +

q m +

v1

Solution

Both particles move under mutual electrostatic repulsion. Let v1 and v2 are final speed (when the separation becomes large). So, vrel = v1 + v2 Since no external force is acting on particles, so by Law of Conservation of Linear Momentum, we have F

v1

rmin Finally



Since no external force is acting on system, total momentum remains conserved. The two charges get close to each other and their separation will be minimum when each will move with equal common speed say v1. When their speeds are equal say v1, then by Law of Conservation of Linear Momentum, we have

mv = mv1 + mv1

To find the minimum separation between the two particles, using Law of Conservation of Energy, Total Initial Energy = Total Final Energy ⇒ ⇒ ⇒

q2 1 1 1 mv 2 = mv12 + mv12 + 2 2 2 4pe 0 rmin 2 q2 1 ⎛ v⎞ mv 2 = m ⎜ ⎟ +  ⎝ 2⎠ 2 4pe 0 rmin

q2 mv 2 = 4 4pe 0 rmin

⇒ rmin =

q

v⎫ ⎧ ⎨∵ v1 = ⎬ 2⎭ ⎩

pe 0 mv

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 108

B m2, q

F

- m1v1 + m2 v2 = 0

By Law of Conservation of Mechanical Energy we have Ui + Ki = U f + K f q2 1 1 ⇒ + 0 + 0 = m1v12 + m2 v22 + 0 4pe 0 r 2 2 p2 p2 q2 + = 2m1 2m2 4pe 0 r

2 ⎛ 2m1 m2 ⎞ q ⇒ p2 = ⎜ ⎟ ⎝ m1 + m2 ⎠ 4pe 0 r 2 ⎛ 2m1 m2 ⎞ q ⇒ p= ⎜ ⎝ m1 + m2 ⎟⎠ 4pe 0 r

⇒ v1 = and v2 =

p 1 ⎛ 2m1 m2 ⎞ q2 = m1 m1 ⎜⎝ m1 + m2 ⎟⎠ 4pe 0 r

p 1 ⎛ 2m1 m2 ⎞ q2 = m2 m2 ⎜⎝ m1 + m2 ⎟⎠ 4pe 0 r

⇒ vr = v1 + v2

2 2

r

⇒ m1v1 = m2 v2 = p ( say )

⇒

v ⇒ v1 = 2

A m1, q

⇒ vr =

2q 2 ⎛ 1 1 ⎞ + ⎜ 4pe 0 r ⎝ m1 m2 ⎟⎠

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Chapter 1: Electrostatics 1.109 Illustration 77

Figure shows a charge +Q fixed at a position in space. From a large distance another charged particle of charge -q and mass m is thrown with a speed v towards +Q with an impact parameter d (it is the perpendicular distance between the velocities of the two particles) as shown. Find the distance of closest approach of the two particles. m

q

v d Q fixed

Solution

Here we can see that when -q moves toward +Q , an attractive force acts on -q toward +Q . Here as the line of action of force passes through the fix charge, no torque act on -q relative to the fix point charge +Q , thus here we can say that with respect to +Q , the angular momentum of -q must remain constant. The charge -q will be closest to +Q when it is moving perpendicularly to the line joining the two charges as shown. –q v′ F

rmin

If the closest separation in the two charges is rmin then, from conservation of angular momentum we have mvd = mv ′rmin …(1) Further, by Law of Conservation of Energy, we have 1 1 1 qQ mv 2 = mv ′ 2 2 2 4pe 0 rmin

From (1), we get

⇒

vd rmin

⎛ d2 ⎞ 1 1 1 qQ mv 2 = mv 2 ⎜ 2 ⎟ …(2) 2 2 4 pe ⎝ rmin ⎠ 0 rmin

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 109

⇒ rmin =

⎤ ⎡ 16p 2 e 02 m2 v 4 d 2 ⎢ 1+ - 1⎥ 2 2 Q q 2pe 0 mv ⎢⎣ ⎥⎦ Qq

2

Illustration 78

A particle of mass 40 mg having a charge of 5 × 10 -9 C is moving directly towards a fixed positive point charge of magnitude 10 -8 C . When it is at a distance of 10 cm from the fixed positive point charge it has a velocity of 50 cms -1 . At what distance from the fixed point charge will the particle come momentarily to rest? Is the acceleration constant during the motion? Solution

Let it come to momentary rest at point P at distance x from O . Then, by Law of Conservation of Energy. 10–8 C O Fixed

u = 50 cms–1

P

5 × 10–9 C 10 cm x



( U + K )initial = ( U + K )final

⇒

Qq Qq 1 1 2 + mu2 = + m(0 ) 4pe 0 r0 2 4pe 0 x 2

–q

Q

v′ =

⎛ 2Qq ⎞ 2 ⇒ rmin +⎜ r - d2 = 0 2 ⎟ min ⎝ 4pe 0 mv ⎠

where Q = 10 -8 C ⇒

q = 5 × 10 -9 C r0 = 10 cm =

10 m 100

m = 40 mg =

40 g = 40 × 10 -6 kg 1000

u = 50 cms -1 = 0.5 ms -1 9 × 109 × 5 × 10 -17 1 ( 2 + 40 × 10 -6 ) ( 0.5 ) = 0.1 2 9 × 109 × 5 × 10 -17 x

⇒ 45 × 10 -7 + 50 × 10 -7 =

45 × 10 -8 x

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1.110  JEE Advanced Physics: Electrostatics and Current Electricity

⇒ x=

45 × 10 -8 95 × 10 -8

Electric potential at point P will be

= 0.473 m

V=

2q 2Q ( ) 4pe 0 BP 4pe 0 ( AP )

⇒ x = 4.73 cm



The force of repulsion increases as the particle approaches the fixed charge. So the acceleration is not constant during the motion.

⎡ 8 × 10 -6 10 -6 ⎤ ⇒ V = 2 × 9 × 109 ⎢ ⎥ 3 2 ⎥ ⎢ 27 + x 2 +x ⎢⎣ 2 ⎥⎦ 2

Illustration 79

8 ⎡ ⇒ V = 1.8 × 10 4 ⎢ 27 ⎢ + x2 ⎣ 2

Four point charges +8 m C , -1 m C, -1 m C and +8 m C are fixed at the points -

27 3 3 m, m, + m 2 2 2

27 m respectively on the Y-axis. A particle of and + 2 mass 6 × 10 -4 kg and charge +0.1 m C moves along the -x direction. Its speed at x → ∞ is v0 . Find the least value of v0 for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Solution



q0 = +0.1 m C = 10 -7 C ,



m = 6 × 10

-4



3 ⎡ 2 1 ⎞ ⎛ 27 ⎛ ⎞ 2 ⎢ E = -1.8 × 10 ( 8 ) ⎜ - ⎟ ⎜ +x ⎟ ⎝ 2⎠ ⎝ 2 ⎠ ⎢⎣

1 3 ( 1 ) ⎛⎜ - ⎞⎟ ⎛⎜ + x 2 ⎞⎟ ⎝ 2⎠ ⎝ 2 ⎠

 So, E = 0 on x-axis either at x = 0 or when

⇒

kg

and Q = 8 m C = 8 × 10 -6 C

3 m A 2 O 3 – m C 2 D 27 m – 2

+Q



=

32

=

( 4 )3 2 ⎛ 27 ⎞ + x2 ⎟ ⎜⎝ ⎠ 2

3 2

⎤ ⎥ ( 2x ) ⎥⎦

1 ⎛3 2⎞ ⎜⎝ + x ⎟⎠ 2

32

1 ⎛3 2⎞ ⎜⎝ + x ⎟⎠ 2

32

v0 P

–q

m q0

x

+Q

AP = CP =

3 + x2 2

BP = DP =

27 + x2 2

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 110

5 m 2 The least value of kinetic energy of the particle at infinity should be enough to take the particle upto

This equation gives x = ±

–q x

Let P be any point at a distance x from origin O . Then

⎛ 27 ⎞ + x2 ⎟ ⎜⎝ ⎠ 2

32

-

⎛3 ⎞ ⎛ 27 ⎞ ⇒ ⎜ + x2 ⎟ = 4 ⎜ + x2 ⎟ ⎝2 ⎠ ⎝ 2 ⎠

y 27 m 2 B

dV , so dx

4

8



⎤ …(1) ⎥ 3 + x2 ⎥ 2 ⎦

Since, electric field at P is given by E = -

In the figure, q = 1 m C = 10 -6 C ,

1

5 5 m because at x = + m , E = 0 , so electro2 2 static force on charge q0 is zero or Fe = 0 . x=+

For at x > x-axis) and for x < x-axis)

5 m , E is repulsive (towards positive 2 5 m , E is attractive (towards negative 2

9/20/2019 10:27:17 AM

Chapter 1: Electrostatics

Now, from equation (1), potential at x = 8 ⎡ V = 1.8 × 10 4 ⎢ 27 5 ⎢ + ⎣ 2 2



x=

From equation (1), potential at origin ( x = 0 ) is

5 m 2

1 ⎤ ⎡ 8 V0 = 1.8 × 10 4 ⎢ = 2.4 × 10 4 V 3⎥ 27 ⎢ ⎥ 2 2 ⎣ ⎦ Let T be the kinetic energy of the particle at origin. Applying energy conservation at x = 0 and at x = ∞

1

⎤ ⎥ 3 5 + ⎥ 2 2⎦

⇒ V = 2.7 × 10 4 V Applying energy conservation at

x→∞

and

5 m 2



1 mv02 = q0V 2

⇒

v0 =

…(2)

2q0V m

Substituting the values v0 =

2 × 10

-7

× 2.7 × 10

4

6 × 10 -4

v0 = 3 ms -1

∴

1.111



T + q0V0 =

1 mv02 2

But

1 mv02 = q0V 2

⇒

T = q0 ( V - V0 )

⇒

T = ( 10 -7 ) ( 2.7 × 10 4 - 2.4 × 10 4 )

⇒

T = 3 × 10 -4 J

{from equation (2)}

Please note here that, E = 0 or F on q0 is zero at x = 0 and x = ±

5 m . Of these x = 0 is stable equi2

librium position and x = ±

Minimum value of v0 is 3 ms -1

rium position.

5 m is unstable equilib2

Test Your Concepts-x

Based on Motion of Charged Particles in Electric Field (Solutions on page H.32) 1. Consider an electron to be released from rest in a uniform electric field whose magnitude is 6 kVm-1. Find, (a) through what potential difference will it have passed after moving 1 cm? (b) how fast will the electron be moving after it has travelled 1 cm? 2. (a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120 V. (b) Calculate the speed of an electron that is accelerated through the same potential difference. 3. Two particles, having charges 20 nC and -20 nC, are placed at the points with coordinates (0, 4) cm and (0, -4) cm, as shown. A particle with charge 10 nC is located at the origin.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 111

20 nC 4 cm 10 nC

3 cm

40 nC

4 cm –20 nC

(a) Find the electric potential energy of the configuration of the three fixed charges. (b) A fourth particle, with a mass of 2 × 10 -13 kg and a charge of 40 nC, is released from rest at the point (3, 0) cm. Find its speed after it has moved freely to a very large distance away.

9/20/2019 10:27:29 AM

1.112  JEE Advanced Physics: Electrostatics and Current Electricity

4. In 1911 Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles from thin sheets of gold. An alpha particle, having charge +2e and mass 6.67 × 10 -27 kg, is a product of certain radioactive decays. The results of the experiment led Rutherford to the idea that most of the mass of an atom is in a very small nucleus, with electrons in orbit around it his planetary model of the atom. Assume an alpha particle, initially very far from a gold nucleus, is fired with a velocity of 2 × 107 ms -1 directly toward the nucleus (charge +79e). How close does the alpha particle get to the nucleus before turning around? Assume the gold nucleus remains stationary. 5. Four identical particles each have charge q and mass m. They are released from rest at the vertices of a square of side L. How fast is each charge moving at the instant when their distance from the center of the square doubles? 6. An electron is released from rest on the axis of a uniform positively charged ring, 0.1 m from the ring’s center. If the linear charge density of the ring is 0.1 mCm-1 and the radius of the ring is 0.2 m, how fast will the electron be moving when it reaches the center of the ring? 7. Four balls, each with mass m, are connected by four non-conducting strings to form a square with side , as shown in Figure. The assembly is placed on a horizontal nonconducting frictionless surface. Balls 3 and 4 each have charge q, and balls 1 and 2 are uncharged. Find the maximum speed of balls 3 and 4 after the string connecting them is cut. 1

2 l

4

3

8. The x-axis is the symmetry axis of a stationary uniformly charged ring of radius R and charge Q. A point charge Q of mass M is located initially at the center of the ring. When it is displaced slightly, the point charge accelerates along the x-axis to

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 112

infinity. Show that the ultimate speed of the point charge is v=

  

Q2 2pe 0 MR Q R

Q

x

v

Uniformly charged ring

9. Figure shows a small ball A of mass m and charge q connected at one end of light rod of length . Another identical ball B is placed on smooth ground at a distance  from the point O, where the rod is clamped. A slight jerk on ball A makes the rod start rotating clockwise. At the instant when A strikes the ground, ball B is moving with a velocity u towards right and is at a position 2  away from the point O. Calculate the kinetic energy of the ball A when it strikes the floor. A l B

O l

10. A charged particle of charge +q, mass m is thrown towards another fixed charge +Q from a very large distance away with speed v0 with an impact parameter . Find the distance of closest approach of the two charges. 11. The potential difference between two large parallel plates is varied as V = at, where a is a constant and t is time. An electron starts from rest at t = 0 from the plate which is at negative potential. If the distance between the plates is  , mass of electron is m and charge on electron is -e then find the velocity of the electron when it reaches the other plate. 12. A point charge -q revolves around a fixed charge +Q in elliptical orbit. The minimum and maximum distance of q from Q are r1 and r2, respectively. The

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Chapter 1: Electrostatics 1.113

mass of revolving particle is m. Q > q and assume no gravitational effects. Find the velocity of q at positions when it is at a distance r1 and r2 from Q. 13. A proton of mass m approaches a free proton from a very large distance with a velocity v0 along the straight line joining their centres. Calculate the

CONDUCTORS AND THEIR PROPERTIES Insulators such as glass or paper are materials in which electrons are attached to some particular atoms and cannot move freely. On the other hand conductors (such as metals) have electrons capable of moving around freely. If a resultant electric field exists in the conductor then free charges will experience a force which will set a current flow. When no current flows, the resultant force and the electric field must be zero. Thus, under electrostatic conditions the value of  E at all points within a conductor is zero. This idea, together with the Gauss’s Law can be used to prove several interesting facts regarding a conductor.

The Electric Field is Zero Inside a Conductor If we place a solid spherical conductor in a constant  external field E0 , the positive and negative charges will move toward the polar regions of the sphere (the regions on the left and right of the sphere in figure shown), thereby inducing an electric field Ei . Inside  E the conductor, i points in the direction opposite to  E0 . Since charges are mobile, they will continue to  move until Ei completely cancels E0 inside the conductor. At electrostatic equilibrium, E must vanish insidea conductor. Outside the conductor, the electric field Ei due to the induced charge distribution corresponds to a dipole field, and the total electric field is    simply the resultant of both, so E = E0 + Ei . The field lines are depicted in figure. Thus if a metal body is placed in an electric field, charge induction on the body surface starts and the continuous flow of electrons takes place inside

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 113

distance of closest approach between the two protons. v0

Rest

m

m

v

m

v

rmin

the metal body (conductor) till the net electric field inside the body becomes zero.

E=0

E = E0 + E1 Placing a conductor in a uniform electric field E

Hence E - Ei = 0 Thus we can say that when all charges in a metal body are at rest, net electric field inside it is zero or the electric field inside the body due to the charges on its surface always balances the external electric field in the body. This is the condition of ELECTROSTATIC EQUILIBRIUM.

Net Excess Charge on a Conductor Resides on Its Outer Surface Consider a charged conductor carrying a charge q and no currents are flowing in it i.e., in electrostatic equilibrium. Now consider a Gaussian  surface inside the conductor everywhere on which E = 0 (because it lies inside the conductor). Since, according to Gauss’s Law,   q E ⋅ dA = enc e0 S   Since E = 0 , so qenc = 0 (because it lies inside the conductor) Thus, the sum of all charges inside the Gaussian ­surface is zero.

∫

9/20/2019 10:27:39 AM

1.114  JEE Advanced Physics: Electrostatics and Current Electricity Gaussian (E = 0) surface



Et = 0 (on the surface of a conductor) En

q

Et b

Conductor a Δx

This Gaussian surface has been taken just inside the surface of the conductor. So, any charge on the conductor must reside on the surface of the conductor. In other words, “Under electrostatic conditions (condition of electrostatic equilibrium) the excess charge on a conductor resides on its outer surface”.

∫

path abcda shown in figure.  Since the electric field E is conservative, the line ­integral around the closed path abcda must be zero.   E ⋅ dl = Et ( Δl ) + En ( Δx ) cos ( 180 ) + ⇒

∫

Δl′

c

d conductor

This implies that the surface of a conductor in electrostatic equilibrium is an equipotential surface. Consider two points A and B on the surface of a conductor as shown, B

The Tangential Component of( E )is Zero on the Surface of a Conductor We have already seen that for an isolated conductor, the electric field is zero in its interior. Any excess charge placed on the conductor must then distribute itself on the surface, as implied by Gauss’s Law.   Consider the line integral E ⋅ dl around a closed

Δl

Δx

A conductor

Since the tangential component Et = 0 , the potential difference is B

  VB - VA = - E ⋅ dl = 0

∫

A

  because E is perpendicular to dl . Thus, points A and B are at the same potential with VA = VB .

abcda

⇒

0 ( Δl ) + En ( Δx ) cos 0 = 0



∫

  E ⋅ dl = Et ( Δl ) = 0

abcda

where Et and En are the tangential and the normal components of the electric field, respectively. We have oriented the infinitesimal segment ab so that it is parallel to Et . Also here we take ad = bc . If we take ad ≠ bc then too since we have taken an infinitesimal loop abcda so in the limit where both Δx and Δx ′ → 0 , we will have Et Δl = 0 . However, since the length element Δl is finite, we conclude that the tangential component of the electric field on the surface of a conductor vanishes.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 114

Electric Field at any Point Close to the Charged Conductor is s /e0 Consider a charged conductor of irregular shape. In general, surface charge density will vary from place to place. At a small surface ΔA , let us assume it to be a constant s . Now construct a Gaussian surface in the form of a cylinder also called as Gaussian Pill Box of cross-section ΔA. One plane face of the cylinder is inside the conductor and other outside the conductor close to it. The surface inside  the conductor does not contribute to the flux as E is zero everywhere inside the conductor. The curved surface outside the conductor also does not contribute to flux as E is always normal to the charged conductor and hence

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Chapter 1: Electrostatics 1.115

parallel to the curved surface. Thus the only contribution to the flux is through the plane face outside the conductor. E ΔA

The Potential of a Charged Conductor Throughout Its Volume is Same

 In any region in which E = 0 at all points, such as the region vary far from all charges or the  interior of a charged conductor, the line integral of E is zero along any path. It means that any two points in the conductor are at the same potential or have zero potential difference or the interior of a charged conductor is an equipotential region.

E=0

So, from Gauss’s Law,   q E ⋅ dA = enc e0 S ( s ) ( ΔA ) ⇒ EΔA = e0

∫

⇒ E = En =

s­ urface. Since the conductor itself has a charge +Q, the amount of charge on the outer surface of the conductor must be Q + q . (Think about the concept of charge conservation to get this as an answer).

Remark(s)

s e0

(This result holds good for a conductor of any arbitrary shape)

(a) Electric field changes discontinuously at the surface of a conductor. Just inside the conductor it s is zero and just outside the conductor it is . In e0 s fact, the field gradually decreases from to zero e0 in a small thickness of about 4 to 5 atomic layers at the surface.

Conductor with Charge Inside a Cavity

E2

Consider a hollow conductor shown in figure. Suppose the net charge carried by the conductor is +Q . In addition, there is a charge q inside the cavity. What is the charge on the outer surface of the conductor? Gaussian surface + + + +

+

– q– + – – – – –

+ +

+

+ + +

+

+

Since the electric field inside a conductor must be zero, the net charge enclosed by the Gaussian surface shown in figure must be zero. This implies that a charge -q must have been induced on the cavity

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 115

++ + + + + + + +

+ + + 2 + + + + ++

+ + + ++

+ + 1 + + + + ++

E1

(b) For a nonuniform conductor the surface charge density (s ) varies inversely as the radius of curvature ( r) of that part of the conductor, i.e.,

+

+ + + Conductor with a cavity

+

++

s∝

1 radius of curvature ( r )



  



For case in the figure, r1 < r2



⇒ s1 > s 2



⇒ E1 > E2 

s ⎫ ⎧ ⎨∵ E = ⎬ e0 ⎭ ⎩

9/20/2019 10:27:54 AM

1.116  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 80

(–7Q + 2Q) = –5Q –2Q +2Q

Consider two concentric conducting spheres. The outer sphere is hollow and initially has a charge -7Q on it. The inner sphere is solid and has a charge +2Q on it. (a) Calculate the charge on the outer and inner ­surface of the outer sphere. (b) If a wire is connected between the inner and outer spheres, after electrostatic equilibrium is established, how much total charge is on the outside sphere? How much charge is on the outer and inner surface of outer sphere? Does the electric field at the surface of the inside sphere change when the wire is connected? (c) Now, if we return to the original condition as specified in (a) and connect the outer sphere to the ground with a wire and then disconnect it. How much total charge will be on the outer sphere? How much charge will be on the inner and outer surface of the outer sphere?

Charge transferred to the ground –5Q

(b)

Total charge on outer surface of outer sphere is -5Q and the total charge on inner surface is zero. The electric field at the surface of the inner sphere goes to zero after connection. Consider a Gaussian surface just on the surface of inner sphere. So, according to Gauss’s Law

ϕE = E ( 4p r 2 ) =



Qenclosed =0 e0

( as Qenclosed = 0 )

⇒ E = 0

Solution

(a) The charges on the inner sphere induce equal magnitude of charge, but opposite in sign, on the inner surface of outer sphere. Sum of all the induced charges is always zero. Therefore, an equal amount of charge must come on the outer surface. Thus outer and inner surface of outer sphere have charges -5Q and -2Q respectively. – – – – – + + + – + – – + + – – + + – – – ++ +2Q + – + – – + – + + – – – – – – –

–

(–7Q + 2Q) on outer surface

–2Q on inner surface

(a)

(b) In electrostatic equilibrium, charge does not reside inside a conductor. So, when outer and inner spheres are connected by a wire, the entire charge is transferred to the outer surface from inner sphere.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 116

(c) When the outer sphere is grounded the charge on the surface is transferred to ground, thus charge is reduced to zero. The final charge distribution is shown in figure (b). Illustration 81

A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge per unit length of l , and the cylinder has a net charge per unit length of 2l . From this information, use Gauss’s Law to find (a) the charge per unit length on the inner and outer surfaces of the cylinder and (b) the electric field outside the cylinder, a distance r from the axis. Solution

(a) Inside surface: consider a cylindrical surface within the metal. Since E inside the conducting shell is zero, the total charge inside the gaussian charge surface must be zero, so the inside = -l length

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Chapter 1: Electrostatics

Outside surface: The total charge on the metal charge cylinder is qout = 2l l + l l so the outside length is 3 l

0 = l l + qin

So

1.117

qin = -l l

⇒ 2l l = qin + qout

(b) E =

3l radially outward 2p ∈0 r

lenc ⎫ ⎧ ⎨∵ E = ⎬ 2pe 0 r ⎭ ⎩

Test Your Concepts-xI

Based on Conductors (Solutions on page H.35) 1. Two identical conducting spheres each having a radius of 0.5 cm are connected by a light 2 m long conducting wire. A charge of 60 mC is placed on one of the conductors. Assume that the surface distribution of charge on each sphere is uniform. Determine the tension in the wire. 2. Two infinite parallel planes carry equal but opposite uniform charge densities sA and sB (shown in figure). Find the field in each of the three regions:

5.

6. (i)

(ii)

(iii)

A

B

σA

σB

(i) to the left of both (ii) between them (iii) to the right of both. Also find the field in the three regions for s A = +s and s B = -s . 3. In a particular region of air at an altitude of 500 m above the ground the electric field is found to be 120 NC -1 directed downwards. At 600 m above the ground the electric field is 100 NC -1 downwards. Calculate the average volume charge density in the layer of air between these two elevations. Is it positive or negative? 4. The electric field on the surface of an irregularly shaped conductor varies from 56 kNC -1 to 112 kNC -1 . Calculate the local surface charge

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 117

7.

8.

density at the point on the surface where the radius of curvature of the surface is (a) greatest and (b) smallest A conducting spherical shell of inner radius a and outer radius b carries a net charge Q. A point charge q is placed at the center of this shell. Determine the surface charge density on (a) the inner surface of the shell and (b) the outer surface of the shell. A hollow conducting sphere is surrounded by a larger concentric spherical conducting shell. The inner sphere has charge -Q, and the outer shell has net charge +3Q. The charges are in electrostatic equilibrium. Using Gauss’s Law, find the charges and the electric fields everywhere. R A positive point charge is at a distance from the 2 center of an uncharged thin conducting spherical shell of radius R. Sketch the electric field lines set up by this arrangement both inside and outside the shell. A solid, insulating sphere of radius a has a uniform charge density r and a total charge Q. Concentric with this sphere is an uncharged, conducting hollow sphere whose inner and outer radii are b and c, as shown in figure. Insulator a

Conductor

b c

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1.118  JEE Advanced Physics: Electrostatics and Current Electricity



(a) Find the magnitude of the electric field in the regions r < a, a < r < b, b < r < c, and r > c. (b) Determine the induced charge per unit area on the inner and outer surfaces of the hollow sphere. 9. A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly distributed throughout its volume. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c, and having a net charge -Q, as shown in Figure –Q 3Q

a



c b

(a) Construct a spherical gaussian surface of radius r > c and find the net charge enclosed by this surface.

CHARGE DISTRIBUTION ON A SYSTEM OF PARALLEL PLATES CASE-1: Consider a large conducting plate that has been given a charge +Q . Due to symmetry, this charge is distributed equally on both sides of its surfaces as shown. If surface area of each face is A , charge density on the surface is given by

s=

Q 2A



(b) Find the magnitude and the direction of electric field at r > c.    (c) Find the electric field in the region with radius r where b < r < c. (d) Construct a spherical gaussian surface of radius r, where b < r < c, and find the net charge enclosed by this surface.   (e)  Construct a spherical gaussian surface of radius r, where a < r < b, and find the net charge enclosed by this surface. (f)  Find the electric field in the region a < r < b.   (g) Construct a spherical gaussian surface of radius r < a, and find an expression for the net charge enclosed by this surface, as a function of r.   (h)  Find the electric field in the region r < a.    (i) Determine the charge on the inner surface of the conducting shell. (j) Determine the charge on the outer surface of the conducting shell.   (k) Make a plot of the magnitude of the electric field versus r.

CASE-2: Now, if another uncharged plate is placed in front of this charged sheet as shown in figure. On the sides facing each other, plate of this plate an equal and Q is induced and on the back suropposite charge 2 Q face + is induced. The electric field, between the 2 s plates is , where s is the charge density of the e0 surface. In figure the direction of electric field in different region is indicated by arrows.

+Q

+Q

A

+

Q 2

+

Q 2

B

The electric field, on both sides of the plate, at points A and B is given by

EA = EB =

s Q = e 0 2 Ae 0

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 118

+Q 2

–Q

A

+Q 2

–Q 2

B

+Q 2

d

Now we find the potential difference between the two plates A and B as

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Chapter 1: Electrostatics 1.119



VA − VB =

σ d ε0

{As electric field is uniform between the plates}

⇒ VA − VB =

Qb

Q ⎫ ⎧ ⎬ ⎨∵ σ = 2A ⎭ ⎩

Q d 2 Aε 0

CASE-3: Consider three parallel metallic plates each of area A which are kept as shown and charges Q1 , Q2 and Q3 are given to them. Let us find the resulting charge distribution on the six surfaces. Neglecting edge effects as usual, we observe that the plate separations do not affect the distribution of charge here. Let the plates 1, 2 and 3 have charges Q1 , Q2 and Q3 . If we assume the charge on side a to be Qa on the side c to be Qc and that on the side e to be Qe , then we have the charges on the left out surfaces given by

Qa

Qb = Q1 − Qa , Qd = Q2 − Qc and Q f = Q3 − Qe . Q1

Q2

Q3

P

Qc

Qd

Q

Qe

Qf

R

From here we may conclude that, half of the sum of all charges appears on each of the two outermost ­surfaces of the system of plates. Further we have a condition,

EQ = 0 ⎡⎣ Qa + ( Q1 − Qa ) + Qc − ( Q2 − Qc ) − Qe − ( Q3 − Qe ) ⎤⎦ =0 2ε 0 A

⇒ Q1 + 2Qc − Q2 − Q3 = 0 ⇒ Qc =

Q2 + Q3 − Q1 2

and Qb = Q1 − Qa =

Q1 − Q2 − Q3 = −Qc 2

Similarly, we can show that, Qd = −Qe . a

b

c

d

e

f

Electric field at point P is zero EP = 0 At P , charge Qa will give an electric field towards right. All other charges Qb , Qc …, etc., will give the electric field towards left. So, ⎡⎣ Qa − ( Q1 − Qa ) − Qc − ( Q2 − Qc ) − Qe − ( Q3 − Qe ) ⎤⎦ =0 2ε 0 A ⇒ 2Qa − Q1 − Q2 − Q3 = 0 Q + Q2 + Q3 ⇒ Qa = 1 2

From here we can find another important result that the pairs of opposite surfaces like b , c and d , e carry equal and opposite charges and the outer surfaces carry equal charges. The properties can often be applied to similar problems, provided additional conditions or constraints are not applied. These conditions (or constraints) can be in the form of some plates being connected together by conducting wires or some plates being held at fixed potentials through the use of batteries. Illustration 82

Consider a system of parallel plates A , B and C . The plates are given charges Q , −2Q and 3Q respectively. Find the distribution of charge on the faces of the three plates. Q

–2Q

3Q

A

B

C

Similarly the condition, ER = 0 will give the result,

Qf =

Q1 + Q2 + Q3 2

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1.120  JEE Advanced Physics: Electrostatics and Current Electricity Solution

Illustration 83

Let us assume that after final distribution of charge on the left face of plate A , charge is q then on right side of it remaining charge will be ( Q - q ) as shown. Always remember that “In a system of large parallel conducting plates, the facing surfaces of plates always carry equal and opposite charges”.

An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E , such that electric field is perpendicular to sheet and covers the entire sheet. Find the charges appearing on the two surfaces of the sheet.

–2Q q

3Q

Q–q –(Q – q)

–Q – q

P

E

(Q + q)

A

B

2Q – q C

So, using this concept, we distribute the charges on the plates as shown in figure. Let us consider a point P inside the plate A . The net electric field at point P must be zero as it is inside a conducting body. This fact helps us to determine the charge q and hence the complete charge distribution. Also, we observe that the electric field at point P is only due to the left face of A and right face of C , because all other charges produces electric field at point P in equal and opposite direction Thus we have electric field at point P as 2Q – q Aε0

P

2Q - q q =0 Ae 0 Ae 0

So, using this concept, we calculate the final distribution of charges on the system of parallel plates shown in figure. 0

A

Let charge on left side of plate be x , then ( Q - x ) is charge on right side of plate. Since inside the plate, at point P , we have EP = 0



x

0

–2Q

B

Q Total initial charge Qi = Q – 2Q + 3Q = 2Q Total final charge Qf = Q – 2Q + 2Q + Q = 2Q ⇒ Qi = 2Q = Qf

2Q

C

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 120

Q–x P

Q–x 2Aε0 1

⇒

x Q-x +E= 2 Ae 0 2 Ae 0

⇒

x Q = -E Ae 0 2 Ae 0

⇒ x=

⇒ q=Q

Q

Solution

q Aε0

Since the point P lies inside a conductor, so net electric field at point P has to be zero. ⇒

Q

x +E 2Aε0 2

Q Q - e o AE and ( Q - x ) = + e o AE 2 2

⎛Q ⎞ So charge on one side is ⎜ - EAe 0 ⎟ and other side ⎝ 2 ⎠ ⎛Q ⎞ ⎜⎝ + EAe 0 ⎟⎠ 2 If Q = 0 i.e. the sheet is neutral, then a charge -e o AE and + e o AE appear on the two surfaces of conducting sheet.

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Chapter 1: Electrostatics 1.121 Illustration 84

Two conducting plates A and B are placed parallel to each other. A is given a charge Q1 and B a charge Q2 . Prove that the charges on the inner facing surfaces are of equal magnitude and opposite sign. Also calculate the charges on the inner and outer surfaces of the plates. Solution

Consider a Gaussian surface as shown in figure. Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The flux through these faces is, therefore, zero. The other parts of the closed surface which are outside the conductor are parallel to the electric field and hence the flux on these parts is also zero. The total flux of the electric field through the closed surface is, therefore zero. From Gauss’s law, the total charge inside this closed surface should be zero. The charge on the inner surface of A should be equal and opposite to that on the inner surface of B . Q2

Q1

E=0

E=0

field due to charge +q is

q (downwards) 2 Ae 0 Q +q and field due to charge Q2 + q is 2 (upwards) 2 Ae 0 field due to charge -q is

The net electric field at P due to all the four charged surfaces is (in the downward direction) Ep =

Q1 - q q q Q +q + - 2 2 Ae 0 2 Ae 0 2 Ae 0 2 Ae 0

As the point P lies inside the conductor so, this field should be zero. Hence, Q1 - q - q + q - Q2 - q = 0 ⇒ q=

Q1 - Q2 2

Problem Solving Technique(s) So, when charged conducting plates are placed parallel to each other, then the outermost surfaces get equal charge (that is exactly half the total charge on the plates) and the plate surfaces facing each other have equal and opposite charge. Q1

E A

q (upwards) 2 Ae 0

B

The distribution of charges is shown in figure. To find the value of q , consider the field at a point P inside the plate A . Suppose, the surface area of the plate q (one side) is A . Since we have E = , So at P 2e 0 A A

Q1 + Q2 2

Q2

Q1 – Q2 2 Q1 – Q2 – 2

Q1 + Q2 2

Half the total charge i.e.

B

Q1 + Q2 2

For Example: Q Q1 – q

q

field due to charge Q1 - q is

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 121

–q

Q2 + q

Q1 - q (downwards) 2 Ae 0

3Q 2

2Q

–

Q 2

Q 2

3Q 2

Half the total charge

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1.122 JEE Advanced Physics: Electrostatics and Current Electricity ILLUSTRATION 85

Three identical metallic plates are kept parallel to one another at a separation of a and b. The outer plates are connected by a thin conducting wire and a charge Q is placed on the central plate. Find the final charges on all the six faces of the plates shown.

a

Further A and C are at same potentials, if E1 be the field between the plates A and B and that between B and C is E2 , then, VB − VA = VB − VC ⇒

E1 a = E2 b

⇒

⎛ q1 ⎞ ⎛ Q − q1 ⎞ ⎜⎝ Aε ⎟⎠ a = ⎜⎝ Aε ⎟⎠ b 0 0

⇒

q1 a = ( Q − q1 ) b

Qb …(2) a+b Electric field inside any conducting plate (say inside C ) is zero. Therefore, ⇒

b

q1 =

q2 q q Q − q1 q1 − Q q − 1 + 1 + + − 3 =0 2 Aε 0 2 Aε 0 2 Aε 0 2 Aε 0 2 Aε 0 2 Aε 0

SOLUTION

Let the charge distribution in all the six faces be as shown in figure. While distributing the charge on different faces, we have used the fact that two opposite faces have equal and opposite charges on them. q2

– q1

1

q1

2

(Q – q1)

3

4

q2 − q3 = 0

E1

B

a

E2

Solving equations, (1), (2) and (3), we get Qb Q , q2 = q3 = a+b 2 Hence, charge on different faces are as follows :

6

b

…(1)

Face

Q 2

1

q2 =

2

− q1 = −

3

q1 =

q2 − q1 + q3 + ( q1 − Q ) = 0 q2 + q3 = Q

Charge

C

Net charge on plates A and C is zero. Hence,

⇒

…(3)

q1 =

q3

5 (q1 – Q)

⇒

Face A

{ A = Area of plates}

Qb a+b

Qb a+b

Charge

Qa

4

( Q − q1 ) = a + b

5

( q1 − Q ) = − a + b

6

q3 =

Qa

Q 2

Test Your Concepts-XII

Based on Charge Distribution (Solutions on page H.37) 1. Figure shows three metallic plates with charges −Q, +3Q and Q respectively. Determine the final charges on all the surfaces.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 122

Q

3Q

–Q

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Chapter 1: Electrostatics 1.123

2. Prove that if an isolated large conducting sheet (isolated means no charges are near the sheet) is given a charge, then the charge distributes equally on its two surfaces. 3. An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such that electric field is perpendicular to sheet and covers the entire sheet. Find the charges appearing on the two surfaces of the sheet. Also calculate the charge appearing on each face of the sheet if the sheet is neutrals. Q

E

4. An isolated conducting sheet of area A and carrying a charge Q is placed in a uniform electric field E, such that electric field is perpendicular to sheet and covers all the sheet. Calculate the resultant electric field on the left and right side of the plate.

METHOD OF IMAGES This method can be used for Boundary Value Problems in which we are asked to find electrostatic potential which results from a given charge configuration in the neighbourhood of an array (arrangement) of grounded conductors in the region outside the conductors. Here, by Grounded Conductors we mean that conductors along the surface of which electrostatic potential V is set to zero.

l

q

Q

E

5. Two uncharged parallel conducting sheets each of area A are placed in a uniform electric field E at a finite distance from each other such that electric field is perpendicular to sheets and covers the entire sheets arrangement. Find out charges appearing on its two surfaces.

E

6. There are two large parallel metallic plates S1 and S2 carrying surface charge densities s1 and s2 respectively ( s 1 > s 2 ) placed at a distance d apart in vacuum. Find the work done by the electric field in moving a point charge q a distance a ( a < d ) from p S1 towards S2 along a line making an angle with 4 the normal to the plates.

Such problems can be solved by reducing to an equivalent simpler problem. We begin by “removing” the conductor(s) and then introducing a new auxiliary system of charges. The auxiliary charges are also called IMAGE charges. The reason behind this nomenclature is that the auxiliary charges are considered as being practically the mirror images of the real charges (at least when the conductor is spherical or planar) in the sense that their positions and relative magnitudes are related in a manner exactly similar to the relations between the positions and relative linear magnification of the object and its corresponding image in Geometrical optics (also called Gaussian ray optics). EXAMPLE If we are asked to find the force between an infinite earthed conductor and a point charge q placed at

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1.124  JEE Advanced Physics: Electrostatics and Current Electricity

perpendicular distance  from the earthed conductor (see figure), than we proceed as follows. Firstly, the conductor being earthed implies V = 0 . So, we redraw the situation in which we replace the conductor and introduce an IMAGE charge –q as shown. The force between the two charges (object charge q and image charge -q ) is the electrostatic force between the infinite grounded conductor and q. So,



F=

⇒ F=

q2 1 4pe 0 ( 2 )2

⇒ W=

∫

⇒ W=⇒ W=

q

l

Electrical image of q V=0

Illustration 86

Calculate the work done in taking the charge q to infinity, when, initially it is at a distance l from an infinite conducting plane. Solution



⇒ F=

Illustration 87

A small ball is suspended over an infinite horizontal conducting plane by means of an insulating elastic thread of stiffness k. As soon as the ball was charged, it descended by x and its new separation from the plane became equal to l. Find the charge of the ball. Solution

When the ball is charged, for the equilibrium of ball, electric force on it must counter balance the excess spring force, exerted, on the ball due to the extension in the string.

+q l

q2 1 4pe 0 ( 2x )2

l

1

4pe 0 ( 4 x 2 )

dW = Fdx cos ( 180° )

⇒ dW = -

q

2

4pe 0 ( 4 x 2 )

Image –q

Thus Fel = Fspr

Let it be further moved through dx away from grounded surface (or towards infinity), then

l

⎤ ⎥ ⎥⎦

q2 16pe 0 l

While taking the charge to infinity let it (at any instant) be at a distance x from the grounded surface. The force with which it is attracted towards grounded surface is given by F=

dx

l

-2 + 1 ∞

q2 ⎡ x ⎢ 16pe 0 ⎢⎣ -2 + 1

-2

q2 is to be 16pe 0 l supplied to the charge (system) to set it free from the electrostatic influence of the grounded conductor. q

l

∫x

So, this much amount of energy i.e.,

–q

l

∞

q2 ⎛ 1 1 ⎞ ⎜ - ⎟ 16pe 0 ⎝ ∞ l ⎠

⇒ W=-

1 q2 (attractive in nature) 4pe 0 4 2

q2 dW = 16pe 0

dx

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 124

The force on the charge q is considered due to attraction by the electrical image ⇒

q2 4pe 0 ( 2l )

2

= kx

⇒ q = 4l pe 0 kx

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Chapter 1: Electrostatics 1.125 Illustration 88 –q

Two point charges, q and -q, are separated by a disL tance L, both being located at a distance from the 2 infinite conducting plane. Find

l –q

 F =

⇒ 

q2

q2 8pe 0 L2

+q

-

4pe 0 L2

(2

4pe 0 ( 2L )

Fq P

–q

l

F–q –q

+q

(b) Also, from the figure, magnitude of electric field strength at P

(2

2 - 1 ) q2

32pe 0 l 2

(It is attractive)

CHARGE DISTRIBUTION ON METAL OBJECT

Fq

L/2



2

2 - 1)

L/2

2 -q ) ( F= + 2 2 4pe 0 ( 2l ) 4pe 0 ( 2 2l )

2q 2

⇒ F=

q2

q l

Solution

     F = 2

q

l l

(a) the modulus of the vector of the electric force acting on each charge; (b) the magnitude of the electric field strength v ­ ector at the mid-point between these charges. (a) Using the concept of electrical image, it is clear that the magnitude of the force acting on each charge,

l

Whenever a charge is given to a metal body then, due to mutual repulsion, it automatically spreads on the outer surface of the body. Let us consider an example for understanding the distribution of charges in detail. Figure shows two conducting metal spheres of radii r1 and r2 at large separation having charges q1 and q2 initially. If the two spheres are connected by a metal wire, charge flow takes place from the sphere at higher potential to the one at lower potential till final potential of the two spheres become equal. After final distribution the charges on the two spheres be ( q1 ) f and ( q2 ) f respectively such that

1 ⎞ q ⎛ E = 2⎜ 1⎟ ⎝ 5 5 ⎠ pe 0 L2

( q1 ) f

4pe 0 r1

Illustration 89

A point charge q is located between two mutually perpendicular conducting half-planes. Its distance from each half-plane is equal to l . Find the modulus of the vector of the force acting on the charge. Solution

=

( q2 ) f 4pe 0 r2

…(1) q2f

q1f

r1

r2

Using the concept of electrical image, it is easily seen that the force on the charge q is,

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1.126  JEE Advanced Physics: Electrostatics and Current Electricity

( q1 ) f

= s 1 4p r12

(

)

If the external electric field on the surface AB is E in the direction shown, then it has two components, one parallel to the surface, other perpendicular to the ­surface, given by

( q2 ) f

= s 2 4p r22

(

)



If after distribution the final surface charge densities on the two spheres are s 1 and s 2 , then we have and

E⊥ = E cos θ and E = E sin θ

Now from equation (1) we get

(

s 1 4p r12

) = s2 (

4pe 0 r2

4pe 0 r1 ⇒ s 1 r1 = s 2 r2 ⇒

4p r22

E

)

E⊥ E

dA θ

s 1 r2 = s 2 r1

A

Thus we can say s ∝

1 r

⇒ s r = constant This shows that for charged metal bodies at same potential, the surface charge density is inversely proportional to the radius of curvature of the body.

r

r

Consider a randomly shaped metal body that has been given some charge as shown, then charge is distributed on the outer surface of this body in such a way that at sharp edges of body (where radius of curvature is less), charge density is more and on the broader parts of body (where radius of curvature is large) charge density is less because we already know that at every point of a metal body potential is same.

B

Due to E , the force on surface is tangential which only stretches the surface and due to E⊥ , the surface experiences an outward force and hence an outward pressure. The outward force on AB due to external electric field is

r

dF = E⊥ dq

⇒ dF = ( s E⊥ ) dA

Thus outward electrostatic pressure Pe is given by

Pe =

dF = s E⊥ dA

MECHANICAL FORCE ON A CHARGED CONDUCTOR Similar charges repel each other, hence the charge on any part of surface of the conductor is repelled by the charge on its remaining part. The surface of the conductor thus experiences a mechanical force.

ELECTRIC PRESSURE ON A CHARGED SURFACE DUE TO EXTERNAL ELECTRIC FIELD Consider a surface uniformly charged with charge density s placed in electric field E . Let us consider a small portion AB having area dA , on the surface. Charge on AB is given by

σ

E1

E2 ΔA

P Q E1

E2

Conductor

dq = s dA

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Chapter 1: Electrostatics 1.127

Also, we have seen that at the boundary of a conductor with uniform charge density s , the tangential component of the electric field is zero, and hence is continuous. But the normal component of the electric s field exhibits discontinuity with En = just outside e0 and zero inside the conductor. Let us now consider a small patch of charge on the surface of conductor as shown. A question comes to our mind “what is the force experienced by this patch”? Since by Newton’s Third Law, the patch cannot exert the force on itself and hence the force on the patch must come only from the remaining part of the conductor (called as remainder). Assume the patch to be a flat surface of area A on the surface of the conductor.  The electric field ( E ) at any point P near the conductor surface is the sum of the electric field due to a small part of the surface (called as Patch) of area say A immediately in the neighbourhood of the point under consideration and due to the remaining  part of  the surface (called as Remainder). Let EPatch and ERemainder be the field intensities due to these parts respectively. Then, total electric field,    E = EPatch + ERemainder n ΔA

Hence, the force experienced by small surface of area A due to the charge on the rest of the surface is ⇒ F = qERemainder = ( s A ) ( ERemainder ) = Thus, the force acting on the Patch is   ⎛ s 2A ⎞ F = qERemainder = ⎜ nˆ ⎝ 2e 0 ⎟⎠

This is precisely the force needed to drive the charges on the surface of a conductor to an equilibrium state where the electric field just outside the conductor has s a value and vanishes inside. The most interesting e0 thing we must note here that irrespective of the sign of s , the force always tends to pull the patch into the field. Using the result obtained above, we may define the electrostatic pressure on the patch as



Force F s 2 1 = = = e 0 E2  Area A 2e 0 2

⇒ Electrostatic Pressure =

Remainder

Patch

 s Now, since E has a magnitude at any point P e0 just outside the conductor and is zero at point Q just inside the conductor. Thus,   s at P , EPatch + ERemainder = and e0   at Q , EPatch - ERemainder = 0   ⇒ EPatch = ERemainder ⇒

  s EPatch = ERemainder = 2e 0

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 127

s ⎫ ⎧ ⎬ ⎨ as E = e 0 ⎭ ⎩

1 s2 e 0 E2 = 2 2e 0

Here, we also note that the pressure is being transmitted via the electric field. Also, this is the value of the energy density ue due to the field. So, ue =

Conductor

s 2A 2e 0

1 s2 e 0 E2 = 2 2e 0

(Please keep in mind that actually pressure and energy density possess the same dimensional formula i.e., ML-1T -2 )

ELECTROSTATIC ENERGY DENSITY (ue) Electrostatic energy density is defined as the energy stored in unit volume in any electric field. Its mathematical formula is given by

Energy density ue =

1 2 eE 2

where E = electric field intensity at that point and

e = e o e r electric permittivity of medium

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1.128  JEE Advanced Physics: Electrostatics and Current Electricity

Calculate the energy stored in an imaginary cubical volume of side a placed in front of an infinitely large nonconducting sheet of uniform charge density s .

To find the total field energy due to this sphere, we consider an elemental spherical shell of radius x and thickness dx as shown. The volume enclosed in this shell is

Solution



Illustration 90

Since electrostatic energy stored is

U=

⇒ U=

∫

1 e 0 E2 dt where dr is small volume 2

1 s e 0 E2 dt , where E = = constant 2 2e 0

∫

1 ⎛ s2 ⎞ s 2 a3 ⇒ U = e 0 ⎜ 2 ⎟ a3 = 2 ⎝ 4e 0 ⎠ 8e 0

dt = ( 4p x 2 ) dx Here we have taken symbol for volume as t (and not V) as V is the symbol for potential.

Thus the field energy stored in the volume of this elemental shell is

ue =

dU ⎛ 1 ⎞ = ⎜ e 0 E2 ⎟ ⎝ ⎠ 2 dt

⇒ dU =

Illustration 91

Using the concept of energy density, show that the total energy stored by a shell having charge Q and Q2 radius R is . 8pe 0 R Solution

Since we have discussed that in space wherever electric field exists, there must be some field energy stored which has energy density, given as

⇒ dU =

⇒ U= Q

R

x

Q2 8pe 0 x 2

2

( 4p x 2 ) dx

dx

Thus total field energy associated with the sphere can be calculated by integrating this expression from surface of sphere to infinity (as electric field inside the sphere is zero). Total field energy in the surrounding of sphere is U=

1 ue = e 0 E2 2

1 ⎛ Q ⎞ e0 2 ⎜⎝ 4pe 0 x 2 ⎟⎠

dx

∫

Q2 dU = 8pe 0

Q ⎡ 1 ⎢8pe 0 ⎣ x 2

⇒ U R→∞ =

∞

1

∫x

2

dx

R

∞

⎤ ⎥ R ⎦

Q2 8pe 0 R

Illustration 92

Here we can see that when the sphere was uncharged, there was no electric field in its surroundings. But when the sphere becomes fully charged, electric field will exist in its surrounding from its surface to infinity. Let us calculate the field energy associated with this charged conducting sphere. We know electric field due to a sphere at outer points varies with distance from centre as E=

Q 4pe 0 x 2

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 128

A non-conducting sphere of radius R has a charge Q distributed uniformly on it. Using the concept of energy density, find the (a) energy stored outside the sphere. (b) energy stored inside the sphere and (c) total energy stored inside and outside the sphere. Solution

(a) We know in outside region of a non-conducting uniformly charged sphere, every point is same as that of a conducting sphere of same radius.

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Chapter 1: Electrostatics 1.129

Thus field energy in the surrounding of this sphere from surface to infinity is also

 

U R→∞ =

Q2 8pe 0 R

(b) As in the case of conducting sphere (where Einside = 0 ), in non-conducting sphere, at interior points E ≠ 0 . Thus field energy also exists in the interior region. This can be calculated by considering an elemental shell of radius x and thickness dx inside the sphere as shown. Q

R x dx

(c) So, total self-energy of this sphere, also called Electrostatic Self Energy is    Uself = U0→ R + U R→∞ Q2 Q2 ⇒   U = + self 40pe 0 R 8pe 0 R 3Q 2 ⇒   Uself = U0→ R + U R→∞ = 20pe R 0 Illustration 93

Consider a thin spherical shell of radius R having charge q distributed uniformly on it. At the centre of shell a negative point charge -q0 is placed. The shell is cut in two identical hemispherical portions along a diametrical section yy′ as shown. Due to mutual repulsion the two hemispherical parts tend to move away from each other but due to the attraction of -q0, they may stay in equilibrium. Find the condition of equilibrium of the hemispherical shells. Calculate minimum value of q0 to attain this equilibrium. y

Here field energy in the volume of this elemental shell is ⎛1 ⎞ dU = ⎜ e 0 E2 ⎟ dt ⎝2 ⎠    where dt = 4p x 2 dx and E = Einside = ⇒ ⇒

 

 

dU =

dU =

1 ⎛ Qx ⎞ e0 2 ⎜⎝ 4pe 0 R3 ⎟⎠ Q

2

8pe 0 R6

2

⇒  U=

∫

dU =

4pe 0 R3

( 4p x 2 dx )

R

8pe 0 R

⎛ R5 ⎞ ⎜ ⎟ 8pe 0 R6 ⎝ 5 ⎠ Q2

Q2 ⇒   U0→ R = 40pe R 0

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 129

∫ x dx 4

6

0

y′

Solution

Let the outward electric pressure at every point of the spherical shell, due to its own charge q be P1 . Then

x 4 dx

Q2

+q

Qx

Total field energy inside the sphere can be given as      U=

R

–q0



P1 =

s2 2e 0 2

⇒ P1 =

q2 q ⎫ 1 ⎛ q ⎞ ⎧  ∵s= = ⎬ ⎜⎝ ⎟ 2 ⎠ 2 4 ⎨ 2e 0 4p R 4p R2 ⎭ 32p e 0 R ⎩

Due to charge -q0 , the electric field on the surface of shell is q0 E=  {radially inwards} 4pe 0 R2 This electric field pulls every point of the shell in an inward direction. Let the inward pressure on the surface of shell due to this negative point charge at centre be P2 . Then

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1.130  JEE Advanced Physics: Electrostatics and Current Electricity



P2 = s E 

{as done already}

  ⇒ Eat P due to + E

qq0 ⎛ q ⎞ ⎛ q0 ⎞ ⇒ P2 = ⎜ = ⎝ 4p R2 ⎟⎠ ⎜⎝ 4pe 0 R2 ⎟⎠ 16p 2 e 0 R 4

 ⇒ E

For equilibrium of hemispherical shell or for the shells not to separate,

⇒

⇒

E

at P due to induced charges

+q

= Eat P due =

2

16p e 0 R

4

≥

q

at P due to induced charges

2

 =0

 = -Eat P due to

tude) and E

qq0

to + q

q 4pe 0 x 2

(in magni-

will have a direction oppo-

site to the direction of field at P due to +q . So, field q due to induced charges is directed from P to 4pe 0 x 2 A i.e., towards +q .

32p 2 e 0 R 4

q 2

⇒ qmin =

+q

at P due to induced charges

P2 ≥ P1

⇒ q0 ≥

at P due to induced charges

q 2

Electrostatic Potential

ELECTROSTATIC FIELD AND POTENTIAL DUE TO INDUCED CHARGES Electrostatic Field Consider the situation in which a metal sphere of radius R is placed at a distance r from the point charge +q . Consider a point P in the sphere at a distance x from +q where we wish to find the electric field and potential due to the induced charges on the surface of sphere. The charge +q at A will induce a negative and a positive charge -qi and +qi as shown at the nearer and farther end respectively. +qi –qi P

C

x r

+q

A

Since, inside the sphere at every point E = 0 and due to this the entire region is equipotential. Hence the potential at the centre of sphere, will be only due to the charge +q , because potential due to induced charges at the centre will be zero (due to symmetrical placement of induced charges about C ). So, net potential at the centre of sphere due to q at A is VC =

q 4pe 0 r

Since the entire sphere is equipotential, so potential at point P must be equal to the potential at point C. However, at P the potential due to induced charges will be non-zero because induced charges are not equidistant from point P . Thus net potential at point P will be VP = Potential at P due to +q at A + Potential due to induced charges ⇒ VP =

q + Vi 4pe 0 x

and since VP = VC =

Now since P lies inside the metal body,   so ( EP )NET = 0 i.e., the total field at E due to the point charge +q (at A ) and due to the charges induced on the sphere must be zero.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 130

q 4pe 0 r

(because the sphere is equipotential inside) Hence

VP =

⇒ Vi =

q q = + Vi 4pe 0 r 4pe 0 x q ⎛ 1 1⎞ ⎜ - ⎟ 4pe 0 ⎝ r x ⎠

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Chapter 1: Electrostatics 1.131

potential zero. Thus the final potential on sphere can be taken as

EARTHING OF CHARGED OR UNCHARGED METAL BODIES As far as electrical behaviour of the earth is concerned, the earth is assumed to be a very large conducting sphere of radius 6400 km. If some charge Q is given to earth, its potential VE becomes

VE =

Q {where R = Radius of Earth} 4pe 0 R

x +q



Here it is obvious that earth has supplied a negative charge so as to develop a negative potential on sphere to nullify the initial positive potential on it due to q .

S

qe

C R

S

Consider a solid uncharged conducting sphere of radius R . A point charge q is placed in front of the sphere at a distance x as shown. Due to q , the potential at sphere is

Always remember whenever a metal body is connected to earth, we consider that earth supplies a charge to it (say qE) to make its final potential zero due to all the charges including the charge on body and the charges in its surrounding.

Problem Solving Technique(s) Earthing a conductor: Potential of earth is often taken to be zero. If a conductor is connected to the earth, the potential of the conductor becomes equal to that of the earth, i.e., zero. If the conductor was at some other potential, charges will flow from it to the earth or from the earth to it to bring its potential to zero.

CURRENT DUE TO MOVEMENT OF CHARGES THROUGH EARTH Consider a metal ball of radius R connected to earth through an ammeter. Let a charge +q be moving towards the ball at speed v , due to which the potential of ball increases continuously. When the charge is at a distance x from ball, the potential of ball due to this charge is given by

V=

q V= 4pe 0 x

Here we have ignore the charges induced due to q because potential at C due to induced charges on the sphere is zero. If we close the switch S , earth ­supplies a charge qE on to the sphere to make its final

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 131

qR x

Remark(s)

C

As R is very large VE becomes a negligible value. Thus for very small bodies whose dimensions are negligible compared to earth we can assume the earth to be at zero potential. If we connect a small body to earth then charge flow takes place between earth and the body till both acquire the same potential i.e., zero potential. The potential of earth will always remain zero, no matter if charge flows into earth or from earth. This implies that if a body at some positive potential is connected to earth, earth will supply some negative charge to this body so that the final potential of body becomes zero.

x

q qE + =0 4pe 0 x 4pe 0 R

⇒ qE = -

R

+q

V=

q 4pe 0 x qE v

+q

x R

A

I

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1.132  JEE Advanced Physics: Electrostatics and Current Electricity

Since the metal ball is connected to the earth, so the earth supplies a charge qE (say) to make the potential of the ball to be zero. This qE is given by (as done earlier)

qE = -

qR x

A

Since, x is continuously decreasing hence the charge on ball must increase continuously. As a result of this the ammeter shows a current I , which is given by



I=

dqE d ⎛ qR ⎞ qR ⎛ dx ⎞ = ⎜⎟= ⎜ ⎟ dt dt ⎝ x ⎠ x 2 ⎝ dt ⎠

⎛ qR ⎞ ⇒ I=⎜ 2 ⎟v ⎝x ⎠ So, in the surrounding of an earthed conducting body whenever one or more charges are in motion, a continuous current flows in the wires connecting body to earth.

Problem Solving Technique(s) Charges appearing on different surfaces of concentric spherical conducting shells: Figure shows three concentric thin spherical conducting shells A, B and C of radii a, b and c. The shells A and C are given charges q1 and q2 and the shell B is earthed. We are interested in finding the charges on inner and outer surfaces of A, B and C. To solve such type of problems following points should keep in mind: C B A a b

q1

q1 q1

q6 q q4 5 q3

q2

c

(a) The entire charge q1 will come on the outer surface of A unless some charge is kept inside A.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 132

Gaussian surface inside A



To understand this, let us consider a Gaussian surface (a sphere) through the material of A. As the electric field in a conducting material is zero. The flux through this Gaussian surface is zero. Using Gauss’s Law, the total charge enclosed must be zero. (b) Similarly if we draw a Gaussian surface through the material of B, then we have q + q = 0 ⇒ q3 = -q1    3 1 and if we draw a Gaussian surface through the material of C, then    q5 + q4 + q3 + q1 = 0 ⇒ q5 = -q4

(c) Since q2 charge was given to shell C, so q5 + q6 = q2 (d) Since B is earthed, so potential of B must be zero. Hence     VB = 0

⇒ 

1 ⎛ q1 q3 + q4 q5 + q6 ⎞ + ⎜ + ⎟ =0 b c ⎠ 4pe 0 ⎝ b



Using the above conditions we can find charges on different surfaces, we can summarise the above points as under (i) Net charge inside a closed Gaussian surface drawn in any conducting shell is zero. (ii) Potential of the conductor which is earthed is zero. (iii) If two conductors are connected, they are at same potential.  (iv) Charge remains constant on all conductors except those which are earthed.  (v) Charge on the inner surface of the innermost shell is zero provided no charge is kept inside it. In all other shells charge resides on both the surfaces.  (vi) Equal and opposite charges appear on surfaces facing each other. Based on this, try to understand the following ILLUSTRATIONS.

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Chapter 1: Electrostatics 1.133 Illustration 94

5Q

The two conducting spherical shells as shown in figure are joined by a conducting wire. After a long time, the wire is cut and the charge stops flowing. Calculate the final charge on each sphere after that.

–2Q

Q

3

2

Q 1 2R 3R

–3Q

R

Solution

Let the charge on the innermost sphere be x . Then finally,

2R

6Q – x

Solution

–2Q

Since the shells had been connected by a conducting wire for a long time, so after the conducting wire is removed then both shells must be at the same potential. If x be the charge that has flown from inner shell to outer shell, then Potential of inner shell is Kx K ( -2Q - x ) k ( x - 2Q ) + = R 2R 2R and potential of outer shell is Vin =



Vout =

Kx K ( -2Q - x ) - KQ + = 2R 2R R –2Q – x x

Since Vout = Vin ⇒

R

- KQ K ( x - 2Q ) = R 2R

⇒ -2Q = x - 2Q ⇒ x=0 So final charge on inner spherical shell is zero and on outer spherical shell is -2Q , i.e. entire charge on inner shell flows to the outer shell.

3

2

x 1

R 2R 3R

Potential of shell 1 = Potential of shell 3 Kx K ( -2Q ) K ( 6Q - x ) Kx K ( -2Q ) K ( 6Q - x ) + + = + + R 2R 3R 3R 3R 3R ⇒ 3 x - 3Q + 6Q - x = 4Q ⇒ 2x = Q ⇒ x=

Q 2

Charge on innermost shell =

Q 2

5Q 2 Charge on middle shell = -2Q

Charge on outermost shell =

Final charge distribution is as shown in figure. Q +3Q/2 –3Q/2 –Q/2 Q/2

Illustration 95

Calculate the final charge on each spherical shell after joining the inner most shell and outer most shell by a conducting wire.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 133

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1.134  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 96

Two conducting hollow spherical shells of radii R and 2R carry charges -Q and 3Q respectively. How much charge will flow into the earth if inner shell is grounded?

q

R 2R

3Q –Q

Solution R

2R

Solution

(a) The charge on the inner surface should be -q, because if we draw a closed Gaussian surface through the material of the hollow sphere the total charge enclosed by this Gaussian surface should be zero. Let q ′ be the charge on the outer surface of the hollow sphere.

When inner shell is grounded, then the potential of inner shell becomes zero.

q

q′ –q

3Q x R

2R

⇒

Kx K ( 3Q ) + =0 R 2R

⇒ x=

-3Q 2

Since the charge has increased by

Δq =

-3Q -Q - ( -Q ) = 2 2

Since, the outer sphere is earthed, its potential should be zero. The potential on it is due to the charges q , -q and q ′ , Hence, V=

q q′ ⎞ 1 ⎛ q + ⎜ ⎟ =0 4pe 0 ⎝ 2R 2R 2R ⎠

  ⇒   q′ = 0

Therefore, there will be no charge on the outer surface of the hollow sphere. (b)

q′ q

Q Hence charge flown into the Earth is 2

R1

Illustration 97

A charge q is distributed uniformly on the surface of a sphere of radius R . It is covered by an earthed concentric hollow conducting sphere of radius 2R . Find the charges on inner and outer surfaces of hollow sphere when (a) thickness of shell is negligible (b) the shell has considerable thickness.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 134

–q

r R2 R 3

P

In this case, we can set V = 0 at any point on the hollow sphere. Let us select a point P a distance r from the centre. Where R2 < r < R3 . So,

VP = 0

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Chapter 1: Electrostatics 1.135





⇒ 

1 ⎛ q q q′ ⎞ - + =0 4pe 0 ⎜⎝ r r R3 ⎟⎠

⇒

q q ’ q ′ 2q + q ′ ⎞ 1 ⎛ q + + ⎜ ⎟ =0 4pe 0 ⎝ 2R 2R 2R 3R 3R ⎠

Solving this equation, we get

⇒   q′ = 0

4 q′ = - q 3

i.e., in this case also there will be no charge on the outer surface of the hollow sphere.



Illustration 98

⇒ 2q + q ′ = 2q -

Figure 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. Find the charges appearing on all the surfaces of A , B and C.

2 4 q= q 3 3 Therefore, charges on different surfaces in tabular form are given below: A

B

C

INNER OUTER INNER OUTER INNER

q

0

4 - q 3

-q

4 q 3

OUTER

2 q 3

Illustration 99 A B C

Figure shows two conducting thin concentric shells of radii r and 3r . The outer shell carries charge q. Inner shell is neutral. Find the charge that will flow from inner shell to earth after the switch S is closed. q

Solution

Since there is no charge inside A . The whole charge q given to the shell A will appear on its outer surface. Charge on its inner surface will be zero. Moreover if a Gaussian surface is drawn on the material of shell B net charge enclosed by it should be zero. Therefore, charge on its inner surface will be -q . Now let q ′ be the charge on its outer surface, then charge on the inner surface of C will be - q ′ and on its outer surface will be, 2q - ( - q ′ ) = 2q + q ′ as total charge on C is 2q. 2q + q′ q′ –q′ –q q R

S

3r

Solution

Let q′ be the charge on inner shell when it is earthed.

Vinner = 0

⇒

1 ⎛ q′ q ⎞ ⎜ + ⎟ =0 4pe 0 ⎝ r 3 r ⎠

⇒ q′ = -

2R

i.e., + 3R

Shell B is earthed. Hence, its potential should be zero.

VB = 0

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 135

r

q 3

q charge will flow from inner shell to earth. 3

FIELD ENERGY DENSITY OF ELECTRIC FIELD An energy is associated with every region where electric field is present. This energy is called the field energy of the electric field. Using the concept of field

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1.136  JEE Advanced Physics: Electrostatics and Current Electricity

energy we can calculate the amount of energy stored in the space where electric field exists. To calculate the Field Energy Density (FED) we have two cases to be taken into account. CASE-1: Field Energy Density of a charged body, also called Self Energy. “Self energy of a charged body is the total field energy, associated with the electric field due to the body itself in its surrounding”. CASE-2: Field Energy Density of a system of charged bodies. Here, we have to take Field Energy Density as the total of self energies associated with the bodies and the interaction energies of the pairs of charged bodies. ⎛ ⇒ U=⎜ ⎜⎝

∑

⎛ self energy ⎞ ⎜ of all charged ⎟ + ⎜ ⎟⎠ ⎜ bodies ⎝

∑

interaction ⎞ en nergy of all ⎟ ⎟ pairs of ⎟ charged bodies ⎠

Electrostatic Self Energy of a Charged Conducting Sphere We have discussed that in space wherever electric field exist, there must be some field energy stored which has energy density, given as

U=

1 e 0 E2 2



E=

Q

R

dx

Here we can see that when the sphere was uncharged, there was no electric field in its surroundings. But when the sphere becomes fully charged, electric field will exist in its surrounding from its surface to infinity. Let us calculate the field energy associated with this charged conducting sphere. We know electric field due to a sphere at outer points varies with distance from centre as

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 136

4pe 0 x 2

To find the total field energy due to this sphere, we consider an elemental spherical shell of radius x and thickness dx as shown. The volume enclosed in this shell is dt = ( 4p x 2 ) dx Thus the field energy stored in the volume of this elemental shell is

⎛1 ⎞ dU = ⎜ e 0 E2 ⎟ dt ⎝2 ⎠

⇒ dU = ⇒ dU =

1 ⎛ Q ⎞ e0 2 ⎜⎝ 4pe 0 x 2 ⎟⎠ Q2 8pe 0 x 2

2

( 4p x 2 ) dx

dx

Thus total field energy associated with the sphere can be calculated by integrating this expression from surface of sphere to infinity (as electric field inside the sphere is zero). Total field energy in the surrounding of sphere is U= ⇒ U=

Volume = dτ

Q

∫

Q2 dU = 8pe 0

Q2 ⎡ 1 ⎢8pe 0 ⎣ x

⇒ U R→∞ =

∞

1

∫x

2

dx

R

∞

⎤ ⎥ R ⎦

Q2 …(1) 8pe 0 R

Alternatively, whenever a system of charges is assembled, some work is done and this work is stored in the form of electrical potential energy of the system. Let consider an example of charging a conducting sphere of radius R. In the process of charging we bring charge to the sphere from infinity in steps of elemental charges dq. The charge on sphere opposes the elemental charge being brought to it. Let us assume that at an instant sphere has a charge q , due to which it has a potential given as

V=

q 4pe 0 R

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Chapter 1: Electrostatics 1.137

If now a charge dq is brought to its surface from infinity, work done in this process can be given as

Q

dU = Vdq

R

⎛ q ⎞ ⇒ dU = ⎜ dq ⎝ 4pe 0 R ⎟⎠

x dx

q

dq R

Total work done in charging the sphere is ⇒ U=

∫

⇒ U R→∞

Q

1 dU = qdq 4pe 0 R

∫ 0

Q2 = …(2) 8pe 0 R

Equation (1) gives the total work done in charging the sphere of radius R. Here we can see that again result in equation (2) is same as that in equation (1) We can conclude by this that whatever work is done in charging a body is stored in its surrounding in the form of its field energy and can be regarded as self energy of that body. Once a body is charged in a given configuration, its self energy is fixed, if the body is now displaced or moved in any manner keeping its shape and charge distribution constant, its self energy does not charge.

As in the case of conducting sphere (where Einside = 0), in non-conducting sphere, at interior points E ≠ 0. Thus field energy also exists in the interior region. This can be calculated by considering an elemental shell of radius x and thickness dx inside the sphere as shown. Here field energy in the volume of this elemental shell is

⎛1 ⎞ dU = ⎜ e 0 E2 ⎟ dV ⎝2 ⎠

where E = Einside = ⇒ dU = ⇒ dU =

METHOD 1: We know in outside region of a non-conducting uniformly charged sphere, every point is same as that of a conducting sphere of same radius. Thus field energy in the surrounding of this sphere from surface to infinity is also

U R→∞

Q2 = 8pe 0 R

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 137

4pe 0 R3

1 ⎛ Qx ⎞ e0 2 ⎜⎝ 4pe 0 R3 ⎟⎠ Q2 8pe 0 R6

2

( 4p x 2 dx )

x 4 dx

Total field energy inside the sphere can be given as U=

R

Q2

∫ dU = 8pe R ∫ x dx 0

⇒ U=

Electrostatic Self Energy of a Uniformly Charged Non-conducting Sphere

Qx

4

6

0

⎛ R5 ⎞ ⎜ ⎟ 8pe 0 R6 ⎝ 5 ⎠ Q2

⇒ U0→ R =

Q2 40pe 0 R

So, total self energy of this sphere is

Uself = U0→ R + U R→∞

⇒ Uself =

Q2 Q2 + 40pe 0 R 8pe 0 R

⇒ Uself = U0→ R + U R→∞ =

3Q 2 20pe 0 R

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1.138  JEE Advanced Physics: Electrostatics and Current Electricity

METHOD 2: Let us assemble the sphere from infinitesimal concentric shells. Q

R

­ on-conducting sphere of radius R1 having charge n Q1 and a charged conducting sphere of radius R2 having charge Q2 respectively separated by a ­distance r .

Q′

x

R1

R2

Q1

Now to add an infinitesimal spherical element of radius x , thickness dx having charge dq on a charged spherical core of radius x having charge q ′ , the work that has to be done is given by

dU =

1 ⎛ Q ′dq ⎞ ⎜ ⎟ 4pe 0 ⎝ x ⎠

(

1 16p 2 r 2 ⇒ U= 4pe 0 3

)

⇒ U=

⇒ U=

1 4pe 0

R

∫

r

The total electrostatic energy of this system, will be

Q ⎛4 ⎞ where Q ′ = ⎜ p x 3 ⎟ r, dq = 4p x 2 dx r and r = ⎝3 ⎠ 4 p R3 3 2 2 1 16p r 4 ⇒ dU = x dx 4pe 0 3

Q2

U = Uself + Uinteraction

⇒ U=

3Q12 Q22 QQ + + 1 2 20pe 0 R1 8pe 0 R2 4pe 0 r

Electrostatic Energy of a System of Concentric Shells Consider two concentric shells of radii R1 and R2 having uniform charges Q1 and Q2 .

x 4 dx

0

⎛ 16p 2 ⎞ ⎛ Q 2 ⎜⎝ ⎟ 3 ⎠ ⎜ 16p 2 6 ⎜ R ⎝ 9

⎞ ⎛ R5 ⎞ ⎟ ⎜⎝ 5 ⎟⎠ ⎟ ⎠

R1

3Q 2 20pe 0 R

Q1

Q2

R2

This work done in assembling the system is stored in the form of electrostatic self energy.

Here the total energy of this system will be

Total Electrostatic Energy of a System of Charged Bodies

⎛ self energy U total = ⎜ of inner ⎜ shell ⎝

Total electrostatic potential energy a system of charges will be given as U = Total of self energy of all charged bodies + Total of interaction energy of all pairs of charged bodies

⇒

nteraction ⎞ ⎛ Total self ⎞ ⎛ Total in ⎜ energy of ⎟ ⎜ energy of all ⎟ ⎟ +⎜ ⎟ U=⎜ ⎜ all charged ⎟ ⎜ pairs of charged ⎟ ⎜⎝ bodies ⎟⎠ ⎜⎝ ⎟⎠ bodies

Let us understand this concept by taking this simple example. Figure shows uniformly charged

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 138

⇒ U total =

⎞ ⎛ self energy ⎟ + ⎜ of outer ⎟ ⎜ shell ⎠ ⎝

⎛ interaction ⎞ ⎜ energy of ⎟ +⎜ ⎟ ⎜ the two ⎠ ⎝ shells

⎞ ⎟ ⎟ ⎟ ⎠

Q22 Q1Q2 ⎞ 1 ⎛ Q12 + + R2 ⎟⎠ 4pe 0 ⎜⎝ 2R1 2R2

The above result can also be obtained by following method. Since, the total electrostatic energy of a system is stored in the form of field energy of the system hence here we can calculate the total electrostatic energy of the system by integrating the field energy density in the space surrounding the shells where the electric field exists.

9/20/2019 10:30:08 AM

Chapter 1: Electrostatics 1.139

⎛ Workdone ⎞ ⎛ Total initial ⎞ ⎛ Total final ⎞ ⎟ = ⎜ energy ⎟ - ⎜ energy of ⎟ ⇒ ⎜ by ⎜ ⎟ ⎜ of system ⎟ ⎜ system ⎟ field ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ R2

R1

x

Now, total initial energy of system is

dx



Ui = SE1 + SE2 + SE3 + IE12 + IE13 + IE23

( SE denotes self energy and IE denotes interaction energy)

r dr

Total field energy in the electric field associated with the system shown in figure can be given as R2

U=

∫

R1

1

∫ 2e



⎛ ( Q1 + Q2 ) ⎞ 2 ⎜ 4pe r 2 ⎟ 4p r dr ⎝ ⎠ 0 2

0

R2



U f = SE1 + SE2 + SE3 + IE12 + IE13 + IE23

⇒ Uf =



⇒ U=

1 ⎛ q12 q22 q32 q1 q3 q1 q2 q2 q3 ⎞ + + + + + ⎜ ⎟ 4pe 0 ⎝ 2a 2b ′ 2c c b′ c ⎠

Work done by field = Ui - U f

Q12 ⎛ 1 1 ⎞ ( Q1 + Q2 ) ⎛ 1 ⎞ + ⎜ ⎜⎝ R ⎟⎠ 8pe 0 ⎝ R1 R2 ⎟⎠ 8pe 0 2

⇒ W=

Q2 Q Q ⎞ 1 ⎛ Q12 + 2 + 1 2⎟ ⎜ 4pe 0 ⎝ 2R1 2R2 R2 ⎠

Illustration 101

2

U=

1 ⎛ q12 q22 q32 q1 q3 q1 q2 q2 q3 ⎞ + + + + + ⎜ ⎟ 4pe 0 ⎝ 2 a 2b 2c c b c ⎠

Total final energy of system

2

1 ⎛ Q1 ⎞ 4p x 2 dx + e0 2 ⎜⎝ 4pe 0 x 2 ⎟⎠ ∞

⇒ Ui =

Illustration 100

Three shells are shown carrying charge q1 , q2 and q3 and of radii a, b and c respectively. If the middle shell expands from radius b to b ′ ( b ′ < c ) . Find the work done by electric field in process.

b a q1

1 ⎛ q22 q22 q1 q2 q1 q2 ⎞ + ⎜ ⎟ 4pe 0 ⎝ 2b 2b ′ b b′ ⎠

Figure shows a shell of radius R having charge q1 uniformly distributed over it. q

R q1

A point charge q is placed at the centre of shell. Find work required to increase radius of shell from R to R1 ( > R ). Solution

b′ c

q2



Work done = Decrease in Potential Energy

⇒ Work = Ui - U f

q3

R1

Solution

Since work done by a conservative field equals the decrease in potential energy, so we have

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 139

R +q q1

9/20/2019 10:30:15 AM

1.140 JEE Advanced Physics: Electrostatics and Current Electricity

Now Ui = ( SE )q + ( SE )q1 + ( IE ) ⇒

Ui = ( SE )q +

and U f = ( SE )q +

q12 qq + 1 8pe 0 R 4pe 0 R q12 q1 q + 8pe 0 R1 4pe 0 R1

⇒

Work done = Ui - U f

⇒

q q⎞ 1 ⎛ q12 q1 q q12 W= + - 1 ⎟ ⎜ 4pe 0 ⎝ 2R R 2R1 R1 ⎠

The electrostatic field energy stored in the volume of this shell is

⎛1 ⎞ dU = ⎜ e 0 E2 ⎟ dV ⎝2 ⎠

⇒

dU =

The total electrostatic energy stored in the above mentioned volume can be obtained by integrating the above expression within limits from a to b as

IllustratIon 102

Find the electrostatic energy stored in a cylindrical shell of length l, inner radius a and outer radius b, coaxial with a uniformly charged wire with linear charge density l .

U= ⇒

U=

solutIon

For this we consider an elemental shell of radius x and width dx. The volume of this shell dV can be given as dV = ( 2p xl ) dx

The electric field due to the wire at the shell is

E=

2

1 ⎛ l ⎞ ( 2p xl ) dx e0 2 ⎜⎝ 2pe 0 x ⎟⎠

∫

b

dU =

l 2l 4pe 0

∫ a

b

2

1 ⎛ l ⎞ ( 2p xl ) dx e0 2 ⎜⎝ 2pe 0 x ⎟⎠

1

∫ x dx a

⇒

U=

l 2l log e x 4pe 0

⇒

U=

l 2l ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠ 4pe 0

(

b a

)

l 2pe 0 x

Test Your Concepts-xIII

Based on Concept of self and Interaction Energy (Solutions on page H.39) 1. A spherical shell of radius R1 with uniform charge q is expanded to a radius R2. Calculate the work performed by the electric forces in this process. 2. A spherical shell of radius R1 with a uniform charge q has a point charge q0 at its centre. Find the work performed by the electric forces during the shell expansion from radius R1 to radius R2.

01_Physics for JEE Mains and Advanced - 2_Part 3.indd 140

3. A spherical shell is uniformly charged with the surface density s. Using the Energy Conservation Law, find the magnitude of the electric force acting on a unit area of the shell. 4. Calculate the work that must be done to charge a spherical shell of radius R to a total charge Q.

9/20/2019 10:30:20 AM

Chapter 1: Electrostatics 1.141

Solved Problems Problem 1

A rectangular tank of mass m0 and charge Q over it is placed over a smooth horizontal floor. A horizontal electric field E exist in the region. Rain drops are falling vertically in the tank at the constant rate of n drops per second. Mass of each drop is m. Find the time taken by tank to reach to half the maximum speed. Solution

Applying Newton’s Second Law, we get QE - v

⇒



dm dv =M dt dt

dv QE - v ( nm ) = ( m0 + nmt ) …(1) dt

When v becomes vmax , then

m0 dm dt

QE

⇒ QE = vmax ( nm ) ⇒ vmax = v0 =

m0 + nmt = m0

⇒

m0 + nmt 2 QE = m0 QE

⇒ 1+

QE nm

⎛ nm ⎞ t=1 ⇒ ⎜ ⎝ m0 ⎟⎠ ⇒ t=

m0 nm

Two small identical balls lying on a horizontal plane are connected by a weightless spring. One ball is fixed at the origin and the other is free. The balls are charged identically as a result of which the spring length increases ’n ’ times. Determine the ratio of new frequency (when balls are charged) to old frequency (when balls were uncharged) when the free ball is displaced slightly from its mean position. 2

⇒

dt = + mnt 0

∫m 0

⇒

1 ⎛ log ⎜ mn ⎝

⇒

1 + nmt = m0



⇒

Solution

1 k , where k 2π m is force constant of the spring and m = mass of the When the balls are uncharged, f0 =

v0 2

dv

∫ QE - nmv 0

mnv ⎞ ⎛ QE m0 + nmt ⎞ 1 ⎜ 2 ⎟ log ⎜ =⎟⎠ ⎟ ⎝ m0 nm QE ⎠ QE ⎛ v⎞ QE - mn ⎜ ⎟ ⎝ 2⎠

m0 = nmt = m0

1

O

QE - nmv ⎛ dv ⎞ =⎜ ⎟ m0 + nmt ⎝ dt ⎠ t

QE ⎫ ⎧ ⎬ ⎨∵ v0 = nm ⎭ ⎩

nmt =2 m0

From (1), we get

2QE  ⎛ QE ⎞ 2QE - mn ⎜ ⎝ mn ⎟⎠

Problem 2

dv =0 dt m

Thrust = v

⇒

2QE ⎛v ⎞ 2QE - mn ⎜ 0 ⎟ ⎝ 2 ⎠

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 141

oscillating ball (ball 1). When charged we have in the equilibrium position of ball 1,



q2 1 = k ( ηl - l ) = kl ( η - 1 ) 4πε 0 ( ηl )2

3 ⇒ l =

q2 4πε 0η2 ( η - 1 ) k

When the ball 1 is displaced by a small distance from the equilibrium position to the right, the unbalanced force to the right is given by

9/20/2019 10:23:48 AM

1.142  JEE Advanced Physics: Electrostatics and Current Electricity

Resultant force to the right is

F=

Solution

Electric field at P due to ring

2

q 1 - k ( ηl + x - l ) 4πε 0 ( ηl + x )2

E1 =

From Newton’s Law, we have m

d2 x dt 2

=

2

1 q 4πε 0 η2 l 2

1 q2 ⇒ m 2 = 4πε 0 η2 l 2 dt d2 x

x⎞ ⎛ ⎜⎝ 1 + ηl ⎟⎠

-2

- kl ( η - 1 ) - kx

2x ⎞ ⎛ ⎜⎝ 1 - ηl ⎟⎠ - kl ( η - 1 ) - kx

1 q2 1 q 2 2x - kl ( η - 1 ) - kx ⇒ m 2 = 4πε 0 η2 l 2 4πε 0 η2 l 2 ηl dt d2 x

⇒ m

⇒ m

⎛ 1 2q 2 ⎞ = ⎜ 4πε 3 3 + k ⎟ x 2 ⎝ ⎠ dt 0 η l

d2 x



qx 1 4πε 0 ( x 2 + R2 )3 2 {charged toward centre as it is negatively}

Electric field at P due to +q E2 =



q 4πε 0 x 2

⇒ Enet =

⎛ 1 ⎞ 2q 2 = + k⎟ x ⎜ 2 2 dt q ⎜ 4πε 0 η3 ⎟ 2 ⎜⎝ ⎟⎠ 4πε 0η ( η - 1 ) k

d2 x

dt

2

=-

⇒ ω2 = ⇒ ⇒

f = f0

+q x

P

3η - 2 k x η m

3η - 2 k η m

1 f = 2π

q ⎛ 1 x ⎞ 4πε 0 ⎜⎝ x 2 ( x 2 + R2 )3 2 ⎟⎠

–q

⎛ 2(η - 1) ⎞ 3η - 2 ⇒ m 2 = -⎜ kx k + k⎟ x = ⎝ ⎠ η η dt d2 x

{away from centre}

Thus net field at P is    Enet = E2 + E1 {vectorially} ⇒ Enet = E2 - E1

d2 x

⇒



For x  R ,

3η - 2 k η m 3η - 2 η

Thus the frequency is increased

Enet

⇒ 3η - 2 times η

Here η = 2 and so frequency increases

2 times.

3 ⎡ R2 ⎞ 2 ⎢ ⎛ = 1- ⎜ 1+ 2 ⎟ x ⎠ 4πε 0 x 2 ⎢⎣ ⎝

Enet =

q

⎡ ⎛ 3 R2 ⎞ ⎤ ⎢1 - ⎜ 1 ⎟⎥ 2 x2 ⎠ ⎦ 4πε 0 x ⎣ ⎝ q

2

 ⇒ Fnet =

⎤ ⎥ ⎥ ⎦

{using Binomial Approximation} 3 qR2 8πε 0 x 4

Problem 3

Problem 4

A point charge q is located at the centre of a thin ring of radius R with uniformly distributed charge -q . Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance x from its centre if x  R .

Find the electric flux crossing the wire frame ABCD of length l width b and whose centre is at a distance OP = d from an infinite line of charge with linear charge density l . Consider that the plane of frame is perpendicular to the line OP.

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 142

9/20/2019 10:24:03 AM

Chapter 1: Electrostatics 1.143 Solution b2

b2

ϕe = 2

∫ dϕ

e

=2

0

Now

l

⎛

⎞

∫ ⎜⎝ 2πε r ⎟⎠ ( l dx ) cosq 0

0

dx = vx has the nearly constant value v dt

⇒ dvy =

where, r = d 2 + x 2 and cos q =

d = r

vy

⇒ vy =

d 2

d +x

2

d

O

P D

⇒ ϕ=2

∫ 0

⇒ ϕ=

l

∞

C

⇒

The vertical velocity component of the moving charge increases according to y

θ

0

q

d Q

⇒ m

-∞

y

-∞

Q

∫ E ( 2π d ) dx = ε y

-∞

0

Q

∫ E dx = 2π dε y

vy =

Solution

dt

∫ E dx …(1)

0

and hence from equation (1), we get

A particle of mass m and charge q moves at high speed along the x axis. It is initially near x → -∞, and it ends up near x → + ∞ . A second charge Q is fixed at the point x = 0, y = - d . As the moving charge passes the stationary charge, its x component of velocity does not change appreciably, but it acquires a small velocity in the y direction. Determine the angle through which the moving charge is deflected.



∫

-∞

ll ⎛ b ⎞ tan -1 ⎜ ⎝ 2d ⎟⎠ πε 0

dvy

∞

∞

Ey dA =

∞

Problem 5

m

q mv

The radially outward component of the electric field varies along the x-axis, and is described by using Gauss’s Theorem

⎛ l ld ⎞ dx ⎜⎝ 2πε ⎟⎠ d 2 + x 2 0

v

∫

dvy =

0

A b B

b2

q Ey dx mv

= Fy

dvy dx = qEy dx dt

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 143

vx

vy x

qQ mv 2π dε 0

The angle of deflection is described by tan q =

vy v

=

qQ 2πε 0 dmv 2

qQ ⎞ ⎛ ⇒ q = tan -1 ⎜ 2 ⎟ ⎝ 2πε 0 dmv ⎠ Problem 6

Two small balls A and B with charges -q and +q respectively are fixed on a horizontal plane at a distance d from each other. A third ball C with charge +Q is suspended from a string. The string makes an angle of 30° with the vertical when the ball C is in equilibrium at a height d vertically above the ball A. When the ball C is in an identical situation above ball B, find the angle which the string now makes with the vertical. Solution

Since ball C is in equilibrium, the sum of torques of all the forces about the point of suspension must be zero.

9/20/2019 10:24:16 AM

1.144  JEE Advanced Physics: Electrostatics and Current Electricity

30° 30 °

1 qQ 4πε 0 2d 2

T

45° 105° Q C mg 1 qQ 4πε 0 d 2 45° –q +q A B d

KqQ ⇒

2d 2 KqQ 2d

2

⇒ Q=

⎛ KqQ ⎞ l sin 75° = ⎜ 2 + mg ⎟ l sin 30° …(1) ⎝ d ⎠ 2d

2

=

mg 2

2

⇒ ( 2 - cos 15° ) =

cos 45° cot q + sin 45° 2

⇒ ( cot q + 1 ) = 2 2 ( 2 - cos 15° ) ⇒ cot q = 2 2 ( 2 - cos 15° ) - 1 Since cos ( 15° ) = cos ( 45° - 30° ) ⇒ cos ( 15° ) = cos 45° cos 30° + sin 45° sin 30° ⇒ cos ( 15° ) =

1 2 2

(

3 + 1)

1 ⎛ ( 3 + 1 ) ⎞⎟ - 1 ⇒ cot q = 2 2 ⎜ 2 ⎝ ⎠ 2 2

mg Kq 2d

⇒ Q=

KqQ

cos 15° -

⎛ 1 - ( cos 15° - 1 ) ⎞ cos 45° cos q + sin 45° sin q ⇒ ⎜ = ⎝ ( cos 15° - 1 ) ⎟⎠ 2 ( cos 15° - 1 ) sin q

⇒ cot q = ( 4 2 - 3 - 1 ) - 1

( cos 15° - 1 ) × 2

⇒ cot q = 4 2 - 3 - 2

2

mgd …(2) Kq ( cos 15° - 1 )

Similarly, in second case,

⇒ q = cot -1 ( 4 2 - 3 - 2 ) Problem 7

KqQ ⎛ KqQ ⎞ l sin ( 135° - q ) ⎜⎝ 2 - mg ⎟⎠ l sin q = d 2d 2

Since sin ( 135° - q ) = cos ( 45° - q )

KqQ ⎛ KqQ ⎞ l cos ( 45° - q ) …(3) ⇒ ⎜ 2 - mg ⎟ l sin q = ⎝ d ⎠ 2d 2

A straight infinitely long cylinder of radius R0 is uniformly charged with charge density σ . The cylinder serves as a source of electrons, with the velocity vector of emitted electrons perpendicular to its surface. What must be the electron velocity to ensure that the electron can move away from the axis of the cylinder to a distance greater than r.

1 qQ 4πε 0 d 2 T2 θ

θ

135°– θ

1 qQ 4πε 0 2d 2

C

45°

45°

–q A

r

+q d

B

mg mg cos ( 45° - q ) ⎛ ⎞ - mg ⎟ sin q = ⎜⎝ ( ο 2 ( cos 15° - 1 ) ⎠ cos 15 - 1 )

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 144

L

mg

Substituting the value of charge Q from equation (2) in (3), we get



R0

Solution

Let us, first of all determine electric field at a distance r from the cylinder. For that we consider a coaxial cylinder as a Gaussian surface, then according to Gauss’s Law, we have

ε 0 ( 2π rL ) E = ( 2π R0 L ) σ

9/20/2019 10:24:29 AM

Chapter 1: Electrostatics 1.145

⇒ E=

σ R0 …(1) ε0 r

Applying Law of Conservation of Energy, we get

1 me v02 - eV0 = - eV …(2) 2

where V0 is the potential of cylinder and V is the potential at a distance r from the cylinder’s axis. dV Employing the relationship E = , we have dr R0σ dV =…(3) ε0 r dr Integrating, we get Rσ V = - 0 log e r + C …(4) εo

where C is constant of integration. Also, from equation (4), we get V0 = -

R0σ log e ( R0 ) + C …(5) ε0

Substituting values of V and V0 in (2), we get



v0 =

2eR0σ ⎛ r ⎞ log e ⎜ ε 0 me ⎝ R0 ⎟⎠

from the large sphere D, r = radius of each small ball, a = each side of the equilateral triangle. Let Q be the charge on the sphere and V be its potential. First consider sharing of charge between the sphere and the first ball, i.e., between D and A. Then



⇒ VA =

Three small identical neutral metal balls are at the vertices of an equilateral triangle. The balls are connected to a large charged sphere held above the plane of the triangle. The first and the second ball acquire charge q1 and q2 respectively. How much is q3 ? The connecting wires are very thin. D

l A

l

a

l

a

C

Solution

When a small ball is connected to a large charged sphere, the potential of the sphere will remain unchanged. Let l be the distance of each small ball

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 145

1 q1 1 Q + 4πε 0 r 4πε 0 l

But potential of the ball is same as that of the sphere ⇒ V=

1 ⎛ q1 Q ⎞ ⎜ + ⎟⎠ …(1) 4πε 0 ⎝ r l

Next consider sharing of charge between ( A + D ) and B . 1 ⎛ q2 q1 Q ⎞ The potential of B is VB = ⎜ + + ⎟⎠ 4πε 0 ⎝ r a l But potential of B is same as that of ( A + D ) , so

⇒

B

⎞ ⎛ Potential ⎞ ⎟ ⎜ ⎟ due to ⎟ +⎜ ⎟ ⎟ ⎜ charge on the ⎟ ⎟⎠ ⎜⎝ sphere D ⎟⎠

VB = V =

1 ⎛ q2 q1 Q ⎞ ⎜ + + ⎟⎠ …(2) 4πε 0 ⎝ r a l

Consider sharing of charge between ( A + D + B ) and C.

Problem 8

a

⎛ Potential ⎛ Potential ⎞ ⎜ ⎜ of small ⎟ = ⎜ due to ⎜ ⎟ ⎜ its own ⎝ sphere A ⎠ ⎜ ⎝ charge

V=

1 ⎛ q3 q1 q2 Q ⎞ + + + ⎟ …(3) ⎜ 4πε 0 ⎝ r a a l ⎠

From equations (1) and (2), we get ⇒

q1 q2 q1 = + r r a r q1 - q2 = a q1

From equations (1) and (3), we get q1 q3 q1 + q2 = + r r a ⇒

r q1 - q3 = a q1 + q2

q1 - q2 q1 - q3 = ⇒   q1 q1 + q2 ⇒ q12 - q22 = q12 - q1 q3 ⇒ q3 =

q22 q1

9/20/2019 10:24:44 AM

1.146  JEE Advanced Physics: Electrostatics and Current Electricity Problem 9

If the string slacks at q = 45° , then T = 0

AB is a vertical rigid thin infinite wire carrying a lin-1 ear charge of density l = 10 μ Cm . A particle having mass m = 2 g and charge is fixed to the wire by means of a light, insulating and inextensible string having length l = 2 2 m. Find the vertical velocity u with which it should be projected under gravity from the shown position so that the string slacks when its angle with vertical becomes 45°. A

u l

lQ sin q mv 2 …(1) + 2πε 0 ( l sin q ) l

Using Work Energy principle, we have ⎛ Work done ⎞ ⎛ Work done ⎞ ⎛ Change ⎞ ⎜⎝ by gravity ⎟⎠ + ⎜ by electrostatic ⎟ = ⎜⎝ in KE ⎟⎠ ⎜⎝ ⎟⎠ forces 1 ⇒ - mgl cos q + 2πε 0

l sin q

∫

⎛ lQ ⇒ - mgl cos q + ⎜ ⎝ 2πε 0

λ

m

⇒ mg cos q =

l

⎞ 1 ( 2 2) ⎟⎠ log e ( sin q ) = 2 m v - u

Using (1), we get

B

-2mgl cos q +

lQ log e ( sin q ) πε 0

Solution

Let us visualise the situation by drawing a diagram showing all the forces acting on the particle. v Fe =

λQ 2πε 0 (l sin θ )

Q ( l dx ) 1 = m ( v 2 - u2 ) 2 x

= mgl cos q ⇒ mu2 = 3 ml cos q - 2 ⇒ u=

θ

3 gl

mg 45°

2

-

lQ - mu2 2πε 0

1 lQ ⎡⎣ 2 log e ( sin q ) + 1 ⎤⎦ 4πε 0

lQ ( - log e 2 + 1 ) 2πε 0 m

⇒ u = 5.7 ms -1 Problem 10

Fe sinθ

v

θ

mv 2 Fe l

θ

T mg cos θ

mg

The electric field, when the particle is at a perpendicular distance l sin q from the wire, is E=

l

=

l ( 2πε 0 l sin q )

2πε 0 ( r⊥ ) So, the electrostatic force Fe on the charge Q is

Fe =

lQ 2πε 0 ( l sin q )

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 146

Two thin concentric rings are placed in a gravity free region in yz- plane, one of radius R carries a charge +Q and the second of radius 4R and charge -8Q distributed uniformly over it. Find the minimum velocity with which a point charge of mass m and charge -q should be projected from a point at a distance 3R from the centre of the ring on its axis so that it will reach to the centre of the rings. –8Q 4R R

Q B P

A

–q

X 2R 3R

9/20/2019 10:24:55 AM

Chapter 1: Electrostatics 1.147 Solution

Note that at a point distant 3R charge -q will have repulsive force but on points close to the centre it will have an attraction force. First we have to find out a point where the electric field is zero because, beyond that, the charge will have attraction force and will be attracted to the centre of the rings. Electric field at a point P distant x from the centre. EP =

1 4πε 0

Qx

(

3 R2 + x 2 2

)

-

On solving for x, we get

1 4πε 0

8Qx

(

3 16 R2 + x 2 2

)

VA =

1 4πε 0

m

r

A

Q 1 ⎡ 8Q ⎤ ⎢ 2 2 2 2 ⎥ 4πε o ⎣ 9R + R 9R + 16 R ⎦

⇒ VA = -

Initially

=0

x = 2R We have to give the charge sufficient minimum KE so that it could reach this point, because for x < 2R the field will be attractive. Potential at A is

c­ ompletely on the rod and can move horizontally on the rod. B is a non-conducting charged small particle of mass 2m and charge Q and connected to A by an inextensible light string of length l . B is released from rest from the position shown in figure. Find velocity of B and tension in the string when the string becomes vertical. Assume gravitational acceleration g and lQ > 0 .



1 4πε 0

The particle of charge q , mass 2m is released from rest (from 1) and it goes to 2, where it has a horizontal velocity v . Also, 2m is attached through string to m (the smooth ring) capable of moving along the rod. So, let the velocity of the ring, when the string becomes vertical be v ′ . Then by Law of Conservation of Linear Momentum (applied only along the horizontal) mv ′ = ( 2m ) v ⇒ v ′ = 2v

⎛ 16 - 10 ⎞ Q ⎜⎝ ⎟⎠ 10 R Q 4 R2 + R2

Finally

-

1 4πε 0

8Q

WA→B = q ( VB - VA )

1 1 Qq ⎡ 6 5 + 10 - 16 ⎤ mv 2 = ⎢ ⎥ 2 4πε 0 R ⎣ 10 ⎦

⇒ vmin =

2m

In the figure shown R is a long smooth fixed nonconducting rod of uniform linear charge density l and radius of cross-section r. A is non-conducting uncharged smooth ring of mass m which fits

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 147

1

v B 2

Further in going from 1 to 2, we have

W = q ( V2 - V1 ) + Work done by  1→ 2  Gravitational Force

⎡ l ⎛ r+L⎞ ⎤ ( ⇒ ΔK = Q ⎢ log e ⎜ + 2m ) gL ⎝ r ⎟⎠ ⎥⎦ 2 πε 0 ⎣ ⎛ Work done ⎞ ⎛ Change ⎞ ⎛ Work done ⎞ = ⎜⎝ in KE ⎟⎠ ⎜ by electrostatic ⎟ + ⎜⎝ by gravity ⎟⎠ ⎜⎝ ⎟ forces ⎠

Qq ( 6 5 + 10 - 16 ) 20πε 0 mR

Problem 11

m A

4 R2 + 16 R2 1 3 Q 4πε 0 5 R

1 1 Qq ⎡ 3 16 - 10 ⎤ 2 Now mv = ⎢ ⎥ 2 4πε 0 R ⎣ 5 10 ⎦ ⇒

v′

=

From Conservation of Energy, we have

B(2m, Q)

Solution

Potential at point B is VB =

L

⇒

1 1 Ql ⎛ r+L⎞ ( mv ′ 2 + ( 2m ) v 2 = log e ⎜ + 2m ) gL ⎝ r ⎟⎠ 2 2 2πε 0

⇒

1 ( 2) 1 Ql ⎛ r+L⎞ + 2mgL m 4v + ( 2m ) v 2 = log e ⎜ ⎝ r ⎟⎠ 2 2 2πε 0

⇒ 3 mv 2 =

Ql ⎛ r+L⎞ log e ⎜ + 2mgL ⎝ r ⎟⎠ 2πε 0

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1.148  JEE Advanced Physics: Electrostatics and Current Electricity

Now time period of particle would be given as

2 gL Ql ⎛ r+L⎞ + log e ⎜ ⎝ r ⎟⎠ 3 6πε 0 m

⇒ v=

Now, tension in the string, when it becomes vertical is



T = ( 2m ) g +

where vnet

(

2 m vnet

L



⎛ l ⎞ ⎛ 1 ⎞ T = 2π ⎜ ⎟ = 2π ⎜ ⎝ 109.8 ⎟⎠ ⎝ g′ ⎠

T = 0.6 s Now from figure, at point A we have

) + QE

l = v + v ′ = 3v and E = ( 2πε 0 r + L )

lQ 9mv 2 ⇒ T = 2mg + + L 2πε 0 ( r + L )

and

mvA2 = TA - mg ′ = TA - ( qE + mg ) …(1) l mvB2 = TB + ( qE + mg ) l

To complete the circle at point B tension TB should be zero, thus we have

Problem 12

A uniform electric field of strength 106 Vm -1 is directed vertically downwards. A particle of mass 0.01 kg and charge 10 -6 coulomb is suspended by an inextensible thread of length 1 m. The particle is displaced slightly from its mean position and released. Calculate the time period of its oscillation. What minimum velocity should be given to the particle at rest so that it completes a full circle in vertical plane without the thread getting slack? Calculate the maximum and minimum tension in the thread in this situation.

mvB2min

= qE + mg …(2) l Using Work Energy Theorem at points A and B , we have 1 1 mvA2 - 2mgl - 2 gEl = mvB2 + UB …(3) 2 2 From equation (2) and (3), we get

vA2 =

5 ( qE + mg ) l m

qE + mg ⎫ ⎧ ⎬ ⎨ where g ′ = m ⎭ ⎩

Solution



The situation is shown in figure Acceleration a due to electric field

⇒ vA = 23.42 ms -1

a=

-6

Now, TA = ( qE + mg ) +

6

qE 10 × 10 = = 100 ms -2 m 0.01

Thus effective acceleration would be

qE g′ = g + = 9.8 + 100 = 109.8 ms -2 m vB

mg′ TB O E

TA A



TA = 6 ( qE + mg ) = 6 ( 1 + 0.098 )



TA = 6.588 N

Problem 13

vA

Solution

The electric field due to dipole is given by

mg



01_Physics for JEE Mains and Advanced - 2_Part 4.indd 148

mvA2 l

A point electric dipole with a moment p is placed in the external uniform electric field whose strength equals E0 with p parallel to E0 . In this case one of the equipotential surfaces enclosing the dipole forms a sphere. Find radius of this sphere.

B

l

= 5 g ′l = 5 × 109.8 × 1

E=

1 p 1 + 3 cos 2 q 4πε 0 R3

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Chapter 1: Electrostatics 1.149

 and its direction with the radius vector r is given 1 by tan a = tan q . Now given that external field 2     E0 is parallel to p. So that Er and Eq are the radial and transverse component of electric field due to the dipole and E0 sin q and E0 cos q are two rectangular components of external field E0. Since at equipotential surface, there is no tangential components of electric field. So E0 sin q must be equal and opposite to E0 sin a E0 sin θ

E0 θ

p

Er = E cos θ α

E0 cos θ E

E sin a sin q

⇒

1 ⎡ p 1 sin q ⎤ × 1 + 3 cos 2 q ⎥ E0 = ⎢ 3 2 4πε 0 ⎣ R ⎦ 1 + 3 cos q sin q



3

2

⇒ q = 45°

 Positive vector r of point P is

(

)

 r = 5 cos 45iˆ + sin 45 ˆj 6 =

(

5 ˆ ˆ i+j 2

(

5 ˆ ˆ i+j 2

)

)

Problem 15

1 ⎫ ⎧ ⎨∵ tan a = tan q ⎬ 2 ⎩ ⎭

E0 4πε 0 R3 = p R=

⇒

1

⇒ sin q =

 ⇒ r=

E0 =

= sin q

5

⇒ 1 + 3 sin 2 q = 5 sin 2 q



⇒

 ⇒

1 + 3 sin 2 q

⇒

iˆ

Hence E0 sin q = E sin a



1 p 1 p sin q 1 + 3 sin 2 q = 3 4πε 0 r 4πε 0 r 2

⇒

Eθ = E sin α

θ

EP = VP



Find the potential of an isolated ball shaped conductor of radius R1 surrounded by an adjacent concentric layer of dielectric with dielectric constant K and outer radius R2 . Solution

Let a charge q be assumed to be given to the conductor

p 4πε 0 E0

∫

Since, V = - E ⋅ dr

Problem 14

A short electric dipole is situated at the origin of coordinate axis with its axis along x-axis and equator along y-axis. It is found that the magnitudes of the electric field intensity and electric potential due to the dipole are equal at a point distant r = 5 m from origin. Find the position vector of this point. Solution

Consider a point P at distance r and angle q from equation. Since y

R1 +q R2

R1 ⎤ ⎡ R2 q q 1 1 ⎢ ⎥ V = dr + dr ⇒ ⎢ 4πε 0 r 2 4πε 0 K r 2 ⎥ R2 ⎢⎣ ∞ ⎥⎦

∫

P r

⇒ V=-

θ x

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 149

∫

R1 ⎡ R2 q ⎢ dr dr + 4πε 0 ⎢ r 2 Kr 2 R2 ⎢⎣ ∞

∫

∫

⎤ ⎥ ⎥ ⎥⎦

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1.150  JEE Advanced Physics: Electrostatics and Current Electricity

⇒ V=

q 4πε 0

⎡ ⎛ 1 ⎞ R2 ⎛ 1 ⎞ R1 ⎤ ⎢ ⎜⎝ - ⎟⎠ + ⎜⎝ ⎟ ⎥ r ∞ Kr ⎠ R2 ⎥⎦ ⎢⎣

⇒ x=

⇒ V=

q 4πε 0

1 1 ⎤ ⎡ 1 ⎢ R + KR - KR ⎥ ⎣ 2 1 2 ⎦

If the discriminant of the quadratic equation is real, we have two points where the field is zero. Discriminant is positive for Q ≥ 8 al .

⇒ V=

q 4πε 0

⎡ KR1 + R2 - R1 ⎤ ⎢ ⎥ KR1 R2 ⎣ ⎦

⇒ V=

q ⎡ R1 ( K - 1 ) + R2 ⎤ ⎥ KR1 R2 4πε 0 ⎢⎣ ⎦

q ⇒ V= 4πε 0

2

1⎛ Q 1⎛ Q ⎞ ⎞ - 2a ⎟ ± - 2a ⎟ - a2 ⎜⎝ ⎜⎝ ⎠ ⎠ 2 2l 4 2l

FOR REGION II In the region (II) the electric field of wire and point charge point in the same direction, positive x-axis. So no point can exist where the field is zero.

1 ⎤ ⎡⎛ K -1⎞ 1 ⎢ ⎜⎝ K ⎟⎠ R + KR ⎥ ⎣ 2 1 ⎦

FOR REGION III Now we take a point to the right of the point charge, at a distance x from it. Resultant field at this point is  ER =

Problem 16

An infinitely long conducting wire of charge density + l and a point charge -Q are at a distance from each other. In which of the three Regions (I, II or III) are there points that lie on the line passing through point charge perpendicular to the conductor and at which the field is zero?

Resultant field is zero if

l Q = ( a + x ) 2x 2

On solving the quadratic in x , we have x=

I

II

III –Q

A

l Q ( ˆ) iˆ + -i 2πε 0 ( x + a ) 4πε 0 x 2

Q Q2 aQ ± + 2 4l 2l 16 l

The negative sign infront of the radical has no meaning because it would mean that the point is to the left of point charge, where field of wire and point charge are added, the magnitudes of the two fields are zero. Problem 17

Solution

FOR REGION I Now we check the region I, take a point to the left of wire at a distance x from it. The resultant field is  ER =

l ( ˆ) Q -i + iˆ 2 2πε 0 x 4πε 0 ( a + x )

The two fields point in the opposite directions, so resultant field can be zero if,



Figure shows two identical beads of mass m and charge q . The beads can slide smoothly on a wire frame kept in a vertical frame.

l Q = 2πε 0 x 4πε 0 ( a + x )2

Q⎞ ⎛ 2 ⇒ x 2 + ⎜ 2a ⎟ x+a = 0 ⎝ 2l ⎠

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 150

O θ

m, q

θ

m, q

(a) Determine angular position q w.r.t. vertical diameter. (b) Now the beads are given a small angular displacement. Show that they perform simple harmonic motion.

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Chapter 1: Electrostatics 1.151 Solution

(a) Each bead is in equilibrium under the action of three forces Coulomb force of repulsion, Weight mg and Normal reaction, N



⎛ mgR cos q + FR sin q ⎞ ⇒   a = -⎜ ⎟⎠ ϕ ⎝ mR2



⎛ mR cos q + FR sin q ⎞ ⇒   ϕ + ⎜ ⎟⎠ ϕ = 0 ⎝ mR2



Comparing with standard equation of SHM 2 i.e., ϕ + ω ϕ = 0 , we get

O N F

N θ θ

m, q

mgR cos q + F m, q

mg

2R sin θ

mg



From conditions of equilibrium, we have



N sin q = F =



and N cos q = mg …(2)



From equations (1) and (2), we have



tan q =

q2 …(1) 4πε 0 ( 2R sin q )

q2 …(3) 4πε 0 mg ( 2R sin q )

(b) After a small angular displacement, we may consider the Coulombic force F to be approximately constant. In equilibrium, net torque about centre of circle is zero. So we have

- mgR sin q + FR cos q = 0 …(4)



After a small angular displacement ϕ of right bead, we have



- mgR sin ( q + ϕ ) + FR cos ( q + ϕ ) = Ia = mR a …(5) 2



⇒   - mgR ( sin q cos ϕ + cos q sin ϕ ) +



  FR ( cos q cos ϕ - sin q sin ϕ ) = mR2a …(6)



For small ϕ , sin ϕ  ϕ and cos ϕ  1 , so equation (6) reduces to



- mgR ( sin q + ϕ cos q ) +







From equations (4) and (7), we get



- ( mgR cos q + FR sin q ) ϕ = mR2a

  FR ( cos q - ϕ sin q ) = mR2a …(7)

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 151



  ω =



⇒  ω=

q2 4πε 0 ( 2R sin q )

2

sin q

mR2

8πε 0 R3 mg sin 2 q + q2 16πε 0 mR2 sin q

Problem 18

In an insulating medium (di-electric constant K = 1) volumetric charge density varies with y-coordinate according to the law ρ = ay . A particle of mass m having positive charge q is at point A ( 0 , y0 ) and  projected with velocity v = v iˆ as shown in figure. 0

Neglecting gravity and frictional resistance of the medium and assuming electric field strength to be zero at y = 0, calculate the slope of the trajectory of the particle as a function of y . y

A

x

Solution

As charge density varies with y-coordinate only, electric field must be directed along y-axis. Consider a thin layer of medium, having thickness dy at a distance y from x-axis as shown in figure. y

y y + dy x

Let strength of electric field at positions y and ( y + dy ) be E and (E + dE) respectively.

9/20/2019 10:25:50 AM

1.152  JEE Advanced Physics: Electrostatics and Current Electricity

Then for area A of the layer, according to Gauss’s Law,

( E + dE ) A - EA =

⇒ dE =

ρ A dy ε0

ρ a dy = y dy ε0 ε0

Integrating above equation with in appropriate limits, we get E



∫ 0

a dE = ε0

y

∫ ydy

Problem 19

A charged particle having a charge q moves along the x-axis with a constant velocity v0. Another particle B with charge q and mass m is lying on the y-axis at y = a. The particle B is constrained to move along the y-axis, while the particle A moves along the x-axis. Assuming that the velocity v0 is very large, find the impulse imparted to B along the y-axis as the particle A moves from -∞ to ∞ assuming that the motion of particle B is negligible. Solution

0

ay 2 ⇒ E= …(1) 2ε 0

First of all, let us determine the angular speed of A relative to B at angular position q shown in figure. y

There is no force on the particle along x-axis, velocity v0 of the particle along x-axis remains constant Due to electric field E , the particle accelerates along positive y-direction with acceleration ay , given by ay =

qE qay 2 = m 2mε 0

Let y-component of velocity of particle be v , then from Newton’s Second Law we get

dv qay 2 v = dy 2mε 0

⇒ v dv =

⇒

∫ 0

⇒ v=

y

∫ y dy

dy = dx

For particle A

dx d = ( a tan q ) = v0 …(1) dt dt dq = v0 dt

Since particle A is moving very fast, we can assume that when it crosses along the x-axis, the y-component of the force on B due to A is

y0

)

qa 3 mε 0 v02

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 152

Fy =

q2 1 × cos q …(3) 2 4πε 0 a sec 2 q

Impulse on B =

dy dy dt v = = Slope of trajectory = dx v0 dx dt ⇒ Slope =

x

A

2

qa y 3 - y03 3 mε 0

(

V0

⎛ dq ⎞ v0 = ⇒ ⎜ cos 2 q …(2) ⎝ dt ⎟⎠ a

qa 2 y dy 2mε 0

qa v dv = 2mε 0

θ

a

⇒ a sec 2 q

Since at point A ( y = y0 ) , y-component of velocity is v=0 v

q, m

B

( y 3 - y03 )

⇒ B=

⇒ B=

∫

Fy dt =

1 q 4πε 0 av0

∫ ⎛⎜ dq ⎞⎟ ⎝ dt ⎠

1 q2 a 4πε 0 a 2 v0 2

Fy dq

cos 3 q

∫ cos q dq 2

π 2

∫

cos q dq

π q=2

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Chapter 1: Electrostatics 1.153

The impulse delivered to B, is given by



ΔPy =

2

2

q 1 2q = …(4) 4πε 0 av0 2πε 0 ( av0 )

Problem 20

A particle of mass m and charge +q is constrained to move along the x-axis. Two charged rings of radius R, charge Q are placed with their centres at ( - R, 0 ) and ( R, 0 ) as shown. (a) Obtain an expression for the potential due to the rings as a function of x for - R < x < R. (b) Show that in this region V ( x ) is minimum at x=0. (c) Show that for x  R , the potential is of the form V ( x ) = V ( 0 ) + a x 2. (d) Derive an expression for the time period of oscillation of the mass m when it is displaced slightly from the origin and released. Q

Q

R

m

X = –R

X = +R

z

1 4πε 0

Q

(

1 R2 + x 2 2

(a) At any point both rings

V(x) =

)

( x, 0, 0 ) the potential due to

1 1 Q ⎡ ⎤ + ⎥ ⎢ 2 2 2 4πε 0 R2 + ( R + x ) ⎦ ⎣ R + (R - x)

(b) For minimum V ( x ) ,





⇒ 

dV ( x ) Q ⎡ 1 2 ( R - x ) ( -1 ) = 3 dx 4πε 0 ⎢ 2 2 2 ⎢ 2 ⎡⎣ R + ( R - x ) ⎤⎦ ⎣ 1 2

2( R + x )

⎤ ⎥ ⎥ ⎦ ⎦

3 ⎡ R 2 + ( R + x )2 ⎤ 2

⎣

dV ( x ) Q ⎡ R-x R+x = ⎢ 3 3 dx 4πε 0 ⎢ ⎡⎣ R2 + ( R - x )2 ⎤⎦ 2 ⎡⎣ R2 + ( R + x )2 ⎤⎦ 2 ⎣

dV ( x ) Q ⎡ R-x R+x = ⎢ 3 3 4πε 0 dx ⎢ ⎡⎣ R2 + ( R - x )2 ⎤⎦ 2 ⎡⎣ R2 + ( R + x )2 ⎤⎦ 2 ⎣

The expression at x = 0 is



     



Thus V ( x ) is minimum at x = 0

dV Q ⎡ R R = ⎢ 3 3 dx 4πε 0 ⎢⎣ ( 2R2 ) 2 ( 2R2 ) 2

     Vmin = V ( 0 ) =





     V ( x ) =

{

d       [ V ( x ) ] = 0 dx

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 153



⎤ ⎥ ⎥ ⎦

Q 2 4πε 0 R

{

1

1

Q ⎡ 2 2 2 2 ⎣ R + ( R - x ) ⎤⎦ 2 + ⎡⎣ R + ( R + x ) ⎤⎦ 2 4πε 0 1

1

1 1 ⎫ ⎧ 2 ⎤- 2 2 ⎤- 2 ⎪ ⎪ ⎡ ⎡ x⎞ Q ⎨ x⎞ ⎛ ⎛ ⎬ ( ) ⇒   V x = 4πε R ⎪ ⎢⎣ 1 + ⎜⎝ 1 - R ⎟⎠ ⎥⎦ + ⎢⎣ 1 + ⎜⎝ 1 + R ⎟⎠ ⎥⎦ ⎪ ⎭ 0 ⎩

Now we apply the binomial approximation, n we have ( 1 + x )  1 + nx when x  1 2





}

}

1 1 ⎫ ⎧ 2 ⎤- 2 2 ⎤- 2 ⎪ ⎪ ⎡ ⎡ x⎞ Q ⎨ x⎞ ⎛ ⎛ ⎢ 1 + ⎜ 1 - ⎟⎠ ⎥ + ⎢ 1 + ⎜⎝ 1 + ⎟⎠ ⎥ ⎬ V ( x ) =  4πε 0 R ⎩⎪ ⎣ ⎝ R ⎦ ⎭⎪ R ⎦ ⎣



⎤ ⎥ ⎥ ⎦

⎤ ⎥=0 ⎥⎦

Q ⎡ 2 2 2 2 V(x) = ⎣ R + ( R - x ) ⎤⎦ 2 + ⎡⎣ R + ( R + x ) ⎤⎦ 2 4πε 0 

Electric potential due to a ring on its axis is given by





x

Solution

V(x) =

⇒ 

(c) For x  R ,

y

R

q



2

x⎞ x⎞ 2x 2x ⎛ ⎛ and ⎜ 1 + ⎟  1 +       ⎜⎝ 1 - ⎟⎠  1 ⎠ ⎝ R R R R



1 1 ⎡ 2x ⎞ 2 ⎛ 2x ⎞ 2 Q ⎢⎛ ⇒   V(x) = ⎜ 2 - ⎟⎠ + ⎜⎝ 2 + ⎟ 4πε 0 R ⎢⎣ ⎝ R R⎠

⎤ ⎥ ⎥⎦

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1.154  JEE Advanced Physics: Electrostatics and Current Electricity









Now we apply the binomial expansion,



x⎞ ⎛       ⎜⎝ 1 - ⎟⎠ R

-

1 2

x⎞ ⎛       ⎜⎝ 1 + ⎟⎠ R

-

1 2

Problem 21

x 3 x2 = 1+ - 2R 8 R 2 = 1+

x 3 x2 + + 2R 8 R 2

where, we ignore the cubic and higher powers of x .





V(x) =

Q 4πε 0

1 1 ⎡ 2 x x ⎛ ⎞ ⎛ ⎞ ⎢⎜ 1- ⎟ + ⎜ 1+ ⎟ 2 ⎝ R⎠ R⎠ 2R ⎢⎣ ⎝

Q

⎡ 3x ⎤ ⎢2+ ⎥ 4 R2 ⎦ 2R ⎣ 2



⇒   V(x) =



Q Q 2 3 x2 ⇒   V ( x ) = 4πε R + 4πε 3 4 2 R 0 o





4πε 0

⎤ ⎥ ⎥⎦

V ( x ) = V0 ( 0 ) + ax 2 , where a =

3Q 4πε 0 4 2R3

(d) The charged particle is displaced from the origin by a distance x (say) For x  R, the potential function is

      V ( x ) = V ( 0 ) + ax dV Since Ex = dx ⇒   Ex = -2 ax



The force experienced by the charge is given by



( )        F = qEx = -Q 2 ax So, acceleration of the charge is



2 aQ a=x        m ⎛ 2Qa ⎞ ⇒   x + ⎜⎝ m ⎟⎠ x = 0



Now, we compare a with standard expression 2 of SHM, a = -ω 2 x , or (x + ω x = 0) , we get



      



m ⇒   T = 2π 2 aQ

ω=

Solution

(a) For simplicity, let’s assume the potential to be zero at infinity, V ( ∞ ) = 0 . Consider an infinitesimal charge element dq = l dz located at a distance z along the z-axis. Its contribution to the electric potential at a point z > d is

l dz 4πε 0 z

Integrating over the entire length of the rod, we obtain z-d

∫

l dz ⎛ z+d⎞ = log e ⎜ ⎝ z - d ⎟⎠ 4πε 0 z

z+d (b) Using the result derived in (a), the electrical potential at z = 4 d is





2 aQ 2π = m T

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 154

       dV =

l       V ( z ) = 4πε 0

2





Athin rod extends along the z-axis from z = - d to z = d . The rod carries a positive charge Q uniformly distribQ uted along its length 2d with charge density l = . 2d (a) Find the electric potential V ( z ) at a point z > d along the z-axis. (b) Find the change in potential energy if an electron moves from z = 4 d to z = 3 d . (c) Assume that the electron started from rest at the point z = 4 d , then find its velocity at z = 3 d .



 V

z= 4d

=

l l ⎛ 4d + d ⎞ ⎛ 5⎞ = log e ⎜ log e ⎜ ⎟ ⎝ 4 d - d ⎟⎠ 4πε 0 ⎝ 3⎠ 4πε 0

V

z= 3d

=

l l ⎛ 3d + d ⎞ log e ⎜ log e 2 ⎟⎠ = ⎝ 4πε 0 3d - d 4πε 0

Similarly, the electrical potential at z = 3 d is

The electric potential difference between the two points is given by ΔV = V

z= 3d

-V

z= 4d

=

l ⎛ 6⎞ log e ⎜ ⎟ > 0 ⎝ 5⎠ 4πε 0

Using the fact that the electric potential difference ΔV is equal to the change in potential energy per unit charge, we have        ΔU = qΔV = -

el ⎛ 5⎞ log e ⎜ ⎟ ⎝ 6⎠ 4πε 0

where q = - e is the charge of the electron.

9/20/2019 10:26:34 AM

Chapter 1: Electrostatics 1.155

(c) If the electron starts out at rest from z = 4 d , then the change in kinetic energy is



1 2        ΔK = mv f 2 By conservation of energy       ΔU + ΔK = 0 el ⎛ 6⎞ log e ⎜ ⎟ > 0 ⇒   ΔK = -ΔU = ⎝ 5⎠ 4πε 0

Similarly, potential on the outermost shell,





2e l ⎛ 6⎞ log e ⎜ ⎟ ⎝ 5⎠ 4πε 0 m



q Q q Q - + =- + …(2) a 2a 3a 3a

⇒ q=

Three concentric conducting shells of A , B and C radii a , 2a and 3a are shown in figure. The charge on the shell A , B and C is Q . When the key K is closed, find the charges on the innermost and outermost shells and ratio of charge densities of the shells.

Charge on innermost shell is - q = -

2a



a

Q 4

Thus charge on outermost shell is q =

3a

q Q q q Q q - + + =- + + …(1) a 2a 3 a 3a 3a 3a

Equation (1) now becomes

Problem 22

1 ⎛ q Q q ⎞ + ⎟ ⎜- + 4πε 0 ⎝ 3 a 3 a 3 a ⎠

Since Va = Vc , so we get

Thus, the magnitude of the velocity at z = 3 d is        v f =

Vc =



σA = σB = σC =

Q 4

Q , so 4

1 ⎛ Q⎞ ⎜- ⎟ 4π a 2 ⎝ 4 ⎠ Q

4π ( 4 a 2 ) +Q

4π ( 9 a 2 )

⇒ σ A : σ B : σ C = -9 : 9 : 1 Solution

When the key is closed, then the innermost and outermost shells will acquire the same potential. Let the charge on the outer shell be q and that on the inner shell be -q , the total charge on inner and outer shells is zero. Potential on innermost shell, Va = Sum of potentials due to -q , Q and q ⇒

Va =

1 ⎛ q Q q ⎞ + ⎟ ⎜- + 4πε 0 ⎝ a 2 a 3 a ⎠

Problem 23

Two concentric shells of radii R and 2R are shown in figure. Initially a charge q is imparted to the inner shell. Now key K1 is closed and opened and then key K 2 is closed and opened. After the keys K1 and K 2 are alternately closed n times each, find the potential difference between the shells. Note that finally the key K 2 remains closed.

2R R

–q Q q K2

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 155

K1

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1.156  JEE Advanced Physics: Electrostatics and Current Electricity Solution

Problem 24

When K1 is closed first time, outer sphere is earthed and the potential on it becomes zero. Let the charge on it be q1′

Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density σ . The spheres are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger sphere.

V1′ = Potential due to charge on inner sphere and that due to charge on outer sphere V1′ =

q′ ⎤ 1 ⎡ q + 1 ⎥=0 ⎢ 4πε 0 ⎣ 2R 2R ⎦

Solution

⇒ q1′ = - q

Let q1 and q2 be the charges on the two spheres before connecting them.

When K 2 is closed first time, the potential V2′ on inner sphere becomes zero as it is earthed. Let the new charge on inner sphere be q2′



0=

1 q2′ 1 ( -q ) + 4πε 0 R 4πε 0 ( 2R )

Now when K1 will be closed second time, charge on q outer sphere will be - q2′ , i.e., 2 After one event involving closure and opening of K1 and K 2 , charge is reduced to half its initial value. Similarly, when K1 will be closed nth time, charge on q outer sphere will be - n -1 as each time charge will 2 be reduced to half the previous value. After closing K 2 nth time, charge on inner shell will be negative of half the charge on outer shell, i.e., ⎛ q ⎞ ⎜⎝ + n ⎟⎠ and potential on it will be zero. 2 For potential of outer shell,

- q ( -1 + 2 ) 4πε 0 2

n+1

R

=

⇒

V0 - Vi =

4πε 0 2n +1 R

-q 4πε 0 2n +1 R

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 156

2

q = q1 + q2 = 20σπ R2 Now, after connecting, the charge is distributed in the ratio of their capacities, which in turn depends on the ratio of their radii ( C = 4πε o R ) . ∴

q ’1 R 1 = = q ’2 2R 2

⇒ q ’1 =

q 20 = σπ R2 3 3

and q ’2 =

2q 40 = σπ R2 3 3

Therefore, surface charge densities on the spheres are:

σ1 =

q ’1 4π R

2

=

( 20 / 3 ) σπ R2 4π R

q ’2

4π ( 2R )

2

=

2

=

5 σ 3

( 40 / 3 ) σπ R2 16π R2

=

5 σ 6

Hence, surface charge density on the bigger sphere is ⎛ 5⎞ σ 2 i.e., ⎜ ⎟ σ . ⎝ 6⎠

-q

Potential difference = V0 - Vi =

q2 = σ ( 4π )( 2R ) = 16σπ R2

and σ 2 =

q ⎞ ⎛ q ⎞ ⎛ + 1 ⎜⎝ 2n ⎟⎠ 1 ⎜⎝ 2n -1 ⎟⎠ + V0 = 4πε 0 2R 4πε 0 2R

⇒ V0 =



) and

Therefore, total charge ( q ) on both the spheres is

q ⇒ q2′ = 2



(

2 Then, q1 = σ 4π R

Problem 25

-q 4πε 0 2n +1 R

-0

A conducting sphere S1 of radius r is attached to an insulating handle. Another conducting sphere S2 of radius R is mounted on an insulating stand S2 is initially uncharged.

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Chapter 1: Electrostatics 1.157

S1 is given a charge Q brought into contact with S2 and removed. S1 is recharged such that the charge on it is again Q and it is again brought into contact with S2 and removed. This procedure is repeated n times.



Similarly,



2 3 ⎡ R ⎛ R ⎞ ⎛ R ⎞ ⎤ q3 = Q ⎢ +⎜ + ⎟ ⎜⎝ ⎟ ⎥ R + r ⎠ ⎥⎦        ⎢⎣ R + r ⎝ R + r ⎠

(a) Find the electrostatic energy of S2 after n such contacts with S1 . (b) What is the limiting value of this energy as n→∞?



and



n R⎡ ⎛ R ⎞ ⎤ 1 ⇒   q = Q ⎢ ⎥ …(2) ⎜ ⎟ n r ⎢⎣ ⎝ R + r ⎠ ⎥⎦

n 2 ⎡ R ⎛ R ⎞ ⎛ R ⎞ ⎤ qn = Q ⎢ +⎜ + ... + ⎟ ⎜⎝ ⎟ ⎥ R + r ⎠ ⎥⎦ ⎢⎣ R + r ⎝ R + r ⎠      

Solution

Capacities of conducting spheres are in the ratio of their radii. Let C1 and C2 be the capacities of S1 and S2 , then



Therefore, charge on S1 will be Q - q1 . Say it is q’1 . q1 q1 C R = = 2 = q ’1 Q - q1 C1 r



⇒ 



It implies that Q charge is to be distributed in S2 and S1 in the ratio of R / r.



⎛ R ⎞ ⇒   q1 = Q ⎜⎝ R + r ⎟⎠ …(1)



In the second contact, S1 again acquires the same charge Q.



Therefore, total charge in S1 and S2 will be





R ⎞ ⎛ Q + q1 = Q ⎜ 1 + ⎟ ⎝ R+r⎠

       This charge is again distributed in the same ratio. Therefore, charge on S2 in second contact, R ⎞⎛ R ⎞ ⎛ q2 = Q ⎜ 1 + ⎟⎜ ⎟ ⎝ R+r ⎠ ⎝ R+r ⎠

   2 ⇒ q = Q ⎢⎡ R + ⎛ R ⎞ ⎥⎤ ⎜ ⎟ 2 ⎣ R+r ⎝ R+r⎠ ⎦

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 157

)







Therefore, electrostatic energy of S2 after n such contacts is

C2 R = C1 r

(a) Charges are distributed in the ratio of their capacities. Let in the first contact, charge acquired by S2 , is q1 .

(

a 1 - r n ⎪⎫ ⎪⎧ ⎬ ⎨∵ Sn = ( 1 - r ) ⎪⎭ ⎩⎪



      

q2 qn2 Un = n = 2C 2 ( 4πε o R )       



qn2 U = ⇒   n 8πε o R



where qn can be written from equation (2).

(b) qn =

QR ⎡ R ⎛ R ⎞ + ...... + .... + ⎜ ⎢1+ ⎝ R + r ⎟⎠ R + r ⎢⎣ R+r

Since, the sum of infinite GP is S∞ =

So, when n → ∞ , then ⎡ QR ⎢ 1 q∞ = ⎢ R+ r ⎢ 1- R ⎣ R+r

n -1

⎤ ⎥ ⎥⎦

a 1- r

⎤ ⎥ ⎥ ⎥ ⎦







⇒   q∞ =

QR ⎛ R + r ⎞ ⎜ ⎟ R+r ⎝ R+r -R⎠



⇒   q∞ =

QR ⎛ R + r ⎞ R ⎜⎝ ⎟⎠ = Q R+r r r



q∞2 Q 2 R2 / r 2 ⇒   U = = ∞ 8πε 0 R 8πε o R



⇒   U∞ =

Q2 R 8πε o r 2

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1.158  JEE Advanced Physics: Electrostatics and Current Electricity

and -2Q are placed at these two points. So, between -3 a < x < 3 a we can find potential by putting y = 0 in equation (1). Therefore,

Problem 26

Two fixed charges -2Q and Q are located at the points with coordinates ( -3 a, 0 ) and ( +3 a, 0 ) respectively in the x -y plane. (a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Give the expression V ( x ) at a general point on the x-axis and sketch the function V ( x ) on the whole x-axis. (c) If a particle of charge +q starts form rest at the centre of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so. Solution

(a) Let P ( x , y ) be a general point of x -y plane. Electric potential at point P would be,

V = ( potential due to Q ) + ( potential due to - 2Q )



⎛ 1 ⇒ V = 4πε ⎜ o ⎜ ⎝

⎞ Q -2Q ⎟ + 2 2 2 2 ⎟ ( 3a - x ) + y ( 3a + x ) + y ⎠

…(1)

Given V = 0



2 2 2⎤ 2 ⎡ ⇒   4 ⎣ ( 3a - x ) + y ⎦ = ( 3a + x ) + y



Solving and rearranging, we get



2      ( x - 5 a ) + y = ( 4 a ) which is the equation of a circle of radius 4a and centre at ( 5 a, 0 ) 2

2

y V=0 (a, 0) O

C(5a, 0)

(9a, 0)

x

4a

Q ⎛ 1 2 ⎞ ⎜⎝ ⎟ , for -3 a < x < 3 a 4πε o 3 a - x 3 a + x ⎠



 V =



For -3 a < x < 3 a, we observe that



     V = 0 at x = a      V → -∞ as x → -3 a and      V → +∞ as x → 3 a



For x > 3 a, there is again a point where potential will become zero so for x > 3 a, we can write Q ⎛ 1 2 ⎞ V= ⎜⎝ ⎟ for x > 3 a       4πε o x - 3 a 3 a + x ⎠



and observe that, V = 0 at x = 9 a



For x < -3 a, we can write V=

Q ⎛ 1 2 ⎞ ⎜ ⎟ for x < -3 a 4πε o ⎝ 3 a - x 3 a - x ⎠



   In this region potential will be zero only when x → -∞



So, we can summarise V ( x ) as under



(i) At x = 3 a, V → +∞



(ii) At x = -3 a, V → -∞



(iii) For x < -3 a , V ( x ) =



(iv) For -3 a < x < 3 a , expression of V ( x ) is same i.e.,     

V(x) =

Q ⎛ 1 2 ⎞ ⎜ ⎟ 4πε o ⎝ 3 a - x 3 a + x ⎠

Q ⎛ 1 2 ⎞ ⎜ ⎟ 4πε o ⎝ 3 a - x 3 a + x ⎠







  (v)  For x > 3 a , V ( x ) =



Potential on x-axis is zero at two places at x = a and x = 9 a . The V -x graph is shown here.

Q ⎛ 1 2 ⎞ ⎜⎝ ⎟ 4πε o x - 3 a 3 a + x ⎠

(b) On x- axis, potential will be undefined (or say ±∞ ) at x = 3 a and x = -3 a, because charge Q

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Chapter 1: Electrostatics 1.159 Solution

V

By Law of Conservation of Energy Total Energy at A = Total Energy at B –3a

a

9a +3a

x

⇒ K A + U A = KB + UB Energy of particle at A EA =

(c) Potential at centre i.e., at x = 5 a will be,



2 ⎞ Q ⎛ 1 Q V = = =       C 4πε o ⎜⎝ 2 a 8 a ⎟⎠ 16πε o a positive

Potential on the circle will be zero. Since, potential at centre > potential on circumference on it, the particle will cross the circle because positive charge moves from higher potential to lower potential. Speed of particle, while crossing the circle would be,      

v=

2q ( ΔV ) = m

Qq 8πε o ma

Here, ΔV is the potential difference between the centre and circumference of the circle.

Qq 1 mv 2 + 2 4πε 0 x

At any point P inside the sphere at a distance r from its centre, the potential energy is given by

UP = qVP

where, Vinside - VP =

Q 8πε 0 R3

( 3R2 - r 2 )

So, energy of particle at B is EB = 0 + ⇒

EB =

⎛ 2 ⎛ R⎞2 ⎞ ⎜ 3 R - ⎜⎝ ⎟⎠ ⎟⎠ 2 8πε 0 R3 ⎝ Qq

⎛ 11R2 ⎞ 11 Qq ⎜ ⎟= 8πε 0 R3 ⎝ 4 ⎠ 32 πε 0 R Q

Q

Problem 27

A positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R . A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point at distance x ( x > R ) from the centre of the sphere. Calculate the minimum velocity v so that it R penetrates a distance inside the sphere. Neglect 2 any resistance other that electrostatic interaction. Assume that the charge on small mass remains constant through out the motion.

01_Physics for JEE Mains and Advanced - 2_Part 4.indd 159

A q, m

v

R/2 x

⇒

Qq 1 11 Qq = mv 2 + 2 4πε 0 x 32 πε 0 R

⇒

Qq ⎡ 11 1 ⎤ 1 mv 2 = 2 4πε 0 ⎢⎣ 8 R x ⎥⎦

⇒ v=

B

C

R

Qq ⎛ 11 1 ⎞ - ⎟ ⎜ 2πε 0 m ⎝ 8R x ⎠

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1.160  JEE Advanced Physics: Electrostatics and Current Electricity

Practice Exercises Single Correct Choice Type Questions This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

An insulating long light rod of length L pivoted at its centre O and balanced with a weight W at a distance x from the left end as shown in figure. Charges q and 2q are fixed to the ends of the rod. Exactly below each of these charges at a distance h a positive charge Q is fixed. Then x is x

q

2q +q m W h

Q

Q

QLq + ε 02 h 2 LW QLq + ε h 2 LW (A) (B) 20 2 ε0h W hW 2

2

4QLq + ε 0 h LW QLq + 4πε 0 h LW (C) (D) 8πε 0 h 2W 8π h 2W Three point charges q1 , q2 and q3 are taken such that when q1 and q2 are placed close together to form a single point charge, the force on q3 at distance L from this combination is a repulsion of 2 unit in magnitude. When q2 and q3 are so combined the force on q1 at distance L is an attractive force of magnitude 4 unit. Also q3 and q1 when combined exert an attractive force on q2 of magnitude 18 unit at same distance L. The algebraic ratio of charges q1 , q2 and q3 is



(A) 1 : 2 : 3 (C) 4 : –3 : 1

3.

In the process, when two bodies are charged by rubbing against the other, one becomes positively charged while the other becomes negatively charged. Then (A) mass of each body remains unchanged (B) mass of each body changes marginally (C) mass of each body changes slightly and hence the total mass (D) mass of each body changes slightly but the total mass remains the same

4.

a0

L O

h

2.

Electric field E exists between the vertical side walls of the elevator. The time taken by the block to come to the lowest point of inclined plane is (assuming the surface to be smooth)

(B) 2 : –3 : 4 (D) 4 : –3 : 2

A small block of mass m, charge +q is kept at the top of a smooth inclined plane of angle 30° placed in an elevator moving upward with an acceleration a0.

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 160

30°

(A) t= (B) t=

(C) t=2

(D) t=

5.

2h g 2h

( g - a0 ) +

qE m

2h

( g + a0 ) -

3 qE m

2h

( g + a0 )

2

⎛ qE ⎞ 2 h -⎜ ⎝ m ⎟⎠

An electron of mass m, initially at rest, moves through a certain distance in a uniform electric field in time t. A proton of mass M, also initially at rest, takes time T to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio T is nearly equal to t

2M m (A) (B) m M M M (C) (D) m m 6.

A proton of mass m and accelerated by a potential difference V gets into a uniform electric field of a

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Chapter 1: Electrostatics 1.161

parallel plate capacitor parallel to plates of length l at mid point of its separation between plates. The field strength in it varies with time as E = at , where a is a positive constant. Find the angle of deviation of the proton as it comes out of the capacitor. (Assume that it does not collide with any of the plates). -1 ⎛

m ⎞ ⎟ 8 eV ⎠

-1 ⎛

al 2 (B) θ = tan ⎜ ⎝ 2V

m ⎞ ⎟ eV ⎠

⎛ al 2 (C) θ = tan -1 ⎜ ⎝ 4V

m ⎞ ⎟ 2eV ⎠

⎛ al 2 (D) θ = tan -1 ⎜ ⎝ V

m ⎞ ⎟ 2eV ⎠

al 2 θ = tan ⎜ (A) ⎝ V

7.

a

(C)

O

8.

a

(B)

a

x

O

(D) a x

9.





∫ -E ⋅ dl,

(D) None of these

11. Two identical point charges are placed at a separation of l. P is a point on the line joining the charges, at a distance x from any one charge. The resultant field at P due to them is E. E is plotted against x for values of x from close to zero to slightly less than l. The curve that best represents the resulting plot of E for x values ranging from very close to zero to slightly less than l is (B) E

L

x

(C) E

O

O

L

x

(D) E

L

x

O

L

x

l =∞

(B) –1 (D) 0

In a regular polygon of n sides, each corner is at a distance a from the centre. Identical charges of

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 161

2σ eR (C) mε 0

O

l=0

(l = 0 being the centre of the ring) in volt is

σ eR σ eR (A) (B) 2mε 0 mε 0

x

O

where in space. The value of the line integral

√3R

E (A)

A non-conducting ring of radius 0.5 m carries a total on charge of 1.11 × 10 -10 C distributed non–uniformly  its circumference producing an electric field E every-

(A) +2 (C) –2

q q ⎛ n - 1⎞ ⎛ n ⎞ (C) (D) ⎜⎝ ⎟ ⎜⎝ ⎟ n - 1 ⎠ 4πε 0 a 2 n ⎠ 4πε 0 a 2

R

x

O

q q ( n - 1) (A) 2 (B) 4πε 0 a 4πε 0 a 2

10. An infinite dielectric sheet having charge density σ has a hole of radius R in it. An electron is released on the axis of the hole at a distance 3R from the centre. The speed with which it crosses the centre of the hole.

Consider two identical positive charges which are fixed on the y-axis, at ( 0, a ) and ( 0, - a ). Let a particle having a negative charge start for the origin O from a point P ( x, 0 ) at a large distance for O, move along the x-axis, passes through O and moves far away from O. Let the acceleration a be taken as positive along its direction of motion. The particle’s acceleration a is plotted against its x-coordinate. Then a vs x curve is given by

(A)

­magnitude q are placed at ( n - 1 ) corners. The field at the centre is

12. A charge +q is fixed at each of the points x = x0 , x = 3 x0 , x = 5x0 ,.... ad infinitum on the xaxis and a charge -q is fixed at each of points x = 2x0 , x = 4 x0 , x = 6 x0 .... ad infinitum. Here x0 is a positive constant. The potential at the origin due to the above system of charges is

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1.162  JEE Advanced Physics: Electrostatics and Current Electricity



(A) zero

(B)

q 8πε 0 x0 log e 2

q log e 2 (C) ∞ (D) 4πε 0 x0 13. A solid spherical region has a spherical having a diameter R (equal to the radius of the spherical region), has a total charge Q. The electric field and potential at a point P at x = 2R as shown is R

P

α a3 ε 0α a 3 (B) (A) ε0

(

)

1 (C) ε 0α a 2 6

(D) zero

16. An electrostatic potential is given by V = kxy , where k is a constant. A particle of charge q0 is first taken from ( 0, 0 ) to ( 0, a ) to ( a, a ) , then directly from ( 0, 0 ) to ( a, a ) and lastly from ( 0 , 0 ) to ( a, 0 ) to ( a, a ) . If W1, W2 and W3 be the work done for the individual paths respectively then (A) W1 = W2 = W3 = - q0 ka 2 W1 = W3 = - 2q0 ka 2 (B)

x – 2R

(C) W1 = W3 > W2

1 ⎛ ρR ⎞ 5 ⎛ ρR ⎞ , (A) ⎜ 27 ⎝ ε 0 ⎟⎠ 12 ⎜⎝ ε 0 ⎟⎠ 2

2 ⎛ ρR ⎞ 5 ⎛ ρR 2 ⎞ , (B) ⎜ 27 ⎝ ε 0 ⎟⎠ 36 ⎜⎝ ε 0 ⎟⎠ ⎛ ρR ⎞ 5 ⎛ ρR 2 ⎞ 3⎜ , (C) ⎝ ε 0 ⎟⎠ 6 ⎜⎝ ε 0 ⎟⎠

(D) W1 > W2 > W3 17. A solid sphere of radius R is charged uniformly. At a point P, a distance r from the surface, the electrostatic potential is observed to be half of the potential at the centre. Then 4R r = 2R (B) r= (A) 3

1 ⎛ ρR ⎞ 1 ⎛ ρR 2 ⎞ , (D) 9 ⎜⎝ ε 0 ⎟⎠ 9 ⎜⎝ ε 0 ⎟⎠

R (C) r = R (D) r= 3

14. An electron enters the region between the plates of a parallel-plate capacitor with same initial velocity at an angle θ to the plates. The plate width is l and the plate separation is d. The electron follows the path shown, just missing the upper plate. Neglect gravity.

18. A line charge of charge density λ lies along the x-axis and let the surface of zero potential passes through ( 0, 5, 12 ) m. The potential at point ( 1, 3, - 4 ) m is Z V=0 (0, 5, 12)

d

θ

Y l

(A) tan θ =

2d 4d (B) tan θ = l l

tan θ = (C)

6d 8d (D) tan θ = l l

15. In a region of space, the electric field isin the x-direction, proportional to x and given by E = α xi, where α is a positive constant. Consider an imaginary cubical volume of edge a, with its edges parallel to the axes of coordinates. The charge inside this cubical volume is

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 162

λ 2λ ⎛ 13 ⎞ ⎛ 13 ⎞ (A) log e ⎜ ⎟ (B) log e ⎜ ⎟ ⎝ 5 ⎠ ⎝ 3 ⎠ 2πε 0 πε 0 λ ⎛ 13 ⎞ (C) log e ⎜ ⎟ ⎝ 5 ⎠ 4πε 0

(D) None of these

19. A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R from the centre of the shell. The electric potential at the centre of shell will be

9/20/2019 10:34:11 AM

Chapter 1: Electrostatics 1.163 y

q q (A) (B) 3πε 0 R 8πε 0 R

e–

q 5q (C) (D) 6πε 0 R 24πε 0 R 20. In a certain charge distribution, all points having zero potential can be joined by a circle C. Points inside C have positive potential, and points outside C have negative potential. A negative charge, which is free to move, is placed inside C. Then the negative charge (A) may move, but will ultimately return to its starting point. (B) will remain in equilibrium. (C) can move inside C, but it cannot cross C. (D) must cross C at some time.



(A) zero

(B)

dε 0 mu2 dε mu (B) σ= 0 el el

(C) σ=

2dε 0 mu2 2dε 0 mu (D) σ= el el 2

q q 2q , and 3 3 3 placed at points A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60° .

24. Consider a system of three charges

y B C 60°

q 2 ( Q1 + Q2 ) q ( Q1 + Q2 ) ( 2 + 1 ) (C) (D) 4πε 0 a 4 2πε 0 a

Q ⎛1 1 ⎞ Q (A) (B) ⎜ + ⎟ 4πε 0 ⎝ r 2R ⎠ 8πε 0 R Q ⎛1 1 ⎞ Q (D) (C) ⎜ ⎟ 4πε 0 ⎝ r 2R ⎠ 4πε 0 r 23. An electron is projected from a distance d and with initial velocity u parallel to a uniformly charged flat conducting plate as shown. It strikes the plate after travelling a distance l along the direction of projection. The surface charge density σ of the conducting plate is

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 163

O

x

A



Conductor

2R

x

(A) σ=

4 2πε 0 a

+Q r O R

σ

l

2 - 1)

22. A point charge Q is placed at a distance r ( < R ) from the centre O of a uncharged spherical shell of inner radius R and outer radius 2R. The electric potential at the centre of the shell is given by

u

d

21. Two identical thin rings, each of radius a metre are coaxial and placed a metre apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q coulomb from the centre of one ring to that of the other is q ( Q1 - Q2 ) (

m



q (A) The electric field at point O is directed 2 8 πε 0R along the negative x-axis (B) The potential energy of the system is zero (C) The magnitude of the force between the charges q2 at C and B is 54πε 0 R2 q (D) The potential at point O is 12πε 0 R

25. Two particles A and B are suspended from the same support. The particles carry negative charges with A having more value of negative charge than B. They diverge and reach equilibrium with A and B making angles α and β with the vertical respectively. (A) α>β (B) α> r

R

q2 ⎛ 1 1 ⎞ 1 q2 (A) ⎜ + ⎟ (B) 4πε 0 ⎝ R r ⎠ 4πε 0 r q2 1 (C) 4πε 0 r ( R + r )

(D) None of these

33. Consider a metal sphere, of radius R that is cut in two along a plane whose minimum distance from the sphere’s centre is h. Sphere is uniformly charged by a total electric charge Q. The force necessary to hold the two parts of the sphere together is

R h m Initially

h m Finally

2 mgh (B) 3 mgh (A) mgh (D) 4 mgh (C)

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Chapter 1: Electrostatics 1.165

(A) F=

Q2 ( 2 R - h2 ) 4πε 0 R 4

(B) F=

Q2 4πε 0 R2

(C) F=

Q2 (R - h) 32πε 0 R3

(D) F=

Q2 ( 2 R - h2 ) 32πε 0 R 4

37. A hollow sphere of radius 2R is charged to a potential V and another smaller sphere of radius R is V charged to a potential . The smaller sphere is now 2 placed inside the bigger sphere without changing the net charge on each sphere. The potential difference between the two spheres is 3V V (A) (B) 2 4 V (C) (D) V 2

34. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to q2

+q1

–q1

(A) q2 (B) only the positive charges (C) all the charges +q1 and -q1 (D) 35. Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is z σ

–2σ

P

z = 3a z=0 x

–σ

z = –a

2σ ˆ 2σ (A) kˆ (B) k ε0 ε0

38. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V . If the shell is now given a charge of -3Q , the new potential difference between the same two surfaces is V (B) 2V (A) (C) 4V (D) -2V a 39. A disk of radius having a uniformly distributed 4 charge 6C is placed in the x -y plane with its centre at ⎛ a ⎞ ⎜⎝ - , 0 , 0 ⎟⎠ . A rod of length a carrying a uniformly dis2 a tributed charge 8C is placed on the x-axis from x = 4 5a to x = . Two point charges -7C and 3C are placed 4 a ⎛a ⎞ ⎛ 3a 3a ⎞ , 0 ⎟ , respectively. at ⎜ , - , 0 ⎟ and ⎜ - , ⎝4 ⎝ 4 4 ⎠ 4 ⎠ Consider a cubical surface formed by six surfaces a a a x = ± , y = ± , z = ± . The electric flux through 2 2 2 this cubical surface is y

x

4σ ˆ 4σ k (C) kˆ (D) ε0 ε0 36. The potential on the Nth shell due to N concentric shells having charges Q, 2Q, 3Q, …….. NQ and radii a, 2a, 3a ………… Na respectively is Q( N + 1) QN ( N + 1 ) (B) (A) 8πε 0 a 8πε 0 a Q (C) 2πε 0 a

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 165

(D) None of these

2C 2C (A) (B) ε0 ε0 10C 12C (C) (D) ε0 ε0 40. A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held

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1.166  JEE Advanced Physics: Electrostatics and Current Electricity together by pressing them with force F (see figure). F is proportional to

F

F

Q= (A)

4 4 πρ0 R5 (B) Q = πρ0 R 4 5 5

(C) Q=

2 2 πρ0 R3 (D) Q = πρ0 R6 3 3

1 1 2 σ R (A) σ 2 R2 (B) ε0 ε0

45. An oil drop is found floating freely between the plates of a parallel plate condenser, the plates being horizontal and the lower plate carrying a charge +Q. The area of each plate is A and the distance of separation between them is D. The charge on the oil drop must be (g is the acceleration due to gravity)

1 σ2 1 σ2 (D) (C) ε0 R ε 0 R2

Ag ε 0 MgA (A) (B) QM Q

41. A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of 81π strength × 10 5 Vm -1 . When the field is switched 7 off, the drop is observed to fall with terminal velocity

AgQ MgAQ (C) (D) ε0 D

2 × 10 -3 ms -1 . Given g = 9.8 ms -2, viscosity of the air = 1.8 × 10 -5 Nsm -2 and the density of oil = 900 kgm -3 , the magnitude of q is 1.6 × 10 -19 C (B) 3.2 × 10 -19 C (A) (C) 4.8 × 10 -19 C (D) 8 × 10 -19 C 42. A thin wire ring of radius r has an electric charge q0 . The increment of the force stretching the wire if a point q charge 0 is placed at the ring’s centre is 2 q2 q02 (A) 2 0 2 (B) 8π ε 0 r 16π 2ε 0 r 2 q2 2q02 (C) 02 2 (D) 4πε 0 r πε 02 r 2 43. Two point charges q1 = 4 μC and q2 = 9 μC are placed 20 cm apart. The electric field due to them will be zero on the line joining them at a distance of (A) 8 cm from q1 (B) 8 cm from q2 80 80 cm from q2 (C) cm from q1 (D) 13 13 44. A spherical region of space has a distribution of charge such that the volume charge density varies with the radial distance from the centre r as ρ = ρ0 r 3 , 0 ≤ r ≤ R where ρ0 is a positive constant. The total charge Q on sphere is

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 166

46. Two similar point charges q1 and q2 are placed at a distance r apart in air. Assume that a slab of thickness one third the separation between the charges is placed between the charges and it is observed that the Coulomb’s repulsive force is reduced in the ratio 25 : 9 . Then the dielectric constant K of such a slab is (A) 1 (B) 4 (C) 9 (D) 25 47. A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by 2 π R2E (B) π R 2E (A)

(

)

(C) π R2 + π L2 E

(D) ZERO

48. Two concentric, thin metallic spheres of radii R1 and R2 ( R1 > R2 ) bear charges Q1 and Q2 respectively. Then the potential at distance r between R1 and R2 1 ⎞ ⎛ will be ⎜ k = ⎟ πε 4 ⎝ 0⎠ ⎛ Q + Q2 ⎞ ⎛Q Q ⎞ (A) k⎜ 1 k⎜ 1 + 2 ⎟ ⎟⎠ (B) ⎝ r R2 ⎠ ⎝ r Q ⎞ ⎛Q ⎛Q Q ⎞ k ⎜ 2 + 1 ⎟ (D) k⎜ 1 + 2 ⎟ (C) R1 ⎠ ⎝ r ⎝ R1 R2 ⎠ 49. A wire of length L is placed along x-axis with one end at the origin. The linear charge density of the wire var⎛ x3 ⎞ ies with distance x from the origin as λ = λ 0 ⎜ ⎟ , ⎝ L⎠ where λ 0 is a positive constant. The total charge Q on the rod is

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Chapter 1: Electrostatics 1.167

(A) Q=

λ 0 L2 λ L3 (B) Q= 0 3 3

(C) Q=

λ 0 L3 λ L4 (D) Q= 0 4 5

50. A flat circular disc has a charge +Q uniformly distributed on the disc. A charge +q is thrown with kinetic energy E, towards the disc along its normal axis. The charge q will (A) hit the disc at the centre (B) return back along its path after touching the disc (C) return back along its path without touching the disc (D)  any of the above three situations is possible depending on the magnitude of E 51. The electric intensity due to a uniformly charged infinite cylinder of radius R, at a distance r ( > R ) , from its axis is proportional to 2

3

(A) r (B) r 1 1 (C) (D) r r2 52. Two identical balls each having a density ρ are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle θ with vertical. Now, both the balls are immersed in a liquid. As a result of immersion in the liquid the angle θ does not change. The density of the liquid is σ . The dielectric constant of the liquid is

σ ρ (A) (B) ρ σ σ ρ (C) (D) σ -ρ ρ -σ 53. The radius of a hollow metallic sphere is R. If the potential difference between its surface and a point at a distance of 3R from its centre is V , then the electric field intensity at a distance of 3R from its centre is V V (A) (B) 2R 3R V V (C) (D) 6R 4R 54. The electric field due to a point charge at a distance R from it is E. If the same charge is placed on a metallic sphere of radius R, the electric field on the surface of the sphere will be E (A) ZERO (B) 2 (C) E (D) 2E

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 167

55. A positive point charge 50 μC is located in the plane  xy at the point with radius vector r0 = 2iˆ + 3 ˆj , where iˆ and ˆj are the unit  vectors of the x and y axes. The electrostatic force F and its magnitude on a charge of  2 μC placed at the point with radius vector r = 8iˆ - 5 ˆj   is (Here r0 and r are expressed in metre) (A) 3 N (B) 9 mN (C) 9 μN (D) 3 mN 56. Two free charges q and 4q are placed at a distance d apart. A third charge Q is placed between them at a distance x from charge q such that the system is in equilibrium. Then 4q 4q d d , x = (B) , x= Q= Q= (A) 9 9 4 3 4q 4q d d (C) Q = - , x = (D) Q=- , x= 9 3 3 4 57. A ring of radius 0.1 m is made out of a thin metallic wire of area of cross-section 10 -6 m 2 . The ring has a uniform charge of π coulomb. The change in the radius of the ring when a charge of 10 -8 coulomb is placed at the centre of the ring is (Young’s modulus of the metal is 2 × 1011 Nm -2 ) (A) ΔR = 2.25 cm (B) ΔR = 2.25 mm (C) ΔR = 22.5 cm (D) ΔR = 22.5 mm 58. Two isolated, charged conducting spheres of radii R1 and R2 produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is R1 R2 (B) (A) R2 R1 R2 R22 (C) 12 (D) R2 R12 59. Two identical metallic blocks resting on a frictionless horizontal surface are connected by a light metallic spring having a spring constant k and an unstretched length L0 . A total charge Q is slowly placed on the system, causing the spring to stretch to an equilibrium 3L length L = 0 , as shown. The value of Q, assuming 2 that all the charge resides on the blocks and assuming the blocks as point charges is

m

L0 k

m

Initially

m

L k

m

Finally

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1.168  JEE Advanced Physics: Electrostatics and Current Electricity

(A) Q = 3 L0 2πε 0 L0 (B) Q = 2L0 3πε 0 L0 (C) Q = 3 L0 3πε 0 L0 (D) Q = 2L0 2πε 0 L0 60. A charged particle of mass m and charge q is released from rest in a uniform electric field E. The kinetic energy of the particle after time t is Eq2 m 2E2t 2 (A) (B) mq 2t 2 E 2 q 2t 2 Eqm (C) (D) 2m 2t 61. Two identical oppositely charged metallic spheres placed 0.5 m apart, attract each other with a force of 0.108 N. When connected to each other by a copper wire for a short while, they begin to repel each other with a force of 0.036 N. The initial charge on each one of them is +1 μC, -3 μC (B) -1 μC, +3 μC (A) (C) -3 μC , +5 μC (D) 5 μC , -

3 μC 5

62. In the electric field of a point charge q, a certain charge is carried from point A to B, C, D and E. Then the work done A

q

B C



E D

63. Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, non-conducting walls, the beads move, and at equilibrium they are a distance R apart (shown in figure). The charge on each bead is R m R

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 168

4πε 0 mg (B) q = 2R 3

(C) q = 2R

πε 0 mg 3

3πε 0 mg

(D) None of these

64. Charges +q, -q, +q and -q are placed at the corners A, B, C and D respectively of a square of side a. The 1 potential energy of the system is times 4πε 0 A(+q)

B(–q)

a

a

a

D(–q)

a

C(+q)

q2 q2 ( -4 + 2 ) (A) ( -4 + 2 ) (B) a 2a 4 q2 4 2q 2 (C) (D) a a 65. Two particles each of charge 10 -7 C and mass 5 g, stay in limiting equilibrium on a horizontal surface. The particles have a separation of 10 cm between them. Assume the coefficient of friction between each particle and the table to be μ . Then μ is (A) 1.8 (B) 0.018 (C) 0.18 (D) 1.08 66. Three charges, each of +4 μC, are placed at the corners B, C, D of a square ABCD of side 1 m. The electric field at the center O of the square is A

(A) is least along the path AB (B) is least along the path AD (C) is zero along any one of the path AB, AC, AD and AE (D) is least along AE

R m

(A) q = 2R

1m

B O

D

C

(A) 7.2 × 10 4 N towards A (B) 7.2 × 10 4 N towards C (C) 3.6 × 10 4 N towards A (D) 3.6 × 10 4 N towards C 67. The magnitude of the electric field E in the annular region of a charged cylindrical capacitor (A) is same throughout (B) is higher near the outer cylinder than near the inner cylinder

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Chapter 1: Electrostatics 1.169



(C) varies as

1 , where r is the distance from the axis r

1 (D) varies as 2 , where r is the distance from the r axis

68. Four equal positive charges, each of charge Q are arranged at the corners of a square of side . A unit positive charge of mass m is placed at point P at height h above the centre of the square. Calculate Q, so that the unit charge is in equilibrium. 3

πε mg ⎛ 2 h 2 ⎞ 2 Q= 0 (A) ⎜ l + ⎟⎠ l ⎝ 2

72. A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is q q (A) (B) ε0 6 L2ε 0 6 qL2 (C) ε0

(D) ZERO

73. Two small conductors A and B are given charges q1 and q2 respectively. Now they are placed inside a hollow metallic conductor C carrying a charge Q. If all the three conductors A, B and C are connected by a conducting wire as shown, the charges on A, B and C will be respectively

3

Q

πε mg ⎛ 2 l 2 ⎞ 2 (B) Q= 0 ⎜ h + ⎟⎠ 2 h ⎝

A q1

3

πε mg ⎛ 2 l 2 ⎞ 2 Q= 0 (C) ⎜ h + ⎟⎠ 4h ⎝ 2

(D) None of these

69. A metallic sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in the figure as



1 2

1 2

3

3

4

4

(A) 1 (C) 3

(B) 2 (D) 4

C q2 B

q 1 + q2 q1 + q2 (A) , , Q 2 2 Q + q + q Q + q1 + q2 Q + q1 + q2 (B) 1 3 , , 3 3 3 q1 + q2 + Q q1 + q2 + Q (C) , , 0 2 2 (D) 0, 0, Q + q1 + q2

70. Three charge q, Q and 4q are placed in a straight line l and l respectively of length l at points distant 0, 2 from one end. In order to make the net force on q zero, the charge Q must be equal to (A) -q (B) -2q

74. Two small equally charged spheres, each of mass m, are suspended from the same point by light silk threads of length l. The separation between the dq spheres is x  l . The rate with which the charge dt leaks off each sphere, if their velocity of approach var6 ies as v = , (where α is a positive constant) is x

-q (C) (D) q 2

2πε 0 mg 3 2πε 0 mg (B) (A) 4 l l 2

71. Three identical spheres each having a charge 2q and radius R are kept such that each touches the other two. Find the magnitude of the electric force on any sphere due to the other two.

9 (C)

3 q2 3 q2 (A) 2 (B) 4πε 0 R 2πε 0 R2 3 q2 3 q2 (C) (D) 8πε 0 R2 4πε 0 R2

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 169

2πε 0 mg 2πε 0 mg (D) 9 3l l

75. The  electric field in a region of space is given by E = 5iˆ + 2 ˆj NC -1 . The electric flux due to this field through an area 2 m 2 lying in the YZ plane, in S.I. units, is 10 (B) 20 (A) (C) 10 2 (D) 2 29

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1.170  JEE Advanced Physics: Electrostatics and Current Electricity 76. Consider an annular thin disc of inner radius a and outer radius b ( > a ). The surface charge density σ varies with the distance r from the centre of the disc σ as σ = 20 where a < r < b and σ 0 is a positive conr stant. The total charge on the disc is q , then Q equals

σ 2σ (A) (B) ρ0 ρ0

⎛ b⎞ ⎛ b⎞ σ 0 log e ⎜ ⎟ (B) 2σ 0 log e ⎜ ⎟ (A) ⎝ a⎠ ⎝ a⎠

81. An electric charge is placed at the centre of a cube of side a. The electric flux through one of its faces will be

⎛ a⎞ ⎛ b⎞ 2πσ 0 log e ⎜ ⎟ (D) 2πσ 0 log e ⎜ ⎟ (C) ⎝ b⎠ ⎝ a⎠ 77. If Ea be the electric field intensity due to a short dipole at a point on the axis and Er be that on the right bisector at the same distance from the dipole, then Ea = Er (B) Ea = 2Er (A) (C) Er = 2Ea (D) Ea = 2Er 78. Two identical conducting spheres carrying different charges attract each other with a force F when placed in air medium at a distance d apart. The spheres are brought into contact and then taken to their original positions. Now the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force. The ratio between initial charges on the spheres is - ( 3 + 8 ) only (A)

79. Equipotential surfaces are shown in figure. Then the electric field strength will be y (cm) O

10

30° 20

20 V

30 V

q q (C) 2 (D) ε 4πε 0 a 0 82. Point charge q moves from point P to point S along the path PQRS (figure shown) in a uniform electric field E pointing coparallel to the positive direction of the X-axis. The coordinates of the points P, Q, R and S are ( a, b , 0 ) , ( 2 a, 0 , 0 ) , ( a, - b , 0 ) and ( 0 , 0 , 0 ) respectively. The work done by the field in the above process is given by the expression Y

S

P Q R

x E

(A) qEa (B) -qEa 2

30

x (cm)

(A) 100 Vm -1 along X-axis 100 Vm -1 along Y-axis (B) 200 Vm -1 at an angle 120° with X-axis (C) 50 Vm -1 at an angle 120° with X-axis (D) 80. A solid sphere of radius R1 and volume charge denρ sity ρ = 0 is enclosed by a hollow sphere of radius r R2 with negative surface charge density σ , such that the total charge in the system is zero. ρ0 is a positive constant and r is the distance from the centre of the R spheres. The ratio 2 is R1

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 170

q q (A) (B) 6ε 0 ε 0 a2

-3 + 8 only

(B)

(C) - ( 3 + 8 ) or ( -3 + 8 ) (D) + 3

10 V

ρ ρ0 (C) 0 (D) 2σ σ

(C) qEa 2 (D) qE ( 2 a ) + b 2 83. A positively charged disc is placed on a horizontal plane. A charged particle is released from a certain height on its axis. The particle just reaches the center of the disc. Select the correct alternative. (A) Particle has negative charge on it. (B)  Total potential energy (gravitational + electrostatic) of the particle first increases then decreases. (C)  Total potential energy of the particle first decreases then increases. (D) Total potential energy of the particle continuously decreases. 84. A simple pendulum has a length l, mass of bob m. The bob is given a charge q . The pendulum is suspended between the vertical plates of charged parallel plate capacitor. If E is the field strength between the plates, then time period T equals

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Chapter 1: Electrostatics 1.171

X

(A) 2π

(C) 2π

89. A charge q is distributed uniformly over the volume of a insulating solid sphere of radius R. It is enclosed by an earthed conducting spherical shell of inner radius R1 and outer radius R2 . The charge on the outer surface of the shell will be

l l 2π (B) qE g g+ m l qE gm

2π (D)

l

⎛ qE ⎞ g +⎜ ⎝ m ⎟⎠ 2

2

85. Two positive point charges are 3 m apart and their combined charge is 20 μC . If the force between them is 0.075 N, the charges are (A) 10 μC, 10 μC (B) 15 μC, 5 μC (C) 12 μC , 8 μC (D) 14 μC , 6 μC 86. Four charges of 6 μC , 2 μC , -12 μC and 4 μC are placed at the corners of a square of side 1 m. The square is in x -y plane and its center at its origin. Electric potential due to these charges is zero everywhere on the line x = y , z = 0 (B) x=0=z (A) (C) x = 0 = y (D) x = z, y = 0 87. Two vertical metallic plates carrying equal and opposite charges are kept parallel to each other like a parallel plate capacitor. A small spherical metallic ball is suspended by a long insulated thread such that is hangs freely in the centre of the two metallic plates. The ball, which is uncharged, is taken slowly towards the positively charged plate and is made to touch that plate. Then the ball will (A) stick to the positively charged plate (B) come back to its original position and will remain there (C) oscillate between the two plates touching each plate in turn (D) oscillate between the two plates without touch them 88. As the electric charge on the surface of a hollow metal sphere increases, the electric field intensity inside the sphere

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 171

(A) decreases (B) increases (C) remains the same (D) may increase or decrease depending on the radius of the sphere

R1

R

R2

R2 (A) +q (B) q R1 R (C) q (D) zero R2 90. A hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. The potential at the centre of the sphere is (A) ZERO (B) 10 V (C) same as at a point 5 cm away from the surface (D) same as at a point 25 cm away from the surface 91. A rod of length L has a total charge q distributed uniformly along its length. It is bent in the shape of a semicircle. The electric potential at the center of the semicircle is q 1 q (B) (A) 4πε o L 4ε o L q 1 q (C) (D) 2ε o L πε o L 92. Two charged conducting spheres of radii R1 and R2 , separated by a large distance, are connected by a long wire. The ratio of the charges on them is R1 R2 (B) (A) R2 R1 R2 R22 (C) 12 (D) R2 R12 93. In PROBLEM 92, the ratio of the electric fields on the surfaces of the two spheres is

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1.172  JEE Advanced Physics: Electrostatics and Current Electricity

R1 R2 (A) (B) R2 R1 R2 R22 (C) 12 (D) R2 R12 94. A spherical volume contains a uniformly distributed charge of density ρ . The electric field inside the sphere at a distance r from the center is

ρr ρr (A) (B) 3ε o 4πε o







(A) points where the electric potential due to the two charges is zero, lie on a circle of radius 4a and center ( 5 a, 0 ) (B) potential is zero at x = a and x = 9 a (C) If a particle of charge +q is released from the center of the circle obtained in part (A) it will eventually cross the circle (D) All of the above

98. In the electric field of a point charge q shown, a charge is carried from A to B and from A to C. Compare the work done B

ρ ρr (C) (D) ε or 4πε o 95. Which one of the following graphs represents the variation of electric field strength E with distance r from the centre of a uniformly charged non-conducting sphere E (B)

E (A)

R

r

E (C)

R





r

99. A particle of mass m and charge -q is projected from the origin with a horizontal speed v into an electric field of intensity E directed downward. Select the incorrect statement. y

96. Four charges +q, +q, -q and -q are placed respectively at the corners A, B, C and D of a square of side a. The potential and field at the centre O of the 1 square are respectively times 4πε 0 A(+q)

a

D(–q)



4q (A) ZERO, 2 a

x



C(–q)



4 2q (B) ZERO, a2

4 2q 4 q 4 2q 4 2q (C) , 2 (D) , a a a a2 97. Two fixed charges -2Q and +Q are located at points ( -3 a, 0 ) and ( +3 a, 0 ) respectively. Then

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 172

E

B(+q) O

a

(A)  work done is greater along the path AC than along AB (B) work done is the same in both the cases (C)  work done is greater along the path AB than along AC (D) work done is zero in both the cases

r

R

r

A

C



E (D)

R

q



(A) The kinetic energy after a displacement y is qEy (B) The horizontal and vertical components of accelqE eration are ax = 0, ay = m 1 ⎛ qEx 2 ⎞ (C) The equation of trajectory is y = ⎜ 2 ⎝ mv 2 ⎟⎠ (D) The horizontal and vertical displacements x and 1 y after a time t are x = vt and y = ay t 2 2

100. A positively charged thin metal ring of radius R is fixed in the X - Y plane with its centre at the origin O. A negatively charged particle P is released from rest at point ( 0 , 0 , z0 ) where z0 > 0 . Then the motion of P is

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Chapter 1: Electrostatics 1.173

(A) simple harmonic for all values of z0 satisfying 0 < z0 < ∞



(B) simple harmonic for all values of z0 satisfying 0 < z0 ≤ R



(C)  approximately z0 a ) . The electric field at a distance R ( a < R < b ) from the centre is Q 3Q (A) (B) 2πε 0 R 2πε 0 R 3Q 4Q (C) 2 (D) 4πε 0 R 4πε 0 R2 103. Two infinitely long parallel wires having linear charge densities λ1 and λ 2 respectively are placed at a distance R metre. The force per length on either 1 ⎞ ⎛ wire will be ⎜ K = 4πε 0 ⎟⎠ ⎝

105. The potential energy of an electric dipole in a uniform electric field is U. The magnitude of the torque acting on the dipole due to the field is N . Then (A)  U is minimum and N is zero when the dipole is parallel to the field. U is zero and N is zero when the dipole is per (B)  pendicular to the field. U is minimum and N is maximum when the (C)  dipole is perpendicular to the field U is minimum and N is zero when the dipole (D)  is anti-parallel to the field 106. Five point charges ( +q each) are placed at the five vertices of a regular hexagon of side 2a. What is the magnitude of the net electric field at the centre of the hexagon? q 1 q (A) 2 (B) 4πε o a 16πε o a 2 2q 5q (C) 2 (D) 4πε o a 16πε o a 2 107. A charge Q is divided into two parts and the two parts are separated by a certain distance. The force between them will be maximum if one of the charges is Q Q (B) (A) 2 3

⎛ 2λ λ ⎞ ⎛ 2λ λ ⎞ (A) K ⎜ 12 2 ⎟ (B) K⎜ 1 2 ⎟ ⎝ R ⎠ ⎝ R ⎠

Q (C) 4

⎛λλ ⎞ ⎛λλ ⎞ K ⎜ 1 2 2 ⎟ (D) K⎜ 1 2 ⎟ (C) ⎝ R ⎠ ⎝ R ⎠

108. The variation of electric field between the two charges q1 and q2 along the line joining the charges is plotted against distance from q1 (taking rightward

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 173

(D) None of these

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1.174  JEE Advanced Physics: Electrostatics and Current Electricity direction of electric field as positive) as shown in the figure. Then the correct statement is

(A) E

E (B)

E

q1

q2

r

x

q1 and q2 are positive and q1 < q2 (A) q1 and q2 are positive and q1 > q2 (B) (C) q1 is positive and q2 is negative and q1 < q2 (D) q1 and q2 are negative and q1 < q2 109. A and B are two spherical conductors of same extent and size. A is solid and B is hollow. Both are charged to the same potential. If the charges on A and B are QA and QB respectively, then (A) QA is less then QB (B) QA is greater than QB but not double (C) QA = QB (D) QA = 2QB 110. Three point charges q, 2q and 8q are to be placed on a straight line 9 cm long. The system possesses minimum potential energy when 2q and q lie at ends with 8q at 3 cm from 2q. (A) (B) 2q and 8q lie at ends with q at 6 cm from 8q. (C) q and 8q lie at ends with 2q at 6 cm from q. (D) 2q and 8q lie at ends with q at 3 cm from 8q. 111. Two large parallel planes charged uniformly with surface charge density σ and -σ are located as shown in the figure. Which one of the following graphs shows the variation of electric field along a line perpendicular to the planes as one moves from A to B ? σ

A

x E (D)

E (C)

x

x

112. An infinite number of charges, each equal to q coulomb, are placed along the x-axis at x (in metres) = 1, 2, 4, 8,....... and so on. The potential and field in SI units at x = 0 due to this set of charges are respec1 times tively 4πε 0 2q , 4q 2q, 4q (B) (A) 3 2q 4 q 4q (C) , (D) 2q, 3 3 3 113. In PROBLEM 112, if the consecutive charges have opposite signs, then the potential and field at x = 0 1 are respectively times 4πε 0 2q 4 q 2q 4 q , (B) , (A) 3 7 5 5 2q 4 q 2q 4 q , (D) , (C) 3 5 5 7 114. Two conducting plates 1 and 2 each having large surface area A (on one side) are placed parallel to each other. The plate A is given a charge q, while the plate 2 is neutral. Then the electric field at a point in between the plates is 1

2

–σ

B

q q (A) (B) 2 Aε 0 Aε 0 2q 3 q (C) (D) Aε 0 2 Aε 0

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Chapter 1: Electrostatics 1.175 115. Two concentric metallic spherical shells are given positive charges of different value. Then (A) the outer sphere is always at a higher potential (B) the inner sphere is always at a higher potential (C) both the spheres are at the same potential (D) no prediction can be made about their potentials unless the actual values of charges and radii are known 116. Two thin infinite parallel plates have uniform charge densities +σ and -σ . The electric field in the space between them is

σ σ (A) (B) 2ε 0 ε0 2σ (C) ε0

120. Equal charges q are placed at the four corners A, B, C, D of a square of side a. The magnitude of the force on the charge at C will be 3 q2 4 q2 (A) 2 (B) 4πε 0 a 4πε 0 a 2 ⎛ 1 + 2 2 ⎞ q2 1 ⎞ q2 ⎛ (C) (D) ⎜⎝ ⎟⎠ ⎜⎝ 2 + ⎟ 2 2 2 ⎠ 4πε 0 a 2 4πε 0 a 121. A uniform electric field of 400 Vm -1 is directed at 45° above the x-axis as shown in the figure. The potential difference VA - VB is given by

(D) ZERO

y(cm)

117. A charge q is placed at O in the cavity in a spherical uncharged conductor. Point S lies outside the conductor. If the charge is displaced from O towards S still remaining within the cavity, then

(0, 2) A

45°

(3, 0) B x(cm)

q O

S

(A) 0 (B) 4V (C) 6.4 V (D) 2.8 V



(A) electric field at S will increase (B) electric field at S will decrease (C) electric field at S will first increase and then decrease (D) electric field at S will not change

118. The electric potential (in volt) in a region is given by V = 6 x - 8 xy 2 - 8 y + 6 yz - 4 x 2 Then electric force acting on a point charge of 2 C placed at the origin will be (A) 2 N (B) 6 N (C) 8 N (D) 20 N 119. The electric field at a point A due to dipole p is perpendicular to p. The angle θ is A p

122. A charge Q is distributed over two concentric hollow spheres of radii r and R ( R > r ) such that the surface densities are equal. The potential at the com1 mon centre is times 4πε 0 Q⎛ r+R ⎞ ⎛ r+R ⎞ Q⎜ 2 (B) (A) ⎜ ⎟ ⎝ r + R2 ⎟⎠ 2 ⎝ r 2 + R2 ⎠ ⎛ r+R ⎞ 2Q ⎜ 2 (C) ⎝ r + R2 ⎟⎠

(D) zero

123. An electric dipole is placed at the origin along the x-axis. The electric field at any point, whose position vector makes an angle θ with the x-axis, makes an angle α (B) θ (A) (C) θ + α (D) θ + 2α

θ

(A) 0° (B) 90° (C) tan -1 ( 2 ) (D) tan -1 ( 2 )

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 175

1 tan θ . 2 1 24. Initially the spheres A and B are at potentials VA and VB respectively. Now sphere B is earthed by closing the switch. The potential of A becomes



with the x-axis, where tan α =

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1.176  JEE Advanced Physics: Electrostatics and Current Electricity

A

B

1 ⎛ 4π A ⎞ 1 ⎛ A⎞ (A) ⎜ (B) 4πε 0 ⎝ r1 ⎠⎟ 4πε 0 ⎝⎜ r1 ⎟⎠ A A (C) (D) πε 0 2ε 0

VA (A) 0 (B)

130. An electric dipole is placed at the origin and is directed along the x-axis. At a point P , far away from the dipole, the electric field is parallel to the y-axis. OP makes an angle θ with the x-axis, then

VA - VB (D) VB (C)

tan θ = 3 (B) tan θ = 2 (A)

125. A charge Q is placed at each of the two opposite corners of a square. A charge q is placed at each of the other two corners. If the resultant force on Q is zero, then

⎛ 1 ⎞ (C) θ = 45° (D) tan θ = ⎜ ⎝ 2 ⎟⎠

Q = 2q (B) Q = - 2q (A) Q = 2 2q (D) Q = -2 2q (C) 126. Electric charges q, q and -2q are placed at the three corners of an equilateral triangle of side l. The magnitude of the electric dipole moment of the system is ql (B) 2ql (A) 3ql (D) 4ql (C) 127. Which one of the following graphs represents, variation of the electric field strength E with distance r from the centre of a charged conducting sphere?

r

E (C)

R

132. An uncharged sphere of metal is placed in uniform electric field produced by two oppositely charged plates. The lines of force will appear as (A)

(B)

(C)

(D)

(B) E

(A) E

R

131. Two identical balls having like charges and placed at certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases 4.5 times in comparison with the initial value. The ratio of the initial charges of the balls is (A) 2 (B) 3 (C) 4 (D) 6

R

r

E (D)

r

R

r

128. A charged particle of mass m and charge q is released from rest from position ( x0 , 0 ) in a uniform electric field E0 ˆj . The angular momentum of the particle about origin (A) is zero (B) is constant (C) increases with time (D) decreases with time 129. An insulating solid sphere of radius R is charged in a non-uniform manner such that volume charge A density ρ = , where A is a positive constant and r r the distance from centre. Electric field strength at any inside point at distance r1 is

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 176

133. Two small identical spheres having charges +10 μC and -90 μC attract each other with a force of F newton. If they are kept in contact and then separated by the same distance, the new force between them is F (B) 16F (A) 6 16 F (C) (D) 9F 9 134. The magnitude of electric field at the centre O of regular hexagon produced by the system of charges will be

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Chapter 1: Electrostatics 1.177

l

–4q

A

+6q l

+3q F

B l

l

O

+4q C l

y

–3q E l

+q

+5q D

q 4πε 0 2

+q O

(–a, 0)

q 2q (A) 2 (B) 4πε 0 4πε 0 2

(2 + 2 ) (C)

139. Two identical point charges q are placed at x = - a and x = a, as shown in figure. The graph showing the variation of E along the x-axis is (assuming positive E to be taken along the +x-axis )

(a, 0)

x

E

(A)

(D) 0 x = –a

135. Two equal and opposite charges are placed at a certain distance, the force between them is F. If 25% of one charge is transferred to other, then the force between them is 9F F (B) (A) 16

O

X

x = +a

X

E

(B)

15 F 4F (C) (D) 16 5

x = +a

x = –a

O

136. The electric field inside a sphere which carries a charge density proportional to the distance from the origin ρ = α r, where ( α is a constant) is

αr3 αr2 (A) (B) 4 ε0 4 ε0 αr2 (C) 3 ε0

(D) None of these x = –a

137. A regular polygon has 20 sides. Equal charges, each Q, are placed at 19 vertices of the polygon and a charge q is placed at the centre of polygon. If the distance of each vertex from the centre O is a, net force experienced by q is 1 20Qq 1 Qq (A) (B) 4πε 0 a 2 4πε 0 a 2 1 19Qq (C) 4πε 0 a 2

E

(C)

O

x = +a

X

E

(D)

x = –a

O

x = +a

X

(D) zero

138. Two particles of masses m and 2 m and charges 2q and 2q are placed in a uniform electric field E and allowed to move for the same time. The ratio of kinetic energies will be

140. The number of electrons that should be removed from a coin of mass 1.6 g, so that it may float in electric field intensity 109 NC -1 directed upwards is

2 : 1 (B) 8 :1 (A)

(A) 9.8 × 107 (B) 9.8 × 10 5

(C) 4 : 1 (D) 1: 4

(C) 9.8 × 10 3 (D) 9.8 × 101

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 177

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1.178  JEE Advanced Physics: Electrostatics and Current Electricity 141. Two identical metallic spheres A and B carry charges +Q and -2Q respectively. The force between them is F Newton, when they are separated by a distance d in air. The spheres are allowed to touch each other and are moved back to their initial positions. The force between them now is F F (B) (A) 2 4 F (C) 8

(D) zero

142. Initially the spring is in unstreched condition. An electric field ( E ) is switched on in right direction. The maximum extension in the spring will be q

K

m

2qE 2qE (A) (B) K K qE qE (C) (D) K K 143. Three wires AB, BC , CD of equal length l are charged uniformly with linear charged density λ and are placed as shown. P is a point which lies at a distance l from the wire BC on its perpendicular bisector. The electric field at P is B

C

qq qq0 qq0 (A) 0 2 × 2 (B) + 2 4πε 0 a 4πε 0 2 a 8πε 0 a 2 qq ⎛ qq ⎛ 1 1⎞ ⎞ (C) 0 2 ⎜ 2 - ⎟ (D) 0 2 ⎜ - 2 ⎟ ⎠ 2⎠ 4πε 0 a ⎝ 4πε 0 a ⎝ 2 145. A neutral sphere of radius r and density ρ is intended to be suspended with an electric field E existing on the earth’s surface directed vertically up. What fraction of the electrons should be removed from the sphere. Given that, the atomic number and atomic mass of the material of the sphere are Z & A respectively. Avogadro number is N A . Assume that the sphere is able to hold the necessary charge without any leakage 2g 3gA (B) (A) eEZAN A eEZ gA 2gA (C) (D) eEZN A eEZN A 146. A wheel having mass m has charges +q and -q fixed at diametrically opposite ends. For the wheel to be in equilibrium on a rough inclined plane in the presence of uniform vertical electric field E, the magnitude of E is E

+q –q 30°

3 mg mg (A) (B) 2q 2q

P

A

3 mg 3 mg (C) (D) 2q 4q

D

λ 2λ (A) ( 2 5 - 1 ) (B) ( 5 - 1 ) 2 5ε 0 l 2 5ε 0 l λ λ (C) ( 2 5 - 3 ) (D) 4 5ε 0l 4 5ε 0l 144. The net force acting on the charge q0 present at the origin is (0, a)q

–q(a, a)

45° q0

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 178

(a, 0) q

147. Seven identical charge particles each having charge q are placed on the vertices of a cube of edge , then the magnitude of electric field at centre of cube is q q (A) 2 (B) 4πε 0 3πε 0 2 q q (C) 2 (D) 8πε 0 6πε 0 2 148. A simple pendulum of mass m and charge q is hanging from roof of a car moving on a circular track of radius r with constant speed v and string makes angle θ with vertical. Magnitude of horizontal electric field required so that string makes same angle θ on other side of vertical is

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Chapter 1: Electrostatics 1.179

4 mv 2 3 mv 2 (A) (B) qr qr 2mv 2 mv 2 (C) (D) qr qr 149. Two identical charges repel each other with a force equal to 10 mg wt, when they are 0.6 m apart in air. The value of each charge is g = 10 ms -2

(

)

2 mC (B) 2 × 10 -7 C (A) (C) 2 nC (D) 2 μC 150. A cavity of radius r is made inside a solid sphere. The volume charge density of the remaining sphere is ρ . An electron (charge e, mass m ) is released from rest inside the cavity from point P as shown in figure. The centre of sphere and centre of cavity are separated by a distance a. The time after which the electron again touches the sphere is

ω

q, m

l

q, m

l

q2 q2 (A) 2 - mω 2 (B) 2 + mω 2 4πε 0 4πε 0 q2 q2 (C) 2 - mω 2 (D) 2 + mω 2 16πε 0 16πε 0 153. A short dipole is placed along x-axis with centre at P, is at a origin. The electric field at a point  which  distance r from origin such that OP = r makes an angle of 45° with x-axis, is directed along a direction making (A) tan -1 ( 0.5 ) with x-axis

P 45° a

π (B) + tan -1 ( 0.5 ) with x-axis 4 π (C) + tan -1 ( 0.5 ) with y-axis 4 (D) tan -1 ( 0.5 ) with y-axis

6 2 rε 0 m 2r ε 0 m (A) (B) eρ a 6 ρa 6 rε m rε 0 m (C) 0 (D) eρ a eρ a 151. A charged soap bubble having surface charge density σ and radius R. If pressure inside soap bubble and pressure outside it is same then the surface tension for soap bubble is

σ 2R σ 2R (B) T= T= (A) 8ε 0 4ε 0 T= (C)

σ 2R σ 2R (D) T= 2ε 0 ε0

152. Two identical point charges each q of mass m are connected by a insulating string of length 2 and the system are rotated about an axis passing through mid point of string and perpendicular to its length as shown. Then neglecting gravity, tension in the string is (Assume only coulombic force between the charges)

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 179

154. A particle of mass m and charge Q is attached to a string of length L. It is whirled into a vertical circle in field E. The speed of the particle at point A, so that the tension in the string at A is ten times the weight of particle, is ( g = acceleration due to gravity)

E L A Q, m

V

QEL (A) 5gL (B) m QE ⎞ ⎛ L ⎜ 9g + (C) ⎟ ⎝ m ⎠

(D) None of these

155. In an experiment to demonstrate Coulomb’s law in electrostatics, the force F between two small charged spheres is measured for various distances r between their centres. A graph is plotted of log e F ( y-axis ) against log e r ( x-axis ). The slope of this graph is

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1.180  JEE Advanced Physics: Electrostatics and Current Electricity

­distance y from a point O. Consider a sphere of radius R with O as centre and R > y . Electric flux through the surface of the sphere is

1 (A) -2 (B) + 2 1 (C) - (D) +2 2 156. Two point like charges Q1 and Q2 of whose strength are equal in absolute value of are placed at a certain distance from each other. Assuming the field strength to be positive in the positive direction of x-axis, the signs of the charges Q1 and Q2 for the graphs (field strength versus distance) shown in figure 1, 2, 3, and 4 are E

E



(A) zero

(B)

2λ R ε0

2λ R 2 - y 2 λ R2 + y 2 (D) (C) ε0 ε0 159. Two charges of -4 μC and +4 μC are placed at points A ( 1, 0, 4 ) and B ( 2, - 1, 5 ) located in an  electric field E = 0.20iˆ Vcm -1 . The torque acting on the dipole is iˆ

Q1

x

Q2 Figure 1

Q1

E

Q1

Q2



   

Q2

x

4πε 0 ( d + 2

(C)

3 a2 2

)

-2qd

4πε 0 ( d 2 +

3 a2 2

)

Figure 4

kˆ (B)

qd

4πε 0 ( d + 2

- qd kˆ (D)

4πε 0 ( d 2 +

3 a2 2

)

3 a2 2

)

kˆ

kˆ

158. A uniformly charged and infinitely long line having a linear charge density λ is placed at a normal

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 180

8 2 × 10 -5 Nm (C)

(D) 2 2 × 10 -5 Nm

4000 (C) coulomb (D) -4000ε 0 coulomb ε0

157. Two point charges q and -q are at positions (0, 0, d ) and (0, 0, -d ) respectively. The electric field at ( a, 0, 0) is 2qd

(B)

(A) -4000 coulomb (B) 4000 coulomb

(A)  Q1 positive; Q2 negative; both positive; Q1 negative; Q2 positive; both negative Q1 negative; Q2 positive; Q1 positive; Q2 neg(B)  ative; both positive; both negative Q1 positive; Q2 negative; both negative; Q1 (C)  negative; Q2 positive; both positive (D) both positive; Q1 positive; Q2 negative; Q1 negative; Q2 positive; both negative

(A)

8 × 10 -5 Nm 2

8 × 10 -5 Nm (A)

160. The inward and outward electric flux from a closed surface are respectively 8 × 10 3 and 4 × 10 3 units . Then the net charge inside the closed surface is

Q1

x

Figure 3



x

Figure 2

   

E



Q2

161. A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is q



(A) zero

(B)

q ε0

q 2q (C) (D) ε0 2ε 0 162. The block shown in the diagram has a mass of 2 microgram and a charge of 2 × 10 -9 coulomb . If the block is given an initial velocity of 2 ms -1 in x direction at t = 0 and an electric field of 2 NC -1 in x direction is switch on at that moment, then R will be g = 10 ms -2

(

)

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Chapter 1: Electrostatics 1.181

u = 2 ms–1

m

E = 2 NC–1 5m

165. A charge Q is distributed uniformly on a ring of radius r. A sphere of equal radius r is constructed with its centre at the periphery of the ring. Find the flux of the electric field through the surface of the sphere Ring

Final position

Sphere O

R

(A) 4 m (B) 1m (C) 2 m (D) 3m 163. A ring of radius R is marked in six equal parts and these parts are charged uniformly each with a charge of magnitude Q but positive and negative alternately as shown. The electric field at centre of ring will be

O′

Q Q (A) (B) 4ε 0 6ε 0 Q Q (C) (D) 3ε 0 2ε 0 166. A charge Q is distributed over a line of length L. Another point charge q is placed at a distance r from the centre of the line distribution. Then the force experienced by q is

R O

Q

r q

L

3Q λ (A) where λ = 4πε 0 r πR

qQ qQ (A) 2 (B) 2 2) ( ( 4πε 0 r - L πε 0 4 r - L2 )

3Q 2λ (B) where λ = πR 4πε 0 r

qQ qQL (C) 2 (D) 4πε 0 r 4πε 0 r 3

3Q 3λ (C) where λ = πR 4πε 0 r

167. Two charges Q1 and Q2 are distance d apart. Two dielectrics of thickness t1 and t2 and dielectric constant k1 and k2 are introduced as shown. Find the force between the charges

3Q 3λ (D) , where λ = πR 2πε 0 R

Q1

164. Some point charges are placed on the circumference of circle at equal distance. (see figure) The direction of electric field at centre O will be along B(q) q

–q O

(q)C

q

–q D(q)

(A) OA (B) OB (C) OC (D) OD

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 181

k2

t1

t2

Q2

Q1Q2 (A) 2 4πε 0 [ d – (t1 + t2 ) + k1t1 + k2t2 ] A(–q)

r

k1



(B) zero

Q1Q2 (C) 4πε 0 d + k1 t1 + k2 t2

(

(D) 4πε 0 ⎡⎣

(

Q1Q2

)

k1 - 1 t1 +

(

)

2

)

k2 - 1 t2 ⎤⎦

2

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1.182  JEE Advanced Physics: Electrostatics and Current Electricity 168. For two infinitely charged parallel sheets the electric field at P will be σ

σ

171. A conducting shell of radius R carries charge -Q. A point charge +Q is placed at the centre of shell. The electric field E varies with distance r (from the centre of the shell) as E (B)

E (A)

x P

r

σ σ σ σ + (A) (B) ( ) ( 2x 2 r - x 2ε 0 x 2π r - x ) ε 0 σ (C) ε0

(D) zero

E (A)

(B) E

x

rA rB

O

rA r B

x

(D) E

E (C)

O

x

r A rB

rA rB

x

170. The minimum value of μ for which small block B of mass m remain at rest on horizontal is (Assume that block A is fixed and small block has identical charge q and separation is r ) q A

r

q B

m

μ

q2 q2 (A) (B) 8πε 0 mgr 2 4πε 0 mgr 2 q2 2q 2 (C) (D) 2πε 0 mgr 2 πε 0 mgr 2

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 182

r

R

r

R

172. Two balls with equal charges are in a vessel with ice at -10 °C at a distance of 25 cm from each other. On forming water at 0 °C , the balls are brought nearer to 5 cm for the interaction between them to be same. If the dielectric constant of water at 0 °C is 80, the dielectric constant of ice at -10 °C is (A) 40 (B) 3.2 (C) 20 (D) 6.4 173. The  electric field varies along the x-direction as E = ( E0 x ) iˆ . Consider an imaginary cubical volume of edge l, with its edges parallel to the axes of coordinates. The charge inside this volume is

O

r

R

E (D)

E (C)

169. Two concentric conducting thin spherical shells A, and B having radii rA and rB ( rB > rA ) are charged to QA and -QB ( QB > QA ). The variation of electric field ( E ) with distance r from centre is best represented by

O

r

R

(B) ε 0E0l 3

(A) zero

1 1 (C) E0l 3 (D) ε 0E0l 2 ε0 6 174. Four point charges, each +q, are fixed at the corners of a square of side a. Another point charge q0 is placed at a height h vertically above the centre of square, assuming the square to be in a horizontal plane. Magnitude of force experienced by q0 is qq0 qq0 1 1 (A) (B) 3 4πε 0 ⎛ 2 a 2 ⎞ 4πε 0 ⎛ 2 a2 ⎞ 2 ⎜⎝ h + ⎟⎠ 2 ⎜⎝ h + ⎟⎠ 2 1 (C) πε 0

qq0 h

3



(D) zero

⎛ 2 a2 ⎞ 2 ⎜⎝ h + ⎟⎠ 2

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Chapter 1: Electrostatics 1.183 175. A hollow cylinder has a charge q coulomb within it. If ϕ is the electric flux in units of voltmeter associated with the curved surface B, the flux linked with the plane surface A in units of voltmeter will be

∞

λ = 1 C/cm

1m

B q

C

A –∞

q ϕ (A) (B) 2ε 0 3 ⎞ q 1⎛ q -ϕ⎟ (C) - ϕ (D) ε0 2 ⎜⎝ ε 0 ⎠ 176. Two small spheres each of mass m and charge q are tied from the same rigid support with the help of silk threads of length L. They make angle θ with the vertical as shown in the figure. If length L is decreased then angle θ with the vertical

(A) ϕ = 100 (B) ϕ=1 1 100 (C) ϕ= (D) ε0 ε0 179. The path of a positively charged particle 1 through a rectangular region of uniform electric field as shown in the figure. The direction of electric field and the direction of deflection of particles 2, 3 and 4 will be Up 2

3

1

4 Down

θ θ

L

q, m





L

q, m

(A) increases (C) unaffected

(B) decreases (D) cannot say

177. A charge is suspended from a light thread in a uniform field as shown. The tension in the thread at equilibrium is

(A) (B) (C) (D)

Up, down, up, down Up, down, down, up Down, up, up, down Down, up, down, down

180. In  a region of space the electric field is given by E = 8iˆ + 4 ˆj + 3 kˆ . The electric flux through a surface of area of 100 units in x -y plane is (A) 800 units (B) 300 units (C) 400 units (D) 1500 units 181. Ratio of electric field intensity at P & Q in the shown arrangement is q 2r

E = 2 × 104 NC–1

Q

3q P

a b

2 × 10–8 C

(A) 8.8 N (B) 8.8 × 10 2 N (C) 8.8 × 10 -4 N (D) 8.8 × 10 -3 N 178. Total flux coming out of the cylindrical Gaussian surface is

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 183

(A) 1 : 2 (B) 2:1 (C) 1 : 1 (D) 4:3 182. An oil drop carrying a charge of two electrons has a mass of 3.2 × 10 -17 kg . It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is (neglect buoyant force) 2 × 10 3 Vm -1 (B) 4 × 10 3 Vm -1 (A) (C) 3 × 10 3 Vm -1 (D) 8 × 10 3 Vm -1

9/20/2019 10:40:43 AM

1.184  JEE Advanced Physics: Electrostatics and Current Electricity 183. A charge q is located at the centre of a cube the electric flux through any face is 4π q πq (B) (A) 6 ( 4πε 0 ) 6 ( 4πε 0 ) q 2π q (C) (D) 6 ( 4πε 0 ) 6 ( 4πε 0 ) 184. A point charge Q is placed at the corner of a square plate of side a then the flux through the square plate is 90° and hence ϕ = 0 Q

2rT (A) 2 (B) 8π r 2rT ε 0 ε0 2rT (C) 8π r rT ε 0 (D) 8π r ε0 186. If a small sphere of mass m and charge q is hanged from a silk thread at an angle θ with the surface of a vertical charged conducting plate, then for equilibrium of sphere the surface charge density of the plate is

ε 0 mg tan θ ε 0 2mg tan θ (B) (A) q q

a

ε 0 mg tan θ (C) ε 0 mgq tan θ (D) 3q

a

187. Two similar point charges q1 and q2 are placed at a distance r, apart in air. The force between them is F1 . A dielectric slab of thickness t ( < r ) and dielectric constant K is placed between the charges. Then the F force between the same charges is F2. The ratio 1 is F2 (A) 1 (B) K

Q Q (A) (B) 8ε 0 4ε 0 Q (C) ε0

(D) zero

185. An isolated and charged spherical soap bubble has a radius r and the pressure inside is atmospheric. If T is the surface tension of soap solution, then charge on drop is

2

2 ⎛ r–t+t K ⎞ r ⎛ ⎞ (C) ⎜⎝ ⎟⎠ (D) ⎜⎝ ⎟⎠ r r–t+t K

Multiple Correct Choice Type Questions This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1.

A charged cork ball of mass 1 g is suspended on a light string in the presence  of a uniform electric field as shown. When E = 3i + 5j × 10 5 NC -1, the ball is in equilibrium at θ = 37° . T is the tension in the string and q is the charge on the ball Take sin 37° = 0.60 and g = 10 ms -2

(

(

)

)

2.

Consider a point charge Q placed at the origin O of    the coordinate system. Let EA , EB and EC be the electric fields at three points A ( 1, 2, 3 ) m, B ( 1, 1, - 1 ) m and C ( 2, 2, 2 ) m due to charge Q. Then     (A) EA ⊥ EB (B) EA  EC     (C) EB = 4 EC (D) EB = 8 EC 3.

θ

E q

(A) q = 11 nC (B) q = 12 nC (C) T = 5.55 × 10 -3 N (D) T = 4.55 × 10 -3 N

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 184



Equipotential surfaces representing a uniform electric field must be (A) plane surfaces. (B) normal to the direction of the field. (C) spaced such that surfaces having equal differences in potential are separated by equal distances. (D) having potentials decreasing in the direction of the field.

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Chapter 1: Electrostatics 1.185 4.

5.

Consider two equipotential surfaces A and B having different potentials. Then A and B cannot intersect. (A)  A and B cannot both be plane surfaces. (B)  (C) In the region between A and B , the field is maximum where equipotential surfaces are closest to each other. (D) A line of force from A to B must be perpendicular to both. An ellipsoidal cavity is carved within a perfect conductor (see figure). A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in figure. Then, A q



6.

7.

B

(A) electric field near A in the cavity = electric field near B in the cavity. (B) charge density at A = charge density at B. (C) potential at A = potential at B. (D) total electric flux through the surface of the cavity q . is ε0 Consider two large, identical, parallel conducting plates have surfaces X and Y, facing each other. The charge per unit area on X is σ 1 and charge per unit surface area on Y is σ 2 . Then, σ 1 = -σ 2 only if a charge is given to either X or Y (A)  (B)  σ 1 = σ 2 = 0 if equal charges are given to both X and Y (C)  σ 1 > σ 2 if X is given a charge more than that given to Y (D)  σ 1 = -σ 2 in all cases A solid metallic sphere is place in a uniform electric field. Which of the curves shown in the figure represent the lines of force correctly? (a) (b) (c) (d)



(A) (a) (C) (c)

8.

X , Y and Z are parallel plates. Y is given some positive charge. Two electrons A and B start with zero initial velocity from X and Z respectively and reach Y in times t and T respectively.



01_Physics for JEE Mains and Advanced - 2_Part 5.indd 185

(B) (b) (D) (d)

X

Y A

2d

Z B

d



(A) t = T (B) t = 2T (C)  2t = T (D) t = 2T

9.

In the set up of PROBLEM 8, an electron E moves from X to Y and a proton P moves from Y to Z. Both particles start from rest. (A) E reaches Y with greater energy than P. P reaches Y with greater energy than E. (B)  P and E reach Y with equal energies. (C)  (D) P reaches Y with greater momentum than E.



10. A charged particle of mass m , charge q moves with a speed v in a circular path of radius r around a long uniformly charged conductor having charge density λ . Then

(A) v ∝ q (B) v∝ λ



(C)  v∝

1 1 (D) v∝ m r

11. Two charges Q1 and Q2 are placed at the points A and B having separation r lying inside and outside the uncharged conducting shell. The force on Q1 is F and that on Q2 is f . Then 1 Q1Q2 (A) F = 0 (B) F= 4πε 0 r 2 1 Q1Q2 (C) f = (D) f =0 4πε 0 r 2 12. A particle of charge q and mass m is moving in an electric field (A) must undergo change in velocity. (B) must undergo change in speed. (C) may not undergo change in velocity. (D) may not undergo change in speed. 13. Two charges Q1 and Q2 have their respective placements at the inside and the outside of a closed surface S. If E be the field at any point on S and ϕ be the flux of E over the surface S, then if

(A)  Q1 changes, both E and ϕ will change.



(B)  Q2 changes, E will change but ϕ will not change. Q1 = 0 and Q2 ≠ 0 then E ≠ 0 but ϕ = 0. (C) 



(D)  Q1 ≠ 0 and Q 2 = 0 then E = 0 but ϕ ≠ 0.

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1.186  JEE Advanced Physics: Electrostatics and Current Electricity 14. A particle having specific charge s starts from rest in a region where the electric field has constant direction but magnitude varying with time t as E = E0t. In time t, it is observed that the particle acquires a velocity v after covering a distance x, then

(A) x =

1 1 x = ( E0 s ) t 3 ( E0 s ) t 3 (B) 2 6

L

1 (C)  v = ( E0 s ) t (D) v = ( E0 s ) t 2 2 15. At a point in space, let the field and potential be having values E and V respectively. Out of the following assertions which one(s) is/are correct? (A) For V = 0, E must be zero (B) For V ≠ 0, E cannot be zero (C) For E ≠ 0, V cannot be zero (D) None of these 2

16. In a parallel plate capacitor, the potential difference between the plates is V . A particle of mass m and charge -Q leaves the negative plate and reaches the positive plate at distance d in time t with a momentum p. Then

p = 2mQV (A) p = mQV (B)



md 2 2md 2 (D) t= QV QV

(C)  t=

17. A ring with a uniform charge Q and radius R, is placed in the yz plane with its centre at the origin O of the coordinates. Let VO and EO be the potential and the field at the origin, V and E be the potential and field at all points on the positive x axis, other than the origin, then

(A) EO = 0



VO = (B) 



(C)  V=



1 (D) E = 4πε 0

E

u

EsL2 L (B) t= 2 u 2u



(A) Δ =



⎛ EsL ⎞ ⎛ Est ⎞ θ = tan -1 ⎜ 2 ⎟ (D) θ = tan -1 ⎜ (C)  ⎝ u ⎟⎠ ⎝ u ⎠

19. Three concentric metallic shells A, B and C are arranged such that shell A is the innermost and shell C is the outermost. Now, A is given some charge. Then the (A) inner surfaces of B and C have the same charge density. (B) inner surfaces of B and C have the same charge. (C) outer surfaces of A, B and C have the same charge density. (D) outer surfaces of A, B and C have the same charge. 20. Two concentric shells A and B have radii R and 3V 2R, charges qA and qB and potentials 2V and 2 respectively. Now, the shell B is earthed and the new charges on them become q′A and qB′ . Then B A

1 Q 4πε 0 R

1 4πε 0

Q 2

x +R

2

Qx

(R

2

+

3 x2 2

)

and is maximum at x =

R 2

18. There is a uniform electric field E in the region between two parallel plates. A charged particle with a specific charge s enters the region with a speed u, in a direction parallel to the plates. The length of each plate is L. The time taken to come out of the plates is t and it suffers a vertical displacement Δ. Also as the particle crosses the plates, its direction of motion changes by an angle θ , then

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 186



q 1 (A) A = qB 2



q′ (B) A = 1 qB′





(C)  Potential difference between A and B after 3V earthing becomes 2 (D)  Potential difference between A and B after V earthing becomes 2

21. In a uniform electric field, (A) all points are at the same potential. (B) no two points can have the same potential.

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Chapter 1: Electrostatics 1.187

(C) pairs of points separated by the same distance must have the same difference in potential. (D) None of the above.

22. Three concentric conducting spherical shells A, B and C have radii r, 2r and 3r and possess charges Q1, Q2 and Q3 respectively. The innermost and the outermost shells are earthed as shown in the figure. Select the mathematical relations between the charges that are correct. Q3 Q2 Q1



(A) positive, along the line joining the two identical charges. (B) positive, perpendicular to the line joining the two identical charges. (C) negative, perpendicular to the line joining the two identical charges. (D) negative, along the line joining the two identical charges.

26. An electron is projected between two plates at separation d = 3 cm each having length l = 11 cm as shown in figure, at a speed of 6 × 106 ms -1 at an angle of 45° in a uniform field E = 2000 Vm -1 directed upwards. The electron strikes the

r E d

2r θ = 45°

3r

l

(A)  Q1 + Q3 = -Q2 (B) Q1 = -



Q Q3 1 (C)  3 = 3 (D) =Q1 Q2 3





(A) upper plate (B) lower plate (C) lower plate at the edge (D) nowhere

(C) Charge -Q1 will appear on the inner surface of B. (D) Charge Q1 + Q2 will appear on the outer surface of B.



(B) Charge Q2 will appear on the outer surface of B.



27. Two identical particles of mass m carry a charge Q each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first particle from a large distance with a speed v. If d is the distance of closest approach then 1 (A) d ∝ Q 2 (B) d∝ 2 v

23. A spherical conductor A lies inside a hollow spherical conductor B. Charges Q1 and Q2 are given to A and B respectively. (A) Charge Q1 will appear on the outer surface of A.



Q2 4



24. A particle of charge Q having a mass m moves rectilinearly under the action of an electric field that varies with distance x from the initial position of rest of the particle E = α - β x, where α and β are positive constants. Then (A) the motion of the particle is oscillatory α (B) the amplitude of the particle is β α (C) the mean position of the particle is at x = β qα (D) the maximum acceleration of the particle is m 25. Two identical charges each having a charge +Q are fixed at a separation 2a. A small particle P having a charge of magnitude q is placed midway between them. Now, P is given a displacement Δ (  a ) , such that it undergoes simple harmonic motion. For this, we have the following ( q, Δ ) combinations.

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 187

(C)  d∝

1 1 (D) d∝ v m

28. An electron moving with a speed of 5 × 106 ms -1 is shot parallel to an electric field of strength 1000 Vm -1, arranged so as to retard its motion. The electron travels a distance s in the field before coming momentarily to rest in time t. It is observed that the electric field ends abruptly after 0.8 cm and due to this the electron loses a percentage fraction f of its initial energy, then

(A) s = 0.14 m (B) s = 0.07 m (C)  t = 0.03 μs (D) f = 11%

29. A positively charged thin metal ring of radius R is fixed in the xy plane, with its centre at the origin O. A negatively charged particle P is released from rest at the point ( 0 , 0 , z0 ), where z0 > 0 . Then the motion of P is (A) periodic, for all value of z0 satisfying 0 < z0 < ∞ . (B) simple harmonic, for all values of z0 satisfying 0 < z0 ≤ R.

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1.188  JEE Advanced Physics: Electrostatics and Current Electricity 34. For spherical symmetrical charge distribution, variation of electric potential with distance from centre is q for r ≤ R0 given in diagram. Given that V = 4πε 0 R0 q for r ≥ R0 and V = 4πε 0 r

(C)  approximately simple harmonic, provided z0  R . (D) such that P crosses O and continues to move along the negative z-axis towards z = -∞. 30. Two infinitely large plane sheets separated by a distance l carry a uniform surface charge densities +σ and -σ . The planes have identical coaxial holes each of radius R ( l  R ). The potential and the field at a point on the axis of the holes at a distance x from the midpoint O between the holes is V and E respectively. Then

V

r r = R0

O



P



x l



V= (A) 



V= (C) 

σ lx 2ε 0 R 2 + x 2

E= (B)

σ lR2 2ε 0 ( R2 + x 2 )

32

σ σ E= (D) 2 2ε 0 x 2ε 0 R5

31. At the corners of a square, four particles having charges of the same magnitude are placed. If the electrostatic potential and field at the centre of the square due to these four charges is V and E respectively, then by suitably selecting the signs of the four charges of identical magnitude, which of the following conditions can be achieved? V = 0, E = 0 (B) V = 0, E ≠ 0 (A)  (C)  V ≠ 0, E = 0 (D) V ≠ 0, E ≠ 0 32. A pendulum bob of mass m = 71 mg , carrying a charge of Q = 2 × 10 -8 C, is at rest in a horizontal, uniform electric field E = 20000 Vm -1. The tension T in the thread of the pendulum is T and the angle it makes with the vertical is θ . Then (A)  θ = 27° (B) T = 880 μN

(C)  T = 803 μN (D) θ = 30°

33. Four identical charges each having a charge q are placed at the points ( a, 0 , 0 ), ( 0 , a, 0 ), ( - a, 0 , 0 ) and ( 0, - a, 0 ). Let V0 and E0 be the potential and the field at the origin, V and E be the potential and field at all points on the z axis, other than the origin, then V0 ≠ 0 (B) E0 = 0 (A) 

(C)  V≠0

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 188

(D) E ≠ 0, along z-axis

Then which option(s) is/are correct (A) Total charge within 2R0 is q. (B) Total electrostatic energy for r ≤ R0 is zero. (C) At r = R0 electric field is discontinuous. (D) There will be no charge anywhere except at r = R0.

35. In a region of space three charged particles are observed to be in equilibrium under their mutual electrostatic forces only. Then the particles (A) must be collinear. (B) cannot possess the same magnitude. (C) cannot possess the same sign. (D) will be in unstable equilibrium. 36. Under the influence of the Coulomb field of charge +Q, a charge -q is moving around it in an elliptical orbit. Find out the correct statement(s). (A)  The angular momentum of the charge -q is constant. (B)  The linear momentum of the charge -q is constant. (C) The angular velocity of the charge -q is constant. (D) The linear speed of the charge -q is constant. 37. At the corners of an equilateral triangle of side l, three point charges are placed such that the only interaction force between them is the electrostatic force. Then this system of three charges will be (A) in equilibrium if the charges have the same magnitude but all do not have the same sign. (B) in equilibrium if the charges have different magnitudes but all do not have the same sign. (C) in equilibrium if the charges rotate about the centroid of the triangle. (D) never in equilibrium. 38. A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that

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Chapter 1: Electrostatics 1.189



(A) Q1 > Q2



Q1 < Q2 (B) 



(C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero



39. Which of the following is/are incorrect statement? (A) Electric field is always conservative (B) Electric field due to a varying magnetic field is non-conservative (C)  Electric field due to a stationary charge is conservative. (D) Electric field lines are always closed loops. 40. A particle of mass m and charge q is fastened to one end of a string of length l. The other end of the string is fixed to the point O. The whole system lies on a frictionless horizontal plane. Initially, the mass is at rest at A. A uniform electric field in the direction shown is then switched on. Then

(A) The inner surfaces of B and C will have the same charge (B) The inner surface of B and C will have same charge density (C) The outer surface of A, B and C will have the same charge (D)  The outer surface of C will have no charge density

42. Two charge Q1 & Q2 placed at (0, 0, 0) and ( a, 0, 0) are ⎛a ⎞ such that x component of field is zero at ⎜ , a, 0 ⎟ . ⎝2 ⎠ Then

⎛a ⎞ (A) y component of field is zero at ⎜ , 0 , a ⎟ ⎝2 ⎠ Q1 2 (B)  = Q2 1 Q (C)  1 = 1 Q2

⎛a ⎞ (D) y component of field is zero at ⎜ , a, 0 ⎟ ⎝2 ⎠

43. An electrically isolated, hollow conducting sphere has a small positively charged ball suspended by an insulating rod from its inside surface, as shown. This causes the appearance of negative charge on the inner surface of the sphere. The electric field outside and far from the conducting sphere is approximately

A

E l

O





41.

60°

B

(A) the speed of the particle when it reaches B is 2qEl m (B) the speed of the particle when it reaches B is qEl m (C) the tension in the string when particles reaches at B is qE (D) the tension in the string when the particle reaches at B is 2qE A, B and C are three concentric metallic shells. Shell A is the inner most and shell C is the outermost. A is given some charge and shell C is earthed. Select the correct statement(s).

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 189



(A) zero (B) the same as if the sphere was not there (C) half what it would be if the sphere was not there Q (D)  , where r is the distance from centre of 4πε 0 r 2 sphere

44. Two identical conducting balls have positive charges q1 & q2 respectively. The balls are brought together so that they touch and then put back in their original positions. The force between the balls may be (A)  The same as it was before the balls had been touched. (B) greater than the force before the balls had been touched (C)  less than the force before the balls had been touched (D) zero

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1.190  JEE Advanced Physics: Electrostatics and Current Electricity 45. Figure shows two dipole moments parallel to each other and placed at a distance x apart (where x is very large compared to size of dipoles) then +q1

+q2 p2

p1 –q1

x

50. In front of an earthed conductor, a point charge +q is placed as shown in figure. Select the correct statement(s).

–q2



(A) they will repel each other (B) they will attract each other



(C) force of interaction is of magnitude of

3 p1p2 4πε 0 x 4



(D) force of interaction is of magnitude of

6 p1p2 4πε 0 x 4

46. A positive point charge is placed at P in front of an earthed metal sheet S. Two points Q and R lie between P and S as shown. If the electric field strength at Q and R are respectively EQ and ER and potential at Q and R are respectively VQ and VR , then

+q





(A)  On the surface of conductor the net charge is always negative. (B) On the surface of conductor at some points the charges are negative and at some points the charges may be positive distributed non uniformly. (C) Inside the conductor electric field due to point charge is non-zero. (D) None of these

P



Q

51. A charged particle q is placed at a distance d from the centre of conducting sphere of radius R ( < d ) , then in 1 ⎞ ⎛ static condition at the centre of sphere ⎜ k = ⎟ , we πε 4 ⎝ 0⎠ have

R S



(A) EQ > ER (B) EQ < ER



(C)  VQ > VR (D) VQ < VR

47. A particle of mass 2 kg and charge 1 mc is projected vertically with a velocity 10 ms -1 . There is a uniform horizontal electric field of 10 4 NC -1. Select the correct statements. (A) The horizontal range of the particle is 10 m. (B) The time of flight of the particle is 2 s. (C) The maximum height reached is 5 m. (D) The horizontal range of the particle is zero. 48. A certain spherical region contains no charge, then (A) Electric field at any point in the sphere must be zero. (B) Electric field at any point in the sphere may not be zero. (C) Electric potential at any point in the sphere must be zero. (D) Electric potential at any point in the sphere may not be zero.

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 190

49. A dipole is placed in the electric field created by a uniformly charged long wire. It is possible that the dipole experiences (A) No net force but a net torque (B) No net force and no net torque (C) A net force but no net torque (D) A net force and a net torque







(A) magnitude of electric field due to induced charge kq is 2 d (B) magnitude of electric field due to induced charge kq is 2 R (C) magnitude of electric field due to induced charge is zero kq (D) magnitude of electric field due to charge q is 2 d

52. We have an infinite non-conducting sheet of negligible thickness carrying a uniform surface charge density -σ and next to it, an infinite parallel slab of thickness D with uniform volume charge density + ρ . All charges are fixed. –σ

D

+ρ

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Chapter 1: Electrostatics 1.191



(A) Magnitude of electric field at a distance h above ρD - σ the negatively charged sheet is 2ε 0 (B) Magnitude of electric field inside the slab at a distance h below the negatively charged sheet

( h < D ) is

σ + ρ ( D - 2h ) 2ε 0



(C) Magnitude of electric field at a distance h below ρD - σ the bottom of the slab is 4ε 0



(D) Magnitude of electric field at a distance h below ρD - σ the bottom of the slab is 2ε 0



55. A conductor A is given a charge +Q and then placed inside a deep metal can B, without touching it. Then (A) 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 is increased

53. A conducting sphere A of radius a, with charge Q, is placed concentrically inside a conducting shell B of radius b. B is earthed. C is the common centre of A and B, then the B

56. A positive charge of q1 is located 3 m to the left of a negative charge -q2 . Magnitude of charge q1 greater than magnitude of negative charge q2. The net electric field intensity is zero at a point on the line passing through charges and 1 m to the right of -q2 . On this line there are also points where the potential is zero. Location of these two points relative to the negative charge is/are

Q

A C

a

(A) field at a distance r from C, where a ≤ r ≤ b , is Q 4πε 0 r 2



(B) potential at a distance r from C, where a ≤ r ≤ b , Q is 4πε 0 r



(C)  potential difference between Q ⎛ 1 1⎞ ⎜ - ⎟ 4πε 0 ⎝ a b ⎠



(D) potential at a distance r from C, where a ≤ r ≤ b , Q ⎛ 1 1⎞ is ⎜ - ⎟ 4πε 0 ⎝ r b ⎠

A

and

B

is

54. Four spheres each of different radii named a, b, c, d are given. Each is solid, insulating and uniformly charged. The variation of electric field with distance r from the centre is given. Straight portion of curve c and d is overlapping and straight portion of curve a and b is overlapping. Select the correct statement(s). E c b d

a r

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 191

(A) at 0.2 m left of negative charge

b



(A) Radius of sphere a > Radius of sphere b (B) Radius of sphere a < Radius of sphere b (C) Volume charged density of c > volume charged density of b (D) Volume charged density of c < volume charged density of b



(B) at 0.2 m right of negative charge

(C) at

3 m left of negative charge 17 3 (D) at m right of negative charge 17

57. Select the correct statement(s). (A) When a positively charged sphere is brought near a metallic sphere, it is observed that a force of attraction exists between two, which means the metallic sphere become negative charged. (B) A given conducting sphere cannot be charged to a potential greater than a certain value. (C) Potential difference between two points lying in a uniform electric field may be equal to zero. (D) Potential difference between two points lying in a non-uniform electric field may be equal to zero. 58. A block having mass m and charge q is connected by spring of force constant k. The block lies on a frictionless horizontal track and a uniform electric field E acts on system as shown. The block is released from rest when spring is unstretched (at x = 0). Then E q, m

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1.192  JEE Advanced Physics: Electrostatics and Current Electricity





2qE k (B) In equilibrium position, stretch in the spring is qE k qE (C) Amplitude of oscillation of block is k 2qE (D) Amplitude of oscillation is k

59. A uniform electric field of magnitude E0 exists in a region at an angle 45° with x-axis. There are two point A ( a, 0) and B (0, b ) having potential VA and VB respectively, then

string makes an angle q with vertical. Its time period of oscillation is T in this position. Then

(A) Maximum stretch in the spring is



(A) tan θ =

σq σq tan θ = (B) ε 0 mg 2ε 0 mg



(C)  T < 2π

l l (D) T > 2π g g

61. Two point charges Q1 and Q2 lie along a line at a distance from each other. Figure show the potential variation along the line of charges V

(A)  VA > VB if a > b (B) VA = VB if a = b (C)  VA > VB if a < b (D) VA < VB if a > b

60. A simple pendulum of length  has a bob of mass m, with a charge q on it. A vertical sheet of charge, with charge σ per unit area, passes through the point of suspension of the pendulum. At equilibrium, the

Q2

Q1 1



2 3

r

(A)  Q1 is positive and Q2 is negative. Q1 is negative and Q2 is positive. (B)  (C) Magnitude of Q1 is more than magnitude of Q2 . (D) Electric field is zero at 3.

Reasoning Based Questions This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A)  Bubble (B)  Bubble (C)  Bubble (D) 

If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.

1.

Statement-1: If there exists coulombic attraction between two bodies both of them may not be charges.  Statement-2: In coulombic attraction two bodies are always oppositely charged. 2.

Statement-1: Two charges q1 and q2 are placed at separation r. Then magnitude of force on each charge qq is F = 1 2 2 . 4πε 0 r

Statement-2: Now a third charge q3 is placed near q1 and q2. Then force on q1 due to q2 remains F. 3.

Statement-1: While going away from a point charge or a small electric dipole, electric field decreases at the same rate in both the cases. Statement-2: Electric field is inversely proportional to square of distance from the point charge.

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 192

4.

Statement-1: The potential of an uncharged conducting sphere of radius R, for a point charge q located at q distance r from its centre ( r > R ) is . 4πε 0 r Statement-2: Electric field intensity inside the conductor is zero therefore potential at each point on conductor is zero. 5.

Statement-1: When a neutral body is charged negatively, its mass increases slightly. Statement-2: When a body is charged negatively, it gains some electrons and electron has finite mass : though quite small 6.

Statement-1: Electric field at a point is always inversely 2 proportional to ( distance ) Statement-2: Electric field due to a line charge at a point is inversely proportional to distance.

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Chapter 1: Electrostatics 1.193 7.

Statement-1: Any charge will move from electric potential V1 to V2 by its own only when V1 < V2.

Statement-2: Electron moves from V1 = 2V towards V2 = 4 V. 8.

Statement-1: Electric field intensity within an isolated conductor will be zero. Statement-2: No net charge can exist within an isolated conductor. 9.

Statement-1: Excess charge on a conductor resides entirely on the outer surface. Statement-2: Like charges repel each another. 10. Statement-1: The surface of a charged conductor is always equipotential. Statement-2: Electric field lines are always perpendicular to the equipotential surface. 11. Statement-1: When a charged body is brought near to an uncharged conducting body, equal and opposite charge is induced on the nearer surface of the conducting body. Statement-2: Net electric field inside the conductor is zero. 12. A charge is launched with a velocity perpendicular to uniform electric field then Statement-1: Initial power delivered by electric field is zero. Statement-2: Path of charged particle is circular. 13. Statement-1: A charge conductor may have charged particle inside it. Statement-2: There can’t exist electric field line inside the conductor. 14. Statement-1: Electric field intensity at surface of uniformly charge spherical shell is E. If shell is punctured E at a point then intensity at punctured point become . 2 Statement-2: Electric field intensity due to spherical charge distribution can be found out by using Gauss Law. 15. Statement-1: For practical purposes, the earth is used as a reference at zero potential in electrical circuits.

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 193

Statement-2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the Q surface is given by . 4πε 0 R 16. Statement-1: Induced charge does not contribute to electric field or potential at a given point. Statement-2: A point charge q0 is kept outside a solid metallic sphere, the electric field inside the sphere is zero. 17. Statement-1: A point charge q is placed in front of a solid conducting sphere. Electric field due to induced charges at the centre of sphere is zero. Statement-2: Electric field at a point inside the solid body of conductor is zero.

r

O

q

R

18. Statement-1: Consider a conducting sphere of radius R. Now a charge q is placed in front of sphere. Electric q potential at point O is . 4πε 0 r

O

r

q

R

Statement-2: Electric potential at the centre of sphere due to induced charges is zero. 19. Statement-1: A metallic shield in form of a hollow shell may be built to block an electric field. Statement-2: In a hollow spherical shield, the electric field inside it is zero at every point. 20. Statement-1: If a point charge q is placed in front of an infinite grounded conducting plane surface, the point charge will experience a force. Statement-2: This force is due to the induced charge on the conducting surface, which is at zero potential.

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1.194  JEE Advanced Physics: Electrostatics and Current Electricity

Linked Comprehension Type Questions This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options)

Comprehension 1



In the classical model of a hydrogen atom, the electron revolves around the proton in a circle of radius r = 0.53 Å. The magnitude of the charge of the electron and proton is e = 1.6 × 10 −19 C. Based on the above facts, answer the following questions. 1. 2.

An electron is injected ( at t = 0 ) horizontally into a uniform field produced by two oppositely charged plates having length l and separation d, as shown in figure. The  par ticle has an initial velocity v0 = v0 iˆ perpendicular to E = Ejˆ . y



(C)  5.76 × 1010 NC −1

3.

The ratio of the magnitudes of the electrical and gravitational force between electron and proton is (A) 2.2 × 10 39 , dependent on r



(B)  2.2 × 10 36 , dependent on r (C)  2.2 × 10 39 , independent of r (D) 2.2 × 10 36 , independent of r

e–

v0 O

P θ

E y1

(D) None of these



The mass of the oil drop is (A)  1.57 × 10 −14 kg (B) 1.57 × 10 −13 kg kg (D) 1.57 × 10

−11

x

Based on the above facts, and the figure provided, answer the following questions.

An oil drop of radius r = 1.64 × 10 −6 m having mass density ρoil = 8.51 × 10 2 kgm −3 is allowed to fall from rest.  It, then enters into a region of constant external field E applied in the downward direction. The oil drop has an unknown electric charge q. The magnitude of the electric  field is adjusted till the gravitational force, Fg = mg = − mg ˆj on  the oil drop is exactly balanced by the electric force, Fe = qE . Suppose this balancing occurs when the value of  electric field is given by E = − Ey ˆj = − ( 1.92 × 10 5 NC −1 ) ˆj , with Ey = 1.92 × 10 5 NC −1. Based on the above facts, answer the following questions.

y2

Screen

6.

Comprehension 2

−12

L

l

The magnitude of the electric field due to the proton at distance r is (A) 5.76 × 1012 NC −1 (B) 5.76 × 1011 NC −1

4.

Comprehension 3

The magnitude of the electric force between the proton and the electron is 8.2 × 10 −7 N (A) 8.2 × 10 −6 N (B) −8 (C)  8.2 × 10 N (D) None of these



(C)  N ( = 4 ) , excess of electrons (D) N ( = 5 ), excess of electrons

The path followed by the electron is a/an (A) straight line (B) ellipse (C) cycloid (D) parabola

7.

The time t1 taken by the electron to leave the plate is t1. Then



(A) t1 =

2lm l t1 = (B) v0 eE



t1 = (C) 

d 2md t1 = (D) eE v0

8.

The velocity of the electron at time t1 when it leaves the plates is



⎛ eE ⎞ l v0 (B) (A) ⎜ ⎟ ⎝ m ⎠ v0



1 m2v04 + e 2l 2E2 (C)  mv0

(D) None of these

9.

The vertical displacement of the electron, y1 after time t1 when it leaves the plates is



(C)  1.57 × 10

5.

The oil drop falls short of electrons or has the excess electrons amounting to N . Then



(A) y1 =

eEl 2 eEl 2 (B) y1 = mv0 mv02



(A) N ( = 4 ) , short of electrons (B)  N ( = 5 ), short of electrons



(C) y1 =

eEl 2 eEl 2 (D) y1 = 2 2 2 2mv0 mv

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 194

kg

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Chapter 1: Electrostatics 1.195 10. The electron makes an angle q with the horizontal, when leaving the plates at time t1, then

(A) sin q =

eEl eEl sin q = (B) mv02 2mv02



tan q = (C) 

eEl eEl tan q = (D) 2 mv0 2mv02

11. The electron hits the screen located a distance L from the end of the plates at a time t2 . The total vertical displacement of the electron from time t = 0 until it hits the screen at t2 is y. Then eEl ⎛ L ⎞ eEl ⎛ l ⎞ y= ⎜ + l ⎟ (B) ⎜ + L ⎟⎠ mv02 ⎝ 2 ⎠ mv02 ⎝ 2



(A) y =



eEl eEl ⎛ ( L + l ) (D) (C) y = y= ⎜L+ 2 2mv02 ⎝ 2mv0

l⎞ ⎟ 2⎠

Comprehension 4 Two parallel plates having charges of equal magnitude but opposite sign are separated by 12 cm. Each plate has a surface charge density of 36 nCm −1. A proton is released from rest at the positive plate. Based on the above facts, answer the following questions. 12. The potential difference between the plates is (A) 288 V (B) 388 V (C) 244 V (D) 488 V 13. The kinetic energy of the proton when it reaches the negative plate is 7.8 × 10 −17 J (A) 3.9 × 10 −17 J (B)

(C)  4.6 × 10 −17 J (D) 6.2 × 10 −17 J

14. The speed of the proton just before it strikes the negative plate is

(A)  106 kms −1 (B) 235 kms −1



(C)  306 kms −1 (D) 216 kms −1

15. The acceleration of the proton is

(A) 3.9 × 1011 ms −2 (B) 4.68 × 1010 ms −2



(C)  2.3 × 1011 ms −2 (D) 1.95 × 1011 ms −2

16. The force acting on the proton

(A) 6.51 × 10 −16 N (B) 7.81 × 10 −17 N (C)  3.84 × 10

−16

N (D) 3.25 × 10

−16

17. The value of v0 is

(A) 

qq0 qq0 (B) 2πε 0 ma πε 0 ma



(C) 

qq0 3πε 0 ma

Two positive point charges each of magnitude q are placed on the y-axis at the points ( 0, a ) and ( 0, − a ).

(D) None of these

18. The value of x is

(A)  3a (B) 4a



(C)  2a (D) 15a

19. The value of v is

v v0 (A)  0 (B) 3 2 v0 (C)  v0 (D) 4

Comprehension 6 A particle of mass m2 carrying a charge Q2 is fixed on the surface of Earth. Another particle, mass m1 and charge Q1, is positioned right above the first one at an altitude h  R , with R the radius of the Earth. The charges Q1 and Q2 have the same sign. Based on the above facts, answer the following questions. 20. The velocity of m1 at a point P very close to Q2 , a distance h1 from the surface of earth, if the initial velocity of m1 was zero and air drag and earth’s electric field being ignored, is

(A)  2 gh −



(C)  2gh

N

Comprehension 5

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 195

A positively charged particle of charge q0 and mass m, when displaced slightly from the origin in the direction of the negative x-axis attains a speed v0 at infinity. When the particle is projected towards the left along the x-axis from a point at a large distance towards the right of the origin with a velocity half the value of v0 , then it comes to rest at a point P ( x, 0 ). Again, if a negatively charged particle from the rest (assume the negatively charged particle to be at very large distance towards the left of origin), then this negatively charged particle crosses the origin with a speed v . Based on the above facts, answer the following questions.

Q1Q2 Q1Q2 (B) 2 gh − 2πε 0 h1m1 4πε 0 h1m1 (D) None of these

21. The magnitude of charge Q2 at which the velocity of m1 , at altitude h2 is zero is given by

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1.196  JEE Advanced Physics: Electrostatics and Current Electricity



(A)  Q2 = (C)  Q2 =

πε 0 m1 ghh2 4πε 0 m1 ghh2 (B) Q2 = Q1 Q1 2πε 0 m1 ghh2 πε m ghh2 (D) Q2 = 0 1 Q1 4Q1

22. At what altitude h3 will object m1 be in equilibrium and what will be the nature of the object’s oscillations if it is disturbed from equilibrium? (A) h3 =

Q1Q2 , periodic 4πε 0 m1 g

(B)  h3 =

Q1Q2 , periodic and oscillatory 2πε 0 m1 g



h3 = (C) 

Q1Q2 , periodic and oscillatory 4πε 0 m1 g



(D) None of these





24. The potential of A is Q Q − (B) 4πε 0 R 12πε 0 R



(A) −



Q (C)  16πε 0 R

(D) None of these

25. The potential of B is

Q 5Q (A)  (B) 4πε 0 R 12πε 0 R



3Q 5Q (C)  (D) 6πε 0 R 48πε 0 R

Comprehension 8 A rod of length L is placed along the x-axis with its left end at the origin and has a non-uniform charge density varying as λ = α x, where α is a positive constant.

Comprehension 7

y B

Figure shows three concentric spherical conductors A, B and C with radii R, 2R and 4R respectively. A and C are connected by a conducting wire and B is having a uniform charge +Q. Q

C A B

b d A

L

x

Based on the above facts and the figure provided, answer the following questions. 26. The dimensional formula for α is

R

2R 4R

Based on the above facts, answer the following questions. 23. Charge on the conductor A is qA and that on C is qC. Then Q 3



(A) M −2T −1A (B) M 0 L−2TA



(C) M 0 L−1TA (D) M 0 L−2T −1A

27. The electric potential at A, for d =

L is given by 4



α α (A)  ( L − log e 5 ) (B) ( L + log e 5 ) 4πε 0 4πε 0



α α (C)  ( 1 − log e 5 ) (D) ( 1 + log e 5 ) 4πε 0 4πε 0



(A) qA = qC = −



qA = (B) 

Q 2Q and qC = 3 3



(C)  qA =

Q Q and qC = − 3 3



αL αL log e ( 3 ) (B) log e ( 6 ) (A)  4πε 0 4πε 0



(D) qA = −

Q Q and qC = 3 3



αL αL (C)  log e ( 3 ) (D) log e ( 6 ) 8πε 0 8πε 0

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 196

28. The electric potential at point B that lies along the 3L perpendicular bisector of the rod a distance b = 2 above the x-axis is given by

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Chapter 1: Electrostatics 1.197

Comprehension 9



(C) 

Ex

Suppose that the electric potential varies along the x-axis as shown in Figure below. V(Volts) b 15

–3 –2 –1

e

1 2

3

1 2

3

x

10 5c –3 –2 –1

–5

1

d 2

3

x(m)

The potential does not vary in the y or z-direction. Of the intervals shown (ignore the behaviour at the end points of the intervals). The field Ex has a maximum absolute value of FB1 Vm −1 in region FB2. Its value in the region CD is FB3 Vm −1. Based on the above facts, answer the following questions. 29. The value that fills FB1 is



25 25 (A)  (B) 2 15 (C)  (D) −25 2

30. The region that fills FB2 is (A) ab (B) de (C)  bc (D) None of these 31. The value that fills FB3 is (A) 1 (C) 3

(B) 2 (D) ZERO

(A) 

–3 –2 –1

x

33. In the V vs x curve, the potential possesses a zero value at 12 5 m (B) x=− m (A) x = − 5 13

x=− (C) 

13 14 m (D) x=− m 5 5

Comprehension 10 A particle of charge +3 × 10 −9 C is in a uniform field directed toward left. It is released from rest and moves a distance of 5 cm after which its kinetic energy is found to be 4.5 × 10 −5 J. Based on the information supplied, answer the following questions. 34. What is the work that is being done by the electrical force? 45 μ J (A) 90 μ J (B) (C)  15 μ J (D) None of the above

32. The plot of Ex vs x is

Ex

(D) 

–10

a





Ex

35. The magnitude of the electric field is

–3 –2 –1



(B) 

1 2

3

x



3 MNC −1 (A) 0.3 MNC −1 (B)



(C)  0.5 MNC −1 (D) 1.5 MNC −1

36. The potential of starting point w.r.t. the end point is (A) 1 kV (B) 10 kV (C) 1.5 kV (D) 15 kV

Ex

Comprehension 11 –3 –2 –1

1 2

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 197

3

x

The nuclear charge ( Ze ) is non-uniformly distributed within a nucleus of radius R. The charge density ρ ( r ) (charge per unit volume) is depended only on the radial

9/24/2019 11:03:36 AM

1.198  JEE Advanced Physics: Electrostatics and Current Electricity distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction. ρ (r)

d

a

37.

R

r

The electric field at r = R is (A) independent of a (B) directly proportional to a (C) directly proportional to a 2 (D) inversely proportional to a

38. For a = 0, the value d (maximum value of ρ as shown in the figure) is

3 Ze 2 3 Ze (A)  3 (B) 4π R π R3



4 Ze Ze (C)  3 (D) 3π R 3π R3

39. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies

R a= (A) a = 0 (B) 2



2R (C)  a = R (D) a= 3

Comprehension 12

41. If an electron is placed at the centre of this arrangement and slightly displaced then it will execute SHM. R is The time period of oscillation assuming x < 2

(A) 2π

ε 0π mR3 15πε 0 mR3 2π (B) 8Qe 8Qe



2π (C) 

7πε 0 mR3 13πε 0 mR2 2π (D) 8Qe 6Qe



2α r ⎛ α R3 r⎞ (A)  2 + ⎜ 1 − ⎟⎠ R 24 r ε 0 3ε 0 ⎝



α R3 2α r ⎛ r⎞ (B)  2 + ⎜ 1 − ⎟⎠ R 8 r ε 0 6ε 0 ⎝



− (C) 

2α r ⎛ α R3 r⎞ + ⎜⎝ 1 − ⎟⎠ 2 R 16 r ε 0 3ε 0



(D) −

2α r ⎛ 3r ⎞ α R3 + ⎜1− ⎟ 4R ⎠ 96ε 0 r 2 3ε 0 ⎝

Comprehension 13 Two spherical cavities, of radii a and b are hollowed out from the interior of a neutral conducting sphere of radius R. At the centre of each cavity, a point charge is placed. Call these charges qa and qb , distance between qa and qb is r.

A region in space contains a total positive charge Q that is distributed spherically such that the volume charge density is given by



R ⎡ ⎢ α for r ≤ 2 ρ(r ) = ⎢ ⎢ 2α ⎛ 1 − r ⎞ for R ≤ r ≤ R ⎟ ⎢⎣ ⎜⎝ 2 R⎠

where α is a positive constant and ρ ( r ) = 0 for r ≥ R. Based on the above facts, answer the following questions. 40. The fraction of total charge contained in the region R r≤ is 2

4 8 (A)  (B) 15 15



7 11 (C)  (D) 15 15

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 198

R < r < R is 2

42. The electric field in a region

qa R qb

Based on the above facts, answer the following questions. 43. The force on charge qa is.

qa ( qa + qb ) qq (A)  a b 2 (B) 4πε 0 r 4πε 0 R2



( qa ) ( qa + qb ) (C)  4πε 0 r 2

(D) zero

44. The field outside the conductor at any point P at a distance r0 ( > R ) from centre of conductor?

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Chapter 1: Electrostatics 1.199



q qa + qb (A)  a 2 (B) 4πε 0 r0 4πε 0 R2





q + qb qb (C)  a (D) 4πε 0 r02 4πε 0 r02

where ρ0 is a constant. Assume ε as the permittivity of the ball, answer the following questions.

r⎞ ⎛ ρ = ρ0 ⎜ 1 − ⎟ ⎝ R⎠

49. The magnitude of the electric field as a function of the distance r outside the ball is given by

45. In each cavity, there is certain electric field say E. If another charge qc were brought near the conductor then E (A) increases. (B) decreases. (C) remains same. (D) depends on the nature of charge qc .



(A) E =

ρ0 R 3 ρ R3 E= 0 2 (B) 2 8ε r 12ε r



E= (C) 

ρ0 R 2 ρ0 R 2 E = (D) 8ε r 3 12ε r 3

Comprehension 14

50. The value of distance rm , at which electric field intensity is maximum, is given by

A small plastic ball of mass m covered with a thin zinc coating, is suspended by means of an insulating thread from a fixed point. There exists an electric field of magnitude E directed along horizontal. Now ultraviolet light is made to incident on the ball due to which it acquires a positive charge. Acceleration due to gravity is g. Based on the above facts, answer the following questions. 46. What is the origin of positive charge acquired by the ball? (A) Excess of protons in plastic ball (B) Deficiency of electrons in zinc coating (C) Deficiency of electrons in plastic ball (D) Some positively charged particles given by ultraviolet light 47. If the string makes an angle q with the vertical, when the ball comes to equilibrium, the number of electrons lost by the ball is mg mg sin q (A)  (B) eE eE

mg tan q mg cot q (C)  (D) eE eE

48. If the charge that suddenly appears on the ball is q, what is the maximum angle by which the string gets deflected?

⎛ mg ⎞ ⎛ 2mg ⎞ (B) tan −1 ⎜ (A) tan −1 ⎜ ⎝ qE ⎟⎠ ⎝ qE ⎟⎠ ⎛ qE ⎞ ⎛ qE ⎞ (C)  2 tan −1 ⎜ tan −1 ⎜ (D) ⎝ mg ⎟⎠ ⎝ mg ⎟⎠

Comprehension 15 A ball of radius R carries a positive charge whose volume charge density depends only on the distance r form the ball’s centre as.

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 199



(A) rm =

R 3R (B) rm = 2 3



rm = (C) 

2R 4R (D) rm = 3 3

51. The maximum electric field intensity is

(A) Em =

ρ0 R ρε Em = 0 (B) 9ε 9R



Em = (C) 

ρ0 R ρR (D) Em = 0 3ε 6ε

Comprehension 16 Two positive point charges A and B have charge +q and 2q, mass m and 2m respectively as shown. A +q, m

B l0

+2q, 2m

Both the charges are released from rest when they are at a distance l apart. Neglect gravity and also assume the only force acting on either charge is the electrostatic force due to each other. Based on the above facts, answer the following questions. 52. The speed of charge A at the instant separation between both charges is 2l0 is

(A) 

q2 q2 (B) 12πε 0 ml 6πε 0 ml



(C) 

q2 q2 (D) 4πε 0 ml 3πε 0 ml

53. The work done by electrostatic force on charge A while the separation between both charges changes from l to 2l is

9/24/2019 11:03:56 AM

1.200  JEE Advanced Physics: Electrostatics and Current Electricity q2 q2 (A)  (B) 12πε 0l 6πε 0l

57. The dipole moment of the ring in Cm is





( 2π R2λ ) iˆ (A) − ( 2π R2 λ ) iˆ (B)



q2 q2 (C)  (D) 4πε 0l 24πε 0l



( 4R2λ ) iˆ (C)  − ( 4 R2 λ ) iˆ (D)

54. Total work done by electrostatic force on charge A + charge B while the separation between both charges changes from l to 2l is

q2 q2 (A)  (B) 12πε 0l 6πε 0l



q2 q2 (C)  (D) 4πε 0l 24πε 0l

Comprehension 18 A metal sphere of radius R, carrying charge q1 is surrounded by a thick concentric metal shell (inner radius a, outer radius b). The shell carries no net charge.

R a

Comprehension 17

b

A thin ring of radius R metre is placed in x -y plane such that its centre lies at origin O. The half ring in region x < 0 carries uniform linear charge density + λ cm −1 and the remaining half ring in region x > 0 carries uniform linear charge density − λ cm −1. y

O +λ

x –λ

Based on the above facts, answer the following questions. 55. The electric potential (in volt) at point P whose coorR⎞ ⎛ dinates are ⎜ 0 , ⎟ m is ⎝ 2⎠

λ (A)  2πε 0



λ λ (C)  (D) 4πε 0 6πε 0

(B) 0

56. The direction of electric field at point P whose coordiR⎞ ⎛ nates are ⎜ 0 , ⎟ m is ⎝ 2⎠

(A) along positive x-direction

(B) along negative x-direction

(C) along negative y-direction

(D) not possible to be determined.

01_Physics for JEE Mains and Advanced - 2_Part 5.indd 200

Based on the above facts, answer the following questions. 58. The surface charge density σ at r = R, r = a and r = b is

σR = (A) 

q q q , σa = , σb = 4π R2 4π a 2 4π b 2



(B)  σR =

q −q q , σa = , σb = 4π R2 4π a 2 4π b 2



(C)  σR =

−q q q , σa = , σb = 4π R2 4π a 2 4π b 2



(D)  σR =

q q −q , σa = , σb = 4π R2 4π a 2 4π b 2

59. The potential at the centre, assuming potential to be zero at infinity is

q ⎛ 1 1⎞ q ⎛ 1 1 1⎞ (A)  ⎜ − ⎟ (B) ⎜ + + ⎟ 4πε 0 ⎝ R a ⎠ 4πε 0 ⎝ R a b ⎠



q q ⎛1 1 1⎞ (D) (C)  ⎜ − + ⎟ 4πε 0 R 4πε 0 ⎝ b a R ⎠

60. Now the outer surface is touched to a grounding wire, which lowers its potential to zero. Now the potential at the centre (Assume at infinity also potential is zero)

q ⎡ 1 1⎤ q ⎡ 1 1 1⎤ − (B) + + (A)  4πε 0 ⎢⎣ R a ⎥⎦ 4πε 0 ⎢⎣ R a b ⎥⎦



q q ⎡1 1 1 ⎤ − + (D) (C)  4πε 0 R 4πε 0 ⎢⎣ b a R ⎥⎦

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Chapter 1: Electrostatics 1.201

Matrix Match/COLUMN MATCH Type Questions Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of bubbles will look like the following: A B C D

1.

p p p p p

q q q q q

s

t

s s s s

t t t t

Match the forces given in COLUMN-I with the properties given in COLUMN-II COLUMN-I

COLUMN-II

(A) Electrostatic force of a nucleus on an electron

(p) Conservative

(B) The force attracting a man towards the centre of the earth

(q) Action-reaction force

(C) Force between earth and sun

(r) Depends on the nature of medium between the interacting objects

(D) Force between two protons

(s) Principle of Superposition is applicable (t) Repulsive

2.

r r r r r

COLUMN-I

COLUMN-II

(A)  Q = +q, displaced along x-axis

(p)  Force on Q before being displaced is zero

(B)  Q = -q, displaced along x-axis

(q) Potential energy of the system is maximum

(C) Q = q, displaced along y-axis

(r) Potential energy of the system is minimum

(D) Q = -q, displaced along y-axis

(s) The charge Q is in stable equilibrium (t) Executes SHM about O

3.

In the arrangement shown, identical charges, each +q are fixed at A and B respectively. O is the midpoint of AB = 2l . A charge Q is placed at O and can be displaced slightly either along x -axis or along y-axis and then released. Assume Q to move only along the axis along which it had been displaced. Now match the options in COLUMN-I to the information in COLUMN-II (regarding Q at O ) y

Match COLUMN-I with COLUMN-II COLUMN-I

COLUMN-II

(A) Force on an electron in an atom

(p) Gravitational force

(B) Force between a proton and a neutron inside nucleus

(q) Strong force

(C) Force between a proton and proton inside nucleus

(r) Coulomb force

(D) Conservative force

(s) Electric force (t) Spring force

+q A

+q O

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 201

B

x

4.

Match the facts given in COLUMN-I with the systems given in COLUMN-II

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1.202  JEE Advanced Physics: Electrostatics and Current Electricity

COLUMN-I

COLUMN-II

COLUMN-I

COLUMN-II

(A) Electric field strength is zero in the volume

(p) A non conducting solid sphere charged uniformly

(A)  E2

(p) decreases

(B)  E4

(q) is constant near it and then decreases

(C)  E1

(r) is discontinuous at O

(D)  E3

(s)  is constant

(B) Electric field strength (q) A conducting is non-zero in the spherical shell volume but zero at charged uniformly the centre (C) Electric field strength (r) A non conducting is maximum on the hollow sphere surface charged uniformly (D) Electric field strength outside the system varies inversely as the square of the distance from the centre 5.

(s) A conducting solid sphere charged uniformly

(t) first increases and then decreases 6.

In figure a conducting spherical shell of inner radius x and outer radius y is concentric with a larger conducting spherical shell of inner radius a and outer radius b . The inner shell has a total charge +3Q and the outer shell has a total charge +5Q . Consider a point P arbitrarily taken in the regions of the system of these two shells. Based on the facts provided, match the contents of COLUMN-I with the locations of the point P mentioned in COLUMN-II.

Figure (A) and (B) show a uniformly charged ring and a uniformly charged disc of large radius R, Figure (C) is a uniformly charged infinite thread and figure (D) shows a uniformly charged infinite sheet. In each figure let us consider the point O as the reference and then move away from O along the +x axis. If E1 , E2 , E3 and E4 be the respective fields, then match the contents of COLUMN-I with those of COLUMN-II. y

y

x

x

σ

O

Ring of finite radius Disc of large radius (A) (B)    y

y

σ

λ

O

x

O

x

Infinite sheet of charge Infinite line of charge (D) (C)   

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 202

a

y

COLUMN-I

λ

O

b xO

COLUMN-II

(A) Electric field strength (p) Outer surface of the larger spherical is zero shell (B) Electric field strength is non-zero

(q) Inner surface of the larger spherical shell

(C) Magnitude of charge on this surface is 3Q

(r) Outer surface of the smaller spherical shell

(D) Charge on this surface is +8Q

(s) Region between inner and outer surface of bigger shell (Continued)

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Chapter 1: Electrostatics 1.203

COLUMN-I

7.

COLUMN-II

COLUMN-I

COLUMN-II

(t) Region between inner and outer surface of smaller shell

(A)  E ≠ 0, V ≠ 0

(p) 

(B)  E = 0, V ≠ 0

(q) 

(C)  E = 0, V = 0

(r) 

(D)  E ≠ 0, V = 0

(s) 

For the electric dipole shown in figure, match the facts in COLUMN-I with the facts in COLUMN-II D –q A

+q a

O

B

C

2a

COLUMN-I

COLUMN-II

(A) Electric potential is zero

(p) Point A

(B) Direction of electric (q) Point C (on the axial field strength is line) opposite to the direction of electric dipole moment of the dipole (C) Direction of electric field strength is the same as that of electric dipole moment

(r) Point O

(D) Electric field strength at C and D for OC  a and OD  a are EC and ED

(s) Any point on the equatorial line of the dipole.

9.

(t)  EC > ED 8.

(t) 

Four charges, all having same magnitude are placed at the corners of a square of side a . At the centre of the square, the electrostatic field and potential due to the system of four charges is E and V respectively. Match the cases in COLUMN-I with the situations in COLUMN-II

Match COLUMN-I with COLUMN-II COLUMN-I

COLUMN-II

(A)  E ∝

1 r

(p) At large distance from the centre of the dipole

(B)  E ∝

1 r2

(q) At large distance from the centre, at the axis of a uniformly charged ring.

(C)  E ∝

1 r3

(r) Due to an infinite charged conducting plate.

(D)  E ∝

1 r0

(s) At large distance from the axis of a thin conducting charged cylinder (Continued)

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1.204  JEE Advanced Physics: Electrostatics and Current Electricity

COLUMN-I

COLUMN-II

COLUMN-I

COLUMN-II

(t) At a distance r from the axis of a thin conducting cylinder

(C)  MLT -3 A -1

(r)  ϕE

(D)  ML2T -3 A -1

(s)  e 0

10. Match COLUMN-I with the standard quantities/symbols in COLUMN-II COLUMN-I

COLUMN-II

(A) Jm-3

(p)  e0

2 -1

(C) C J m

-1

(D) farad

(r) 

U V2

(s) 

V E

(t) 

1 e 0 E2 2

a b

11. A particle of charge -q , mass m moves in a region of space between two plates from a plate at potential -V to the plate at potential +V . The plate separation is d. If K , U , T and E be the respective kinetic energy, potential energy, total mechanical energy of the particle and E be the electric field between the plates, then match the facts in COLUMN-I with those in COLUMN-II COLUMN-I

COLUMN-II

(A)  K

(p) constant

(B)  U

(q) first increase and then decrease

(C)  T

(r) increase

(D)  E

(s) decrease (t) other than those in (p), (q), (r) or (s)

12. Match the dimensional formula in COLUMN-I with the quantities in COLUMN-II COLUMN-I

COLUMN-II

(A)  M -1L-3T 4 A 2

(p)  V

(B)  ML3T -3 A -1

(q)  E

σ 2e 0

13. A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d as shown in figure. The inner shell has total charge +2q and the outer shell has charge -4q . Match the column for the electric field (magnitude and direction) in term of q and distance r from the common centre of two shells.

(q)  Pe

(B) metre

(t) 

c

d

COLUMN-I

COLUMN-II

(A)  r < a

(p)  E = 0

(B)  b < r < c

(q)  E =

(C)  c < r < d

(r)  radially inward

(D)  r > d

(s)  radially outward

q 2πe 0 r 2

14. A charged sphere is enclosed by a concentric neutral conducting sphere as shown. The three region are marked as I, II and III. Region I is inside hollow charged sphere, region II is outside charged sphere and inside uncharged (outer) sphere, while region III is outside to both. Match the entries of COLUMN-I with the entries of COLUMN-II. All answers have to be given in steady state. q III

II

I

S

(Continued)

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Chapter 1: Electrostatics 1.205

COLUMN-I

COLUMN-II

COLUMN-I

COLUMN-II

(A) When S is open, then in region II

(p) Electric field intensity is zero

(C) When S is closed, then in region I

(r) Electric field intensity is non zero

(D) When S is closed, then in region II

(s) Electrical potential is non zero

(B) When S is open, then (q) Electric potential is in region III zero

15. In each situation of COLUMN-I, some charge distributions are given with all details explained. The electrostatic potential energy and its nature is given situation in COLUMN-II. Then match situation in COLUMN-I with the corresponding results in COLUMN-II. COLUMN-I

COLUMN-II

(A) A thin shell of radius a and having a charge -Q uniformly distributed over its surface as shown

(p) 

Q2 in magnitude 8πe 0 a

(q) 

3Q 2 in magnitude 20πe 0 a

(r) 

2Q 2 in magnitude 5πe 0 a

–Q a

5a and having a charge -Q uniformly distributed 2 over its surface and a point charge -Q placed at its centre as shown

(B) A thin shell of radius

–Q 5a 2

(C) A solid sphere of radius a and having a charge -Q uniformly distributed throughout its volume as shown –Q a

(D) A solid sphere of radius a and having a charge -Q uniformly distributed throughout its volume. The solid sphere is surrounded by a concentric thin uniformly charged spherical shell of radius 2a and carrying charge -Q as shown

(s)  Positive in sign

–Q

a

–Q 2a

(t)  Negative in sign

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1.206  JEE Advanced Physics: Electrostatics and Current Electricity 16. In COLUMN-I various configuration of charges are given and in COLUMN-II Information about electric field E and electric potential V is given. Match COLUMN-I with COLUMN-II. COLUMN-I

COLUMN-II

(A) Square

(p) Potential at any point on z-axis is zero.

y

q

q

Centre

COLUMN-I

COLUMN-II

(A) Hollow hemisphere, at its centre

(p) 

λ 2 2πe 0 x

(q) 

Q 8πe 0 R2

(r) 

3Q 8πe 0 R2

(s) 

λx ⎛ 3x 2 ⎞ 1 ⎜ ⎟ 2e 0 R2 ⎝ 2R 2 ⎠

Q R

E

x –q

(B) Solid hemisphere, at its centre

–q

(B)  Equilateral triangle y –2q

q

Centroid x

y

+q

x

(D)  Semi-infinite wire

–q

(s)  z-component of electric field intensity at any point on z-axis is zero.

y

+q

+q

λ

x

E

18. In a certain region of x -y plane, potential function is given in COLUMN-I and corresponding electric lines of force are given in COLUMN-II. Match the potential function of COLUMN-I with their respective field line representation in COLUMN-II.

+q +q x

Square

E

Centre

(D) Both squares are equidistant from origin

+q

R

–q

–q

+q

Q

(r) Potential energy of system is negative.

(C)  Regular hexagon

+q

E

(C) On the axis of ring, near its centre

q

+q

R

Q

(q)  y-component of electric field intensity at any point on z-axis is zero.

+q +q Square

(t) Potential is zero but electric field intensity is non-zero at z-axis. 17. Match the column for electric field given in COLUMN-II for the uniformly charged bodies in COLUMN-I.

COLUMN-I (Potential Function) (A) 

V = x2 - y 2

COLUMN-II (Electric Lines of Force) (p) 

y

x

(Continued)

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Chapter 1: Electrostatics 1.207

COLUMN-I (Potential Function)

COLUMN-II (Electric Lines of Force)

(B)  V = xy

(q) 

COLUMN-I

COLUMN-II

(A)  S1

y

(B)  S2

x

(C)  S3 (r) 

(C)  V = x 2 + y 2

(p) 

3q 8e 0

(q) 

q 2e 0

(r) 

q 6e 0

(s) 

q 4e 0

y

(D)  S4 x

(D)  V =

(s) 

x y

20. Two spherical shells shown in figure have uniformly distributed charge q1 and q2 . If r is the distance of a point from common centre and

y

x













E1 is electric field for r < R1 , E2 is electric field for R1 < r < R2 , V1 is electric potential for r < R1 and

V is electric potential for R1 < r < R2 , 2 then match the quantities in COLUMN-I with their respective property(ies) in COLUMN-II. q2

19. Consider the imaginary surfaces S1 , S2 , S3 and S4 drawn near a point charge as shown in figure.

q1

Infinite sheet

a

R1 O

q

a 2

a S1

a

q

S2

R2

a 15 a S3

COLUMN-I

S2

q

a 15 a

S4

q

S3

q

21. COLUMN-I gives different surfaces and COLUMN-II corresponding electric flux. Match the entries of COLUMN-I to COLUMN-II.

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 207

q

COLUMN-II

(A)  E1

(p) is constant for q2 and varies for q1.

(B)  V1

(q) is zero for q2 and varies for q1

(C)  V2

(r)  is constant

(D)  E2

(s)  is zero

Infinite sheet

a

S4

q

n identical drops of a liquid each having charge q and radius r coalesce. Match the unfilled places in COLUMN-I with their respective fills given in COLUMN-II.

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1.208  JEE Advanced Physics: Electrostatics and Current Electricity

COLUMN-I

COLUMN-II

COLUMN-I

COLUMN-II

(A)  Ebig = ....Esmall

(p)  n1 3

(B) 

(q)  E

(B)  Vbig = ....Vsmall

(q)  n3

(C)  C big = ....Csmall

(r)  n2 3

(D)  σ big = ....σ small

(s)  n

+Q

(C) 

r

+Q

(r)  E

-1 3

r

22. Entries in COLUMN-I show four hollow metal spheres each with internal radius a and external radius b . Match each charge distribution in COLUMN-I with the corresponding graph of electric field versus radial distance r from centre as shown in COLUMN-II. COLUMN-I (A) 

(D) 

+2Q +Q

(s)  E r

COLUMN-II –Q

(p)  E

+Q

r

(Continued)

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s). 1.

A ring of radius 0.1 m is made out of a thin metallic wire of area of cross-section 10 -6 m 2 . The ring has a uniform charge of π coulomb. The change in the radius of the ring, when a charge of 10 -8 coulomb is placed at the centre of the ring is x × 10 -5 m. Find x, if Three identical small balls, each of mass 0.1 g, are suspended from one point by silk threads each having a length of l = 20 cm. What charge, in nC, must be imparted to each ball so as to make each thread form

Four point charges each of magnitude 1 μC (each positive) are fixed at the vertices of a square ABCD. The diagonals of the square intersect at O. Length of each diagonal is 2 m. A particle of mass 2 g having a charge +2 μC is released from a point 1 cm from O and 99 cm from A. The motion of the particle is SHM with

an angle α = 30° with the vertical?

period

the young’s modulus of the metal is 2 × 1011 Nm -2 . 2.

3.

the force of friction between the drop and the air to be proportional to the velocity of the drop, find the time t, in millisecond during which the drop travels the same distance after the field is switched off.

In classical experiments performed to measure the charge of an electron, a charged drop of oil is placed between the horizontal plates of a plane capacitor. Under the action of an electrostatic field, the drop moves uniformly upward, covering a certain distance during the time t1 = 6 ms, or downward, when the sign of the charge on the plates changes, covering the same distance during the time t2 = 3 ms. Assuming

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 208

4.

5.

π x s. Find the value of x. (Neglect gravity). 3

An electron is projected with an initial speed u = 4 × 106 ms -1 into the uniform field between the parallel plates in figure. The direction of the field is vertically downward, and the field is zero except in the space between the plates. The electron enters the field at a point midway between the plates. If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field in NC -1.

9/20/2019 10:34:40 AM

Chapter 1: Electrostatics 1.209 2 cm u

1 cm

E

6.

7 .

A uniform surface charge density 8σ exists over the entire x -y plane except for a circular hole of radius a centred at the origin. The electric field at a point σ P ( 0, 0, 3 a ) on the z-axis is found to be x . Find e0 the value of x . Two fixed, equal, positive charges, each of magnitude 5 × 10 -5 C are located at points A and B separated by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge when it reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 joules. The distance of the farthest point D, in metre, where the negative charge will reach before returning towards C is

x . Find x . A

+q 3m

O D

4m

C

3m B

8.

+q

Positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R. A point mass having charge +q and mass m is fired towards the centre of the sphere with velocity v from a point 5R A at distance from the centre of the sphere. The 2 R disminimum velocity v so that it can penetrate 2 Qqx tance of the sphere is . Neglect any resistance 2πe 0 m other than electric interaction. Charge on the small mass remains constant throughout the motion. Find the integral value of x to the nearest approximation.

9.

A particle A having a charge q is fixed on a vertical (insulated) wall. A second particle B of mass m, charge Q is suspended by a silk thread of length l from a point P on the wall that is at a distance l above the A. Calculate the angle in degree made by the thread with the vertical, when B stays in equilibrium for Qq = 4πe 0 mgl 2.

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 209

10. Two positively charged particles each of mass 1.7 × 10 -27 kg having charge of 1.6 × 10 -19 C are placed at separation r. Calculate r, (in cm) to the nearest integer if each one experiences a repulsive force equal to its weight. Take g = 10 ms -2 .

(

)

11. The bob of a simple pendulum has a mass of 40 g and a positive charge 4 × 10 -6 C. It makes 20 oscillation in 45 s. A vertical electric field pointing upward and of magnitude 2.5 × 10 4 NC -1 is switched on. How much time, in seconds will it now take to complete 20 oscillations. 12. A thin fixed ring of radius 1 meter has a positive charge 1 × 10 -5 coulomb uniformly distributed over it. A particle of mass 0.9 g and having a negative charge of 1 × 10 -6 coulomb is placed on the axis at a distance of 1 cm from the centre of the ring. Show that the motion of the negatively charged particle is approximately π simple harmonic. The time period of oscillations is . k Find k . 13. An inclined plane making an angle 30° with the horizontal is placed in a uniform horizontal electric field E of 100 Vm -1 as shown. A particle of mass 1 kg and charge 0.01 C is allowed to slide down from rest from a height of 1 m. If the coefficient of friction is 0.2, find the time, in millisecond it will take the particle to reach the bottom. (Take g = 10 ms -2 ) E = 100 vm–1

q 1m

30°

14. A ball of mass m with a charge q can rotate in a vertical plane at the end of a string of length l = 3.6 m in a uniform electrostatic field whose lines of force are directed upwards. What horizontal velocity, in ms -1, must be imparted to the ball in the upper position so that the tension in the string in the lower position of the ball is 15 times than the weight of the ball? Given 6 mg and g = 10 ms -2 . E= 5q 15. Two equally charged identical metal spheres A and B repel each other with a force 2 × 10 -5 N. Another identical uncharged sphere C is touched to B and then placed at the midpoint between A and B. What is the net electric force, in μN on C ? 16. Two identically charged spheres are suspended by strings of equal length. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 gmcc -1 the angle remains same. What is the dielectric constant of liquid. Density of sphere = 1.6 gcc -1.

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1.210  JEE Advanced Physics: Electrostatics and Current Electricity If this particle moves with a uniform horizontal speed 0.02 ms -1 , find the number of electrons in it. 30°

22. Two circular wire loops of radii 0.05 m and 0.09 m, respectively are put such that their axes coincide and their centres are 0.12 m apart. Charge of 10 -6 C is spread uniformly on each loop. Find the potential difference, in volt between the centres of the loops.

15°

17. Two identical pith balls are charged by rubbing against each other. They are suspended from a horizontal rod through two strings of length 20 cm each. The separation between the suspension points being 5 cm. In equilibrium the separation between the balls is 3 cm. Find the mass m in gram, nearest to an integer, of each ball and the tension in millinewton, nearest to two digit integer, in the strings. The charge on each ball has magnitude 2 × 10 -8 C . 3

⎛ 11 ⎞ 2 1 8. A charge Q1 = +11 ⎜ ⎟ × 10 -9 C is located at the ori⎝ 3⎠ gin in free space and another charge Q at ( 2, 0 , 0 ). If the x-component of the electric field at ( 3 , 1, 1 ) is zero, calculate the value of Q in nC.

19. Two similar helium filled spherical balloons tied to a 5 g weight with strings each carrying an electric charge q float in equilibrium as shown in figure. q

A

B

q

B

A r1 R1

C2

C1

R2 r2 d = 0.12 m

23. The radii of internal and external spheres of concentric spherical air capacitor are 1 cm and 4 cm respectively. A potential difference of 3000 V is applied between the spheres. A velocity x × 10 5 ms -1 will be imparted to an electron when it approaches from a distance of r1 = 3 cm to r2 = 3 cm as measured from the centre of spheres. Then find x closest to three digit integer. 24. Three point charges of 1C, 2C and 3C are placed at the corners of an equilateral triangle of side 1 m. The work required to move these charges to the corners of a smaller equilateral triangle of sides 0.5 m as shown in Figure is x × 109 J. Find x 1C

m 1m

(a) Find the magnitude of q , in μC , assuming that the charge on each balloon as if it were concentrated at its centre. (b) Find the volume of each balloon, in cc (neglect the weight of the unfilled balloons and assume that the density of air 1.29 kgm -3 and density of helium inside the balloon 0.2 kgm -3 ). 5 20. A charge Q = nC is distributed over two concen100 tric hollow spheres of radii r = 3 cm and R = 6 cm such that the surface densities are equal. Find the potential, in volt, at the common centre.

25. Three point charges of 0.1 C are placed at the corners of an equilateral triangle with side L = 1 m. If this system is supplied energy at the rate of 1 kW, how much time, in hours will be required to move one of the charges on to the mid point of the line joining the two charges.

21. A charged dust particle of radius 5 × 10 -7 m is located in a horizontal electric field having an intensity of 6.28 × 10 5 Vm -1. The surrounding medium is air with coefficient of voscosity η = 1.6 × 10 -5 Nsm -2.

26. An electron enters a region between the plates of a parallel plate capacitor at a point equidistant from either plate. The capacitor plates are 2 × 10 -2 m apart and 10 -1 m long. A potential difference of 300 V is



01_Physics for JEE Mains and Advanced - 2_Part 6.indd 210

0.5 m 2C

3C

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Chapter 1: Electrostatics 1.211 maintained across the plates. Assuming that the initial velocity of the electron is parallel to the capacitor plates, the largest value of the velocity of the electrons so that they do not fly out of the capacitor at the other end is x × 10 5 ms -1. Taking, me = 9 × 10 -31 kg and e = 1.6 × 10 -19 coulomb, find the value of x closest to three digit integer. 27. Two plane parallel conducting plates 1.5 × 10 -2 m apart are held horizontal one above the other in air. The upper plate is maintained at positive potential of 1.5 kV while the other plate is earthed. (a) Calculate the number of electrons which must be attached to a small oil drop of mass 4.9 × 10 -15 kg between the plates to maintain it at rest, assuming that the density of air is negligible in comparison with that of oil. (b) If the potential of above plate is suddenly changed to -1.5 kV and what is the initial acceleration, in ms -2 , of the charged drop? (c) Also obtain the terminal velocity of the drop, in μms -1 if its radius is 5 × 10 -6 m and the coefficient of viscosity of air is 1.8 × 10 -5 Nsm -2 .

to 0.2 m from the conductor. Give your answer closest to the two digit integer, in this case. 30. Two concentric spherical shells have radii 0.15 m and 0.3  m. The inner sphere is given a charge -6 × 10 -2 μC. An electron escapes from the inner sphere with negligible speed. Assuming that the region between the spheres is vacuum, the speed with which it will strike the outer sphere is ∗ × 106 ms -1, where ∗ is not readable. Find the value of ∗. ⎛ e ⎞ for electron = 1.76 × 1011 Ckg -1 ⎟ . ⎜⎝ ⎠ m 31. A non-conducting sphere of radius R = 5 cm is centred at origin of coordinate system. It has a spherical cavity of radius 1 cm having centre at ( 0 , 3 ) cm . The material of sphere has a uniform charge density 10 -6 ρ= Cm -3. Calculate the electric potential at point π P ( 4 cm, 0 ) to the nearest two digit integer. Y

( Take g = 10 ms-2 ) X

28. A positively charged sphere of mass m = 5 kg is attached by a spring of force constant K = 10 4 Nm -1. The sphere is tied with a thread so that the spring is in its natural length. Another identical, negatively charged sphere is fixed to the floor, vertically below the positively charged sphere as shown in figure. Initial separation between sphere is x0 = 0.5 m. Now thread is burnt and maximum elongation of the spring is 0.1 m, calculate the charge, in μC on each sphere. Take g = 10 ms -2 .

32. Two small identical balls P and Q, each of mass 0.866 g, carry identical charges and are suspended by two threads of equal length of 50 cm each from the same point. At equilibrium, they are separated by a distance of 50 cm. Find the charge on each ball in nC. (Take g = 10 ms -2 ).

x0

33. A particle having a charge of q = 8.85 μC is placed on the axis of a circular ring of radius R = 30 cm at a point P at a distance of a = 40 cm from centre of ring. Calculate the electric flux passing through the ring in kNC -1.

Initial position

29. An infinite long straight conductor is uniformly charged, the charge per unit length being 6 × 10 4 Cm -1 . (a) What is the force experienced by a proton, in μN, at a perpendicular distance 0.5 m from it? (b) How much work in mJ will be required to bring the proton at a perpendicular distance from 0.5 m

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 211

O

P

34. A particle of mass 2 μg and carrying a charge of 1 × 10 -9 C is moving directly towards a fixed positive point charge of magnitude 10 -8 C. When it is at a distance of 10 cm from the fixed point charge it has a velocity of 1 ms -1 . At what distance from the fixed point charge will the particle come momentarily to rest? (in mm)

9/20/2019 10:35:29 AM

1.212  JEE Advanced Physics: Electrostatics and Current Electricity 35. A copper atom consists of copper nucleus surrounded by 29 electrons. The atomic weight of copper is 63.5 gmol -1. Let us now take two pieces of copper each weighing 10 g. Let us transfer one electron from one piece to another for every 1000 atoms in that piece. The coulomb force between the two pieces after the transfer of electrons if they are 1 cm apart is x × 106 N . Find x, close to the nearest integer. (Avogadro number N A = 6 × 10 23 gmole -1, charge on an electron = -1.6 × 10 -19 coulomb) 36. It is required to hold four equal point charges each 8 having a charge Q = ( 1 - 2 2 ) C in equilibrium at 7 the corners of a square. Find the point charge, in coulomb, that will do this if placed at the centre of the square. 37. A radioactive source in the form of a metal sphere of radius 10 -2 m , emits beta particles (electrons) at the rate of 5 × 1010 particles per second. The source is electrically insulated. How long, in μs will it take for its potential to be raised by 2 V assuming that 40% of the emitted beta particles escape the source. 38. A small ball of mass 2 × 10 -3 kg having a charge of 1 μC is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity in cms -1 which should be imparted to the lower ball so that it can make complete revolution.

q

u q, m

39. The centres of two metal spheres of radius 10 cm are 50 cm apart on the x-axis. The spheres are initially neutral, but a charge Q is transferred from one sphere to the other, creating a potential difference 100 V. A proton is released from rest from the surface of the positively charged sphere and travels to the negatively charged sphere. At what speed, in kms -1 does it strike the negatively charged sphere? 40. A simple pendulum with a bob of mass m = 1 kg, charge q = 5 μC and string length l = 1 m is given a horizontal velocity u in a uniform electric field E = 2 × 106 Vm -1 at its bottommost point A, as shown in figure. It is given that the speed u is such that the

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 212

particle leaves the circle at point C. Find the speed u to the nearest integer. Take g = 10 ms -2

(

)

C

60° B l A

u

E

41. Two insulating spheres have radii 0.3 cm and 0.5 cm, masses 0.1 kg and 0.7 kg, and uniformly distributed charges of -2 μC and 3 μC . They are released from rest when their centers are separated by 1 m. (a) How fast in cms -1 will each be moving when they collide? (b)  If the spheres were conductors, would the speeds be greater or less than those as calculated in part (a)? Give reasonable explanation to your answer. 42. A particle having mass m = 1 kg and charge q = 2 μC is thrown from a horizontal ground with a speed u = 20 ms -1 that makes an angle α = 45° with the horizontal. A horizontal electric field E = 2 × 107 Vm -1 exist in this space. Find the range, in metre on the horizontal ground of the projectile thrown. 43. A ball of mass 2 kg, charge 1 μC is dropped from top of a high tower. In space electric field exist in horizontal direction away from tower which varies as E = ( 5 - 2x ) × 106 Vm -1 . Find maximum horizontal distance that the ball can go from the tower. 44. A particle having a charge of 1.6 × 10 -19 C enters midway between the plates of a parallel plate condenser. The initial velocity of particle is parallel to the plates. A potential difference of 300 volts is applied to the capacitor plates. If the length of the capacitor plates is 10 cm and they are separated by 2 cm, calculate the greatest initial velocity, in kms -1 for which the particle will not be able to come out of the plates. The mass of the particle is 12 × 10 -24 kg . 45. A uniform electric field E is created between two parallel charged plates as shown in figure. An electron enter the field symmetrically between the plates with a speed u. The length of each plate is l. Find the angle of deviation, in degree of the path of the electron as it comes out of the field for u2 =

3Ele . m

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Chapter 1: Electrostatics 1.213 m

q

L

P

l

46. A particle having charge q = +2 μC and mass m = 0.01 kg is connected to a string that is L = 1.5 m long and is tied to the pivot point P in figure. The particle, string and pivot point all lie on a frictionless horizontal table. The particle is released from rest when the string makes an angle θ = 60° with a uniform electric field of magnitude E = 300 Vm -1. Determine the speed of the particle in cms -1 when the string becomes parallel to the electric field (point A in figure).

θ

A

E

47. A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ = kr a , where k and a are constants and r is the distance from its cenR 1 tre. If the electric field at r = is times that at r = R, 2 8 find the value of a .

ARCHIVE: JEE MAIN 1. [Online April 2019]  The bob of a simple pendulum has mass 2 g and a charge of 5.0 μC . It is at rest in a uniform horizontal electric field of intensity 2000 Vm -1 . At equilibrium, the angle that the pendulum makes with the vertical is (take g = 10 ms -2 ) (A) tan -1 ( 0.2 ) (B) tan -1 ( 5.0 ) (C) tan

-1 (

-1 (

2.0 ) (D) tan 0.5 )

2. [Online April 2019]  A solid conducting sphere, having a charge Q , is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of -4Q , the new potential difference between the same two surfaces is -2V (B) V (A) (C) 2V (D) 4V 3. [Online April 2019]   The electric field in a region is given by E = ( Ax + B ) iˆ , where E is in NC -1 and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = -5 is V2 , then V1 - V2 is

mass m . It is kept in a uniform electric field E . If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is 2qE qE (B) 2 (A) md md qE qE (D) (C) 2md md 5. [Online April 2019]  A positive point charge is released from rest at a distance r0 from a positive line charge with uniform ­density. The speed ( v ) of the point charge, as a function of instantaneous distance r from line charge, is proportional to

r0

r

+ ⎛ r⎞ (A) v ∝ ln ⎜ ⎟ (B) v ∝ e r0 ⎝ r0 ⎠

(A) 180 V (B) -520 V (C) 320 V (D) -48 V

⎛ r⎞ ⎛ r⎞ (C) v ∝ ln ⎜ ⎟ (D) v∝⎜ ⎟ ⎝ r0 ⎠ ⎝ r0 ⎠

4. [Online April 2019]  An electric dipole is formed by two equal and opposite charges q with separation d . The charges have same

6. [Online April 2019]  A system of three charges are placed as shown in the figure

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 213

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1.214  JEE Advanced Physics: Electrostatics and Current Electricity D +q

–q d

E , as shown in figure. Its bob has mass m and charge q . The time period of the pendulum is given by

Q

If D  d , the potential energy of the system is best given by 1 ⎛ q2 qQd ⎞ 1 ⎛ q2 qQd ⎞ (A) ⎜ - - 2 ⎟ (B) ⎜ + + 2 ⎟⎠ 4πe 0 ⎝ d 4πe 0 ⎝ d D ⎠ D 1 ⎛ q2 2qQd ⎞ 1 ⎛ q2 qQd ⎞ (C) ⎜ - + (D) ⎟ ⎜- ⎟ 2 4πe 0 ⎝ d 4πe 0 ⎝ d 2D2 ⎠ D ⎠ 7. [Online April 2019]  Four point charges -q , +q , +q and -q are placed on y-axis at y = -2d , y = - d , y = + d and y = +2d , ­respectively. The magnitude of the electric field E at a point on the x-axis at x = D , with D  d , will behave as E∝ (A)

1 1 (B) E∝ D D3

E∝ (C)

1 1 E∝ 2 (D) D4 D

8. [Online April 2019]  A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centred at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4 a . The minimum value of v such that it crosses the origin is 1

1

2 ⎛ 1 q2 ⎞ 2 2 ⎛ 1 q2 ⎞ 2 (A) ⎜ (B) m ⎝ 15 4πe 0 a ⎟⎠ m ⎜⎝ 5 4πe 0 a ⎟⎠ 1

1

2 ⎛ 4 q2 ⎞ 2 2 ⎛ 2 q2 ⎞ 2 (C) ⎜ (D) ⎜ ⎟ m ⎝ 15 4πe 0 a ⎠ m ⎝ 15 4πe 0 a ⎟⎠

9. [Online April 2019]  In free space, a particle A of charge 1 μC is held fixed at a point P . Another particle B of the same charge and mass 4 μg is kept at a distance of 1 mm from P . If B is released, then its velocity at a distance of 9 mm 1 from P is [Take = 9 × 109 Nm 2 C -2 ] 4πe 0

q, m

(A) 2π

2π (C)

1 0. [Online April 2019]  A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 214

L L 2π (B) qE ⎞ qE ⎞ ⎛ ⎛ ⎜⎝ g ⎜⎝ g + ⎟ ⎟ m⎠ m⎠ L ⎛ qE ⎞ g2 + ⎜ ⎝ m ⎟⎠

2

2π (D)

L g2 -

q 2E2 m2

1 1. [Online April 2019]   A point dipole p = - p0 xˆ is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively (Take V = 0 at infinity)     p p p -p (B) , (A) 2 , 3 2 4πe 0 d 4πe 0 d 3 4πe 0 d 4πe 0 d   -p p (C) 0, (D) 0, 4πe 0 d 3 4πe 0 d 3 1 2. [Online April 2019]  Shown in the figure is a shell made of a conductor. It b , and carries has inner radius a and outer radius  charge Q . At its centre is a dipole P as shown. In this case

P



2.0 × 10 3 ms -1 (B) 3.0 × 10 4 ms -1 (A) (C) 1.5 × 10 2 ms -1 (D) 6.3 × 10 4 ms -1

L

E

(A)  Surface charge density on the outer surface  depends on P (B)  Surface charge density on the inner surface is ­uniform and equal to



⎛ Q⎞ ⎜⎝ ⎟⎠ 2

4π a 2 (C) Electric field outside the shell is the same as that of point charge at the centre of the shell (D) Surface charge density on the inner surface of the shell is zero everywhere

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Chapter 1: Electrostatics 1.215 1 3. [Online April 2019]  Let a total charge 2Q be distributed in a sphere of radius R , with the charge density given by ρ ( r ) = kr , where r is the distance from the centre. Two charges A and B , of -Q each, are placed on diametrically opposite points, at equal distance, a , from the centre. If A and B do not experience any force, then a= (A)

3R 1 24

(C) a=2

-

R (B) a= 3

1 4R

-

1

(D) a = 8 4R

1 4. [Online January 2019]  Three charges +Q , q , +Q are placed respectively, at d distance, 0, and d from the origin, on the x-axis . 2 If the net force experienced by +Q , placed at x = 0 , is zero then value of q is

a 1 a Q ⎞ ⎞ ⎛ ⎛ (A) log (B) log ⎜ 1 ⎟ ⎜ ⎟ ⎝ Q 2π aA ⎠ 2 2 ⎜⎝ 1 ⎟⎠ 2π aA Q ⎞ 1 ⎛ ⎞ ⎛ (C) a log ⎜ 1 a log ⎟ (D) ⎜ ⎝ Q ⎟ 2π aA ⎠ ⎜⎝ 1 ⎟ 2π aA ⎠ 1 8. [Online January 2019]  Two electric  dipoles, A , B with respective dipole moments dA = -4 qaiˆ and dB = -2qaiˆ are placed on the x-axis with a separation R , as shown in the figure. R A

B

X

The distance from A at which both of them produce the same potential is

Q Q + (B) (A) 2 2

2R 2R (A) (B) 2 -1 2 +1

Q Q (C) - (D) + 4 4

R R (C) (D) 2 -1 2 +1

1 5. [Online January 2019]  For a uniformly charged ring of radius R , the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is

1 9. [Online January 2019]  A charge Q is distributed over three concentric spherical shells of radii a , b , c ( a < b < c ) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a , would be

R R (A) (B) 2 5 R (D) R 2 (C) 16. [Online January 2019]

 Two point charges q1 ( 10 μC ) and q2 ( -25 μC ) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in Vm -1 ) at a point y = 3 m 1 ⎡ ⎤ = 9 × 109 Nm 2 C -2 ⎥ on y-axis is, ⎢ Take 4πe 0 ⎣ ⎦

Q( a + b + c ) Q (A) (B) 4πe 0 ( a + b + c ) 4πe 0 ( a 2 + b 2 + c 2 ) Q ( a2 + b 2 + c2 ) Q ab + bc + ca (C) (D) 12πe 0 abc 4πe 0 ( a 3 + b 3 + c 3 ) 2 0. [Online January 2019]  Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), ( 4 , - 2 ) and ( 0 , - 2 ) . The work required to put a fifth charge Q at the origin of the coordinate system will be

( ) (B) ( 81iˆ - 81ˆj ) × 102 (C) ( -81iˆ + 81ˆj ) × 102 (D) ( -63iˆ + 27 ˆj ) × 102

1 ⎞ Q2 Q2 ⎛ 1+ (B) (A) ⎟ 4πe 0 ⎜⎝ 3⎠ 2 2πe 0

1 7. [Online January 2019]  Charge is distributed within a sphere of radius R with 2r A volume charge density ρ ( r ) = 2 e a , where A and r a are constants. If Q is the total charge of this charge distribution, the radius R is

2 1. [Online January 2019]  Charges -q and +q located at A and B , respectively, constitute an electric dipole. Distance AB = 2 a , O is the mid point of the dipole and OP is perpendicular to AB . A charge Q is placed at P where OP = y and

(A) 63iˆ - 27 ˆj × 10 2

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 215

1 ⎞ Q2 Q2 ⎛ 1+ (C) (D) ⎟ 4πe 0 4πe 0 ⎜⎝ 5⎠

9/20/2019 10:36:57 AM

1.216  JEE Advanced Physics: Electrostatics and Current Electricity y  2 a . The charge Q experiences an electrostatic force F . If Q is now moved along the equatorial line ⎛ y⎞ to P ′ such that OP ′ = ⎜ ⎟ , the force on Q will be ⎝ 3⎠ ⎛y ⎞ close to ⎜  2 a ⎟ ⎝3 ⎠ P

A

Q

–q

O

P′ B +q

-9 × 10 -20 J (B) -10 × 10 -29 J (A) (C) -7 × 10 -27 J (D) -20 × 10 -18 J 2 5. [Online January 2019]  There is a uniform spherically symmetric surface charge density at a distance R0 from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V ( R ( t ) ) of the distribution as a function of its instantaneous radius R ( t ) is (A) V(R(t))

(B) V(R(t))

F (A) (B) 9F 3 (C) 27F (D) 3F 2 2. [Online January 2019]  The given graph shows variation (with distance r from centre) of

R(t)

R0

(C) V(R(t))

R(t)

R0



(A) (B) (C) (D)

r

Potential of a uniformly charged spherical shell Electric field of a uniformly charged sphere Electric field of uniformly charged spherical shell Potential of a uniformly charged sphere

R0

R(t)

2 6. [Online January 2019]  Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure y –2q

2 3. [Online January 2019]  Three charges Q, +q and +q are placed at the vertices of a right-angle isosceles triangles as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is

l

+q

Q

R(t)

(D) V(R(t))

r0

r0

R0

l l

+q

x

⎛ ˆj - iˆ ⎞ (A) 2qljˆ (B) 3 ql ⎜ ⎝ 2 ⎟⎠ +q

+q

- 2q (B) (A) +q 2 +1 -q (C) -2q (D) 1+ 2 2 4. [Online January 2019]  An electric field of 1000 Vm -1 is applied to an electric dipole at angle of 45° . The value of electric dipole moment is 10 -29 Cm . What is the potential energy of the electric dipole?

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 216

⎛ iˆ + ˆj ⎞ (C) - 3qljˆ (D) ql ⎜ ⎝ 2 ⎟⎠ 2 7. [Online January 2019]  A parallel plate capacitor with plates of area 1 m 2 each, are at a separation of 0.1 m . If the electric field between the plates is 100 NC -1 , the magnitude of charge on each plate is (Take e 0 = 8.85 × 10 -12 C 2 N -1m -2 ) 8.85 × 10 -10 C (B) 9.85 × 10 -10 C (A) (C) 6.85 × 10 -10 C (D) 7.85 × 10 -10 C

9/20/2019 10:37:13 AM

Chapter 1: Electrostatics 1.217 28. [2018] Three concentric metal shells A, B and C of respective radii, a, b and c ( a < b < c ) have surface charge densities +σ , -σ and +σ respectively. The potential of shell B is ⎞ ⎞ σ ⎛ a2 - b 2 σ ⎛ a2 - b 2 (A) ⎜ + c ⎟ (B) + c⎟ ⎜⎝ ⎠ e0 ⎝ a e b ⎠ 0 ⎞ ⎞ σ ⎛ b2 - c2 σ ⎛ b2 - c2 + a ⎟ (D) + a⎟ (C) ⎜ ⎜⎝ ⎝ ⎠ ⎠ e0 b e0 c 29. [Online 2018] a above the centre A charge Q is placed at a distance 2 of the square surface of edge a as shown in the figure. The electric flux through the square surface is P a/2

a

Q Q (A) (B) 3e 0 6e 0 Q Q (C) (D) e0 2e 0 30. [Online 2018] A solid ball of radius R has a charge density ρ given r⎞ ⎛ by ρ = ρ0 ⎜ 1 - ⎟ for 0 ≤ r ≤ R. The electric field out⎝ R⎠ side the ball is

ρ R3 4 ρ0 R 3 (A) 0 2 (B) 12e 0 r 3e 0 r 2 3ρ R3 ρ0 R 3 (C) 0 2 (D) 4e 0 r e0r 2 31. [Online 2018] Two identical conducting spheres A and B , carry equal charge. They are separated by a distance much larger than their diameters and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B and then removed. As a result, the force between A and B would be equal to

3F F (A)  (B) 8 2



3F (C)  (D) F 4

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 217

32. [Online 2018] A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x-direction about its equilibrium position, taken to be at x = 0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct? 1 1 q 2E2 (A) The total energy of the system is mω 2 A 2 . 2 2 k

2qE (B) The new equilibrium position is at a distance k from x = 0. qE (C) The new equilibrium position is at a distance 2k from x = 0. 1 1 q 2E2 . (D) The total energy of the system is mω 2 A 2 + 2 2 k

33. [2017]   An electric dipole has a fixed dipole moment p, to x-axis. When which makes angle θ with respect  ˆ , it experiences a = subjected to an electric field E Ei 1  torque T1 = τ kˆ . When subjected to another electric field    E = 3E ˆj it experiences a torque T = -T . The angle 2

θ is

1

2

1



(A) 30° (B) 45°



(C)  60° (D) 90°

34. [Online 2017] There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between the limits 589 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60° with the direction of the field? (A) 589.2 V (B) 589.6 V (C) 589.5 V (D) 589.4 V 35. [Online 2017] Four closed surfaces and corresponding charge distributions are shown below. q –q q q

2q S1

S2

5q q

8q q

S3

3q

–2q –4q S4

Let the respective electric fluxes through the surfaces be Φ1, Φ 2, Φ 3 and Φ 4 . Then

9/20/2019 10:37:35 AM

1.218  JEE Advanced Physics: Electrostatics and Current Electricity

Φ1 > Φ 3 ; Φ 2 < Φ 4 (A) Φ1 = Φ 2 = Φ 3 = Φ 4 (B)



Φ1 > Φ 2 > Φ 3 > Φ 4 (D) Φ1 < Φ 2 = Φ 3 > Φ 4 (C) 

36. [2016] The region between two concentric spheres of radii ’a ’ and ’b ’, respectively (see figure), has volume charge A density ρ = , where A is a constant and r is the r distance from the centre. a Q

b

At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is

Q Q (A)  2 (B) 2π a 2π ( b 2 - a 2 )



2Q 2Q (C)  2 (D) 2) ( π a2 π a -b

39. [2015] A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞) on its surface. For this sphere the equipotential surfaces with 3V0 5V0 3V0 V0 , , and have radius potentials 4 2 4 4 R1 , R2 , R3 and R4 respectively. Then

(A) R1 = 0 and R2 > ( R4 - R3 )



R1 ≠ 0 and ( R2 - R1 ) > ( R4 - R3 ) (B) 



R1 = 0 and R2 < ( R4 - R3 ) (C) 



(D) 2R < R4

40. [2015] A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge -σ in the lower half. The electric field lines around the cylinder will look like figure given in (figures are schematic and not drawn to scale) (A)

(B) 

(C)

(D) 

37. [Online 2016]  The potential (in volts) of a charge distribution is given by

V ( z ) = 30 - 5 z 2 for z ≤ 1 m



V ( z ) = 35 - 10 z for z ≥ 1 m V ( z ) does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ0 (in units of e 0 ) which is spread over a certain region, then choose the correct statement (A) ρ0 = 20e 0 in the entire region.



(B)  ρ0 = 10e 0 for z ≤ 1 m and ρ0 = 0 elsewhere



(C)  ρ0 = 20e 0 for z ≤ 1 m and ρ0 = 0 elsewhere



(D) ρ0 = 40e 0 in the entire region



38. [Online 2016] Within a spherical charge distribution of charge density ρ ( r ), N equipotential surfaces of potential V0 , V0 + ΔV , V0 + 2 ΔV , ….. V0 + N ΔV ( ΔV > 0 ) , are drawn and have increasing radii r0 , r1, r2 , …., rN , respectively. If the difference in the radii of the surfaces is constant for all values of V0 and ΔV then 1 r2



ρ(r ) ∝ (A) ρ ( r ) = constant (B)



1 (C)  ρ ( r ) ∝ (D) ρ(r ) ∝ r r

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 218

41. [Online 2015] A thin disc of radius b = 2 a has a concentric hole of radius a in it (shown in figure). It carries uniform surface charge σ on it. If the electric field on its axis at height h ( h  a ) from its centre is given as Ch then value of C is



σ σ (A)  (B) ae 0 2 ae 0



σ σ (C)  (D) 4 ae 0 8 ae 0

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Chapter 1: Electrostatics 1.219 42. [Online 2015] Shown in the figure are two point charges +Q and -Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If σ 1 is the surface charge on the inner surface and Q1 net charge on it and σ 2 the surface charge on the outer surface and Q2 net charge on it then +Q –Q

σ 1 ≠ 0 , Q1 ≠ 0 ; σ 2 ≠ 0 , Q2 ≠ 0 (A) 



σ 1 ≠ 0 , Q1 = 0 ; σ 2 ≠ 0 , Q2 = 0 (B) 



σ 1 ≠ 0 , Q1 = 0 ; σ 2 = 0 , Q2 = 0 (C) 



σ 1 = 0 , Q1 = 0 ; σ 2 = 0 , Q2 = 0 (D) 

43. [Online 2015]  A wire, of length L ( = 20 cm ), is bent into a ­semi-circular arc. If the two equal halves of the arc, were each to be uniformly charged with charges ±Q, ( Q = 10 3 e 0 Coulomb where e 0 is the permittivity (in SI units) of free space) the net electric field at the centre O of the semi-circular arc would be Y

X

O



( × 103 NC -1 ) ˆj (B) ( 25 × 103 NC -1 ) iˆ (A) 50



( × 103 NC -1 ) ˆj (D) ( 50 × 103 NC -1 ) iˆ (C) 25

44. [Online 2015]  An electric field E = 30iˆ + 30 ˆj NC -1 exists in a region of space. If the potential at the origin is taken to be zero then the potential at x = 2 m , y = 2 m is

(

)



-120 JC -1 (A) -130 JC -1 (B)



-140 JC -1 (D) -110 JC -1 (C) 

45. [2014]   Assume that an electric field E = 30 x 2iˆ exists in space. Then the potential difference VA - VO , where VO is the potential at the origin and VA the potential at x = 2 m is

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 219

(A)  120 V (B) -120 V



(C)  -80 V (D) 80 V

46. [2013] Two charges, each equal to q , are kept at x = - a and x = a on the x-axis. A particle of mass m and charge q q0 = is placed at the origin. If charge q0 is given a 2 small displacement ( y  a ) along the y-axis, the net force acting on the particle is proportional to -y (A) y (B)



O



1 1 (C)  (D) y y

47. [2013] A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is O

L

A

L

B



Q ln 2 3Q (A)  (B) 8πe 0 L 4πe 0 L



Q Q ln 2 (D) (C)  4πe 0 L ln 2 4πe 0 L

48. [2012] This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. An insulating solid sphere of radius R has a uniformly positive charge density ρ . As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point outside the sphere. The electric potential at infinity is zero. Statement 1: When a charge q is taken from the centre to the surface of the sphere, its potential energy qρ . changes by 3e 0 Statement 2: The electric field at a distance r ( r < R ) ρr from the centre of the sphere is . 3e 0

(A) Statement 1 is true, Statement 2 is false. (B) Statement 1 is false, Statement 2 is true. (C) Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation of Statement 1. (D) Statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation of Statement 1.

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1.220  JEE Advanced Physics: Electrostatics and Current Electricity 49. [2012] In a uniformly charged sphere of total charge Q and radius R, the electric field E is plotted as a function of distance from the centre. The graph which would correspond to the above will be (B) E

(A) E

R

r

(C) E

R

r

(D) E

­distance from the origin. The electric field at a distance r ( r < R ) from the origin is given by

ρ r⎛ 5 r ⎞ 4πρ0 r ⎛ 5 r ⎞ (A)  0 ⎜ - ⎟ (B) ⎜ - ⎟ ⎠ ⎝ 3e 0 4 R 3e 0 ⎝ 3 R ⎠



ρ r⎛5 r⎞ 4 ρ0 r ⎛ 5 r ⎞ (C)  0 ⎜ - ⎟ (D) ⎜ - ⎟ 4e 0 ⎝ 3 R ⎠ 3e 0 ⎝ 4 R ⎠

54. [2010] A thin semi-circular ring of radius r has a positive q distributed uniformly over it. The net field charge  E at the centre O is j

R

r

R

r

50. [2011] Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d ( d  l ) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v . Then as a function of distance x between them

v ∝ x -1 (A) v ∝ x -1 2 (B)



(C)  v ∝ x1 2 (D) v∝x

51. [2011] The electrostatic potential inside a charged spherical ball is given by ϕ = ar 2 + b where r is the distance from the centre; a, b are constants. Then the charge density inside the ball is

-6 ae 0 r (A) -24π ae 0 r (B)



-24π ae 0 (D) -6 ae 0 (C) 

52. [2010]  Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 g cm -3 , the angle remains the same. If density of the material of the sphere is 1.6 g cm -3 , the dielectric constant of the liquid is (A) 1 (B) 4 (C) 3 (D) 2 53. [2010] Let there be a spherically symmetric charge distribu⎛5 r⎞ tion with charge density varying as ρ ( r ) = ρ0 ⎜ - ⎟ ⎝ 4 R⎠ upto r = R and ρ ( r ) = 0 for r > R , where r is the

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 220

O

i



q q ˆj (A)  2 2 ˆj (B) 2 2π e 0 r 4π e 0 r 2



(C)  -

q q ˆj (D) - 2 2 ˆj 4π 2e 0 r 2 2π e 0 r

55. [2009] A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then the Q equals q

(A) -2 2 (B) -1



1 1 (D) (C)  2

56. [2009] Two points P and Q are maintained at the potentials of 10 V and -4 V respectively. The work done in moving 100 electrons from P to Q is

(A) -9.60 × 10 -17 J

(B) 9.60 × 10 -17 J

(C)  -2.24 × 10 -16 J

(D) 2.24 × 10 -16 J 57. [2009] Q r be the charge density distribution for π R4 a solid sphere of radius R and total charge Q. For a point ’ p ’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is

Let ρ ( r ) =

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Chapter 1: Electrostatics 1.221



Q (A) 0 (B) 4πe 0 r12



Qr 2 Qr12 (C)  1 4 (D) 4πe 0 R 3πe 0 R 4

58. [2009] This question contains Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement 1: For a charged particle moving from point P to point Q, the net work done by an electrostatic

field on the particle is independent of the path connecting point P to point Q Statement 2: The network done by a conservative force on an object moving along a closed loop is zero. (A) Statement 1 is true, Statement 2 is false. (B) Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation of Statement 1. (C) Statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation of Statement 1. (D) Statement 1 is false, Statement 2 is true.

ARCHIVE: JEE advanced Single Correct Choice Type Problems

R2 P a

This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. [JEE (Advanced) 2019]  A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the ­potential at its surface is V0 . A hole with a small area α 4π R2 ( α  1 ) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct? (A) The potential at the centre of the shell is reduced by 2αV0 .





(B) The magnitude of electric field at the centre of the αV0 shell is reduced by . 2R (C) The magnitude of electric field at a point, located on a line passing through the hole and shell’s centre, on a distance 2R from the centre of the αV0 spherical shell will be reduced by . 2R (D) The ratio of the potential at the centre of the shell 1 to that of the point at R from centre towards the 2 1-α . hole will be 1 - 2α

2. [JEE (Advanced) 2015] Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution, a spherical cavity of radius R2 , centred at P with distance OP = a = R1 - R2 (see figure) is made. If the elec   tric field inside the cavity at position r is E ( r ), then the correct statements is/are

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 221

R1

O



 (A) E is uniform, its magnitude is independent of R2  but  its direction depends on r (B)  E is uniform, its magnitude depends on R2 and  its  direction depends on r (C)  E is uniform, its magnitude is independent of a  but its direction depends on a  (D) E is uniform and both its magnitude and direc tion depend on a

3. [JEE (Advanced) 2015] The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density λ are kept parallel to each other. In their resulting electric field, point charges q and -q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statements is/are λ

λ

+q



λ

x

λ

–q

x

(A) both charges execute simple harmonic motion (B) both charges will continue moving in the direction of their displacement

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1.222  JEE Advanced Physics: Electrostatics and Current Electricity (C) charge +q executes simple harmonic motion while charge -q continues moving in the direction of its displacement (D) charge -q executes simple harmonic motion while charge +q continues moving in the direction of its displacement





4. [JEE (Advanced) 2014] Charges Q, 2Q and 4Q are uniformly distributed R in three dielectric solid spheres 1, 2 and 3 of radii , 2 R and 2R respectively, as shown in figure. If ­magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are E1 , E2 and E3 respectively, then P R

P

Q

R 4Q

R/2 Sphere-1

Sphere-2



(C)  E2 > E1 > E3 (D) E3 > E2 > E1

5. [IIT-JEE 2012] Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field E ( r ) and the electric potential V ( r ) with the distance r from the centre, is best represented by which graph?



0

R

(A) E(r)

V(r)

R

(B) E(r)



1 × 10 -5 V (B) 1 × 10 -7 V (A) 



(C)  1 × 10 -9 V (D) 1 × 10 -10 V

z

r

(a, 0, a)



y

x



(A) 2E0 a 2 (B) 2E0 a 2



E0 a 2 (C)  E0 a 2 (D) 2

8. [IIT-JEE 2010]  A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is made of two hemispherical shells, held together by pressing them with force F (see figure). F  is proportional to

V(r)

R

(C) E(r)

0

(a, a, a)

(0, 0, 0) (0, a, 0)

F 0

r

6. [IIT-JEE 2012] Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45° to the vertical just after release. Then X is nearly

Sphere-3

(A) E1 > E2 > E3 (B) E3 > E1 > E2

0

V(r)

2R





(D) E(r)

7. [IIT-JEE 2011]   Consider an electric field E = E0 xˆ , where E0 is a ­constant. The flux through the shaded area (as shown in the figure) due to this field is

P

2Q



r V(r)

R

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F

r



1 1 2 σ R (A)  σ 2 R2 (B) e0 e0



1 σ2 1 σ2 (C)  (D) e0 R e 0 R2

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Chapter 1: Electrostatics 1.223 9. [IIT-JEE 2010] A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of 81π × 10 5 Vm -1 . When the field is switched strength 7 off, the drop is observed to fall with terminal velocity

12. [IIT-JEE 2008]

q q 2q , and 3 3 3 placed at points A, B and C, respectively, as shown in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60° .

Consider a system of three charges

2 × 10 -3 ms -1 . Given g = 9.8 ms -2, viscosity of the air -5

-2

y

-3

= 1.8 × 10 Nsm and the density of oil = 900 kgm , the magnitude of q is

B

(A)  1.6 × 10 -19 C (B) 3.2 × 10 -19 C (C)  4.8 × 10

-19

C

-19

C (D) 8 × 10 C

10. [IIT-JEE 2009] Three concentric metallic spherical shells of radii R, 2R and 3R are given charges Q1, Q2 and Q3 , respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3 , is

(A)  1 : 2 : 3 (B) 1: 3 : 5



(C)  1 : 4 : 9 (D) 1 : 8 : 18

11. [IIT-JEE 2009] a having a uniformly distributed A disk of radius 4 charge 6C is placed in the x -y plane with its centre at ⎛ a ⎞ ⎜⎝ - , 0 , 0 ⎟⎠ . A rod of length a carrying a uniformly dis2 a tributed charge 8C is placed on the x-axis from x = 4 5a to x = . Two point charges -7C and 3C are placed 4 a ⎛a ⎞ ⎛ 3a 3a ⎞ at ⎜ , - , 0 ⎟ and ⎜ - , , 0 ⎟ , respectively. ⎝4 ⎝ 4 4 ⎠ 4 ⎠ Consider a cubical surface formed by six surfaces a a a x = ± , y = ± , z = ± . The electric flux through 2 2 2 this cubical surface is y

x

2C 2C (B) e0 e0



(A) -



10C 12C (C)  (D) e0 e0

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 223

60°

O

x

A





(A) The electric field at point O is

q directed 8πe 0 R2

along the negative x-axis (B) The potential energy of the system is zero (C) The magnitude of the force between the charges q2 at C and B is 54πe 0 R2 (D) The potential at point O is

q 12πe 0 R

13. [IIT-JEE 2007] Positive and negative point charges of equal magnia⎞ a⎞ ⎛ ⎛ tude are kept at ⎜ 0 , 0 , ⎟ and ⎜ 0 , 0 , - ⎟ , respec⎝ ⎠ ⎝ 2 2⎠ tively, The work done by the electric field when another positive point charge in moved from ( - a, 0 , 0 ) to ( 0, a, 0 ) is

(A) positive (B) negative (C) zero (D) depends on the path connecting the initial and final positions

14. [IIT-JEE 2007] Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then (A)  negative and distributed uniformly over the ­surface of the sphere (B) negative and appears only at the point on the sphere closest to the point charge (C) negative and distributed non-uniformly over the entire surface of the sphere (D) zero

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1.224  JEE Advanced Physics: Electrostatics and Current Electricity 15. [IIT-JEE 2007] A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

c­alculating the flux of the electric field over the ­spherical surface, the electric field will be due to q2

+q1

–q1



(A) zero everywhere (B) non-zero and uniform (C) non-uniform (D) zero only at its centre

16. [IIT-JEE 2007] A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. (A) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder (B) A potential difference appears between the two cylinders when a charge density is given to the outer cylinder (C) No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders (D) No potential difference appears between the two cylinders when same charge density is given to both the cylinders 17. [IIT-JEE 2005] Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is σ

P

(A) q2



(B) only the positive charges (C) all the charges (D) +q1 and -q1

19. [IIT-JEE 2004] Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charge is possible for, P, Q, R, S, T and U respectively? P

U

Q

R

O

T

S



(A) + , - , + , - , - , + (B) + , -, +, -, +, -



(C)  + , + , - , + , - , - (D) -, + , +, -, +, -

z = 3a

20. [IIT-JEE 2003] A metallic shell has a point charge q kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces?

z=0



(A) 

(B)



(C) 

(D)

z –2σ



x –σ

z = –a



2σ ˆ 2σ (A)  kˆ (B) k e0 e0



4σ 4σ ˆ k (C)  kˆ (D) e0 e0

18. [IIT-JEE 2004]  Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When

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Chapter 1: Electrostatics 1.225 21. [IIT-JEE 2002] Two equal point charges are fixed at x = - a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

x2 (A) x (B)



1 (C)  x (D) x

Q

+q

3

22. [IIT-JEE 2002] Two equal point charges are fixed at x = - a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

(A) x (B) x2



1 (C)  x 3 (D) x



(C) 



-q -2q (A)  (B) 1+ 2 2+ 2



-2q (D) +q (C) 



(B)

(A) 

+q

26. [IIT-JEE 1998] A charge +q is fixed at each of the points x = x0 , x = 3 x0 , x = 5x0 ,.... ad infinitum on the x axis and a charge -q is fixed at each of points x = 2x0 , x = 4 x0 , x = 6 x0 .... ad infinitum. Here x0 is a positive constant. The potential at the origin due to the above system of charges is q (A) zero (B) 8πe 0 x0 log e 2

23. [IIT-JEE 2001] Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in

a

q log e 2 (C)  ∞ (D) 4πe 0 x0

(D)

27. [IIT-JEE 1997] An electron of mass me, initially at rest, moves through a certain distance in a uniform electric field in time t1. A proton of mass mp, also, initially at rest takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t2 is nearly equal to t1 1



⎛ mp ⎞ 2 (A) 1 (B) ⎜⎝ m ⎟⎠



⎛ m ⎞2 (C)  e ⎜⎝ mp ⎟⎠

e

1

24. [IIT-JEE 2001] A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm . Then the potentials at the points A , B and C satisfy

(A)  VA < VB (B) VA > VB



(C)  VA < VC (D) VA > VC

25. [IIT-JEE 2000] Three charges Q, +q and +q are placed at the vertices of a right angle triangle (isosceles triangle) as shown. The net electrostatic energy of the configuration is zero, if Q is equal to

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 225

(D) 1836

28. [IIT-JEE 1997] A non-conducting ring of radius 0.5 m carries a total charge of 1.11 × 10 -10 C distributed non-uniformly  on its circumference producing an electric field E everywhere in space. The value of the line integral l=0





∫ -E ⋅ dl, (l = 0 being the centre of the ring) in volt is

l =∞



(A) +2 (C) –2

(B) –1 (D) 0

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1.226  JEE Advanced Physics: Electrostatics and Current Electricity 29. [IIT-JEE 1996] A metallic sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in the ­figure as



1 2

1 2

3

3

4

4

(A) 1 (C) 3



(B) 2 (D) 4

30. [IIT-JEE 1995]  Two point charges +q and -q are held fixed at ( - d, 0 ) and ( d, 0 ) respectively of a x -y co-ordinate system. Then: (A) the electric field E at all points on the x-axis has the same direction (B) work has to be done in bringing at a test charge from ∞ to the origin (C) electric field at all point on y-axis is along x-axis (D) the dipole moment is 2qd along the x-axis 31. [IIT-JEE 1992] Two identical thin rings, each of radius a metre are coaxial and placed a metre apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q coulomb from the centre of one ring to that of the other is (B)

q ( Q1 - Q2 ) ( 2 - 1 )



(A) zero



q 2 ( Q1 + Q2 ) q ( Q1 + Q2 ) ( 2 + 1 ) (C)  (D) 4πe 0 a 4 2πe 0 a

4 2πe 0 a

32. [IIT-JEE 1989] A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V . If the shell is now given a charge of -3Q , the new potential difference between the same two surfaces is V (B) 2V (A)  (C)  4V (D) -2V 33. [IIT-JEE 1987] A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 226



(A) -

Q Q (B) 2 4



(C)  +

Q Q (D) + 4 2

34. [IIT-JEE 1984] Two equal negative charges -q are fixed at points ( 0, a ) and ( 0, - a ) on the y-axis. A positive charge Q is released from rest at the point ( 2 a, 0 ) on the x-axis. The charge Q will (A) Execute simple harmonic motion about the origin (B) Move to the origin and remains at rest (C) Move to infinity (D)  Execute oscillatory but not simple harmonic motion. 35. [IIT-JEE 1983] A hollow metal sphere of radius 5 cm is charged so that the potential on its surface is 10 V. The potential at the centre of the sphere is (A) ZERO (B) 10 V (C) same as at a point 5 cm away from the surface (D) same as at a point 25 cm away from the surface 36. [IIT-JEE 1981]  An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of

1 Å (B) 10 -10 cm (A) 



(C)  10 -12 cm (D) 10 -15 cm

Multiple Correct Choice Type Problems This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1. [JEE (Advanced) 2019]  A charged shell of radius R carries a total charge Q. Given ϕ as the flux of electric field through a closed cylindrical surface of height h , radius r and with its centre same as that of the shell. Here, centre of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct? ( e 0 is the permittivity of free space)

3R Q then ϕ = 5 5e 0 8R 3R (B) If h < and r = then ϕ = 0 5 5 (A) If h > 2R and r =

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Chapter 1: Electrostatics 1.227



(C) If h > 2R and r > R then ϕ =



(D) If h > 2R and r =

4.

Q e0

4R Q then ϕ = 5 5e 0

2. [JEE (Advanced) 2019] p  An electric dipole with dipole moment 0 iˆ + ˆj is 2 held fixed at the origin O in the presence of an uniform electric field of magnitude E0 . If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are ( e 0 is permittivity of free space. R  dipole size)

(

)

y

45

°

B

O



R



Q = 4σπ r02 (A) 

(B) r0 =

λ 2πσ

⎛r ⎞ ⎛r ⎞ (C)  E1 ⎜ 0 ⎟ = 2E2 ⎜ 0 ⎟ ⎝ 2⎠ ⎝ 2⎠

⎛r ⎞ ⎛r ⎞ E2 ⎜ 0 ⎟ = 4E3 ⎜ 0 ⎟ (D) ⎝ 2⎠ ⎝ 2⎠

A

45°

[JEE (Advanced) 2014]

Let E1 ( r ), E2 ( r ) and E3 ( r ) be the respective electric fields at a distance r from a point charge Q, and infinitely long wire with constant linear charge density λ , and an infinite plane with uniform surface charge density σ . If E1 ( r0 ) = E2 ( r0 ) = E3 ( r0 ) at a given distance r0 , then

5. [JEE (Advanced) 2013] Two non-conducting spheres of radii R1 and R2 and carrying uniform volume charge densities + ρ and - ρ , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region

x

 (A) Total electric field at point B is EB = 0  (B) Total electric field at point A is EA = 2E0 iˆ + ˆj

(

)

ρ

–ρ

1

⎛ p0 ⎞ 3 (C) R=⎜ ⎝ 4πe 0E0 ⎟⎠

R1

(D) The magnitude of total electric field on any two points of the circle will be same

3. [JEE (Advanced) 2017] A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the fi ­ gure. Which of the following statements is/are correct? +Q

R



(A) The electric flux passing through the curved surQ ⎛ 1 ⎞ face of the hemisphere is ⎜1⎟. 2e 0 ⎝ 2⎠



(B) The component of the electric field normal to the flat surface is constant over the surface (C) Total flux through the curved and the flat surfaces Q is e0 (D)  The circumference of the flat surface is an equipotential





01_Physics for JEE Mains and Advanced - 2_Part 6.indd 227



R2

(A) the electrostatic field is zero (B) the electrostatic potential is constant (C) the electrostatic field is constant in magnitude (D) the electrostatic field has same direction

6. [JEE (Advanced) 2013] Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities ρ1 and ρ2 respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere, along the line joining the centre of the spheres, is zero. ρ The ratio 1 can be ρ2 32 (A) -4 (B) 25 32 (C)  (D) 4 25 7. [IIT-JEE 2012] Six point charges are kept at the vertices of a regular hexagon of side L and centre O as shown in the 1 q ­figure. Given that K = , which of the following 4πe 0 L2 statements(s) is(are) correct.

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1.228  JEE Advanced Physics: Electrostatics and Current Electricity

L F +q

P O T

S

A +2q

R

+q

C

B



E –q

D –2q



–q

(A) The electric field at O is 6K along OD (B) The potential at O is zero (C) The potential at all points on the line PR is same (D) The potential at all points on the line ST is same

8. [IIT-JEE 2012] A cubical region of side a has its centre at the origin. It a ⎛ ⎞ encloses three fixed point charges, -q at ⎜ 0 , - , 0 ⎟ , ⎝ 4 ⎠ a ⎛ ⎞ +3q at (0, 0, 0) and -q at ⎜ 0 , + , 0 ⎟ . Choose the ⎝ 4 ⎠ correct option(s). z a –q –q

(B) The Gauss’s law can be used to calculate the field distribution around an electric dipole (C) If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same (D) The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VB is ( VB - VA )

10. [IIT-JEE 2011] A spherical metal shell A of radius RA and a solid metal sphere B of radius RB ( < RA ) are kept far apart and each is given charge +Q. Now they are connected by a thin metal wire. Then

(A) EAinside = 0



σ R (C)  A = B σ B RA



(B) QA > QB (D) EAon surface < EBon surface

11. [IIT-JEE 2010] A few electric field lines for a system of two charges Q1 and Q2 fixed at two different points on the x-axis are shown in the figure. These lines suggest that

y

3q

x







a 2 is equal to the net electric flux crossing the plane a x=2 a (B) The net electric flux crossing the plane y = + is 2 more than the net electric flux crossing the plane a y=2 q (C) The net electric flux crossing the entire region is e0 a (D) The net electric flux crossing the plane z = + 2 is equal to the net electric flux crossing the plane a x=+ 2 (A) The net electric flux crossing the plane x = +

9. [IIT-JEE 2011] Which of the following statement(s) is/are correct? (A) If the electric field due to a point charge varies as r -2.5 instead of r -2 , then the Gauss’s law will still be valid

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 228



(A) Q1 > Q2



(B) Q1 < Q2



(C) at a finite distance to the left of Q1 the electric field is zero (D) at a finite distance to the right of Q2 the electric field is zero



12. [IIT-JEE 2009] Under the influence of the coulomb field of charge +Q, a charge -q is moving around it in an elliptical orbit. Find out the correct statement(s). (A)  The angular momentum of the charge -q is constant (B)  The linear momentum of the charge -q is constant (C)  The angular velocity of the charge -q is constant (D) The linear speed of the charge -q is constant

9/20/2019 10:40:37 AM

Chapter 1: Electrostatics 1.229 13. [IIT-JEE 2006] For spherical symmetrical charge distribution, variation of electric potential with distance from centre is q given in diagram. Given that V = for r ≤ R0 4πe 0 R0 q for r ≥ R0 and V = 4πe 0 r V



at the point ( 0 , 0 , z0 ) where z0 > 0 . Then the motion of P is (A) periodic for all values of z0 satisfying 0 < z0 < ∞ (B) simple harmonic for all values of z0 satisfying 0 < z0 ≤ R (C) approximately simple harmonic provided z0  R (D) such that P crosses O and continues to move along the negative z-axis towards z = -∞

Reasoning Based Problems r r = R0



Then which option(s) is/are correct (A) Total charge within 2R0 is q.



(B) Total electrostatic energy for r ≤ R0 is zero. (C) At r = R0 electric field is discontinuous.



(D) There will be no charge anywhere except at r = R0 .

14. [IIT-JEE 1999] An ellipsoidal cavity is carved within a perfect conductor (see figure). A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in figure. Then, A q



B

(A) electric field near A in the cavity = electric field near B in the cavity. (B) charge density at A = charge density at B. (C) potential at A = potential at B. (D) total electric flux through the surface of the cavity q . is e0

15. [IIT-JEE 1998] A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its centre (A) increases as r increases for r < R (B) decreases as r increases for 0 < r < ∞ (C) decreases as r increases for R < r < ∞ (D) is discontinuous at r = R 16. [IIT-JEE 1998] A positively charged thin metal ring of radius R is fixed in the x -y plane with its centre at the origin O . A negatively charged particle P is released from rest

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 229

This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE. 1. [IIT-JEE 2008]  Statement-1: For practical purposes, the earth is used as a reference at zero potential in electrical circuits.  Statement-2: The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by

Q . 4πe 0 R

Linked Comprehension Type Problems This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options)

Comprehension 1 Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match

9/20/2019 10:41:02 AM

Chapter 1: Electrostatics 1.231 List I with List II and select the correct answer using the code given below the lists.

z

+q

L

D

(0, b) a A

Q1

Q2

Q3

Q4

(–2a, 0)

(–a, 0)

(+a, 0)

(+2a, 0)

List I

List II

P.  Q1, Q2, Q3, Q4 all positive

1.  +x

Q.  Q1, Q2 positive; Q3, Q4 negative

2.  -x

R.  Q1, Q4 positive; Q2, Q3 negative

3.  +y

S.  Q1, Q3 positive; Q2, Q4 negative

4.  -y

3 C 2a y

B

x

2. [IIT-JEE 2012] An infinitely long solid cylinder of radius R has a uniform volume charge density ρ . It has a spherical cavity R of radius with its centre on the axis of the cylinder, 2 as shown in the figure. The magnitude of the electric field at the point P, which is at a distance 2R from the 23 ρR axis of the cylinder, is given by the expression . 16 ke 0 The value of k is z

Codes (A) P-3, Q-1, R-4, S-2 (B) P-4, Q-2, R-3, S-1 (C) P-3, Q-1, R-2, S-4 (D) P-4, Q-2, R-1, S-3

R

2R

1. [JEE (Advanced) 2015] An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the y-axis in 3 the y -z plane at z = a (see figure). If the magni2 tude of the flux of the electric field through the rectangular surface ABCD lying in the x -y plane with its

λL ( e 0 = permittivity of free ne 0 space), then the value of n is centre at the origin is

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 231

y

x

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s).

P

R/2

3. [IIT-JEE 2011] Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side a . The surface tension of the soap film is γ . The system of charges and planar film are in equilibrium, and 1

⎛ q2 ⎞ N a = k ⎜ ⎟ , where k is a constant. Then N is. ⎝ γ ⎠ 4. [IIT-JEE 2009] A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ = kr a , where k and a are constants and r is the distance from its cenR 1 tre. If the electric field at r = is times that at r = R, 2 8 find the value of a .

9/20/2019 10:41:12 AM

1.232  JEE Advanced Physics: Electrostatics and Current Electricity

Answer Keys—Test Your Concepts and Practice Exercises Test Your Concepts-I (Based on Coulomb’s Law)

3 q2 4.  4πe 0 R2

1. K = 4 2. 2 × 1016 N

q2 5.  4πe 0 a 2

λ L2 3.  0 3

6. (a)  zero

ρ 4.  ρ -σ

   (b)  F =

5. 4 × 10 42 , 10 36, 8.6 × 10 -11 Ckg -1 6. 9 × 10

-4

(

⎛ Qq ⎞ 7. 2π 2 2 ⎜ ⎟ ⎝ πe 0 ma 3 ⎠

)

6iˆ - 8 ˆj N , 9 × 10 -3 N

(

)

7.  -21iˆ - 28 ˆj + 84 kˆ × 10 -5 N

qQ 8.  4πe 0 a 2

8. 2.25 mm

9. 10 N ( parallel to AB )

9. 33 × 10 -9 C 11. 2π

4πe 0 mL3 Qq

Test Your Concepts-III (Based on Electric Field)

12. Q = 2L 4πe 0 k ( L - L0 )

q 1.  4πe 0 a 2

13. 3 μC and -1 μC 14. 38 μC or 12 μC

Qx 2.  2 8π e 0 R3

15.  4.1 × 1016 rads -1 , 1.5 × 10 -16 s

3. 0.2 N (due North), 0.5 N (due South)

1

⎛ 4πe 0 mgR2 ⎞ 2 17.  ⎜ ⎟⎠ ⎝ 3

6q 4.  4πe 0 2

18.  μ = 0.18

5. E =

19. 1

Test Your Concepts-II (Based on Principle of Superposition) 1. 

   

qQ 4πe 0 a 2

1 1 ⎤ ˆ ˆ ˆ Q2 ⎡ 1+ + i+ j+k 2 3 3 ⎥⎦ 4πe 0 a 2 ⎢⎣

(

)

7. (a)  Q2 is negative and Q1 is positive.    (b)     (c) 

2

3Q ⎛ 1 1 ⎞ + ⎜ 1+ ⎟ MAGNITUDE 2 3 3⎠ 4πe 0 a 2 ⎝

mgπe 0 ⎛ 2  ⎞ 2.  ⎜ h + ⎟⎠ 2 h ⎝ 2

32

3. Radially outwards ±

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 232

Q1 ⎛  + a ⎞ =⎜ ⎟ Q2 ⎝ a ⎠

2

1 ⎛ Q1 ⎞ ⎝⎜ Q ⎠⎟

13

-1

2

8. f = λ RE0 along positive x-axis. 9. (a) 

Qq0 a , 2 3 3πe 0 a 2

mg q

2Qqx0

(

4πe 0 R2 + x02

)3 2



{towards the centre}

   (b)  The motion of the bead is periodic between ± x0

9/20/2019 10:41:27 AM

Chapter 1: Electrostatics 1.233

   (c)  m

2Qqx0 d2x =dt 2 4πe 0 R2 + x02

(

)

32

with

dx = 0 at x = ± x0 dt

       and x = x0 at t = 0

λ 21.  0 8πe 0 x0 22. Ex =

   (d)  x = x0 cos ( ωt ) and v = - x0ω sin ( ωt ), where

4π 3e 0 mR3 2Qq

   (e) 

10. 2π

Test Your Concepts-IV (Based on Dipoles and Dipole Moment)

2Qq 4πe 0 mR3

        ω =

1. 2 p cos

mRπ 4 qE

⎛ 1⎞    at an angle tan -1 ⎜ ⎟ with the z axis. ⎝ 4⎠ 3. T = 2π

mgd

4. (a) 2π

q 2 - d2 

13. T = 2π

2

⎛ qE ⎞ ⎛ qE ⎞ g +⎜ cos β - 2g ⎜ ⎝ m ⎟⎠ ⎝ m ⎟⎠ 2

2md not a simple harmonic motion but periodic in qE nature.

14. 2

15. 

λ 2 -2iˆ + 14 ˆj 4πe 0

(

)

⎛ 3λ ⎞ ˆ j 16.  - ⎜ ⎝ πe 0 ⎟⎠ mg 17. (a)  q = ( A cot θ + B ) mgA    (b)  T = ( A cos θ + B sin θ ) 18. (a) 

2

mv qR

   (b) 

(R

+

md 2λ LE

   (b)  (i)  λ LdE      (ii)  2λ LdE 5. (a)  π

m λ sin 2 θ 0E

   (b)  8 λ R2Esin 2 θ 0 6. (a) zero  λp    (b)  2πe 0 r 2  λp    (c)  2πe 0 r 2 7. π R2 λ 0 8. (a) 

qp 4πe o d 2

   (b) 

pq ˆ i 2πe o d 3

q ⎛  ⎞ 1.  ⎜ 1 2 2 ⎟ e0 ⎝ R + ⎠

Qx 2

4πe 0Ι σ

Test Your Concepts-V (Based on Flux)

mv 2 2qR

19. 5.25 μC 1 20.  4πe 0

θ 2

 2. p = 17 qa

λ 11.  2 2πe 0 R 12. 

2λ λ λ and Ey = with Enet = 4πe 0b 4πe 0b 4πe 0b

3 x2 2

)

iˆ

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 233

q ⎛ L ⎞ 2.  ⎜ e 0 ⎝ L2 + 4 R2 ⎟⎠

9/20/2019 10:41:41 AM

1.234  JEE Advanced Physics: Electrostatics and Current Electricity λR 3.  2e 0

1 5.  kx 2 3e 0

⎛ 3⎞ q 4. ⎜ 1 + ⎟ ⎝ 2 ⎠ 2e 0

6. (a) 

q 2e 0

    (b) 

q 2e 0

5. 2HRE 6. E ( 2RH cos θ + π R2 sin θ ) ch 7.  2

Q - 6q Q 7.  , q= 6 6e 0

2

2

8. (a)  858 Nm C

-1

     (b)  0 2

   (c)  428.75 Nm C

-1

8. (a) 

Qr 3 e 0 a3

    (b) 

Q e0 ΦE

   (c) 

9. 180 kN m 2 C -1

Q ε0

10. 1 MNC -1 11. (a)  -2.34 kNm 2 C -1    (b)  +2.34 kN m 2 C -1

0

     (c)  0 12. +1.87 kNm 2 C -1 13. ERh 14. (a) +    (b)  -

Q 2e 0 Q 2e 0

Test Your Concepts-VI (Based on Gauss’s Law) q e0

3. 4πe 0α R 4.  ϕ

0

9. When R ≤ d , Φ E =

qenc =0 e0

    When R > d , Φ E =

2λ R 2 - d 2 e0

10. (a)

ρ0 R 5 5e 0 r 2

   (b) 

ρ0 r 3 5e 0

⎡ ρ0 x 3 ˆ for x > 0 i ⎢ 3e 0 ⎢     (b)  ⎢ ρ x3 ⎢ - 0 iˆ for x < 0 ⎢⎣ 3e 0

q     (b)  6e 0

λa ε0

r

11. (a)  ⎡ ρ d 3 d 0 for x > iˆ ⎢ 24 2 e 0 ⎢ ⎢ ρ d3 d ⎢ - 0 iˆ for x < 2 ⎢⎣ 24e 0

1. zero 2. (a) 

a

12.  – λε v

λv ε0

0

a v

2a v

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 234

t

ρ0 2 e0

⎛ GME ⎞ 13. (b)  g = ⎜ r ⎝ RE3 ⎟⎠  ρa 15.  2e 0

9/20/2019 10:41:54 AM

Chapter 1: Electrostatics 1.235

Test Your Concepts-VII (Based on Electrostatic Potential and Energy) 2.50 2.00

1. (a)  V (x) Q 4πε0a

10. (a)  4.5 × 10 -3 J    (b)  -4.5 × 10 -3 J ⎛ a⎞ 11. -Q ⎜ ⎟ ⎝ b⎠

1.50

12. (a)  6 m

1.00

0.50 0.00 –4 –3 –2 –1 0 1 2 3 4 x/a

   (b)  -2 μC 13. r = 0.5 m      r = 0.25 m

   (b) 

5 ⎛ q2 ⎞ 14.  ⎜ 9 ⎝ 4πe 0 d ⎟⎠

10 8 6 4 2

V (y) Q 4πε0a

–5 –4 –3 –2 –1

15. -0.553 1 2 3 4 –2 –4 –6 –8 –10

y/a

5

Q 4πe 0 R

16. -q0E0 a

λ 17.  ( π + 2 log e 3 ) 4πe 0 x ⎛ ⎛ πσ 0 ⎞ ⎡ 2 2 2 18. ⎜ ⎢ R R + x + x log e ⎜ ⎝ 4πe 0 ⎟⎠ ⎣ ⎝ R + R2 + x 2 19. (i) 

2q q , 4πe 0 3πe 0

4. 9 × 10 3 V

  (ii) 

q ⎡ 4q ⎤ 1 ⎛ 2q ⎞ ⎜ ⎟, 4πe 0 ⎝ 3 ⎠ 4πe 0 ⎢⎣ 5 ⎥⎦

5. (a) 

2QE k

20. 1 m

    (b) 

QE k

21. 

2. -7.62 × 10 -2 J 3. 1.35 MJ

7Qq 4πe 0 m

⎛ 1 q2 ⎞ ⋅ ⎟ 22. 5.824 ⎜ ⎝ 4πe o a ⎠

m    (c)  2π k    (d) 

⎞⎤ ⎟⎠ ⎥ ⎦

2 ( QE - μ k mg ) k

⎛ a⎞ 23. V ⎜ ⎟ ⎝ 3t ⎠

1/3

4

6. 1.8 × 10 V 7. 40 kV

Test Your Concepts-VIII (Based on Equipotential Surfaces)

1 2Qqx 2 8.  4πe 0 a 3

1. +260V

9. 0.4 ms -1

   (b)  200 NC -1 downwards

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 235

2. (a)  EA > EB

9/20/2019 10:42:06 AM

1.236  JEE Advanced Physics: Electrostatics and Current Electricity A

   (c) 

2. (a)  1.52 × 10 5 ms -1    (b)  6.5 × 106 ms -1 3. (a)  -4.5 × 10 -5 J

8 2

4

   (b)  3.46 × 10 4 ms -1

6

0

4. 27.3 fm 1 ⎞ 1 q2 ⎛ 5.  ⎜ 1 + ⎟ ⎝ 8 ⎠ 4πe 0 mL

B

Test Your Concepts-IX (Based on Relation Between Electrostatic Field and Potential)

6. 1.45 × 107 ms -1 7. 

1 q2 4πe 0 3 m

8. 

Q2 2πe 0 MR

1. 5 2 NC -1 3

(

2

2

2. Ex = 3E0 a xz x + y +

5 z2 2

)

(

    Ey = 3E0 a 3 yz x 2 + y 2 + z 2

(

)

-

    Ez = E0 + E0 a 3 2 z 2 - x 2 - y 2

5 2

)( x

2

+ y 2 + z2

)

-

5 2

Qq Qq ⎞ ⎛ + ⎜ + 2 4πe 0 mv02 ⎝ 4πe 0 mv02 ⎟⎠ 2

4. 3 V

2 ⎛ ae ⎞ ⎛ 6 m ⎞ 3 11.  ⎜ ⎟ ⎜ ⎝ 2m ⎠ ⎝ ae ⎟⎠

5. (a)  1 V  (b) 1 V

(

6. (a)  -2a xiˆ - yjˆ

)

12. v1 =

   (b)  - ayiˆ - axjˆ  7. E = - 2 axiˆ + 2 ayjˆ + 2bzkˆ

(

)

     E = 2 a 2 x 2 + y 2 + b 2 z 2

(

)

   An ellipsoid of revolution with semi-axis V , a

2

10. 

3. -2iˆ - 3 ˆj + kˆ

   

q2 1 1 (1 - 2 ) 9. mg - mu2 + 2 4πe 0 2

V , a

V . b

8. - ( axy + byz ) + C

13. 

Qqr2 , v2 = 2πe 0 mr1 ( r1 + r2 )

e2 πe 0 mv02

Test Your Concepts-XI (Based on Conductors) 1. 2 N 2. -

1 1 1 (σ A + σB ) , (σ A - σB ) , (σ A + σB ) 2e 0 2e 0 2e 0

Test Your Concepts-X (Based on Motion of Charged Particles in Electric Field)

    EI = EIII = 0 , EII =

1. (a)  60 V

4. (a)  496 nCm -2

   (b)  4.6 × 106 ms -1

   (b)  992 nCm -2

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 236

Qqr1 2πe 0 mr2 ( r1 + r2 )

σ e0

3. 1.77 pCm -3

9/20/2019 10:42:19 AM

Chapter 1: Electrostatics 1.237

5. (a)

-q 4π a 2

   (b) 

Q+q 4π b 2

    (i)  -3Q     (j)  +2Q   (k)  E

6. Charge on the outer surface of the sphere is -Q    Charge on the inner surface of the shell is +Q    Charge on the outer surface of the shell is +2Q

a

r

b c

   Inside the sphere within the material of the shell, E = 0    Between the sphere and shell E =    Outside the shell E =

Q 4πe 0 r 2

2Q 4πe 0 r 2

Test Your Concepts-XII (Based on Charge Distribution) Q 2

1. 3Q 2

–

5Q 2

7.  –

Q 2

5Q 2

3Q 2

⎛Q ⎞ ⎛Q ⎞ 3. ⎜ - e 0 AE ⎟ , ⎜ + e 0 AE ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠     - e 0 AE , + e 0 AE 8. (a) For r < a , E =

ρr 3e 0

      For a < r < b and c < r , E =       For b ≤ r ≤ c , E = 0 Q     (b)  4π c 2

Q 4π r 2e 0

Q Q , Eright = E + 2 Ae 0 2 Ae 0

5. AEe 0 , -AEe 0

( σ 1 - σ 2 ) qa

6. 

2 eo

Test Your Concepts-XIII (Based on Concept of Self and Interaction Energy)

9. (a)  +2Q    (b) 

4. Eleft = E -

2Q 4πe 0 r 2

q2 1.  8πe 0

    (c)  E = 0     (d)  0

⎡ 1 1 ⎤ ⎢ ⎥ R R 2 ⎦ ⎣ 1

q⎞ ⎛ q ⎜ q0 + ⎟ ⎝ 2⎠ 2.  4πe 0

    (e)  +3Q 3Q     (f)  4πe 0 r 2 ⎛r ⎞     (g)  +3Q ⎜ 3 ⎟ ⎝a ⎠ 3

⎛ 3Q ⎞     (h)  ⎜ r ⎝ 4πe 0 a 3 ⎟⎠

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 237

⎡ 1 1 ⎤ ⎢ ⎥ R R 2 ⎦ ⎣ 1

σ2 3.  2e 0 1 Q2 4.  4πe 0 2R

9/20/2019 10:42:33 AM

1.238  JEE Advanced Physics: Electrostatics and Current Electricity

Single Correct Choice Type Questions   1. D

  2. D

  3. D

  4. C

  5. C

  6. C

  7. D

  8. A

  9. A

 10. B

 11. B

 12. D

 13. B

 14. B

 15. A

 16. A

 17. D

 18. A

 19. D

 20. D

 21. B

 22. D

 23. C

 24. C

 25. C

 26. D

 27. A

 28. B

 29. B

 30. B

 31. D

 32. A

 33. D

 34. C

 35. B

 36. A

 37. B

 38. A

 39. A

 40. A

 41. D

 42. B

 43. A

 44. D

 45. B

 46. C

 47. D

 48. C

 49. C

 50. D

 51. C

 52. D

 53. D

 54. C

 55. B

 56. C

 57. B

 58. A

 59. A

 60. C

 61. B

 62. C

 63. C

 64. A

 65. C

 66. A

 67. C

 68. B

 69. D

 70. A

 71. D

 72. A

 73. D

 74. D

 75. A

 76. D

 77. B

 78. C

 79. C

 80. C

 81. A

 82. B

 83. C

 84. D

 85. B

 86. C

 87. C

 88. C

 89. D

 90. B

 91. B

 92. A

 93. B

 94. A

 95. C

 96. B

 97. D

 98. B

 99. A

100.  C

101.  D

102.  C

103.  B

104.  A

105.  A

106.  B

107.  A

108.  A

109.  C

110.  B

111.  B

112.  D

113.  C

114.  A

115.  B

116.  B

117.  D

118.  D

119.  D

120.  C

121.  D

122.  A

123.  C

124.  C

125.  D

126.  C

127.  D

128.  C

129.  D

130.  B

131.  A

132.  C

133.  C

134.  D

135.  B

136.  B

137.  B

138.  A

139.  C

140.  A

141.  C

142.  A

143.  A

144.  C

145.  C

146.  B

147.  B

148.  C

149.  D

150.  A

151.  A

152.  D

153.  B

154.  C

155.  A

156.  B

157.  C

158.  C

159.  C

160.  D

161.  A

162.  D

163.  D

164.  A

165.  C

166.  B

167.  D

168.  D

169.  A

170.  B

171.  A

172.  B

173.  B

174.  C

175.  D

176.  A

177.  C

178.  D

179.  A

180.  B

181.  C

182.  A

183.  A

184.  D

185.  B

186.  B

187.  C

Multiple Correct Choice Type Questions 1.  A, C

2.  A, C

3.  A, B, C, D

4.  A, C, D

6.  A, C

7.  D

8.  D

9.  C, D

5.  C, D 10.  A, B, C

11.  A, C

12.  A, B

13.  A, B, C

14.  B, D

15.  D

16.  B, D

17.  A, B, C, D

18.  A, B, C, D

19.  B, D

20.  A, D

21.  D

22.  A, B, C

23.  A, C, D

24.  A, B, C, D

25.  A, C

26.  B, C

27.  A, B, D

28.  B, C, D

29.  A, C, D

30.  A, B

31.  A, B, C, D

32.  C, D

33.  A, B, C, D

34.  A, B, C, D

35.  A, B, C, D

36.  A

37.  D

38.  A, D

39.  A, D

40.  B, D

41.  A, D

42.  A, C

43.  B, D

44.  A, B

45.  A, C

46.  A, C

47.  A, B, C

48.  B, D

49.  A, C, D

50.  A, B, C

51.  A, D

52.  A, B, D

53.  A, C, D

54.  B, C

55.  A, B, C

56.  B, C

57.  B, C, D

58.  A, B, C

59.  B, C, D

60.  A, D

61.  A, C, D

Reasoning Based Questions 1.  C

2.  B

3.  D

4.  C

5.  A

6.  D

7.  D

8.  A

9.  B

10.  A

11.  D

12.  C

13.  B

14.  B

15.  B

16.  D

17.  D

18.  A

19.  A

20.  A

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 238

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Chapter 1: Electrostatics 1.239

Linked Comprehension Type Questions 1.  C

2.  B

3.  C

4.  A

5.  D

6.  D

7.  B

8.  C

9.  C

10.  C

11.  B

12.  D

13.  B

14.  C

15.  A

16.  A

17.  B

18.  D

19.  C

20.  A

21.  B

22.  C

23.  D

24.  C

25.  D

26.  B

27.  A

28.  C

29.  B

30.  A

31.  D

32.  D

33.  C

34.  B

35.  A

36.  D

37.  A

38.  B

39.  C

40.  A

41.  B

42.  D

43.  D

44.  C

45.  C

46.  B

47.  C

48.  C

49.  B

50.  C

51.  A

52.  D

53.  B

54.  C

55.  B

56.  A

57.  C

58.  B

59.  D

60.  A

Matrix Match/Column Match Type Questions 1. A → (p, q, r, s)

B → (p, q, s)

C → (p, q, s)

D → (p, q, r, s, t)

2. A → (p, r, s, t)

B → (p, q)

C → (p, q)

D → (p, r, s, t)

3. A → (p, r, s)

B → (p, q)

C → (p, q, r, s)

D → (p, r, s, t)

4. A → (q, r, s)

B → (p)

C → (p, q, r, s)

D → (p, q, r, s)

5. A → (q, r)

B → (s)

C → (t)

D → (p)

6. A → (s, t)

B → (p, q, r)

C → (q, r)

D → (p)

7. A → (r, s)

B → (p, r, s)

C → (q)

D → (t)

8. A → (q, s)

B → (p)

C → (t)

D → (r)

9. A → (s, t)

B → (q)

C → (p)

D → (r, s)

10. A → (q, t)

B → (s)

C → (p)

D → (r)

11. A → (r)

B → (s)

C → (p)

D → (p)

12. A → (s)

B → (r)

C → (q, t)

D → (p)

13. A → (p)

B → (q, s)

C → (p)

D → (q, r)

14. A → (r, s)

B → (r, s)

C → (p, s)

D → (p, s)

15. A → (p, s)

B → (q, s)

C → (q, s)

D → (s)

16. A → (p, r, s, t)

B → (p, r, s, t)

C → (p, q, r, s)

D → (q)

17. A → (q)

B → (r)

C → (s)

D → (p)

18. A → (s)

B → (r)

C → (p)

D → (q)

19. A → (r)

B → (q)

C → (p)

D → (s)

20. A → (s)

B → (r)

C → (p)

D → (q)

21. A → (p)

B → (r)

C → (p)

D → (p)

22. A → (q)

B → (r)

C → (p)

D → (s)

Integer/Numerical Answer Type Questions 1. 225

2. 33

3. 12

4. 2

6. 12

7. 72

8. 1

9. 60

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 239

5. 2275 10. 12

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1.240  JEE Advanced Physics: Electrostatics and Current Electricity 11. 52

12. 5

13. 1320

14. 24

15. 20

16. 2

17. 8

18. 33

19.  (a) 55, (b) 2293

20. 9

21. 30

22. 7080

23. 154

24. 99

25. 50

26. 365

27.  (a) 3, (b) 20, (c) 58

28. 100

29.  345, 16

30. 8

31. 35

32. 373

33. 100

34. 47

35. 2

36. 2

37. 700

38. 586

39. 758

40. 6

41.  1080, 155

42. 200

43. 5

44. 10

45. 30

46. 30

47. 2

ARCHIVE: JEE MAIN 1.  D

2.  B

3.  A

4.  A

5.  A

6.  A

7.  C

8.  D

9.  D

10.  C

11.  C

12.  C

13.  D

14.  C

15.  A

16.  A

17.  A

18.  B

19.  A

20.  D

21.  C

22.  A

23.  A

24.  C

25.  D

26.  C

27.  A

28.  B

29.  B

30.  A

31.  A

32.  D

33.  C

34.  A

35.  A

36.  B

37.  B

38.  C

39.  C, D

40.  A

41.  C

42.  C

43.  B

44.  D

45.  C

46.  A

47.  D

48.  B

49.  B

50.  A

51.  D

52.  D

53.  C

54.  D

55.  A

56.  D

57.  C

58.  D

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems 1.  D

2.  D

3.  C

4.  C

5.  D

6.  C

7.  C

8.  A

9.  D

10.  B

11.  A

12.  C

13.  C

14.  D

15.  B

16.  A

17.  B

18.  C

19.  D

20.  C

21.  B

22.  B

23.  C

24.  B

25.  B

26.  D

27.  B

28.  A

29.  D

30.  C

31.  B

32.  A

33.  B

34.  D

35.  B

36.  C

Multiple Correct Choice Type Problems 1.  A, B, C

2.  A, C

3.  A, D

4.  C

6.  B, D

7.  A, B, C

8.  A, C

9.  C, D

11.  A, D

12.  A

13.  A, B, C, D

14.  C, D

5.  C, D 10.  A, B, C, D 15.  A, C

16.  A, C

Reasoning Based Problems 1.  B

Linked Comprehension Type Problems 1.  A

2.  B

3.  A

4.  B

5.  C

Matrix Match/Column Match Type Questions 1.  A

Integer/Numerical Answer Type Questions 1.  6

2.  6

3.  3

01_Physics for JEE Mains and Advanced - 2_Part 6.indd 240

4.  2

9/20/2019 10:42:34 AM

CHAPTER

2

Capacitance and Applications

Learning Objectives After reading this chapter, you will be able to: After reading this chapter, you will be able to understand concepts and problems based on: (a) Capacitors (b) Capacitance

(c) Energy Concepts (d) Effect of Dielectric Insertion

(e) Capacitor Circuits (f) Spherical Capacitor and Cylindrical Capacitor

All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main and Advanced) are also given.

CAPACITORS: INTRODUCTION A capacitor is a device which stores electric charge. Capacitors vary in shape and size, but the basic configuration involves two conductors carrying equal but opposite charges (shown in figure). Capacitors have many important applications in electronics. Some examples include storing electric potential energy, delaying voltage changes when coupled with resistors, filtering out unwanted frequency signals, forming resonant circuits and making frequency dependent and independent voltage dividers when combined with resistors.

to the other one, giving one conductor a charge +Q, and the other one a charge −Q as a result of which a potential difference ΔV is created, with the positively charged conductor at a higher potential than the negatively charged conductor. Note that whether charged or uncharged, the net charge on the capacitor as a whole is zero. The simplest example of a capacitor consists of two conducting plates of area A, which are parallel to each other, and separated by a distance d (  A ), as shown in figure (called as a Parallel Plate Capacitor). +Q

d +Q

In the uncharged state, the charge on either one of the conductors in the capacitor is zero. During the charging process, a charge Q is moved from one conductor

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 1

A

–Q

–Q

Experimentally it has been verified that the amount of charge Q stored in a capacitor is linearly proportional to ΔV, the electric potential difference between the plates. Thus, we may write Q = C ΔV

9/24/2019 11:13:55 AM

2.2  JEE Advanced Physics: Electrostatics and Current Electricity

⇒ C=

Q DV

where C is a positive proportionality constant called capacitance. Physically, capacitance is a measure of the capacity of storing electric charge for a given potential difference DV. The SI unit of capacitance is the farad ( F ).

1 F = 1 farad = 1 coulomb volt -1 = 1 CV -1

Figure (a) shows the symbol which is used to represent capacitors in circuits. For a polarized fixed capacitor which has a definite polarity. Figure (b) is rarely used. + (a)

  

– (b)

CALCULATION OF CAPACITANCE As discussed, the method for the calculation of capacitance involves integration of the electric field between two conductors or the plates which are just equipotential surfaces to obtain the potential difference DV. Thus, a

  DV = - E ⋅ dr

∫



b

Since, by definition, we have C = ⇒ C=

Q = DV

Q DV

Q a

  - E ⋅ dr

∫



1 μF = 10 -6 F



1 nF = 10 -9 F

1 pF = 10 -12 F = 1 μμF CGS unit of capacitance is statfarad (SF) and

1 farad = 9 × 1011 statfarad

Dimensional formula for capacitance is,

[ C ] = M -1L-2T 4 A 2

ENERGY STORED IN A CHARGED CAPACITOR Whenever any conductor is charged, the charge produces an electric field outside it. Now if, further similar charge is to be given to the conductor some work has to be done by the external agency against the electrostatic forces of repulsion. This work done is stored as the electrostatic potential energy which is stored in the electrostatic field of the conductor. Let us consider that charge is given in a stepwise manner to the conductor. If V be the potential of the conductor at the instant when it has a charge q, then q V= C If dq is the additional infinitesimal charge that is to be given to the conductor and dW is the corresponding work done, then

dW = Vdq

q dq C Total work done to charge the conductor to Q is

⇒

dW =

b

DEFINITION OF C Q …(1) DV If DV = 1 volt, then C=

C = Q   (numerically)

i.e., capacitance of a conductor is numerically equal to the charge that raises its potential by 1 volt. Practically farad is a very big unit for capacitance and generally smaller units such as μF (microfarad), nF (nanofarad), pF (picofarad) are used.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 2

W= ⇒

W=

∫

Q

1 1 q2 dW = q dq = C C 2

∫ 0

Q 0

Q2 2C

If V0 is the potential that develops on the conductor finally on account of charge Q then Q V0 = C ⇒

W=

1 Q2 1 CV02 = = QV0 2 2C 2

9/20/2019 10:41:38 AM

Chapter 2: Capacitance and Applications 2.3 Illustration 1

Prove that while charging a capacitor half of the energy supplied by the battery dissipates in the form of heat.

the charge now starts flowing from the conductor at higher potential to the conductor at lower potential till both the conductors acquire the same potential. Conducting wire

Solution

C1

C2

C1

C2

When the switch S is closed, the charge stored in the capacitor is

V1

V2

V

V



C + –

S

q V V

q

Charge transferred from the battery is also q. So, energy supplied by the battery = qV = ( CV )( V ) = CV 2 . 1 However the capacitor stores only CV 2 which is 2 exactly half this energy. So, the remaining half or 50% 1 or CV 2 must have been dissipated as heat. 2

CAPACITANCE OF AN ISOLATED SPHERE Let a conducting sphere of radius R acquire a potential V when a charge Q is given to it. The potential acquired by the sphere is

⇒

V=

Q 4πε 0 R

C=

Q = 4πε 0 R V

In CGS system

q2 = C2V2

q1′ = C1V

Initially C



q1 = C1V1

q = CV

Total initial charge = q1 + q2 = C1V1 + C2V2 If V is the common potential after charge sharing takes place then Total final charge = q1′ + q2′ = C1V + C2V By Law of Conservation of Charge Total Initial Charge = Total Final Charge ⇒

C1V1 + C2V2 = ( C1 + C2 ) V

⇒

V=

Hence, in cgs system the capacitance of an isolated sphere equals its radius.

C1V1 + C2V2 …(1) C1 + C2

Further after sharing the new potential is the same. So, q1′ C1V C1 = = q2′ C2V C2

i.e., charges are shared in the ratio of the capacitances after common potential ( V ) is acquired. Further, there is always a loss in energy during the sharing process as some energy gets converted to heat. Loss = - DU = Total Initial Energy - Total Final Energy ⇒

1 =1 4πε 0

So, C = R (numerically).

q2′ = C2V Finally

1 ⎛1 ⎞ Loss = - DU = ⎜ C1V12 + C2V22 ⎟ ⎝2 ⎠ 2 1 ⎛1 2 2⎞ ⎜⎝ C1V + C2V ⎟⎠ 2 2

1 1 1 C1V12 + C2V22 - ( C1 + C2 ) V 2 2 2 2 From (1) put value of V above we get ⇒

Loss = - DU =

Loss = - DU =

1 ⎛ C1C2 ⎞ ( V1 - V2 )2 ⎜ ⎟ 2 ⎝ C1 + C2 ⎠

CHARGE SHARING BETWEEN TWO CHARGED CONDUCTORS



Consider two charged conductors having capacitance C1 and C2 at potential V1 and V2 respectively. Let the two be connected by a conducting wire such that

A capacitor of capacitance C0 is charged to a potential V0 and then isolated. A small capacitor C is then charged from C0 , discharged and then charged again.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 3

Illustration 2

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2.4  JEE Advanced Physics: Electrostatics and Current Electricity

If the process is being repeated n times, the potential of the large capacitor has now fallen to V. Calculate C. Solution

Let V1 be the potential of C0 after first charging, then

( C + C0 ) V1 = C0V0

⇒ V1 =

C0V0 C + C0

C0V1 = ( C + C0 ) V2

⎛ C0 ⎞ ⎛ C0 ⇒  V2 = ⎜ ⎝ C + C0 ⎟⎠ ⎜⎝ C + C0 so on

⎛ 2 ⎞( ) 9 = 6 μC and q2′ = ⎜ ⎝ 1 + 2 ⎟⎠ The common potential is given by

2

⎞ ⎛ C0 ⎞ ⎟⎠ V0 = ⎜⎝ C + C ⎟⎠ V0 and 0



n

A capacitor of 20 μF charged to 500 V, is connected in parallel to another capacitor of 10 μF charged to 200 V, find the common potential.

Two isolated spherical conductors have radii 6 cm and 12 cm respectively. They have charges of 12 μ C and -3 μ C . Find the charges after they are connected by a conducting wire. Also find the common potential after redistribution.

Solution

The common potential,



Solution

–

–

– –

–

–

– – –3 μ C +

+ + + +

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 4

–

+ +

+

+ + + q′1



–

R2

–

V +

+ +

Vcommon =

C1V1 + C2V2 C1 + C2

Here C1 = 20 μF = 20 × 10 -6 F , V1 = 500 V

Net charge = ( 12 - 3 ) μ C = 9 μ C

V

( 9 × 10 -6 ) ( 9 × 109 ) ( 18 × 10 -2 )

Illustration 4

Illustration 3

+ + +

( 9 × 10 -6 ) q1 + q2 = C1 + C2 4πε 0 ( R1 + R2 )

⇒ V = 4.5 × 10 5 V

1 ⎤ ⎡ ⎛ V0 ⎞ n ⎢ ⎥C ⇒ C= ⎜ 1 0 ⎢⎣ ⎝ V ⎟⎠ ⎦⎥

+ + + + + + R1 + + + + + 12 μ C

V=

⇒ V=

⎛ C0 ⎞ ⇒ Vn = V = ⎜ V0 ⎝ C + C0 ⎟⎠

+ + +



q1′ R1 6 1 = = = q2′ R2 12 2

⎛ 1 ⎞( ) ⇒ q1′ = ⎜ 9 = 3 μC ⎝ 1 + 2 ⎟⎠

Let V2 be the potential of C0 after second charging then

Since the charge is distributed in the ratio of their capacitance (or radii in case of spherical conductors), i.e.,

+ + q′2

⇒

C2 = 10 μF = 10 × 10 -6 F , V2 = 200 V

⇒ Vcommon =

20 × 10 -6 × 500 + 10 × 106 × 200 20 × 10 -6 + 10 × 10 -6

⇒ Vcommon = 400 V

Remark(s) CAPACITOR OR CONDENSER An arrangement which has capability of collecting (and storing) charge and whose capacitance can be varied is called a capacitor (or condenser).

9/20/2019 10:41:57 AM

Chapter 2: Capacitance and Applications 2.5

In a capacitor two charged conductors are placed near each other so that the potential of each conductor is determined not only by its own charge but also by the magnitude and sign of the charge on the neighbouring conductor. The capacitance of a capacitor depends (a) directly on the size of the conductors of the capacitor. (b) directly on the dielectric constant K of the medium between the conductors. (c) inversely on the distance of separation between the conductors. The capacitance of a capacitor also depends on the presence of conductors present in neighbourhood and is independent of the material of metallic conductor.

Sharing Of Charges between capacitors And Common Potential Just like conductors, when two isolated capacitors (either both of them are charged or one of them is charged) are connected to each other, then redistribution of charge takes place due to potential difference between them. The charge flows from higher potential to lower potential till both of them acquire the same potential called common potential. Consider two capacitors, C1 and C2 charged to potentials, V1 and V2 . Let the capacitors be connected such that CASE-1: Their similar plates are connected to each other (i.e. positive plate of one capacitor is connected to positive plate of other capacitor and so on). C1, V

C1, V1

K

C2, V

Consider the part of circuit inside dotted loop. This part is isolated from other part. Therefore, total charge of this part remains constant. Hence

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 5

C1V1 + C2V2 = C1V + C2V

⇒ V=

C1V1 + C2V2 C1 + C2

During this redistribution of charge, there is a loss of energy in the form of heat and electromagnetic radiations, given by Loss of energy is - DU = Ui - U f . ⇒ - DU =

1 1 1 C1V12 + C2V22 - ( C1 + C2 ) V 2 2 2 2

Substituting V = Loss = - DU =

C1V1 + C2V2 in above equation, we get C1 + C2

1 ⎛ C1C2 ⎞ ( V1 - V2 )2 ⎜ ⎟ 2 ⎝ C1 + C2 ⎠

CASE-2: Their dissimilar plates are connected together (i.e. positive plate of one capacitor is connected to the negative plate of other capacitor and so on). C1, V1

C 1, V K is closed

C2, V2

K

C2, V

Again applying the Law of Conservation of Charge, we get

C1V1 - C2V2 = C1V + C2V

⇒ V=

C1V1 - C2V2 C1 + C2

So, loss in energy is given by Loss = - DU =

K is closed C2, V2



1 C1C2 ( V1 + V2 )2 2 C1 + C2

Illustration 5

A parallel plate capacitor of capacitance C has been charged, so that the potential difference between its plates is V . Now, the plates of this capacitor are

9/20/2019 10:42:03 AM

2.6  JEE Advanced Physics: Electrostatics and Current Electricity

c­ onnected to another uncharged capacitor of capacitance 2C. Find the common potential acquired by the system and loss of energy.

In this problem, we have

( C1 + C2 ) V = Q1 + Q2 = Q0

Q0 960 μ C ⇒   V = C + C = 12 μF = 80 V 1 2

C1V1 + C2V2 C1 + C2 CV + 0 V = = 3C 3

VCommon =

⇒   Q1 = 640 μ C , Q2 = 320 μ C The total energy stored is given by

1 C × 2C 1 2 and Loss = - DU = × ( V - 0 ) = CV 2 2 3C 3 Illustration 6

In the circuit shown, C1 = 8 μF , C2 = 4 μF and V0 = 120 V. When the switch S is closed, the charged capacitor C1 gets connected to an uncharged capacitor C2 . Calculate the Q0

1 1 1 U = Q1V + Q2V = Q0V 2 2 2      1 -6 ⇒   U = 2 ( 960 × 10 ) ( 80 V ) = 0.038 J The energy after closing the switch is less than the original energy of 0.058 J. This difference has been converted to energy of some other form (e.g. heat energy or energy radiated as electromagnetic waves). Illustration 7

V0

C1

Q1 = C1V …(1) Q2 = C2V …(2)

    

C1 = C , C2 = 2C , V1 = V and V2 = 0 Common potential is given by

VCommon



Adding equation (1) and (2), we get

Solution



In the steady state, we have

S

C2

(a) charge on each capacitor after switching S. (b) total energy after switching S. Solution

Two spheres 1 and 2 of radii R and 2R having Q charges Q and , respectively are connected with 2 a cell of potential difference V as shown in Figure. When the switch Sw is closed, calculate the final charge on each sphere.

(a) Before switching on S, Q0 = C1V0 = 960 μ C and Energy stored U =



U=

1 Q0V0 2

1( 960 × 10 -6 C ) ( 120 V ) 2

–Q1

C1

(b) After we close switch S, we have



Q0 = Q1 + Q2

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 6

V Sw

1

Q 2

2R 2

Solution

U = 0.058 J +Q1

R

Q

+Q2

When the switch is closed, the potential difference between the spheres should be V . Let charge q flow from the sphere of radius R to sphere of radius 2R. Then

( q2 )final ( q1 )final

C2



C2

-

C1

=V

C2 = 4πε 0 ( 2R ) and C1 = 4πε 0 R

9/20/2019 10:42:17 AM

Chapter 2: Capacitance and Applications

R 1

Sw

2R

V

Finally

(q1)f = (Q – q)

⇒

3 q = 8πε 0 RV +

⇒

q=

2 (q2)f =

Q+q 2

2.7

3Q 2

8πε 0 RV Q + 3 2

So the final charges on each of the spheres are ⇒

⎛Q ⎞ ⎜⎝ + q ⎟⎠ (Q - q ) = V 2 ( ) 4πε 0 2R 4πε 0 ( R )

⇒

Q + q - 2Q + 2q = 4πε 0 ( 2R ) V 2

and

( q1 ) f

=Q-q=

( q2 ) f

=

Q 8πε 0 RV 2 3

8πε 0 RV Q +q = Q+ 2 3

Test Your Concepts-I

Based on General Capacitance 1. A capacitor of 20 μF charged to 500 V, is connected in parallel to another capacitor of 10 μF charged to 200 V, find the common potential. 2. Calculate the heat generated when a condenser of 100 μF capacity and charged to 200 V is discharged through a 2 Ω resistance. 3. The capacitance of a variable radio capacitor can be changed from 50 pF to 950 pF by turning the dial from 0° to 180°. With the dial set at 180°, the capacitor is connected to a 400 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned at 0°. (a) What is the potential difference across the capacitor when the dial reads 0°? (b) How much work is required to turn the dial, if friction is neglected? 4. A battery of 10 V is connected to a capacitor of capacitance 0.1 F. The battery is now removed and this capacitor is connected to a second uncharged capacitor. If the charge distributes equally on these two capacitors, find the total energy stored in the

PRINCIPLE OF A PARALLEL PLATE CAPACITOR Consider a conducting plate A which is given a charge Q (as shown in Figure 1(a)) such that its potential rises to V. Then

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 7

(Solutions on page H.126) two capacitors. Further compare this energy with the initial energy stored in the capacitors. 5. Consider two conducting spheres with radii R1 and R2 separated by a distance much greater than radius of either. A total charge Q is shared between the spheres, subject to the condition that the electric potential energy of the system has the smallest possible value. The spheres being very far apart, you can assume that the charge of each is uniformly distributed over the surface of each. (a) Determine the values of charges on the spheres in terms of Q, R1 and R2. (b) Find the potential difference between the surfaces of the spheres. 6. A capacitor of capacitance C0 is charged to a potential V0 and then isolated. A small capacitor C is then charged from C0, discharged and charged again, the process being repeated 10 times. The potential of the capacitor C0 has now fallen to V. Calculate C.

C=

Q V

Let us place another identical conducting plate B parallel to it such that charge is induced on plate B (as shown in Figure 1(b)).

9/20/2019 10:42:23 AM

2.8  JEE Advanced Physics: Electrostatics and Current Electricity

A + + + + + + + + +

+ + + + + + + + Q +

A + + + + + + + + +

 

Figure 1(a)

A

B – – – – – – – – –

+ + + + + + + + +

+ + + + + + + + +

 

+ + + + + + + + +

Figure 1(b)

+ + + + + + + + +

B – – – – – – – – –

+ + + + + + + + +

σ σ = ε Kε0

E=

⇒

A+ + + + + + + +

Figure 1(c)

If V- is the potential at A due to induced negative charge on B and V+ is the potential at A due to induced positive charge on B, then C′ =

Q Q = V ′ V + V+ - V-

Since V ′ < V (as the induced negative charge lies closer to the plate A in comparison to induced positive charge). ⇒

C′ > C

Further, if B is earthed from the outer side (as shown in Figure 1(c)) then V ′′ = V - V- as the entire positive charge flows to the earth. So ⇒

C ′′ =

⇒

{

q V =  d Kε0 A

C=

∵E=

– – – – – – – –

+ + + + + + + +

}

B σ E=σ ε = kε 0

q Kε0 A = V d

If medium between the plates is air or vacuum, then

K=1

⇒

C0 =

ε0 A d

Remark(s)

Q Q = V ′′ V - V-

C ′′  C

So, if an identical earthed conductor is placed in the viscinity of a charged conductor then the capacitance of the charged conductor increases appreciably. This is the principle of a parallel plate capacitor.

Suppose charges Q1 and Q2 are given to the two plates. The resulting charge distribution on the various surfaces is shown in the figure and is derived using the concept that no field exists in the thickness of the conductor.

EA = 0 q

y

Q1 – q

PARALLEL PLATE CAPACITOR It consists of two metallic plates A and B each of area A at separation d. Plate A is positively charged and plate B is earthed. If K is the dielectric constant of the material medium and E is the field that exists between the two plates, then A

q V and σ = d A

B

A

B (Q2 – q′)

⇒

x

q′

1 ⎡ q - ( Q1 - q ) - ( Q2 - q ′ ) - q ′ ⎤⎦ = 0 2ε 0 A ⎣

⇒ q - Q1 + q - Q2 + q ′ - q ′ = 0 –σ

+σ

A = Area of plate d = Separation between the plates.

⇒ 2q - Q1 - Q2 = 0 ⇒ q=

Q1 + Q2 2

d

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 8

9/20/2019 10:42:35 AM

Chapter 2: Capacitance and Applications 2.9

ELECTROSTATIC FORCE BETWEEN THE PLATES OF A PARALLEL PLATE CAPACITOR

Similarly EB = 0 ⇒

1 ⎡ q + ( Q1 - q ) + ( Q2 - q ′ ) - q ′ ⎤⎦ = 0 2ε 0 A ⎣

⇒ Q1 + Q2 - 2q ′ = 0 ⇒ q′ =

Q1 + Q2 2

The plates of the capacitor each carry equal and opposite charges, hence they must attract each other with a force, say F. At any instant let the plate separation be x, then

C=

2Q - Q1 - Q2 Q1 - Q2 ⇒ Q1 - q = 1 = 2 2

Also U =

⎛ Q + Q2 ⎞ and Q2 - q ′ = Q2 - ⎜ 1 ⎟ ⎝ 2 ⎠

⇒

2Q2 - Q1 - Q2 Q2 - Q1 ⎛ Q - Q2 ⎞ = = -⎜ 1 ⎟ ⎝ 2 2 2 ⎠

⇒ Q2 - q ′ =

So, the distribution of charges is as shown below Q1 + Q2

–

2

Q1 – Q2 2

ε0 A x Q2 2C

⎛ Q2 ⎞ U=⎜ x ⎝ 2ε 0 A ⎟⎠

Let the plates be moved towards each other through dx, such that the new separation between the plates is ( x - dx ). If U f is the final potential energy, then



Uf =

Q2 Q2 ( x - dx ) = 2C ′ 2ε 0 A +Q

Q1 – Q2 2

–Q

Q1 + Q2 2

The charges on the inner surfaces are equal and opposite and the field between the plates is,



E=

σ ( q1 - q2 ) = ε0 2 Aε 0

If dU is the change in potential energy, then

Thus, the potential difference between the plates is,



V = Ed =

( q1 - q2 ) d 2 Aε 0

q In order to find C = we must now use the value V ( q1 - q2 ) for q because this is the charge responsi2 ble for creating the field between the two plates. The outer charges

( q1 + q2 )

do not come in the picture.

2 ε0 A Thus, we get C = as before. d

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 9



dU = U f - Ui Q2 Q2 ( x - dx ) x 2ε 0 A 2ε 0 A

⇒

dU =

⇒

dU = -

Q2 dx 2ε 0 A

Further since ⇒ 

F=F=

dU dx

⎛ σ2 ⎞ Q2 ⎞ ⎛1 =⎜ A = ⎜ ε 0 E2 ⎟ A ⎠ ⎝2 2ε 0 A ⎝ 2ε 0 ⎟⎠

σ ⎫ ⎧ ⎨∵ Q = σ A, E = ⎬ ε 0 ⎭ ⎩

9/20/2019 10:42:44 AM

2.10  JEE Advanced Physics: Electrostatics and Current Electricity

From here too, we observe that electrostatic pressure 1 σ2 Pe = ε 0 E2 = 2 2ε 0

(c) CS has a value smaller than the least capacitance of the circuit, provided all capacitors are connected in series.

CAPACITORS IN SERIES

Illustration 8

In the circuit shown in figure, find

+Q –Q +Q –Q +Q –Q A

C2

C1 V1

C3 V2 V

A

B

+Q –Q

V3

V

B

CS

(a) the equivalent capacitance (b) the charge stored in each capacitor and (c) the potential difference across each capacitor. 4 μF

6 μF

In this arrangement of capacitors the charge has no alternative path(s) to flow. (a) The charges on each capacitor are equal i.e., Q = C1V1 = C2V2 = C3V3 …(1) (b) The total potential difference across AB is shared by the capacitors in the inverse ratio of the capacitances.

100 V

Solution

(a) The equivalent capacitance

V = V1 + V2 + V3 …(2)

If CS is the net capacitance of the series combination, then  

Q Q Q Q 1 1 1 1   ⇒  = + + = + + CS C1 C 2 C3 CS C1 C2 C3

Further V1 =



     C=

C1C2 C1 + C2

( 4 )( 6 ) ⇒   C = 4 + 6 = 2.4 μF

Q Q and V = C1 CS

1 V1 C1 ⇒ = 1 V CS 1 ⎛ ⎜ C1 ⇒ V1 = ⎜ ⎜ 1 + 1 + 1 ⎜⎝ C C C3 1 2

6 μF

+ – q V1

+ – q V2

100 V

⎞ ⎟ ⎛ CS ⎞ V ⎟V = ⎜ ⎝ C1 ⎟⎠ ⎟ ⎟⎠

1 ⎛ ⎜ C2 Similarly, V2 = ⎜ ⎜ 1 + 1 + 1 ⎜⎝ C C C3 1 2 1 ⎛ ⎜ C3 and  V3 = ⎜ ⎜ 1 + 1 + 1 ⎜⎝ C C C3 1 2

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 10

4 μF

⎞ ⎟ ⎛ CS ⎞ V ⎟V = ⎜ ⎝ C2 ⎟⎠ ⎟ ⎟⎠

⎞ ⎟ ⎛ CS ⎞ V ⎟V = ⎜ ⎝ C3 ⎟⎠ ⎟ ⎟⎠

(b) The charge q, stored in each capacitor is,

 q = CV = ( 2.4 × 10 -6 ) ( 100 ) C

⇒   q = 240 μ C (c) In series combination, V ∝

1  C

{∵ q = constant }

V1 C2 ⇒  V = C 2 1 ⎛ C2 ⎞ ⎛ 6 ⎞ ⇒   V1 = ⎜ C + C ⎟ V = ⎜⎝ 4 + 6 ⎟⎠ ( 100 ) = 60 V ⎝ 1 2 ⎠ So, V2 = V - V1 = 100 - 60 = 40 V

9/20/2019 10:42:54 AM

Chapter 2: Capacitance and Applications 2.11

CAPACITORS IN PARALLEL +Q1

2 μF 4 μF

–Q1

6 μF

C1 A

+Q2

–Q2

B

A

+Q2

–Q2

C2 +Q3

100 V

V

–Q3 V

B

Cp

C3

Solution

In such an arrangement of capacitors the charge has an alternative path(s) to flow.

(a) The capacitors are in parallel. Hence the equivalent capacitance is,

  Q = Q1 + Q2 + Q3 …(3) If Cp is the net capacitance for the parallel combination of capacitors then

q : q : q = C1 : C2 : C3 = 2 : 4 : 6     1 2 3 2 ⎛ ⎞ ⇒   q1 = ⎜⎝ 2 + 4 + 6 ⎟⎠ × 1200 = 200 μ C

  CPV = C1V + C2V + C3V ⇒  CP = C1 + C2 + C3

4 ⎛ ⎞ ⇒   q2 = ⎜⎝ 2 + 4 + 6 ⎟⎠ × 1200 = 400 μ C

Further Q = CP V ⇒ 

Q1 = C1V ,

Q2 = C2V ,

Q3 = C3V ,

Q1 C1 = Q CP

⎛C ⇒  Q1 = ⎜ 1 ⎝ CP

⇒   C = ( 2 + 4 + 6 ) = 12 μF ( b) Total charge drawn from the battery q = CV = 12 × 100 μ C = 1200 μ C      This charge will be distributed in the ratio of their capacitances. Hence,

6 ⎛ ⎞ × 1200 = 600 μ C and q3 = ⎜ ⎝ 2 + 4 + 6 ⎟⎠

Problem Solving Technique(s)

⎞ ⎟⎠ Q

⎛C Similarly Q2 = ⎜ 2 ⎝ CP



C = C1 + C2 + C3

(a) The potential difference across each capacitor is same and equals the total potential applied. i.e., V = V1 = V2 = V3 …(1) Q Q Q ⇒  V = 1 = 2 = 3 …(2) C1 C2 C3 (b) The total charge Q is shared by each capacitor in the direct ratio of the capacitances.

⎞ ⎟⎠ Q

⎛ C3 ⎞   Q3 = ⎜⎝ C ⎟⎠ Q P (c) CP has a value greater than the greatest capacitance of the circuit. Illustration 9

In the circuit shown in figure, find (a) the equivalent capacitance and (b) the charge stored in each capacitor.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 11

(a) If C1, C2, C3,… are capacitors connected in series and if total potential across all is V, then potential across each capacitor is ⎛ ⎜ V1 = ⎜ ⎜ ⎜⎝

1 C1 1 CS

⎛ 1 ⎞ ⎜ C2 ⎟ ⎟ V ; V2 = ⎜ ⎜ 1 ⎟ ⎜⎝ C ⎟⎠ S



 



and so on, where

⎛ 1 ⎞ ⎜ C3 ⎟ ⎟ V ; V3 = ⎜ ⎜ 1 ⎟ ⎜⎝ C ⎟⎠ S

⎞ ⎟ ⎟V ⎟ ⎟⎠

1 1 1 1 1 = + + + ... + C S C1 C2 C3 Cn

(b) If C1, C2, C3,… are capacitors connected in parallel and if Q is total charge on the combination, then charge on each capacitor is

9/20/2019 10:43:05 AM

2.12  JEE Advanced Physics: Electrostatics and Current Electricity





In this situation except two extreme plates each plate is common to two adjacent capacitors. (i) n plates arranged as shown in figure constitute ε A ( n - 1) capacitors in series each of value ⎛⎜ 0 ⎞⎟ , ⎝ d ⎠ so that

⎛C ⎞ ⎛C ⎞ ⎛C ⎞ Q1 = ⎜ 1 ⎟ Q; Q2 = ⎜ 2 ⎟ Q; Q3 = ⎜ 3 ⎟ Q ⎝ CP ⎠ ⎝ CP ⎠ ⎝ CP ⎠   and so on, where CP = C1 + C2 + C3 + ... + Cn

(c) The current through a capacitor is zero as long as the voltage across it does not change with time (or as long as capacitor is in the steady state i.e., charge on the capacitor has build up to attain a constant maximum value). (d) It is not possible to change the voltage across a capacitor by a finite amount in zero time as this process requires an infinite amount of current to flow through the capacitor. (e) A capacitor can store a finite amount of energy in it even if the current flowing through it is zero when voltage across the capacitor is constant. (f) A true mathematical (theoretical) model of a capacitor should never dissipate energy but can only store it whereas a physical (practical) model never does so. (g) A parallel plate capacitor must have plates of large area (A) in comparison to the distance of separation (d) between the plates to avoid Fringing Effect. Fringing Effect is the bending of field lines at the corners of a capacitors with large d (see figure). A

B



CS =



+ + + + + –







+ + + + +

ε0 A d – – – – –

+ + + + +

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 12

– – – – –

+ + + + +

– – – – –

+ + + + +

– – – – –

– – – – – –

++ ++ ++ ++ ++

– – – – –

– – – – –

+ + + + +

+ + + + +

+

– – – – –

Illustration 10

An air-dielectric capacitor is formed by two nonparallel plates, each of area A. An edge view of the arrangement is shown in figure. Note that the top plate is tilted relative to the bottom plate so that on one edge the plate separation is d + Dd , while on the other edge it is d - Dd . Assuming that Dd  d and that d is small compared with the length of the plate,

Fringing effect for a capacitor with large d.

CP = ( n - 1)

– – – – – –

In this situation except two extreme plates each plate is common to two adjacent capacitors.

show that C =

(h) n plates arranged as shown in figure constitute ( n - 1) capacitors in parallel, each of value ⎛ ε0 A ⎞ ⎜⎝ ⎟ , so that d ⎠

ε0 A d ( n - 1)

2 ε0 A ⎡ 1 ⎛ Dd ⎞ ⎤ ⎢ 1 + ⎜⎝ ⎟ ⎥. d ⎣ 3 d ⎠ ⎦

Δd Δd

Solution

Consider an infinitesimal element of thickness dx at a distance x from O. Then the capacitor is thought of being made from such infinitesimal elements having their capacitances in parallel. If dC be the capacitance of such an ­element, then



dC =

ε 0 ( dA ) y

9/20/2019 10:43:12 AM

Chapter 2: Capacitance and Applications 2.13

Since, Dd  d , so

AN ≅ xθ

⇒ dC =

2Δd

ε 0 ( b dx )

θ l

d + xθ N O

θ

A

Δd

y

x

dx

where b = breadth of the plate and l = length of the plate So, total capacitance C is given by

C = sum of capacitances of infinitesimal elements l 2

⇒ C=

ε 0 b dx

∫ dC = ∫ d + xθ -

l 2

⇒ C=

3 ε 0 b ⎡ ⎛ lθ ⎞ 1 ⎛ lθ ⎞ ⎤ ⎢ 2 ⎜⎝ ⎟⎠ + 2 ⎜⎝ ⎟⎠ ⎥ 3 2d ⎦ θ ⎣ 2d

⇒ C=

ε 0 b ⎡ lθ 1 l 3θ 3 ⎤ ⎢ + ⎥ θ ⎣ d 12 d 3 ⎦

⇒ C=

ε 0 lb ⎡ 1 l2 2 ⎤ θ ⎥ ⎢1+ d ⎣ 12 d 2 ⎦

But

ε 0 lb ε 0 A = d d

and θ 

2Dd l

⇒ C=

2 ε0 A ⎡ 1 l 2 ⎛ 4 ( Dd ) ⎞ ⎤ ⎢1+ ⎜ ⎟⎥ d ⎣ 12 d 2 ⎝ l 2 ⎠ ⎦

⇒ C=

2 ε0 A ⎡ 1 ⎛ Dd ⎞ ⎤ 1 + ⎢ ⎜ ⎟ ⎥ d ⎣ 3⎝ d ⎠ ⎦

l

2 ε b ⇒ C = 0 log e ( d + xθ ) l θ -

2

ε b⎡ lθ ⎞ lθ ⎞ ⎤ ⎛ ⎛ ⇒ C = 0 ⎢ log e ⎜ d + ⎟ - log e ⎜ d - ⎟ ⎥ ⎝ ⎝ 2⎠ 2⎠⎦ θ ⎣ ⇒ C=

ε0 b ⎡ lθ ⎞ lθ ⎞ ⎤ ⎛ ⎛ log ⎜ 1 + ⎟ - log ⎜ 1 - ⎟ ⎥ ⎝ ⎝ θ ⎢⎣ 2d ⎠ 2d ⎠ ⎦

Using log e ( 1 + x ) = x -

x2 x3 x4 + - .......... 2 3 4

⎛ ⎞ x2 x3 x4 and log e ( 1 - x ) = - ⎜ x + + + + ...... ⎟ ⎝ ⎠ 2 3 4 2 3 ⎤ ε b ⎧⎪ ⎡ lθ 1 ⎛ lθ ⎞ 1 ⎛ lθ ⎞ ⇒ C = 0 ⎨⎢ - ⎜ + ⎜ + .... ⎥ + ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ 3 2d θ ⎩⎪ ⎣ 2d 2 2d ⎦



⎡ lθ 1 ⎛ lθ ⎞ 2 1 ⎛ lθ ⎞ 3 ⎤ ⎫⎪ ⎢ + ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ + ..... ⎥ ⎬ 3 2d ⎣ 2d 2 2d ⎦ ⎭⎪

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 13

Illustration 11

A parallel plate square capacitor has the space between the plates filled with a medium whose dielectric constant increases uniformly with distance x from one of its plate as K = K1 + α x . If d is distance between the plates and K1 and K 2 are dielectric constants of the medium at the two square plates, find the capacity of the capacitor. Solution K2 = K1 + α d K = K1 + α x

d dx x A

K1

Since K = K1 + α x and at x = d , K = K 2

9/20/2019 10:43:26 AM

2.14  JEE Advanced Physics: Electrostatics and Current Electricity

Since, these elementary capacitors are joined in ­parallel, so

K 2 - K1 d The capacitance of the elementary capacitor is ⇒ α=



dC =

C=

Aε 0 ( K1 + α x )



dx

∫

d

∫ 0

dx 1 ⎛ ln ( K1 + α x ) ⎞ = ⎜ ⎟⎠ α ( K1 + α x ) Aε 0 ⎝

ε 0 A ⎛ K1 + K 2 ⎞ ⎜ ⎟⎠ d ⎝ 2

0

ELECTROSTATIC ENERGY DENSITY (uE)

Illustration 12

A parallel plate square capacitor has the space between the plates filled with a dielectric whose dielectric constant increases linearly with distance x from one edge to the other as K = K1 + α x , where K1 and K 2 are dielectric constants of the medium at the two edges of the square plates, find the capacity of the capacitor.

x

dx l

x=l

Since K = K1 + α x and at x = l , we have K = K2

K - K1 ⇒ α= 2 l The capacitance of the elementary capacitor is

dC =

ε 0 ( ldx ) ( K1 + α x ) d

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 14

1 CV 2 2

1 ε0 A 2 2 E d 2 d

U=

⇒

⎛1 ⎞ U = ⎜ ε 0 E2 ⎟ ( Ad ) ⎝2 ⎠

⇒

U=

1 ε 0 E2t 2

where t ( = Ad ) is volume of the capacitor Electrostatic Energy 1 U σ2 = uE = = ε 0 E2 = t 2 2ε 0 Volume

This energy is stored in the capacitor in the form of electrostatic field. Since,

Electrostatic Energy = Electrostatic Pressure Volume

⇒

Electrostatic Pressure =

Solution



ε0 A σ and V = Ed ,   where E = ε0 d

⇒

⇒

x=0

1 ⎛ K 2 - K1 ⎞ ⎤ ⎡ ⎟⎠ l ⎥ ⎢ K1 + 2 ⎜⎝ l ⎣ ⎦

For a parallel plate capacitor, we have

Since, U =

K2 = K1 + α l

αl ⎞ ⎛ ⎜⎝ K1 + ⎟⎠ 2

⇒ C=

C=

K1

0

ε0 l2 d

d

Aε 0 ( K 2 - K1 ) ⎛K ⎞ d ln ⎜ 2 ⎟ ⎝ K1 ⎠

d

∫ ( Kl + α x ) dx =

ε0 l2 d

1 1 d ⎛ K + αd ⎞ ⎛K ⎞ = = ⇒ ln ⎜ 1 ln ⎜ 2 ⎟ ⎟ C Aε 0α ⎝ K1 ⎠ Aε 0 ( K 2 - K1 ) ⎝ K1 ⎠ ⇒ C=

l

⇒ C=

Since these elementary capacitors are joined in series, so 1 1 1 = = C dC Aε 0

∫

ε l dC = 0 d

1 σ2 ε 0 E2 = = uE 2 2ε 0

Also we note that if we take dt as the volume ­element, then

uE =

∫ u dt E

I, have taken a new symbol ( t ) for the volume, because we may confuse V for potential.

9/20/2019 10:43:39 AM

Chapter 2: Capacitance and Applications 2.15

Remark(s)



CP = C1 + C2 + C3 + ...

The potential energy U of a charged conductor or a capacitor is stored in the electric field. The energy per unit volume is called the energy density (uE). Energy density (uE) in a dielectric medium is given by,

⇒

1 1 1 1 CPV 2 = C1V 2 + C2V 2 + C3V 2 + ... 2 2 2 2

⇒

U P = U1 + U 2 + U 3 + ...

1 uE = ε 0KE 2 2

This relation shows that the energy stored per unit volume depends on E2. If E is the electric field in a space of volume dt, then the total stored energy in an electrostatic field is given by,

∫

1 U = ε 0K E 2 dt 2

and if E is uniform throughout the volume (electric field between the plates of a capacitor is uniform), then the total stored energy can be given by,

1 U = K ε 0 E 2t 2

i.e., if U1 is energy stored in capacitor C1, if U 2 is energy stored in capacitor C2 and so on, and U P is total energy for parallel combination of C1, C2 , C3 ,… then too



⎛ Total Energy ⎜ for the parallel ⎜⎝ combination

⎞ ⎛ Sum of individual ⎞ ⎟ = ⎜ energy stored in ⎟ ⎟⎠ ⎜⎝ each capacitor ⎟⎠

Problem Solving Technique(s) So, to conclude, if capacitors are connected in series or in parallel the total energy across the combination (any one) is equal to the sum of the individual energies stored in each capacitor.

ENERGY FOR SERIES AND PARALLEL COMBINATIONS

CALCULATING THE NET CAPACITANCE OF CIRCUITS

Series Combination

To calculate the net capacitance of a circuit/network, the steps given below should be followed.

For a series combination of capacitors Q = constant and 1 1 1 1 = + + + ... CS C1 C2 C3 Q2 Q2 Q2 Q2 ⇒ = + + + ... 2CS 2C1 2C2 2C3 ⇒

US = U1 + U 2 + U 3 + ...

i.e., if U1 is energy stored in capacitor C1 , if U 2 is energy stored in capacitor C2 and so on, and US is total energy for series combination of C1 , C2 , C3 ,… then, ⎛ Total Energy ⎞ ⎛ Sum of individual ⎞ ⎜ for the series ⎟ = ⎜ energies stored in ⎟ ⎜⎝ combination ⎟⎠ ⎜⎝ each capacitor ⎟⎠

Parallel Combination For a parallel combination of capacitors V =constant and

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 15

STEP-1: Identify the two points across which the equivalent capacitance is to be calculated. STEP-2: Imagine a battery to be connected between these points. STEP-3: Start solving the circuit from the reference point which is farthest from the points between which the equivalent capacitance has to be calculated. This point is likely to be not a node.

Simple Circuits Analyse the circuit carefully to conclude which pair of capacitors are in series and which are in parallel (This all should be done keeping in mind the points across which net capacitance has to be calculated). Find their net capacitance and again draw an equivalent diagram to apply the above specified technique repeatedly so as to get the total capacitance between the points A and B as Illustrated below.

9/20/2019 10:43:48 AM

9 μF

Series 9 = 3μF 3

9 μF

9 μF

A 6 μF

6 μF

2.16  JEE Advanced Physics: Electrostatics and Current Electricity

(i)

3×6 = 2 μF 3+6 3 μF

6 μF

3 μF

B



B

3 μF

9 μF

9 μF

9 μF

A

9 μF

3 μF

6 μF

6 μF

6 μF

A

9 μF

9 μF

B A

9 μF

Parallel 6 + 3 = 9 μF

9 μF

B

3 μF

9 μF A 2 μF

9 μF

CAB = 2 + 3 = 5 μ F B

A

(ii)

B

3 μF

6 μF

By similar process CAB = 3 μF Illustration 13

6 μF ⇒

3 μF

6 μF

6 μF A

2 μF

6 μF

6 = 3μF B 2

3 μF A

2 μF

9 μF

6 μF

Find C if the equivalent capacitance between the points A and B in the circuit shown is 1 μF. All the capacitances are in μF.

B

A

C

1 6

4

8

⇒

AB

= 3 + 2 = 5 μF

6 μF

(iii)

2 μF 9 μF

B

3 μF 6 μF

B

A

9 μF

2 μF

9 μF

B

9 μF

6 μF

9 μF

9 μF

9 μF

9 μF

9 μF

B

Solution

After a brief simple analysis, we conclude the following information (a) 2 and 2 are in parallel to give 4 μF (b) 6 and 12 are in series to give

A 6 μF

Series 9 = 3μF 3

9 μF

B

9 μF 9 μF

6 μF 9 μF 9 μF

A 6 μF B

6 μF

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 16

9 μF

9 μF

9 μF

6 × 12 = 4 μF 6 + 12

(c) This 4 μF (net capacitance of 6 and 12) is in parallel combination to another 4, so as to give us 8 μF So, using the above information, we redraw the above circuit Further, we conclude C

A 6 μF

12

2

3 μF ⇐

6 μF

A

2

3 + 3 = 6 μF

6 = 3μF 2 C

A 8

1 8

4

9 μF

B

(Continued) 3 μF Parallel 6 + 3 = 9 μF

9/20/2019 10:43:53 AM

Chapter 2: Capacitance and Applications 2.17

(d) 1 and 8 are in series, so we get

1× 8 8 = μF 1+ 8 9

(e) 8 and 4 are in series, so we get

8 × 4 32 = μF 8 + 4 12

(ii) A

A

32 9

A

(iii)

23C = 32

⇒

C=

C

C A

B

C

B

B C

A

B B

C

C

A

B

C

32 μF 23

+ – CAB = 3C

Circuits With Extra Wires

(iv)

C

If there is no capacitor or resistor in any branch of a circuit, then every point of this branch will be at same potential. Suppose equivalent capacitance is to be determined in following cases.

A

C

C

C

B

C

C C

C

A

C

C B

A B

CAB = 3C

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 17

A

C

Parallel C + C = 2C

A

2C B

C

B

Series 2C 2C × C = 2C + C 3

C

C

C

C

C

2C/3

B

C

B

C

Parallel 2C + C = 5C 3 3

B

A

C B

A

A B

B

C A

A

B

C

A

C

C CAB = 2C

C

⇒

A

B

A

A

32C = 32 + 9C

A

B

C

⇒

(i)

C

B

A A B C No p.d. across vertical branch so it can removed

32 C 9 =1 32 +C 9

A

C

C

A

⇒

B

A

A

Finally

C

C C

8 32 (f) μF and μF are in a parallel combination to 9 12 8 32 32 give net capacitance + = μF 9 12 9 32 μF and C are in series between A and B, 9 so as to give 1 μF. So, the final equivalent circuit is shown in figure.

C

C A

B

B

Hence equivalent capacitance between A and B 5C . is 3

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2.18  JEE Advanced Physics: Electrostatics and Current Electricity

(v) Since there is no capacitor in the path APB, the points A, P and B are electrically same i.e., the input and output points are directly connected (short circuited). C

1 C 3 A

C

2

2C

2C C C

C 4

P

C

P

B

C

LOS A

B

C

P

Thus, entire charge will prefer to flow along path APB. It means that the capacitors connected in the circuit will not receive any charge for storing. Thus equivalent capacitance of this circuit is zero.

Now, the concept of Line of Symmetry makes our job easy to calculate capacitance across AP. (1) and (2) are in parallel further in series with (3), whose resultant capacitance is in parallel with (4). Resultant of (1) and (2) is 3C Resultant of 3C and (3) is

3C 7C and (4) is . 4 4 So total capacitance across AB is Resultant of

Concept Of Line Of Symmetry Line of symmetry (L.O.S.) is an imagination of our mind to divide a highly symmetric circuit into two equal halves such that the points of the circuit through which LOS passes are at equal potential.

3C . 4

⇒

CAB =

CAP 2

CAB =

7C 8

Balanced Wheatstone Bridge Illustration 14

Find the net capacitance of the circuit shown between the points A and B.

If in a circuit, five capacitors are arranged as shown in following figure, the circuit is called Wheatstone Bridge type circuit.

C

C1

C

C

C C

A C

A

Solution

This circuit is highly symmetric and so we can consider the line of symmetry to pass through the circuit to divide it into two equal (identical) halves. If line of symmetry passes through a branch possessing a capacitor, then on each side of Line of Symmetry (LOS) the capacitance will become 2C ( 2C and 2C in series will gives C ), as shown.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 18

B

C5

B C2

C

C3

C4

It is called a Balanced Wheatstone Bridge (BWSB), when C1 C3 = C2 C 4

In this situation we observe that no charge flows through C5 and hence C5 can be removed from the circuit to get the equivalent capacitance between A and B.

9/20/2019 10:44:07 AM

Chapter 2: Capacitance and Applications 2.19 C

Conceptual Note(s) C1 C3 = , then no charge exists in the branch conC2 C 4 taining C5 (also called fifth branch) and hence it can be omitted ( just neglect it as if it was never present).

A

C

If

As a consequence of this we can say that C1 and C1C3 , C2 and C4 are in C3 are in series to give C1 + C3 C2 C 4 and both in parallel. So, net series to give C2 + C 4 capacitance Cnet =

C1C3 CC + 2 4 C1 + C3 C2 + C4

Other shapes of Balanced Wheatstone Bridge are indicated below. C1

C5

A E

C3

A

C4

B

C1 C2

C3 C4

C3

C1 C2

C3 A B

C

E

C

G

C

A

B

F

C

H

CAB C

2C 3

C1 C2

C3 C4

In the following infinite circuits, the equivalent capacitance between A and B is to be calculated. Suppose the effective capacitance between A and B is X . Since the network is infinite, so even if we remove one repetitive unit of capacitors from the chain, then too the remaining network would still have infinite pair of capacitors, i.e., effective capacitance between P and Q will still be X P

A

C1

C4

C1

A X B

C2

C1 C2

∞

Q

Extended Wheatstone Bridge The given figure consists of two Wheatstone Bridges connected together. One bridge is connected between points AEGHFA and the other is connected between points EGBHFE.

C1

C2

B

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 19

H

C

Infinite Chain Of Capacitors

C1

C5

C

C

C2 C5

B

This circuit is known as Extended Wheatstone Bridge and it has two branches EF and GH to the left and right of which symmetry in the ratio of capacitances can be seen. C C C = = =1 C C C It can be seen that ratio of capacitances in branches AE and EG is same as that between the capacitances of the branches AF and FH . Thus, in the bridge AEGHFA; the branch EF can be removed. Similarly in the bridge EGBHFE branch GH can be removed.

B

C4

C

G C

F

C

C2

D

C

E

Parallel (C2 + X) A

P X

C2

Q

C1 X

Series C1 (C2 + X) C1 + C2 + X

(C2 + X)

B

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2.20  JEE Advanced Physics: Electrostatics and Current Electricity

Hence equivalent capacitance between A and B CAB =



⇒ CAB =

C1 ( C2 + X ) C1 + C2 + X C2 2

⎛ C2 ⎞ (a) Potential difference across C1 is ⎜ ( V1 - V2 ) ⎝ C1 + C2 ⎟⎠

=X

⎤ ⎡ ⎛ C1 ⎞ - 1⎥ ⎢ ⎜ 1+ 4 ⎟ C2 ⎠ ⎢⎣ ⎝ ⎥⎦

(V1 - V2) and potential difference across C2 is ⎛ C1 ⎞ ⎜⎝ C + C ⎟⎠ ( V1 - V2 ) 1 2

Illustration 15

For what value of C0 in the circuit shown below will the net effective capacitance between A and B be independent of the number of sections in the chain? A

C1

C1

C2

C2

C1 C2

C1

C C0

C2

B

D

Solution

Suppose there are n sections between A and B and the network is terminated by C0 with equivalent capacitance X . A

C

X

C0

B

Network With More Than One Cell

C1

V1

V1 – V2

C1

(b) Consider the arrangement shown. A

C1

V0

A

V0

C1



C0 =

C1 + C2 + C0

⇒ C02 + C2 C0 - C1C2 = 0 ⎤ C ⎡ ⎛ C ⎞ On simplification C0 = 2 ⎢ ⎜ 1 + 4 1 ⎟ - 1 ⎥ 2 ⎢⎣ ⎝ C2 ⎠ ⎥⎦

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 20

B

C2

B

V

Then, potential difference between the ends of this arrangement is ( V - V0 ) . (V – V0)

Now if we add one more sections to the network between D and C (as shown in figure), the equivalent capacitance of the network X will be independent of number of sections if the capacitance between D and C still remains C0 i.e., C1 × ( C2 + C0 )

C2

This arrangement is connected across a battery of voltage V (> V0 ) as shown.

C0

D

C2

V2

Parallel C2 + C0 C2

C1

C2

C1

C2

Case Of Compound Dielectrics If several dielectric medium filled between the plates of a parallel plate capacitor in different ways as shown.

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Chapter 2: Capacitance and Applications 2.21

(a) In the arrangement shown, these two capacitors are in parallel and 1 A/2

K1

A/2

K2



d/2

K1

d/2

K2

K1 ε 0 A K ε A and C2 = 2 0 2d 2d

C2

K + K2 = 1 2

where C1 = K1

(b) The system can be assumed to be made up of two capacitors C1 and C2 which may be said to connected in series



2 C1

A K1

K2



C2

d

C3

⎛ K1 + K 2 ⎞ ε 0 A ⎟⎠ ⋅ ⇒   Ceq = ⎜⎝ 2 d

1

K3

C1

Since, Ceq = C1 + C2

⇒   Keq

A/2

The equivalent capacitor circuit diagram for this arrangement is shown in figure.

C2

d



A/2

C1

2

C1 =

(c)

   

C2 = K 2 C3 = K 3

So, Ceq =

ε0 ( A 2 )

(d 2)

ε0 ( A 2 ) d2

ε0 ( A 2 ) d

A

Keq

⎛ ε A⎞ = K1 ⎜ 0 ⎟ ⎝ d ⎠

⎛ ε A⎞ = K2 ⎜ 0 ⎟ ⎝ d ⎠ ⎛ ε A⎞ = K3 ⎜ 0 ⎟ ⎝ 2d ⎠

K ⎞ ε A⎛ K K C1C2 + C3 = 0 ⎜ 1 2 + 3 ⎟ C1 + C2 d ⎝ K1 + K 2 2 ⎠

⎛ ε A⎞ Ceq = Keq ⎜ 0 ⎟ ⎝ d ⎠ d/2

Since,

d/2

C1 =

K1 ε 0 A K ε A , C2 = 2 0 d2 d2

1 1 1 = + Ceq C1 C2

⇒   Ceq

⎛ 2 K1 K 2 ⎞ ε 0 A =⎜ ⎝ K1 + K 2 ⎟⎠ d

2 K1 K 2 ⇒   Keq = K + K 1 2

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 21

K3 K1 K 2 ⇒   Keq = K + K + 2 1 2 (d) To calculate equivalent capacitance in this case, we have A/2 K1

d/2 d/2

A/2

K2

K3

to proceed as follows by redrawing the arrangement

9/20/2019 10:44:34 AM

2.22  JEE Advanced Physics: Electrostatics and Current Electricity

A/2

A/2

d/2

K1

K1

d/2

K2

K3

where C1 = K1



   

C2 = K 2 C3 = K 3

So, Ceq =

ε0 ( A 2 )

(d 2)

ε0 ( A 2 )

(d 2)

ε0 ( A 2 )

(d 2)

C1

C1

C2

C3

d

A Keq

⎛ ε A⎞ C=⎜ 0 ⎟ ⎝ d ⎠

Now plate-1 connected to positive terminal of battery and part of one capacitor ⎛ ε AV ⎞ q1 = + ⎜ 0 ⎝ d ⎟⎠ Since plate 4 is connected to negative terminal and is common between two capacitors in parallel, so

⎛ ε A⎞ = K1 ⎜ 0 ⎟ ⎝ d ⎠

⎛ ε A⎞ = K2 ⎜ 0 ⎟ ⎝ d ⎠



⎛ ⎡ ε AV ⎤ ⎞ q4 = qC + qD = -2q1 = ⎜ -2 ⎢ 0 ⎝ ⎣ d ⎥⎦ ⎟⎠

⎛ ε A⎞ = K3 ⎜ 0 ⎟ ⎝ d ⎠



⎛ 2ε AV ⎞ q4 = - ⎜ 0 ⎟ ⎝ d ⎠

Illustration 17

CC C1C2 + 1 3 C1 + C2 C1 + C3

ε 0 A ⎡ K1 K 2 K1 K 3 ⇒   Ceq = d ⎢ K + K + K + K 2 1 3 ⎣ 1

⎤ ⎛ ε0 A ⎞ ⎥ = Keq ⎜⎝ d ⎟⎠ ⎦

K1 K 3 K1 K 2 ⇒   Keq = K + K + K + K 1 3 1 2

Two large parallel metal plates are oriented horizontally and separated by a distance 3d. A grounded conducting wire joins them, and initially each plate carries no charge. Now a third identical plate carrying charge Q is inserted between the two plates, parallel to them and located a distance d from the upper plate, as in Figure. d

Illustration 16

Five identical capacitor plates each of area A are arranged in such a way that adjacent plates are at a distance d apart. The plates are connected to source of emf V . Calculate the magnitude and nature of charge on plate 1 and 4 respectively. 2

4

1

5

3 A

B

C

D

2d

(a) What induced charge appears on each of the two original plates? (b) What potential difference appears between the middle plate and each of the other plates? Each plate has area A. Solution

V

Solution

These five plates constitute four identical capacitors in parallel each of capacitance C given by

Imagine the centre plate is split along its midplane and pulled apart. We have two capacitors in parallel, supporting the same DV and carrying total charge Q. ε A The upper has a capacitance C1 = 0 and the lower d ε0 A has a capacitance C2 = . 2d Charge flows from ground onto each of the outside plates of capacitors, so that

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 22

Q1 + Q2 = Q …(1)

9/20/2019 10:44:43 AM

Chapter 2: Capacitance and Applications 2.23 Solution

and ΔV1 = ΔV2 = ΔV

Charges on the surfaces facing each other will be equal and opposite (from Gauss’s Law). Let the distribution be as shown in the following diagram and the charges on the six surfaces are as shown.

d 2d

q1

⇒

Q1 Q2 = C1 C2

⇒

Q1 d Q2 2d = ε0 A ε0 A

As electric field inside the conductor (let at point P) is zero.

2Q2 + Q2 = Q

(a) Q2 =



⇒ q1 =

Q 3

So, on the lower plate the charge is −

q1 = Q3 + Q2 + Q1 − q1

Q3 + Q2 + Q1 2 Since, the outer plates are connected hence, their potential will be same.

Q 3

Q1 =

Q3 + Q2 + Q1 – q1

P

From (1), we get

⇒ Q2 =

Q2 + q –(Q2 + q)

⇒ Q1 = 2Q2

–q

q

2Q 3

So, on the upper plate the charge is − Q 2Qd ( b) ΔV = 1 = C1 3ε 0 A

Q 3 2Q 3

⇒

Q2 + q q d+ d=0 ε0 A Kε0 A

⇒ Q2 + q +

⇒ KQ2 + ( K + 1 ) q = 0 ⇒ q=

Illustration 18

Three large conducting plates are placed a distance d apart in air. The space between the first two plates is completely filled with a dielectric slab of dielectric constant K = 2 as shown in Figure. The plates are given charges Q1 = 7Q , Q2 = 3Q and Q3 = 2Q respectively. The outer two plates are now connected with a conducting wire. Find the charges on all the six surfaces. d

K

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 23

d

q =0 K

− KQ2 = −2Q  K +1

{∵ Q2 = 3Q }

So, the final charge configuration on the six surfaces will be as shown here. 6Q

–2Q +2Q

+Q

–Q

6Q

Illustration 19

Each of the three plates shown in Figure has 200 cm 2 area on one side, and the gap between the adjacent plates is 0.2 mm. The emf of the battery is 20 V. Calculate the distribution of charge on various surfaces of the plates and the equivalent capacitance of the system between the terminal points?

9/24/2019 11:11:36 AM

2.24 JEE Advanced Physics: Electrostatics and Current Electricity

The charge on plate b is negative on both faces. Thus, the charge on faces a and c is

20 V

a b C1 a

b

c

c b C2

SOLUTION

As the potentials of a and c are equal, the capacitors Cab and Cbc are in parallel. Therefore, equivalent capacitance is 2ε A C = Cab + Cbc = 0 d



where, Cab = Cbc = ⇒

C=

ε 0 A ε 0 × 2 × 10 -2 = = 100ε 0 d 0.2 × 10 -3

2 × ε 0 × 2 × 10 -2 0.2 × 10

-3

= 200ε 0

20 V



qa = qc =

ε0 A V d

⇒

qa = qc = 100ε 0 × 20

⇒

qa = qc = 2000ε 0

⇒

Q = 2 × 2000ε 0

⇒

Q = 4000ε 0

Test Your Concepts-II

Based on Series and Parallel Combination of Capacitors 1. Two identical parallel-plate capacitors, each with capacitance C, are charged to potential difference DV and connected in parallel. Then the plate separation in one of the capacitors is doubled. (a) Find the total energy of the system of two capacitors before the plate separation is doubled. (b) Find the potential difference across each capacitor after the plate separation is doubled. (c) Find the total energy of the system after the plate separation is doubled. (d) Reconcile the difference in the answers to parts (a) and (c) with the Law of Conservation of Energy. 2. Two identical parallel plate capacitors are first connected in series and then in parallel. In each case, the plates of one capacitor are brought closer by a distance Dd and the plates of the other capacitor are moved away by same distance Dd. How does

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 24

(Solutions on page H.127) the total capacitance of the system change in both cases by the above displacement of plates? 3. A 10 μF capacitor is charged to 15 V. It is next connected in series with an uncharged 5 μF capacitor. The series combination is finally connected across a 50 V battery, as shown in Figure. Find the new potential differences across the 5 μF and 10 μF capacitors. 5 μF

10 μ F ΔVi = 15 V

50 V

4. Each plate of a parallel plate capacitor has an area A. What amount of work has to be performed

9/20/2019 10:45:02 AM

Chapter 2: Capacitance and Applications 2.25

to slowly increase the distance between the plates from x1 to x2, if (a) the charge of the capacitor, q is kept constant in the process. (b) the voltage across the capacitor, V, is kept constant in the process. 5. Two capacitors A and B are connected in series across a 100 V supply and it is observed that the potential difference across them are 60 V and 40 V. A capacitor of 2 μF capacitance is now connected in parallel with A and the potential difference across B rises to 90 V. Determine the capacitance of A and B. 6. Two capacitors are in parallel and the energy of the combination is 0.1 J, when the differences of potential between terminals is 2 V. With the same two condensers now connected in series, the energy is 1.6 × 10 -2 J for the same difference of potential across the series combination. Calculate the capacitances of the capacitors. 7. If the area of each plate is A and the successive separation are 3d, 2d and d. Then find the equivalent capacitance across A and B. 3d

A 2d

B d

d

b a

10. A capacitor has square plates, each of side a making an angle θ with each other as shown in figure. Show that for small θ , the capacitance C is given by



C=

ε 0 a2 ⎛ aθ ⎞ ⎜ 1- ⎟⎠ d ⎝ 2d a θ

d O

a

11. Find the effective capacitance of the following arrangement of four identical metallic plates between P and Q. The area of each plate is A and separation between successive plates is d. P

1

Q

2 3

8. Two capacitors A and B each having a dielectric of dielectric constant 2 are connected in series. When they are connected across a 230 V D.C. supply it is found that the potential difference across A is 130 V and that across B is 100 V. If the dielectric in the smaller capacitor is replaced by another dielectric of dielectric constant 5, what will be the new values of the potential difference across each? 9. Two capacitors are joined in series shown in figure. The central rigid H-shaped part is movable. Find the equivalent capacitance of the combination and hence show that it is independent of position of the central H-shaped part. The area of each plate is A.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 25

4 ARRANGEMENT 1 1

P

Q

2 3 4 ARRANGEMENT 2

12. A radio capacitor of variable capacitance is made of n plates each of area A and separated from each other by a distance d. The alternate plates are connected together. One group of alternate plates is

9/20/2019 10:45:05 AM

2.26  JEE Advanced Physics: Electrostatics and Current Electricity

fixed while the other is movable. Find the maximum capacitance of the capacitor. 13. (a) Find the equivalent capacitance of the combination shown in figure, where C1 = 6 μF, C2 = 4 μF and C3 = 8 μF. C2

C1

15. The circular plates A and B of a parallel plate air capacitor have a diameter of 0.1 m and are 2 mm apart. The plates C and D, of a similar capacitor have a diameter 0.12 m and are 3 mm apart. Plate A is earthed and the plates B and D are connected together. Plate C is connected to the positive pole of a 120 V battery whose negative terminal is earthed. Calculate 2 mm

C3

3 mm 120 V



(b)  If a battery is connected between a and b, determine the charge on each capacitor and the potential difference across each, if Vab =12 V.

14. Find the capacitance of the following combinations between P and Q. Area of the each plate is A and separation of successive plates is d. Q

P ARRANGEMENT 1 P

Q

A B Diameter = 0.1 m



C D Diameter = 0.12 m



(a) the combined capacitance of the arrangement and (b)  the energy stored in it.



Take

1 = 36π × 109 Nm2 C -2 ε0

16. Using the concept of energy density, find the total energy stored in a (a)  parallel plate capacitor (b)  charged spherical conductor

ARRANGEMENT 2

DIELECTRIC SLAB INSERTED IN A PARALLEL PLATE CAPACITOR When the space between the parallel plate capacitor is partly filled with a dielectric of thickness t ( < d ) If no slab is introduced between the plates of the σ capacitor, then a field E0 given by E0 = , exists in ε0 a space d. E E= 0 K

E0 K exists inside the slab of thickness t and a field E0 exists in remaining space ( d - t ). If V is total potential then On inserting the slab of thickness t , a field E =

⇒ But

+σ

t d

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 26

V = E0 ( d - t ) + Et ⎡ ⎛ E⎞ ⎤ V = E0 ⎢ d - t + ⎜ t ⎝ E0 ⎟⎠ ⎥⎦ ⎣ E0 = K = Dielectric Constant E

⇒

V=

σ ε0

⇒

V=

q Aε 0

t ⎤ ⎡ ⎢⎣ d - t + K ⎥⎦ t ⎤ ⎡ ⎢⎣ d - t + K ⎥⎦

9/20/2019 10:45:12 AM

Chapter 2: Capacitance and Applications 2.27

⇒

C=

q = V

ε0 A 1⎞ ⎛ d -t⎜ 1- ⎟ ⎝ K⎠

⇒

So, on introducing a dielectric slab of thickness t and dielectric constant K the capacitance increases by the same amount as the effective air spacing between the 1⎞ ⎛ plates is made t ⎜ 1 - ⎟ ⎝ K⎠

CONDUCTING SLAB INSERTED IN A PARALLEL PLATE CAPACITOR When the space between the parallel plate capacitor is partly filled by a conducting slab of thickness t ( < d ). If no conducting slab is introduced between the σ plates, then a field E0 = exists in a space d. If C0 ε0 be the capacitance (without the introduction of conducting slab), then

ε A C0 = 0 d On inserting the slab, field inside it is zero and so a σ field E0 = now exists in a space ( d - t ) ε0

+σ

E=0

t d

⇒

V = E0 ( d - t )

⇒

V=

σ (d -t) ε0

⇒

V=

q (d -t) Aε 0

⇒

q ε A C= = 0 V d-t

⇒

C=

ε0 A t⎞ ⎛ d⎜ 1- ⎟ ⎝ d⎠

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 27

E0

C=

C0 t⎞ ⎛ ⎜⎝ 1 - ⎟⎠ d

Since d - t < d ⇒

C > C0

i.e., Capacitance increases on insertion of conducting slab between the plates of capacitor. If t = d , then C → ∞ i.e., if a conducting slab occupies the complete space between the plates of the capacitor, then C → ∞.

CHARGE INDUCED ON A DIELECTRIC and GAUSS’S LAW FOR DIELECTRICS Consider a parallel plate capacitor which has a surface charge density σ and -σ on each plate. If E0 is the electric field between the plates of air capacitor, then E0 =

σ ε0

Let a dielectric of dielectric constant K be now introduced between the plates of the capacitor. If σ p is induced surface charge density (see figure) due to the already existing field, then electric field due to induced charge is Ep =

σp

ε0 So, resultant dielectric field within the plates is

E = E0 - Ep E=

1 ( σ - σ p ) …(1) ε0

Also E =

σ …(2) Kε0

⇒

+qp

–qp +q

–q

E0 Ep E = E0 – EP

9/20/2019 10:45:22 AM

2.28  JEE Advanced Physics: Electrostatics and Current Electricity

1 σ ( σ - σ p ) = Kε ε0 0

⇒ ⇒ ⇒

1⎞ ⎛ σp = σ ⎜ 1- ⎟ ⎝ K⎠ qp A

=

q⎛ 1⎞ ⎜⎝ 1 - ⎟⎠ A K

1⎞ ⎛ qp = q ⎜ 1 - ⎟ ⎝ K⎠

Since according to Gauss’s Law, for dielectric ⇒ ⇒ ⇒ ⇒

∫

  1 1 E ⋅ dA = ( Σqenc ) = ( q - qp ) ε0 ε0

∫

  q 1 ⎡ 1⎞⎤ ⎛ E ⋅ dA = q - q⎜ 1- ⎟ ⎥ = ⎢ ⎝ ⎠ ε0 ⎣ K ⎦ Kε0

∫

  q q E ⋅ dA = = Kε0 ε

{∵ ε = ε r ε 0 = K ε 0 }

  K ε 0 E ⋅ dA = q

∫   D ⋅ dA = q ∫

Regarding dielectrics, it is worth noting that: (a) These are non-conductors upto a certain value of field depending on its nature. If the field exceeds this limiting value called dielectric strength, dielectric loses its insulating property and begins to conduct. (b) These have either permanent dipole moment (polar-dielectrics, e.g., water) or acquire induced dipole moment (non-polar dielectrics) when placed in an electric field. (c) The dielectric constant of polar dielectric depends on its temperature and due to thermal agitation with rise in temperature decreases.

Illustration 20

Two parallel conducting plates of area A charge +q and -q are as shown. A dielectric slab of dielectric constant K, thickness 2d and a conducting plate of thickness d is inserted between them. Taking x = 0 at positive plate and x = 7 d at negative plate, plot E vs x and V vs x graphs. Here E is the electric field and V is the potential and consider the potential at the positive plate to be V0. Dielectric

  where D = K ε 0 E is called the Electric Displacement Vector.

d

Problem Solving Technique(s) 1⎞ ⎛ Since qp = q ⎜ 1- ⎟ ⎝ K⎠ For a conductor K → ∞. Hence, qp = q , σ p = σ and E = 0

Hence, we may conclude the above discussion as under q σ (a) E vacuum = E0 = = ε 0 Aε 0 (b) Edielectric

E = 0 {K = dielectric constant} K

(c) Econductor = 0 {as K → ∞ }

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 28

Conductor

Compare (1) and (2), we get

2d

2d

d

d

Solution

The electric field in air/vacuum is E0 =

q σ = . ε 0 Aε 0

E0 and in conductor field K is zero. Hence, the E-x graph is as shown in figure. In dielectric, field is E =

E

E0 E0 k O

d

3d

5d 6d

7d

x

9/20/2019 10:45:34 AM

Chapter 2: Capacitance and Applications 2.29

V

E0 =

V0 V1 = V0 – E0d V2 = V0 – E0d – 2Ed

EFFECT OF INSERTION OF DIELECTRIC IN A PARALLEL PLATE CAPACITOR

E σ and E = 0 ε0 K

V3 = V0 – 3E0d – 2Ed V4 = V0 – 4E0d – 2Ed O

d

3d

5d

6d

7d

x

Using V = Ed (in uniform field) the V -x graph is shown (slope of this graph gives electric field in that region). Illustration 21

A parallel plate capacitor with air as dielectric has a plate area of 200 cm 2 and plate separation of 4 mm. Calculate the percentage change in the capacitance if a layer of varnish ( k = 3 ) of thickness 0.1 mm is given on the inside of both plates. Solution

C=

ε0 A

t d-t+ k

=

C0  t t 1- + d kd

{

∵ C0 =

ε0 A d

}

where t = 2 ( 0.1 mm ) = 0.2 mm, d = 4 mm and K = 3

The effect of insertion of a dielectric on other physical quantities such as charge, potential difference, electric field and energy associated with a capacitor depends on the fact that whether the charged capacitor is isolated (i.e., charge is kept constant) or is attached to the battery (i.e. potential is kept constant). If q0 , C0 , V0, E0 and U0 represent the charge, capacitance, potential difference, electric field and energy associated with charged air capacitor respectively. On introduction of a dielectric slab of dielectric constant K between the plates let the respective quantities becomes q, C , V , E and U. CASE-1: When charge is kept constant OR Capacitor is Isolated OR Battery is Disconnected (a) Charge remains unchanged, i.e., q = q0, as in an isolated system charge is conserved. (b) Capacitance increases and becomes C = KC0, because due to the presence of a dielectric, the capacitance becomes K times. (c) Potential difference between the plates decreases ⎛V ⎞ and becomes V = ⎜ 0 ⎟ ⎝ K⎠

Putting values, we get C0 = 44.25 pF Now, C =

⇒ C=

⇒ C=

44.25 0.2 0.2 1+ ( 4 3 )( 4 )



q0 + + + + + + + +

44.25 pF 1 1 1+ 20 60 44.25 × 60 2655 = 60 - 3 + 1 58

⇒ C = 45.77% So, %age change is DC ⎛ 45.77 - 44.25 ⎞ × 100 = ⎜ ⎟⎠ × 100 = 3.4% ⎝ C0 44.25

So, percentage increase in the value of capacitance is 3.4%

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 29

q q0 ⎛ ⎞ ⎜⎝ ∵ V = C = KC , as q = q0 and C = KC0 ⎟⎠ 0 q + + + + + + + +

C0, V0, E0, U0

(A)

C, V, E, U

K

(B)

  

(d) Field between the plates decreases and becomes

⎛E ⎞    E = ⎜ 0 ⎟ ⎝ K⎠

{

}

V V V V0 E0 ∵V = 0 and E0 = 0 ⇒   E = d = Kd = K K d  (e) Energy stored in the capacitor decreases and ⎛U ⎞ becomes U = ⎜ 0 ⎟ ⎝ K ⎠

9/20/2019 10:45:46 AM

2.30  JEE Advanced Physics: Electrostatics and Current Electricity

Since U = ⇒ U=

Illustration 22

q2 2C

q02 U = 0  2KC0 K

{∵ q = q0

and C = KC0 }

CASE-2: When potential is kept constant OR b ­ attery remains attached (a) Potential difference remains constant, i.e., V = V0, because battery is a source of constant potential difference. (b) Capacitance increases and becomes C = KC0 , because by presence of a dielectric capacitance becomes K times. q0 + + + + + + + +

(A)

V

C, V, E, U

K

(c) Charge on capacitor increases, i.e., q = Kq0, because



q = CV

⇒   q = ( KC0 ) V = Kq0 

{∵ q0 = C0V }

(d) Electric field remains unchanged, so E = E0 E=

{

V V V0 = = E0  ∵ V = V0 and 0 = E0 d d d

}

(e) Energy stored in the capacitor increases and becomes U = KU0 Since, U =

1 1 2 CV 2 = ( KC0 )( V0 ) 2 2

1 ⇒   U = 2 KU0 

{

1 ∵C = KC0 and U0 = C0V02 2

Problem Solving Technique(s) While solving problems of this type always keep in mind that, whenever a battery is disconnected then, q = constant and if battery remains attached, then V = constant

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 30

Initially charge on the capacitor Qi = 10 × 12 = 120 μ C When dielectric medium is filled, so capacitance becomes k times, i.e., new capacitance is C ′ = kC = 5 × 10 = 50 μ C Final charge on the capacitor Q f = 50 × 12 = 600 μ C



DQ = Q f - Qi = 480 μ C

Illustration 23 (B)

 

Solution

Hence additional charge supplied by the battery is

q + + + + + + + +

C0, V0, E0, U0

V0

An air capacitor of capacity C = 10 μF is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant k = 5. Calculate the charge that flows from battery to the capacitor.

}

The capacitance of a variable radio capacitor can be changed from 50 pF to 950 pF by turning the dial from 0° to 180°. With the dial set at 180°, the capacitor is connected to a 400 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned at 0°. (a) What is the potential difference across the capacitor when the dial reads 0°? (b) How much work is required to turn the dial, if friction is neglected? Solution

When the dial is at 0°, the capacitance of the capacitor is given by C1 = 50 pF = 50 × 10 -12 F When dial is at 180°, the capacitance is given by C2 = 950 pF = 950 × 10 -12 F The potential difference across capacitor C2 , is given by V2 = 400 V Charge on capacitor C2 is

q = C2V2 = 950 × 10 -12 × 400

⇒ q = 380 × 10 -9 C

9/20/2019 10:46:02 AM

Chapter 2: Capacitance and Applications 2.31

(a) When battery is disconnected the charge remains the same and so, q = constant. Let V1 be the potential difference across capacitor when dial reads 0°. Then

C

C

C

KC

q

q

q′

q′

q = C1V1

   

⇒   380 × 10 ⇒   V1 =

-9

V

= 50 × 10

380 × 10 -9 50 × 10 -12

-12

× V1

V

Net capacitance after inserting the slab is ⎛ K ⎞ C′ = ⎜ C ⎝ K + 1 ⎟⎠     

= 7600 V

(b) Work required to turn the dial from 180° to 0° is W = Gain in energy of capacitor

⎛ K ⎞ ⇒   q ′ = CV ⎜⎝ K + 1 ⎟⎠

q2 q2 q2 ⎛ 1 1 ⎞ W = = ⇒   ⎜ 2C1 2C2 2 ⎝ C1 C2 ⎟⎠

Final electric field E′ is given by

W=

⇒

⇒  W=

q2 ( C2 - C1 )

    

2C1C2

( 380 × 10 ) ( 950 - 50 ) × 10 -9 2

Potential Difference Separation between the plates

⎛ 1 ⎞ V⎜ ⎝ K + 1 ⎟⎠ V ⇒   E′ = ( d K + 1) d So, electric field decreases by a factor

-12

2 × 50 × 10 -12 × 950 × 10 -12

-3 ⇒   W = 1.368 × 10 J

Illustration 24

Two parallel capacitors each of capacitance C are connected in series to a battery of emf V. Now, one of the capacitors is filled completely with a dielectric of dielectric constant K. (a) Calculate the ratio of the electric field strength in the capacitor to the electric field when the dielectric is introduced. Does the field increase or decrease after insertion of dielectric? (b) Find the amount of charge that flows through the battery. Solution

(a) Net capacitance without inserting the slab is,

E′ =

C 2

⎛ C⎞ ⇒   q = ⎜⎝ 2 ⎟⎠ V

⎛ V⎞ ⎜⎝ ⎟ K +1 E 2d ⎠ = = V 2 ⎞    E ′ ⎛⎜ ⎝ ( K + 1 ) d ⎟⎠ (b) Charge that flows through the battery is q ′ - q   

Charge Flowing =

Illustration 25

Figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric. S

Initial electric field E is given by ⎛V⎞ ⎜⎝ ⎟⎠ V Potential Difference E= = 2 = separation between the plates d 2d

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 31

CV ( K - 1 ) 2( K + 1)

V

A

C

B

C

9/20/2019 10:46:12 AM

2.32  JEE Advanced Physics: Electrostatics and Current Electricity Solution

A

With the switch S closed, the potential difference across capacitors A and B is same. So, QA QB = C C The initial charges on the capacitors are given by V=

QA = QB = CV When dielectric is introduced, the new capacitance of either capacitor.

C ′ = KC = 3C

QB = CV = C ′V ′

⇒ V′ =

C V V= volt C′ 3

Initial energy of both capacitors 1 1 CV 2 + CV 2 = CV 2 2 2 Final energy of both capacitors Ui =



Uf =

⇒ Uf = ⇒

1 1 C ′V 2 + C ′V ′ 2 2 2 2

1 5 1 V ( 3C ) V 2 + ( 3C ) ⎛⎜ ⎞⎟ = CV 2 ⎝ 3⎠ 2 2 3

Ui 3 CV 2 = = 5 Uf CV 2 5 3

FORCE ON A DIELECTRIC SLAB BEING INSERTED BETWEEN CAPACITOR PLATES When a dielectric slab is being inserted between the plates of an initially charged capacitor, then due to the electric field between the plates of the capacitor, bound charges of opposite nature are induced inside the dielectric. Also it is observed that while inserting the dielectric slab, fringing of electric field lines at the edges of the plates takes place. (Fringing Effect is the bending of field lines at the corners of a capacitor as shown).

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 32

  Fringing effect for a capacitor with large d.

A component of the electric field along the surface of the dielectric pulls the dieletric slab inside the capacitor. Thus force on the dielectric slab is given by F=

Now, when the switch S is opened, let the potential difference across capacitor A be V volt and that across the capacitor B be V ′ volt. When dielectric is introduced with the switch open (i.e., battery disconnected) then, the charge on capacitor B remains unchanged, so

B

dU dx

Let us discuss the two cases under which the dielectric slab can be introduced between the plates of the capacitor. CASE-1: FOR CAPACITOR CONNECTED TO BATTERY (i.e. V = Constant) Consider a dielectric slab inserted partially between the plates of a capacitor. Due to positive charge on the upper plate, some negative charge is induced on the top surface of the dielectric. Similarly, some positive charge is induced at the bottom surface of the slab. The negative induced charge is attracted towards positive charge on upper plate and the positive induced charge is attracted towards negative charge on lower plate as shown. As a result, the net force on the slab acts inwards and the slab is attracted into the capacitor. To calculate this force, let us consider a parallel plate capacitor having plates of length l, width b and separation d. Let the capacitor be connected to a battery of voltage V. Let us find the force on a slab of thickness d and dielectric constant K when a portion of length x ( < l ) lies inside the plates. F

+q V

Dielectric

– qi +qi

E

–q

F

d

V x l

9/20/2019 10:46:22 AM

Chapter 2: Capacitance and Applications 2.33

The capacitance of the capacitor so formed is calculated by assuming the capacitor in this position to be made up of two capacitors in parallel, one without a dielectric (of length ( l - x ), separation d and width b) and the other with a dielectric (of length x, separation d and width b). The capacitance of the portion of capacitor without dielectric is,

This work done dW is equal to the increase in the 1 energy dU = ( dC ) V 2 stored in the capacitor 2 (assuming that no heat losses are produced while inserting the dielectric inside the capacitor). So, we have

ε b(l - x ) C1 = 0 d and that of capacitor with dielectric is

⇒ Fdx =

K ε 0 bx d So, the net capacitance of the capacitor in this ­position is, C2 =



C = C1 + C2

ε b ⇒ C = 0 [ l + x ( K - 1 ) ] …(1) d The electric field inside the capacitor attracts the dielectric slab with a force F (say). For the dielectric slab to slowly enter the capacitor, an external force Fext = F must be applied opposite to F. Due to this external force, the dielectric slab doesnot gain any kinetic energy while entering the capacitor plates. Let the slab move further inside the capacitor by an infinitesimal distance dx. Due to this the capacitance of the capacitor also increases from C to C + dC . Since the capacitor is connected to the battery, so potential difference V remains constant and hence the battery has to supply a charge ( dQ ) to the capacitor given by dQ = V ( dC )



1 ( dC ) V 2 = ( dC ) V 2 - Fdx 2

⇒ F=

1 ( dC ) V 2 2

1 2 ⎛ dC ⎞ V ⎜ ⎝ dx ⎟⎠ 2

From (1), we get dC = ⇒

ε0 b ( K - 1 ) dx d

dC ε 0 b ( K - 1) = dx d

⇒ F=

ε 0 bV 2 ( K - 1 ) 2d

Thus, the electric field attracts the dielectric into the capacitor with the force F given by

ε 0 bV 2 ( K - 1 ) 2d Also, we note that this force is constant and is independent of the distance x . F=

CASE-2: FOR AN ISOLATED CAPACITOR (i.e., q = Constant) When the battery is removed after charging the capacitor then q remains constant. In that case,

U=

q2 2C

During this process work done by the battery is

⎤ ⎡ q2 d ⇒ dU = d ⎢ ⎥ ⎣ 2ε 0 b { l + x ( K - 1 ) } ⎦

dWbattery = VdQ = ( dC ) V 2 Also work done by the external force F during this displacement dx is

⇒ dU = -

dWexternal = Fdx cos ( 180° ) = - Fdx Total work done on the capacitor is

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 33

2

Here, since work done by battery is zero and so only the electrostatic force does the work. Hence, F = -

dW = dWbattery + dWexternal

⇒ dW = ( dC ) V 2 - Fdx

q2 d ( K - 1 ) ⎛ 1 ⎞ ⎜⎝ ⎟ dx l + x ( K - 1) ⎠ 2ε 0 b

⇒ F=

dU dx

q2 d ( K - 1 ) ⎛ 1 ⎞ ⎜⎝ ⎟ l + x ( K - 1) ⎠ 2ε 0 b

2

9/20/2019 10:46:37 AM

2.34  JEE Advanced Physics: Electrostatics and Current Electricity

This force also acts inwards, but in this case it is not constant and is a function of x . Illustration 26

In the situation shown in figure the length of the plates is l, area of the plates is A and separation between then is d. A dielectric slab of dielectric constant K, mass m and thickness d is released from rest. Prove that the slab will execute periodic motion and find its time period. Assume friction to be absent between inner portion of the plates and the outer surface of the slab. Also see that initially the slab has its length x inserted in the plates of the capacitor.

d

K

V x

So, acceleration of slab, a = ⎛ A⎞ ε0 ⎜ ⎟ V 2 ( K - 1 ) ⎝ l ⎠ ⇒ a= 2md ⇒ a=

As derived and seen earlier that the electric forces pull the slab inwards with a constant force, given by

F=

ε 0 bV ( K - 1 ) 2d

ε 0 AV 2 ( K - 1 ) 2mld

At the instant when the slab is fully inside the plates, we get the state of equilibrium for the slab. The slab will execute periodic motion in the phases as shown in figure. So, time taken to reach from position (a) to position (b) equals t (say). Then, t =

T 4

Since, s =

1 2 at 2

⇒ t=

2s  a

l

Solution

⇒ t=

2

ε 0 AV ( K - 1 ) 2mld



=

4 ( l - x ) mld

ε 0 AV 2 ( K - 1 )

( l - x ) mld ε 0 AV 2 ( K - 1 )

DIELECTRIC BREAKDOWN

x (b)

  

x

  

where, b = width of plate =

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 34

2

s = l - x}

The desired time period is T = 4t = 8

(c)

{∵

2(l - x )

2d

(a)

F m

A l

(d)

When a dielectric material is subjected to a strong electric field then there comes a breakdown field strength at which it loses its insulating or dielectric ability and the dielectric becomes a conductor. This occurs when the electric field is so strong that electrons are ripped loose from their molecules. This maximum electric field magnitude that a material can withstand without the occurrence of breakdown is called its Dielectric Strength. The dielectric strength of dry air is about 3 × 106 Vm -1.

9/20/2019 10:46:46 AM

Chapter 2: Capacitance and Applications 2.35

ELECTRIC ENERGY DENSITY OF DRY AIR

(b) For glass, breakdown voltage is given by

The breakdown field strength at which dry air loses its insulating ability and allows a discharge to pass through is Eb = 3 × 106 Vm -1. At this field strength, the electric energy density is



 

uE =

2 1 1 ε 0 E2 = ( 8.85 × 10 -12 ) ( 3 × 106 ) = 40 Jm -3 2 2

Illustration 27

Four identical plane capacitors with an air dielectric are connected in series. The intensity of the field at which the air is punctured is Ea = 3 × 10 4 Vcm -1 . The distance between the plates d = 0.7 cm. (a) What maximum voltage can be fed to this battery of capacitors? (b) What will this maximum voltage be if one of the capacitors is replaced by a similar one in which glass is used as a dielectric? The permitivity of glass K = 7 and the puncturing field intensity of glass Eg = 9 × 10 4 Vcm -1.



Vg = 6.3 × 10 4 V B

A C0

C C0

D C0

    

q = C0 ( DVAB ) = C0 ( DVB C ) =

C0 ( DVCD ) = 7 C0 ( DVDE )

⇒   DVAB = DVBC = DVCD = K DVDE ⇒   DVAB = DVBC = DVCD = 7 DVDE OPTION 1 Take DVAB = 2.1 × 10 4 V and find DVDE From the above equality, we get DVDE =   

DVAB = 0.3 × 10 4 V 7

OPTION 2

(a) Potential across each capacitor = V0 = Ed

Take DVDE = 6.3 × 10 4 V ­voltage of glass), then

⇒   V0 = ( 3 × 10 × 10 Vm 2

-1

) ( 0.7 × 10

-2

m)

4 ⇒   V0 = 2.1 × 10 V

a

b

c

d

e

Now DV = DVab + DVbc + DVcd + DVde 4 4 4 4 ⇒   DV = 2.1 × 10 + 2.1 × 10 + 2.1 × 10 + 2.1 × 10

KC0

If the glass capacitor is connected between the points D and E , then

Solution

4

E

(The

breakdown

DV = DVBC = DVCD = 7 ( 6.3 × 10 4 V ) AB ⇒   DVAB = DVBC = DVCD 5 4    DVAB = 4.41 × 10 V > 2.1 × 10 V So, as per OPTION 2, all the first three capacitors are going to breakdown, so, we have to accept



4 DVDE = 0.3 × 10 V (for glass capacitor)

4 4 ⇒   ( DV )Max = 4 ( 2.1 × 10 ) = 8.4 × 10 V

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 35

9/20/2019 10:46:57 AM

2.36 JEE Advanced Physics: Electrostatics and Current Electricity

Test Your Concepts-III

Based on Dielectrics and Breakdown (Solutions on page H.132) 1. A capacitor is filled with two dielectrics of same dimensions but of dielectric constant 2 and 3 respectively. Find the ratio of capacities in the two arrangements.

d

C1 K1

(d/2)

C2 K2

(d/2)

(A)

d

A/2

A/2

K1

K2

C3

C4 (B)

2. A parallel-plate capacitor of plate separation d is charged to a potential difference DV0. A dielectric slab of thickness d and dielectric constant κ is introduced between the plates while the battery remains connected to the plates. (a) Show that the ratio of energy stored after the dielectric is introduced to the energy stored U = κ . Give a physin the empty capacitor is U0 ical explanation for this increase in stored energy. (b) What happens to the charge on the capacitor? 3. Two parallel capacitors each of capacitance C are connected in series to a battery of emf V. Now, one of the capacitors is filled completely with a dielectric of dielectric constant K. (a) Calculate the ratio of the electric field strength in the capacitor to the electric field when the dielectric is introduced. Does the field increase or decrease after insertion of dielectric? (b) Find the amount of charge that flows through the battery. 4. A parallel plate capacitor with air as dielectric has a plate area of 200 cm3 and plate separation of 4  mm. Calculate the percentage change in the capacitance if a layer of varnish ( k = 3 ) of thickness 0.1 mm is given on the inside of both plates. 5. A thin conducting plate is inserted in halfway between the plates of a parallel plate capacitor.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 36

d/2

d/2

(a) Compute the capacitance before the plate was inserted. (b) What is the capacitance after the plate was inserted? (c) What does the value of capacitance becomes when the thin plate and the upper plate are shortened (i.e., connected by a conducting wire)? Area of each plate is A. 6. A capacitor is composed of three parallel conducting plates. All three plates are of the same area A. The first pair of plates are kept a distance d1 apart, and the space between them is filled with a medium of a dielectric K1. The corresponding data for the second pair are d2 and K2, respectively. d1

d2

K1

K2

V

The middle plate is connected to the positive terminal of a constant voltage source V, and the external plates are connected the other terminal of V. (a) Find the capacitance of the system. (b) What is the surface charge density on the middle plate? (c) Compute the energy density in the medium K1.

9/20/2019 10:46:59 AM

Chapter 2: Capacitance and Applications 2.37

7. Find the capacitance of a system of three parallel plates, each of area A, separated by d1 and d2. The space between them is filled with dielectrics of relative dielectric constants K1 and K2.

10. A capacitor is constructed from two square plates of sides  and separation d. A material of dielectric constant κ is inserted a distance x into the capacitor, as shown in Figure. Assume that d is much smaller than x. l

K1

K2 κ

A

8. A parallel plate capacitor is constructed using three different dielectric materials as shown in figure. The parallel plates across which a potential difference is applied, are of area A = 1 cm2 and are separated by a distance d = 2 mm. If K1 = 4, K 2 = 6 and K 3 = 2, find capacitance across P and Q. (Take

ε 0 = 8.8 × 10 -12 C2N-1m-2 ) P

+Q C2, K2

d/2

C3, K3

d/2

C1, K1

l

d

d2

d1

d

x

l

–Q

9. Figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 37

A

C

(a)  Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor, letting DV represent the potential difference.

(c) Find the direction and magnitude of the force exerted on the dielectric, assuming a constant potential difference DV. Ignore friction. (d) Obtain a numerical value for the force assuming that  = 5 cm, DV = 2000 V, d = 2 mm and the dielectric is glass (κ = 4.5). 11. A vertical parallel-plate capacitor is half filled with a dielectric for which the dielectric constant is 2. When this capacitor is positioned horizontally, what fraction of it should be filled with the same dielectric in order for the two capacitors to have equal capacitance?

(a)

S

V



B

   

(b)

C

9/20/2019 10:47:02 AM

2.38  JEE Advanced Physics: Electrostatics and Current Electricity

KIRCHHOFF’S LAWS FOR CAPACITOR CIRCUITS Kirchhoff’s Junction Law (KJL) According to KJL, charge can never accumulate at a junction i.e., at the junction

∑

qjunction = 0 KJL is based on Law of Conservation of Charge.

Problem Solving Technique(s) This law is helpful in determining the nature of charge on an unknown capacitor plate. +q1 – q1

+q2 – q2

A

B 1 2

C

(b) In a loop, across a battery, if we travel from positive terminal of the battery to the negative terminal then there is a potential fall and a –ve sign is applied with voltage of the battery. (c) In a loop, across a capacitor, if we go from negative plate to the positive plate of the capacitor then there is a potential rise and a +ve sign is to be taken with potential difference across the q capacitor i.e., DV = + . C (d) In a loop, across a capacitor, if we go from positive plate to the negative plate of the capacitor then there is a potential fall and a –ve sign is to be taken with the potential difference across the q capacitor i.e., DV = - . C Illustration 28

Three uncharged capacitors having capacitances 60 μF, 30 μF and 10 μF are connected to each other as shown. Calculate the potential at the point O.

Charge on capacitor C can be determined by using this rule. As no charge must accumulate at the junction O, so if x is charge on plate 1 of C, then

6V 60 μ F

-q1 + q2 + x = 0

⇒ x = q1 - q2

30 μ F

O

10 μ F

i.e., plate 1 has a charge ( q1 - q2 ) and plate 2 has a

charge - ( q1 - q2 ) .

3V

2V

Solution

Kirchhoff’s Loop Law (KLL) In a closed loop/circuit (a closed loop is the one which starts and ends at the same point), the algebraic sum of potential differences across each element of a closed loop/circuit is zero. ⇒

∑V = 0

Conventions Followed to Apply Loop Law (a) In a loop, across a battery, if we travel from negative terminal of battery to the positive terminal then there is a potential rise and a +ve sign is applied with voltage of the battery.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 38

According to Kirchhoff’s Junction Law (KJL), charge cannot accumulate at the junction ( O ), so

Σq0 = 0 , where q = C DV

Let V0 be the potential of junction O, then

60 ( 6 - V0 ) + 10 ( 2 - V0 ) + 30 ( 3 - V0 ) = 0

⇒ 360 + 20 + 90 = ( 60 + 10 + 30 ) V0 ⇒ V0 =

470 100

⇒ V0 = 4.7 V

9/20/2019 10:47:10 AM

Chapter 2: Capacitance and Applications 2.39 Illustration 29

Find the equivalent capacitance between the point A and B in figure. C2

C1

C3

A

C2

⇒ VA - VB =

B

The first equation may be written as

C1

⇒

Let us connect a battery between the points A and B. The charge distribution is shown in figure. Suppose the positive terminal of the battery supplies a charge +Q and the negative terminal a charge -Q. The charge Q is divided between plates a and e. C1

Q1 a

–Q1 Q – Q1 b

D

(2Q1 – Q) –(2Q1 – Q) e Q – Q1

C2

c

f E g –(Q – Q1) Q1

C2

–(Q – Q1)

⎛ 1 1 ⎞ Q VA - VB = Q1 ⎜ + ⎝ C1 C2 ⎟⎠ C2 C C1C2 ( VA - VB ) = Q1 + 1 Q …(3) C2 - C1 C2 - C1

The second equation may be written as

⇒

d

i C3 j

⎛ 1 1 ⎞ Q VA - VB = 2Q1 ⎜ + ⎝ C1 C3 ⎟⎠ C3 C1 C1C2 Q …(4) ( VA - VB ) = Q1 2 ( C1 + C3 ) 2 ( C1 + C3 )

Subtracting (4) from (3) B

⎤ ⎡ C1C2 C1C3 ⎥= ⎣ C2 - C1 2 ( C1 + C3 ) ⎦

( VA - VB ) ⎢

h C1

Q1 2Q1 - Q Q1 + + …(2) C1 C3 C1

We have to eliminate Q1 from these equations to get Q the equivalent capacitance . ( VA - VB )

Solution

A

Also, VA - VB = ( VA - VD ) + ( VD - VE ) + ( VE - VB )

–Q1

Let a charge Q1 goes to the plate a and the rest Q - Q1 goes to the plate e. The charge -Q supplied by the negative terminal is divided between plates d and h. Using the symmetry of the figure, charge -Q1 goes to the plate h (as it has a capacitance C1 ) and -(Q - Q1 ) to the plate d (as it has a capacitance C2 ). This is because if we look into the circuit from A or from B, the circuit looks identical. The division of charge at A and at B should, therefore, be similar. The charges on the other plates may be written easily. The charge on the plate i is 2Q1 - Q which ensures that the total charge on plates b, c and i remains zero as these three plates form an isolated system.

⎤ ⎡ C1 C1 + ⎢ ⎥Q ⎣ C2 - C1 2 ( C1 + C3 ) ⎦

⇒

( VA - VB ) ⎡⎣ 2C1C2 ( C1 + C3 ) - C1C3 ( C2 - C1 ) ⎤⎦ = C1 ⎡⎣ 2 ( C1 + C3 ) + ( C2 - C1 ) ⎤⎦ Q

 ⇒ C=

2C C + C2 C3 + C3 C1 Q = 1 2 VA - VB C1 + C2 + 2C3

Illustration 30

Find the charges on the three capacitors shown in figure. 4 μF

We have VA - VB = ( VA - VD ) + ( VD - VB ) Q Q - Q1 ⇒ VA - VB = 1 + …(1) C2 C1

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 39

8 μF

12 μ F

10 V

20 V

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2.40  JEE Advanced Physics: Electrostatics and Current Electricity

FLOW OF CHARGE

Solution

Let the charges in three capacitors be as shown in figure B

4 μF

8 μF

C

q1 q3

12 μ F A

10 V

D

q2

F

20 V

E

Charge supplied by 10 V battery is q1 and that from 20 V battery is q2. Then, q1 + q2 = q3 …(1) This relation can also be obtained in a different manner. The charges on the three plates which are in contact add to zero. Because these plates taken together form an isolated system which can’t receive charges from the batteries. Thus, q3 - q1 - q2 = 0 ⇒ q3 = q1 + q2 Applying Second Law in Loops BCFAB and CDEFC, we have -

q1 q3 - + 10 = 0 4 12

⇒ q3 + 3 q1 = 120 …(2) and



20 q1 = μC , 3 280 μ C and q3 = 100 μ C 3

Thus, charges on different capacitors in μ C, are shown in figure. 20 μ C 3



qflowing through = a point/branch

∑q -∑q i

f

ENERGY SUPPLIED/CONSUMED BY THE BATTERY We must take a note that whenever a charge +q leaves the positive terminal of the battery of emf V , then work is done by the battery or the battery supplies an energy given by

Esupplied = Wby the battery = qV V

V

Solving the above three equations, we have

q2 =

STEP-1: Find the charges on different capacitors for the initial and final positions/modes of the switch. STEP-2: The charge flowing through a point/branch is calculated by the difference between the charges in the initial and the final positions/modes of the switch.

q

q2 q - 20 + 3 = 0 8 12

⇒ 3 q2 + 2q3 = 480 …(3)



Consider a circuit having a switch, say S . Whenever the switch S in the circuit is opened or closed (or transferred from one mode to the other), then a charge flow takes place through certain points/branches of the circuit. To solve such like problems, we may follow the following steps.

280 μ C 3

q

However, if a charge +q enters the positive terminal of the battery then work is done on the battery or the battery consumes an energy given by Econsumed = Won the battery = qV While solving numerical problems we must keep in mind that the energy supplied by the battery will be taken as positive and the energy consumed by the battery will be taken as negative.

GENERATION OF HEAT 100 μ C

10 V

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 40

20 V

Again, let us consider a circuit having a switch S (as discussed earlier). Now, whenever the switch S in the circuit is opened or closed (or transferred from

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Chapter 2: Capacitance and Applications 2.41

open mode to the close mode), then some amount of heat will be generated in the circuit. If Wsupplied be the energy supplied by the battery/batteries, Wconsumed be the energy consumed by the battery/batteries, ΣUi be the initial energy stored in all the capacitors and ΣU f be the final energy stored in all the capaci-

CV (b) Since a charge flows into the battery i.e., a 2 CV charge enters the positive terminal of the 2 battery so energy is consumed by the battery and CV 2 is given by Wbattery = ( Dq ) Vbattery = . 4

tors, then heat generated is given by

(c) Finally, change in energy of capacitor is

(∑ ∑ )

DH = ( Wsupplied - Wconsumed ) + Ui Uf    by the battery/batteries

for the capacitors

because, we have

Wbattery = DUC + DH

DUc = Ufinal - Uinitial      2

1 1 ⎛V⎞ 2 ⇒   DUc = 2 C ⎜⎝ 2 ⎟⎠ - 2 CV 1 3 1 2 2 2 ⇒   DUc = 8 CV - 2 CV = - 8 CV

Illustration 31

A capacitor of capacitance C which is initially charged upto a potential difference V is connected V with a battery of emf such that the positive ter2 minal of battery is connected with positive plate of capacitor. After a long time, calculate the (a) charge flow through the battery (b) work done by battery (c) heat dissipated in the circuit during the process of charging Solution

(a) Initial charge on the capacitor is qi = CV CV Final charge on the capacitor is q f = 2 Charge flown = Dq = q f - qi = -

Since Wbattery = DUc + DH ⇒   DH = Wbattery - DUc CV 2 ⎛ 3 2⎞ ⇒   DH = - 4 - ⎜⎝ - 8 CV ⎟⎠ 3 CV 2 2 ⇒   DH = CV 8 4 ⇒   DH =

Illustration 32

What charges will flow after shorting of the switch S in the circuit illustrated in figure through sections 1 and 2 in the directions indicated by the arrows?

CV 2

+qi C –qi

+q E

R

–q 1

CV

CV

C1

S

i.e. charge must have flown into the battery. S

CV 2 8

–q

+q

C2

E

2

qi = CV Sw P C

S

V 2

R

qf qf

qf = P

Sw

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 41

Solution

Q

V 2

CV 2

The charge distribution, when the switch is closed, is shown here. In figure, the two capacitors are connected in series and hence their equivalent capacitance is given by

Q



CS =

C1C2 ( C1 + C2 )

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2.42  JEE Advanced Physics: Electrostatics and Current Electricity

K +q2

E

C1

L

–q2 P

+q1

C1 = 2 μ F

M

–q1

C2

1 O2

Sw

E A

N

⎛ CC ⎞ q=⎜ 1 2 ⎟E ⎝ C1 + C2 ⎠

q2 = 0 C2

{∵ - DV = 0 }

⇒ q2 = C2 E …(1) For closed loop MLONM, applying Kirchhoff’s Loop Law, we get



-



q1 C2 E + -E = 0 C1 C2

6 μF 2+3 5 Since both in series, so charge on both will be the same, say q0 . So,



⎛ q0 = ( Ceq ) Vtotal = ⎜ ⎝ +q0

6⎞ ⎟ ( 30 + 60 ) = 108 μ C 5⎠ +q0

J

–q0 C2

C

A 30 V

60 V

When the switch Sw is closed then let q1 and q2 be the charges on capacitors C1 and C2 (as shown). +q1 –q1

+q2

J

C1

Dq1 = C2 E

–q2 C2

Sw

and the charge that flows through section 2 is given by



–q0

=

C1

So, charge flowing through 1 is given by

⎛ CC ⎞ Dq2 = - q1 - q = - ⎜ 1 2 ⎟ E ⎝ C1 + C2 ⎠

( 2 )( 3 )

capacitance is Ceq =

q1 q2 + - E = 0 …(2) C1 C2

⇒ q1 = 0



V2 = 60 V

When the switch Sw is open, then both capacitors 2 μF and 3 μF are in series. So, their equivalent

Substituting the value of q2 from equation (1) in equation (2), we get -

C

Solution

For the closed loop LKPOL, applying Kirchhoff’s Loop Law, we get E-

B V1 = 30 V

If q is the charge on either of the capacitors, then



C2 = 3 μ F

B

A 30 V

C 60 V

Now, q1 = C1V1 = 60 μ C and q2 = C2V2 = 180 μ C

Illustration 33



In the circuit shown, calculate the charge that flows through A, B and C (in the directions shown), when the switch Sw is closed. Also calculate the loss in energy.

Now initial charge on left plate of C1 is q0 = 108 μ C

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 42

and final charge on left plate of C1 is q1 = 60 μ C , which makes us conclude that a charge of -48 μ C is

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2.43

Chapter 2: Capacitance and Applications

being sent (via A) to left plate of capacitor C1. Hence charge flowing from A in the direction shown is ⇒

Dq1 = ( q1 )final - ( q1 )initial

Energy supplied = Dq2 V2 = ( 72 × 10 -6 ) ( 60 )

Dq1 = 60 - 108 = -48 μ C

⇒

Similarly, final charge on right plate of C2 is - q2 = -180 μ C whereas initial charge on right plate of C2 was - q0 = -108 μ C . This simply means that a charge -72 μ C flows (via C ) to right plate of capacitor C2 . Hence charge flowing through A in the diection shown is

Dq2 = ( q2 )final - ( q2 )initial

⇒

Dq2 = -180 - ( -108 ) = -72 μ C

However, final charge at the junction J is ( q2 - q1 ) q2 - q1 = 180 - 60 = 120 μ C

So, finally charges flowing in the circuit are shown in figure. q2 = 180 μ C –180 μ C q1 = 60 μ C –60 μ C J C1

C2

Sw

Δq1 = 48 μ C

B

30 V

48 μ C

60 V

48 μ C 72 μ C

⇒

Energy consumed = 1.44 mJ

Initial energy stored by capacitors is ΣUinitial =

1 1⎛ 6 ⎞ 2 2 CeqVtotal = ⎜ × 10 -6 ⎟ ( 90 ) ⎝ ⎠ 2 2 5

ΣUinitial = 4.86 mJ

Final energy stored by the capacitors is ⇒

ΣUfinal =

1( 1 2 2 2 × 10 -6 ) ( 30 ) + ( 3 × 10 -6 ) ( 60 ) 2 2

ΣUfinal = 6.3 mJ

So, loss in energy is given by

Loss = ( Esupplied - Econsumed ) + ( ΣUi - ΣU f

⇒

Loss = ( 4.32 - 1.44 ) + ( 4.86 - 6.3 )

⇒

Loss = 1.44 mJ

)

72 μ C = Δq2

120 μ C A

Energy supplied = 4.32 mJ Energy consumed = Dq1 V1 = ( 48 × 10 -6 ) ( 30 )

⇒

Initial charge at the junction J is + q0 - q0 = zero ⇒

Now, we observe that energy is supplied by V1 = 60 V battery and consumed by V2 = 30 V battery. So,

C 72 μ C

Test Your Concepts-IV

Based on Capacitor Circuits, kirchhoff’s Laws, Charge Flown and Generation of Heat (Solutions on page H.135) 1. Find the equivalent capacitance between A and B. C A

2C B

2C

2. A circuit has section AB shown in figure. The emf of the source equals E = 10 V, the capacitor capacitances are equal to C1 = 1 μF and C2 = 2 μF , and the potential difference VA - VB = 5 V. Find the voltage across each capacitor. A

2C

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 43

C

B C1

E

C2

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2.44  JEE Advanced Physics: Electrostatics and Current Electricity

3. In the circuit shown, find the charges on 6 μF and 4 μF capacitors. 3 μF

C1

5V

6 μF

5V

7. Calculate the potential difference between points A and B in the circuit shown in figure.

2 μF

4 μF

C3

10 V

A

+q

E1

E2

C

C C

C

C B

5. Determine the potential at point 1 of the circuit shown in figure, assuming the potential at the point O to be equal to zero. Using the symmetry of the formula obtained, write the expressions for the potential at points 2 and 3. E1

C

V

+q –q C2

V2

8. Determine the capacitance CAB of the battery of identical capacitors shown in Figure.

C1 –q

B

V1

4. Find the charges on capacitor C1 and C2 in the ­circuit shown in figure.

C2

A

9. Find the charges on the three capacitors shown in figure. 2 μF

4 μF

6 μF

C1 1 C2

E2

O

2 E3 3

6. Calculate the potential difference between points A and B in the circuit shown in Figure. Under what condition is it equal to zero? C1

A

C2

C3

B

C4

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 44

20 V

10. What amount of heat will be generated in the circuit shown in Figure after the switch S is shifted from position 1 to position 2?

C3

V

10 V

C

S 2

V1

1

V2

11. The arrangement shows a capacitor circuit with a switch S.

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Chapter 2: Capacitance and Applications 2.45

C1 = 3 μ F V1 = 200 V

V2 = 0 V

S a

C3 = 6 μ F



C2 = 6 μ F

b



(i) Calculate the potential difference between a and b when the switch S is open and when S is closed. (ii) Also calculate the charge that flows through the switch when it is closed.

C4 = 3 μ F

SPHERICAL CAPACITOR

To Find C2

It consists of two concentric spherical conductors of radii a and b ( a < b ) respectively. The space between the two conductors is filled with dielectric material of dielectric constant K.

Imagine A to be made open circuited (i.e. made non conducting), then

B

b

C1

a

C2

C2 = 4πε 0 Kb …(2) CASE-1: When battery is connected to B and A is earthed. Then C1 and C2 are in parallel ⇒

C = C1 + C2

⇒

⎛ ab ⎞ + 4πε 0 Kb C = 4πε 0 K ⎜ ⎝ b - a ⎟⎠

⇒

⎛ b2 ⎞ C = 4πε 0 K ⎜ ⎝ b - a ⎟⎠

A

The inner conductor is given a charge q . Depending upon which conductor (inner or outer) is earthed, we get the net capacitance as discussed in the following cases. Let C1 be the capacitance in between the two conductors and C2 be capacitance outside both.

CASE-2: When battery is connected to A , then C1 and C2 are in series. ⇒

1 1 1 = + C C1 C2

To Find C1

⇒

Imagine the outer surface of B to be earthed. Then -q is the charge induced on the inner surface of B.

1 b-a 1 1 = + C ab 4πε 0 K 4πε 0 Kb

⇒

1 1 ⎛ b-a ⎞ = + 1⎟ ⎜ ⎠ C 4πε 0 Kb ⎝ a

⇒

1 1 ⎛ = ⎜ C 4πε 0 Kb ⎝

⇒

C = 4πε 0 Ka

If V is the potential difference between the two ­surfaces, then V=

q -q + 4πε 0 Ka 4πε 0 Kb

⇒

V=

q ⎛ 1 1⎞ ⎜ - ⎟ 4πε 0 K ⎝ a b ⎠

⇒

C1 =

q ⎛ ab ⎞ = 4πε 0 K ⎜ …(1) ⎝ b - a ⎟⎠ V



02_Physics for JEE Mains and Advanced - 2_Part 1.indd 45

b⎞ ⎟ a⎠

CASE-3: When battery connected to A and B is earthed. Then C2 can be omitted as it will not receive any charge. So, C = C1

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2.46  JEE Advanced Physics: Electrostatics and Current Electricity

⇒

Check Point

⎛ ab ⎞ C = 4πε 0 K ⎜ ⎝ b - a ⎟⎠

CASE-4: When battery connected to B and A is open circuited (or made non conducted) then C1 can be omitted (as it is open circuited). So, ⇒

C = C2 C = 4πε 0 Kb

Let us check the limiting value of C when κ 1, κ 2 → 1. In this case, the above expression reduces to C=

4πε 0 abc 4πε 0 abc 4πε 0 ac = = c(b - a) + a(c - b ) b(c - a) (c - a)

which agrees with equation for a spherical capacitor of inner radius a and outer radius c .

Illustration 34

Remark(s)

Consider a conducting spherical shell with an inner radius a and outer radius c . Let the space between two surfaces be filled with two different dielectric materials so that the dielectric constant is κ 1 between a and b and κ 2 between b and c , as shown in ­figure. Determine the capacitance of this system.

(a) Capacitance of a two sphere capacitor of radii a and b when spheres are placed at large distance is

κ1 a +Q

b

κ2

⎛ ab ⎞ C = 4πε 0 ⎜ , for   a,   b ⎝ a + b ⎟⎠ (b) For a spherical capacitor of inner radius a and outer radius b, we have



⎛ ab ⎞ C = 4πε 0 ⎜ ⎝ b - a ⎟⎠

–Q a

l

b

c

Solution

The system can be treated as two capacitors connected in series, since the total potential difference across the capacitors is the sum of potential differences across individual capacitors. The equivalent capacitance for a spherical capacitor of inner radius r1 and outer radius r2 filled with dielectric with dielectric constant κ is given by



⎛ rr ⎞ C = 4πε 0κ ⎜ 1 2 ⎟ ⎝ r2 - r1 ⎠



If both a and b are made extremely large but b - a = d is kept fixed, then 2 2    ab  a ( or b )



⎛ 4π a2 ⎞ ε 0 A , ⇒  C = ε 0 ⎜ = ⎝ d ⎟⎠ d



where A is area of sphere (  4π a2  4π b2 )



The above result is the capacitance of a parallel plate capacitor of plate area A and plate separation d.

Thus, the equivalent capacitance of this system is



κ c ( b - a ) + κ 1a ( c - b ) 1 1 1 = + = 2 C 4πε 0κ 1 ab 4πε 0κ 2 bc 4πε 0κ 1κ 2 abc (b - a) (c - b)

⇒ C=

4πε 0κ 1κ 2 abc ( κ 2c b - a ) + κ 1a ( c - b )

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 46

Illustration 35

In case of two conducting spherical shells having radii a and b ( > a ) calculate the capacity of the system if (a) shells are concentric and inner is given a charge while outer earthed

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Chapter 2: Capacitance and Applications 2.47

(b) shells are concentric and the outer is given a charge while the inner is earthed (c) shells carry equal and opposite charges and are separated by a distance d. Can you try to solve this part, using the concept of self energy and interaction energy?

(ii) As a and b both become very large such that b - a has a finite and small value, say d, then 4π ab  4π a 2  4π b 2 = A and so equation (1) ⎛ ε A⎞ reduces to C = ⎜ 0 ⎟ and hence we observe ⎝ d ⎠ that a spherical capacitor behaves as a parallel plate capacitor when its spherical surfaces have large radii and are close to each other:

Solution

(a) Since the outer sphere is earthed hence, its potential will be zero, i.e., VB = 0. If q ′ is the charge induced on shell B, then

( q + q′ ) = 0

4πε 0 b      ⇒   q′ = -q

(b) In this situation VA = 0 so if q ′ is the charge induced on shell A, then

B

a ⇒   q′ = - b q

A b



a q P

Since no field will exist at P due to B as P lies inside the conductor, so the field at a point P, between the shells is

V=0

q′

So electric field at a point P between the shells is 1 q E = EA + EB = {as EB = Ein = 0 } 4πε 0 r 2 - dV 1 q dV as E = ⇒   dr = 4πε 2  dr 0 r

{

0

q ⇒   - dV = 4πε 0

∫ V

b



1 q′ 4πε 0 r 2 B A q′ V=0

q

dr

∫r

2

a

q V

4πε 0 ab ⇒   C = b - a …(1) From this it is clear that (i) As b → ∞ equation (1) reduces to C → 4πε 0 a. So, we can say that, a spherical conductor is a spherical capacitor with its other plate of infinite radius.

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 47



E = EA + EB =

}

q ⎛ 1 1⎞ ⇒   V = 4πε ⎜⎝ a - b ⎟⎠ 0 Since, C =

1 ⎛ q′ q ⎞ ⎜ + ⎟ =0 4πε 0 ⎝ a b ⎠

dV 1 ⎛ a ⎞ 1 ⇒   - dr = 4πε ⎜⎝ - b q ⎟⎠ 2  r 0

{

∵ E=-

 V

q a ⇒   - dV = - 4πε b 0

∫ 0

b

dV a and q ′ = - q dr b

}

dr

∫r

2

a

q (b - a) ⇒   V = 4πε b2 0 So, C1 =

q 4πε 0 b 2 = V (b - a)

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2.48  JEE Advanced Physics: Electrostatics and Current Electricity

b { as b > a } …(2) ⇒   C1 = a C(> C )    Also, we note that in this situation qB ≠ - qA and hence the system is not a capacitor. However, we observe that C1 =

⎛ a⎞ ⎜⎝ ⎟⎠ q a ⎤ b = = ⎥ ( a b - a) ⎦ B q - ⎛⎜ ⎞⎟ q ⎝ b⎠



  

q 4πε 0 r

2

+

q 4πε 0 ( d - r )

2

q ⎡ 1 1 dV ⇒   - dr = 4πε ⎢ 2 + ( d - r )2 0 ⎣r VB

⇒  -

∫

VA

r

EB

P EA

+q

q dV = 4πε 0

( d -b )

∫ a

{

dV ⎤ ⎥ as E = - dr ⎦

1 ⎡ 1 ⎢ r2 + ( 2 d-r) ⎣

⎤ ⎥ dr ⎦

Since C =

}

1 ⎤ ⎡ 1 ⎢⎣ - r + ( d - r ) ⎥⎦

q ⇒   VA - VB = 4πε 0

1 1 ⎤ ⎡1 1 ⎢⎣ a + b - ( d - a ) - ( d - b ) ⎥⎦

a

For d  a and d  b , we get q ⎡1 1 2⎤ ⇒   VA - VB = V = 4πε ⎢⎣ a + b - d ⎥⎦ 0

–q

q DV

q 4πε 0 ⇒   C2 = V = 1 1 2 …(3) ⎛ ⎞ ⎜⎝ + - ⎟⎠ a b d Using Concept of Self Energy and Interaction Energy The self energy of a spherical conductor is 1 1 ⎡ q2 ⎤ qV = ⎢ ⎥ 2 2 ⎣ 4πε 0 R ⎦

q ⎫ ⎧ ⎨ as V = ⎬ πε 4 0R ⎭ ⎩

while interaction energy of two point charges separated by a distance r U I = qV =

q1 q2 4πε 0 r

So total potential energy of the system is n⎞ ⎛ Self ⎞ ⎛ Self ⎞ ⎛ Interaction ⎜ ⎟ ⎜ ⎟ ⎜ U = Energy + Energy + Energy ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ of A ⎠ ⎝ of B ⎠ ⎝ of AB ⎠

⇒ U = U A + UB + U AB ⇒ U=

q2 ( -q ) + q ( -q ) + 8πε 0 a 8πε 0 b 4πε 0 d

⇒ U=

1 q2 ⎛ 1 1 2 ⎞ ⎜ + - ⎟ 2 4πε 0 ⎝ a b d ⎠

2

( d -b )

q ⇒   VA - VB = 4πε 0

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 48

d

US =

and the system becomes equivalent to two capacitors in parallel having a common potential V. (c) To find the capacitance in this situation, let us consider a point P at a distance r from centre of A on the line joining the centres of two spheres. If E be the net field at this point P, then E=

A

4πε 0 b 2 4πε 0 ab = + 4πε 0 b = CAB + CB (b - a) (b - a)

This system is equivalent to a spherical capacitor (of inner radius a and outer radius b and a spherical conductor of radius b connected in parallel. This is because the charge q given to the ⎛ a⎞ outer shell distributes in such a way that ⎜ ⎟ q ⎝ b⎠ remains on its inner side while the remaining ⎡ ⎛ a⎞ ⎤ ⎢ q - ⎜⎝ b ⎟⎠ q ⎥ lies on its outer side. Hence, ⎣ ⎦ ⎡ qin ⎢q ⎣ out

B

⎛ q2 ⎞ Since, we have U = ⎜ ⎝ 2C ⎟⎠ ⇒ C2 =

4πε 0 ⎛ 1 1 2⎞ ⎜⎝ + - ⎟⎠ a b d

This result is identical to the result obtained earlier, using the method of electric field.

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Chapter 2: Capacitance and Applications 2.49

Now, if d → ∞ then C2′ =

4πε 0 1 1⎞ ⎛ ⎜⎝ + ⎟⎠ a b

⇒ C2′ < C2

1 1 1 1 1 Also, = + = + C2′ 4πε 0 a 4πε 0 b C A CB and hence the given system becomes equivalent to two capacitors CA ( = 4πε 0 a ) and CB ( = 4πε 0 b ) in series. Also, we observe that in this ILLUSTRATION, (a) and (c) represent capacitors (as here the two conductors have equal and opposite charges) while arrangement given in (b) is not a capacitor (because qB is not equal to -qA ).

Substituting



qmax =

C2 from equation (1), we get C1 Qq Q-q

Illustration 37

Three concentric conducting shells A, B and C of radii a, b and c are as shown in figure. Find the capacitance of the assembly between A and C. Now if, the space between A and B is filled with a dielectric material of dielectric constant K, then find the new capacitance between A and C. C A a b

Illustration 36

An isolated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge Q. If q is the charge on the conductor after first operation prove that the maximum charge which can be given to the Qq . conductor in this way is Q-q Solution

Let C1 , be the capacitance of the plate and C2 that of the conductor. After first contact charge on conductor is q. Therefore, charge on plate will remain Q - q. As the charge redistributes in the ratio of capacitances, so

Q - q C1 = …(1) q C2

Let qmax be the maximum charge which can be given to the conductor. Then flow of charge from the plate to the conductor will stop when, both get equal potential. So, ⇒

Vconductor = Vplate qmax Q = C2 C1

⎛C ⎞ ⇒ qmax = ⎜ 2 ⎟ Q ⎝ C1 ⎠

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 49

B

c

Solution

Consider a positive charge q on A and a negative charge -q on C. Find the potential difference V between A and C. The desired capacitance will be C=

q V

While calculating V, notice that net charge on B is zero. Moreover, we can apply the generator principle while calculating V.



V = VA - VC =

q ⎛ 1 1⎞ q ⎛ c-a⎞ ⎜ ⎟ ⎜ - ⎟= 4πε 0 ⎝ a c ⎠ 4πε 0 ⎝ ac ⎠

So, the desired capacitance is,

C=

q ⎛ ac ⎞ = 4πε 0 ⎜ ⎝ c - a ⎟⎠ V

Now when the dielectric is filled between A and B, the electric field will change in this region. Therefore the potential difference and hence the capacitance of the system will change. So, first find the electric field E ( r ) in the region a ≤ r ≤ c . Then find the potential difference ( V ) between A and C and finally the capacitance of the system will be,

9/20/2019 10:49:16 AM

2.50  JEE Advanced Physics: Electrostatics and Current Electricity

C=

q V

where, q = charge on A



q ⎧ ⎪ 4πε Kr 2 ⎪ 0 E( r ) = ⎨ q ⎪ ⎪⎩ 4πε 0 r 2

Since DV =





⎛ 1 1⎞ ⎤ ⎜⎝ - ⎟⎠ ⎥ r b ⎦

for b ≤ r ≤ c

⇒ V=

Q 4πε 0

1 ⎞ ⎛ 1 1 ⎞⎤ ⎡1⎛ 1 ⎢ r ⎜⎝ K - K ⎟⎠ + ⎜⎝ K a - K b ⎟⎠ ⎥ 2 1 1 2 ⎣ ⎦





c

q

∫ 4πε Kr 0

2

dr -

q

∫ 4πε r b

0

2

dr

q ⎡ 1 ⎛ 1 1⎞ ⎛ 1 1⎞ ⎤ ⎜ - ⎟ +⎜ - ⎟ = 4πε 0 ⎢⎣ K ⎝ a b ⎠ ⎝ b c ⎠ ⎥⎦ q ⎡ (b - a) (c - b) ⎤ + 4πε 0 ⎢⎣ Kab bc ⎥⎦ q [ c ( b - a ) + Ka ( c - b ) ] 4πε 0 Kabc

Q 1 ⎞ ⎛ 1 1 ⎞⎤ ⎡1⎛ 1 ⇒ C = = 4πε 0 ⎢ ⎜ +⎜ ⎟ ⎟⎥ V r K K K a K 1⎠ ⎝ 1 2b ⎠ ⎦ ⎣ ⎝ 2

q 4πε 0 Kabc = ( V Ka c - b ) + c ( b - a )

It consists of two coaxial metallic cylinders of inner radius a and outer radius b. The outer surface of the cylinder of radius b (outer one) is earthed. The space between the two cylinders is filled with material of dielectric constant K. Let inner cylinder be given a ⎛ q⎞ charge per unit length of λ ⎜ = ⎟ . A charge -q is ⎝ l⎠ induced on length l at inner surface of outer cylinder

⇒

Illustration 38

A capacitor is formed of two concentric spherical conducting shells of radii a and b with b > a . If the medium between the spherical shells has a dielectric constant K1, from a to c and K 2 from c to b, find the capacitance of the spherical capacitor. Solution

E= -

λ for a < r < b 2πε 0 r

λ dV = dr 2πε 0 Kr

outer surface

⇒

∫

inner surface

⇒

λ dV = 2πε 0 K

Vouter - Vinner = surface

Let the charge on inner sphere be +Q and that on the outer sphere be -Q .

b

surface

a

r=b

∫

r=a

dr r

λ ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠ 2πε 0 K

a q–q

K2 K1

-1

CYLINDRICAL CAPACITOR

The desired capacitance is, C=

b

⎡ 1 ⎛ 1 1⎞ 1 ⎢ K ⎜⎝ a - r ⎟⎠ + K ⎣ 1 2

a

⇒ V=

∫

Q 4πε 0

b

⇒ V=

∫

⎤ dr ⎥ K2 r 2 ⎥ ⎦

⇒ V=

∫ dV = -∫ E ⋅ dr

V = VA - VC = -

∫

r

for a ≤ r ≤ b

So, the potential difference between A and C is

⎡a -Q ⎢ dr V = - Edr = + 4πε 0 ⎢ K1 r 2 ⎣r

l

b a +q

–q

c b Gaussian surface

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Chapter 2: Capacitance and Applications 2.51

Since, inner surface is at higher potential and outer at lower potential, so ⇒

Vinner - Vouter = surface

Vinner - Vouter = surface

⇒

surface

C=

Vinner

surface

surface

⇒

C=

required to add an additional charge dq to the system is dW = Vdq. Thus, the total work is

λ ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠ 2πε 0 K

W=

q ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠ 2πε 0 lK

q - Vouter

surface

=

∫

2πε 0 lK ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠

+λ

1 Q2 ε 0 E2 = 2 32π 2 ε 0 r 4

outside the sphere, and zero inside it. Since the electric field is non-vanishing outside the spherical shell, we must integrate over the entire region of space from r = a to r → ∞. In spherical coordinates the infinitesimal volume element dt = 4π r 2 dr , we have

∫

∫ a

⎛ Q2 ⎜ ⎝ 32π 2 ε 0 r 4

⎞ Q2 2 = π r dr 4 ⎟ 8πε 0 ⎠

∞

∫r

2

a

Q is the electric potential on the sur4πε 0 a

face of the shell, with V ( ∞ ) = 0. We can readily verify that the energy of the system is equal to the work done in charging the sphere. To show this, suppose at some instant the sphere has q charge q and is at a potential V = . The work 4πε 0 a

02_Physics for JEE Mains and Advanced - 2_Part 1.indd 51

E

b–r a

Let the charges on the two wires of length l be +Q and -Q. The charge per unit length is + λ and - λ Q where λ = . The electric field at point P due to l both wires is from positively charged wire towards the negatively charged wire and is given by E=

λ λ + 2πε 0 r 2πε 0 ( b - r )

The potential difference between the two wires is

∫

V = - Edr = -

dr

1 Q2 ⇒ U= = QV 8πε 0 a 2 where V =

b

P

–λ

The corresponding energy density is

∞

a +Q

–Q

⎧ Q ˆ r r>a  ⎪ 2 E = ⎨ 4πε 0 r  ⎪ r C1 ) connected in parallel is 25 times the effective capacitance when they are con6 C nected in series. The ratio 2 is C1 4 3 (A) (B) 2 3 5 25 (C) (D) 3 6 92. In the given system a capacitor of plate area A is charged upto charge q. The mass of each plate is m2 . The lower plate is rigidly fixed. Find the value of m1 so that the system is in equilibrium

Aε Aε 0 (A) 0 (B) 2d d 4 Aε 3 Aε 0 (C) 0 (D) d d 95. Four metallic plates each with a surface area of one side A, are placed at a distance d from each other. The two outer plates are connected to one point A and the two other inner plates to another point B as shown in the figure. Then the capacitance of the system is

A

B

ε0 A 2ε 0 A (A) (B) d d

q ++++ ––––

m2 + (A)

B

m1

q2 m2 (B) ε 0 Ag

q2 + m2 (C) 2 Aε 0 g

(D) None of these

93. Figure shows two capacitors C1 and C2 connected with 10 V battery and terminal A and B are earthed. The graph shows the variation of potential as one C moves from left to right. Then the ratio of 1 is C2

3ε A 4ε 0 A (C) 0 (D) d d 96. Separation between the plates of a parallel plate capacitor is 5 mm. This capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3 mm and dielectric constant K = 10 is placed between the plates, as shown. Potential difference between the plates after the dielectric slab has been introduced is

10 V

A

C1

C2

B

Potential 3 mm

10 V 4V

5 mm r

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2.80  JEE Advanced Physics: Electrostatics and Current Electricity 6 μF

(A) 18.5 V (B) 13.5 V

a

3 μF

(C) 11.5 V (D) 6.5 V 97. The capacitance between the inner and outer curved cylindrical conductor surface as shown in figure is (Space between conductor surface is filled with dielectric of K = 5.5 )

3 μF

b

V=0 6 μF

(A) 300 µC (B) 400 µC (C) 500 µC (D) 100 µC

K = 5.5 30° 60 20 mm

S

+200 V

101. Two charged capacitors A and B have their outer plates fixed and inner plates connected by a spring of force constant k . The charge on each capacitor is q . The extension in the spring at equilibrium is

5 mm

A

B K

6.86 pF (B) 1.86 pF (A) (C) 3.26 pF (D) 12.63 pF 98. If area of each plate is A and the successive separations are d, 2d and 3d, then equivalent capacitances across A and B is A d 2d 3d B

ε0 A ε0 A (B) (A) 6d 4d 3ε A ε0 A (C) 0 (D) 4d 3d 99. An air capacitor is charged up to potential V1 . It is connected in parallel to an identical uncharged capacitor filled with a dielectric medium. After redistribution of charge is final potential difference of this combination is V , then the dielectric constant of the substance will be (Assume that both the capacitors with air medium has same capacitance) V ( V1 - V ) (B) V ( V1 - V ) (A) 2

V ( V1 - V ) (C) (D) V V V ( 1 ) 100. Find the amount of charge that will flow through the switch when it is closed

02_Physics for JEE Mains and Advanced - 2_Part 2.indd 80

q2 q2 (A) (B) 2A ε0k 4 A ε0k q2 (C) A ε0 k

(D) zero

102. Two spherical conductors A1 and A2 of radii r1 and r2 and carrying charges q1 and q2 are connected in air by a copper wire as shown in the figure. Then the equivalent capacitance of the system is A2 A1 Q1

r1

Q2 r2

4πε r r 4πε 0 ( r1 + r2 ) (A) 0 1 2 (B) r2 - r1 4πε 0 r2 (D) 4πε 0 r1 (C) 103. A parallel plate capacitor is charged with a battery. Charging is completed, battery is disconnected. Now the separation between plates is decreased then which of following is correct? (A) Electric field will not be same (B) Potential differences between plates increases (C) Capacitance will decrease (D) Stored energy will decrease

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Chapter 2: Capacitance and Applications 2.81 104. Two identical capacitors are connected in parallel across a potential difference V. After they are fully charged, the positive of first capacitor is connected to negative of second and negative of first is connected to positive of other. The loss in energy will be

3ε A (A) 0 5 d

1 CV 2 (B) CV 2 (A) 2

108. In the given figure, the work done by battery after the switch S is closed will be

CV 2 (C) 4

5ε A (C) 0 3 d



(B)

5 ε0 A 4 d



(D)

4 ε0 A 5 d

10 μ F

(D) zero

105. In the given circuit, if point C is connected to the earth and a potential of 12000 V is given to point A. The potential at B is

10 μ F

4V

10 μ F 10 μ F 5 μF A

S C

B 10 μ F

(A) 100 µ J (B) -100 µ J

(A) 1500 V (B) 1000 V (C) 400 V (D) 500 V 106. A capacitor is half filled with dielectric ( K = 2 ) as shown in Arrangement-1. If the same dielectric is to be filled in the same capacitor as shown in Arrangement-2, then the thickness of dielectric so that capacitor still has same capacitance is a A/2

(C) 80 µ J (D) -80 µ J 109. In the figure shown the plates of a parallel plate capacitor have unequal charges. Its capacitance is C. P is a point outside the capacitor and close to the plate of charge -Q . The distance between the plates is d select incorrect alternative 2Q

–Q

a P

A/2

K

d

d t

b Arrangement – 1

K b Arrangement – 2

2d 3d (A) (B) 2 3 3d 4d (C) (D) 3 4 107. Five identical plates of equal area A are placed parallel to and at equal distance d from each other as shown in figure. The effective capacity of the system between the terminals A and B is



(A) A point charge at point P will experience electric force due to capacitor (B) The potential difference between the plates will 3Q be 2C



(C)  The energy stored in the electric field in the 9Q 2 region between the plates is 8C



(D) The force on one plate due to the other plate is Q2 2Cd

110. The net capacitance between A and B is C

C

A B C A

B C

02_Physics for JEE Mains and Advanced - 2_Part 2.indd 81

C

C C

9/20/2019 10:48:10 AM

2.82  JEE Advanced Physics: Electrostatics and Current Electricity 2C (A) 6C (B) 5 2C (C) (D) None of these 3 111. Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of V0 volt. If one of the capacitors is completely immersed in a liquid with dielectric constant K, the potential difference between the plates of the other capacitor will change to K +1 K V0 (A) V0 (B) K K +1 K +1 2K (C) V0 (D) V0 2K K +1 112. The potential at point A in the circuit is ( N point is grounded. Grounding means that potential of that point is zero.) 10 V C

C

C

N

(C) 480 µC (D) 24 µC 116. Potential difference across C1 and C2, when VA - VB equals 5 V is

A

C

(A) 10 V (B) 7.5 V (C) 5 V (D) 2.5 V

C2

1 μF

10 V

2 μF

2ε 0 AV 2 (D) W=0 (C) d 114. A parallel plate capacitor having a plate separation of 2 mm is charged by connecting it to a 300 V supply. The energy density is -3

10 Jm (B) 1.0 Jm (A) (C) 0.1 Jm -3 (D) 0.01 Jm -3 115. An air capacitor of capacity C = 10 µF is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The (additional) charge that flows now from battery to the capacitor is

B

(C) 10 V and 5 V (D) 7.5 V and 7.5 V 117. Two identical capacitors are connected in parallel across a potential difference V. After they are fully charged the battery is removed and the positive of first capacitor is connected to negative of second and negative of first is connected to positive of other. The loss in energy will be 1 CV 2 (B) CV 2 (A) 2 (D) zero

118. Three capacitors of capacitors C1 , C2 , C3 are connected as shown in the figure. The points A , B and C are at potential V1 , V2 and V3 respectively. Then the potential at O will be A C1 C2

ε 0 AV 2 ε 0 AV 2 (A) (B) 2d d

02_Physics for JEE Mains and Advanced - 2_Part 2.indd 82

E

(A) 5 V and 10 V (B) 15 V and 15 V

113. A parallel plate capacitor has plate area A and separation d and is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart until their final separation is 2d . The work done in this process is

-3

C1

CV 2 (C) 4 B

A

120 µC (B) 600 µC (A)

B

O

C3 C

V1 + V2 + V3 (A) 2 V1V2 + V2V3 + V3V1 (B) V1 + V2 + V3 V1C1 + V2C2 + V3C3 (C) C1 + C2 + C3

(D) zero

119. A circuit element is placed in a closed box. At time t = 0 , a constant current generator supplying a current of I amp is connected across the box. Potential difference across the box varies according to graph as shown in the figure. The element in the box is

9/20/2019 10:48:29 AM

Chapter 2: Capacitance and Applications 2.83 122. What fraction of the energy drawn from the charging battery is stored in a capacitor?

volts (V) 8

(A) 75% (B) 100% (C) 25% (D) 50%

2 3

Time t(sec)

a resistance of 2 Ω a battery of emf 6 V an inductance of 2 H a capacitance



(A) (B) (C) (D)

120.

A , B , C , D , E and F are conducting plates each of area A and any two consecutive plates separated by a distance d. The net energy stored in the system after the switch S is closed is A B

CD

123. The work done in inserting the dielectric slab inside one of the capacitors as shown in diagram. C V

K

E F

C

CV 2 ⎛ K – 1 ⎞ CV 2 ⎛ K – 1 ⎞ (B) (A) ⎜ 2 ⎝ K + 1 ⎟⎠ 4 ⎜⎝ K + 1 ⎟⎠ CV 2 ⎛ K + 1 ⎞ CV 2 ⎛ K + 1 ⎞ (D) (C) ⎜ ⎟ 4 ⎝ K –1 ⎠ 2 ⎜⎝ K – 1 ⎟⎠ 124. The two spherical shells are at large separation one of them has radius 10 cm and has 1.25 µC charge. The other is of 20 cm radius and has 0.75 µC charge. If they are connected by a conducting wire of negligible capacity, the charge on the shells are

3ε 0 A 2 5ε 0 A 2 (A) V (B) V 2d 12d

ε0 A 2 ε0 A 2 (C) V (D) V 2d d 121. Two long coaxial cylindrical metal tubes (inner radius a , outer radius b ) stand vertically in a tank of dielectric oil (of mass density ρ , dielectric constant K ). The inner one is maintained at potential V and the outer one is grounded. The equilibrium height ( h ) to which the oil rise in the space between the tubes is [Assume this height ( h ) as a equilibrium height] 2ε V 2 ( K - 1 ) (A) 0 ⎛ b⎞ ρ g ( b 2 - a 2 ) log e ⎜ ⎟ ⎝ a⎠

2 4 µC , µC 1 µC , 1 µC (B) (A) 3 3 4 2 µC , µC (D) (C) 0.25 µC , 0.25 µC 3 3 125. Three identical metal plates of area A are at distance d1 and d2 from each other. Metal plate A is uncharged, while plate B and C have respective charges +q and -q . If metal plates A and C are connected by switch K through a consumer of unknown resistance, the energy given by the consumer to its surrounding Assume d1 = d2 = d A B

ε V 2 ( K - 1) (B) 0 ⎛ b⎞ ρ g ( b 2 - a 2 ) log e ⎜ ⎟ ⎝ a⎠ 4ε V 2 ( K - 1 ) (C) 0 ⎛ gρ ( b 2 - a 2 ) log e ⎜ ⎝

–q

K

b⎞ ⎟ a⎠

6ε V 2 ( K - 1 ) (D) 0 ⎛ b⎞ ρ g ( b 2 - a 2 ) log e ⎜ ⎟ ⎝ a⎠

02_Physics for JEE Mains and Advanced - 2_Part 2.indd 83

C

+q

q2 d q2 d (A) (B) 4ε 0 A ε0 A q2 d 2q 2 d (C) (D) 2ε 0 A ε0 A

9/20/2019 10:48:50 AM

2.84  JEE Advanced Physics: Electrostatics and Current Electricity 126. A closed body, whose surface F is made of metal foil, has an electrical capacitance C with respect to an uniformly distant point. The foil is now dented in such a way that the new surface F * is entirely inside or an the original surface as shown in the figure. Then

A l v

b

x

F

b

B F∗



dC/dt (A)

dC/dt (B)

(A) Capacitance of F * > capacitance of F *



(B) Capacitance of F < capacitance of F



(C) Capacitance of F * = capacitance of F (D) Nothing can be concluded from given

127. A capacitor is composed of three parallel conducting plates. All three plates are of same area A. The first pair of plates are kept a distance d1 apart and the space between them is filled with a medium of a dielectric ε1 . The corresponding data for the second pair are d2 and ε 2 respectively. The surface charge density on the middle plate is d1

d2

ε1

ε2

v

(C) dC/dt

v dC/dt (D)

v

v

130. Two capacitor C1 and C2 , charged with q1 and q2 are connected in series with an uncharged capacitor C , as shown in figure. As the switch S is closed C

C1

+ q1 –

+ C – 2

q2

V0

S

(A) C gets charged in any condition

ε ⎞ ε ⎞ ⎛ε ⎛ε ε 0V ⎜ 1 + 2 ⎟ (B) - ε 0V ⎜ 1 + 2 ⎟ (A) ⎝ d1 d2 ⎠ ⎝ d1 d2 ⎠

(B) C gets charged only when q1C2 > q2C1

ε ⎞ ε ⎞ ⎛ε ⎛ε 2ε 0V ⎜ 1 + 2 ⎟ (D) -2ε 0V ⎜ 1 + 2 ⎟ (C) ⎝ d1 d2 ⎠ ⎝ d1 d2 ⎠

(D) C gets charged when q1C2 ≠ q2C1

128. A capacitor of capacity C1 is charged upto potential V volt and then connected in parallel to an uncharged capacitor of capacity C2 . The final potential difference across each capacitor will be

(C) C gets charged only when q1C2 < q2C1

131. Initially K1 is closed, now if K 2 is also closed, the heat dissipated in the connecting wires is C

C

CV C1V (B) (A) 2 C1 + C2 C1 + C2 ⎛ ⎛ C2 ⎞ C2 ⎞ (C) ⎜⎝ 1 + C ⎟⎠ V (D) ⎜⎝ 1 – C ⎟⎠ V 1

1

129. Let a dielectric slab be inserted inside the parallel plate capacitor with a speed v . Then the variation of dC rate of change of capacitance with respect to v is dt

02_Physics for JEE Mains and Advanced - 2_Part 2.indd 84

C V

C

K2 K1

1 2 (A) CV 2 (B) CV 2 3 2 1 1 (C) CV 2 (D) CV 2 3 4

9/20/2019 10:49:07 AM

Chapter 2: Capacitance and Applications 2.85 132. Half of the space of a parallel plate capacitor is filled with dielectric as shown. Potential difference V0 is applied across the plates. The electric field between the plates in the space with dielectric is E1 and in the space without dielectric is E2 then

K

(A) E1 = E2 (B) E1 > E2 (C) E1 < E2

(D) None of these

133. The charge on 6 µF capacitor is A 90 V

9 μF

6 μF

12 μ F

B

(A) 540 µC (B) 180 µC (C) 324 µC (D) 810 µC 134. A thin metallic wire is joined between both the plates of parallel plate capacitor as shown in diagram. The capacitance of the system will be

(A) K=

4b d and < b ≤ d 2b - d 3

(B) K=

2b d and < b ≤ d 2b - d 2

(C) K=

2b d and ≤ b ≤ 2d 2b - d 2

(D) K=

2b d and ≤ b ≤ d 2b - d 4

136. Capacitance of an isolated sphere is increased n times when it is enclosed by an earthed concentric sphere. The ratio of their radii is n2 n (A) (B) n-1 n-1 2n 2n + 1 (C) (D) n+1 n+1 137. A capacitor of capacity C1 = 1 µF can with stand the maximum voltage V1 = 16 KV while another capacitor C2 = 2 µF can with stand the maximum voltage V2 = 4 KV. The maximum voltage of capacitance ­connected in series will be (A) 10 KV (B) 12 KV (C) 6 KV (D) 4 KV 138. The equivalent capacitance between points A and B 3C

3C

A

a

3C

C B

(A) C (B) 2C (C) 3C (D) 4C

b

C (A) (B) 2C 2

(C) 0

(D) Infinite

135. A capacitor consists of two parallel metal plate of area A separated by a distance d. A dielectric slab of area A , thickness b and dielectric constant k is placed inside the capacitor. If CK is the capacitance of capacitor with dielectric, how must K and b be selected so that CK = 2C , where C is capacitance without dielectric?

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 85

139. An arrangement of three capacitors is connected to the battery. C1 and C2 has fixed capacitances. If C3 is increased and is made 10 µF then C3 = 1 μ F C2 = 2 μ F C1 = 6 μ F 9V

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2.86  JEE Advanced Physics: Electrostatics and Current Electricity



(A) charge on C1 increases and that on C2 also increases (B) charge on C1 decreases and that on C2 decreases



(C) charge on C1 increases but that on C2 decreases



(D) charge on C1 decreases but that on C2 increases



140. Consider an arrangement of three plates X , Y and Z each of area A and separation d. The energy stored, when the plates are fully charged is X

d

Y

V

ε 0 AV 2 ε 0 AV 2 (A) (B) 2d d

142. A capacitor of 2 µF is charged at a steady rate from zero to 5 coulomb. The graph correctly representing the variation of potential difference across its plates with respect to charge on the condenser is (B) V (× 106 V) 5

5 (Coulomb)

Q

(C) V (× 106 V)

5 (Coulomb)

Q

(D) V (× 106 V)

5

2.5

5 (Coulomb)

Q

5 (Coulomb)

Q

143. The potential difference between points a and b C1 = 1 μ F a C2 = 1.5 μ F

9 × 106 V (B) 4.5 × 106 V (A) (C) 1.8 × 107 V (D) 13.6 × 106 V

Q0 , (C)

141. From a supply of identical capacitors rated 8 mF , 250 V the minimum number of capacitors required to form a composite 16 mF , 1000 V is (A) 2 (B) 4 (C) 16 (D) 32

2.5

144. Two metallic charged sphere whose radii are 20 cm and 10 cm respectively, each having 150 µC positive charge. The common potential after they are connected by a conducting wire is

3Q0 , 3V0 , 3E0 (B) Q0 , 3V0 , 3E0 (A)

2ε 0 AV 2 3ε 0 AV 2 (C) (D) d d

(A) V (× 106 V)

(C) -13 V (D) 13 V

145. A capacitor when filled with a dielectric K = 3 has charge Q0 , voltage V0 and Electric field E0 . If the dielectric is replaced with another one having K = 9, the new value of charge, voltage and field will be respectively

d

Z

(A) 5 V (B) 9V

V0 V E Q0 , 0 , 0 , 3E0 (D) 3 3 3

146. A capacitor with plate separation d is charged to V volts. The battery is disconnected and a dielecd tric slab of thickness and dielectric constant 2 is 2 inserted between the plates. The potential difference across the terminals becomes (A) V (B) 2V 4V 3V (C) (D) 3 4 147. A parallel plate capacitor C0 is charged to a potential V0 . The energy stored in the capacitor when the charging battery is kept connected and the plate separation is doubled is U1 . The energy stored in the capacitor when the charging battery is disconnected and the separation between plates is doubled is U 2 , U then 1 is U2

(A) 4

(B)

1 4



(C) 2

(D)

1 2

148. A capacitor is made of a flat plate of area A and a second plate having a stair like structure as shown in diagram. The capacitance of the arrangement is A/3 d A/3

b C3 = 2.5 μ F C4 = 0.5 μ F

A/3 d

d Area = A

30 V

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Chapter 2: Capacitance and Applications 2.87 18 ε 0 A ε0 A (A) (B) d 11 d 11 ε A ε0 A (C) 0 (D) 18 d 3d 149. Calculate the reading of voltmeter between X and Y then ( Vx - Vy ) is equal to 1F 2F

X

A

V Y

3F

1F

B

6F

A B

(A) 32 µF (B) 2 µF (C) 8 µF (D) 16 µF 153. A capacitor stores 50 µC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 µC flows through the battery. The dielectric constant of the material is (A) 2.5 (B) 2 (C) 4 (D) 3 154.

20 V

(A) 10 V (B) 13.33 V

n conducting plates are placed face to face as shown. Distance between any two plates is d. Area of the A A A , … ( n -1 ) plates is A , 2 4 2 A

(C) 3.33 V (D) 10.33 V 150. Two similar conducting spheres having charge +q and -q are placed at d separation from each other in air. The radius of each ball is r and the separation between their centre is d ( d  r ) . The capacitance of the two ball system is +q

–q

r

r

A 2

d



A 4

d

The equivalent capacitance of system is

ε0 A ε0 A (A) (B) 2n d (2n - 1)d

d

4πε 0 r (B) 2πε 0 r (A) ⎛ ε r⎞ ⎛ r⎞ (C) 4π log e ⎜ 0 ⎟ (D) 4π log e ⎜ ⎟ ⎝ d ⎠ ⎝ d⎠ 151. Potentials of A and B are respectively

ε0 A ε0 A (C) (D) (2n - 2)d (2n -1 - 1)d 155. Four condensers are joined as shown in fig. The capacity of each is 8 µF . The equivalent capacity between points A and B will be

10 V A A

2C

C

C

B B

(A) +6V , -4V (B) -6V , +4V

(A) 32 µF (B) 2 µF

(C) +10V , 0V (D) +4V , +2V

(C) 8 µF (D) 16 µF

152. Four condensers are joined as shown in the adjoining figure. The capacity of each is 8 µF . The equivalent capacity between the points A and B will be

156. The ratio of potential difference between 1 µF and 5 µF capacitors is

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 87

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2.88  JEE Advanced Physics: Electrostatics and Current Electricity 1 μF 2 μF

ε0 A 2ε 0 A (A) (B) d d

2 μF

3 μF

3ε 0 A ε0 A (C) (D) 4d d

5 μF

10 V

1 : 2 (B) (A) 3 :1

160. A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, the time rate of change of electrostatic energy of capacitor is proportional to

1 : 5 (D) 10 : 1 (C) 157. Each capacitor in the circuit has capacitance 4 µF. When the switch S is closed, the charge that flows through AB is S

V d V0

(A) x -2

(B) x

-1

(C) x (D) x2 161. In the circuit shown, the 40 V

12 V

(A) 320 µC (B) 213 µC (C) 107 µC

C1 = 2 μ F

(D) None of these

158. A finite ladder is constructed by connecting several sections of 2 µF , 4 µF capacitor combinations as shown in the figure. It is terminated by a capacitor of capacitance C . The value of C , such that equivalent capacitance of the ladder between the point A and B becomes independent of the number of sections in between is 4 μF 4 μF

4 μF

A

C2 = 8 μ F 6V



(A) charge on C2 is greater than C1



(B) charge on C1 is greater than C2



(C) potential difference across C1 and C2 are same



(D) potential difference across C1 is greater than C2

162. A capacitor of capacitance 1 µF is filled with two dielectrics of dielectric constants 4 and 6. What is the new capacitance?

C

2 μF 2 μF 2 μF B

d

(A) 4 µF (B) 2 µF (C) 18 µF (D) 6 µF 159. Six plates of equal area A and plate separation as shown are arranged equivalent capacitance between A and B is d A

2d 3d 2d d

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 88

1 2 3 4

(A) 10 µF (B) 5 µF (C) 4 µF

163. The resultant capacitance between (A) and (B) in the following figure is A

B 5 6

(D) None of these

3 μF 2 μF

3 μF 2 μF

3 μF

3 μF

3 μF

3 μF

3 μF

B

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Chapter 2: Capacitance and Applications 2.89 C1 = 1 μ F

(A) 1 µF (B) 3 µF

C3 = 3 μ F

(C) 2 µF (D) 1.5 µF 164. A condenser of capacitance 10 µF has been charged to 100 V . It is now connected to another uncharged condenser in parallel. The common potential becomes 40 V . The capacitance of another condenser is

C2 = 2 μ F

S2

C4 = 4 μ F

S1

(A) 15 µF (B) 5 µF V = 12 Volt

(C) 10 µF (D) 16 µF 165. Two parallel plate capacitors of capacitance C and 3C are connected in parallel and charged to a potential difference V with the help of a battery. The battery is disconnected and the space between these two capacitors are filled with dielectric of constant K and K respectively. The voltage across the combination 2 will be 4V 6V (A) (B) 3K 3K 8V 3V (C) (D) 5K 5K 166. Two insulated metallic spheres of 3 µF and 5 µF capacitances are charged to 300 V and 50 V respectively. The energy loss, when they are connected by a wire, is 0.012 J (B) 0.0218 J (A) (C) 0.0585 J (D) 3.75 J 167. The ratio of equivalent capacitances between points P and Q in the following cases is P C C

P

Q

C

C

C P

C C

Q

C

C Q

1 14 (A) (B) 2 15 15 2 (C) (D) 14 1 169. Capacitor C1 of capacitance 1 µF and capacitor C2 of capacitance 2 µF are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at t = 0 . Then (A) at t = 0 , the value of current in the circuit containing 1 µF is more than current in the circuit containing 2 µF (B) at t = 0 , the current in 2 µF capacitor circuit is more than current in 1 µF capacitor circuit

(C)  1 µF capacitor losses 50% charge sooner than 2 µF capacitor



(D) 2 µF capacitor losses 50% charge sooner than 1 µF capacitor

170. A number of capacitors each of capacitance 1 µF and each one of which get punctured if a potential difference just exceeding 500 volt is applied, are provided. Then an arrangement suitable for giving a capacitor of 2 µF across which 3000 volt may be applied requires at least (A) 18 component capacitors (B) 36 component capacitors (C) 72 component capacitors (D) 144 component capacitors 171. Two capacitors are joined in series as shown in figure. The area of each plate is A . The equivalent of the combination is

(A) 1 : 2 : 3 (B) 3 : 2:1

d

(C) 1 : 1 : 1 (D) 1:1: 2 168. Initially only S1 is closed. After a long time S2 is also closed. The ratio of charge on C1 when only S1 is closed to the charge on C1 when both S1 and S2 is closed is

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 89

a

b d

9/20/2019 10:59:48 AM

2.90  JEE Advanced Physics: Electrostatics and Current Electricity ε0 A ε0 A (B) (A) d1 - d2 a-b 1⎞ ⎛ 1 1⎞ ⎛ 1 (C) ε 0 A ⎜ - ⎟ (D) ε0 A ⎜ - ⎟ ⎝ a b⎠ ⎝ d1 d2 ⎠

176. There are two metallic plates of a parallel plate capacitor. One plate is given a charge +q while the other is earthed as shown . Points P, P1 and P2 are taken as shown in adjoining figure. Then the electric intensity is not zero at

172. The charge on capacitor C2 if a potential difference V is applied between A and B is

P2

C3

A

B

C2 C4

CV (A) CV (B) 3 CV CV (C) (D) 2 4 173. A parallel plate capacitor is charged to a definite potential and the charging battery is disconnected. Now if the plates of the capacitor are moved apart, then (A) the stored energy of the capacitor increases (B) Charge on the capacitor increases (C) Voltage of the capacitor decreases (D) The capacitance increases

D

E

C

F

P

–

+

–

+

–

+

–

(C) P2 only

(D) P, P1 and P2

177. A capacitor of capacitance C1 is charged to a potential V0 . The electrostatic energy stored in it is U 0 . It is connected to another uncharged capacitor of capacitance C2 in parallel. The energy dissipated in the process is C C1 (A) 2 U 0 (B) U0 C1 + C2 C1 + C2 2

⎛ C1 – C2 ⎞ C1C2 U0 (C) ⎜⎝ C + C ⎟⎠ (D) 2( C 1 + C2 ) 1 2 178. Charge on capacitor C0 is C0 = 2 μ F 2 μF

4 μF 4 μF

20 V

H

P1

(B) P1 only

3 μF

–20 μ C B

–

(A) P only

174. A charge -20 µC is given to side GH of the negative plate of the capacitor. The charge on positive plate at CD side is A

–

+ +

(assume C1 = C2 = C3 = C4 = C ) C1

+

3 μF

G

(A) +10 µC (B) +20 µC

(A) 10 µC (B) 20 µC

(C) -10 µC

(C) 30 µC (D) 0 µC

(D) None of these

175. The capacitor that will get minimum amount of charge if a battery is connected between Q and S (all capacitors are identical) is 2

P 1

3 Q 4 R

5

179. A capacitor of capacity C1 is charged to the potential of V0 . On disconnecting with the battery, it is connected with a capacitor of capacity C2 as shown in the figure. The ratio of energies before and after the connection of switch S will be

S

S C1, V0



(A) 1 (C) 3

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 90

C2

(B) 2 (D) 6

9/20/2019 10:59:58 AM

Chapter 2: Capacitance and Applications 2.91 C1 ( C1 + C2 ) (B) (A) C1 ( C1 + C2 ) C1 (C) C1C2 (D) C2

182. The outer cylinders of two cylindrical capacitors of capacitance 2 µF each are kept in contact and the inner cylinders are connected through a wire. A battery of emf 8 V is connected. The total charge supplied by battery

180. In the given circuit the charge passing through the battery is 48 µC . The potential difference of capacitor C is

8V

3 μF C 6 μF 6 μF 2 μF

(A) q = 2 µC (B) q = 8 µC (C) q = 16 µC (D) q = 32 µC

4 μF

(A) 6 V (B) 8V

183. Capacitor C1 ( 10 µF ) and C2 ( 20 µF ) are connected in series across a 3 kV supply as shown. The charge on capacitor C1 is

(C) 3 V (D) 4.5 V

3 kV

12 V

181. A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates at any time, then the time rate of change of electrostatic energy of capacitor is proportional to (A) x -2

(B) x

(C) x -1 (D) x2

C1 = 10 μ F

C2 = 20 μ F

(A) 45000 µC (B) 20000 µC (C) 15000 µC (D) 10000 µC

Multiple Correct Choice Type Questions This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1.

A parallel plate capacitor is filled with two dielectrics having dielectric constants K1 and K 2 as shown in the figure. The capacitance of the system is C and that of the two capacitors forming the system is C1 and C2. Then we have

C1 and C2 are in series (C) C= (D) 2.

Plate area = A K1

d/2

K2

d/2

2K1ε 0 A d 2K 2ε 0 A C2 = (B) d

C1 = (A)

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 91



2ε 0 A ⎛ K 1 K 2 ⎞ d ⎜⎝ K1 + K 2 ⎟⎠

An air filled parallel plate capacitor having capacitance C0 is connected to a battery of emf E and then disconnected from it. A dielectric slab having a dielectric constant K , which can just fill the air column in the capacitor, is now inserted in it. Then it is observed that the 1⎞ 1 ⎛ (A) change in energy is C0E2 ⎜ 1 - ⎟ ⎝ 2 K⎠ 1 C0 E 2 ( K - 1 ) 2 (C)  energy stored in the capacitor decreases by a ­factor of K (D) potential difference between the plates decreases by a factor of K (B) change in energy is

9/20/2019 11:00:07 AM

2.92  JEE Advanced Physics: Electrostatics and Current Electricity 3.

Three large, parallel conducting plates P, Q and R are placed horizontally such that the plates P and R are rigidly fixed and earthed and the plate Q is given some charge. Under the influence of the electrostatic and gravitational forces, the plate Q may stay in equilibrium

field in the region between the plates, U be the electrostatic energy stored in the capacitor and ΔV be the potential difference between the plates of the capacitor, then on increasing the separation between the plates we observe that U increases (A)

(B) F decreases

P

(C) ΔV decreases

(D) E remains constant

Q

8.

R



(A) (B) (C) (D)

midway between P and R if it is closer to P than to R if it is closer to R than to P which is always unstable

A parallel plate capacitor has a dielectric slab that completely fills the empty space of the capacitor. The capacitor is charged by a battery and then battery is disconnected. Now the slab is pull out slowly at t = 0. If at time t , capacitance of the capacitor is C , potential difference across is V , and energy stored in it is U, then which of the following graphs are correct?

(A) V

U (B)

4.

Consider a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 4 d . Now a dielectric slab of thickness 2d is inserted in the capacitor such that the slab is equidistant from both the plates. The capacitor is given some charge. As x increases from 0 to 4d , we observe that the (A) magnitude of the electric field does not change (B) direction of the electric field does not change (C) electrostatic potential increases continuously (D)  electrostatic potential first increases, then decreases and then again increases 5.

A parallel plate capacitor having capacitance C has two plates X and Y each having a charge Q . The plate X is now connected to the positive terminal of a battery and the plate Y to the negative terminal of the Q same battery of emf E = . Then it is observed that C



(A) an energy CE2 is supplied by the battery

6.

7.

V (C)

A parallel plate capacitor is charged from a cell and then isolated from it. If F be the force of attraction between the plates of the capacitor, E be the electric

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 92

C (D)

t

C

9.

(B)  Q amount of charge flows, through the capacitor, from the positive terminal to the negative terminal of the battery (C) the total charge on the plate X is 2Q (D) the total charge on the plate Y is zero Consider a parallel plate capacitor that has been charged from a battery and then disconnected from it. Now the separation between the plates is doubled, then the (A) external agent has to do some work on the plates (B) energy stored in the capacitor doubles (C) potential difference between the plates doubles (D) field between the plates does not alter

V

t

Consider two identical dielectric slabs that are inserted into two identical capacitors A and B connected to the battery as shown in figure. Now the slab of capacitor B is pulled out from it with the battery still remaining connected. Then during the process of removing the dielectric, we observe that the following statement(s) to be correct. A

B F

a



b

(A) Charge flows from a to b . (B) Final charge on capacitor B will be less than that on capacitor A . (C)  Work is done by the external force F always appears a significant heat in the circuit. (D) Internal energy of the battery increases.

10. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of

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Chapter 2: Capacitance and Applications 2.93

s and -s, area A, charges Q and -Q. The electrostatic force due to the electric field created by the plate b is Fe and the spring gets an extension x . Then

the capacitor so as to fill the space between the plates. If Q , E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in questions, in the process of inserting the slab, then Q= (A)

k a b Capacitor connected to a spring

ε 0 AV ε KAV Q= 0 (B) d d

V 1⎞ ε AV 2 ⎛ E= (D) W= 0 (C) ⎜ 1 - ⎟⎠ Kd 2d ⎝ K

(A) Fe =

QE sQ (B) x= 2ε 0 k

11. A large vessel is filled with water having dielectric constant K = 81 . Two square plates each of length 10 cm and separated by 1 cm are held parallel to each other with their lower ends just touching the surface of water. The plates are charged by applying a voltage V and then disconnected from the voltage source. The liquid rises by 1 cm between the plates. The capacitor has capacitances C0 and C without and with water respectively. Then

(C) x=

Q2 Q2 (D) Fe = 2ε 0 kA 2kAε 0

14. In an arrangement of capacitors shown C1 = C2 = C3 = C. The emf of the battery connected in the circuit is E . When the switch S is closed, then S

C1

C3 C2

10 cm E 1 cm 1 cm



(A) some charge enters the positive terminal of the battery (B) some charge flows out of the positive terminal of the battery 4 CE flows through the battery 3 (D) a charge CE flows through the battery

C0 = 8.85 µF (B) (A) C = 0.689 nF



(C) V = 1110 V (D) C = 80 µF



12. A conducting sphere of radius R, carrying charge Q, lies inside an uncharged conducting shell of radius 2R. When joined by a metal wire,

15. Two large, parallel conducting plates are placed close to each other. The inner surfaces of the two plates have surface charge densities +s and -s . The outer surfaces are without charge. The electric field has a magnitude of





Q (A)  amount of charge will flow from the sphere to 3 the shell 2Q (B)  amount of charge will flow from the sphere 3 to the shell Q amount of charge will flow from the sphere to (C)  the shell

Q2 amount of heat will be produced (D) 16πε 0 R 13. Consider an air filled parallel plate capacitor with one plate connected to a spring having a force constant k and the other plate is held fixed. The system rests on a frictionless table top. The plates have charge density

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 93

(C) a charge

Q1

Q2

C

A B X

D Y

s (A) in the region between the plates. ε0 s (B) in the region outside the plates. ε0

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2.94  JEE Advanced Physics: Electrostatics and Current Electricity

2s (C) in the region between the plates. ε0

(D) zero in the region outside the plates.



16. In the figure shown, the charges on capacitors C1 , C2 and C3 are Q1 , Q2 and Q3 respectively. Then

19. In a parallel plate capacitor of plate area A , plate separation d and charge q , the force of attraction between the plates is F

C2 = 2 μ F

C1 = 4 μ F

C3 = 3 μ F

F ∝ q2 (B) (A) F∝d F∝ (C)

9V

Q1 = 20 µC (B) Q2 = 12 µC (A) Q2 = 8 µC (D) Q3 = 12 µC (C) 17. In the diagram shown, we have three large, identical, parallel conducting plates A , B and C placed such that the switches S1 and S2 are open initially and they can be used to earth the plates A and C by just closing them. A charge +Q is given to the plate B . It is observed that a charge of amount A

B

C

S2

S1







(A)  Q will pass through S1 , when S1 is closed and S2 is open. (B)  Q will pass through S2 , when S2 is closed and S1 is open.

1 1 (D) F∝ A d

20. An air filled parallel plate capacitor has capacitance C0 . Now the air column in the capacitor is completely replaced by a dielectric slab having a dielectric constant K and the capacitor is connected to a battery of emf E and then the slab is taken out. In the process of taking out the dielectric, it is observed that

(A)  the energy stored in the capacitor reduces by C0 E 2 ( K - 1 ) 1 C0E2 ( K - 1 ) has to be done by the exter2 nal agent to take the slab out



d

2d

(A) no charge will flow between C and D (B) some charge will flow between C and D (C)  equivalent capacitance between C and D will not change (D)  the equivalent capacitance between C and D will change

(B) a work

(C) an energy C0E2 ( K - 1 ) is absorbed by the cell



(D) a charge C0E ( K - 1 ) flows through the cell

21. In the circuit shown the potential difference between the points ab and ef are Vab and Vef respectively. Then 23 V 15 μ F

Q 2Q (C)  will pass through S1 , will pass through 3 3 S2 , when S1 and S2 are closed together.

A 0.75 μ F

Q 4Q (D)  will pass through S1 , - will pass through 3 3 S2 , when S1 and S2 are closed together.

5 μF

18. In the circuit shown, some potential difference is applied between A and B. If C is joined to D 4 μF

C

8 μF

A

B

8 μF

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 94

D

16 μ F

a

5 μF

e

B

b

15 μ F

15 μ F D 0.75 μ F

f

5 μF

Vab = 5 V (B) (A) Vef = -5 V Vab = 0 (D) Vef = 0 (C) 22. Two large, parallel conducting plates X and Y each having an area A are placed close to each other. The plate X is given a charge Q whereas the plate Y has no charge on it. Three points A , B and C are taken in   the different regions of the plates as shown. If EA , EB  and EC be the respective fields at these points, then

9/20/2019 11:00:36 AM

Chapter 2: Capacitance and Applications 2.95

Q1

Q3

C

B

A

Q2 X

Y

  Q 1⎛ Q ⎞ (A) EB = (B) EB = ⎜ ε0 A 2 ⎝ ε 0 A ⎟⎠ (C)

   EA = EB = EC

  (D) EA = - EC

23. In an attempt to double the separation between the plates of a parallel plate capacitor while it still being connected to a battery, we observe that (A) the electric field in the region between the plates becomes half (B) some energy is absorbed by the battery (C) the external agent has to do some work on the plates (D) the charge on the capacitor becomes half 24. Consider a situation where a charge of amount Q is given to an isolated metal plate X of surface area A such that its surface charge density becomes s 1 . Now an isolated identical plate Y is brought close to X such that the surface charge density on X becomes s 2 . When the plate Y is earthed, the new surface charge density becomes s 3 , then which of the following relations seem to be the most appropriate and correct?

s1 = (A)

Q (B) s 2 = s1 A

(C) s1 =

Q 1⎛ Q⎞ s3 = ⎜⎝ ⎟⎠ (D) A 2 A

Q4

(A) Q2 = C ΔV (B) Q1 + Q2 = C ΔV (C) ( Q1 + Q4 ) + ( Q2 - Q3 ) = C ΔV (D) Q3 = C ΔV 27. In the arrangement of the capacitors shown, select the correct options +90 V C1 = 20 μ F C2 = 30 μ F C3 = 15 μ F



(A) Total charge in this series combination is 600 µC



(B) The potential difference between the plates of C1 is 30 V (C) The potential difference between the plates of C2 is 20 V (D) The potential difference between the plates of C3 is 40 V



28. Consider a conducting sphere A having radius a , charge Q placed concentrically inside a conducting shell B having radius b ( > a ) and earthed. Let C is the common centre of the sphere A and the shell B. If E and V be the respective electrostatic field and potential at a distance r ( a ≤ r ≤ b ) from C and ΔV be the electrostatic potential difference between the two shells, then B A

25. Consider a capacitor C that is charged by a battery of emf V to a potential V. The capacitor is then disconnected from the battery and then connected again with it but now with its polarity reversed. Then the

Q C

a

b



(A) work done by the battery is CV 2 (B) total charge that passes through battery is 2CV (C) initial and final energy of the capacitor is same

1 ⎛Q⎞ (A) E = 0 (B) E= ⎜ ⎟ 4πε 0 ⎝ r 2 ⎠



(D) work done is by the battery is 2CV 2

(C) ΔV =

26. Consider an isolated parallel plate capacitor having capacitance C. The four surfaces of the capacitor have charges Q1 , Q2 , Q3 and Q4 , as shown. If ΔV is the potential difference between the plates then

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 95

Q ⎛ 1 1⎞ Q ⎛ 1 1⎞ V= ⎜ - ⎟ (D) ⎜ - ⎟ 4πε 0 ⎝ a b ⎠ 4πε 0 ⎝ r b ⎠

29. Consider two large parallel conducting plates placed close to each other. It is observed that the inner surfaces of the two plates have surface charge densities +s and -s whereas the outer surfaces possess no

9/20/2019 11:00:47 AM

2.96  JEE Advanced Physics: Electrostatics and Current Electricity



 charge. If E be the electric field between the plates of the capacitor then in the region  s (A) between the plates, E = ε0  2s (B) between the plates, E = ε0  (C) outside the plates, E = 0  s (D) outside the plates, E = ε0



34. Identical dielectric slabs are inserted into two identical capacitors A and B . These capacitors and are connected as shown in figure. Now the slab of capacitor B is pulled out with battery remaining connected. Select the correct statement(s).

30. Two identical sheets of a metallic foil are separated by d and capacitance of the system is C0 and charged to a potential difference V0 . Keeping the charge constant, the separation is increased by l . Then the new capacitance and potential difference are C and V respectively. Then l⎞ ⎛ (B) C= C = C0 ⎜ 1 + ⎟ (A) ⎝ l⎞ d⎠ ⎛ ⎜⎝ 1 + ⎟⎠ d C0

⎛ (C) V = V0 ⎜ 1 + ⎝



(A) the charge on each capacitor is 4.8 × 10



(B) the potential difference across C1 is 60 V



(C) the potential difference across C2 is 240 V



(D) the energy stored in the system is 7.2 × 10 -2 J

C

32. A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q0 , V0 , E0 and U 0 respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q , V , E and U are related to the previous ones as Q > Q0 (B) V > V0 (A) E > E0 (D) U > U0 (C) 33. In the given circuit, the potential difference across the 7 µF capacitor is 6 V. Then

3.9 μ F 7 μF

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 96

3 μF

a





31. Two capacitors, C1 = 2 µF and C2 = 8 µF are connected in series across a 300 V source. Then -4

A



l⎞ V0 V= ⎟⎠ (D) l⎞ d ⎛ ⎜⎝ 1 + ⎟⎠ d

12 μ F

(A) the potential difference across the 12 µF capacitor is 10 V (B) the potential difference across the 3.9 µF capacitor is 10 V (C) the charge on the 3 µF capacitor is 42 µC (D) the emf of the battery is 30 V

B

b

(A) During this process the charge flows from a to b . (B) Final charge on capacitor B will be less than on capacitor A. (C) During this process, work is done by external force F which appears as heat in the circuit. (D) During this process, the battery receives energy.

35. Three capacitors each having capacitance C = 2 µF are connected with a battery of emf 30 V as shown in figure. Select the correct statement(s), when the switch S is closed. S

C C C 30 V



(A) The amount of charge flown through the battery is 20 µC (B) The heat generated in the circuit is 0.6 mJ (C) The energy supplied by the battery is 0.6 mJ (D) The amount of charge flown through the switch S is 60 µC

36. A parallel plate capacitor has a dielectric slab in it. The slab just fills the space inside the capacitor. The capacitor is charged by a battery and then battery is disconnected. Now the slab is started to pull out slowly at t = 0 . If at time t, the capacitance of the capacitor is C and potential difference between the plates of a capacitor is V then which of the following graphs is/are correct

9/20/2019 11:00:56 AM

Chapter 2: Capacitance and Applications 2.97 C (B)

C (A)



t

(C) V

t

(D) V

C

37. A parallel plate air capacitor is connected to a battery. If plates of the capacitor are pulled further apart, then which of the following statement(s) is/are correct? (A)  Strength of electric field inside the capacitor remain unchanged, if battery is disconnected before pulling the plate. (B) During the process, work is done by an external force applied to pull the plates whether battery is disconnected or it remain connected. (C) Potential energy in the capacitor decreases if the battery remains connected during pulling plates apart. (D) None of the above 38. A 2 µF capacitor is charged to a potential of 15 V and a 3 µF is charged to a potential of 10 V and the capacitors are connected such that positive plate of one is connected to the negative plate of the other capacitor and negative plate of one is connected to the positive plate of the other capacitor. Select the correct statement(s) about the final circuit. (A) Final charge on each capacitor is zero (B) Final total electrical energy of the capacitor will be non zero (C) Total charge flown from A to D is 30 µC (D) Total charge flown from A to D is -30 µC 39. In the circuit shown in the figure, switch S is closed at time t = 0 . Select the correct statements 2R

2C

R

E



S

(A) Rate of increase of charge is same in both the capacitors

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 97

(D) Steady state charge in capacitors C and 2C are in the ratio of 1 : 2

40. In the circuit shown in figure, C1 = C2 = 2 µF . In steady state, the charge stored in

C

C

(B) Ratio of charge stored in capacitors C and 2C at any time t would be 1 : 2 (C) Time constants of both the capacitors are equal

1Ω

2Ω

C1

C2

2Ω

1Ω

3Ω

3Ω

120 V



(A) capacitor C1 is zero



(B) capacitor C2 is zero



(C) both capacitors is zero



(D) capacitor C1 is 40 µC

41. The two plates x and y of a parallel plate capacitor of capacitance C are given a charge of amount Q each. x is now joined to the positive terminal and y to the Q negative terminal of a cell of emf V = , then select C the correct statements.



(A) A charge of amount Q will flow from the positive terminal to the negative terminal of the cell through the capacitor (B) The total charge on the plate x will be 2Q (C) The total charge on the plate y will be zero



(D) The energy supplied by the cell is CV 2

42. The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now a dielectric of dielectric constant K is inserted to fill the whole space between the plates with voltage source remaining connected to the capacitor. Select the correct statement(s). (A) The energy stored in the capacitor will become K-times the initial energy. (B) The electric field inside the capacitor will decrease K-times the initial field. (C) The force of attraction between the plates will become K 2 -times the initial force. (D) The charge on the capacitor will become K-times the initial charge.

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2.98  JEE Advanced Physics: Electrostatics and Current Electricity

Reasoning Based Questions This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A)  If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B)  If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C)  If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. Bubble (D)  If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE. 1.

Statement-1: A spherical conductor charged upto 50 V is placed at the centre of a conducting shell which is charged upto 100 V and connected by a wire. All the charge of the shell flows to the sphere.

Statement-2: The positive charge always flows from higher to lower potential. 2.

Statement-1: A single isolated conductor is always equivalent to a capacitor.

Statement-2: The second plate of an isolated conductor can be assumed at infinity. 3.

Statement-1: The two adjacent conductors carrying same charge can be at different potentials.

Statement-2: The potential of a conductor depends on the charge on it, shape and size of its the surrounding charged bodies and the relative separation between them. 4.

Statement-1: A parallel plate capacitor is charged using a battery and then a dielectric slab is inserted completely filling space between plates without disconnecting battery. Electric field between plates of capacitor will decrease.

Statement-2: When the battery remains connected, then charge on plates of capacitor increases. 5.

Statement-1: If two concentric conducting sphere which are connected by a conducting wire. No charge can exist on inner sphere.

Statement-2: When charge on outer sphere will exist then potential of inner shell and outer shell will be same 6.

Statement-1: Two concentric spherical shell of different radius are at potential VA and VB . If outer shell is earthed then potential difference will not be changed.

Statement-2: Potential difference between the surfaces of two concentric spherical shells does not depends on the charge on the outer shell. 7.

Statement-1: If three capacitors of capacitance C1 < C2 < C3 are connected in parallel then their equivalent capacitance Cparallel > Cseries

Statement-2:

1 1 1 1 = + + Cparallel C1 C2 C3

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 98

8.

Statement-1: A charged capacitor is disconnected from a battery. Now if its plates are separated further, the potential energy will fall.

Statement-2: Energy stored in a capacitor is equal to the work done in charging it. 9.

Statement-1: There cannot be a potential difference between two adjacent conductors that carry the same amount of positive charge.

Statement-2: Potential of a conductor can be found by ⎛ Q⎞ ⎜⎝ ⎟⎠ ratio and capacitance C depends on the geoC metrical parameters like size, area etc. 10. Statement-1: When a capacitor is charged by a battery, both the plates receive charges of equal magnitude, irrespective of their sizes. Statement-2: The charge distribution on the plates of capacitor is in accordance with charge conservation principle. 11. Statement-1: The capacitance of any capacitor is always constant for any charge. Statement-2: If the charge on a capacitor increases, its Q capacitance increases as C = . V 12. Statement-1: A parallel plate capacitor is connected across battery through a key. A dielectric slab of constant K is introduced between the plates. The energy which is stored becomes K times.  Statement-2: The surface density of charge on the plate remains constant. 13. Statement-1: When one plate of a charge parallel plate capacitor is connected to the earth, its capacitance increases.  Statement-2: Electric potential difference between the plates always decreases. 14. Statement-1: If the distance between parallel plates of a capacitor is halved and dielectric constant is made three times, then the capacitance becomes 6 times.  Statement-2: Capacitance does not depend upon the nature of material of the capacitor plates.

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Chapter 2: Capacitance and Applications 2.99

Linked Comprehension Type Questions This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options)

Comprehension 1

6 μF

A parallel plate capacitor having capacitance C0 consists of two metal plates of area A and separation d. A slab of thickness t ( < d ) is inserted between the plates with its faces parallel to the plates and having the same surface area as that of plates as shown.

3 μF

600 μ C S2 S1

150 μ C S3

200 V

d

Based on the above facts, answer the following questions. 4.

It is observed that an isolated system is formed by plate(s) (A) a and d separately (B) b and c separately

t

Based on the above facts, answer the following questions. 1.

The capacitance of the system for the slab to be dielect tric having dielectric constant 2 and d = is 3

3 3 C0 (A) C0 (B) 2 4 5 6 C0 (C) C0 (D) 6 5 2.

The ratio

t for which the capacitance of system will d

⎛ 3⎞ be ⎜ ⎟ times that of the conductor with air filling the ⎝ 2⎠ full space is 2 3 (A) (B) 3 2 (D)

1 3



(C) 1

3.

The ratio of energy in two cases mentioned in PROBLEM 2, is

2 3 (A) (B) 3 2 1 1 (C) (D) 3 2

Comprehension 2 Two capacitors of capacity 6 μF and 3 μF are charged to 100 V and 50 V separately and are then connected as shown. Now all the three switches S1 , S2 and S3 are closed.

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 99

(C) a and d jointly 5.

(D) b and c jointly

In steady state, the charge on 6 μF and 3 μF capacitor respectively is

400 μC, 400 μC (B) 700 μC, 250 μC (A) (C) 800 μC, 350 μC (D) 300 μC, 450 μC 6.

Suppose Q1 , Q2 and Q3 be the magnitudes of charges that flows from switches S1 , S2 and S3 after they are closed. Then

(A) Q1 = Q3 and Q2 = 0 (B) Q1 = Q3 =

Q2 2

(C) Q1 = Q3 = 3Q2 (D) Q1 = Q2 = Q3

Comprehension 3 In the given set up the parallel plate capacitor AB has vertical plates with plate separation 50 mm and capacitance C0 . From the plate A a small conducting ball of mass m , capacitance C hangs from a non-conducting silk thread of length 100 mm. The ball initially touches the plate A which is connected to a power supply of voltage V0 for a short time by closing the switch S and then opening it again. The plate B is grounded. The motion of conducting ball is now observed. It is observed that due to the charge deposited on plates and ball, the ball swings across, touches the plate B , returns, touches A and finally swings out again such that it almost touches plate B . Taking g = 10 ms −1 and based on the above facts, answer the following questions.

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2.100  JEE Advanced Physics: Electrostatics and Current Electricity A

B

1 : 1 (B) 1: K (A) (C) K : 1 (D) 1: K

S V0

11. After introduction of dielectric slab in B , the ratio of potential differences across A and B will be

C1

(A) 1 : 1 (B) 1: K (C) K : 1 (D) 1: K d

7.

The angle made by ball and string with the plate A when the ball is in its final position is

12. The ratio of potential differences across A before and after the introduction of dielectric slab in B will be (A) 1 : 1 (B) 1: K

30° (B) 45° (A)

( K + 1 ) : 2 (D) ( K + 1 ) : 2K (C)

(C) 60° (D) 0°

13. The ratio of potential differences across B before and after the introduction of dielectric slab in B will be

8.

The ball initially touches plate B, it loses its charge to B, swings back and touches A and then finally again swings out and almost touches plate B. The new common potential V is given by

CV C0V0 (A) 0 (B) C + C0 C + C0 C ⎞ ⎛ V0 (D) (C) ⎜⎝ 1 + C ⎟⎠ V0 0 9.

The required power supply voltage V0 is given by

C ⎞ ⎛ (A) ⎜⎝ 1 + C ⎟⎠ 0

m C0 ⎞ m ⎛ (B) ⎜⎝ 1 + ⎟ C ⎠ 2 3C 3C

C ⎞ m C0 ⎞ ⎛ ⎛ (C) ⎜⎝ 1 + ⎟ ⎜⎝ 1 + C ⎟⎠ 2 3C (D) C ⎠ 0

5m 3C

(A) 1 : 1 (B) K :1

( K + 1 ) : 2 (D) ( K + 1 ) : 2K (C) 14. The ratio of energy stored in capacitors A and B after the introduction of dielectric slab in B is (A) 1 : 1 (B) 1: K 2

( K + 1) : K2 K : 1 (D) (C)

Comprehension 5 Two spherical conductors A and B have capacitance 2 μF and 3 μF . A is given a charge Q and allowed to share with B by means of a conducting wire. Based on these facts, answer the following questions. 15. The ratio of radii of spheres A and B is (A) 2 : 3 (B) 3:2

Comprehension 4 Two identical capacitors A and B , each of capacitance C are connected in series. The combination is connected to a battery of emf E . A dielectric slab of dielectric constant K is slipped between the plates of capacitor B to cover entire space between the plates.

16. The charge on A after sharing is Q 4Q (A) (B) 5 5 2Q 3Q (C) (D) 5 5

A E K B

Based on these facts, answer the following questions. 10. After introduction of dielectric slab in B , the ratio of capacitance of A and B is

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 100

(C) 4 : 9 (D) 9:4

17. The ratio of electrostatic energies of A before and after sharing is 2 3 (A) (B) 5 5 4 9 (C) (D) 25 25

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Chapter 2: Capacitance and Applications 2.101 18. The ratio of final electrostatic energy of system to initial electrostatic energy of system is 1 : 1 (B) 2:5 (A) (C) 3 : 5 (D) 2:3

Comprehension 6 Two metal plates form a parallel plate capacitor. The separation between the plates is d . Now half of the separation between the plates is filled with a dielectric of relative permittivity 2 and of same area.

air

K=2

d/2

d/2

The dielectric slab is placed inside A as shown in Figure (a). A is then charged to a potential difference of 110 V. The capacitance of A is C A and the energy stored in it is U A . The battery is disconnected and then the dielectric slab is removed from A . The work done by the external agency in removing the slab from A is W . The same dielectric slab is now placed inside B , filling it completely. The two capacitors A and B are then connected as shown in Figure (c). The energy stored in the system is U . Based on the above facts, answer the following questions. 22. The value of C A is (A) 0.18 nF (C) 3 nF



(B) 2 nF (D) 1 nF

23. The value of U A approximately is 10 μ J (B) 11 μ J (A)

The new arrangement gets punctured when a potential difference just exceeding 1200 V is applied. Based on these facts, answer the following questions.

(C) 12 μ J (D) 13 μ J

19. The ratio of capacitance before and after insertion of dielectric is

(A) 48.4 μ J (B) 38.4 μ J

1 : 2 (B) 2:1 (A) (C) 3 : 4 (D) 4:3 20. The maximum potential difference that can exist across the dielectric is (A) 300 V (B) 400 V (C) 600 V (D) 800 V 21. The ratio of electrostatic energy stored by air and dielectric parts of capacitor is 1 : 2 (B) 2:1 (A) (C) 3 : 4 (D) 4:3

24. The value of W is (C) 28.4 μ J (D) 18.4 μ J 25. The value of U is 10 μ J (B) 11 μ J (A) (C) 12 μ J (D) 13 μ J

Comprehension 8 Three capacitors are connected to a battery of voltage V as shown in Figure. Their capacitances are C1 = 3C , C2 = C and C3 = 5C . Based on the information and the diagram provided, answer the following questions. C1

Comprehension 7 Two parallel plate capacitors A and B have the same separation d = 8.85 × 10 −4 m between the plates. The plate areas of A and B are 0.04 m 2 and 0.02 m 2 respectively. A slab of dielectric constant (relative permittivity) K = 9 has dimensions such that it can exactly fill the space between the plates of capacitor B . A

B

A

B

C2

26. The equivalent capacitance of this set of capacitors is 23C 4C (B) (A) 6 2C (C)

110 V (a)

(b)

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 101

(c)

C3

(D) None of these

27. The ordering of the capacitors according to the charge they store, from largest to smallest is

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2.102  JEE Advanced Physics: Electrostatics and Current Electricity C1 > C2 > C3 (B) C3 > C2 > C1 (A)

Comprehension 10

C3 > C1 > C2 (D) C2 > C1 = C3 (C)

Given here are four SITUATIONS. Read them carefully to answer the questions that follow.

28. Rank the capacitors according to the potential difference across them, from largest to smallest. C1 > C2 > C3 (B) C1 < C2 = C3 (A) C1 > C2 = C3 (D) C1 < C2 < C3 (C) 29. Now if, C3 is increased then the charge on (A) C1 and C2 increase, C3 decreases (B) C1 and C3 increase, C2 decreases (C) C1 and C2 decrease, C3 increases (D) C1 and C3 decrease, C2 increases

Comprehension 9 Four parallel metal plates P1 , P2 , P3 and P4 , each of area 7.5 cm 2 are separated successively by a distance d = 1.19 mm as shown in Figure. P1 is connected to the negative terminal of a battery and P2 to the positive terminal. The battery maintains a potential difference of 12 V. Based on the information and the diagram provided, answer the following questions. P2 P3 P4

P1

Situation 1: A 1 μF capacitor and a 2 μF capacitor are connected in series across a 300 V supply line. The charge on each capacitor is found to be Q1 and Q2 and potential across each is V1 and V2 . Situation 2: The charged capacitors are now disconnected from the line and reconnected with their positive plates together and the negative plates together, with no external voltage being applied. The respective charges and potentials across capacitors are found to be Q1′ , Q2′ and V1′, V2′ . Situation 3: Again the system is reset to the initial conditions and the charged capacitors are disconnected from the line and reconnected with plates of opposite polarity connected together, with no external voltage being applied. The respective charges and potentials across capacitors are found to be Q1′′ , Q2′′ and V1′′ and V2′′ . Situation 4: The charged capacitors in situation 2 are disconnected and reconnected with the plates of opposite polarity together. The respective charges and potentials across the capacitors are found to be Q1′′′, Q2′′′ and V1′′′ , V2′′′ . Based on the Situations discussed above, answer the following questions. 34. The ( Q , V ) values in μC and volt respectively for

12 V

SITUATION 1 are (A) Q1 = Q2 = 200 ; V1 = V2 = 100 d

d

d

30. If P3 is connected to the negative terminal, the capacitance of the three-plate system P1P2 P3 is

(A) 12.1 pF (C) 5.58 pF

(B) 11.2 pF (D) 16.7 pF

31. The charge on P2 , in pC is (A) 34 (B) 67 (C) 134 (D) None of these 32.



P4 is now connected to the positive terminal of the battery. The capacitance of the four plate system P1P2 P3 P4 is (A) 12.1 pF (B) 11.2 pF (C) 5.58 pF (D) 16.7 pF

33. The charge on P4 , in pC is (A) 34 (B) 67 (C) 134 (D) 268

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 102

Q1 = Q2 = 100 ; V1 = 100 ; V2 = 200 (B) Q1 = 100 ; Q2 = 200 ; V1 = V2 = 100 (C) Q1 = Q2 = 200 ; V1 = 200 ; V2 = 100 (D) 35. The ( Q , V ) values in μC and volt respectively for SITUATION 2 are (A) V1′ = V2′ = 200 ; Q1 = Q2 = 200 (B) V1′ = V2′ =

200 400 ; Q1′ = Q2′ = 3 3

(C) V1′ = V2′ =

400 400 800 ; Q1′ = ; Q2′ = 3 3 3

(D) V1′ =

200 400 400 800 ; V2′ = ; Q1′ = . Q′2 = 3 3 3 3

36. The ( Q , V ) values in μC and volt respectively for SITUATION 3 are

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Chapter 2: Capacitance and Applications 2.103 41. Energy lost by the system of capacitors, after the switches are closed is (A) 20 mJ (B) 5 mJ (C) 15 mJ (D) 10 mJ

V1′′= V2′′= 0 ; Q1′′= Q2′′ = 0 (A) (B) V1′′≠ V2′′= 0 ; Q1′′= Q2′′ ≠ 0 V1′′= V2′′≠ 0 ; Q1′′≠ Q2′′ = 0 (C) (D) ′′= ( V1 V2′′) ≠ 0 ; ( Q1′′= Q2′′) ≠ 0

Comprehension 12

37. The ( Q , V ) values in μC and volt respectively for

2k A B

k

SITUATION 4 are (A) V1′′= V2′′=

200 800 ; Q1′′′= Q2′′′ = 9 9

(B) V1′′′= V2′′′=

400 400 800 ; Q1′′′= ; Q2′′′ = 9 9 9

(C) V1′′′=

400 800 400 ; V2′′′= ; Q1′′′= Q2′′′ = 9 9 9

(D) V1′′′=

200 400 200 400 ; V2′′′= ; Q1′′′= ; Q2′′′ = 9 9 9 9

Comprehension 11 Two capacitors C1 = 1 μF and C2 = 3 μF are each charged to a potential V0 = 100 V but with opposite polarity, as shown. Switches S1 and S2 are now closed. Based on the above facts, answer the following questions. e a

S1

d

S2

c

38. What is the potential difference between points ’e ’ and ’ f ’ ?

(A) 25 V (C) 75 V

(B) 50 V (D) 100 V

39. What is the final charge on C1 ? 50 μC (B) 75 μC (A) (C) 150 μC (D) 300 μC 40. What is the charge on C2 ? 50 μC (B) 75 μC (A) (C) 150 μC (D) 300 μC

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 103

In the given circuit, two identical parallel conducting plates A and B are connected to a 50 V battery by metal springs of spring constants 2k and k respectively. Initially, the switch S is open and the plates are uncharged. In fact, the two plates A and B form a capacitor. When the switch S is closed, the separation between the plates becomes 2 mm and this is one fourth of the initial separation between plates. Also the electrostatic energy stored in the capacitor is found to be 10 mJ . Based on the information provided, answer the following questions. 42. Initially, when the plates were uncharged capacitance of the capacitor was (C) 2 μF (D) 4 μF

C2

f

S

(A) 0.5 μF (B) 1 μF

C1 b

50 V

43. Extension in the spring connected to plate B is (A) 4 mm 3 mm (B) (C) 5 mm (D) 6 mm 44. Spring constant of the spring connected to plate A is 625 Nm −1 (B) 1250 Nm −1 (A) (C) 2500 Nm −1 (D) 5000 Nm −1

Comprehension 13 Inside a parallel plate capacitor, there is a dielectric slab parallel to the outer plates whose thickness is equal to 70% of the gap width. The dielectric constant of material of the slab is 7. When the slab is absent, the capacitance of the capacitor equals 10 μF. The capacitor was connected to a 10 V battery. Now the capacitor is disconnected from the battery and the slab is removed from the gap. After removal of the slab, the positively charged plate of the capacitor is reconnected with negative terminal of the same battery and

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2.104  JEE Advanced Physics: Electrostatics and Current Electricity negatively charged plate with positive terminal of the battery. Based on the information provided, answer the following questions.

50. At a given moment plates are unfixed. What is relative velocity of plates when their distance is reduced to d (A) 9Q

45. Energy stored in capacitor before reconnection is 500 μ J (B) 1250 μ J (A) (C) 3215 μ J (D) 4250 μ J

3 Q 1 (B) mC 2 mC

2Q 1 9Q 1 (C) (D) 3 3 mC 2 3 mC

46. Energy stored in the capacitor after reconnection is

Comprehension 15

(A) 500 μ J (B) 1250 μ J

A fuel gauge uses a capacitor to determine the height of the fuel in the tank. The effective dielectric constant K eff changes from a value of 1 (when the tank is empty) to a value of K , where K is dielectric constant of the fuel (when the tank is full). The appropriate electronic circuitry can determine the effective dielectric constant of the combined air and fuel between the capacitor plates.

(C) 3215 μ J (D) 4250 μ J 47. Heat generated in connecting wires after reconnection with the battery is 500 μ J (B) 1250 μ J (A) (C) 3215 μ J (D) 4250 μ J

battery

Comprehension 14 A parallel plate capacitor has two plates of area A each. The mass of the left plate is m and its electric charge is Q. The mass of the right plate is 2m and its electric charge is −2Q . Initially the plates are fixed and the separation between the plates is 3d . If C is the capacitance of the capacitor when plates would have been lying at the separation d , then based on the information provided, answer the following questions. Q

A

–2Q

A

2m

m 3d

L

W h

Each of the two rectangular plates has a width W and a length L . The height of the fuel between the plates is h . (Ignore fringing effect). Based on the information provided, answer the following questions. 51. The electronic gadget which senses the fuel height is called (A) level tester (B) strain gauge (C) transducer (D) capacitor

48. Potential difference across plates of capacitor is

52. The value of K eff as a function of h is

3Q 9Q (A) (B) 2C 2C

( K − 1) h ( K − 1) h (A) + 1 (B) L L

6Q 3Q (C) (D C C

( K − 1) h ( K + 1) h (C) − 1 (D) − 1 L L

49. Energy possessed by the electric field between the plates is

53. When the tank is

27 Q 2 9 Q2 (B) (A) 4C 8C

3K K −1 (A) (B) 4 4

27 Q 2 81Q 2 (D) (C) 8C 8C

( K − 1) (C)

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 104

1 th empty, then the value of K eff is 4

3 K +1 (D) 4 4

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Chapter 2: Capacitance and Applications 2.105

Comprehension 16

54. Applied potential difference V0 is

Consider a series combination of capacitor C1, C2 and C3. Capacitor C1 and C3 have air as the dielectric medium between the plates while C2 has a medium of dielectric constant K = 3 filled uniformly between the plates. A potential difference V0 is applied across the combination. It is observed that potential difference across C3 is 12 V . The dielectric material ( K = 3 ) between the plates of C2 is now removed so that its capacity now becomes C2′ . In this situation, the potential difference across C3 is found to be 9 V . It is also observed that the potential differences across C1 and C2′ are equal. If the capacitance of C2 is 6 μF , then based on the information provided, answer the following questions.

(A) 10 V (B) 20 V (C) 30 V (D) 36 V 55. Capacitance C1 is (A) 8 μF (B) 6 μF (C) 3 μF (D) 2 μF 56. Capacitance C3 is (A) 8 μF (B) 6 μF (C) 3 μF (D) 2 μF

Matrix Match/Column Match Type Questions Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of bubbles will look like the following:

1.

p

q

r

s

t

A

p

q

r

s

t

B

p

q

r

s

t

C

p

q

r

s

t

D

p

q

r

s

t

Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. COLUMN-I

COLUMN-II

(A) Energy stored in a capacitor carrying charge Q . (potential difference = V between the plates)

(p)  CV 2

COLUMN-I (C) Energy loss when plates of a charged capacitor are connected together.

COLUMN-II (r) 

C1C2 ( V1 − V2 )2 2 ( C1 + C2 )

(D) Energy loss 1 (s)  CV 2 when charges 2 of two parallel plates capacitors of capacitances C1 and C2 at potential differences V1 and V2 respectively are shared.

(B) Energy supplied Q2 to a capacitor by a (q)  2C battery of potential difference V .

(Continued)

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 105

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2.106  JEE Advanced Physics: Electrostatics and Current Electricity 2.

In the circuit shown C1 = C , C2 = 2C , C3 = 3C , C4 = 4C . Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. C3 C2

C1

C4

V

COLUMN-I

COLUMN-II

(A) Charge across the capacitor C3

(p) 

2 23

(q) 

6 23

4.

    (multiple of CV ) (B) Potential difference across the capacitor C4     (multiple of V )

(C) Energy stored across the 1 (r)  capacitor C3 66     (multiple of CV 2 ) (D) Energy stored across the 1 (s)  capacitor C4 88

COLUMN-I

COLUMN-II

(B) Electric field strength between the plates will reduce if

(q) Separation between the plates is increased

(C) Electric energy stored in the capacitor will decrease if

(r) A dielectric with K > 1 is filled between the plates of capacitor

(D) Electric energy density will decrease if

(s) Separation between the plates is reduced

Three capacitors of capacitances 2 µF , 3 µF and 6 µF are connected in series with a 12 V battery. All the connecting wires are disconnected. The three positive plates are connected together and three negative plates are connected together. The charges on three capacitors after reconnection are Q1 , Q2 and Q3 respectively. The charge supplied by battery is Q . Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. COLUMN-I

COLUMN-II

(A)  Q1

(p)  12 µC

(B)  Q2

(q) 

216 µC 11

(C)  Q3

(r) 

72 µC 11

(D)  Q

(s) 

108 µC 11

2

    (multiple of CV ) 3.

A parallel plate capacitor with air between its plates is charged using a battery and then disconnected from the battery. Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. COLUMN-I

COLUMN-II

(A) Potential difference between the plates will decrease if

(p) Separation between the plates is K increased to 2 times the initial value and space between the plates after the separation has increased, is completely filled with a dielectric (here K is the dielectric constant)

(Continued)

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 106

5.

Capacitors with capacitances C , 2C , 3C and 4C are charged to the voltage V , 2V , 3V , 4V respectively. Circuit is closed. Assume voltages across capacitors in equilibrium are V1 , V2 , V3 and V4 respectively. Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. C, V – + q1 4 C, 4V

+ –

– +

2 C, 2V

+ – 3 C, 3 V

9/20/2019 11:02:33 AM

Chapter 2: Capacitance and Applications 2.107

6.

COLUMN-I

COLUMN-II

COLUMN-I

COLUMN-II

(A)  V1

(p) 

2V 5

(A) Charge on plate 1 ε 0 AV ⎞ ⎛ ⎜⎝ multiple of ⎟ d ⎠

(p)  -2

(B)  V2

(q) 

7V 5

(B) Charge on plate 4

(q) 1

ε 0 AV ⎞ ⎛ ⎜⎝ multiple of ⎟ d ⎠

19V 5

(C)  V3

(r) 

(D)  V4

14V (s)  5

(C) Potential difference between the plates 2 and 3 (multiple of 2 V) (D) Potential difference between the plates 1 and 5 (multiple of 2 V)

In the diagram shown, q1 , q2 and q3 are the charges on the capacitors C1 , C2 and C3 . The potential difference across C3 is V , where q1 , q2 and q3 are in µC and V is in volt. Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. 8.

C1 = 2 μ F C2 = 4 μ F

C3 = 6 μ F

10 V

7.

20 V

COLUMN-I

COLUMN-II

(A)  q1

(p) 50

(B)  q2

(q) 

10 3

(C)  q3

(r) 

140 3

(D)  V

(s) 

25 3

2

4

A

B

C

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 107

D

1 2

In case of two conducting spherical shells having radii a and b ( > a ) . Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. COLUMN-I

COLUMN-II

(A) Shells are concentric and inner is given a charge while outer is earthed

(p)  C = 4πε 0 ( a + b )

(C) Shells carry equal and opposite charges and are separated by a distance d (D) Shells are connected by a conducting wire 9.

5

3

(s) 

(B) Shells are 4πε 0 ab concentric and the (q)  C = b - a outer is given a charge while inner is earthed

Five identical capacitor plates each of area A are arranged such that adjacent plates are at d distance apart. Plates are connected to a source of emf V as shown. Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II. 1

(r) 0

V

(r)  C =

(s)  C =

4πε 0b 2 b-a

4πε 0 1 1 2 + a b d

A parallel plate capacitor with air between its plates is charged using a battery and remains connected to the battery. Match the quantities in COLUMN-I with their respective answer(s) in COLUMN-II.

9/20/2019 11:02:39 AM

2.108  JEE Advanced Physics: Electrostatics and Current Electricity

10.

COLUMN-I

COLUMN-II

(A) Electric field strength between the plates will reduce if

(p) Separation between the plates is increased

(B) Electric energy stored in the capacitor will increase if

(q) Separation between the plates is reduced

(C) Electric energy density will increase if

(r) A dielectric with K > 1 is filled between the plates

(D) Charge in the capacitor will decrease if

(s) Separation between the plates is K increased to 2 times the initial value and space between the plates, after the separation has increased, is completely filled with a dielectric (here K is the dielectric constant and K = 3 )

COLUMN-I

COLUMN-II

(A) If distance between plates of an isolated capacitor decreases

(p) Potential difference across plate is decreased

(B) If dielectric is inserted between plates of capacitor whose plates are connected with battery

(q) Capacitance of the capacitor will increase

(C) If area of plates of an isolated capacitor is increased

(r) Energy of capacitor will increase

(D) If distance between plates of capacitor is decreased when the capacitor is connected with battery

(s) Force between the plates will decrease

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 108

11. A capacitor of capacitance C is charged to a potential V. Now it is connected to a battery of e.m.f E as shown in the figure. Match the conditions in COLUMN-I with the arguments in COLUMN-II. V

E

COLUMN-I

COLUMN-II (p) non-zero charge is supplied by the +ve terminal of battery to +ve plate of capacitor.

(A) If V = E , then

(q) Zero charge is supplied by +ve terminal of battery to +ve plate of capacitor.

(B) If V > E

(r) non-zero thermal energy will be dissipated in the circuit.

(C) If V < E , then

(s) outer surfaces of the plates of capacitor have zero charge.

(D) If V ≠ E , then

12. Match the columns for the capacitance of systems in COLUMN-I to their respective values of capacitance in COLUMN-II. COLUMN-I (A) 

COLUMN-II

b

c

a A

(p) 

4πε 0c 2 c-a

(q) 

4πε 0 ac c-a

B

(Concentric shells) capacitance CAB

(B)  c

a

B

A (Concentric shells) Capacitance CAB

(Continued)

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Chapter 2: Capacitance and Applications 2.109

COLUMN-I

COLUMN-II

(C) 

B R

A

(r)  d

2πε 0 ⎛ d⎞ log e ⎜ ⎟ ⎝ R⎠

(Concentric cylinders) Capacitance CAB per unit length

(D) 

A d

R

(s) 

R B (Parallel wires of radius R at separation d) capacitance CAB per unit length)

πε 0 ⎛ d⎞ log e ⎜ ⎟ ⎝ R⎠

13. Match the Column COLUMN-I

COLUMN-II

(A) 

(p)  Ceq between a and b is 2C

a C

C C C

C b

(B) 

(q)  Ceq between a and b is C

a 2C

2C

C 2C

14. Match the entries of COLUMN-I with the entries of COLUMN-II. COLUMN-I

(A) When a dielectric (p) The electric slab is gradually potential energy inserted between the of the system plates of an isolated decreases parallel plate capacitor (B) When a dielectric gradually inserted between the plates of a parallel plate capacitor connected to battery

(q) Work done by external agent is positive

(C) When the plates of a parallel plate capacitor are pulled apart for a capacitor connected to battery

(r) Work done by battery is positive

(D) When the plates of a parallel plate isolated capacitor are pulled apart

(s) Work done by external agent is negative

15. In the given figure, the separation between the plates of C1 is slowly increased to double of its initial value, then match the following

2C

b

(C) 

(r)  Ceq between a and b is 3C

a C 2C 2C C b

(s)  Ceq between a and b is 4C

a

(D)  C

C

C

C b

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 109

COLUMN-II

C1

C2

2 μF

4 μF V

COLUMN-I

COLUMN-II

(A) The potential difference across C1

(p) increases

(B) The potential difference across C2

(q) decreases

(C) The energy stored in C1

(r) increases by a 6 factor of 5

(D) The energy stored in C2

(s) decreases by a factor of

18 25

9/20/2019 11:02:45 AM

2.110  JEE Advanced Physics: Electrostatics and Current Electricity

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s). 1.

A capacitor has rectangular plates of length a and width b. The top plate is inclined at a small angle as shown in figure. The plate separation varies from d = y0 at the left to d = 2 y0 at the right where y0  a and y0  b . The capacitance of the system is found to ε ab be C = 0 log e a . Then find value of a . y0

b

axis has a diameter of 0.2 mm. The dielectric strength of the gas between the central wire and the cylinder is 1.2 × 106 Vm -1 . Calculate the maximum potential difference, in volt that can be applied between the wire and the cylinder before breakdown occurs in the gas. 6.

A radioactive source in the form of a metal sphere of diameter 10 -3 m emits β particles at a constant rate of 6.25 × 1010 particles per second. If the source is electrically insulated, how long ( in ns ) will it take for its potential to rise by 1 V, assuming that 80% of emitted β particles escape from the surface.

7.

A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced between the plates; in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.

8.

The capacitance of all the capacitors shown in figure are in µF



(a) What is the equivalent capacitance, in µF , between x and y ? (b) If the charge on 5 µF capacitor is 120 µC , what is the potential difference, in volt, between x and z?

2y0

y0

a

2.

3.

4.

A 10 µF capacitor has plates with vacuum between them. Each plate carries a charge of magnitude 1000 µC . A particle with charge -3 µC and mass 2 × 10 -16 kg is fired from the positive plate toward the negative plate with an initial speed of 2 × 106 ms -1. Does it reach the negative plate? If so, find its impact speed in kms -1 . If not, what fraction of the way across the capacitor does it travel? Capacitors C1 = 6 µF and C2 = 2 µF are charged as a parallel combination across a 250 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge, in µC, on each capacitor.

3 C1

a

4 μF

2 μF

8 μF

4

b

A detector of radiation called a Geiger tube consists of a closed, hollow, conducting cylinder with a fine wire along its axis. Suppose that the internal diameter of the cylinder is 2.5 cm and that the wire along the

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 110

4 3 C4

9. 4 μF

C2

5

x

Calculate the equivalent capacitance, in µF, between the points a and b in figure.

2 μF

5.



y

4 z C3

2

A capacitor of capacitance C1 = 1 µF withstands the maximum voltage V1 = 0.6 kV while another capacitor of capacitance C2 = 2 µF withstands the maximum voltage V2 = 4 kV. What maximum kilovolt will the system of these two capacitors withstands if they are connected in series?

10. A parallel capacitor of plate area 0.2 m 2 and spacing 10 -2 m is charged to 10 3 V and is then disconnected from the battery. How much work is required if the

9/20/2019 11:02:53 AM

Chapter 2: Capacitance and Applications 2.111 plates are pulled apart to double the plate spacing? Calculate the final voltage, in kilo volt, of the capacitor 1 if ε 0 = Fm -1. 36π × 107

S 2

11. Find the potential difference in volt, between points M and N of the system shown, if the e.m.f. is equal to C E = 110 V and the capacitance ratio 2 = η = 2. C1 C1

3 C1

1 E

C2

2

1

q

p 15 μ F 3 μF 10 μ F

M

6 μF

4 C2

(a)  Calculate the charge, in µC. (b) A second condenser has a capacitance of 20 µF and is charged to a potential of 300 V . If after charging, the two condensers are connected in parallel by wires of negligible capacitance, how much energy, in mJ, is dissipated?

13. A parallel plate capacitor of plate area A = 10 -3 metre 2 and plate separation d = 10 -2 m is charged to V0 = 100 V. Then after removing the charging battery, a slab of insulating material of thickness b = 0.5 × 10 -2 m and dielectric constant K = 7 is inserted between the plates. Calculate the free charge, in pC, on the plates of the capacitor, electric field intensity, in kVm -1 , in air, electric field intensity, in Vm -1 , in the dielectric, potential difference, in volt, between the plates and capacitance, in pF, with dielectric present. 14. The distance between the parallel plates of a charged condenser is d = 5 cm and the intensity of the field E = 300 V cm -1 . A slab of dielectric constant K = 5 and 1 cm wide is inserted parallel to the plates. Determine the potential difference, in volt, between the plates, before and after the slab is inserted. If the slab is replaced by a metal plate so that the final potential difference remains unchanged, what be the thickness, in mm, of the plate? 15. The circuit shown in Figure is in steady state. The potential difference between points a and b is 10 V. Find the capacitance, in µF, of the capacitor C.

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 111

5 μF 6 μF a 3 μF b

16. In the circuit shown in Figure. Find the charge, in µC, on each capacitor. B

A

3 μF

3 μF

3 μF D

12. A condenser has capacitance 10 µF and is charged to a potential 150 V.

C

r

N



60 V

C 6 μF

E = 10 V

17. Consider the circuit shown in Figure, where C1 = 6 µF, C2 = 3 µF and ΔV = 20 V. Capacitor C1 is first charged by closing the switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by closing the switch S2 . Calculate

C1

ΔV

S1

C2

S2



(a)  the initial charge acquired by C1 , in µC.



(b)  the final charge on each capacitor, in µC.

18. The circuit in figure consists of two identical parallel metal plates connected by identical metal springs to a 100 V battery. With the switch open, the plates are uncharged, are separated by a distance d = 8 mm and have a capacitance C = 2 µF . When the switch is closed, the distance between the plates decreases by a factor of half. k

d

k

S ΔV

9/20/2019 11:03:01 AM

2.112  JEE Advanced Physics: Electrostatics and Current Electricity



(a)  How much charge, in µC, collects on each plate?



(b) What is the spring constant, in Nm -1, for each spring?

19. A certain storm cloud has a potential of 1 × 108 V relative to a tree. If, during a lightning storm, 50 C of charge is transferred through this potential difference and 1% of the energy is absorbed by the tree, how much sap, in kg (nearest to two digit integer) in the tree can be boiled away? Assume that the “sap” is “water initially at 30 °C ”. Water has a specific heat of 4186 Jkg -1 °C , a boiling point of 100 °C , and a latent heat of vaporisation of 2.26 × 106 Jkg -1 20. Each capacitor in the combination shown in figure has a breakdown voltage of 16 V. What is the breakdown voltage, in V, of the combination? 20 μ F

20 μ F 10 μ F

20 μ F

C1

M

C2

C2

E

N

E

22. A 20 pF parallel plate capacitor with air as medium is charged to 200 V and then disconnected from the battery. What is the energy U i of the capacitor? The plates are then slowly pulled apart (in a direction normal to the plate area) so that the plate separation is doubled. What is the mechanical work done, in nJ, in the process? 23. A 3 µF parallel plate capacitor is connected to a battery of 400 V. The plates are then pulled apart as in above problem, so that the capacitance value becomes 1 µ F. This operation is carried out while the capacitor is still connected to the battery of 400 V. Calculate the mechanical work done, in mJ. Account for the loss of energy of the capacitor.

20 μ F

21. Find the potential difference, in volt, between points M and N of the system shown, if the emf is equal to C E = 110 V and the capacitance ratio 2 is 2. C1

ARCHIVE: JEE MAIN 1. [Online April 2019] Voltage rating of a parallel plate capacitor is 500 V. Its dielectric can withstand a maximum electric field of 106 Vm -1 . The plate area is 10 -4 m 2 . What is the dielectric constant if the capacitance is 15 pF? (given ε 0 = 8.86 × 10 -12 C2 Nm -2 )

(A) 3.8 (C) 8.5

(B) 4.5 (D) 6.2

2. [Online April 2019] A parallel plate capacitor has 1 µF capacitance. One of its two plates is given +2 µC charge and the other plate, +4 µC charge. The potential difference developed across the capacitor is 5 V (B) 1V (A) (C) 3 V (D) 2V

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 112

3. [Online April 2019] A capacitor with capacitance 5 µF is charged to 5 µC . If the plates are pulled apart to reduce the capacitance to 2 µF, how much work is done? 2.55 × 10 -6 J (B) 6.25 × 10 -6 J (A) (C) 3.75 × 10 -6 J (D) 2.16 × 10 -6 J 4. [Online April 2019] The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V . When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is L

9/20/2019 11:03:09 AM

Chapter 2: Capacitance and Applications 2.113

( n + 1 )V V (A) (B) (K + n) K+n

1 μF

4 μF

nV (C) V (D) K+n 5. [Online April 2019] Figure shows charge ( q ) versus voltage ( V ) graph for series and parallel combination of two given capacitors. The capacitances are q( μ C)

5 μF 3 μF 10 V

(A) 5.4 µC (B) 9.6 µC (C) 13.4 µC (D) 24 µC

A 500 B 80 V (Volt)

10 V

8. [Online January 2019] A parallel plate capacitor is made of two square plates of side a, separated by a distance d ( d  a ). The lower triangular portion is filled with a dielectric of dielectric constant K , as shown in the figure. Capacitance of this capacitor is

(A) 40 µF and 10 µF (B) 20 µF and 30 µF (C) 60 µF and 40 µF (D) 50 µF and 30 µF 6. [Online April 2019] Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K 2 and K 3 . The first capacitor is filled as shown in figure i, and the second one is filled as shown in figure ii. If these two modified capacitors are charged by the same potential V , the ratio of the energy stored in the two, would be ( E1 refers to capacitor (I) and E2 to capacitor (II)) K1 K2

K1

K2

K

a

Kε a2 Kε 0 a2 ln K (B) ln K (A) 0 d( K - 1) d Kε a2 1 Kε 0 a2 (D) (C) 0 2d ( K + 1 ) 2 d 9. [Online January 2019] A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K 2 , K 3 , K 4 arranged as shown in the figure. The effective dielectric constant K will be

K3

K3 (i)

(ii)

E1 ( K1 + K 2 + K 3 ) ( K 2 K 3 + K 3 K1 + K1K 2 ) (A) = E2 K 1K 2 K 3

( K 1 + K 2 + K 3 ) ( K 2 K 3 + K 3 K 1 + K 1K 2 )

K1

K2 L/2

K3

K4 L/2

d/2 d/2

E1 9 K 1K 2 K 3 (C) = E2 ( K1 + K 2 + K 3 ) ( K 2 K 3 + K 3 K1 + K1K 2 )

( K1 + K 2 ) ( K 3 + K 4 ) 2 ( K1 + K 2 + K 3 + K 4 ) ( K + K2 )( K3 + K4 ) (B) K= 1

E1 K 1K 2 K 3 (D) = E2 ( K1 + K 2 + K 3 ) ( K 2 K 3 + K 3 K1 + K1K 2 )

(C) K=

E (B) 1 = E2

9 K 1K 2 K 3

7. [Online April 2019] In the given circuit, the charge on 4 µF capacitor will be

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 113

d

K= (A)

K1 + K 2 + K 3 + K 4

(D) K=

( K1 + K 3 ) ( K 2 + K 4 ) K1 + K 2 + K 3 + K 4

( K1 + K 4 ) ( K 2 + K 3 )

2 ( K1 + K 2 + K 3 + K 4 )

9/20/2019 11:03:18 AM

2.114  JEE Advanced Physics: Electrostatics and Current Electricity 1 0. [Online January 2019] A parallel plate capacitor is of area 6 cm 2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10 , K 2 = 12 and K 3 = 14 . The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be K1



K2

K3

(A) 36 (C) 12

(C)

(D)

3 mm

(B) 14 (D) 4

1 1. [Online January 2019] A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is 560 pJ (B) 692 pJ (A)

1 4. [Online January 2019] In the figure shown, after the switch S is turned from position A to position B, the energy dissipated in the circuit in terms of capacitance C and total charge Q is A B S ε

C

3C

508 pJ (D) 600 pJ (C) 1 2. [Online January 2019] In the figure shown below, the charge on the left plate of the 10 µF capacitor is ... The charge on the right plate of the 6 µF capacitor is 6 μF

10 μ F

2 μF 4 μF

(A) +18 µC (B) -12 µC (C) +12 µC (D) -18 µC 1 3. [Online January 2019] Seven capacitors, each of capacitance 2 µF, are to be connected in a configuration to obtain an effective ⎛ 6 ⎞ capacitance of ⎜ ⎟ µF . Which of the combinations, ⎝ 13 ⎠ shown in figures below, will achieve the desired value? (A)

3 Q2 1 Q2 (A) (B) 8 C 8 C 5 Q2 3 Q2 (C) (D) 8 C 4 C 1 5. [Online January 2019] In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 µF . All values in the circuit are in µF . C

2

A 2

1

2 2

2

2 B

7 (A) µF (B) 4 µF 11 6 7 (C) µF (D) µF 5 10

(B)

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 114

1 6. [Online January 2019] The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure

9/20/2019 11:03:26 AM

Chapter 2: Capacitance and Applications 2.115

6 5 4 q( μ C) 3 2 1 0

as shown in the figure. The sum of final charges on C2 and C3 is (1)

(2)

B 2

6

4

C2

60 V

8

C1

C3

t(s)



What is the value of current at t = 4 s ?

(A) 2 µA



(B) zero

(C) 3 µA (D) 1.5 µA 17. [2018] A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material 5 is inserted between the of dielectric constant K = 3 plates, the magnitude of the induced charge will be (A) 1.2 nC (B) 0.3 nC (C) 2.4 nC (D) 0.9 nC 18. [Online 2018] The equivalent capacitance between A and B in the circuit given below, is 6 μF A 5 μF

2 μF 5 μF

4 μF

2 μF

B

(A) 5.4 µF (B) 4.9 µF

(A) 20 C (C) 36 C

(B) 40 C (D) 54 C

21. [2017] A capacitance of 2 µF is required in an electrical circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is (A) 2 (B) 16 (C) 24 (D) 32 22. [Online 2017] The energy stored in the electric field produced by a metal sphere is 4.5 J . If the sphere contains 4 µC charge, its radius will be



1 ⎡ 9 2 -2 ⎤ ⎢ Take 4πε = 9 × 10 Nm C ⎥ 0 ⎣ ⎦ (A) 32 mm (C) 16 mm

(B) 20 mm (D) 28 mm

23. [Online 2017] A combination of parallel plate capacitors is maintained at a certain potential difference.

(C) 3.6 µF (D) 2.4 µF 19. [Online 2018] A parallel plate capacitor with area 200 cm 2 and separation between the plates 1.5 cm, is connected across a battery of emf V . If the force of attraction between the plates is 25 × 10 -6 N, the value of V is approximately 2 ⎞ ⎛ -12 C ⎜⎝ ε 0 = 8.85 × 10 ⎟ Nm 2 ⎠



(A) 150 V (C) 250 V

(B) 100 V (D) 300 V

20. [Online 2018] A capacitor C1 = 1 µF is charged up to a voltage V = 60 V by connecting it to battery B through switch (1). Now C1 is disconnected from battery and connected to a circuit consisting of two uncharged capacitors C2 = 3 F and C3 = 6 F through switch (2),

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 115

C1 A

D

C

C2

C3 E

B

When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab (A) 6 (B) 5 (C) 4 (D) 3 24. [2016] A combination of capacitors is set up as shown in the figure. The magnitude of the electric field due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal

9/20/2019 11:03:34 AM

2.116  JEE Advanced Physics: Electrostatics and Current Electricity 3 μF

4 μF

(A) Charge

(B) Charge

Q2

9 μF

Q2

2 μF 1 μF

3 μF

C

(C) Charge 8V

(A) 240 NC

-1

1 μF

(D) Charge

Q2

Q2

-1

(B) 360 NC

(C) 420 NC -1 (D) 480 NC -1 25. [Online 2016] Three capacitors each of 4 µF are to be connected in such a way that the effective capacitance is 6 µF . This can be done by connecting them (A) all in series (B) all in parallel (C) two in parallel and one in series (D) two in series and one in parallel

1 μF

3 μF

3 μF

C

2 μF

a

3 μF

S 3 μF

b 2 μF

10 V

(B) 20 µC from a to b

A 6

(C) 5 µC from a to b

4

2

1 μF

(A) 5 µC from b to a

1 8

C

28. [Online 2015] In figure is shown a system of four capacitors connected across a 10 V battery. Charge that will flow from switch S when it is closed is

26. [Online 2016] Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 µF is C

C

3 μF

2

12 B

32 31 µF (A) µF (B) 23 23 33 34 µF (C) µF (D) 23 23 27. [2015] In the given circuit, charge Q2 on the 2 µF capacitor changes as C is varied from 1 µF to 3 µF . Q2 as a function of C is given properly by (figures are drawn schematically and are not to scale) 1 μF C 2 μF

(D) zero

29. [2014] A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 10 4 V/m , the charge density of the positive plate will be close to (A) 6 × 10 -7 C/m 2 (B) 3 × 10 -7 C/m 2 (C) 3 × 10 4 C/m 2 (D) 6 × 10 4 C/m 2 30. [2013] Two capacitors C1 and C2 are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then 5C1 = 3C2 (B) 3C1 = 5C2 (A)

E

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 116

(C) 3C1 + 5C2 = 0 (D) 9C1 = 4C2

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Chapter 2: Capacitance and Applications 2.117

ARCHIVE: JEE advanced Single Correct Choice Type Problems This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. [JEE (Advanced) 2015] A parallel plate capacitor having plates of area A and plate separation d , has capacitance C1 in air. When two dielectrics of different relative permittivities ( ε1 = 2 and ε 2 = 4 ) are introduced between the two plates as shown in the figure, the capacitance becomes C C2 . The ratio 2 is C1 d/2

ε2

A/2

A/2

3. [IIT-JEE 2011] A 2 µF capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is 1

2

8 μF

2 μF



(A) 0% (C) 75%

(B) 20% (D) 80%

4. [IIT-JEE 2002] Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2 . The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

(

)

(

1 1 C V12 + V22 (A) C V12 - V22 (B) 4 4

ε1

d

6 5 (A) (B) 5 3 7 7 (C) (D) 5 3 2. [IIT-JEE 2012] In the given circuit, a charge of +80 µC is given to the upper plate of the 4 µF capacitor. Then in the steady state, the charge on the upper plate of the 3 µF capacitor is

1 1 2 2 (C) C ( V1 - V2 ) (D) C ( V1 + V2 ) 4 4 5. [IIT-JEE 2001] Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is q + + + + + +

+80 μ C 4 μF

– – – – – –

S

A

2 μF

3 μF

)



(A) zero

B

(B)

q 2

(C) q (D) 2q

(A) +32 µC (B) +40 µC (C) +48 µC (D) +80 µC

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 117

6. [IIT-JEE 2000] A parallel plate capacitor of area A , plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants k1 , k2 and k3 as shown. If a single dielectric material is to be

9/20/2019 11:03:52 AM

2.118  JEE Advanced Physics: Electrostatics and Current Electricity used to have the same capacitance C in this capacitor then its dielectric constant k is given by

d 2

A 2

A 2

k1

k2 d k3

1 1 1 1 1 1 1 (A) = + + (B) = + k k1 k2 2k3 k k1 + k2 2k3 kk kk k k 1 k= 1 3 + 2 3 (C) = 1 2 + 2k3 (D) k k1 + k2 k1 + k3 k2 + k3 7. [IIT-JEE 1999] Two identical metal plates are given positive charges Q1 and Q2 ( < Q1 ) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C , the potential difference between them is ( Q + Q2 ) (B) ( Q1 + Q2 ) (A) 1 2C C

10. [IIT-JEE 1995] A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V . Another capacitor of capacitance 2C is similarly charged to potential difference 2V . The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration

(A) ZERO

(B)

3 CV 2 2

25 9 (C) CV 2 (D) CV 2 6 2 11. [IIT-JEE 1990] Seven capacitors each of capacitance 2 µF are to be connected in a configuration to obtain an effective ⎛ 10 ⎞ capacitance of ⎜ ⎟ µF. Which of combination(s) ⎝ 11 ⎠ shown in figure, will achieve the desired results? (A)

( Q - Q2 ) (D) ( Q1 - Q2 ) (C) 1 2C C 8. [IIT-JEE 1999] For the circuit shown, which of the following statements is true? V2 = 20 V

V1 = 30 V s1

C1 = 2pF

(B)

s3

C2 = 3pF

s2



(A) With S1 closed, V1 = 15 V , V2 = 20 V



(B) With S3 closed, V1 = V2 = 25 V



(C) With S1 and S2 closed, V1 = V2 = 0



(D) With S1 and S3 closed, V1 = 30 V , V2 = 20 V

9.

[IIT-JEE 1996]

(C)

(D)

The magnitude of the electric field E in the annular region of a charged cylindrical capacitor

(A) is same throughout (B) is higher near the outer cylinder than near the inner cylinder 1 (C) varies as , where r is the distance from the axis r 1 (D) varies as 2 , where r is the distance from the r axis

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 118

Multiple Correct Choice Type Problems This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

9/20/2019 11:04:00 AM

Chapter 2: Capacitance and Applications 2.119 1. [IIT-JEE 2014] A parallel plate capacitor has a dielectric slab of dielec1 tric constant K between its plates that covers of 3 the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is C1 . When the capacitor is charged, the plate area covered by the dielectric gets charge Q1 and the rest of the area gets charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2 . Choose the correct option/ options, ignoring edge effects. Q1

E1 E2

Q2



(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 at first, then decreases and again increases

4. [IIT-JEE 1991] A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q , E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in questions, in the process of inserting the slab, then Q= (A)

ε 0 AV d

E1 E1 1 = 1 (B) = (A) E2 E2 K

Q= (B)

ε 0 KAV d

Q 3 C 2+K (C) 1 = (D) = Q2 K C1 K

E= (C)

V Kd

2. [JEE (Advanced) 2013] In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to charge the capacitor C2 . After some time, S2 is released and then S3 is pressed. After some time

W= (D)

S1 2V0



(A) (B) (C) (D)

S2 C1

S3 C2

V0

the charge on the upper plate of C1 is 2CV0 the charge on the upper plate of C1 is CV0 the charge on the upper plate of C2 is 0 the charge on the upper plate of C2 is -CV0

3. [IIT-JEE 1998] A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at x = 0 and positive plate is at x = 3 d . The slab is equidistant from the plates. The capacitor is given some charge. As x goes from 0 to 3d

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 119

1⎞ ε 0 AV 2 ⎛ ⎜ 1 - ⎟⎠ 2d ⎝ K

5. [IIT-JEE 1987] A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved farther apart by means of insulating handles (A) the charge on the capacitor increases (B) the voltage across the plates increases (C) the capacitance increases (D)  the electrostatic energy stored in the capacitor increases 6. [IIT-JEE 1985] A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q0 , V0 , E0 and U 0 respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities now given by Q , V , E and U are related to the previous ones as Q > Q0 (B) V > V0 (A) E > E0 (D) U > U0 (C)

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2.120  JEE Advanced Physics: Electrostatics and Current Electricity

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s). 1. [JEE (Advanced) 2019] A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The region between the plates is filled with N dielectric layers, d parallel to its plates, each with thickness δ = . The N

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 120

m⎞ ⎛ dielectric constant of the mth layer is K m = K ⎜ 1 + ⎟ . ⎝ N⎠ For a very large N ( > 10 3 ) , the capacitance C is ⎛ Kε A ⎞ a ⎜ 0 ⎟ . The value of a will be ________. ⎝ d ln 2 ⎠ (ε 0 is the permittivity of free space)

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Chapter 2: Capacitance and Applications 2.121

Answer Keys—Test Your Concepts and Practice Exercises Test Your Concepts-I (Based on General Capacitance)

8. 78.7 V , 151.3 V .

1. 400 V

9. 

ε0 A a-b

10. 

ε 0 a2 ⎡ aθ ⎤ 1⎢ d ⎣ 2d ⎥⎦

2. 2 J 3. (a)  7600 V  

(b)  1.368 × 10 -3 J 1 2

4. 5 J ,

⎛ R1 ⎞ ⎛ R2 ⎞ 5.  (a)  ⎜ Q, ⎜ Q ⎝ R1 + R2 ⎟⎠ ⎝ R1 + R2 ⎟⎠  

11.  (a)  Arrangement 1,

3 ⎛ ε0 A ⎞ ⎜ ⎟ 2⎝ d ⎠

 

2 ⎛ ε0 A ⎞ ⎜ ⎟ 3⎝ d ⎠

(b)  Arrangement 2,

12.  ( n - 1 )

(b)  0

1 ⎤ ⎡ ⎛ V0 ⎞ 10 ⎢ ⎥C 1 6.  ⎜ ⎢⎣ ⎝ V ⎠⎟ ⎥⎦ 0

Test Your Concepts-II (Based on Series and Parallel Combination of Capacitors) 1. (a)  C ( ΔV ) (b) 

 

(c)  4C

13. (a)  4 µF  

(b)  V1 = 8 V, Q1 = 48 µC

     

V2 = 4 V, Q2 = 16 µC

     

V3 = 4 V, Q3 = 32 µC

14.  Arrangement 1,

2ε 0 A d

 

3ε 0 A d

2

4 ΔV 3

 

( ΔV )2 3

2. (a)  Net capacitance remains same   (b)  Net capacitance increases.

  Arrangement 2,

15. (a)  17 pF  

(b)  1.224 × 10 -7 J

16.  (a) 

q2 2C

 

q2 8πε 0 R

3. Across the 5 µF capacitor 23.3 V  

ε0 A d

  Across the 10 µF capacitor 26.7 V

(b) 

4. (a) 

q2 ( x2 - x1 ) 2ε 0 A

Test Your Concepts-III (Based on Dielectrics and Breakdown)

 

1 1⎞ ⎛ 1 ε 0 AV 2 ⎜ 2 ⎝ x2 x1 ⎟⎠

1. 

(b) 

24 25

5. 0.24 µF , 0.16 µF

2. (b)  It becomes κ times the initial charge

6. 40 mF and 10 mF

⎛ K + 1⎞ 3. (a)  Electric field decreases by a factor ⎜ ⎝ 2 ⎟⎠

7. 

ε0 A 4d

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 121

 

(b) 

CV ( K - 1 ) 2( K + 1)

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2.122  JEE Advanced Physics: Electrostatics and Current Electricity 4. 3.4%

5. V1 =

ε A 5. (a)  0 d  

ε A (b)  0 d-t

 

(c)

 

2ε 0 A d

 

K ⎞ ⎛K 6. (a)  ε 0 A ⎜ 1 + 2 ⎟ ⎝ d1 d2 ⎠  

 

K ⎞ ⎛K (b)  ε 0V ⎜ 1 + 2 ⎟ ⎝ d1 d2 ⎠

1 ⎛ ε K V2 ⎞ (c)  ⎜ 0 12 ⎟ 2 ⎝ d1 ⎠

7. 

ε 0 K 1K 2 A d1K 2 + d2 K1

V2 = V3 =

E1 ( C2 + C3 ) - E2C2 - E3C3 C1 + C2 + C3 E2 ( C1 + C3 ) - E1C1 - E3C3 C1 + C2 + C3 E3 ( C1 + C2 ) - E1C1 - E2C2 C1 + C2 + C3

C1C4 - C2C3

6. 

( C1 + C2 ) ( C3 + C4 )

7. 

V,

C1 C3 = C2 C 4

V2C2 - V1C1 C1 + C2 + C3

8. 2C 9. Charge on 2 µF capacitor is

8. 1.54 pF 9. 

3 5

10.  (a)   

ε0 2 [  + x ( κ - 1 ) ] d

2 1 ⎛ ε ( ΔV ) ⎞ 2 (b)  ⎜ 0 ⎟⎠ [  + x (κ - 1 ) ] 2⎝ d 2 ε 0 ( ΔV )

 (κ - 1 )

 

(c) 

 

(d)  1.55 × 10 -3 N

11. 

2d

2 3

10 1.  C 7 2. V1 = 10 V, V2 = 5 V

4. 

140 µC 3

 

 Charge on 4 µF capacitor is

 

 Charge on 6 µF capacitor is 50 µC 1 2 V2 C 2

10. 

11. (i) 

200 V, 0 V;  (ii) 300 µC 3

Test Your Concepts-V (Based on Spherical and Cylindrical Capacitors) 1. 2 pF

Test Your Concepts-IV (Based on Capacitor Circuits, Kirchhoff’s Laws, Charge Flown and Generation of Heat)

3. 10 µC and

10 µC 3

40 µC 3

E2 - E1 C1C2 C1 + C2

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 122

2. (a)  2.5 × 10 4 V  

 (b)  +1 µC 2

5

3. 

C R V U Q = , = N3, = N3 = N, c r v u q

4. 

4πε 0b 2 b-a

5. +24 nC 6. 

πL ( K1 + K 2 ) ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠

9/20/2019 11:04:32 AM

Chapter 2: Capacitance and Applications 2.123

Single Correct Choice Type Questions 1.  A

2.  B

3.  B

4.  A

5.  D

6.  D

7.  B

8.  C

9.  B

10.  C

11.  B

12.  B

13.  B

14.  C

15.  D

16.  B

17.  D

18.  B

19.  C

20.  A

21.  D

22.  D

23.  D

24.  D

25.  B

26.  B

27.  C

28.  B

29.  C

30.  C

31.  C

32.  C

33.  A

34.  C

35.  C

36.  B

37.  C

38.  B

39.  B

40.  B

41.  B

42.  C

43.  C

44.  A

45.  D

46.  C

47.  C

48.  C

49.  B

50.  C

51.  D

52.  B

53.  A

54.  A

55.  B

56.  A

57.  A

58.  B

59.  C

60.  C

61.  B

62.  B

63.  D

64.  A

65.  C

66.  C

67.  B

68.  C

69.  B

70.  A

71.  B

72.  B

73.  D

74.  A

75.  C

76.  D

77.  A

78.  B

79.  C

80.  A

81.  C

82.  D

83.  D

84.  B

85.  A

86.  B

87.  C

88.  C

89.  C

90.  A

91.  A

92.  C

93.  B

94.  A

95.  B

96.  C

97.  A

98.  B

99.  D

100.  A

101.  A

102.  B

103.  D

104.  B

105.  D

106.  A

107.  C

108.  C

109.  D

110.  B

111.  B

112.  B

113.  A

114.  C

115.  C

116.  C

117.  B

118.  C

119.  D

120.  C

121.  B

122.  D

123.  B

124.  A

125.  A

126.  B

127.  A

128.  B

129.  B

130.  D

131.  C

132.  A

133.  C

134.  D

135.  B

136.  B

137.  B

138.  B

139.  C

140.  B

141.  D

142.  A

143.  C

144.  A

145.  D

146.  D

147.  A

148.  C

149.  C

150.  B

151.  A

152.  A

153.  D

154.  C

155.  A

156.  C

157.  B

158.  A

159.  A

160.  A

161.  D

162.  A

163.  A

164.  A

165.  C

166.  C

167.  C

168.  C

169.  C

170.  C

171.  B

172.  B

173.  A

174.  A

175.  A

176.  A

177.  A

178.  B

179.  A

180.  A

181.  A

182.  D

183.  B

Multiple Correct Choice Type Questions 1.  A, B, C, D

2.  A, C, D

3.  B, D

4.  B, C,

6.  A, B, C, D

7.  A, D

8.  A, B, C, D

9.  A, D

5.  A, B, C, D 10.  A, C, D

11.  A, C, D

12.  C, D

13.  A, B, C,

14.  B, C,

15.  A, B, C, D

16.  A, C, D

17.  A, B, C,

18.  A, C,

19.  A, D

20.  B, C, D

21.  A, B,

22.  B, C, D

23.  A, B, C, D

24.  B, C, D

25.  B, C, D

26.  A, D

27.  A, B, C, D

28.  B, C, D

29.  A, C,

30.  B, C,

31.  A, D

32.  A, D

33.  A, C, D

34.  A, D

35.  A, C, D

36.  A, C,

37.  A, B, C,

38.  A, C,

39.  B, C, D

40.  B, D

41.  A, B, C, D

42.  A, C, D

Reasoning Based Questions 1.  D

2.  A

3.  A

4.  D

11.  C

12.  C

13.  C

14.  B

5.  A

6.  A

7.  C

8.  D

9.  D

10.  A

Linked Comprehension Type Questions 1.  D

2.  A

3.  B

4.  D

5.  B

6.  D

7.  A

8.  B

9.  C

10.  B

11.  C

12.  D

13.  C

14.  C

15.  A

16.  C

17.  C

18.  B

19.  C

20.  B

21.  B

22.  B

23.  C

24.  A

25.  B

26.  C

27.  A

28.  C

29.  B

30.  B

31.  C

32.  D

33.  B

34.  D

35.  C

36.  A

37.  B

38.  B

39.  A

40.  C

41.  C

42.  C

43.  B

44.  C

45.  B

46.  A

47.  D

48.  B

49.  C

50.  D

51.  C

52.  A

53.  D

54.  D

55.  D

56.  C

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 123

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2.124  JEE Advanced Physics: Electrostatics and Current Electricity

Matrix Match/Column Match Type Questions 1. A → (q, s)

B → (p)

C → (q, s)

D → (r)

2. A → (q)

B → (p)

C → (s)

D → (r)

3. A → (p, r, s)

B → (p, r)

C → (p, r, s)

D → (p, r)

4. A → (s)

B → (q)

C → (r)

D → (p)

5. A → (r)

B → (p)

C → (q)

D → (s)

6. A → (q)

B → (r)

C → (p)

D → (s)

7. A → (q)

B → (p)

C → (s)

D → (r)

8. A → (q)

B → (r)

C → (s)

D → (p)

9. A → (p, s)

B → (q, r, s)

C → (q, r, s)

D → (p)

10. A → (p, q)

B → (r, q)

C → (p, q, s)

D → (r, q)

11. A → (q, s)

B → (r, s)

C → (p, r, s)

D → (p, r, s)

12.  A → (q)

B → (p)

C → (r)

D → (s)

13.  A → (p)

B → (p)

C → (r)

D → (q)

14. A → (p, s)

B → (r, s)

C → (p, q)

D → (q)

15. A → (p, r)

B → (q)

C → (s, q)

D → (q)

Integer/Numerical Answer Type Questions 1.  a = 2

2.  Yes, 1000

3.  q6 mF = 750 mC, q2 mF = 250 mC

4.  3

5. 579

6. 6950

7.  5

9. 9

10. 2

11. 10

8.  (a) 5, (b) 64

12.  (a) 7500, (b) 75

13.  890,  10,  1430,  57,  16 14.  1500, 1260, 8

15. 9

16.  Charge  on  6 mF  capacitor  is  24 mC,  Charge  on  3 mF capacitor is 18  mC

18.  (a) 400, (b) 2500

19. 10

20. 24

23. 160

21. 550

22. 400

ARCHIVE: JEE MAIN 1.  C

2.  B

3.  C

4.  A

5.  A

6.  C

7.  D

8.  A

9.  (*)

10.  C

11.  C

12.  A

13.  A

14.  A

15.  A

16.  B

17.  A

18.  D

19.  C

20.  B

21.  D

22.  C

23.  B

24.  C

25.  D

26.  A

27.  B

28.  A

29.  A

30.  B

5.  A

6.  D

7.  D

8.  D

9.  C

10.  B

5.  B, D

6.  A, D

* No given option is correct

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems 1.  D

2.  C

3.  D

4.  C

11.  A

Multiple Correct Choice Type Problems 1.  A, D

2.  B, D

3.  B, C

4.  A, C, D

Integer/Numerical Answer Type Questions 1.  1

02_Physics for JEE Mains and Advanced - 2_Part 3.indd 124

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CHAPTER

3

Electric Current and Circuits

Learning Objectives After reading this chapter, you will be able to: After reading this chapter, you will be able to understand concepts and problems based on: (a) (b) (c) (d) (e)

Current Ohm’s Law Resistance Resistance Combinations Kirchhoff ’s Laws

(f) Nodal Analysis (g) Cells (h) Cell Combinations (i) Ammeter ( j) Voltmeter

(k) Heating Effects of Current (l) Wheatstone Bridge (m) Post Office Box (n) Potentiometer (o) RC Circuit and Applications

All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main and Advanced) are also given.

ELECTRIC CURRENT Whenever a charge moves (i.e. a charge moving with respect to an observer) we get current. So, moving charges (positive and negative) constitute electric current. Flow of electric charge is a direct measure of electric current. Suppose a collection of charges is moving perpendicular to a surface of area A , as shown in figure. I A

Charges moving through a cross section

The electric current is defined as the rate at which charges flow across any cross-sectional area. If an amount of charge ΔQ passes through a surface in a time interval Δt, then the average current Iavg is given by ΔQ I avg = …(1) Δt

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 1

The SI unit of current is the ampere (A), where 1 A = 1 coulomb sec −1 . Common currents range from mega-amperes (in lightning) to nano-amperes (in your nerves). In the limit Δt → 0, the instantaneous current I may be defined as I = lim t→0

ΔQ dQ = Δt dt

…(2)

Since flow has a direction, so we have implicitly introduced a convention that the direction of current corresponds to the direction in which positive charges are flowing. This current is called the Conventional Current. The flowing charges inside wires are negatively charged electrons that move in the opposite direction of the current and is called ‘Electronic Current’. Electric currents flow in conductors : solids (metals, semiconductors), liquids (electrolytes, ionized) and gases (ionized), but the flow is impeded in non-conductors or insulators.

9/24/2019 11:16:36 AM

3.2  JEE Advanced Physics: Electrostatics and Current Electricity

According to its magnitude and direction we categorise the current in two types. (a) Direct Current (DC): If the magnitude and direction of current does not vary with time, it is said to be direct current (DC). This type of current is provided by a Cell, a battery or DC dynamo. (b) Alternating Current (AC): If a current is periodic (with constant amplitude) and has half cycle positive and half negative, it is said to be alternating current (AC). This type of current is generally sinusoidal in nature. An AC dynamo provides this type of current.

I=

DQ Nq = = nqA Dt Dt

n=

N ADx

   (c) If there are n particles per unit volume in the conductor (each having a charge q ) moving with velocity v , then   

So, the current through cross-sectional area A is Δx q v

v A

Remark(s) (a) In all electric circuits, the current shown (or the arrows drawn) must represent (or indicate) the “Conventional Current”. (b) A negative charge moving to the right is equivalent to a positive charge moving to the left. So current due to this negative charge is towards the right. (c) If a charge Q revolves in a circle of radius R with angular frequency w or frequency f, then equivalent current,



vΔ t

  

I=

DQ Nq Dx = = nqA = nqvA Dt Dt Dt

Due to Circular Motion of Charge If a point charge q is moving in a circle of radius r with 2π ⎛ 2π r ⎞ 1 speed v , then its time period T = =⎜ ⎟= . w ⎝ v ⎠ f v

Charge Circulating I= Time to Complete one Revolution Q Qw Qv I= = = T 2π 2π R

DIFFERENT SITUATIONS PRODUCING CURRENT

q r

So through a given hypothetical cross-section (­perpendicular to motion), the current is

Due to Translatory Motion of Charges (a) If N particles, each having a charge q , pass through a given area in time t , the current is given by DQ Nq = Dt t    (b) If n particles (each having a charge q ) pass per second per unit area, then I=

N n= ADt    So, the current associated with cross-sectional area A is

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 2



I=

⇒ I=

Charge Circulating Period of One Revolution q qv qw = qf = = 2π r 2π T

where w is the angular velocity of the charge.

Due to Convectional Motion of Charge It is possible that a charged body is transported from one place to another. The convectional current is the current which is developed due to the transportation of charge or the mechanical transfer of a charge.

9/20/2019 11:06:35 AM

Chapter 3: Electric Current and Circuits 3.3

Conceptual Note(s) Instantaneous current through a cross-section is dq I= dt Charge passed through the cross-section in the interval t to t + dt dq = Idt Total charge in the interval t1 to t2 is given by t2

∫

Dq = Idt = Area under I versus t graph in the t1  interval t1 to t2 as shown in figure.

I

t2

t1

t

Average current in the interval t1 to t2 is

Dq Dq Iav = = Dt t2 - t1

∫ Idt

Dq Area under I versus t graph t = 1 = t2 - t1 t2 - t1 Time interval

Illustration 1

The current in a wire varies with time according to the relation I = a + bt 2 , where current I is in ampere and time t is in second and a = 4 A , b = 2 As -2 .

(b) I e =

(a) Dq =

∫ 5

I dt =

∫(

)

4 + 2t 2 dt

5

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 3

2 ( 1000 - 125 ) 3

Dq 603.33 = = 120.67 A Dt 10 - 5

Illustration 2

In the Bohr model of hydrogen atom, the electron is pictured to rotate in a circular orbit of radius 5 × 10 -11 m , at a speed of 2.2 × 106 ms -1 . What is the current associated with electron motion ? Solution

The time taken to complete one rotation is 2π r T= v Therefore, the current is

q e ev 1.6 × 10 -19 × 2.2 × 106 I= = = = = 1.12 mA t T 2π r 2 × 3.14 × 5 × 10 -11

Illustration 3

In a neon discharge tube 2.9 × 1018 Ne + ions move to

Solution

Since current is actually due to the flow of positive charges. Also we know that negative charge moving to the left is equivalent to positive charge moving to the right, so Net current i = i+ + i- = e–

( n+ ) ( q+ ) ( n- ) ( q- ) t

+

t

Ne– i

⇒ i=

Solution 10

5

= 4 ( 10 - 5 ) +

   Dq = 603.33 C

(a) How many coulomb pass a cross-section of the wire in the time interval between t = 5 s and t = 10 s ? (b) What constant current could transport the same charge in same time interval ?

10

10

the right each second while 1.2 × 1018 electrons move to the left per second. Electron charge is 1.6 × 10 -19 C. Calculate the current in the discharge tube.

t2

⇒ Iav =

2 ⇒   Dq = 4t + t 3 3

( n+ ) e + ( n- ) e t

⇒ i = ( 2.9 × 10

t

18

+ 1.2 × 1018 ) × 1.6 × 10 -19

⇒ i = 0.66 A

9/20/2019 11:06:42 AM

3.4 JEE Advanced Physics: Electrostatics and Current Electricity IllUSTRATION 4

A long cylinder with uniformly charged surface and cross-sectional radius a = 1 cm moves with a constant velocity v = 10 ms -1 along its axis. An electric field strength at the surface of the cylinder is equal to E = 0.9 kVcm -1 . Find the resulting convection current, that is, the current caused by mechanical transfer of a charge. SOlUTION

The convection current is dq …(1) dt where, dq = l dx , l is the linear charge density I=

But, from the Gauss’s Law, electric field at the surface of the cylinder, E=

l 2πε 0 a

Substituting the value of l and subsequently of dq in equation (1), we get I=

dq l dx = dt dt

⇒

I=

2Eπε 0 adx dt

⇒

I = 2πε 0 Eav

{

∵

dx =v dt

}

Substituting the given values, we get I = 0.5 μ A

Test Your Concepts-I

Based on Current Definition 1. In the Bohr model of the hydrogen atom, an electron in the lowest energy state follows a circular path 5.29 × 10 -11 m from the proton. (a) Show that the speed of the electron is 2.19 × 106 ms -1. (b) What is the effective current associated with this orbiting electron? 2. The quantity of charge q (in coulombs) that has passed through a surface of area 2 cm2 varies with time according to the equation q = 4t 3 + 5t + 6, where t is in seconds. (a) What is the instantaneous current through the surface at t = 1 s? (b) What is the value of the current density? 3. An electric current is given by the expression I ( t ) = 100 sin( 120π t ) , where I is in ampere and t is in second. What is the total charge carried by the ⎛ 1 ⎞ s? current from t = 0 to t = ⎜ ⎝ 240 ⎟⎠ 4. A very small earthed conducting sphere is at a distance a from a point charge q1 and at a distance b from a point charge q2 ( a < b ) . At a certain instant, the sphere starts expanding so that its radius grows according to the law R = vt . Determine the time dependence I(t) of the current in the earthing conductor, assuming that the point charges and the

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 4

(Solutions on page H.198) centre of the sphere are at rest, and in due time the initial point charges get into the expanding sphere without touching it (through small holes). 5. The current in a wire varies with time according to the relation ⎛ A⎞ I = ( 3A ) + ⎜ 2 ⎟ t ⎝ s⎠ (a) How many coulomb of charge passes a crosssection of the wire in the time interval between t = 0 and t = 5 s? (b) What constant current would transport the same charge in the same time interval? 6. The gap between two plane plates of a capacitor equal to d is filled with a gas. One of the plates emits n0 electrons per second, which while moving in an electric field, ionize the gas molecules. This way each electron produces a new electrons (and ions) along a unit length of its path. Find the electronic current at the opposite plate, neglecting the ionization of gas molecules by the ions so formed. Take charge on an electron as e. 7. A long conductor of charge q, with charge density l is moving with a velocity 2v parallel to its own axis. Find the convectional current due to motion of conductor.

9/20/2019 11:06:49 AM

Chapter 3: Electric Current and Circuits 3.5

 CURRENT DENSITY ( J ) The current density at any point inside a conductor is defined as a vector quantity whose magnitude is equal to current per unit infinitesimal area at that point, the area held normal to the direction of flow of current. A θ

I

I An = A cos θ

θ

 Current density J points along the direction of ­current flow. If An is small area normal to current I , then Current density J =

I An

If the plane of the small area A is not normal to ­current, but makes an angle θ with the to current, then



J=

I I = An A cos θ

The unit of current density is Am -2 . From above, we have   I = JA cos θ = J ⋅A

RESISTIVITY (r), CONDUCTIVITY (s) and CONDUCTANCE (G) For a given conductor of uniform cross-section A and length l , it has been observed that the electrical resistance R is (a) directly proportional to length l and (b) inversely proportional to cross-sectional area A

⇒  R ∝

rl …(1) A where r is a constant of proportionality called the specific resistance or resistivity of the metal of the conductor at a given temperature. Since RA r= l If l = 1 m , A = 1 m 2 , then r = R (numerically). So, the specific resistance of the material of the conductor is defined as the resistance offered by the conductor of length 1 m and cross sectional area 1 m 2 , when current flows normal to the area. Alternatively, the specific resistance of a material is defined as the resistance offered by the opposite faces of a cube of that material of side 1 m. The unit of resistivity is ohm metre ( Wm ) .

⇒  R =

OHM’S LAW AND ELECTRICAL RESISTANCE On applying a potential difference across a conductor, a current I is set up in the conductor. According to Ohm’s Law “Under given physical conditions the potential difference ( V ) applied across a conductor produces a proportionate amount of current ( I ) in the conductor” i.e.,

V ∝I

⇒ V = IR where the constant R is called the electrical resistance of the given conductor. The unit of resistance R is volt/ampere.

1 VA -1 = 1 ohm ( W )

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 5

l A

I

I

2

ρ=R l = 1m

A

=

1m

The reciprocal of resistivity ( r ) is called the conductivity ( s ) ⇒

s=

1 r

The unit of conductivity is mho metre -1 ( W -1 m -1 ) Ohm’s Law in alternative form may be expressed as J = sE where J = Current Density and E = Electric Field Strength.

9/20/2019 11:07:02 AM

3.6  JEE Advanced Physics: Electrostatics and Current Electricity

Conductance (G) is the reciprocal of resistance ⇒

Consider a thin disk of radius r at a distance x from the left end. From the figure shown, we have

1 R

G=

Solution

Unit of conductance is siemen or mho ( = W -1 ) .

Illustration 5

A 3000 km long cable consists of seven copper wires, each of diameter 0.73 mm, bundled together and surrounded by an insulating sheath. Calculate the resistance of the cable. Use 3 × 10 -6 W cm for the resistivity of the copper.

b-r b-a = x h

x ⇒ r = (a -b) + b h

b

r

a

Solution

The resistance R of a conductor is related to the rl resistivity r by R = , where l and A are the A length of the conductor and the cross-sectional area, respectively. Since the cable consists of N = 7 copper wires, the total cross sectional area is A = Nπ r 2 = N

π d2 π ( 0.073 cm ) =7 4 4

2

h

Since resistance R is related to resistivity r by rl R = , where l is the length of the conductor and A A is the cross section, the contribution to the resistance from the disk having a thickness dy is

The resistance then becomes R=

rl ( 3 × 10 -6 Wcm ) ( 3 × 108 cm ) = = 3.1 × 10 4 W 2 A 7π ( 0.073 cm )

dR =

∫ 0



Illustration 6

Consider a material of resistivity r in a shape of a truncated cone of altitude h , and radii a and b , for the right and the left ends, respectively, as shown in the figure.

⇒ R=

=

A truncated cone

Assuming that the current is distributed uniformly throughout the cross-section of the cone, what is the resistance between the two ends?

( a - b )x ⎤ ⎡ π ⎢b + ⎥⎦ h ⎣

r dx

( a - b )x ⎤ ⎡ π ⎢b + ⎥⎦ h ⎣

2

2

=

rh π ab

rh π ab du

a

r dx

where we have used

h

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 6

πr

2

h

R=

b

r dx

Straight forward integration then yields

4

⇒ R = 31 kW

θ

x

∫ (au + β )

2

=-

1 a (au + β )

Note that if b = a , we get R =

r h rl = π a2 A

Illustration 7

Consider a hollow cylinder of length L and inner radius a and outer radius b, as shown in figure. The material has resistivity r .

9/20/2019 11:07:13 AM

Chapter 3: Electric Current and Circuits 3.7

Since, m = volume × density = Ald …(2) m ⇒ A = ld

L

a ρ b A hollow cylinder

(a) Suppose a potential difference is applied between the ends of the cylinder and produces a current flowing parallel to the axis. What is the resistance measured? (b) If instead the potential difference is applied between the inner and outer surfaces so that current flows radially outward, what is the resistance measured? Solution

(a) When a potential difference is applied between the ends of the cylinder, current flows parallel to the axis. In this case, the cross-sectional area is A = π ( b 2 - a 2 ) , and the resistance is given by

rL rL R= = 2 A π ( b - a2 )    (b) Consider a differential element which is made up of a thin cylinder of inner radius r and outer radius r + dr and length L . Its contribution to the resistance of the system is given by   

dR =

r dl r dr = A 2π rL

where A = 2π rL is the area normal to the direction of current flow. The total resistance of the system becomes b

R=   

rdr

r

a

RECASTING A WIRE OF GIVEN MASS Consider a conductor of mass m , density d , resistivity r to be recast to a wire of length l , cross-sectional area A (or radius r ), then we have

R=

rl …(1) A

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 7



rl ⎛ rd ⎞ 2 =⎜ ⎟ l …(3) m ld ) ⎝ m⎠ (

R=

As r , d , m are constants, so we have

R ∝ l2

⇒

R2 ⎛ l2 ⎞ = …(4) R1 ⎜⎝ l1 ⎟⎠

2

So, we conclude that if a given mass of wire is stretched (or drawn) to n times its original length, then the new resistance is n2 times the original resistance of wire. Hence Rnew = n2 Rold m Further if we put value of l from (2) i.e. l = in (1), Ad then we get

rm ⎛ mr ⎞ ⎛ 1 ⎞ =⎜ ⎟⎜ ⎟ AdA ⎝ d ⎠ ⎝ A 2 ⎠

R=



⇒ R∝

1 A2

…(5)

Since A = π r 2 ⇒ R∝

1 r4

Remark(s) Thus if given mass of material is stretched to make its length l, then R ∝  2 and if the given mass of material is stretched to make its radius r, then

⎛ b⎞

∫ 2π rL = 2π L log ⎜⎝ a ⎟⎠ e

Then from (1), we get

or

R∝

1 r4

R2 ⎛ r1 ⎞ = R1 ⎜⎝ r2 ⎟⎠

4

So, we conclude that if a given mass of wire is stretched (or draw) such that new radius of wire is n times its old radius, then

Rnew =

1 n4

Rold

9/20/2019 11:07:24 AM

3.8  JEE Advanced Physics: Electrostatics and Current Electricity

FRACTIONAL CHANGE IN RESISTANCE The resistance of a wire depends on its length L , radius r and cross-sectional area A (where A = π r 2 ). If the wire is stretched by a small amount, its resistance R changes. CASE-1: Suppose the length L increases by a small amount DL (  L ) . Then, we can write



R=

⎛ L2 ⎞ rL = r⎜ ⎟  ⎝V⎠ A {as AL = volume, V = constant}

Hence fractional change in resistance with a small change in length is DR DL =2 R L

⇒

∫

(

rL ⎛ V ⎞ = r⎜ 2 ⎟ ⎝A ⎠ A

DR DA = -2 R A

)

DQ = q nADx . Suppose that the charge carriers move with a speed vd , then the displacement in a time interval Dt will be Dx = vd Dt . So, by definition of average current, we get

CASE-2: Suppose the wire is stretched so that the cross-sectional area A reduces by small amount DA (  A ) . Then we can write R=

Let the total current through a surface be written as   I = J ⋅ dA …(1)  where J is the current density (the SI unit of current density are Am -2 ). If q is the charge of each carrier, and n is the number of charge carriers per unit volume (also called “Charge Carrier Density”), the total amount of charge in this section is then

I avg =

DQ = nqAvd …(2) Dt

The speed vd (along with the random speed) at which the charge carriers are moving is known as the drift speed. Physically, vd is the average speed of the charge carriers inside a conductor when an external electric field is applied. Actually an electron inside the conductor does not travel in a straight line. Instead, its path is rather erratic, as shown in figure. vd

CASE-3: Suppose the wire is stretched so that the radius reduces by Dr (  r ) . Then, we can write R= ⇒

rL V ⎛ LA ⎞ = r⎜ 2 ⎟ = r ⎝A ⎠ A π r2

( )

2

⎛ rV ⎞ 1 =⎜ 2 ⎟ 4 ⎝π ⎠r

DR Dr = -4 R r

CURRENT DENSITY and DRIFT VELOCITY To relate current, a macroscopic quantity, to the microscopic motion of the charges, let’s examine a conductor of cross-sectional area A , as shown in figure. J vd q A

Δx = vd Δt

A microscopic picture of current flowing in a conductor

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 8

E Motion of an electron in a conductor

From the above equations (1) and (2), the current ­density J can be written as   J = nqvd …(3)   Thus, we see that J and vd point in the same direction for positive charge carriers, in opposite directions for negative charge carriers. To find the drift velocity of the electrons, we first note that an electron  in the conductor experiences an electric force Fe = - eE which gives an acceleration   eE  Fe a= =me me Let the velocity of a given electron immediate after  a collision be u . The velocity of the electron immediately before the next consecutive collision is then given by

9/20/2019 11:07:36 AM

Chapter 3: Electric Current and Circuits 3.9

     ⎛ eE ⎞ v = u + at = u + ⎜ t ⎝ me ⎟⎠

where t is the time between two consecutive colli sions. The average of v over all time intervals is  eE   v = u t me  This average, v is equal to the drift velocity vd . Since in the absence of electric field, the velocity of the electron is completely random, it follows that   u = 0 . If t = t is the average characteristic time between successive collisions (the average relaxation time or the mean free time), we have  ⎛ eE ⎞   vd = v = - ⎜ t ⎝ me ⎟⎠ The current density, from equation (3) becomes   ⎛ eE ⎞ ⎛ ne 2t ⎞   J = -nevd = -ne ⎜ E …(4) t = ⎝ m ⎟⎠ ⎜⎝ m ⎟⎠ e

e

  Note that J and E will be in the same direction for either negative or positive charge carriers.   Also, we compare (4) with J = s E , then we get

s=

⇒ r=

ne 2t me me

ne 2t

Remark(s)

  (a) J and vd point in the same direction for positive charge carriers, in opposite directions for negative charge carriers.   (b) J and E will be in the same direction for either negative or positive charge carriers.

OHM’S LAW: REVISITED In many materials, the current density is linearly  dependent on the external electric field E with their relation expressed as   J = sE where s is called the conductivity of the material. The above equation is known as the Microscopic

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 9

Ohm’s Law. A material that obeys this relation is said to be ohmic, otherwise, the material is non-ohmic. Also, we have seen that the conductivity s is given by

s=

ne 2t me

To obtain a more useful form of Ohm’s Law for practical applications let us consider a segment of straight wire of length l and cross-sectional area A , as shown in figure. Va

l J A Vb

E

A uniform conuctor of lenght l and potential difference ΔV = Vb – Va

Suppose a potential difference DV = Vb - Va is applied across the ends of the wire, creating an electric field  E and hence  make a current I to flow through it. Assuming E to be uniform, then b

  DV = Vb - Va = - E ⋅ dl = El

∫

a The current density can then be written as



⎛ DV ⎞ J = sE = s ⎜ ⎝ l ⎟⎠

where J =



I . The potential difference becomes A

DV =

where

l ⎛ l ⎞ I = RI J=⎜ ⎝ s A ⎟⎠ s

R=

DV l = , is the resistance of the sA I

conductor. The equation DV = IR is the “macroscopic” version of the Ohm’s Law. The SI unit of R is the ohm ( W , the Greek letter Omega), where 1 W≡

1V 1A

Once again, a material that obeys the above relation is ohmic, and non-ohmic if the relation is not obeyed.

9/20/2019 11:07:48 AM

3.10  JEE Advanced Physics: Electrostatics and Current Electricity

Most metals, with good conductivity and low resistivity, are ohmic. We shall focus mainly on ohmic materials. I

I I l ΔV R ΔV

ΔV

Ohmic vs. Non-ohmic behaviour

The resistivity r of a material is defined as the reciprocal of conductivity

r=

m 1 = 2e s ne t

From the above equations, we see that r can be related to the resistance R of an object by



DV RA E r= = l = I l J A

⇒ R=

m rl where r = 2e A ne t

MOBILITY Conductivity arises from mobile charge carriers. These mobile charge carriers are electrons in case of metal, electrons and positive ions in case of an ionized gas and both positive and negative ions in case of an electrolyte. In semiconductor’s material such as germanium and silicon, conduction is partly due to electrons and partly due to electron vacancies called holes. Holes are sites of missing electrons which acts like positive charges. The mobility μ of a charge carrier is defined as the drift velocity per unit electric field.

μ=

vd vd = E E

Mobility is positive for both electrons and holes, although their drift velocities are opposite to each other. The electric conductivity for a semiconductor containing electrons and holes as charge carriers can be expressed as

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 10



s = e ( ne μe + nh μ h )

where μ e and μ h are electron and hole mobilities and ne and nh are electron and hole concentrations (in charge carriers per unit volume) also called as electron and hole density. qt E  m qt v ⇒ μ= d = E m

⎧ ⎛ qE ⎞ ⎫ ⎨∵ vd = at = ⎜⎝ ⎟t⎬ m⎠ ⎭ ⎩

Also, vd =

Hence, μ e =

et e et and μ h = h me mh

t e and t h are relaxation time for electrons and holes, respectively. me and mh refer to mass of electron and holes respectively. Charge on either of the carriers is e. Illustration 8

The resistivity of sea water is about 25 W cm . The charge carriers are chiefly Na + and Cl - ions, and of each there are about 3 × 10 20 cm -3 . If we fill a plastic tube 2 m long with sea water and connect a 12 V ­battery to the electrodes at each end, what is the resulting average drift velocity of the ions, in cms -1 ? Solution

The current in a conductor of cross sectional area A is related to the drift speed vd of the charge carriers by I = neAvd where n is the number of charges per unit volume. We can then rewrite the Ohm’s Law as

⎛ rl ⎞ V = IR = ( neAvd ) ⎜ ⎟ = nevd ( rl ) ⎝ A⎠

⇒ vd =

V nerl

Substituting the values, we get vd =

( 6 × 1020

12 V cm

-3

) ( 1.6 × 10 -19 C )

( 25 W cm )( 200 cm )

⇒ vd = 2.5 × 10 -5

Vcm = 2.5 × 10 -5 cms -1 CW

9/20/2019 11:07:59 AM

3.11

Chapter 3: Electric Current and Circuits IllUSTRATION 9

Show that the total amount of charge at the junction 1 ⎞ ⎛ 1 of the two materials in figure is ε 0 I ⎜ , where ⎝ s 2 s 1 ⎟⎠ I is the current flowing through the junction, and s 1 and s 2 are the conductivities for the two materials. σ1 > σ2

I

σ1

E1 J

⇒

⎛s ⎞ E2 = ⎜ 1 ⎟ E1 ⎝ s2 ⎠

Let the charge on the interface be qenc , then from the Gauss’s Law, we have   q E ⋅ dA = ( E2 - E1 ) A = enc ε0

∫

Layer of positive charge J

E2

σ2

⇒ I

Charge at a junction

SOlUTION

In a steady state of current flow, the normal component of the current density J must be the same on both sides of the junction. Since J = s E , we have s 1E1 = s 2 E2

E2 - E1 =

qenc Aε 0

Substituting the expression for E2 from above then 1 ⎞ ⎛s ⎞ ⎛ 1 qenc = ε 0 AE1 ⎜ 1 - 1 ⎟ = ε 0 As 1E1 ⎜ - ⎟ ⎝ s 2 s1 ⎠ ⎝ s2 ⎠ Since the current is I = JA = ( s 1E1 ) A . So, the amount of charge on the interface becomes 1 ⎞ ⎛ 1 qenc = ε 0 I ⎜ ⎝ s 2 s 1 ⎟⎠

Test Your Concepts-II

Based on Resistance, Resistivity and Ohm’s law 1. A close analogy exists between the flow of energy by heat because of a temperature difference and the flow of electric charge because of a potential difference. The energy dQ and the electric charge dq can both be transported by free electrons in the conducting material. Consequently, a good electrical conductor is usually a good thermal conductor as well. Consider a thin conducting slab of thickness dx, area A, and electrical conductivity s, with a potential difference dV between opposite faces. dq Show that the current I = is given by the equadt tion on the left below: Charge Conduction

Thermal Conduction

dq dV = sA dt dx

dQ dT = kA dt dx

In the analogous thermal conduction equation on dQ the right, the rate of energy flow (in SI units of dt joules per second) is due to a temperature gradient

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 11

(Solutions on page H.199) dT , in a material of thermal conductivity k. State dx analogous rules relating the direction of the electric current to the change in potential, and relating the direction of energy flow to the change in temperature. 2. Material with uniform resistivity r is formed into a wedge as shown in figure. Show that the resistance between face A and face B of this wedge is Face A Face B y1

y2 L

R=r

w

⎛y ⎞ log e ⎜ 2 ⎟ w ( y2 - y1 ) ⎝ y1 ⎠ L

3. A piece of wire of uniform cross section has a resistance 8 W. If the length of the wire is doubled and its area of cross section is increased four times,

9/20/2019 11:08:07 AM

3.12  JEE Advanced Physics: Electrostatics and Current Electricity

calculate the new resistance. Neglect the temperature variation of resistance. 4. (a) If a copper wire is stretched to make it 0.1% longer, keeping volume constant, what is the percentage change in its resistance? (b) What is percentage change in its resistance if radius is increased by 1% and area is increased by 1%? 5. A steady current passes through a cylindrical conductor. Is there an electric field inside the conductor?

VARIATION OF RESISTANCE WITH TEMPERATURE The resistance of a conductor varies with temperature. The graph of variation of resistance of pure metal with temperature is shown. R

O

T

Mathematically the dependence of ( R ) on temperature ( T ) is expressed as

6. The region between two concentric conducting spheres with radii a and b is filled with a material with resistivity r. (a) Show that the resistance between the spheres is given by R=

r ⎛ 1 1⎞ ⎜ - ⎟ 4π ⎝ a b ⎠

  (b)  Find the current density as a function of radius, in terms of the potential difference Vab between the spheres. So, the temperature coefficient of resistance is given by

a=

R2 - R1 R1T2 - R2 T1

The resistivity of a material actually varies with temperature T . For metals, the variation is linear over a large range of T

rT = r0 ( 1 + a T ) where a is the temperature coefficient of resistivity. Typical values of r , s and a (at 20 °C ) for different types of materials are given in the table below. Material

RT = R0 ( 1 + a T + βT 2 ) …(1)

Resistivity Conductivity Temperature coefficient r (Wm) s (Wm)–1 a (°C)–1

where a and β are Temperature Coefficients of Resistance (TCR). The values of a and β vary from metal to metal. If temperature T is not sufficiently large (as in most practical cases), then equation (1) may be expressed as

Elements

Aluminium

2.82 × 10

RT = R0 ( 1 + a T ) …(2)

Tungsten Iron



The constant a is called the Temperature Coefficient of Resistance (TCR) of the material. a is positive for metals and negative for semiconductors and electrolytes. If R1 and R2 are the resistances of the same specimen at temperatures T1 °C and T2 °C , then

Silver

1.59 × 10-8

6.29 × 107

0.0038

Copper

1.72 × 10-8

5.81 × 107

0.0039

-8

7

3.55 × 10

0.0039

5.6 × 10-8

1.8 × 107

0.0045

10 × 10-8

1 × 107

0.0050

10.6 × 10-8

1 × 107

0.0039

Brass

7 × 10-8

1.4 × 107

0.002

Platinum Alloys



R1 = R0 ( 1 + a T1 ) …(3)

Manganin

44 × 10-8

0.23 × 107

1 × 10-5



R2 = R0 ( 1 + a T2 ) …(4)

Nichrome

100 × 10-8

0.1 × 107

0.0004

(Continued)

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 12

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Chapter 3: Electric Current and Circuits 3.13

Material

Resistivity Conductivity Temperature coefficient r (Wm) s (Wm)–1 a (°C)–1

Semiconductors 3.5 × 10-5

2.9 × 104

-0.0005

Germanium (pure)

0.46

2.2

-0.048

Silicon (pure)

640

1.6 × 10-3

-0.075

Carbon (graphite)

Insulators

Solution

Glass

1010 - 1014

10-14 - 10-10

Sulfur

1015

10-15

Quartz (fused)

75 × 1016

1.33 × 10-18

and r =

m ne 2t A m

ne 2t The number density of free electrons, n, is practically independent of temperature for most metals. However, an increase of temperature increases the amplitude of vibration of atoms and also the average speed of the free electrons. Therefore the average relaxation time t decreases and hence resistivity increases with temperature. If r0 and rT are the values of resistivity of a material at 0°C and T°C respectively then over a temperature range that is not too large, we have approximately,

(a) a = ⇒

1 dr r dT dr dT = a dT = - a r T

Let r = r0 at T = T0 , then

Variation of Resistivity of Metals with Temperature Since, R =

(a) Assume that a is not constant and is given by a a = - , where T is the absolute temperature T where a is a constant, show that the resistivity r b is given by r = n , where b is another constant. T (b) Using the values r = 3.5 × 10 -5 Wm and -1 a = -5 × 10 -4 ( C° ) for graphite at 293 K , determine a and b . (c) Using your result from part (b), determine the resistivity of graphite at -196 °C and 300 °C . (Remember to express T on absolute scale).

rT = r0 ( 1+ a T )

Illustration 10

The temperature coefficient of resistivity a is given ⎛ 1 ⎞ dr by a = ⎜ ⎟ , where r is the resistivity at temper⎝ r ⎠ dT ature T.

r

∫



r0

⇒

⎛T ⎞ ⎛ r ⎞ ⎛ T ⎞ log e ⎜ = - a log e ⎜ ⎟ = log e ⎜ 0 ⎟ ⎝ T ⎠ ⎝ r0 ⎟⎠ ⎝ T0 ⎠

⇒

r = ( r0 T0a )

T0

1 T

a

=

a

b Ta

Here, b = r0 T0a (b) Given that r0 = 3.5 × 10 -5 Wm and





a 0 = -5 × 10 -4 ( °C )

-1

at T0 = 293 K

a = -a 0 T0 

{

as a = -

a T

}

a ≈ 0.15 ⇒ 0.15 and b = r0 T0a = ( 3.5 × 10 -5 ) ( 293 ) ⇒ b = 8 × 10 -5 W mK 0.15  {be careful while taking units of a and b } (c) Now at T = -196 + 273 = 77 K ,

r=

b Ta

=

8 × 10 -5

( 77 )0.15

≈ 4.3 × 10 -5 Wm

  Similarly at T = 300 + 273 = 573 K  

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 13

∫

T

dr dT = -a r T

r=

8 × 10 -5

( 573 )0.15

≈ 3 × 10 -5 Wm

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3.14  JEE Advanced Physics: Electrostatics and Current Electricity

COLOUR CODE FOR CARBON RESISTANCES Carbon resistances find their wide use in electronic circuits at low voltages due to following reasons. (a) They may have values ranging from few ohms to 100 MW . (b) They are made up of small handy sizes. (c) They are quite cheap. A Colour Code is used to indicate the resistance and its percentage reliability. The carbon resistor has a set of concentric rings of various colours on it with their significance in the table. Colour

Number

Multiplier 0

% Tolerance

Black (B)

0

10 = 1

Gold 5%

Brown (B)

1

101

Silver 10%

Red (R)

2

102

No Colour 20%

Orange (O)

3

103

Yellow (Y)

4

104

Green (G)

5

105

Blue (B)

6

10

6

Violet (V)

7

107

Grey (G)

8

108

White (W)

9

109

The colour bands are formed from left to right. The first three bands ( A, B and C ) give the value of resistance. The colours of first and second bands indicate the first and second significant digits while the colour of third band gives decimal multiplier (i.e., the number of zeros which follow the first two digits). The colour of fourth band represents its tolerance. Absence of any colour means a tolerance of 20%.

The table may be memorised by the mnemonic BB ROY of Great Britain has a Very Good Wife. Capital letters represent the first letter of colour (from Darkest to the lightest colour). Also you may try to remember the following mnemonic Black Bears Roar, Orangutans Yell, Goats Bleat Violently, Go Weep. Another way of representation is the Dot-Body convention in which the body of the resistance is given one colour, the ends are given the other second same colour and a dot is marked over the body. A ring (silver or gold) on one side indicates the tolerance. The colour code table remains the same. Body Dot (Green) (Red)

End (Yellow)

Ring (Gold)

End (Yellow)

Colour of body describes the first significant digit. Identical end colours give the second significant figure. Colour of dot gives the decimal multiplier. The colour of the ring gives the tolerance. So, for the colours specified in the figure the resistance has a value

R = 54 × 10 2 W ± 5%

THERMISTOR A thermistor is a heat sensitive resistor usually made of semiconductor material. The temperature coefficient of a thermistor is negative but is unusually -1 large, of the order of -0.04 ( °C ) . The V -I curve of a thermistor is unusual and is shown in Figure. I

ABC D R = AB × 10c ± Tolerance (in %)

e.g. if A is red, B is green, C is grey and there is no colour at D, then

V

R = 25 × 108 W ± 20%

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 14

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Chapter 3: Electric Current and Circuits

Uses of Thermistor (a) They are used for resistance thermometers in very low temperature measurement of the order of 10 K .

3.15

(b) They may be used to safeguard electronic circuits against current jumps because initially (when cold) thermistor has high resistance and its resistance drops appreciably when it warms up.

Test Your Concepts-III

Based on Variation of Resistance with Temperature (Solutions on page H.200) 1. When the temperature of a heating element changes, its resistance also changes and so does the temperature coefficient. Suppose the temperature varies linearly with time and is given by T ( °C ) = ( 20 + 10t ) where t is time in second. The temperature coefficient of the material is -1 0.0065 ( °C ) at 0°C. If the initial resistance of the heating element is 2r, then express the resistance as a function of time. 2. Obtain a general relationship between a0 and aT, the respective temperature coefficients at 0°C and T°C. 3. Obtain a general relationship between a1 and a2, the respective temperature coefficients at T1°C and T2°C. 4. Obtain the expression for the ratio of resistances of a coil at two temperatures T1 and T2 assuming that the only other given quantity is the temperature coefficient a0 at 0°C. 5. It has been experimentally found that the resistivity of conducting materials such as copper and aluminium varies linearly with temperature. Obtain a relation between resistivities of a metal r1 and r2 at two temperatures T1 and T2 respectively.

ElECTRICAl ENERGY AND POWER Consider a circuit consisting of a battery and a resistor with resistance R . b ΔV a R

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 15

I

ρ

Metal 2 (Aluminium) Metal 1 (Copper)

ρ2 ρ1

T1 T2

T

6. When a metal rod is heated, not only its resistance but also its length and area of cross-section changes. Find the percent change in R, l and A of a copper wire for a temperature rise of 1°C. Coefficient of linear expansion for copper is -1 1.7 × 10 -5 ( °C ) and its thermal coefficient of -1 resistance is 3.9 × 10 -3 ( °C ) .

7. Two coils connected in series have resistance of 600 W and 300 W and temperature co-efficient -1 -1 of 0.001 ( °C ) and 0.004 ( °C ) respectively at 20°C. Find resistance of the combination at a temperature of 50°C. What is the effective temperature co-efficient of combination?

Let the potential difference between two points a and b be DV = Vb - Va > 0 . If a charge Dq is moved from a through the battery, its electric potential energy is increased by DU = Dq DV . On the other hand, as the charge moves across the resistor, the potential energy is decreased due to collisions with atoms in the resistor. If we neglect the internal resistance of the battery and the connecting wires, upon returning to a the potential energy of Dq remains unchanged. Thus, the rate of energy loss through the resistor is given by

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3.16  JEE Advanced Physics: Electrostatics and Current Electricity



P=

like road maps. Some frequently used symbols are shown here for the purpose of convenience.

DU ⎛ Dq ⎞ =⎜ ⎟ DV = I DV Dt ⎝ Dt ⎠

This is precisely the power supplied by the battery. Using DV = IR , the above equation can be rewritten as

P = I 2R =

( DV )2

Introduction to Electrical Circuits Electrical circuits connect power supplies to loads such as resistors, motors, heaters, or lamps.

R2 +

R2

R1 +

–

–

(a)

(b)

Elements connected (a) in parallel and (b) in series

The connection between the supply and the load is made with the help of wires (that are often called leads), or with many kinds of connectors and ­terminals. Energy is delivered from the source to the user on demand at the flick of a switch. Sometimes many circuit elements are connected to the same lead, which is called a common lead for those elements. Various parts of the circuits are called circuit elements, which can be in series or in parallel, as we have already seen in the case of capacitors. Elements are said to be in parallel when they are connected across the same potential difference. Generally, loads are connected in parallel across the power supply. On the other hand, when the elements are connected one after another, so that the current passes through each element without any branches, the elements are in series. There are pictorial diagrams that show wires and components roughly as they appear, and s­ chematic diagrams that use conventional symbols, somewhat

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 16

+ –

Resistor Switch

R

R1

Voltage source

Often, in circuits we observe a switch to be connected in series. When the switch is open the load is disconnected and when the switch is closed, the load is connected. One can have closed circuits, through which current flows, or open circuits in which there are no currents. Usually by accident, wires may touch, causing a short circuit. Most of the current flows through the short, very little will flow through the load. This may burn out a piece of electrical equipment such as a transformer. To prevent damage, a fuse or circuit breaker is put in series. When there is a short the fuse blows, or the breaker opens. In electrical circuits, a point (or some common lead) is chosen as the ground. This point is assigned an arbitrary voltage, usually zero, and the voltage V at any point in the circuit is defined as the voltage difference between that point and ground.

Active and Passive Elements in Electric Circuits An electric circuit is an arrangement of active and passive elements, having one or more closed paths for electric current to flow. It is due to the active elements (such as cells, batteries, dynamos, generators, etc.) that the current is forced to flow in a closed path. These supply energy to the circuit. The active element in a dc circuit is a cell or a battery. Its circuit symbol is + – E, r

On the contrary, the passive elements either consume or store electric energy. There are three basic passive elements.

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Chapter 3: Electric Current and Circuits 3.17

1 Passive element (property)

2

3

Resistor Capacitor Inductor (Resistance) (Capacitance) (Inductance)

Symbol

R

C

L

Units

W

F

H

Circuit Symbol

or

V-I Relationship Energy consumed/ Stored

V = IR I 2 Rt

I=C

dV dt

1 CV 2 2

Opposes Characteristic Converts variations in property electric energy into voltage heat

V=L

+

A simple circuit consisting of a battery and a resistor

dI dt

1 2 LI 2 Opposes variations in current

Assuming that the battery has no internal resistance, the potential difference DV (or terminal voltage) between the positive and the negative terminals of the battery is equal to the emf E . To drive the current around the circuit, the battery undergoes a discharging process which converts chemical energy to emf. The current I is found by noting that no work is done in moving a charge q around a closed loop due to the conservative nature of the electrostatic force. So,

W=q

We are aware of the fact that energy has to be supplied to maintain a constant current in a closed circuit. The source of energy is commonly referred to as the electromotive force, or the emf (symbol E ). The ability of a cell or a battery to make the current flow in the circuit is measured in terms of its electromotive force or emf. Batteries, solar cells and thermocouples are some examples of emf source. They can be thought of as a “charge pump” that moves charges from lower potential to the higher potential. Mathematically emf is defined as E=

W …(1) q

So, emf is the work done to move a unit positive charge in the direction of the applied external filed or from higher potential to the lower potential. SI unit for E is volt ( V ) . Consider a simple circuit consisting of a battery as the emf source and a resistor of resistance R , as shown in figure.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 17





∫ E ⋅ dl = 0 …(2)

Consider the loop abcda and the point a in figure be the point of start.

ELECTROMOTIVE FORCE (EMF)



–

c

R

I

E b

+ –

d

a I

When crossing from the negative to the positive terminal, the potential increases by E . On the other hand, as we cross the resistor, the potential decreases by an amount IR , and the potential energy is converted into thermal energy in the resistor. Assuming that the connecting wire carries no resistance so, upon completing the loop the net change in potential difference must be zero. Hence, we must have

E - IR = 0 …(3)

⇒ I=

E …(4) R

However in reality a battery always carries an internal resistance r and the potential difference across the

9/20/2019 11:08:46 AM

3.18  JEE Advanced Physics: Electrostatics and Current Electricity V I

e

d

R

I E a

b

+ –

E

ε IR

r

R

Ir

r c

e a b c d (b) (a) (a) Circuit with an emf source having an internal resistance r and a resistor of resistance R. (b) Change in electric potential around the circuit.

battery terminals, also called as the Terminal Potential Difference (TPD), becomes

Series Combination The two resistors R1 and R2 in figure are connected in series to a voltage source DV . By current conservation, the same current I is flowing through each resistor. The total voltage drop from a to c across both elements is the sum of the voltage drops across the individual resistors DV = IR1 + IR2 = I ( R1 + R2 )



a

R1

b

R2

a

c

I

Req

c

DV = E - Ir …(5)

Since there is no net change in potential difference around a closed loop abcdea , we have

RESISTORS IN SERIES AND IN PARALLEL

I ΔV

E - Ir - IR = 0 …(6)

+

(a) Figure (b) depicts the change in electric potential as we traverse the circuit clockwise. From the ­figure, we see that the highest voltage is immediately after the battery. The voltage drops as each resistor is crossed. Note that the voltage is essentially constant along the wires. This is because the wires have a negligibly small resistance compared to the resistors. (b) For a source with emf E, the power or the rate at which energy is delivered is P = IE = I ( IR + Ir ) = I2R + I2 r That the power of the source emf is equal to the sum of the power dissipated in both the internal and load resistance is required by energy conservation. (c) V (potential difference across the terminals of the cell) may be less than E, equal to E or greater than E. (i)  V < E, when the circuit draws current I from the cell and in that case V = E - Ir . (ii)  V = E, when the circuit draws no current from the cell i.e. when I = 0. (iii) V > E, when the cell is being charged and in that case V = E + Ir .

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 18

ΔV

–

+

(a) Resistors in series

E ⇒ I= …(7) R+r

Remark(s)

I –

(b) Equivalent circuit

The two resistors in series can be replaced by one equivalent resistor Req (figure (b)) with the identical voltage drop DV = IReq which implies that Req = R1 + R2 The above argument can be extended to N resistors placed in series. The equivalent resistance is just the sum of the original resistances, N

Req = R1 + R2 + .... =

∑R

i

(a) Notice that if one resistance R1 is much larger than the other resistances Ri , then the equivalent resistance Req is approximately equal to the largest resistor R1. (b) The current flowing in each resistor is same. (c) The total potential difference V across the combination is equal to the sum of the potential difference across individual resistances i.e., i =1

V = V1 + V2 + V3 such that V1 : V2 : V3 ≡ R1 : R2 : R3 (d) The equivalent or effective resistance ( RS ) of the combination is equal to the sum of individual resistances i.e.,

RS = R1 + R2 + R3

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Chapter 3: Electric Current and Circuits 3.19

Parallel Combination

(e) Net Conductance (G) is 1 1 1 1 = + + GS G1 G2 G3

(f) Net Resistance Rs has a value greater than the largest resistance of the combination. ⎛R ⎞ (g) V1 = ⎜ 1 ⎟ V ⎝ RS ⎠



⎛R ⎞ V2 = ⎜ 2 ⎟ V ⎝ RS ⎠ ⎛R ⎞ V3 = ⎜ 3 ⎟ V ⎝ RS ⎠

where RS = R1 + R2 + R3

Next let’s consider two resistors R1 and R2 that are connected in parallel across a voltage source DV . By current conservation, the current I that passes through the voltage source must divide into a current I1 that passes through resistor R1 and a current I 2 that passes through resistor R2 . Each resistor individually satisfies Ohm’s Law, DV1 = I1 R1 and DV2 = I 2 R2 . However, the potential across the resistors are the same, DV1 = DV2 = DV . Current conservation then implies I = I1 + I 2 =



I1

R1

I2

R2

I

ΔV

ΔV

+ – (a) Two resistors in parallel

+ –

Voltage Divider Circuit Consider a voltage source Vin that is connected in series to two resistors, R1 and R2 .

a

R1 Vin R2

Vout

Voltage divider

The voltage difference, Vout , across resistor R2 will be less than Vin . This circuit is called a voltage divider. From the loop rule, Vin - IR1 - IR2 = 0 So the current in the circuit is given by Vin I= R1 + R2

Thus the voltage difference, Vout , across resistor R2 is given by Vout = IR2 =

R2 Vin R1 + R2

Note that the ratio of the voltages characterizes the voltage divider and is determined by the resistors

Vout R2 = R1 + R2 Vin

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 19

DV DV 1 ⎞ ⎛ 1 + = DV ⎜ + R1 R2 ⎝ R1 R2 ⎟⎠ Req

I b

(b) Equivalent resistance

The two resistors in parallel can be replaced by one equivalent resistor Req with DV = IReq . Comparing these results, the equivalent resistance for two resistors that are connected in parallel is given by 1 1 1 = + Req R1 R2

This result easily generalizes to N resistors c­ onnected in parallel



1 1 1 1 = + + + .... = Req R1 R2 R3

N

∑R

1

i =1

i

(a) The potential difference ( V ) across each resistor is the same. (b) The current is different in different resistances such that the total current flowing in the combination is shared by the individual resistances in the inverse ratio of their resistances i.e.,

I = I1 + I 2 + I 3

such that I1 : I 2 : I 3 ≡

1 1 1 : : R1 R2 R3

9/20/2019 11:09:10 AM

3.20  JEE Advanced Physics: Electrostatics and Current Electricity

(c) The equivalent or effective resistance (RP) of the combination is given by 1 1 1 1 = + + RP R1 R2 R3

Current Divider Consider two resistors connected in parallel as shown. From figure, we have

(d) Effective conductance GP is the sum of conductance of individual resistors i.e., (e) Net Resistance RP has a value smaller than the least resistance of the combination. ⎛ ⎞ ⎛ (f) I1 = ⎜ ⎟⎠ I , I 2 = ⎜⎝ ⎝ 1 1 1 1 = + + RP R1 R2 R3

RP R2

⎞ ⎛ ⎟⎠ I and I 3 = ⎜⎝

RP R3

⎞ ⎟⎠ I where

(g) When one resistance R1 is much smaller than the other resistances, then the equivalent resistance Req is approximately equal to the smallest resistor R1 . In the case of two resistors, RR RR = 1 2 ≈ 1 2 = R1 R1 + R2 R2

Req This means that almost all of the current that enters the node point will pass through the branch containing the smallest resistance. So, when a short develops across a circuit, all of the current passes through this path of nearly zero resistance. Hence SHORT CIRCUITING a branch implies taking the resistance of that branch to be zero or close to zero. Shown here is an electric circuit whose two points are directly connected by a conducting wire. R1

R2 E

Here we say that the resistance R1 in the circuit is short circuited. Under such condition, both points are at same potential and hence the potential difference across R1 is zero. Hence, no current will flow through R1 and the current through R2 E is I = . R2

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 20

R1

I2

R2

I

GP = G1 + G2 + G3

RP R1

I1



I = I1 + I 2 …(1)



I1 R1 = I 2 R2 …(2)

On solving equations (1) and (2), we get

⎛ R2 ⎞ ⎛ R1 ⎞ I and I 2 = ⎜ I I1 = ⎜ ⎟ ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎟⎠

So, we observe that the division of current in the branches of a parallel circuit is inversely proportional to their resistances. Similarly, consider three resistors connected in parallel as shown.

I

I1

R1

I2

R2

I3

R3

I

From figure, we have

I = I1 + I 2 + I 3



I1 R1 = I 2 R2 = I 3 R3

and Req =

R1 R2 R3 R1 R2 + R2 R3 + R3 R1

R2 R3 ⎤ ⎡ I1 = I ⎢ ⎥ R R R R R R + + 2 3 3 1 ⎦ ⎣ 1 2 1 ⎛ ⎞ ⎛ 1 ⎜ ⎜ R1 ⎟ R1 ⇒ I1 = ⎜ ⎟I=⎜ ⎜ 1 + 1 + 1 ⎟ ⎜ 1 ⎜⎝ R R ⎟ R3 ⎠ ⎝ RP 1 2

⎞ ⎟ ⎟I ⎟ ⎠

9/20/2019 11:09:19 AM

Chapter 3: Electric Current and Circuits 3.21

R1 R3 ⎤ ⎡ I2 = I ⎢ ⎥ R R R R R R + + 2 3 3 1 ⎦ ⎣ 1 2 1 ⎛ ⎞ ⎛ 1 ⎜ ⎜ R2 ⎟ R2 ⇒ I2 = ⎜ ⎟I=⎜ ⎜ 1 + 1 + 1 ⎟ ⎜ 1 ⎜⎝ R R ⎟⎠ R ⎝ RP 1 2 3 R1 R2 ⎡ ⎤ I3 = I ⎢ ⎥ R R R R R R + + 2 3 3 1 ⎦ ⎣ 1 2 1 ⎛ 1 ⎛ ⎞ ⎜ R3 ⎜ ⎟ R3 ⇒ I3 = ⎜ ⎟I=⎜ ⎜ 1 ⎜ 1 + 1 + 1 ⎟ ⎜⎝ R ⎜⎝ R R ⎟⎠ R 1 2 3 P

Solution

In this problem, our aim is to calculate only the current through 6 W resistor, we can do it easily by using current divider concept.

⎞ ⎟ ⎟I ⎟ ⎠

7Ω Ix

⎞ ⎟ ⎟I ⎟ ⎟⎠

RA IB RB

Illustration 11

In the given network, calculate the current through 6 W resistor. 7Ω 12 Ω

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 21

6Ω

60 V

6Ω

6Ω

(b)

12 Ω

3Ω

60 V

   

(c)

We first calculate the current I supplied by the ­ attery. For this we rearrange the network and simb plify it step by step as follows: (a) We interchange the position of 6 W and 12 W resistors. This will not make any difference, as the two resistors are in parallel [Fig. (a)]. (b) We then combine the two 12 W resistors to give a single resistor of 6 W [Fig. (b)] (c) We again combine the two 6 W resistors to give a single resistor of 3 W [Fig. (c)]. (d) The current I is clearly given as I =

RD RC

Remember, parallel resistor must be the resistors ­connected between the same pair of terminals.

60 V

7Ω I

Ix

RB is wrong. RA + RB

IS

12 Ω

I1

7Ω I

Be careful in applying current division. Without parallel resistors, there is no way that the current division can be applied. To say

IA

12 Ω

(a)

Word of Advice



6Ω

60 V

It is easy to remember the expressions for I1 , I 2 and I 3 just by noticing which resistance is missing in the term at the numerator.

IA = IS

I1

60 =6A 7+3

(e) Using Fig. (b), and the current divider concept, we get the current through 6 W resistor as ⎛ 6 ⎞ =3A Ix = I ⎜ ⎝ 6 + 6 ⎟⎠   

NETWORK ANALYSIS (a) The first step in analyzing a circuit is to simplify it to simplest equivalent configuration. In ­figure, each of the circuits has the same equivalent resistance between A and B , in each version, R1 , R2 and R3 are in parallel.

9/20/2019 11:09:28 AM

3.22  JEE Advanced Physics: Electrostatics and Current Electricity

R3

R3

R2

R2

R2

R3 A

R1

R1

R1

B

(a)

   

A

B

(b)

   

R3

R1

R2

R2

A

(c)

B

(d)

(e)

  

(d) If the same current passes through every resistor in a given branch, respective of the presence of sources in that branch, the resistors are in series even though they are not directly connected to each other. Same is true about capacitors.

R3

R1 A

B

(d)

   

A

(e)

–

B

R1

A

+ – + – V1

+ R2

(b) All the resistors in figure (a) are in parallel arrangement. All the resistors in figure (b) are in series arrangement.

– – + V2

+

I + – V3

R3

– R4 + B

(a) I –

V1 – V2 – V3

+

A + – R1 + R2 + R3 + R4

B

(b) (a)

(b)

  

(c) Figure shows several circuits in which the circuit elements are neither in series nor in parallel.

+ –

+ (a)

– –

+

+

–

EARTHING OR GROUNDING IN AN ELECTRICAL CIRCUIT If a certain point in an electrical circuit is earthed, its potential is taken as zero. Consider the following circuit. When only one point is earthed, then there is no effect on current flowing in the circuit. A

I E = 10 V

I

I

(b)

   

1Ω

I

I

B (c)

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 22

E

C I

3Ω

I

6Ω

I

D

Earthing

9/20/2019 11:09:32 AM

Chapter 3: Electric Current and Circuits 3.23

Using KVL in ABCDEA,

Applying KJL to closed loop ABCDEFA, we get

⇒ I =1A

-4 I + 0 - 5I - ( 1 ) I + 10 = 0 ⇒ I =1A

Now, VC = 0 V (because C is grounded)

Now, VA - VB = 4 I = 4 V

Using KVL , VB - 3 I = VC

⇒ VA = 4 V



-3 I - 6 I - I + 10 = 0

Also, VC - VD = 5I

⇒ VB = VC + 3 I

⇒ VD = -5 V

⇒ VB = 3 V Similarly VC - 6 I = VD

DELTA TO STAR OR DELTA-STAR TRANSFORMATION

⇒ VD = -6 V Hence, earthing point gives us the reference potential i.e., zero potential and potentials of different points can then be calculated with respect to this new reference. Illustration 12

Suppose we are given three resistances R12 , R23 and R13 connected in delta fashion between terminals 1, 2 and 3 as in figure (a). So far as the respective terminals are concerned, these three given resistances can be replaced by the three resistances R1 , R2 and R3 connected in star as shown in figure (b). 1

In the electrical circuit shown, points B and C are earthed. Find the potentials of points A and D . 10 V

R13

1Ω

3 4Ω

10 Ω B

A

R1

1 R12 R23

2

C



VB = VC = 0

R1 =

R12 R13 R12 + R23 + R13

⇒ VB - VC = 0



R2 =

R23 R12 R12 + R23 + R13

⇒ Current through 10 W is zero

and R3 =

R13 R23 R12 + R23 + R13

10 V

1Ω

I

E I

I A

B 10 Ω C 4Ω I

I

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 23

I

5Ω

I

R2 2

(b)

These two arrangements will be electrically equivalent because resistance as measured between any pair of terminals is the same in both the arrangements.

D

Since B and C both are earthed, so

I

N

3

(a)

5Ω

Solution

F

R3

D

Problem Solving Technique(s) How to Remember (Delta to Star)? Resistance of each arm of the star is given by the product of the resistances of the two delta sides that meet at its end divided by the sum of the three delta resistances.

9/20/2019 11:09:46 AM

3.24  JEE Advanced Physics: Electrostatics and Current Electricity

R1

4Ω

A

C

1 R13 3 R3



16/9 Ω

R12 R23

12/9 Ω

2 R2

R12R13 R12 + R23 + R13

R2 =

R23R12 and R12 + R23 + R13

R3 =

R13R23 R12 + R23 + R13

35/9

4Ω B

Suppose we are given three resistances R1, R2 and R3 connected in star fashion between terminals 1, 2 and 3 as shown in figure (a). So far as the respective terminals are concerned, these three resistances can now be replaced by the three resistances R12, R23 and R13 connected in the delta network as shown in figure (b). 1 1

R1

4Ω C

D

R13 6Ω

8Ω

R3

8Ω

N

R2

3

E 4Ω

2

3

(a)

R12 R23

2

(b)

These two arrangements will be electrically equivalent because resistance as measured between any pair of terminals is the same in both the arrangements.

B

Solution

For finding RAB , we will convert the delta CDE of figure into its equivalent star as shown in figure.



E

STAR TO DELTA OR STAR-DELTA TRANSFORMATION

Find the input resistance of the circuit between the points A and B of figure.



16/9

B

4Ω

C

24/9 Ω

D

Illustration 13

A

S

8Ω

R1 =

4

A

RCS = 8 ×

4 16 6 24 = W ; RES = 8 × = W 18 9 18 9

RDS = 6 ×

4 12 = W 18 9

R12 = R1 + R2 +

R1 R2 R3

R23 = R2 + R3 +

R2 R3 R1

and R13 = R1 + R3 +

R1 R3 R2



The two parallel resistances between S and B can be reduced to a single resistance of As seen from figure

RAB

35 W 9

⎛ 16 ⎞ ⎛ 35 ⎞ 87 = 4+⎜ ⎟ +⎜ ⎟ = W ⎝ 9 ⎠ ⎝ 9 ⎠ 9

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 24

Problem Solving Technique(s) How to Remember (Star to Delta)? The equivalent delta resistance between any two terminals is given by the sum of star resistances between those terminals plus the product of these two star resistances divided by the third star resistance.

9/20/2019 11:09:52 AM

Chapter 3: Electric Current and Circuits 3.25

When combined together, the final circuit is as shown in figure (c).

R1 1 R13 R23

18 Ω

R23 = R2 + R3 +

R2R3 and R1

R13 = R1 + R3 +

R1R3 R2

Ω

R1R2 , R3

Ω .5



R12 = R1 + R2 +

13



R2

1.5

R3

2

9Ω

3



A

R12

9Ω B

C 1Ω (b) A

6Ω

27/20 Ω

Illustration 14

A network of resistances is formed as follows: A

B

9/10 Ω

C

9Ω

6Ω

(c)

D

4Ω B

(a) As seen, there are two parallel paths across points A and B .

1.5 Ω

1Ω

3Ω

(i) First, directly from A to B having a resistance of 6 W , and C

(a)

AB = 9 W ; BC = 1 W ; CA = 1.5 W forming a delta and AD = 6 W , BD = 4 W and CD = 3 W forming a star. Compute the network resistance measured between (a) A and B , (b) B and C , and (c) C and A .

(ii) the other via C having a total resistance of ⎛ 27 9 ⎞ + ⎟ = 2.25 W ⎜⎝ 20 10 ⎠ 6 × 2.25 18 ⇒   RAB = ( 6 + 2.25 ) = 11 W

(b) RBC

9 ⎛ 27 ⎞ ×⎜6+ ⎟ 10 ⎝ 20 ⎠ 441 W = = 27 ⎞ 550 ⎛ 9 +6+ ⎜⎝ ⎟⎠ 10 10

(c) RCA

27 ⎛ 9 ⎞ ×⎜6+ ⎟ 20 ⎝ 10 ⎠ 621 W = = 27 ⎞ 550 ⎛ 9 +6+ ⎜⎝ ⎟⎠ 10 20

Solution

The star of figure (a) may be converted into the equivalent delta and combined in parallel with the given delta ABC. The three equivalent delta resistances of the given star become as shown in figure (b).

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 25

9/20/2019 11:10:01 AM

3.26 JEE Advanced Physics: Electrostatics and Current Electricity

Test Your Concepts-IV

Based on Series and Parallel Combination of Resistances (Solutions on page H.201) 1. Four copper wires of equal length are connected in series. Their cross-sectional areas are 1 cm2, 2 cm2, 3 cm2, and 5 cm2. A potential difference of 120 V is applied across the combination. Determine the voltage across the wire having the area 2 cm2. 2. Consider the circuit shown in figure. Find

1 5Ω 2 200 V

25 Ω

25 V a

10 Ω

3 b

10 Ω 5Ω

10 Ω

20 Ω

5Ω

6. In the circuit shown, calculate the net resistance between the points A and B. 2Ω

A B

(a) the current in the 20 W resistor and (b) the potential difference between points a and b. 3. A 6 V battery supplies current to the circuit shown in figure. When the double-throw switch S is open, as shown in the figure, the current in the battery is 1 mA. When the switch is closed in position 1, it is observed that the current in the battery is 1.2 mA and when the switch is closed in position 2, it is observed that the current in the battery is 2 mA. Find the resistances R1, R2 and R3.

8Ω

15 Ω

20 Ω 30 Ω

10 Ω 40 Ω

7. In the circuit shown, find the current supplied by the battery. 8Ω

24 V

R2

R1

12 Ω

R2 1

6V

2

S

R3

4. Two resistors connected in series have an equivalent resistance of 690 W. When they are connected in parallel, their equivalent resistance is 150 W. Find the resistance of each resistor. 5. In the circuit shown, find the magnitude and polarity of the voltages mentioned below: (i) V1, (iv) V3- 2 ,

10 Ω

(ii) V2 ,

(iii) V3 ,

(v) V1- 2 ,

(vi) V1- 3 .

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 26

8. Find the current through the battery and the equivalent resistance of the network shown. 8Ω

24 V

4Ω

6Ω

12 Ω

9. Find the total equivalent circuit resistance and current through the battery in the circuit shown.

9/20/2019 11:10:05 AM

Chapter 3: Electric Current and Circuits 3.27

8Ω

A Rc

6Ω

24 V

Ra

D

10. Compute the value of battery current I shown in figure. 6Ω 4Ω 4Ω

12 V

C

B

12 Ω

I

Rb Rd

13. Resistances R1 and R2, each 60 W, are connected in series. The potential difference between points A and B is 120 V. Find the reading of a voltmeter connected to points C and D if its internal resistance is r = 120 W. A

8Ω

B

2Ω R1

11. Find the equivalent resistance between A and B in the following circuits:

C

4Ω

Ω

6

2Ω

A

Ω

B

14. Show by diagram, how can we use a rheostat as the potential divider? 15. In the figure shown, determine the current through each 5 W resistor and the battery.

4Ω (a)

A 10 Ω 6Ω

A 4Ω

5Ω

5Ω

10 Ω

10 Ω B

2Ω (b)

12. Three wires having resistance Ra, Rb and Rc are joined in the form of a triangle as shown. Another wire is joined at A and it makes a free sliding contact with BC. The resistance of this wire is Rd. Find the maximum resistance of the circuit (assume all wires to be uniform). The arc BC is a part of a circle whose centre is at A.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 27

10 Ω

20 V

8Ω B

D

V

4Ω 8

R2

16. In figure shown we wish to determine current in one of the 4 W resistors in the circuit. 4Ω 12 V

4Ω

4Ω

4Ω

4Ω

4Ω 4Ω

4Ω

9/20/2019 11:10:07 AM

3.28  JEE Advanced Physics: Electrostatics and Current Electricity

SHORT AND OPEN CIRCUITS When two points of a circuit are connected together by a conducting wire, they are said to be short ­circuited. The connecting wire is assumed to have zero ­resistance. No voltage can exist across a short, secondly current through it is very large (­theoretically infinity).

Short circuit across R3 shorts R1 and R2 as well, short across one branch in parallel means short across all branches. There is no current in shorted resistors. The shorted components are not damaged, they will function normally when short circuit is removed. A R1

V

Remaining Short circuit circuit

VAB = 0 Remaining Open I = 0 AB ISC circuit circuit RAB → ∞ ISC = IAB → ∞

Two points are said to be open circuited when there is not direct connection between them, a break in the continuity of circuit exists. Due to this break the resistance between the two points is infinite and there is no flow of current between the two points. Please note that the entire applied voltage is felt across the open, i.e., across terminals A and B , so, VAB = 100 V . R1 A A Open V 100 V circuit B VAB = 100 V

100 V = V

Short circuit R2 R3

R1

C

B

B

In figure shown, short circuit across R3 may short out R2 but not R1 , because it is protected by R4 . R4 R1

D

In figure shown, if a battery is connected between points A and B , emf E = 18 V , what is current through it? 18 Ω

4/5 Ω 3 Ω 1/5 Ω C

V V ⇒ I= = and VB = VD Req R1 Bulb B1 acts as open circuit, bulb B1 will not glow. However, other bulb B2 remains connected across the voltage supply, it will operate normally.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 28

B2

9

A

Req = R1

Open filament

R3 Short circuit

R2

Illustration 15

V

B1

B

C

In the circuit shown if B and D are shorted, then

220 V

Short circuit

D

R2



R3

R2

B

Ω

3Ω 6Ω

E

Solution

The resistors at C and D are redundant, they are open branches, therefore no current through them. The point E is grounded, but no current will leak to ground, no return path for the current. Points a and d are at the same potential, similarly points b and c are at the same potential, because these points are connected by conducting wires without any circuit element.

9/20/2019 11:10:15 AM

Chapter 3: Electric Current and Circuits 3.29

d

18 Ω

3Ω

c

Flip A

B 9/

4/5 Ω 3 Ω

4 Ω

a

6Ω

3Ω

A

Notice the 6 W and 3 W resistors, similarly 18 W , 9 W resistors, these pairs are a parallel arrangement. 4

2Ω

3Ω

4/5 Ω

B

(Parallel of 3 Ω and 6 Ω)

Thus the equivalent resistance of the circuit between A and B is 1.8 W . The current supplied by battery.

10 V

Find the equivalent resistance across AB of the circuit given. 5Ω

10 Ω

3Ω

20 Ω

5Ω

Solution

The circuit is redrawn to make it easily evaluable. 5Ω

A

B

10 V

5Ω C

10 Ω

1Ω

3Ω

5Ω

A B

Wires are not in contact

A

C

25 Ω

Illustration 16

In figure shown, determine the current provided by the battery.

10 V

Illustration 17

⎛ 18 ⎞ I=⎜ A = 10 A ⎝ 1.8 ⎟⎠

2Ω

10 V

B

6Ω

2Ω



1Ω 3Ω

10 V

2Ω A

2Ω

1Ω

2Ω

b

3Ω

Wires are not in contact

25 Ω

5Ω 20 Ω

B

The left block is equivalent to 15 W and 25 W in ­ arallel, and its resistance is given as p R1 =

25 × 15 = 9.375 W 25 + 15

Solution



Just flip the circuit about the dotted line shown in the figure. Now the circuit is simple with equivalent ⎛ 10 ⎞ resistance 2 W and current I = ⎜ ⎟ A = 5 A . ⎝ 2 ⎠

The right block is equivalent to 10 W and 20 W in parallel, and its resistance is given as

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 29



R2 =

10 × 20 200 = = 6.667 W 10 + 20 30

9/20/2019 11:10:22 AM

3.30  JEE Advanced Physics: Electrostatics and Current Electricity

The circuit now reduces to two resistors in series, as shown. 9.375 Ω A

6.667 Ω C

R1

B

R2

Hence, the total resistance,

R = R1 + R2 = 9.375 + 6.667 = 16.042 W

The current flow is not a mirror image in branches ab and bc because the flow is in same direction. This is called asymmetric condition. The special thing about this asymmetry is that current incoming at b is equal to outgoing current, similar situation exists at b and d also. Thus resistors in branches be and de are ineffective. In the following figure there is a symmetry along the xy. X

PRINCIPLE OF SYMMETRY

R

If a network is symmetrical on both sides of a line, all the points lying on the line will have same potential. Hence, no current will flow in a resistance connected between two such points. Such resistances can be ignored while calculating the equivalent resistance. This simplifies the network. Or, else all the points on this line may be treated as shorted.

SYMMETRICAL CIRCUITS When a circuit is symmetrical about a line (By symmetry we mean that two parts are mirror images of each others), then the potential and current must also be symmetrical. b R A

a R

R

R

c R

I3

B

B

R

R Y (a)

The current I2 reaching O is equal to outgoing ­current, which simply means there is no mingling of current from upper branch and lower branch into middle branch and hence we can remove the connection of the respective upper and lower branches as shown.

O

I2

I1

I3

I3

(b)

Therefore, currents in ab and ad are same. Currents in dc and bc are same. Potentials of the points b , e and d are same. The equivalent circuit is redrawn, 2 the equivalent resistance is R . Note that there is no 3 current in branches be and ef . Another symmetry is visible along line bd . R

b

R

R

e

R

R

f

R

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 30

I1

R

(a)

(b)

R I3

R

R

d

A

R I 2 O R

R

I I1 I2

R

R e

A

R

R I1 I R 2

The resulting circuit is simple enough and the 4R . ­equivalent resistance is 5 Illustration 18

Find the current in the branch AB of the circuit. Also find the current in the branches BF and EA . 12 V F

A

2Ω B

C

6Ω G

4Ω D

4Ω

6Ω B

2Ω E

12 V

9/20/2019 11:10:29 AM

Chapter 3: Electric Current and Circuits 3.31 Solution

Referring to figure we see that symmetry demands that current only circulates in outer branch. Points A and B are at the same potential because the circuit is symmetrical. Therefore, no current can go across the resistors in that branch. The current through BF and EA is 2 A.

⇒ Req =

R R 2R + = 3 3 3

Illustration 20

Calculate the effective resistance between points A and C , by applying symmetry principle.

Illustration 19

B

Calculate effective resistance between points A and C for the networks shown. A R

B

R

R

R

R

R

R

R

D

C

A

R

O

R

R

R

R D

C

R

Solution

Solution

The two sides of the line BOD , are symmetrical. Thus, if the points A and C are connected to a source of emf, points B , O and D of the network will have same potential. Therefore, no current flows through RBO and RDO . These resistances can be ignored. The network then reduces to that shown in figure.

AE

B

R

A

Break the branch AC into two resistors in series, each R , and consider the dotted line passing through B , 2 D and E . The network on the two sides of this line is symmetrical. Hence, one can short-circuit the points B , D and E and calculate ( Req ) .

B

R R

O

D

R

R R A

C

R

Thus, between points A and C , we now have three branches in parallel, each having resistance of 2R . 2R 3 Alternately, since the points B , O and D are at same potential, we can treat them shorted. ⇒ Req =

R A

R

B

R

R

O R R

D

R

A C

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 31

R

R

B, O

R

R

D

R

R/2 E

R R/2

R

R C

R B D

A

R/2

E

R/2

C

R are in 2 parallel across A and E . The parallel combination of R R R R and R gives . This in parallel with gives 2 2 2 R . Hence, 4

Now, we find that resistances R , R and

R

R

R

D

R

R

C

⇒

( Req )AE = 4

R

R

( Req )AC = 2 ( Req )AE = 2 ⎛⎜⎝ 4 ⎞⎟⎠ = 2

9/20/2019 11:10:43 AM

3.32  JEE Advanced Physics: Electrostatics and Current Electricity

IDENTICAL POTENTIAL POINTS

Illustration 22

In some networks, you may not find symmetry on the two sides of a line. But, you may find that the network contains some set of points having identical potential. Such a set of points can be joined together to make the network simple.

Consider the circuit shown in figure. For a given resistance R0 , what must be the value of R1 so that the equivalent resistance between the terminals is equal to R0 ? R1

Illustration 21

Twelve equal wires of resistances R each are joined up to form the edges of a cube, as shown in figure. The cube is connected into a circuit across the diagonal AG . Find the equivalent resistance of the network. D

C

A

B H E

R1

The equivalent resistance, R ′ , due to the three resistors on the right is

Let us search the points of identical potential. Since the three edges of the cube from A , viz., AB , AD and AE are identical in all respects in the circuit, the points B , D and E are at the same potential. Similarly, for the point G , the sides GC , GH and GF are symmetrical and the points C , H and F are at the same potential. C

B E

H F

G

Next, to simplify the circuit, we bring together the points, B , D and E and also C , H and F . Now, it becomes obvious that the resistance between A and D = R/3 the resistance between D and C = R/6 the resistance between C and G = R/3 Thus, the circuit is equivalent to three resistances of value. R/3, R/6 and R/3, in series, and hence the net resistance between A and G is

RAG =

R R R 5R + + = 3 6 3 6

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 32

R1 ( R0 + R1 ) R0 + 2R1

Since R′ is in series with the fourth resistor R1, the equivalent resistance of the entire configuration becomes

F

D

R0 + 2R1 1 1 1 = + = R ′ R1 R0 + R1 R1 ( R0 + R1 )

⇒ R′ =

Solution

A

R0

Solution

G

R1

Req = R1 +

R1 ( R0 + R1 )

If Req = R0 , then

R0 + 2R1

=

3 R12 + 2R1 R0 R0 + 2R1

R0 ( R0 + 2R1 ) = 3 R12 + 2R1 R0

⇒ R02 = 3 R12 ⇒ R1 =

R0 3

INFINITE LADDER AND GRID Some networks make a ladder (or a grid) and extend to infinity. To reduce such networks we use the ­following steps. Step-1: Let us assume the total resistance of the infinite network to be X (say). Step-2: Now just retain one basic repetitive unit and we observe the remaining circuit to be the same as the original circuit. So, resistance of this left out circuit must be X. Step-3: Now the equivalent circuit, is the ­combination of basic unit and original repetitive circuit of resistance X, such that the net resistance of the entire circuit is X.

9/20/2019 11:10:56 AM

Chapter 3: Electric Current and Circuits 3.33

The following illustrations are done on the basis of these three steps.

X=

Illustration 23

(a) Find effective resistance between points A and B of an infinite chain of resistors joined as shown in Figure 1. R1

A

And as resistance can not be negative, we have

C

R1

R2 B

R1

R2

R2

To infinity

Figure 1

(b) For what value of R0 in the circuit shown in Figure 2 will the net effective resistance is independent of the number of cells in the chain? R1

However, if R1 = R2 = R we get   

D

R1

  

X=

R1 R2

R ( 1 + 5 ) = 1.6 R 2

(b) Suppose there are n sections between points A and B and the network is terminated by R0 with equivalent resistance X . Now, if we add one more to the network between the C and D, the equivalent resistance of the network X will be independent of number of cells if the resistance between points C and D still remains R0 (or X ). So, the circuit reduces to an equivalent circuit as shown. R1

C R2

4R ⎞ R1 ⎛ 1+ 1+ 2 ⎟ ⎜ 2 ⎝ R1 ⎠

R1

R1

C

R0

R2

D

R2

D

Figure 2

C

R0

R1

Solution R2

(a) Suppose the effective resistance between A and B is X . Applying the steps discussed, we get ⇒

R2 X X = R1 + [ R2  X ] = R1 + ( R2 + X )

⇒

X 2 - R1 X - R1 R2 = 0

⇒

X=

A

( R12 + 4R1R2 ) )

(

1 R1 ± 2

R1 C R1 R2

B

R2 D

A

X

R1

R2

R2

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 33

D

R2 R0 = R0 (R2 + R0 )

⇒

R1 +

⇒

X = R0 =

4R ⎞ 1⎛ 1+ 1+ 2 ⎟ ⎜ 2⎝ R1 ⎠

If R1 = R2 = R , then we get

R1 To infinity



X = R0 =

R ( 1 + 5 ) = 1.6 R 2

Problem Solving Technique(s)

C

B

R0 = X

X

Whenever each element of the circuit is multiplied by a factor, say k, then the equivalent resistance is also multiplied by the same factor. 

D

9/20/2019 11:11:06 AM

3.34  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 24

The circuit diagram shown consists of a large number of elements (each element has two resistors R1 and R2 ). The resistances of the resistors in each sub1 sequent element differs by a factor of k = from the 2 resistances of the resistors in the previous elements. Find the equivalent resistance between A and B shown in figure. A

R1

kR1

R2

kR2

k 2R1 k2R2

k 3R1 k 3R2

BALANCED WHEATSTONE BRIDGE LIKE SITUATIONS Condition of balance is

P R = Q S

So, to conclude B

k 4R1 A

∞

k 4R2

B

R

I

When each element of circuit is multiplied by a factor k then equivalent resistance also becomes k times. Let the equivalent resistance between A and B be X . R1 R2

kR1

k2R1

kR2

k2R2

R1

A

+

R2 B

B X

kR1

k2R1

kR2

k2R

2

kX

k3R1 k3R

S

E

(a) When battery and galvanometer arms of a Wheatstone’s Bridge are interchanged, the ­balance position remains undisturbed while sensitivity of bridge changes. (b) When Wheatstone’s Bridge is balanced, the resistance in arm BD may be ignored while calculating the equivalent resistance of bridge between A and C . So, P and Q are in series, R and S are in series and both in parallel to each other. 1 1 1 = + Rnet P + Q R + S

Three other common forms of balanced Wheatstone’s Bridge are shown here. R

So, equivalent circuit becomes A X

R1

R2

I

D + –

⇒   2

C

G

Solution

A

Q

P

A

kX

G I

B

I

S

P Q

B

Q

P

1 For k = 2 ⇒

X=

P

G

Q A

R

( R1 - R2 ) +

R12 + R22 + 6 R1 R2 2

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 34

B S

E

and

S

R G

9/20/2019 11:11:13 AM

Chapter 3: Electric Current and Circuits 3.35 Illustration 25

Illustration 26

Two resistors with temperature coefficients of resistance a 1 and a 2 have resistances R01 and R02 at 0 °C. Find the temperature coefficient of the compound resistor consisting of the two resistors connected, (a) in series (b) in parallel

A rod of length L and cross-section area A lies along the x-axis between x = 0 and x = L. The material obeys Ohm’s Law and its resistivity varies along the rod according to,

Solution

In Series: At 0°C

R01

R0 = R01 + R02

R02

At t°C R01(1 + α 1 t) R02(1 + α 2 t)



R0(1 + α t)

R01 ( 1 + a 1t ) + R02 ( 1 + a 2 t ) = R0 ( 1 + a t )

⇒ R01 ( 1 + a 1t ) + R02 ( 1 + a 2 t ) = ( R01 + R02 ) ( 1 + a t ) ⇒ R01 + R01a 1t + R02 + R02a 2 t = R01 + R02 + ( R01 + R02 ) a t

R a + R02a 2 ⇒ a = 01 1 R01 + R02

(a) Find the total resistance of the rod and the c­ urrent in the wire. (b) Find the electric potential V ( x ) in the rod as a function of x . Solution

(a) Resistance of elementary section dx at x = x is, ⇒

R02

⇒

1 1 1 = + R0 ( 1 + a t ) R01 ( 1 + a 1t ) R02 ( 1 + a 2 t )

R01 + R02 1 1 = + R01 R02 ( 1 + a t ) R01 ( 1 + a 1t ) R02 ( 1 + a 2 t )

Since a t  1, so by using the Binomial expansion, we get

dR =

L

R R R = 01 02 R01 + R02

∫

R=

At t °C ,

dR =

{

r ( x ) dx  A r0 e

-

x L

∵ R=r

⇒

0

x

L

x=0

dx

x

∫ 0

r L⎛ 1⎞ R = 0 ⎜ 1- ⎟ A ⎝ e⎠ I=

(b) dV = IdR

V0 V0 A ⎛ e ⎞ = ⎜ ⎟ r0 L ⎝ e - 1 ⎠ R

dV =

V0 A ⎛ e ⎞ r0 e - x L ⋅ dx ⎜ ⎟ r0 L ⎝ e - 1 ⎠ A

dV =

V0 ⎛ e ⎞ - x L dx ⎜ ⎟e L ⎝ e -1⎠

V( x )

∫

V0 ⎛ e ⎞ ⎜ ⎟ L ⎝ e -1⎠

x

∫e

⇒

a 1 ⎞ a1 ⎛ 1 ⇒ at ⎜ + = t+ 2 t ⎟ R R R R ⎝ 01 02 ⎠ 01 02

⇒

⎛ e ⎞( 1 - e-x L ) V ( x ) - V0 = V0 ⎜ ⎝ 1 - e ⎟⎠

⇒

V(x) =

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 35

x=L

r r L dR = 0 e L dx = 0 ( 1 - e -1 ) A A

1 1 1 1 ( 1 - at ) + ( 1 - at ) = ( 1 - a 1t ) + ( 1 - a 2 t )    02 R R01 R01 R02

a 1 R02 + a 2 R01 R01 + R02

}

Current in the wire is given by

1 1 1 1 ( 1 - at ) + ( 1 - at ) = ( 1 - a1t ) + ( 1 - a 2t ) R02 R01 R01 R02

⇒ a=

l A

dx

A Since all such elements are in series, hence

In Parallel: R01

r ( x ) = r0 e - x L The end of the rod at x = 0 is at a potential V0 and it is zero at x = L

dV =

V0

-x L

dx

0

V0 ( e - x L - e -1 ) 1 - e -1

9/20/2019 11:11:29 AM

3.36 JEE Advanced Physics: Electrostatics and Current Electricity

Test Your Concepts-V

Based on Series and Parallel Combination of Resistances 1. The space between two coaxial cylinders, whose radii are a and b (where a < b) is filled with a conducting medium. The specific conductivity of the medium is s. (a) Compute the resistance between the cylinders in the radial direction. Assume that the cylinders are very long compared to their radii, i.e., L  b, where L is the length of the cylinders. (b) Calculate the resistance, assuming s varies as s s (r ) = 0 r σ

a

b

(Solutions on page H.203) (b) If the above network were generalised so that there are n ( n = even ) resistors each of resistance R placed along the sides of a regular n sided polygon and if each terminal point of a resistor were connected by (n - 2) insulated wires each of resistance R, directly to the (n  -  2) terminals, other than its, nearest terminals, find the equivalent resistance across any two terminals of the network (i.e., current entering at one of the two terminals and leaving by the other. 3. Calculate the equivalent resistance of the triangular bipyramid between the points specified. Assume the resistance of each branch to be R. D

L C

A B

2. (a) A network of 12 resistors each of value R = 6 W are interconnected as shown in figure, being placed along the sides of a regular dodecagon. Each of the terminals 1, 2, 3, …, 12 has been connected to each of the 9 terminals (other than nearest) directly by insulated wires each of resistance R, there being 9 wires from each terminal making 108 wire connections totally. [Only one set of 9 wires, from terminal 1 have been shown]. Find the equivalent resistance of the network when the current enters at the terminal 1 and leaves at terminal 2. R

3

4

R

2 R 1 R 12 R

R

R R R R R R R R

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 36

P

O

A 6

R

11

(a) A and C (b) D and E 4. A hemispherical network of radius a is made by using a conducting wire of resistance per unit length l. Calculate the equivalent resistance across OP.

B 5

R R

E

R 7 R 8

C

D

5. In the circuit shown, all the side resistances are of value 2 W and all inner resistances are of value 4 W. Find the current via branch AB.

R 10

R

9

9/20/2019 11:11:31 AM

Chapter 3: Electric Current and Circuits 3.37

R′

2Ω

4Ω 2Ω

4

Ω

Ω

4

4Ω

4Ω 2Ω



2Ω

4Ω

4

2Ω

Ω

2Ω

2Ω

2Ω

(a) Show that the effective resistance between A and B is 2 W (b) What is the current that passes through the 2 W resistance nearest to the battery? 9. Nine wires each of resistance r are connected to make a prism as shown in figure. Find the equivalent resistance of the arrangement across. E

A

20 V

D

6. A square pyramid, having vertex at M is formed by joining 8 equal resistances R across the edges. The square base of the pyramid has the corners at A, B, C and D. Calculate the current in the edge M D

C

A

B

F C

(a) AD (b) AB 10. In figure, determine the current through 6 W and 9 W resistor.

B

9Ω

C

E



(a)  MC if an ideal cell of emf E is connected across the adjacent corners A and B. (b)  MA if an ideal cell of emf E is connected across the opposite corners A and C. 7. A length 4a of uniform wire having resistance per unit length l, is bent in the form of a square and the opposite angular points are formed with straight pieces of the same wire which are in contact at the intersection. A given current enters at the intersection of diagonals and leaves at an angular point. Find the resistance of the whole network. 8. An infinite ladder network of resistances is constructed with 1 W and 2 W resistances, as shown in figure. (1987; 7M) The 6 volt battery between A and B has negligible internal resistance. A1Ω

1Ω

1Ω

1Ω

2Ω

2Ω

2Ω

2Ω

6V

B

2Ω 7Ω D

8Ω

2Ω

2Ω

4Ω

E

A

6Ω

12 V

11. Determine equivalent resistance between A and B. 3 r

r

1 r A

r r

6

r

r 4

r B

r r

2

7 r

r 5

12. In the figure shown, the resistances specified are in ohms. Determine the equivalent resistance between points A and D.

B

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 37

9/20/2019 11:11:33 AM

3.38  JEE Advanced Physics: Electrostatics and Current Electricity

A

B

D

F

2 D

E

B

2

F

1

A

C

1

C

E

16. Find the equivalent resistance between A and B in the following circuit: 10 Ω

Ω

10 Ω

10 Ω

5

Ω

B

5

13. In the network shown in figure all the resistances are equal. Determine equivalent resistance between A and E.

Ω

2

1

1

1

5

1

5

Ω

R

R A

R

F

R

R GR

E R R

H

R

B

17. Find the equivalent resistance between A and B in the following circuit:

R

R

10 Ω

A

C

R

R

R

D

14. Twelve equal resistors each R W are connected to form the edges of a cube. Find the equivalent resistance of the network:

R

R

R

R R

R

D

A

R

R

R

C′

R

B′ A

18. Find the equivalent resistance between A and B in the following circuit:

C A′

Ω

B

2Ω

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 38

1Ω

2Ω

(a) when current enters at A and leaves at A′. (b) when current enters at A and leaves at B′. (c) when current enters at A and leaves at B. 15. What will be the change in the resistance of a circuit between A and F consisting of five identical conductors if two similar conductors added as shown by the dashed line in figure.

A

2Ω

D′

2

B

B

2Ω 1Ω

1Ω 2Ω

9/20/2019 11:11:35 AM

Chapter 3: Electric Current and Circuits 3.39

19. An oceanographer is studying the variation of the ion concentration in sea water with depth. This is done by lowering into the water a pair of concentric metallic cylinders at the end of a cable and taking data to determine the resistance between these electrodes as a function of depth. The water between the two cylinders forms a cylindrical shell of inner radius ra, outer radius rb and length L much larger than rb. A potential difference DV is applied between the inner and outer surfaces, producing an outward radial current I. Let r represent the resistivity of the water. Find the

by half tends to infinity. The side AB has a resistance R0. Find the equivalent resistance between A and B.

A

B

22. In the circuit shown in figure, determine the ­current and potential drop across resistor R1.

rb

f ra

R1 = 6 Ω

g

R2 = 4 Ω

17 Ω

21 Ω 2Ω

L

12 V

6V

(a) resistance of the water between the cylinders in terms of L, r, ra and rb. (b) resistivity of the water, r in terms of the measured quantities L, ra , rb , DV and I. 20. The cross section area and length of a cylindrical conductor are A and l, respectively. The specific 1 , where x is conductivity varies as s ( x ) = s 0 x the distance along the axis of the cylinder from one of its ends. (a) Compute the resistance of the system along the cylindrical axis. (b) Compute the current density if the potential drop along the cylinder is V0. What is the electric field at each point in the cylinder in the case described? 21. A frame made of thin homogeneous wire is shown in figure. Assume that the number of successively embedded equilateral triangle with sides decreasing

KIRCHHOFF’S CIRCUIT RULES In analyzing circuits, there are two fundamental (Kirchhoff’s) rules

Junction Rule At any point where there is a junction between various current carrying branches, by current conserva-

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 39

d

6V

h

c 1Ω



e

1Ω

a

b 12 V

23. Determine the current through the 8 W resistor in figure. 2Ω 2Ω 8Ω

6Ω 2Ω 2Ω 7Ω

6V

4Ω 12 Ω

10 Ω

12 V

5Ω

15 Ω

6V

12 V

tion the sum of the currents into the node must equal the sum of the currents out of the node (otherwise charge would build up at the junction)

∑

∑

I in = I out In other words this law states that current or charge cannot accumulate at a junction. This law is in accordance with the Law of Conservation of Charge.

9/20/2019 11:11:38 AM

3.40  JEE Advanced Physics: Electrostatics and Current Electricity I2 I1

Problem Solving Technique(s) Applying Kirchhoff’s Rules

a

In this chapter we have seen how Kirchhoff ’s rules can be used to analyze multiloop circuits. The steps are summarized below:

I3 Kirchhoff’s junction rule

According to the junction rule, the three currents shown here are related by I1 = I 2 + I 3



Loop Rule The sum of the voltage drops DV , across any circuit elements that form a closed circuit is zero.

∑

DV = 0

closed loop



This rule is simply saying that there cannot be a potential difference between a point and itself and is also in accordance with the Law of Conservation of Energy. The rules for determining DV across a resistor and a battery with a designated travel direction are shown below: Travel direction

ΔV = Vb – Va = –IR

a

a

–

E

+

b

Travel direction

Higher V Higher V

ΔV = Vb – Va = +E

Higher V

ΔV = Vb – Va = +IR

Travel direction Lower V

I

b

a

+

E

–

Lower V

ΔV = Vb – Va = –E

I2

(I1 – I2)

(d) Mark the polarity of the voltage across each ­element. In a resistor, the polarity depends upon the assumed direction of current. The end into which the current enters is marked positive. Assumed direction of current

b

Convention for determining ΔV

Note that the choice of travel direction is arbitrary. The same equation is obtained whether the closed loop is traversed clockwise or counter clockwise. Also, remember that a closed loop must start and end at the same point.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 40

C

O B

Lower V Lower V b

I1

A

Travel direction

Higher V I a

(a) Draw a circuit diagram, and label all the quantities, both known and unknown. The number of unknown quantities is equal to the number of linearly independent equations we must look for. (b) Assign a direction to the current in each branch of the circuit. (If the actual direction is opposite to what you have assumed, your result at the end will be a negative number). (c) Apply KCL to each junction. Often, it is found more convenient to apply KCL while marking the currents in the branches. This reduces the number of unknowns to be solved. For example, if we have assumed currents in the branches AO and OB as I1 and I2, we need not mark the current in branch OC as I3. Instead we apply KCL and say that the current in branch OC is ( I1 - I2 ) .

I

+

R V(= IR)

–

(e) Apply the loop rule to the loops until the number of independent equations obtained is the same as the number of unknowns. For example, if there are three unknowns, then we must write down three linearly independent equations in order to have a unique solution. Traverse the loops using the convention below for DV

9/20/2019 11:11:41 AM

Chapter 3: Electric Current and Circuits 3.41



Resistor Travel direction Higher V a

Lower V

I

ΔV = Vb – Va = –IR

b

Travel direction Lower V a



I Higher V

ΔV = Vb – Va = +IR

b

EMF Source Travel direction Lower V a

–

E

+

Higher V

ΔV = Vb – Va = +E

b

Travel direction Higher V a



+

E

–

Lower V

ΔV = Vb – Va = –E

b

(f) If necessary, choose other loops and repeat above steps until you get as many independent equations as the unknowns. The set of equations obtained will be independent provided each new loop equation contains a voltage change not included in a previous equation. The easiest way to ensure the independency of equations is to select all meshes ( just like the panes of a window) as the loops for writing KVL. You may have to select loops as per the number of variables to be calculated. Another thing we must keep in mind while selecting loops is that we must select the loops independently. Suppose we require three loops to be selected and we select a bigger loop along with two loops (these two loops collectively form the bigger loop) and write three equations. We will notice that actually we may not be getting three equations but instead we shall be getting only two equations. So avoid taking such type of loops and try to identify and take independent loops. Suppose, in this circuit, we take the currents as shown. Hence we observe that at least three equations are required to get I, I1 and I2. I1

Capacitor P

Travel direction Lower V a

–q

+q

b

I



The same equation is obtained whether the closed loop is traversed clockwise or counterclockwise. Try to take as many equations as the number of variables. Start from any point and mentally go around the loop in the designated direction. Write down directly the voltage of each element whose positive (+) terminal is entered, and write down the negative of every voltage whose negative (-) terminal is entered.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 41

c

d

S (I – I1 + I2) g

E

Lower V

ΔV = Vb – Va = –q/C



R I – I1

h –q

Q

Higher V

Travel direction +q

(I1 – I2) I2

a

ΔV = Vb – Va = +q/C

Higher V

b



Please be careful and take loops as abda, bcdb (or abcda) because abcda = abda + bcdb. So these are just two loops. The third loop must be abcgha (or adcgha or adbcgha or abdcgha). So, while taking the loops, keep in mind to take unique loops. Avoid taking big loops made from smaller loops (for which equations have already been taken) because the big loop made from the combination of small loops will again fetch you the identical equations which cannot be used as such. (g) Solve the simultaneous equations to obtain the solutions for the unknowns.

9/20/2019 11:11:42 AM

3.42  JEE Advanced Physics: Electrostatics and Current Electricity

Always remember the signs of emf are independent of the current. Hence, we have

Illustration 27

In the arrangement shown, find the R1 = 5 Ω

a

b R2 = 5 Ω c

r1 = 1 Ω

E2 = 4 V d

E1 = 12 V

r2 = 1 Ω e



- IR1 - IR2 - E2 - Ir2 - IR3 + E1 - Ir1 = 0



I=

E1 - E2 R1 + R2 + R3 + R4 + R5

Note that if E2 is greater than E1 , we get a negative value of I , which shows that assumed direction of current is wrong. 12 - 4 = 0.5 A 5 + 5 + 4 + 1+ 1 ( b) Now we determine the potential at each labelled point in the circuit.

I=



Vg = V f + E1 = 0 + 12 = 12 V



Va = Vg - Ir1 = 12 - ( 0.5 ) ( 1 ) = 11.5 V

Solution



(a) First we assume that current is clockwise. Now we apply KVL in the assumed direction of current.

Vb = Va - IR1 = 11.5 - ( 0.5 )( 5 ) = 9 V



Vc = Vb - IR2 = 9 - ( 0.5 )( 5 ) = 6.5 V



Vd = Vc - E2 = 6.5 - 4 = 2.5 V



Ve = Vd - Ir2 = 2.5 - ( 0.5 ) ( 1 ) = 2 V



V f = Ve - IR3 = 2 - ( 0.5 ) ( 4 ) = 0

f

R3 = 4 Ω

(a) current I in the circuit, ( b) potential at each of the labelled points a, b, c, d, e, assuming potential at f to be zero, (c) power input and output in the circuit.

I

a

R1 = 5 Ω +

b

–

+

R2 = 5 Ω c – +

r1 = 1 Ω g

+

E1 = 12 V

d

E2 = 4 V r2 = 1 Ω

Note that, here we have assigned f to be at zero potential however, we can choose any point of the circuit to be at zero potential and then determine the potentials of the other points relative to it. The zero potential point is indicated by the

f

R3 = 4 Ω

ground symbol at point f . (c) Power delivered by source of emf E1 ,

e

Circuit element

Changes in potential

Sign

R1 (a → b)

Drop IR1

minus

R2 (b → c)

Drop IR2

minus

E2 (c → d)

Drop E2

minus

r2 (d → e)

Drop Ir2

minus

R3 (e → f )

Drop IR3

minus

E1 ( f → g)

Gain E1

plus

r1 (g → a)

Drop Ir1

minus

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 42



PE1 = E1 I = ( 12 ) ( 0.5 ) = 6 W V 12 10 8 6 4 2 f g a b c def

Power dissipated in resistors,

PR = I 2 ( R1 + R2 + R3 + r1 + r2 )



2 PR = ( 0.5 ) ( 5 + 5 + 4 + 1 + 1 ) = 4 W

9/20/2019 11:11:51 AM

Chapter 3: Electric Current and Circuits 3.43

The power consumed by battery 2 in getting charged,  

PE2 = E2 I = 4 ( 0.5 ) = 2 W

Note that battery E1 is discharging, the current comes out of its positive terminal. The terminal voltage across its terminals is

V1 = E1 - Ir1

⇒

V1 = 12 - ( 0.5 × 1 )

⇒

V1 = 11.5 V

Writing KVL for loop I, we get xR1 + E3 + yR3 - E1 = 0 ⇒ 5x + 10 + 4 y – 6 = 0 ⇒ 5x + 4 y = -4 …(1) Writing KVL for loop II, we get ⇒

( x – y ) R2 + E2 – yR3 – E3 = 0 ( x – y ) 10 + 8 – 4 y – 10 = 0

⇒ 10 x – 14 y = 2 …(2)

The battery E2 is charging, the current goes in from its positive terminal. The terminal voltage across it V2 = E2 + Ir2 = 4 + ( 0.5 ) ( 1 ) = 4.5 V . The terminal voltage is greater than the emf of the battery. Illustration 28

Calculate the currents through each resistance in the given circuit. Also calculate the potential difference between points a and b . a

5Ω

4Ω

x=-

8V

b

Solution

Let the current through R1 be x and through R3 be y , as shown. Applying KCL at junction a , the current through R2 will be ( x - y ) .

The signs indicate that the directions of x and y were assumed incorrectly, whereas the assumed direction of current x -y through R2 was correct. Potential difference between a and b , Vab = Potential of a with respect to b is Va - Vb . So, Va - Vb = Potential rise when we go from b to a

10 Ω (x – y)

R1 6V

E1

I

y

R2 10 V II E2 4Ω

E3 R3 b

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 43

8V

5 90 × 4 + 10 = V 11 11

Illustration 29

Two dissimilar cells of emfs E1 and E2 and internal resistances r1 and r2 respectively are connected in parallel across a load resistance R , as shown. Find the emf and internal resistance of the equivalent cell of this combination. E1

E2

According to the assumed direction of currents, we mark the polarity of voltage drops across different resistances. 5Ω x a

24 5 1 A , y=A and x - y = A 25 11 55

⇒ Va - Vb = yR3 + E3 = -

10 Ω 10 V

6V

Solving equations (1) and (2), we get

r1

r2 R Load

Solution

Let I1 and I 2 be the currents supplied by the two cells. Let the current through load resistance R be I . According to KCL at junction L ,

I = I1 + I 2 …(1)

9/20/2019 11:12:09 AM

3.44  JEE Advanced Physics: Electrostatics and Current Electricity I1 E1

A

r1

C

I M

B

Eeq

O

L

(a) If in the circuit battery E2 is reversed E2 ⇒ -E2 , equation (7) reduces to

D

r2

I2 E2

Problem Solving Technique(s)

req

N

R (a)

⎡ rr ⎤ E r -E r req = ⎢ 1 2 ⎥ and Eeq = 1 2 2 1 ( r1 + r2 ) ⎣ r1 + r2 ⎦

R

(b) If large number of dissimilar cells are joined in parallel, in the light of equation (6), we have

(b)

For loop ALMNOBA , writing KVL equation,

IR + I1 r1 – E1 = 0

E - IR ⇒ I1 = 1 …(2) r1 Similarly, for loop CLMNODC , the KVL equation,

IR + I 2 r2 – E2 = 0

⇒ I2 =

E2 - IR …(3) r2



-1

and Eeq

⎡ ⎛ 1 1⎞ ⇒ I ⎢R+ ⎜ + ⎟ ⎝ r1 r2 ⎠ ⎢⎣

-1

2Ω

5Ω

3Ω

Y

Solution

There are two closed circuits or loops. Let the current in the two loops be I1 and I 2 as shown. a

)

So, comparing equations (4) and (5), we get -1

⎡1 1⎤ ⎡E E ⎤⎡ 1 1 ⎤ req = ⎢ + ⎥ and Eeq = ⎢ 1 + 2 ⎥ ⎢ + ⎥ ⎣ r1 r2 ⎦ ⎣ r1 r2 ⎦ ⎣ r1 r2 ⎦  …(6) r1 r2 E r + E2 r1 and Eeq = 1 2 …(7) r1 + r2 r1 + r2

X

I2 5Ω

I2

4V c

Y

I2

f

no current can flow through branch cd , not a part of any closed loop. The whole goes round the left-hand loop, and whole goes round the right-hand loop.

I1 =

2 V = 0.4 A 2 W+3 W

and I 2 =

4 V = 0.5 A 3 W+5 W

∴

e

3Ω

I1

Note that since it is current I1 current I 2

4V

d

3Ω

2V

b

⇒ I R + req = Eeq …(5)

2Ω

1Ω

- IR – Ireq + Eeq = 0

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 44

∑

-1

4V

X

3Ω

2V

resistance of a single cell as shown in Fig. (3.139(b)), then according to loop rule, we have

⇒ req =

∑

1⎤ ⎥ ri ⎦

4V

⎤ ⎡ E E ⎤ ⎡ 1 1 ⎤ -1 ⎥ = ⎢ 1 + 2 ⎥ ⎢ + ⎥ …(4) ⎥⎦ ⎣ r1 r2 ⎦ ⎣ r1 r2 ⎦

-1

Ei ⎤ ⎡ ⎥⎢ ri ⎦ ⎣

1Ω

Now if Eeq and req are the equivalent emf and internal

(

⎡ =⎢ ⎣

In the given network, determine the potential of point X with respect to the point Y .

⎡1 1⎤ ⎡E E ⎤ I = ⎢ 1 + 2 ⎥ - IR ⎢ + ⎥ ⎣ r1 r2 ⎦ ⎣ r1 r2 ⎦

⎡ ⎛ 1 1 ⎞⎤ ⎡E E ⎤ ⇒ I ⎢1+ R⎜ + ⎟ ⎥ = ⎢ 1 + 2 ⎥ ⎝ r1 r2 ⎠ ⎦ ⎣ r1 r2 ⎦ ⎣



∑

1⎤ ⎥ ri ⎦

Illustration 30

Substituting the value of I1 and I 2 from equation (2) and (3) into (1), we get



⎡ req = ⎢ ⎣

9/20/2019 11:12:24 AM

Chapter 3: Electric Current and Circuits 3.45

The potential of point X with respect to point Y is written as VXY . If we traverse any path starting from point Y and end at point X , the net potential rise will be VXY . Now, let us find VXY , by traversing the path YdcX ,

VXY = 3 I 2 + 1 ( 0 ) – 4 – 3 I1



VYX = -VXY = - ( -3.7 ) V = 3.7 V

It means point Y is 3.7 V higher in potential than the point X . Note that whichever path we traverse from Y to X, the result should be the same, that is, -3.7 V . Let us check if by traversing the path YfedcbaX, VXY = 0 – 5I 2 + 4 + 1 ( 0 ) – 4 + 0 – 2 + 2I1



VXY = 0 – 5 ( 0.5 ) + 4 + 1 ( 0 ) – 4 + 0 – 2 + 2 ( 0.4 )



VXY = -2.5 + 4 - 4 - 2 + 0.8 = 4.8 - 8.5 = -3.7 V

I = I1 + I 2 …(1)

Applying Kirchhoff’s Second Law in Loop ABEFA ,

VXY = 3 ( 0.5 ) + 0 - 4 - 3 ( 0.4 ) = 1.5 - 4 - 1.2 = - 3.7 V It means that point X is 3.7 V lower in potential than the point Y . We can also write



Applying Kirchhoff’s First Law (Junction Law) at junction B ,

-4 I + 4 - 2I + 2 = 0 …(2)

⇒ I1 = 6 A Applying Kirchhoff’s Second Law in Loop BCDEB ,

-2I1 - 6 - 4 I1 - 4 = 0 …(3)

⇒ I1 = -

10 5 =- A 6 3

Using (1), we get I1 = -

5 A 3

Here, negative sign of I1 implies that current I1 is in the direction opposite to that assumed by us in the figure i.e., it is from C to B (and not from B to C ). A

Illustration 31

Find currents in different branches of the electric ­circuit shown in figure. Also find the potential difference between points F and C . A

4Ω

2Ω

B

C

4V

2V F

2Ω

E

6V

4Ω

D

Solution

In this Illustration there are three wire segments EFAB , BE and BCDE and two junctions at B and E . Therefore we have three unknowns I , I1 and I 2 and hence we require three equations. One equation will be obtained by applying Kirchhoff’s Junction Law (either at B or at E ) and the remaining two equations, we get from the Second Law (Loop Law). Consider three loops ABEFA , ACDFA and BCDEB. But we have to choose any two of them. Further, we can choose any arbitrary directions of I , I1 and I 2 .

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 45

2V

H I H

4Ω

L

L F

L I

B I1 H

L

H

H

2Ω

I2

L

2Ω

4V

C H L

H L E I1 4 Ω

6V D 4Ω

To find the potential difference between any two points of a circuit let us reach from one point to the other via any path of the circuit. It is advisable to choose a path in which we come across the least number of resistors preferably a path which has no resistance. Let us reach from F to C via A and B ,

VF + 2 - 4 I - 2I1 = VC

⇒ VF - VC = 4 I + 2I1 - 2 Substituting, I = 1 A and I1 = VF - VC = -

5 A , we get 3

4 V 3

Here negative sign implies that VF < VC or F is at a lower potential than C .

9/20/2019 11:12:45 AM

3.46  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 32

Parallel E

In the arrangement shown, find the 1 kΩ

E

F

1 kΩ A

C 12 V

H

(a) equivalent resistance between C and G . (b) current provided by the source. (c) voltage across points G and E . Solution

(a) Imagine the wires to be flexible, lift up the inside square with the resistors along with the source attached and then following figures (a), (b), (c) and (d) to arrive at equivalent circuit. The equivalent resistance is 3 kW . 1 kΩ

F

1 kΩ A

1 kΩ

1 kΩ

1 kΩ

1 kΩ

1 kΩ

D

C 12 V

H

1 kΩ Series

Series

B

1 kΩ

G

(c)

(d)

(b) Since V = IReq , I =

G

1 kΩ

E

⇒ ⇒

VG + 12 - IR = VE VE - VG = 12 - IR

VE - VG = 12 - ( 4 × 10 -3 × 2 × 10 3 ) = 4 V

THE PRINCIPLE OF SUPERPOSITION OR SUPERPOSITION THEOREM Complex network problems can sometimes be solved easily by using the principle of superposition. This principle essentially states that when a number of emf’s act in a network, the solution is the same as the superposition of the solutions for one emf acting at a time, the others being shorted. Figure shows a network with two loops. 21.6 V 8 Ω 24 Ω 4Ω

Parallel 1 kΩ 2 kΩ

A

2 kΩ

2 kΩ

2 kΩ

C 12 V (b)

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 46

V = 4 mA Req

(c) Start at point G , assign it a potential VG , proceed toward E along any path. When you reach point E after adding potential drops and gains you get potential of E

(a) E

12 V

G Equivalent circuit

G

1 kΩ

1 kΩ

D

3 kΩ

12 V

1 kΩ

1 kΩ

2 kΩ C

2 kΩ

B

1 kΩ

1 kΩ

2 kΩ

C

G

16 Ω 28.8 V

The currents in various branches can be calculated using Kirchhoff’s Laws. We can get the same solution by considering only one battery at a time and then superposing the two solutions. If a battery has an internal resistance, it must be left in place when the emf of the battery is removed. Figure shows how the superposition principle can be applied to the present problem.

9/20/2019 11:12:51 AM

Chapter 3: Electric Current and Circuits 3.47

The current values in figure (a) and (b) are easily verified. For example when the 21.6 V battery alone is acting, the total resistance in the circuit is, 8+



24 × 16 + 4 = 21.6 W 24 + 16

21.6 V 8 Ω

8Ω

1A 1A

24 Ω

0.4 A 4Ω

1.2 A 0.4 A

16 Ω

16 Ω

24 Ω

0.8 A

0.6 A 4Ω

(a)

(b)

28.8 V

21.6 V = 1 A . This 21.6 W ­current splits between 16 W and 24 W in the ratio 3 : 2. Similarly, the total resistance when only the 28.8 V battery is acting is, This makes the total current



16 +

24 × 12 = 24 W 24 + 12

Therefore, the total current is

28.8 V = 1.2 A 24 W

The superposition principle shows that there is no current in the 24 W resistance. 21.6 V 1.8 A 4Ω

8Ω

16 Ω 28.8 V

Only a current of 1.8 A flows through the outer loop. All these conclusions can also be verified by analyzing the circuit using Kirchhoff’s Laws.

Problem Solving Technique(s) (a) If there are a number of emfs acting simultaneously in any linear network, then each emf acts independent of the others, i.e., as if the other emfs did not exist. The value of current in any conductor is the algebraic sum of the currents due to each emf. Similarly, voltage

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 47

NODAL ANALYSIS IN CIRCUITS Node is a point at which two or more elements are joined together. For two nodes to be different, their voltages must be different. A conductor with a substantially zero resistance is considered to be a node for the purpose of analysis. So, in circuit diagrams where connections are done using ideal wires (i.e. wires with zero resistance), a node may consist of the entire section of wire between elements, not just a single point. a Node

b Ideal wire (Full section ab of wire can be called as a node)

1.8 A

24 Ω

(c)

across any conductor is the algebraic sum of the voltages which each emf would have produced while acting singly. (b) This theorem is applicable only to linear networks where current is linearly related to voltage as per Ohm’s Law. (c) Please note that while applying the superposition theorem only the current sources (or batteries) and the currents have to be superimposed. Do not superimpose the resistances, because then you will be getting inaccurate results.

Junction is a point where three or more branches meet together as shown in figure. Node

R1

Junction

R2

Node

R3 Node

Conceptual Note(s) A junction can also be called a node but a node may or may not be a junction. Think about two wires/branches meeting at a point, which makes it a Node, whereas think about three wires meeting at a point which may be called a Node (two or more branches meeting at a point) as well as a Junction (three or more branches meeting at a point).

9/20/2019 11:12:55 AM

3.48  JEE Advanced Physics: Electrostatics and Current Electricity E1

a

E2

f

In this step, for each node, we assume that all branch currents are either leaving from the node (or entering the node), and then we describe the branch current(s) in term of node voltages. Always keep in mind that a Junction may be called a node but a node may or may not be a junction. Step-4: Solve the system of linear equations obtained to find the unknown node voltage(s) and hence the current through different branches of the circuit.

e

R3 b

R1

c

d

R2

In the above figure we can say that points a, b, c, d, e, f are Nodes and point c and f are called Junctions. (or can also be called as Nodes)

Conceptual Note(s) A current through a resistor can be calculated by using node-voltages as follows: V1

I

R

Node 1

I=

V1 - V2 (when V1 > V2 ) R V1

I

R

Node 1

I=

V2(< V1) Node 2

NODAL ANALYSIS METHOD To solve problems on the basis of nodal analysis, we use the following steps. Step-1: Assign potential to every node of circuit. For doing this, we assign zero potential to any one of the node of the circuit. This Node (which is assigned zero potential) is called as Reference Node OR Reference Junction. For example, in the circuit shown, we assign zero potential to the point c.

V2(> V1) Node 2

+20 V

The nodal analysis is a systematic way of applying Kirchhoff ’s Circuit Laws (KCL) at each essential node of a circuit and represents the branch current in terms of the node voltages. This will give us a set of equations that we solve together to find the node voltages. Once we find the node voltages, we can use them to calculate any other currents or voltages of interest. Step-1: Identify all the essential nodes and select one of them as a reference node i.e. the node at zero potential i.e. 0 V. This reference node (can generally be taken as the ground) has usually most elements tied to it. Step-2: Now assign voltages (with respect to the reference node) to all the nodes except the reference node. Step-3: Apply the KCL at each node except the reference node and write the equations.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 48

10 Ω

f I1

30 Ω

e

5V

d

x I2

–5 V

20 Ω

V2 - V1 (when V2 > V1 ) R

Problem Solving Technique(s)

a

I3 +20 V

b

20 V

c 0V

–5 V

Then, Vc = 0 V , Vb = +20 V , Vd = -5 V . Since ab and de are perfectly conducting sections of the circuit. So, Va = Vb = +20 V , Ve = Vd = -5 V . Let the potential of node/junction f be x . Step-2: Now at junction f , apply KJL i.e., ΣI = 0 DV where I = . Assuming all currents I1 , I 2 and I 3 R to enter the junction, we get ⇒

I1 + I 2 + I 3 = 0 20 - x -5 - x 0 - x + + =0 10 30 20

⇒ 6 ( 20 - x ) + 2 ( -5 - x ) + 3 ( - x ) = 0 ⇒ 120 - 6 x - 10 - 2x - 3 x = 0 ⇒ 11x = 110 ⇒ x = 10 V

9/20/2019 11:13:08 AM

Chapter 3: Electric Current and Circuits 3.49

So, potential of junction f is V f = x = 10 V ⇒ I1 =

6V f

20 - x 20 - 10 = =1A 10 10

1 -5 - x -5 - 10 ⇒ I2 = = =- A 30 30 2



(Negative sign with I 2 implies that its direction is opposite to the direction as assigned in the circuit i.e., actually from f to e ) ⇒ I3 = -

10 1 x ==- A 20 20 2

I2 =

So, current through 6 V battery is

( I1 - I 2 ) =

5 A (from e to f ) 6

I1 = I 28 W =

1 A (from f to a ) 2

I 2 = I 54 W =

1 A (from f to e ) and 3



Illustration 33



Using Nodal Analysis method, calculate the current through the resistors 28 W , 54 W and 6 V battery.

Illustration 34

54 Ω

I 6 V battery = I1 + I 2 =

5 A (from c to f ) 6

Using Nodal Analysis method, calculate the current through 5 W , 4 W , 10 W resistors. Also calculate potential difference between the points a and b .

6V

5Ω

8V

4Ω

12 V 9V

Let us redraw the circuit and take a zero reference point in the circuit. So let potential of the point c be taken as zero. Then with respect to this point potential of point b is -8 V , point d is -12 V , point f is 6 V . a

6

28 Ω

1 I1 = A 2 I +I 1 2

54 Ω

e

1 I2 = A 3

–12

c b

12 V

0

8V

d

–12

Also sections ab and de are perfectly conducting, so, Va = Vb = -8 V , Ve = Vd = -12 V . A careful observation of this circuit reveals that –8 V a



28 Ω

6 - ( -8 ) 1 I1 = = A 28 2

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 49

I1

6V f

12 V

10 V a

Solution

Let Va = 0 and the circuit is redrawn as shown. 9

f

5Ω

I1

9

x b

I2 10 Ω

c

e

12

4Ω

I3 g 10 10 V a 0

9V

6V –8

10 Ω

b

Solution

–8

–12 V e

6 - ( -12 ) 1 = A 54 3

1 ⇒ I 3 = A ( from f to c ) 2 Also, we observe that I1 = I 2 + I 3

28 Ω

54 Ω I2

12 V d

0

Let x be the potential at junction b . f 5Ω 9V

I1

x b

I2 10 Ω

c 12 V

4Ω I3 g 10 V

9/20/2019 11:13:23 AM

3.50 JEE Advanced Physics: Electrostatics and Current Electricity

Applying KJL at b, we have ΣI b = 0

⇒

I1 =

1 A (from b to f ) 5

⇒

I1 + I 2 + I 3 = 0

⇒

I2 =

⇒

9 - x 12 - x 10 - x + + =0 5 10 4

12 - x 12 - 10 1 = = A 10 10 5

⇒ ⇒ ⇒ ⇒

4 ( 9 - x ) + 2 ( 12 - x ) + 5 ( 10 - x ) = 0 36 - 4x + 24 - 2x + 50 - 5x = 0 11x = 110 x = 10 V

⇒

I2 =

1 A (from c to b ) 5

⇒

I3 =

10 - x 10 - 10 = =0A 4 4

⇒

9 - x 9 - 10 1 I1 = = =- A 5 5 5

Since no current flows through the branch ab , so Va - Vb = Vab = 10 V

Test Your Concepts-VI

Based on Kirchhoff’s laws and Nodal Analysis (Solutions on page H.211) 1. A dead battery is charged by connecting it to the live battery of another car with jumper cables. Determine the current in the starter and in the dead battery. 1Ω

0.01 Ω + –

+ –

12 V

Live battery

10 Ω

2. For the circuit shown in figure, calculate (a) The current in the 2 W resistor and (b) The potential difference between points a and b.

A

R1

4Ω

2Ω

R2

b

a 6Ω

3. Calculate the potential difference between points a and b in figure and identify which point is at the higher potential.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 50

b

4. A circuit has a section ABC as shown in figure. If the potentials at points A, B and C are V1, V2 and V3 respectively, calculate the potential at point O.

Dead battery

8V

a

4Ω

12 V

0.06 Ω Starter

10 V

12 V

4V

2Ω

B

O

R3 C

5. In the circuit shown in figure, calculate the currents I1, I2 and I3. Also find out the potential differences between the points A and B and B and C.

9/20/2019 11:13:30 AM

Chapter 3: Electric Current and Circuits 3.51

2Ω

3Ω

B

A A

1Ω

I2

I1

1Ω

C

6. Find the current in each branch of the circuit shown in figure. 4Ω

A

5V

6Ω 8Ω

E

1Ω

+ – E3

B

5Ω

21 V

r2 r3

I3

5Ω

+ – E1 + – E2

R

6V

8V

r1

(a) Find the potential difference between the points A and B and the currents through each branch. (b) If r2 is short circuited and the point A is connected to point B, find the currents through E1, E2, E3 and the resistor R. 10. Calculate the steady state current in the 2 W resistor shown in the circuit (shown in figure). The internal resistance of the battery is negligible and the capacitance of the condenser C is 0.2 μF. 2Ω

B

C

3Ω 16 Ω

D

0.2 μ F

7. Figure shows part of a circuit. Calculate the power dissipated in 3 W resistance. What is the potential difference VC - VB ? 12 V

1Ω C

D 2Ω

3V

E

5A

4Ω 2A B

6A 3Ω

6Ω

A

11. In the circuit shown in figure E, F, G, H are cells of emf 2 V, 1 V, 3 V and 1 V respectively, and their internal resistances are 2 W, 1 W, 3 W and 1 W respectively. Calculate (a) the potential difference between B and D and (b) the potential difference across the terminals of each cells G and H.

4Ω

B

H

F B

10 Ω

9. In the circuit shown in figure E1 = 3 V, E2 = 2 V, E3 = 1 V and R = r1 = r2 = r3 = 1 W

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 51

E

A

2Ω

8Ω 6Ω

2.8 Ω

6V

8. Find the equivalent resistance between A and B in the following circuit: 2Ω

4Ω

C

2V

D

G

C

12. A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances etc., is shown in the figure. Calculate the energy stored in the capacitor C ( 4 μF ) .

9/20/2019 11:13:32 AM

3.52  JEE Advanced Physics: Electrostatics and Current Electricity

1A 4V

3Ω

a 2A

C

4 μF

b

3Ω

1Ω 4Ω

3A

2A 1Ω

5Ω

3A

C 3V

2Ω

6A

13. In the given circuit:



E1 = 3E2 = 2E3 = 6 V, R1 = 2R4 = 6 W

14. Determine the current in each branch of the circuit shown in Figure. 3Ω 5Ω

Find the current in R3 and the energy stored in the capacitor.

PRIMARY AND SECONDARY CELLS/BATTERIES A battery consists of a series or parallel combination of two or more similar cells. Cells or batteries can be divided into primary and secondary types. The secondary cell is rechargeable, whereas the primary is not. The battery used in a car is of secondary type, since it can be recharged. But the cells used in a torch are primary type, as they cannot be recharged. The resistance offered by the electrolyte to the flow of current through the cell is called the internal resistance of the cell. The internal resistance of a cell depends on (a) the distance between the electrodes ( r ∝ d ) ⎡ ⎛ 1⎞⎤ (b) the area of the electrodes ⎢ r ∝ ⎜ ⎟ ⎥ ⎝ A⎠ ⎦ ⎣ (c) the concentration of the electrolyte ( r ∝ C ) (d) the absolute temperature of the electrolyte ⎡ ⎛ 1⎞⎤ ⎢ r ∝ ⎜⎝ T ⎟⎠ ⎥ ⎦ ⎣ Internal resistance is different for different types of cells and even for a given type of cell it varies from cell to cell.

R4

E3

R3 = 2R2 = 4 W, C = 5 μF

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 52

R3

E2 R2

1A



E1

R1 3Ω

8Ω

1Ω + –

4V

1Ω + –

12 V

IDEAL VOLTAGE SOURCE An ideal voltage source is one that can maintain constant voltage across its terminals whatever be the current drawn from it. The internal resistance of an ideal cell is zero. In practice, no cell can be ideal.

PRACTICAL VOLTAGE SOURCE It is due to the internal resistance that the source does not behave as an ideal voltage source. It is usual practice to show a practical voltage source as consisting of an ideal voltage source of voltage E in series with a resistance r . E is the emf of the cell and r is its internal resistance. If such a cell is connected to load of resistance R , the current supplied by the cell is I=

total emf in the circuit E = total resistance R+r

Therefore, the potential difference that is now existing between the terminals A and B of the cell is given as

V = E – Ir

This quantity V is called the terminal voltage of the cell.

9/20/2019 11:13:36 AM

Chapter 3: Electric Current and Circuits 3.53

DISCHARGING OF CELL

SHORT-CIRCUITING OF A CELL

When current is drawn from a cell, the potential difference across it is less than the emf of the cell and is given by

The terminals A and D of a cell are connected by a conducting wire of zero resistance. Using KVL in ABCDA , - Ir + E = 0

V = E - Ir If the load resistance i.e. R → ∞ (i.e., when the terminals A and B are open-circuited), then V → E. Thus, the emf of a cell is nothing but its terminals voltage when its terminals are open-circuited. Suppose a battery of emf E and internal resistance r is connected to a box which contains sources with greater emfs.

⇒ E = Ir

I

A

r

V

R

Load

Then the current I may flow into the positive terminal of the battery (instead of its flowing out of the battery). The battery does not supply energy to the box; on the other hand the box supplies energy to the battery. Now, the terminal voltage is given as V = E + Ir

CHARGING OF CELL When a cell is being charged, the voltage across its terminals is greater than its emf and is given by V = E + Ir A

E

I

r

I

D I

I

I

C



⎛ E⎞ VAD = E - Ir = E - ⎜ ⎟ r ⎝ r⎠

⇒ VAD = 0

B

Cell

A

Terminal potential difference across the cell is

E



E r

B

Practical voltage source



⇒ I=

INTERNAL RESISTANCE OF CELL (r) The internal resistance of a cell is the resistance offered by the electrolyte of the cell to the flow of current between its electrodes. It is denoted by r. The internal resistance of a cell (a) varies directly as concentration of the electrolyte solution of the cell. (b) varies directly as the separation between electrodes i.e., length of electrolyte solution between electrodes. (c) varies inversely as the area of immersed electrodes. (d) is independent of the material of electrodes.

I

Remark(s) r

Box

E

The internal resistance of a circuit is always in series with external resistance of the circuit.

B

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 53

9/20/2019 11:13:40 AM

3.54  JEE Advanced Physics: Electrostatics and Current Electricity

TERMINAL POTENTIAL DIFFERENCE ACROSS A CELL If a cell of e.m.f. E , internal resistance r sends a current I , through an external resistance, then terminal potential difference

V = IR = E - Ir

In general the terminal potential difference (T.P.D.) is the potential difference across the external resistance of the circuit. where I =

E R+r r

I

For P to be MAXIMUM R-r = 0 ⇒ R=r So, we observe that for maximum power consumption across the external resistance, we must have External Resistance = Internal Resistance of the battery. ⇒ Pmax =

E2 E2 = 4R 4r

This is an example of “impedance matching”, in which the variable resistance R is adjusted so that the power delivered to it is maximized. The behaviour of P as a function of R is depicted in figure below.

E

P Pmax

R

During the process of charging of a battery current is driven into a battery in the reverse direction. In such a case positive charge enters the battery at the positive terminal and leaves the battery from the negative terminal. So, when a cell is being charged by an external battery, then terminal potential difference

V = E + Ir

Illustration 35

Shown here are two resistors with resistances R1 and R2 connected in parallel and in series. The battery has a terminal voltage of E .

MAXIMUM POWER TRANSFER THEOREM The power consumption across the external resistance of the circuit is maximum when the net external resistance of the circuit equals the net internal resistance. Consider the circuit shown. The power ­consumed E across R is I 2 R , where I = . R+r R

E, r

⇒ P= ⇒ P=

E2 R

( R + r )2 E2 R

( R - r )2 + 4Rr

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 54

R

r

a

I1

R1

I2

R2

I

E

E

+ –

+ –

a

b

b

I

c

I

Suppose R1 and R2 are connected in parallel, then (a) find the power delivered to each resistor. (b) show that the sum of the power used by each resistor is equal to the power supplied by the battery. Suppose R1 and R2 are now connected in series, then (c) find the power delivered to each resistor. (d) show that the sum of the power used by each resistor is equal to the power supplied by the battery. Which configuration, parallel or series, uses more power?

9/20/2019 11:13:46 AM

Chapter 3: Electric Current and Circuits 3.55 Solution

2

(a) When two resistors are connected in parallel, the current through each resistor is E E I1 = and I 2 = R1 R2 and the power delivered to each resistor is given by P1 = I12 R1 =

E2 E2 and P2 = I 22 R2 = R1 R2

The results indicate that the smaller the resistance, the greater the amount of power delivered. If the loads are the light bulbs, then the one with smaller resistance will be brighter since more power is delivered to it. (b) The total power delivered to the two resistors is PR = P1 + P2 =

E2 E2 E2 + = R1 R2 Req

1 1 1 where = + Req R1 R2 ⇒

Req =

R1 R2 R1 + R2

is the equivalent resistance of the circuit. On the other hand, the total power P supplied by the battery is P = IE , where I = I1 + I 2 , as seen from the figure. Thus, ⎛ E⎞ ⎛ E ⎞ P = I1E + I 2 E = ⎜ E+⎜ E ⎝ R1 ⎟⎠ ⎝ R2 ⎟⎠

P=

E2 E2 E2 + = = PR R1 R2 Req

The above results are in accordance with Law of Conservation of Energy. (c) When the two resistors are connected in series, the equivalent resistance becomes Req ′ = R1 + R2 and the currents through the resistors are I1 = I 2 = I =

E ⎞ ⎛ and P2 = I 22 R2 = ⎜ R2 R + ⎝ 1 R2 ⎟⎠ Contrary to what we have seen in the parallel case, when connected in series, the greater the resistance, the greater the fraction of the power delivered. Once again, if the loads are light bulbs, the one with greater resistance will be brighter. (d) The total power delivered to the resistors is 2



PR′ =

E2 E2 = R1 + R2 Req ′

On the other hand, the power supplied by the battery is E ⎞ E2 E2 ⎛ E= P ′ = IE = ⎜ = ⎟ R1 + R2 Req ′ ⎝ R1 + R2 ⎠

Again, we see that P ′ = PR′ , as required by energy conservation Comparing the results obtained in (b) and (d), we see that P=

E2 E2 E2 + > = P′ R1 R2 R1 + R2

⇒ P > P ′ which means that the parallel connection uses more power because of the fact that the equivalent resistance of two resistors connected in parallel is always smaller than that connected in series.

IDENTICAL CELLS IN SERIES In this arrangement the positive terminal of one cell is connected to negative terminal of the other in succession. Figure represents n cells, each of e.m.f. E and internal resistance r connected in series and an external resistance R is connected across the combination. E

E R1 + R2

Therefore, the power delivered to each resistor is

2

E ⎞ E ⎞ ⎛ ⎛ PR′ = P1 + P2 = ⎜ R1 + ⎜ R2 ⎟ ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎟⎠

E r

E r

r

I

2



E ⎞ ⎛ P1 = I12 R1 = ⎜ R1 ⎝ R1 + R2 ⎟⎠

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 55

R

9/20/2019 11:13:52 AM

3.56  JEE Advanced Physics: Electrostatics and Current Electricity

Potential difference between terminals A and C is

Since all the cells are in series, so Net e.m.f. = nE Similarly all resistances are in series, so Net internal resistance = nr ⇒ Total resistance of circuit = R + nr



⇒ VAC = ( E1 + E2 ) - I ( r1 + r2 ) …(1)

We can replace the above combination by a single cell of emf Eeq and internal resistance req as shown below.

Net e.m.f. ⇒ Current I = Net resistance I=

VAC = VA - VC = VA - VB + VB - VC

nE R + nr

(a) If R >> nr , then I=n

E = n ( current due one cell ) R

(b) If R > net internal resistance, then to get the maximum current, the cells must be connected in series for maximum current, the cells should be connected in series, when net external resistance >> net internal resistance.



Eeq I

A

A

r1

B

I

r2

C

then, VAC = E1 - Ir1 - E2 - Ir2 ⇒

VAC = ( E1 - E2 ) - I ( r1 + r2 )

⇒

VAC = Eeq - Ireq

Thus, equivalent emf Eeq = E1 - E2 and internal resistance req = r1 + r2

NON-IDENTICAL CELLS IN SERIES E1 and E2 are emf of two cells used in series as shown in figure. If r1 and r2 are their respective internal resistances and I is the current flowing from C to A , potentials at points A , B and C respectively be VA , VB and VC , then

(c) If out of N identical cells, n are wrongly connected then Eeq = ( N - n ) E - nE = ( N - 2n ) E and req = Nr



VAB = VA - VB = E1 - Ir1 and

Illustration 36



VBC = VB - VC = E2 - Ir2

In the circuit shown in figure, E1 = 10 V , E2 = 4 V , r1 = r2 = 1 W and R = 2 W . Find the potential difference across battery 1 and battery 2.

I A

E1 r1

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 56

I B

E2 r2

I C

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Chapter 3: Electric Current and Circuits 3.57

Similarly all internal resistances are in parallel, so

E2 r 2

E1 r1

Net internal resistance = Rint = R

⇒ Total resistance of circuit = R + ⇒ Current I =

Solution

Net emf of the circuit = E1 - E2 = 6 V Total resistance of the circuit = R + r1 + r2 = 4 W

(a) If R > R

E ⎛ r⎞ R+⎜ ⎟ ⎝ m⎠

r m

r , then m

E ≈ current due to one cell . R

So, when Net Internal Resistance >> Net External Resistance, then to get maximum current, the cells must be connected in parallel.

and V2 = E2 + Ir2 = 4 + ( 1.5 ) ( 1 ) = 5.5 V

NON-IDENTICAL CELLS IN PARALLEL

IDENTICAL CELLS IN PARALLEL

Consider two cells of emf E1 and E2 connected in parallel across AB .

In this arrangement the positive terminals of all cells are connected to one point and negative terminals to the other point. Figure represents m cells, each of e.m.f. E and internal resistance r , connected in parallel and an external resistance R is connected across the combination. Since all the cells are in parallel, so net e.m.f. equals to the e.m.f. due to a single cell. E I

r r

E

r

E

r

E R

⇒ Net e.m.f. = E

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 57

I

I1 A

I

E1 I1 I

r1 E2 I2

r2

B

I2

If I1 and I 2 are the currents leaving positive electrodes of both cells, then total current from B to A .

I = I1 + I 2 …(1)

VA and VB are potentials at A and B respectively, then potential difference between A and B , can be written as

V = VA - VB = E1 - I1 r1 …(2)

and V = VA - VB = E2 - I 2 r2 …(3) Using equation (1), (2) and (3), we get

I = I1 + I 2

9/20/2019 11:14:12 AM

3.58  JEE Advanced Physics: Electrostatics and Current Electricity Solution

E1 - V E2 - V ⎛ E1 E2 ⎞ ⎛ 1 1⎞ + =⎜ + -V ⎜ + ⎟ r1 r2 ⎝ r1 r2 ⎠ ⎝ r1 r2 ⎟⎠

⇒ I= ⇒ V=

E1 r2 + E2 r1 ⎛ rr ⎞ - I ⎜ 1 2 ⎟ …(4) r1 + r2 ⎝ r1 + r2 ⎠

We can replace the given combination by a single cell of emf equation and internal resistance req across AB as shown in the figure. We have,

VAB = VA - VB = Eeq - Ireq …(5)

Comparing (4) and (5), we have

Eeq

A

Eeq req

=

I



where,

A

10 V 6V

Solution

The given combination consists of two batteries in parallel and resultant of these two in series with the third one. 6V

Calculate equivalent emf of four different cells ­connected in parallel as shown.

Eeq

2 V, 1 Ω

4 V, 2 Ω

B

E=3V 1Ω

r=2Ω

For parallel combination we can apply,

1 V, 1 Ω

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 58

3V 1Ω

Illustration 37

3 V, 1 Ω

2Ω 4V

r and Eeq = E n

A

2Ω

1Ω

1 1 1 1 = + + ..... + req r1 r2 rn

req =

B

Find the emf and internal resistance of a single ­battery which is equivalent to a combination of three ­batteries as shown in figure.

E E E = 1 + 2 + ...... + n r1 r2 rn

(b) For identical cells, r1 = r2 = ....... = rn = r (say) and E1 = E2 = ...... = En = E (say), then

2 4 V, Ω 7 7

Illustration 38

(a) If n cells of emf E1 , E2 ,……, En and internal resistances r1 , r2 ,……, rn are in parallel, then req

The equivalent cell arrangement between A and B is shown for reference.

B

E1 E2 1 1 1 + , where = + r1 r2 req r1 r2

Eeq

⇒ Eeq

1 2 3 4 + - + 2 4 =1 1 1 2 = = V 1 1 1 1 ⎛ 7⎞ 7 + + + ⎜ ⎟ 1 1 1 2 ⎝ 2⎠

req

rr = 12 r1 + r2

and req ⇒

E1 r2 + E2 r1 r1 + r2

Eeq =

E1 E2 E3 E4 + + + r r2 r3 r4 Since Eeq = 1 1 1 1 1 + + + r1 r2 r3 r4

Further,

E1 E2 10 4 r r2 = 1 = 2 2 =3V 1 1 1 1 + + 2 2 r1 r2 1 1 1 1 1 = + = + =1 req r1 r2 2 2

⇒ req = 1 W

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Chapter 3: Electric Current and Circuits 3.59

Now this resistance is in series with the third resistance so the equivalent emf of these two is ( 6 - 3 ) V = 3 V and the internal resistance will be ( 1 + 1 ) = 2 W .

Conceptual Note(s) The equivalent emf can also be found by another method. Suppose we wish to find the equivalent emf of the circuit. Here we shall be using the fact that when no current is drawn by the cell then, we must have E = V However we may note that the current in the internal circuit may be non zero. This current is, Ι=

10 + 4 14 = A 2+1 3

For path ACDB, we have ⎛ 14 ⎞ Now, VA + 4 - 1⎜ ⎟ - VB = 0 ⎝ 3⎠ ⇒ VA - VB =

14 2 -4= V 3 3

Internal resistance of the equivalent battery is found by the same routine procedure. For example, here 2 W and 1 W resistances are in parallel. Hence, their combined resistance is

1 1 1 3 = + = r 1 2 2

⇒ r =

2 W 3

MIXED GROUPING OF IDENTICAL CELLS In this arrangement a total of N cells are used. This arrangement comprises of m rows of cells in parallel, each row containing n cells in series. So, N = mn. The e.m.f. of each cell is E and internal resistance of each cell is r . The combination is connected to external resistance R . Net e.m.f. = nE nr Net internal resistance Rint = m

I

r

I

2 V 3

⇒ E =

E

m rows

10 V

I

F 2Ω

A I=0 4V

R B I=0

1Ω C

⇒

E A

r B

D

The identical result can also be obtained for the path AEFB for which we will get

n cells

E

2 ⇒ E = VA - VB = V = V 3

⎛ 14 ⎞ VA - 10 + 2 ⎜ ⎟ + VB = 0 ⎝ 3⎠

2 ⇒ E = VA - VB = V 3 Further, since, VA - VB is positive, i.e., VA > VB . So, we conclude that A is connected to the positive terminal of the battery and B to the negative.

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 59

nr m NE nE = ⇒ Current I = nr mR + nr R+ m

Net resistance of circuit = R +

⇒ I=

NE

(

mR - nr ) + 2 mnRr 2

For maximum current ⇒

R=

nr m

Rext = Rint

Thus for maximum current, the cells should be connected in mixed grouping when external resistance

9/20/2019 11:14:26 AM

3.60  JEE Advanced Physics: Electrostatics and Current Electricity

R = net internal resistance i.e., Rext = Rint



and I max =

maximum when R = 0 (i.e., the battery is shorted), but PR will be maximum, when

nr = m

R = r

1 mE 1 nE = 2 r 2 R

⇒  External Resistance = Internal Resistance

Illustration 39

Conceptual Note(s) Please note that, for the current in the circuit to be maximum or for maximum power to be consumed by the external resistance R, following cases exist. (a) In the first case, we arrange the cells in such a manner that current and power in the circuit should be maximum. And this happens when we arrange the cells in such a manner that total internal resistance comes out to be equal to total external resistance i.e., load resistance. (b) In the second case, when a battery of emf E, internal resistance r is connected in series with a load of resistance R, then R

In the circuit shown in figure, the emfs of batteries are E1 and E2 which have internal resistances R1 and R2 . At what value of the resistance R will the thermal power generated in it be the highest? What it is?

E1 R1



Ι=

E R+r 2





⎛ E ⎞ PR = Ι 2 R = ⎜ R ⎝ R + r ⎟⎠

(i) Now if r can be varied keeping E and R are fixed, then both I and PR are maximum when r = 0. (ii) However, if R can be varied keeping E and r are fixed. Then, current in the circuit is

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 60

R2

Solution

The two batteries are in parallel. Thermal power generated in R will be maximum when, Total internal resistance = total external resistance ⇒ R=

R1 R2 R1 + R2

⇒ Eeq

⎛ E1 E2 ⎞ ⎜⎝ R + R ⎟⎠ ⎛ E R + E2 R1 ⎞ 1 2 =⎜ 1 2 = 1 1 ⎛ ⎞ ⎝ R1 + R2 ⎟⎠ ⎜⎝ R + R ⎟⎠ 1 2

I

E, r

E2

R

Since, Rnet = ⇒ I=

Eeq Rnet

2R1 R2 R1 + R2 =

E1 R2 + E2 R1 2R1 R2

Maximum power through R .



Pmax = I 2 R =

( E1R2 + E2 R1 )2 4 R1 R2 ( R1 + R2 )

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Chapter 3: Electric Current and Circuits

3.61

Test Your Concepts-VII

Based on EMF, Internal Resistance and Combination of Cells 1. An automobile battery has an emf of 12.6 V and an internal resistance of 0.08 W. The headlights together present equivalent resistance 5 W (assumed constant). What is the potential difference across the headlight bulbs when (a) they are the only load on the battery and (b) the starter motor is operated, taking an additional 35 A from the battery? 2. Under what circumstances can the terminal potential difference of a battery exceeds its emf? 3. Two batteries having the same emf E but different internal resistances r1 and r2 are connected in series with an external resistor R. For what value of R does the potential difference between the terminals of the first battery become zero? 4. The emf of a storage battery is 90 V before charging and 100 V after charging. When charging began the current was 10 A. What is the current at the end of charging if the internal resistance of the storage battery during the whole process of charging may be taken as constant and equal to 2 W? 5. A battery is made by joining m rows of identical cells in parallel each row consists n cells joined in series. This battery sends a maximum current I in a given external circuit. Now the cells are so arranged that instead of m rows, n rows are joined in parallel and each row consists of m cells joined in series. Show that current through the same 2mn I. external circuit is now given by I ′ = 2 m + n2

AMMETER Measurement of Current: The instrument used for measuring current in a circuit is called ammeter. It is basically a moving coil galvanometer in which a low resistance wire (shunt) is connected in parallel with the coil. The shunt is connected for two reasons. (a) To increase the range of current measurement: The moving coil galvanometer is a very sensitive device and its coil gets a full scale deflection for a very small current ( ~ 1 mA ) . By connecting a

03_Physics for JEE Mains and Advanced - 2_Part 1.indd 61

(Solutions on page H.215) 6. We are made available with 4 batteries each of emf 2 V and supplying a current of 2 A. Show by a circuit diagram how these can be arranged to supply 4 A of current at 4 V. 7. The potential difference across a battery is 8.5 V when there is a current of 3 A in the battery from the negative to the positive terminals. When the current is 2 A in the reverse direction, the potential difference becomes 11 V. What is the emf and the internal resistance of the battery ? 8. The output voltage can be reduced in the output circuits of generators as desired by means of an attenuator designed as the voltage divider shown in figure. U1 U0

R1

U2 R1

R2

U3 R1

R2

Un–2 R1

R2

R2

Un–1

Un

R1 R2

R3

A special selector switch makes it possible to connect the output terminal either to the point with a potential U0 produced by the generator, or to any of the points U1, U2 …, Un, each having a potential k times smaller ( k > 1) than the previous one. The second output terminal and the lower ends of the resistances are earthed. Find the ratio between the resistances R1 : R2 : R3 with any number of cells in the attenuator.

shunt of suitable resistance, the excess current passes through it, without damaging the galvanometer. (b) To reduce the error of measurement: Generally the galvanometer coil has a fairly high resistance (10 to 1000 Ω) and therefore it will give error in the measurement of current-the measured current will be less than the actual current. By connecting a low resistance wire in parallel with the coil, the effective resistance of the ammeter can be made small and hence the error is reduced.

9/20/2019 11:14:31 AM

3.62  JEE Advanced Physics: Electrostatics and Current Electricity

SHUNT RESISTANCE TO CONVERT A GALVANOMETER INTO AN AMMETER OF DESIRED RANGE

Illustration 40

The galvanometer shown in figure has resistance 50 W and current required for full scale

If I is the range of the ammeter, I g is the galvanometer current for full scale deflection, G is the resistance of the coil and S is the shunt resistance then, clearly,

G

R1 I

Ig

G

G



( I - I g ) S = I gG

⇒ S=

I gG I - Ig



For the range 0.1 A , R1 , R2 and R3 are in series combination.



RA  S

i.e., ammeter is a low resistance device and hence it is always connected in series in a circuit.

Remark(s) (a) The reading of an ammeter is always lesser than the actual current in the circuit. If V is the potential difference across a resistance R, the true cur⎛V⎞ rent is I = ⎜ ⎟ . However, when an ammeter of ⎝ R⎠ resistance r is used to measure it, the reading will be V I′ = ( R+r) which is less than the true current I (b) Smaller the resistance of an ammeter, the more accurate will be its reading. An ammeter is said to be ideal if its resistance (r) is zero. However, ideal ammeter cannot be realized in practice.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 62

Ig

G

R1

R2

I

Generally S  G , so G+S G

0.1 A Z

Y

Solution

GS G+S



X

deflection is 1 mA . Find the resistances R1 , R2 and R3 required to convert it into ammeter having ranges as indicated.

The resistance of the ammeter so obtained is RA =

R3

1A

10 A A

(I – Ig)

S Ammeter

R2

R3

(I – Ig) A

Z

So, equating potential difference across galvanometer and series combination of R1 , R2 and R3 , we get

I g G = ( I - I g ) ( R1 + R2 + R3 )

Hence R1 + R2 + R3 =

I gG

=

10 -3 × 50

( I - I g ) ( 0.1 - 10 ) -3

=

50 W 99

…(1) For range 1 A, ( R1 + R2 ) is in parallel to ( G + R3 )

I

Ig

G

R1

R2

R3

(I – Ig) A

Y

⇒ I g ( G + R3 ) = ( I - I g ) ( R1 + R2 ) ⇒ I g ( G + R1 + R2 + R3 ) = I ( R1 + R2 )

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Chapter 3: Electric Current and Circuits 3.63

Substituting for I g , G , I and have



R1 + R2 = 10

-3

( R1 + R2 + R3 ) ,

we

50 ⎞ ⎛ 50 + ⎜ 99 ⎟ = 5 W …(2) ⎜⎝ ⎟ 1 ⎠ 99

For range 10 A, R1 is in parallel to ( G + R2 + R3 )

I

Ig

G

R1

R2

R3

(I – Ig) A

X

So, I g ( G + R2 + R3 ) = ( I - I g ) R1 ⇒ R1 =

I g ( G + R1 + R2 + R3 ) I

50 ⎞ ⎛ 10 -3 ⎜ 50 + ⎟ ⎝ 5 1 99 ⎠ ⇒ R1 = = W= W …(3) 10 990 198 So, R2 =

5 5 1 = W 99 990 22

⇒ R3 =

50 - ( R1 + R2 ) 99

small potential difference. However, if a large resistance is connected in series with the coil, any desired potential difference can be measured. (b) To reduce the error of measurement: The voltmeter is connected in parallel with that part of the circuit across which the potential difference is to be measured. Therefore its own resistance should be very high, otherwise it will change the current in (and hence the potential difference across) that part of the circuit. The measured potential difference will be less than the actual value. By connecting a high resistance in series with the coil, we can minimize this error.

SERIES RESISTANCE TO CONVERT A GALVANOMETER INTO A VOLTMETER OF DESIRED RANGE If V is the desired range and R is the additional series resistance, then V = Ig (G + R )



⇒ R=

V -G Ig

Voltmeter is a high resistance device and hence it is always connected in parallel across a circuit whose voltage is to be measured. V

50 5 45 ⇒ R3 = = 99 99 99 ⇒ R3 =

Ig

G

15 W 33

Hence, R 1 =

1 1 15 W , R2 = W and R3 = W 198 22 33

VOLTMETER Measurement of Potential Difference: The most commonly used instrument for measurement of potential difference is voltmeter. It is a moving coil galvanometer to which a high resistance is connected in series. The high resistance is connected for two reasons. (a) To increase the range of measurement: Since the galvanometer has a full scale deflection for very small currents and hence it can measure only a

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 63

G

R

Voltmeter

Remark(s) (a) The reading of a voltmeter is always lesser than the true value. If a current I0 is passing through a resistance R, the true value V = IoR. However, when a voltmeter having resistance r is connected across R, the current through R will become I′ =

r Io (R + r )







and so V ′ = I ′R =

V [1+ (R / r )]

9/20/2019 11:07:04 AM

3.64  JEE Advanced Physics: Electrostatics and Current Electricity



(b) If the voltmeter of range V and resistance G is to be converted into an ammeter of range I, then shunt S to be connected in parallel is

When the voltmeter is connected across R, its reading will also be V′ which is less than V. I

I0

I′

R

V

R

+ V r –

I

(b) Greater the resistance of voltmeter, the more accurate will be its reading. A voltmeter is said to be ideal if its resistance r is infinite. An ideal voltmeter draws no current from the circuit element for its operation.



S=



VG IG - V

Illustration 42

A galvanometer has an internal resistance of 50 W and current required for full scale deflection is 1 mA. Find the series resistances required (as shown in ­figure) to use it as a voltmeter with different ranges, as indicated in figure. G

R2

R1

R3

Illustration 41

How can we make a galvanometer with G = 20 W and I g = 1 mA into a voltmeter with a maximum range of 10 V?

10 V 100 V

Solution

Solution

For range 1 V , galvanometer and R , are in series

V Using R = -G Ig We have, R =

10

- 20 = 9980 W

10 -3 Thus, a resistance of 9980 W is to be connected in series with the galvanometer to convert it into the voltmeter of desired range. Please note that at full scale deflection current through the galvanometer, the voltage drop across the galvanometer Vg = I g G = 20 × 10 -3 V = 0.02 V and the voltage drop across the series resistance R is, V = I g R = 9980 × 10 -3 V = 9.98 V



1V



Ig =

V ( G + R1 )

⇒ 10 -3 =

1

( 50 + R1 )

⇒ 50 + R1 = 1000 ⇒ R1 = 1000 - 50 = 950 W For range 10 V , galvanometer and R2 , R3 are in series.

10 -3 =

10 ( G + R1 + R2 )

⇒ G + R1 + R2 =

10 10

-3

= 10 × 10 3

⇒ R2 = 10000 - ( 50 + 950 ) = 9000

Conceptual Note(s) (a) If an ammeter of range I and resistance G is to be converted into a voltmeter of range V, then the resistance R to be connected in series is



V R = -G I

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 64

R2 = 9 kW and for range 100 V, galvanometer, R1 , R2 and R3 are in series. 10 -3 =

100

( G + R1 + R2 + R3 )

⇒ G + R1 + R2 + R3 =

100 10

-3

= 100 × 10 3

9/20/2019 11:07:14 AM

Chapter 3: Electric Current and Circuits 3.65

⇒

R3 = 100 × 10 3 - ( G + R1 + R2 )

Solution

Let E be the emf of the battery

⇒ R3 = 100 × 10 3 - 10 × 10 3

R

⇒ R3 = 90 × 10 3 = 90 kW

R2 A

Illustration 43

The scale of galvanometer is divided into 150 equal divisions. The galvanometer has the current sensitivity of 10 divisions per mA and the voltage sensitivity of 2 divisions per mV. How can you convert this galvanometer into a/an

R1 V

R2 A

B

8I 3 C I 3

V

R1 D

F

3I

I A

E

E

G

In the first case let I be the current in the circuit, then

(a) ammeter of 6 A per division (b) voltmeter of 1 V per division



Calculate the galvanometer resistance G , value of shunt S to be connected to convert into ammeter and value of series resistance R to be connected to convert into voltmeter.

In the second case main current increases three times I while current through voltmeter will reduce to . 3 I 8I Hence, the remaining 3 I - = passes through R 3 3 as shown in figure.

Solution

⎛I⎞ ⎛ 8I ⎞ VC - VD = ⎜ ⎟ R1 = ⎜ ⎟ R ⎝ 3⎠ ⎝ 3⎠

Since galvanometer resistance is

⇒ R1 = 8 R

Full scale voltage G= Full scale current Also, full scale voltage =

Applying Kirchhoff’s Second Law in Loop ABFGA ,

No. of divisions 150 = = 75 V Voltage sensitivity 2

and full scale current =

No. of divisions 150 = = 15 mA Current sensitivity 10

= 15 × 10 -3 A 75 ⇒ G= = 5000 W 15 × 10 -3



⇒ S= ⇒ R=

I gG I - Ig

=

15 × 10 -3 × 5000 150 × 6 - 15 × 10 -3



R ⎞ ⎛ ⎛I⎞ E = 3 I ( R2 ) + ⎜ ⎟ ( R1 ) = I ⎜ 3 R2 + 1 ⎟ …(2) ⎝ 3⎠ ⎝ 3 ⎠

From equations (1) and (2), we get

R1 + R2 = 3 R2 +

⇒ 2R2 =

= 83 mW

150 V -G = - 5 = 5000 W Ig 15 × 10 -3

Illustration 44

A voltmeter of resistance R1 and an ammeter of resistance R2 are connected in series across a battery of negligible internal resistance. When a resistance R is connected in parallel to voltmeter, reading of ammeter increases three times while that of voltmeter reduces to one third. Find R1 and R2 in terms of R .

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 65

E = I ( R1 + R2 ) …(1)

⇒ R2 =

R1 3

2R1 3

R1 8 R = 3 3

Illustration 45

A battery of emf 1.4 V and internal resistance 2 W is connected to a resistor of 100 W . In order to measure the current through the resistance and the potential difference across its ends, an ammeter is connected in series with it and a voltmeter is connected across its 4 ends. The resistance of the ammeter is W and that 3 of the voltmeter is 200 W . What are the readings of the two instruments? What would be their reading if they were ideal instruments?

9/20/2019 11:07:27 AM

3.66  JEE Advanced Physics: Electrostatics and Current Electricity Solution

Let RA and RV be the resistances of the ammeter and voltmeter respectively. Then the total resistance across the emf E is



RRV 100 × 200 4 + + 2 = 70 W = + RA + r = 100 + 200 3 R + RV

Req

I + E–

r

1.4 V

2Ω

R 100 Ω V

A

Connecting the ammeter changes the current. Let RA is the resistance of the ammeter, then I′ =

E R + RA + r

Clearly I ′ < I . So an ammeter reads less than actual value. Error in the reading = I ′ - I

error in the measurement by voltmeter I0

Consider the same circuit as before. Now, we wish to find out the potential drop across the external resistance R . To measure the potential difference, a voltmeter of resistance RV is connected between A and B.

RA

Rv

I

Therefore, the current Io =

R

A

V

1.4 E = = 0.02 A Req 70

This is the current through ammeter. Hence, the reading of ammeter is 0.02 A Reading of voltmeter is the potential difference across its terminals. So reading of voltmeter is ⎛ RRV ⎞ ⎛ 100 × 200 ⎞ V = Io ⎜ = 0.02 ⎜ = 1.33 V ⎝ 100 + 200 ⎟⎠ ⎝ R + RV ⎟⎠

If the ammeter and the voltmeter were ideal, RA = 0 and RV → ∞ . Then, 1.4 E The reading of ammeter = = = 0.0137 A r + R 2 + 100 The reading of voltmeter = I o R =

Current through the cell is I=

E R+r

Now, V = VAB = IR ⇒ V = VAB =

Now, R ′ =

1.4 × 100 = 1.37 V 102

R

R

I=

A

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 66

E

E 1+

r R

V R

B

V′

r

E R+r

=

R × RV R = Vrated, then Pconsumed > Prated (c) Resistance of electrical appliance: If variation of resistance with temperature is neglected then resistance of any electrical appliance can be

9/20/2019 11:08:08 AM

Chapter 3: Electric Current and Circuits 3.73

calculated by rated power and rated voltage i.e. V2 by using R = R . PR (d) It is observed that for a bulb Brightness ∝ Pconsumed (e) Power consumed and illumination: An electrical appliance (Bulb, heater, …. etc.) consume rated power PR only when the applied voltage VA is equal to the rated voltage VR. So

(i) If VA = VR, then Pconsumed = PR



V2 = A R

(ii) If VA < VR, then Pconsumed

Since, R =

VR2 and hence PR

⎛ V2 ⎞ ⇒ Pconsumed = ⎜ A2 ⎟ PR ∝ ( Brightness ) ⎝ VR ⎠

c­ onsumption across the parallel combination of the appliances will be Pp = where



V2 , Rp

1 1 1 1 = + + and Rp R1 R2 R3

R1 =

V2 V2 V2 , R2 = and R3 = P1 P2 P3

1 1 ⎞ ⎛ 1 ⇒ Ps = V 2 ⎜ + + ⎝ R1 R2 R3 ⎟⎠ ⇒ Ps = P1 + P2 + P3

COMBINATION OF BULBS Series Combination P1, V

P2, V

POWER TRANSFORMATION RULE Case-1: Bulbs Connected in Series When bulbs/appliances having rated voltage V , and rated powers P1 , P2 , P3 , …. are connected in series across a voltage supply V , then the power consumption across the series combination of the appliances will be V2 Ps = , Rs

where Rs = R1 + R2 + R3 and

R1 =

⇒ Ps =

⇒

V2 V2 V2 , R2 = and R3 = P1 P2 P3 V2 ⎛ V2 V2 V2 ⎞ ⎜⎝ P + P + P ⎟⎠ 1 2 3

1 1 1 1 = + + Ps P1 P2 P3

Supply V

(a) Total power consumed

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 73

Ptotal

=

(b) If n bulbs are identical, Ptotal =

1 1 + + ...... P1 P2 P n

(c) Pconsumed (Brightness) ∝ V ∝ R ∝

1 Prated

i.e. in series combination, the bulb of lesser wattage will give more bright light and potential difference appeared across it will be more.

Parallel Combination (a) Total power consumed

Ptotal = P1 + P2 + P3 ...... + Pn P1, V

Case-2: Bulbs Connected in Parallel When bulbs/appliances having rated voltage V, and rated powers P1 , P2 , P3 , …. are connected in parallel across a voltage supply V , then the power

1

P 2, V Supply V

9/20/2019 11:08:19 AM

3.74  JEE Advanced Physics: Electrostatics and Current Electricity

(b) If n identical bulbs are in parallel, then Ptotal = nP

Conceptual Note(s) (a) If VApplied < VRated then percentage drop in output power of electrical device is %age drop =

(b) Different bulbs

25 W 220 V

(PR - Pconsumed ) × 100 PR 100 W 220 V

1000 W 220 V



⇒ Resistance

R25 > R100 > R1000



⇒ Thickness of filament

t1000 > t100 > t40



⇒ Brightness

B1000 > B100 > B25

(c) Necessary series resistance to glow a bulb of rated power PR, if VApplied > VRated is ⎛ VApplied - VRated ⎞ R=⎜ ⎟⎠ × VR P ⎝

R (d) When some potential difference applied across the conductor then collision of free electrons with ions of the lattice result’s in conversion of electrical energy into heat energy (e) If a heating coil of resistance R, (length l) consumed power P, when voltage V is applied to it then by keeping V constant if it is cut in n equal R parts then resistance of each part will be and n 1 from Pconsumed ∝ , power consumed by each part

P ’ = nP .

r

R

(f) In series a device of higher power rating consumes less power. (g) Consider that n bulbs are connected in series across V volt supply. If one bulb gets fused and

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 74

Heater

(n - 1) bulbs are again connected in series across same supply, the illumination will be more with (n - 1) bulbs than with n bulbs, but risk of fusing of bulbs will increases. (h) When a heavy current appliance such us motor, heater or geyser is switched on, it will draw a heavy current from the source so that terminal voltage of source decreases. Hence power consumed by the bulb decreases, so the light of bulb becomes less.

1 (c) Pconsumed (Brightness) ∝ PR ∝ i ∝ R i.e. in parallel combination, bulb of greater wattage will give more bright light and more current will pass through it.

K

Illustration 50

A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to this current, the temperature of the wire is raised by ΔT in time t . A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L . The temperature of wire is raised by same amount ΔT in the same time t . Find the value of N. Solution

Since we have

Heat = mcΔT = i 2 Rt

Case-1: Length ( L ) , Resistance R and mass m Case-2: Length ( 2L ) , Resistance 2R and mass 2m ⇒

m1S1 ΔT1 i12 R1t1 = m2 S2 ΔT2 i22 r2 t2

⇒

i 2 Rt mSΔT = 21 2mSΔT i2 2Rt

⇒ i1 = i2 ⇒

(3E)2 ( NE)2 = 12 2R

⇒ N=6

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Chapter 3: Electric Current and Circuits 3.75

TIME RELATION FOR SERIES and PARALLEL COMBINATION Let an electrical appliance having rated power P1 can generate energy E in time t1 , then E = P1t1 . Similarly another electrical appliance having rated power P2 can generate same energy E in time t2 , then E = P2 t2 . Now when these appliances are connected across the same voltage source Case-1: In Series Then power generated is 1 1 1 = + PS P1 P2

So time required to generate the same energy E is tS = ⇒

E E E E E = + = + = t1 + t2 PS P1 P2 ⎛ E ⎞ ⎛ E ⎞ ⎜⎝ t ⎟⎠ ⎜⎝ t ⎟⎠ 1 2

tS = t1 + t2

Case-2: In Parallel Then power generated is PP = P1 + P2 So, time required to generate the same energy E is tP = ⇒

tt E E E = = = 12 E E t1 + t2 PP P1 + P2 + t1 t2

1 1 1 = + tP t1 t2

Illustration 51

An electric tea kettle has a multi-position switch and two heating coils. When only one of the coils is switched on, the well-insulated kettle makes a full pot of water to a boil in a time interval Δt . When only the other coil is switched on, it requires a time interval of 2Δt to boil the same amount of water. Find the time interval required to boil the same amount of water if both coils are switched while being used (a) in a parallel connection and (b) in a series connection.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 75

Solution

A certain quantity of energy ΔEint = P ( time ) is required to raise the temperature of the water to 100°C. For the power delivered to the heaters we have P = I ΔV =

( ΔV )2

where ΔV is a constant. R Thus comparing coils 1 and 2, we have for the energy H=



( ΔV )2 Δt R1

=

( ΔV )2 2Δt R2

⇒ R2 = 2R1 (a) When connected in parallel, the coils present equivalent resistance

2R 1 1 = = 1 1 1 1 1 3 + + R1 R2 R1 2R1

Rp =

and H =

( ΔV )2 Δt R1

=

( ΔV )2 Δtp 2R1 3

2 Δt 3 ( b) For the series connection, Rs = R1 + R2 = R1 + 2R1 ⇒ Δtp =

= 3R1 and

( ΔV )2 Δt R1

=

( ΔV )2 Δts 3 R1

⇒   Δts = 3 Δt

CHARACTERISTICS OF A FUSE Fuse is used with the main electrical circuit for the safety of electrical appliances. A fuse wire must have high resistance and low melting point. Hence, generally it is made of tin-lead alloy. Let R be the resistance, ρ resistivity, l length, a cross-sectional area and I ampere be its current carrying capacity. When the fuse is safe, then for its steady state temperature, heat produced per second must be equal to heat radiated by it per second. Heat produced in fuse wire per second

H= H=

Q ⎛ ρl ⎞ = I 2R = I 2 ⎜ ⎟ ⎝ A⎠ t I 2 ρl

π r2

joule sec -1 …(1)

9/20/2019 11:08:42 AM

3.76  JEE Advanced Physics: Electrostatics and Current Electricity

If e is emissivity of fuse material of radius r and T is the excess safe temperature of wire above surroundings then according to Newton’s Law of Cooling, the energy radiated per second

H = e ( 2π rl ) T …(2)

For steady state

e ( 2π rl ) T = I 2

⇒ T=

ρl π r2

I 2ρ

…(3) 2π 2 er 3 So, we observe that the steady state temperature of a fuse is independent of its length. Hence length is immaterial for an electric fuse. For a given material of a fuse wire

I 2 ∝ r3

Illustration 52

What amount of heat will be generated in a coil of resistance R due to a charge q passing through it if the current in the coil decreases down to zero (a) uniformly during a time interval t0 ? (b) halving its value every t0 seconds?

Here, I0 is unknown, which can be obtained by using the fact that area under I -t graph gives the flow of charge. Hence, 1 q = ( t0 )( I 0 ) 2 2q I0 = ⇒ t0 Substituting in (1), we get, I=





⇒

2q ⎛ t ⎞ 1- ⎟ t0 ⎜⎝ t0 ⎠

⎛ 2q 2qt ⎞ I=⎜ - 2 ⎟ t0 ⎠ ⎝ t0

Now at time t , heat produced in a short interval dt is,





⇒

dH = I 2 R dt 2

⎛ 2q 2qt ⎞ dH = ⎜ - 2 ⎟ Rdt t0 ⎠ ⎝ t0 t0

So, total heat produced is H =

0

t0



⇒

H=

Solution

H = I 2 Rt We can directly use this formula provided I is constant. Here, I is varying. So, first we will calculate I at any time t , then find a small heat dH in a short interval of time dt . Then by integrating it with proper limits we can obtain the total heat produced. (a) The corresponding I - t graph will be a straight line with I decreasing from a peak value (say I 0 ) to zero in time t0 . I - t equation will be as

⇒

∫ 0

Heat generated in a resistance is given by,

⎛I I = I0 - ⎜ 0 ⎝ t0

∫ dH



⇒

H=

2

⎛ 2q 2qt ⎞ ⎜⎝ t - t 2 ⎟⎠ R dt 0 0

4 q2 R 3 t0

(b) Here, current decreases from some peak value (say I 0 ) to zero exponentially with half life t0 . I -t equation in this case will be - lt I = I 0 e log e ( 2 ) Here, l = t0



∞

Now, q =

⎞ ⎟⎠ t    ( y = - mx + c ) …(1)

∞

∫ I dt =∫ I e 0

0

0

- lt

⎛I ⎞ dt = ⎜ 0 ⎟ ⎝ l⎠

I I0

I I0

t0 t t0 t

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 76



⇒

I0 = l q

9/20/2019 11:08:57 AM

Chapter 3: Electric Current and Circuits



I = ( l q ) e - lt

⇒

2 2 -2 l t

2

dH = I Rdt = l q e

⇒

∞



⇒

H=

∫

Rdt

V1 = V30 W

∞

∫

dH = l 2 q2 R e -2 lt dt = 0

0

q2 l R 2

A

log e ( 2 ) Substituting l = , we have t0



B

IllustRatIon 53

A circuit shown in the figure has resistances 20 W and 30 W. At what value of resistance Rx will the thermal power generated in it be practically independent of small variations of that resistance? The voltage between points A and B is supposed to be constant in this case. A

20 Ω

30 Rx ⎛ ⎞ ⎜ 30 + Rx ⎟ 30 Rx ⎞ ⎛ =⎜ V ⎟V = ⎜ 30 R ⎝ 50 Rx + 600 ⎟⎠ x ⎜ ⎟ + 20 ⎜⎝ 30 + R ⎟⎠ x

20 Ω

20 Ω

A

30 Rx V 30 + Rx 1

Rx V

30 Ω

V

q2 R log e ( 2 ) H= 2t0

3.77

Original circuit

B

Equivalent circuit

Now, power generated in Rx is P=

900 RxV 2 V12 = Rx ( 50 Rx + 600 )2

For P to be constant, we have

30 Ω

Rx



dP =0 dRx

( 50Rx + 600 )2 ( 900V 2 ) B

⇒

( 1800 )( 50 ) ( RxV 2 ) ( 50 Rx + 600 )

( 50Rx + 600 )4

solutIon



V = Constant

Now, voltage across Rx is equal to the voltage across the 30 W resistor, because both are in parallel. Hence, if V1 is the voltage across Rx , then

⇒

50 Rx + 600 - 100 Rx = 0

⇒

Rx = 12 W

=0

Test Your Concepts-IX

Based on Heating Effects and Power Consumption 1. Calculate the power delivered to each resistor in the circuit shown. 2Ω

3Ω

18 V

4Ω

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 77

1Ω

(Solutions on page H.220) 2. An electric teakettle has a multiposition switch and two heating coils. When only one of the coils is switched on, the well-insulated kettle makes a full pot of water to a boil in a time interval Δt. When only the other coil is switched on, it requires a time interval of 2Δt to boil the same amount of water. Find the time interval required to boil the same amount of water if both coils are switched while being used

9/20/2019 11:09:06 AM

3.78  JEE Advanced Physics: Electrostatics and Current Electricity

(a) in a parallel connection and (b) in a series connection. 3. When two unknown resistors are connected in series with a battery, the battery delivers total power Ps and carries a total current of I. For the same total current, a total power Pp is delivered when the resistors are connected in parallel. Determine the values of the two resistors. 4. Two resistors R1 and R2 are in parallel with each other. Together they carry total current I. (a) Determine the current in each resistor. (b) Prove that this division of the total current I between the two resistors results in less power delivered to the combination than any other division. In other words prove the general principle that current in a direct current circuit distributes itself so that the total power delivered to the circuit is a minimum. 5. Consider an ideal battery with emf V0 and two resistors with resistances R1 and R2. For the power absorbed by resistors to be maximum, should they be connected in series or in parallel, as in figure. R1

R2



(b) A fuse wire of radius 0.1 mm blows when a current of 10 A passes through it. What should be the radius of a fuse wire made of the same material which will blow at a current of 20 A? 9. In the circuit shown in figure abcd is a square. All the wires forming the square and its diagonals are homogeneous and have same cross-section. Find the ratio of power dissipated in resistors ab and cd. c

d

a

b V

10. The variable capacitor in figure is connected to a battery of emf V and internal resistance r. The current in the circuit is kept constant by changing the capacitance. (Assume that we are able to increase the capacitance indefinitely in order to accomplish this). C

V0

V0 Series connection

R1

R2

Parallel connection

6. Two resistors are made of identical material, have the same length but one has the diameter twice that of the other. Determine the ratio of powers dissipated in the two resistors when (a) each resistor is being connected across the same voltage source i.e., in parallel (b) they are connected in series 7. A 100 W, 110 V light bulb has a filament made of an alloy having a temperature coefficient of 0.0055 °C -1 at 0°C. The normal operating temperature of the bulb is 2000°C. How much current will the bulb draw at the instant it is turned on when the room temperature is 20°C and 2000°C? 8. (a) Two wires made of same tinned copper alloy having equal cross-sectional area but of different lengths are to be used as fuses. Show that the fuses will melt at the same value of current which is independent of the length of the wires.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 78

Vc I

Battery r S V



(a) Calculate the power supplied by the battery. (b) Compare the rate at which energy is supplied by the battery with the rate of change of the energy stored in the capacitor. 11. In the circuit shown in figure, find the power 4V

10 V I 3Ω



(a) supplied by 10 V battery (b) consumed by 4 V battery and (c) dissipated in 3 W resistance.

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Chapter 3: Electric Current and Circuits 3.79

12. In the circuit shown in figure, find the heat developed across each resistance in t = 2 s. 6Ω

3Ω

3Ω

5Ω

20 V

13. In which branch of the circuit shown, a 11 V battery be inserted so that it dissipates minimum power. What will be the current through the 2 W resistance for this position of the battery?

2Ω

4Ω

6Ω

14. In a circuit shown in figure if the internal resistances of the sources are negligible then at what value of resistance R will the thermal power generated in it be the maximum. What is its value? R

15. The current I through a rod of a certain metallic oxide is given by I = 0.2 V 1 2 , where Vis the potential difference across the rod. The rod is connected in series with a resistance to a 6 V battery of negligible internal resistance. What value should the series resistance have so that: (a) the current in the circuit is 0.4 A. (b) the power dissipated in the rod is twice that dissipated in the resistance. 16. A storage battery with emf 2.6 V loaded with external resistance produces a current 1 A. In this case the potential difference between the terminals of the storage battery equals 2 V. Find the thermal power generated in the battery and the power developed in it by electric forces. 17. The circuit considered shown in figure is connected for a duration of 2 minute. (a) Find the energy delivered by each battery. (b) Find the energy delivered to each resistor. (c) Identify the types of energy transformations that occur in the operation of the circuit and the total amount of energy involved in each type of transformation.

10 V

6V

3Ω 5Ω

3Ω 8Ω

6Ω

1Ω 4V

WHEATSTONE BRIDGE: CONDITION OF BALANCE The Wheatstone’s Bridge is shown in figure. B

A

I1

(I – I1) I

P

IG

Q I1 – IG C

G R

(I – I1 + IG) S

D E

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 79

I

1Ω 12 V

P, Q, R and S are four resistances, G is galvanometer and E is a battery. The Wheatstone’s Bridge is said to be balanced when no current flows in galvanometer. So, Potential of B = Potential of D . The condition of balance is best achieved by applying Kirchhoff’s Laws Applying Kirchhoff’s 2nd Law to loop ABDA

- I1 P - IG G + ( I - I1 ) R = 0 …(1)

Applying Kirchhoff’s 2nd Law to loop BCDB

- ( I1 - IG ) Q + S ( I - I1 + IG ) + IG G = 0 …(2)

9/20/2019 11:09:11 AM

3.80  JEE Advanced Physics: Electrostatics and Current Electricity

For balance IG = 0, so that (1) and (2) give

Illustration 54



I1 P = ( I - I1 ) R and



I1Q = ( I - I1 ) S …(4)

…(3)

Dividing (3) by (4), we get Condition of balance is

An electrical circuit is shown in figure. Calculate the potential difference across the resistor of 400 W as will be measured by the voltmeter V of resistance 400 W either by applying Kirchhoff’s rules or otherwise. V 400 Ω

P R = Q S



100 Ω

100 Ω

200 Ω

100 Ω

Remark(s) So, to conclude (a) When battery and galvanometer arms of a Wheatstone’s Bridge are interchanged, the balance position remains undisturbed while sensitivity of bridge changes. (b) When Wheatstone’s Bridge is balanced, the resistance in arm BD may be ignored while calculating the equivalent resistance of bridge between A and C. So, P and Q are in series, R and S are in series and both in parallel to each other. 1 1 1 = + ⇒  Rnet P + Q R + S Three other common forms of balanced Wheatstone’s Bridge are shown here.

The given circuit actually forms a balanced Wheatstone bridge (including the voltmeter) as shown in figure. Rv = 400 Ω V 400 Ω 100 Ω I2

(I – I1 + IG) S

B

10 V (a)



(ii) from B to D if

P R < Q S

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 80

= P

Ω 20

0 B

10 V (b)

Here, we see that

10

0

(c) If the bridge is not balanced, then the current will flow P R (i) from D to B if > Q S

Ω

G

=

R = 100 Ω

Ω

S

S

I2

0 10

R

B

I1

=

and

A

R

E

P = 100 Ω 100 Ω B A S = 200 Ω

R

Q

Q = 200 Ω

0

P

Q

20

P G

A

=

(I1 – IG) Q

Q

Ω

I

200 Ω

S

P

100 Ω

I1 100 Ω

I′

IG G

I1 I

Solution

R

I – I1 A

10 V

10 V (c)

P R =  Q S

{Bridge is balanced}

Therefore, resistance between A and B can be ignored and equivalent simple circuit can be drawn as shown in figure (c).

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Chapter 3: Electric Current and Circuits 3.81

The voltmeter will read the potential difference across resistance Q . So, currents I1 and I 2 are given by 10 1 I1 = I 2 = = A 100 + 200 30 ⇒ Potential difference across voltmeter is 20 ⎛ 1 ⎞ ΔV = I1Q = ( 200 ) ⎜ V= V ⎝ 30 ⎟⎠ 3 So, reading of voltmeter is

20 V 3

Solution

Let AJ = l cm , then JB = ( 100 - l ) cm At zero deflection of Galvanometer R1 RAJ l = = R2 RJB 100 - l 15 l ⇒ = 10 100 - l ⇒

3 l = 2 100 - l

⇒ 300 - 3l = 2l

THE METRE BRIDGE

⇒ 5l = 300

The metre bridge is the practical application of the Wheatstone network principle in which the ratio of two of the resistances, say R and S , is deduced from the ratio of their balancing lengths. AC is a 1 m long uniform wire. If AD = l cm , then DC = ( 100 - l ) cm Since Resistance ∝ Length

⇒ l = 60 cm

P

G R A

l

B

Q

Illustration 56

In a meter bridge, null point is found to be at 20 cm. When the known resistance R is shunted by 10 W resistance, null point is found to be shifted by 10 cm. Find the unknown resistance X. Solution R

J

S C

(100 – l)

D

G A

⇒

P l = Q 100 - l

If P is known then Q can be determined. At the balancing point galvanometer G gives no deflection at all. At A and C, the galvanometer must have deflections in opposite direction because, then only zero deflection can be expected when the jockey ( J) attached to the galvanometer is moved from A to C. Illustration 55

⇒

R1 = 15 Ω



R2 = 10 Ω

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 81

80 cm

B

R X = 20 80 R 1 = …(1) X 4

⎛ 10 R ⎞ ⎜⎝ ⎟ 10 + R ⎠ X = 10 90

10 4 = 10 + R 9 ⇒ 90 = 40 + 4 R ⇒

G

J

20 cm

Now shunting R by 10 W , i.e., 10 W connected in 10 R parallel to R , then new resistance becomes 10 + R (which is less than R ) so new balance length is 20 - 10 = 10 cm . Hence

In the meter bridge circuit shown in figure. Calculate the length AJ for null deflection in galvanometer.

A

X

B

⇒ R = 12.5 W ⇒ X = 4 R = 50 W

9/20/2019 11:09:32 AM

3.82  JEE Advanced Physics: Electrostatics and Current Electricity Illustration 57

Illustration 58

If resistance R1 in resistance box is 300 W , then the balanced length is found to be 75 cm from end A . The diameter of unknown wire is 1 mm and length of the unknown wire is 31.4 cm . Find the specific resistance of the unknown wire.

When we use 100 W and 200 W in place of R and X we get null point deflection at l = 33 cm . On interchanging the resistors, the null point length is found to be 67 cm . Calculate the end corrections x1 and x2.

Solution

Solution

R l Since, = X 100 - l ⎛ 100 - l ⎞ ⇒ X=⎜ R ⎝ l ⎟⎠ ⎛ 100 - 75 ⎞ ( ⇒ X=⎜ 300 ) = 100 W ⎝ 75 ⎟⎠

ρl ρl = A ⎛ π d2 ⎞ ⎜⎝ ⎟ 4 ⎠ π d2 X ⇒ ρ= 4l ⎛ 22 ⎞ ( -3 )2 ( 100 ) ⎜⎝ ⎟⎠ 10 ⇒ ρ= 7 ( 4 ) ( 0.314 )



⇒ x1 =

END CORRECTIONS IN METRE BRIDGE In meter bridge, some extra length (under the metallic strips) comes at points A and C. Therefore, some additional length x1 and x2 should be included at the ends. So, x1 and x2 are called the end corrections. Hence, in place of l we use l + x1 and in place of ( 100 - l ) we use 100 - l + x2 . To find x1 and x2, we use known resistors R1 and R2 in place of R and X and suppose we get null point length equal to l1 Then, R1 l1 + x1 = …(1) R2 100 - l1 + x2 Now, we interchange the positions of R1 and R2 and suppose the new null point length is l2 . Then, R2 l2 + x1 = …(2) R1 100 - l2 + x2 Solving equations (1) and (2), we get

x1 =

and x2 =

R2 l1 - R1l2 R1 - R2

R2 l1 - R1l2 R1 - R2

( 200 )( 33 ) - ( 100 )( 67 ) 100 - 200

⇒ x1 = 1 cm

where X =

⇒ ρ = 2.5 × 10 -4 Wm

x1 =



x2 =

⇒ x2 =

R1l1 - R2 l2 - 100 R1 - R2

( 100 )( 33 ) - ( 200 )( 67 ) 100 - 200

- 100

⇒ x2 = 1 cm

POST OFFICE BOX Post office box also works on the principle of Wheatstone’s bridge. In a Wheatstone’s bridge cirP R = , then the bridge is said to be balcuit, when Q X anced. So, unknown resistance X is given by

X=

Q R P B P A

Q C

G R

X D

It is observed that P and Q are set in arms AB and Q BC , where we can have the ratio can be set by P using 10 W , 100 W or 1000 W resistances.

R1l1 - R2 l2 - 100 R1 - R2

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Chapter 3: Electric Current and Circuits 3.83

A

P

B

1000 100 10

X

1

Q 1

10

10

20

Suppose at R = 4 W , we get deflection towards left and at R = 5 W , we get deflection towards right. Then, we can say that for balanced condition, R should lie between 4 W to 5 W . Q 10 Now, X = R = R = R = 4 W to 5 W . P 10 To get closer value of X , in the second o ­ bservation, Q 1 P = 100 ⎛ ⎞ let us choose = i.e., ⎜ ⎝ Q = 10 ⎟⎠ P 10 Suppose, now at R = 42 we get deflection towards left and at R = 43 W deflection is towards right. So R lies between 42 W and 43 W 1 Q 10 Now, X = R = R= R P 100 10

C

100 1000

∞

D

1

2

2

5

20

50 R

E

5000 2000 2000 1000 500 200 200 100

G

K2

K1

These arms are called ratio arm, initially we take Q Q = 10 W and P = 10 W to set = 1 . The unknown P resistance ( X ) is connected between C and D , the battery is connected across A and C . Now, adjust resistance in part A to D such that the bridge gets balanced. For this, keep on increasing the resistance with 1 W interval, check the deflection in galvanometer by first pressing key K1 then galvanometer key K 2 .

S. No.

Resistance in ratio arm AB (is P ohm)

1.

2.

3.

BC (is Q ohm)

10

100

1000

10

10

10

⇒ R lies between 4.2 W and 4.3 W

Q 1 Now, to get further closer value take = and P 100 so on. The observation table is shown below

Resistance in arm AD (is R ohm)

Direction of deflection

4

Left

5

Right

40

Left (large)

50

Right (large)

42

Left

43

Right

420

Left

424

Left

425

No deflection

426

Right

Unknown resistance Q X = × R (ohm) P

4 to 5

4.2 to 4.3

4.25

So, the correct value of X is 4.25 W

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3.84  JEE Advanced Physics: Electrostatics and Current Electricity

Circuit Diagram

Conceptual Note(s) To locate the null point, deflection battery key (K1) is pressed slightly earlier than the galvanometer key (K2) because if galvanometer key K2 is pressed first, then just after closing the battery key K1, the current suddenly increases and due to self-induction, a large back emf is generated in the galvanometer, which may damage the galvanometer.

Potentiometer consists of a long resistive wire AB of length L (about 6 m to 10 m long) made up of manganin or constantan and a battery of known ­ voltage e and internal resistance r called supplier battery or driver cell. Connection of these two forms primary circuit. e, r Primary circuit

Illustration 59

Secondary circuit

Calculate the maximum and minimum values of unknown resistance X , which can be determined using the post office box shown in the figure? A X

P

Q

1000 100 10

∞

D G

B

1

2

2

10 5

10

C

20

50 R

E

5000 2000 2000 1000 500 200 200 100 K2

K1

Solution

QR P So, maximum value of x is given by X=



Xmax =

Qmax Rmax Pmin

1000 ( 11110 ) 10 = 1111 kW

⇒ Xmax = ⇒ Xmax

Xmin =

⇒ Xmin =

Qmin Rmin Pmax

( 10 ) ( 1 )

1000 ⇒ Xmin = 0.01 W

J

A E

B

G

J = Jockey K = Key RAB = R = Resistance of potentiometer wire ρ = Specific resistance of potentiometer wire = Variable resistance which controls the Rh current through the wire AB

100 1000

20

Rh

K

One terminal of another cell (whose emf E is to be measured) is connected at one end of the main circuit and the other terminal at any point on the resistive wire through a galvanometer G . This forms the secondary circuit. Other details are as follows : (a) The specific resistance ( ρ ) of potentiometer wire must be high but its temperature coefficient of resistance ( α ) must be low. (b) All higher potential points (terminals) of primary and secondary circuits must be connected together at point A and all lower potential points must be connected to point B or jockey. (c) The value of known potential difference must be greater than the value of unknown potential difference to be measured. (d) The potential gradient must remain constant. For this the current in the primary circuit must remain constant and the jockey must not slide while in contact with the wire. (e) The diameter of potentiometer wire must be uniform everywhere.

Potential Gradient (x)

POTENTIOMETER

Potential difference (or fall in potential) per unit length of wire is called potential gradient i.e.,

Potentiometer is a device mainly used to measure emf of a given cell and to compare emf’s of cells. It is also used to measure internal resistance of a given cell.



03_Physics for JEE Mains and Advanced - 2_Part 2.indd 84

x=

V volt where L m

9/20/2019 11:10:07 AM

Chapter 3: Electric Current and Circuits 3.85

In balanced condition E = xl

e ⎛ ⎞ V = IRAB = ⎜ R ⎝ R + Rh + r ⎟⎠

So, potential gradient x is given by x=

⇒ E = xl =

V IR I ρ e R = = = L L A ( R + Rh + r ) L

V IR e ⎛ ⎞R l= l=⎜ ×l L L ⎝ R + Rh + r ⎟⎠ L

If V is constant then L ∝ l

(a) The potential gradient directly depends upon the ⎛ R⎞ (i) resistance per unit length ⎜ ⎟ of potentiom⎝ L⎠ eter wire. (ii) radius of potentiometer wire (i.e., Area of cross-section) (iii) specific resistance of the material of potentiometer wire (i.e., ρ ) (iv)  current flowing through potentiometer wire ( I ) (b) Potential gradient indirectly depends upon the (i) emf of battery in the primary circuit (i.e., e) (ii) resistance of rheostat in primary circuit (i.e., Rh )

⇒

x1 L1 l1 = = x2 L2 l2

Standardization of Potentiometer The process of determining potential gradient experimentally is known as standardization of potentiometer. e, r

J

A E0

Working

Rh

K

B

G

E

Suppose the jockey is made to touch a point J on wire. Then potential difference between A and J will be V = xl . At this length ( l ) , two potential difference values are obtained. (a) V due to battery e and (b) E due to unknown cell

⇒ x= Rh

K

e, r

J1

J J2

B

G G G

E l

If V > E , then current will flow in galvanometer ­circuit in one direction, shown as If V < E , then current will flow in galvanometer ­circuit in opposite direction, shown as If V = E , then no current will flow in galvanometer circuit this condition to known as null deflection position, length l is known as balancing length, shown as

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 85

E0 l0

Sensitivity of Potentiometer

()

A

Let the balancing length for the standard emf E0 is l0 , then by the principle of potentiometer we have E0 = xl0

A potentiometer is said to be more sensitive, if it measures a small potential difference more accurately. (a) The sensitivity of potentiometer is assessed by its potential gradient. The sensitivity is inversely proportional to the potential g ­ radient. (b) In order to increase the sensitivity of potentiometer, the (i) resistance in primary circuit will have to be decreased. (ii)  length of potentiometer wire will have to be increased so that the length may be measured more accuracy.

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3.86  JEE Advanced Physics: Electrostatics and Current Electricity

Conceptual Note(s) (a) In the potentiometer arrangement shown, E1 1

A

l

A

J l1

2

E 2 r2

D

l2 R



r1

B

3

A

a R

S



VAJ = VDC



⇒ I1RAJ = E2 - I2 r2



⇒ I1l  = E2 - I2 r2 …(1)

where, l is the resistance per unit length of potentiometer wire AB. The length l is called the balance point length. Currents I1 and I2 are independent with each other. Current I2 = 0, if switch is open. (b) From equation (1), the null point length is given by =

r

C

under balanced condition (when IG = 0) Loop ➀ and Loop ➂ are independent of each other. All problems in this condition can be solved by a single equation. i.e.,



A constant voltage V0 is applied to a potentiometer of resistance R connected to an ammeter. A constant resistor r is connected to the sliding contact of the potentiometer and the fixed end of the potentiometer. How will the reading of ammeter vary as the sliding contact is moved from one end of the potentiometer to the other. The resistance of ammeter is assumed to be negligible.

G





Illustration 60

E2 - I2 r2 I1l



Now, suppose E1 is increased then I1 will also increase and null point length l will decrease. (c) Under balanced condition, a part of potential difference of E1 is balanced by the lower circuit. So, normally E2 < E1 for taking balance point length. Similarly, we observe that VAJ = VDC. Also, VA = VD and VJ = VC, so we conclude that positive terminals of both batteries should be on same side and negative terminals on the other side. (d) If we do not get any balanced condition ( IG ≠ 0 ), then the given circuit is simply a three loop problem, which can be solved by applying Kirchhoff ’s Laws.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 86

V0

Solution

If x is the resistance of the potentiometer between point a and the sliding contact, the total resistance rx between a and the sliding contact is , while the r+x rx resistance of the entire circuit is ( R - x ) + . The r+x current supplied by the source is I=

V0 R-x+

rx r+x

The potential difference between the sliding contact and point a is V=

V0 rx

( R - x ) ( r + x ) + rx

=

V0 rx

Rx - x 2 + Rr The current passing through ammeter is, V0 r …(1) Rx - x 2 + Rr To find the extremum (maximum, minimum or constant)

Ia =



dI a =0 dx

- R + 2x ⎡ ⇒ V0 r ⎢ 2 2 ⎣ ( Rx - x + Rr ) ⇒ x=

⎤ ⎥=0 ⎦

R 2

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Chapter 3: Electric Current and Circuits 3.87

Substituting x in equation (1) we find the minimum current I min =

V0 r R⎞ ⎛ R⎜ r + ⎟ ⎝ 4⎠

Thus, as the sliding contact is moved the current through the ammeter passes through a minimum and the smaller the r the deeper the minimum. At x = 0 V and x = R , a current of I max = 0 passes through the R I x ammeter. The a versus curves for several valR I max r are shown in figure. ues of R 1

Ia Imax

0.6 0.4

r = 0.2 R

0.2 O

1

0.5

x R

APPLICATIONS OF POTENTIOMETER To Determine the Internal Resistance of a Primary Cell (a) Initially, in the secondary circuit the key K ′ remains open and balancing length l1 = lkey open is obtained. Since cell E is in open circuit so it’s emf balances for a length l1 i.e., E = xl1 …(1) (b) Now key K ′ is closed so cell E comes in closed circuit. If the process of balancing repeated again then potential difference V balances on length l2 = lkey closed i.e.,

V = xl2 …(2) Rh

K

e, r

J

A E R

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 87

G K′





⎛E ⎞ r = ⎜ - 1⎟ R ⎝V ⎠



⇒

⎛ lkey open ⎞ ⎛ l -l ⎞ r=⎜ 1 2⎟R=⎜ - 1⎟ R ⎝ l2 ⎠ ⎝ lkey closed ⎠

Illustration 61

In a potentiometer circuit to find the internal resistance r of the cell of emf E , when K is open, the balance point is obtained at 60 cm . When K is closed, the balance point is obtained at 50 cm . What is the value of r ? If the cell is shorted by a 10 W resistor, then calculate the internal resistance of the cell. Solution

r=R r = 0.5 R

0.8

(c) Since we know that the internal resistance is given by

l1 = 60 cm , when K is open and l2 = 50 cm , when K is closed ⎛ lkey open ⎞ Since r = ⎜ - 1⎟ R l ⎝ key closed ⎠ 1 ⎛l ⎞ ⎛ 60 ⎞ - 1 ⎟ 10 W = × 10 W = 2 W ⇒ r = ⎜ 1 - 1⎟ R = ⎜ ⎝ ⎠ 50 5 ⎝ l2 ⎠ Illustration 62

When a resistor of 5 W is connected across a cell its terminal potential difference is balanced by 150 cm of potentiometer wire and when a resistor of 10 W is connected across the cell, then the terminal potential difference is balanced by 175 cm of the potentiometer wire. Calculate the internal resistance of the cell. Solution

Since the same cell is used in both cases, so lkey open = l will remain the same for both. For the first case, we have lkey closed = l1 and R = R1 . So

B

⎛ lkey open ⎞ ⎛ l ⎞ r=⎜ - 1 ⎟ R = ⎜ - 1 ⎟ R1 …(1) ⎝ l1 ⎠ ⎝ lkey closed ⎠

For the second case, we have lkey closed = l2 and R = R2 . So

⎛ lkey open ⎞ ⎛ l ⎞ r=⎜ - 1 ⎟ R = ⎜ - 1 ⎟ R2 …(2) l l ⎝ ⎠ ⎝ key closed ⎠ 2

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3.88  JEE Advanced Physics: Electrostatics and Current Electricity

From (1), we get

Illustration 63

⎛ l ⎞ r ⎜ 1 ⎟ = l - l1 …(3) ⎝ R1 ⎠

From (2), we get

⎛ l ⎞ r ⎜ 2 ⎟ = l - l2 …(4) ⎝ R2 ⎠

Subtracting (4) from (3), we get

Solution

E1 = 1.5 V , l1 = 30 cm E2 = ? , l2 = 50 cm

⎛ l l ⎞ r ⎜ 1 - 2 ⎟ = l2 - l1 ⎝ R1 R2 ⎠

⇒ r=

In a potentiometer arrangement a cell of emf 1.5 V gives a balance point at 30 cm length of wire. Now, when the cell is replaced by another cell, the balance point shifts to 50 cm. Calculate the emf of second cell?

l2 - l1 175 - 150 25 = = =2 ⎛ l1 l2 ⎞ 150 - 175 12.5 ⎜⎝ R - R ⎟⎠ 5 10 1 2

⇒ r=2W

Using the formula for comparison of emf of cells by potentiometer, we have

⇒ E2 =

Comparison of Emf’s of Two Cell Let l1 and l2 be the balancing lengths with the cells E1 and E2 respectively then E1 = xl1 and E2 = xl2 e, r

Rh

K

E1 E2

Comparison of Resistances Let the balancing length for resistance R1 (when XY is connected) is l1 and let balancing length for resistance R1 + R2 (when YZ is connected) is l2 .

X

2 i

⇒

+

E2

–

( E1 + E2 ) = xl1 ( E1 - E2 ) = xl2 E1 + E2 l1 = E1 - E2 l2

E1 l1 + l2 ⇒ = E2 l1 - l2

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 88

+

E1 –

–

E2

+

Y

G

B Z

R2

R1 K1

Let E1 > E2 and both are connected in series. If balancing length is l1 , when cells assist each other and is l2 , when they oppose each other (as shown), then



J

A

E1 l1 ⇒ = E2 l2

E1 –

Rh

B

G

1

l2 50 × E1 = × 1.5 V = 2.5 V l1 30

K

J

A

+

E2 l2 = E1 l1

Rh1

Then IR1 = xl1 and I ( R1 + R2 ) = xl2 ⇒

R2 l2 - l1 = R1 l1

Illustration 64

One of the circuits for the measurement of resistance using a potentiometer is shown. The galvanometer is connected and zero deflection is observed at length AJ = 30 cm . In second case the secondary cell is changed. Take ES = 10 V and r = 1 W in 1st reading and ES = 5 V and r = 2 W in 2nd reading.

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Chapter 3: Electric Current and Circuits 3.89

In second case, the zero deflection is observed at length AJ = 10 cm . Calculate the resistance R (in ohm). Ep

J

A

B

G

R

r

Es

(b) For the calibration of an ammeter, 1 W standard resistance coil is specifically used in the secondary circuit of the potentiometer, because the potential difference across 1 W is equal to the current flowing through it. So, from Ohm’s Law we get V = I (c) If the balancing length for the emf E0 is l0 then

E0 = xl0



x=

⇒

e

Solution

At zero deflection, we have

I2 =

K1

B

E1 1Ω

ES1

1

2

3

G A

r1 + R 10 1+ R

⇒ I1 =



A

V I1 R I 2 R …(1) = = l 30 10

Now I1 =

E0 (Process of standardisation) l0

K2

ES2 r2 + R

5 2+R Substituting in (1), we get ⇒ I2 =

10 R

( 1 + R ) × 30

=

5R

( 2 + R ) × 10

⇒ 4 + 2R = 3 + 3 R ⇒ R=1W

Calibration of Ammeter Checking the correctness of ammeter readings with the help of potentiometer is called calibration of ammeter. (a) In the process of calibration of an ammeter the current flowing in a circuit is measured by an ammeter and the same current is also measured with the help of potentiometer. By comparing both the values, the error(s) in the ammeter readings can be determined.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 89

(d) Let I ′ be the current that flows through 1 W resistance, thus giving the potential difference as V ′ = I ′ ( 1 ) = xl1 where l1 is the balancing length. So error can be found as

⎛E ⎞ ΔI = I - I ′ = I - xl1 = I - ⎜ 0 ⎟ l1 ⎝ l0 ⎠

Calibration of Voltmeter (a) Practical voltmeters are not ideal, because these do not have infinite resistance. The error of such practical voltmeter can be found by comparing the voltmeter reading with calculated value of potential difference by potentiometer. (b) If l0 is balancing length for E0 , the emf of standard cell (by connecting 1 and 2 of b ­ i-directional E0 key), then x = . l0 (c) The balancing length l1 for unknown potential difference V ′ is given by (by closing 2 and 3), ⎛E then V ′ = xl1 = ⎜ 0 ⎝ l0

⎞ ⎟⎠ l1

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3.90  JEE Advanced Physics: Electrostatics and Current Electricity e

Solution

Rh

K1

Length of potentiometer wire = 10 m A

E0

2

V

3

RB

B

C

1

K2

G

So, fall in potential per meter (also known as poten2 tial gradient) = = 0.2 Vm -1 10 Since the cell gives a null point at 6.9 m, hence its emf is,

Rh

If the voltmeter reading is V then the error will be ( V - V ′ ) which may be positive, negative or zero.

DIFFERENCE BETWEEN VOLTMETER AND POTENTIOMETER Voltmeter

Potentiometer

It’s resistance is high but finite.

It’s resistance is infinite.

It draws some current from source of emf.

It does not draw any current from the source of unknown emf.

The potential difference measured by it is lesser than the actual potential difference.

The potential difference measured by it is equal to actual potential difference.

Its sensitivity is low.

Its sensitivity is high.

It is a versatile instrument.

It measures only emf or potential difference.

It is based on deflection method.

It is based on zero deflection method.



E = ( 0.2 )( 6.9 ) = 1.38 V

Resistance of potentiometer wire = 11.5 × 10 = 115 W . When a resistance of 5 W is connected, total resistance becomes Rnet = 115 + 5 = 120 W and Current, I =

2 A 120

So, new potential gradient = (current) (resistance per unit length) 23 ⎛ 2 ⎞( ⇒ Potential Gradient = ⎜ 11.5 ) = Vm -1 ⎝ 120 ⎟⎠ 120 Let l ′ be the position of new null point, then

⎛ 23 ⎞ ⎜⎝ ⎟ l ′ = E = 1.38 120 ⎠

⇒ l′ =

1.38 × 120 = 7.2 m 23

Illustration 66

In a potentiometer circuit, the emf of driver cell is 2 V and internal resistance is 0.5 W . The potentiometer wire is 1 m long. It is found that a cell of emf 1 V and internal resistance 0.5 W is balanced against 60 cm length of the wire. Calculate the resistance of potentiometer wire. Solution

Illustration 65

A cell having a steady emf of 2 V is connected across the potentiometer wire of length 10 m. The potentiometer wire is of manganin and having resistance of 11.5 Wm -1 . An another cell gives a null point at 6.9 m. If a resistance of 5 W is put in series with the potentiometer wire, find the new position of the null point.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 90

Let resistance of potentiometer wire be RAB = R . At balance, we have

I=

2 R + 0.5

⇒ VAB = IRAB =

2R …(1) R + 0.5

Since, VAJ = 1 V {∵ There is no deflection} Across 60 cm length of potentiometer wire, potential difference is 1 V

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Chapter 3: Electric Current and Circuits 3.91

So, across 1 cm of potentiometer wire, potential dif1 ference is V 60

I1

2 V, 0.5 Ω

A

I

A

60 cm J

40 cm I

20 V

I1

B

6Ω

Hence, across 100 cm of potentiometer wire, potential difference is 5 ⎛ 1 ⎞ = ⎜ ⎟ 100 = V …(2) ⎝ 60 ⎠ 3

where E = 8 V , r = 2 W ⎛ 20 ⎞ =2A I1 = ⎜ ⎝ 2 + 8 ⎟⎠

2l ⎛R ⎞ ⎛ 8 ⎞ RAJ = ⎜ AB ⎟ lAJ = ⎜ ⎟⎠ l = ⎝ 100 25 ⎝ lAB ⎠

8 =1A 2+6 Substituting these values in equation (1), we get I2 =

Equating (1) and (2), we get 5 2R = 3 R + 0.5

⇒ 5R + 2.5 = 6 R



⇒ R = 2.5 W

⇒

Illustration 67

In the figure shown, wire AB has a length of 100 cm and resistance 8 W . Find the balance point length l . 20 V

I1

J

2 8 V, 2 Ω G IG = 0 C D I2 3 I2 I2

B

1 V, 0.5 Ω

VAB

I1

1

G IG = 0



2Ω

2Ω

⎛ 2l ⎞ 2 ⎜ ⎟ = 8 - ( 1 )( 2 ) ⎝ 25 ⎠ 4l =6 25

⇒ 4l = 150 ⇒ l=

150 = 37.5 cm 4

Illustration 68

A l A

J

B

8 V, 2 Ω G I = 0 G

6Ω

A potentiometer wire of length 100 cm has a ­resistance of 10 W . It is connected in series with a resistance R and a cell of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. What is the value of R ? Solution

Solution

VAJ = VCD

⇒ I1 RAJ = E - I 2 r …(1)

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 91

2V

I

For condition of balance, we have IG = 0 . So loops ➀ and ➂ are independent of each other. Hence we observe that A

40 cm J

10 mV

G IG = 0

R

B lAB = 100 cm RAB = 10 Ω

9/20/2019 11:11:24 AM

3.92 JEE Advanced Physics: Electrostatics and Current Electricity

For the condition of balance i.e., when IG = 0, we have I=

⇒

2 2 = R + RAB R + 10

VAB = IRAB

For a length of 1 cm, potential difference is ⎛ 20 ⎞ 1 =⎜ ⎝ R + 10 ⎟⎠ 100

For a length of 40 cm , potential difference is

⎛ 20 ⎞ =⎜ V ⎝ R + 10 ⎟⎠

Since VAJ = 10 mV = 10 × 10 -3 V = 10 -2 V

…(1)

For a length of 100 cm , potential difference across AB is 20 VAB = R + 10



VAJ =

20 × 40 100 ( R + 10 )

⇒

20 ⎤ ⎡ -2 ⎢⎣ 10 ( R + 10 ) ⎥⎦ 40 = 10

⇒ ⇒

800 = R + 10 R = 790 W

{∵ of (1)}

Test Your Concepts-X

Based on Wheatstone Bridge and Potentiometer 1. Figure shows a potentiometer using a cell C of emf 2 V and internal resistance 0.4 W connected to a resistor wire AB. A standard cell of constant emf of 1.02 V gives a balance point at 67.3 cm length of the wire. A very high resistance R = 600 kW is put in series with the standard cell. This resistance is shorted by inserting switch S when close to the balance point. The standard cell is then replaced by a cell of unknown emf E and the null point turns out to be 82.3 cm length of the wire.

(Solutions on page H.226) (f) Would the circuit work well for determining extremely small emfs of the order of a few millivolts? What modification do you suggest in the circuit. 2. Figure shows a potentiometer circuit for determining the internal resistance of a cell. When switch S is open, the balance point is found to be at 76.3 cm of the wire. When switch S is closed and the value of R is 4 W, the balance point shifts to 60 cm. Find the internal resistance of cell C′.

C

C

A

S

J

B G

C1

J′ J

A C′

B

G

R

(a) What is the value of E? (b) What is the purpose of using the high resistance R? (c) Is the null point affected by this high resistance? (d) Is the null point affected by the internal resistance of the cell C? (e) Would this method work if: (i) the internal resistance of cell C were higher than the resistance of wire AB and (ii) the emf of cell C were 1 V instead of 2 V?

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 92

R S

3. Figure shows a metre bridge consisting of two resistances X and Y together in parallel with a metrelong constantan wire AC of uniform cross-section. D is a movable contact that can slide along the wire AC. The resistors X, Y and resistances of segments AD and DC of the wire constitute the four arms of the bridge. The length of wire AC is 100 cm. X is a standard 4 W resistor and Y is a coil of wire. With Y immersed in melting ice the null point is found to be at a distance of 40 m from point A. When the

9/20/2019 11:11:30 AM

Chapter 3: Electric Current and Circuits 3.93

coil Y is heated to 100°C, a 100 W resistor has to be connected in parallel with Y in order to keep the bridge balanced at the same point. Calculate the temperature coefficient of resistance of the coil. X=4Ω

Y

G A

l= 40 cm

(100 – l) = 60 cm

C

4. 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 5 W a balance is obtained when the cell is connected across 0.4 m of the wire. Find the internal resistance of the cell. 5. A battery of emf 4 V is connected across a 10 m long potentiometer wire having a resistance per unit length 1.6 Wm-1. A cell of emf 2.4 V is connected so that its negative terminal is connected to the low potential end of the potentiometer wire and the other end is connected through a galvanometer to a sliding contact along the wire. It is found that the no-deflection point occurs against the balancing length of 8 m. Calculate the internal resistance of the 4 V battery. E

r L

A

L–x G

B x E1 Cell 2.4 V

6. A thin uniform wire AB of length 1 m, an unknown resistance X and a resistance of 12 W are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 93

B

A

C

D



B

D

12 Ω

X

(a) Are there positive and negative terminals on the galvanometer? (b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points. (c)  After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance X. 7. R1, R2, R3 are different values of R. A, B, C are the null points obtained corresponding to R1, R2 and R3 respectively. For which resistor, the value of X will be the most accurate and why? X

R G B

A

C

8. Figure illustrates potentiometer circuit by means of which we can vary a voltage V applied to a certain device possessing a resistance R. The potentiometer wire has a length L and a resistance R0. A voltage V0 is applied to the terminals of wire. Find the voltage V fed to the device as a function of distance x. Analyse the case R  R0 . V0 R0, L x R

9/20/2019 11:11:32 AM

3.94  JEE Advanced Physics: Electrostatics and Current Electricity

9. For the potentiometer arrangement to measure the internal resistance of a cell, when the switch is open in lowermost loop the balance point length is 60 cm. When the switch is closed with a known resistance of R = 4 W, the balance point length decreases to 40 cm. Calculate the internal resistance of the unknown battery. 10. For the potentiometer arrangement shown in the figure, the balance point is obtained at a distance 75 cm from A when the key K is open.

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 94

E0 = 2 V

A

B E1 = 1.5 V

G

r 6Ω



K

The second balance point is obtained at 60 cm from A when the key K is closed. Find the internal resistance (in W) of the battery E1.

9/20/2019 11:11:32 AM

Chapter 3: Electric Current and Circuits 3.95

RC Circuit and Applications RC CIRCUIT

Travel direction

a

Consider the circuit shown below. The capacitor is connected to a DC voltage source of voltage V . At time t = 0 , the switch S is closed. The capacitor initially is uncharged,

R

C +q

R

I 2 S

V

V 4

3 (a) (b) RC circuit diagram for t < 0   Circuit diagram for t < 0

In particular, for t < 0 , there is no voltage across the capacitor so the capacitor acts like a short circuit. At t = 0 , the switch is closed and current begins to flow and has a value

I0 =

V R

At this instant, the potential difference from the battery terminals is the same as that across the resistor. This initiates the charging of the capacitor. As the capacitor starts to charge, the voltage across the capacitor increases in time. At any instant, the voltage across the capacitor is

VC ( t ) =



03_Physics for JEE Mains and Advanced - 2_Part 2.indd 95

b

+q

–q

Lower V

ΔV = Vb – Va = –

q⎞ dq 1 ⎛ = ⎜V - ⎟ ⎝ dt R C⎠

This equation can be solved by the method of separation of variables. The first step is to separate terms involving charge and time, (this means putting terms involving dq and q on one side of the equality sign and terms involving dt on the other side), dq

q =0 C

dq q ⇒ V -R - = 0 dt C

q C

Since I must be the same in all parts of the series ­circuit, the current across the resistance R is equal to the rate of increase of charge on the capacitor plates. The current flow in the circuit will continue to decrease because the charge already present on the capacitor makes it harder to put more charge on the capacitor. Once the charge on the capacitor plates reaches its maximum value q0 the current in the circuit will drop to zero. Thus, the charging capacitor satisfies a first order differential equation that relates the rate of change of charge to the charge on the capacitor

q(t ) C

Applying Kirchhoff’s loop rule to 12341 shown in ­figure (b), we obtain V - IR -

Higher V

b q C Kirchhoff’s rule for capacitors

–q

S

ΔV = Vb – Va = +

Higher V a

1

+q

Travel direction

q ( initial ) = q ( at t = 0 ) = 0 C

–q

Lower V

Charging a Capacitor

⇒

q⎞ ⎛ ⎜⎝ V - ⎟⎠ C

=

1 dt R

dq 1 = dt CV - q RC

Integrating both sides of the above equation, we get dq ⎫ ⎧ ⎬ ⎨∵ I = dt ⎭ ⎩

q



∫ 0

t

dq 1 = dt CV - q RC

∫ 0

9/20/2019 11:11:40 AM

3.96  JEE Advanced Physics: Electrostatics and Current Electricity

The current in the charging circuit decreases exponentially in time, I ( t ) = I 0 e - t RC . This function is

t ⎛ CV - q ⎞ ⇒ log e ⎜ =⎝ CV ⎟⎠ RC ⇒ q ( t ) = CV ( 1 - e - t RC ) = q0 ( 1 - e - t RC ) where q0 = CV is the maximum amount of charge stored on the plates. The time dependence of q ( t ) is plotted in figure below:

often written as I ( t ) = I 0 e - t t where t = RC is called the time constant. The SI unit of t is second, as can be seen from the dimensional analysis ⎛ [V ] ⎞ ⎛ [ C ] ⎞ ⎝ [ A ] ⎟⎠ ⎜⎝ [ V ] ⎟⎠

[ t ] = [ RC ] = [ W ][ F ] = ⎜

q



CV CV(1 – e–1)

τ = RC

t

τ

Once we know the charge on the capacitor we also can determine the voltage across the capacitor, VC ( t ) =

At that time, the voltage across the capacitor is equal to the applied voltage source and the charging process effectively ends,

VC =

C

=

q0 = V (Battery Voltage) C

The current that flows in the circuit is given by I (t ) =



dq ⎛ V ⎞ - t RC =⎜ ⎟e = I 0 e - t RC dt ⎝ R ⎠

The coefficient in front of the exponential ( I 0 ) is equal to the initial current that flows in the circuit when the switch was closed at t = 0 . The graph of current as a function of time is shown in figure below: I I0 = V R I0 e τ

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 96

I ( t + t ) = I ( t ) e -1 Vc V

q ( t → ∞ ) = CV = q0

q(t → ∞ )

[C ] [C ] = = [s] [ A ] [C ] [ s]

The time constant t is a measure of the decay time for the exponential function. This decay rate satisfies the following property

q(t ) = V ( 1 - e - t RC ) C

The graph of voltage as a function of time has the same form as shown in figure. From the figure, we see that after a sufficiently long time the charge on the capacitor approaches the value

[t ] =

t

V(1 –

e–1)

τ = RC

τ

t

which shows that after one time constant t has elapsed, the current falls off by a factor of e -1 = 0.368 , as indicated in figure above. Similarly, the voltage across the capacitor (shown in figure) can also be expressed in terms of the time constant t VC ( t ) = V ( 1 - e - t t ) Notice that initially at time t = 0 , VC ( t = 0 ) = 0 . After one time constant t has elapsed, the potential difference across the capacitor plates has increased by a factor ( 1 - e -1 ) = 0.632 of its final value

VC ( t ) = V ( 1 - e -1 ) = 0.632 V

Discharging a Capacitor Suppose initially the capacitor has been charged to some value q0 . For t < 0 , the switch is open and the potential difference across the capacitor is given by q VC = 0 . On the other hand, the potential difference C across the resistor is zero because there is no current flow, that is, I = 0 . Now suppose at t = 0 the switch is closed (shown in figure). The capacitor will begin to discharge.

9/20/2019 11:11:52 AM

Chapter 3: Electric Current and Circuits 3.97 R –q0 +q0

R

1 S

C

–q

t0

3

Discharging the RC circuit

The charged capacitor is now acting like a voltage source to drive current around the circuit. When the capacitor discharges (electrons flow from the negative plate through the wire to the positive plate), the voltage across the capacitor decreases. The capacitor is losing strength as a voltage source. Applying the Kirchhoff’s loop rule to 12341, the equation that describes the discharging process is given by



∫

q0

t

dq 1 =dt q RC

∫ 0

t ⎛ q ⎞ ⇒ log e ⎜ ⎟ = q RC ⎝ 0⎠ ⇒ q ( t ) = q0 e - t RC The voltage across the capacitor is then

VC ( t ) =

q ( t ) ⎛ q0 ⎞ - t RC =⎜ ⎟e ⎝ C⎠ C

A graph of voltage across the capacitor vs. time for the discharging capacitor is shown in figure. I

q IR - = 0 C

I0 =

q0 RC

The current that flows away from the positive plate is proportional to the charge on the plate,

I=-

dq dt

dq q +R =0 C dt

The current also exponentially decays in the circuit as can be seen by differentiating the charge on the capacitor I=-

Vc

A graph of the current flowing in the circuit as a function of time also has the same form as the voltage graph depicted in figure.

At t = 0 , switch S is closed. The charge on the capacitor is varying with time as Q = Q0 ( 1 - e -α t ) . Obtain the value of Q0 and α in the given circuit parameters. R1

q0 C S

V0 e

dq 1 =dt q RC

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 97

C

R2

V τ



dq ⎛ q0 ⎞ - t RC =⎜ ⎟e dt ⎝ RC ⎠

Illustration 69

This equation when integrated by using the method of separation of variables

V0 =

t

τ

The negative sign in the equation is an indication that the rate of change of the charge is proportional to the negative of the charge on the capacitor. This is due to the fact that the charge on the positive plate is decreasing as more positive charges leave the positive plate. Thus, charge satisfies a first order differential equation

I0 e

t

Solution

Q0 is the steady state charge stored in the capacitor.

9/20/2019 11:11:58 AM

3.98  JEE Advanced Physics: Electrostatics and Current Electricity

Q0 = C (Potential difference across capacitor in steady state) R1

V

R2

C

(a) What is the time constant before the switch is closed? (b) What is the time constant after the switch is closed? (c) Find the current through the switch as a function of time after the switch is closed. Solution

⇒ Q0 = C (Steady state current through R2 ) ( R2 ) ⎛ V ⎞ ⇒ Q0 = C ⎜ R2 ⎝ R1 + R2 ⎟⎠ ⇒ Q0 =

CVR2 R1 + R2

Since, α =

1 1 = t C CRnet

t = Req C = ( R1 + R2 ) C The amount of charge stored in the capacitor is

where, Rnet is the equivalent resistance across ­capacitor after short circuiting the battery. So, R1

R2



Rnet =

⇒ α=

R1 R2 {∵ R1 and R2 are in parallel} R1 + R2 R + R2 = 1 CR1 R2 ⎞ R1 R2 ⎟ R1 + R2 ⎠ 1

⎛ C⎜ ⎝

(a) Before the switch is closed, the two resistors R1 and R2 are in series with the capacitor. Since the equivalent resistance is Req = R1 + R2 , the time constant is given by

q ( t ) = CE ( 1 - e - t t ) ( b) After the switch is closed, the closed loop on the right becomes a decaying RC circuit with time constant t ′ = R2 C . Charge begins to decay according to q ′ ( t ) = CEe - t t ′ = CEe - t R2C (c) The current passing through the switch consists of two sources : (i)  the steady current I1 from the left circuit E given by I1 = and R1 (ii) the decaying current I 2 from the RC ­circuit, given by I′(t ) =

  The negative sign in I ′ ( t ) indicates that the direction of flow is opposite of the charging process. Since both I1 and I ′ move downward across the switch, the total current is given by

Illustration 70

In the circuit in figure, suppose the switch has been open for a very long time. At time t = 0 , it is suddenly closed. R1

E

S

C

R2

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 98

dq ′ ⎛ CE ⎞ - t t ′ ⎛ E ⎞ - t R2C = -⎜ e = -⎜ e ⎝ t ′ ⎟⎠ dt ⎝ R2 ⎟⎠



I ( t ) = I1 + I ′ ( t ) =

E ⎛ E ⎞ - t R2C + e R1 ⎜⎝ R2 ⎟⎠

Problem Solving Technique(s) (a) An uncharged capacitor offers zero resistance to the current in a circuit, i.e., the branch containing the uncharged capacitor can be assumed to be short circuited in terms of the capacitance. That is the capacitor can just be thought of being absent initially.

9/20/2019 11:12:11 AM

Chapter 3: Electric Current and Circuits 3.99

(b) A fully charged capacitor offers an infinite resistance to the current and hence no current will pass through the branch that contains a fully charged capacitor. (c) While solving problems, whenever we come across the branch that contains a fully charged capacitor then we can omit that branch to calculate the net resistance of the circuit. (d) Please be alert that no current will pass through the branch of the circuit that contains a fully charged capacitor However, this does not imply that potential across the capacitor is zero. The Q potential across the capacitor will be ΔV = ± . C For instance, consider the circuit shown. 4Ω S 2Ω

10 V

2Ω

Step-1: Short-circuit the battery i.e., just remove the battery and join the ends across which it was connected. Step-2: Find net resistance across the capacitor (say it Rnet), Just think the capacitor is not there and the two plates of the capacitor had been converted to the ends across which we wish to find the net equivalent resistance Rnet of the circuit. OR For doing this just remove the capacitor and make the points across which the capacitor was connected as the current inlet and outlet points. Step-3: Then, the capacitive time constant is t = ( Rnet ) C

12 R

The circuit at time t = 0, when switch is closed, becomes as shown in figure.

At t = 0

2Ω 1A

2A

2Ω 1A

at t = 0

When the capacitor becomes fully charged i.e., when t → ∞, the circuit in the figure becomes as if no current passes through the branch containing the capacitor. 4Ω I A

C I=

10 = 5 6 3A

are in parallel, so their combined resistance is ( 12R ) ( 6R ) = 4R. Now this 4R is in series with the 12R + 6R remaining 2R. 12 R 2R

2Ω I

B As t → ∞ , the capacitor C blocks the current flow in the branch AB

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 99

6R

B A

I 2Ω

10 V

6R

C

10 V

I

2R

V

4Ω



To find the equivalent time constant of a circuit, follow the steps mentioned here :

For example, in the circuit shown in figure, after short circuiting the battery 6R and 12R

5 μF



Equivalent Time Constant

Step-1: Battery Shorted Step-2: Capacitor removed to find equivalent resistance across A and B. Hence, Rnet = 4R + 2R = 6R ⇒ t = ( Rnet ) C = 6RC

9/20/2019 11:12:14 AM

3.100  JEE Advanced Physics: Electrostatics and Current Electricity

Alternate Method of Finding Current in the Circuit and Charge on the Capacitor at Any Time t: In a complicated RC circuit it is easy to find current in the circuit and charge stored in the capacitor at time t = 0 and t → ∞ by conditions which we have already discussed. But to find the current and charge as function of time t following steps may be followed. (a) Find equivalent time constant (t) of the circuit. (b) Find steady state charge q0 (at time t → ∞) on the capacitor. (c) Charge on the capacitor at any time t is, q = q0 ( 1- e - t t ) By differentiating it w.r.t. time we can find current through the capacitor at time t. Then by using Kirchhoff ’s Laws we can calculate currents in other parts of the circuit also. Otherwise we can also find the current in the circuit as shown and calculated in the following ILLUSTRATION because, it may not always be easy and convenient for us to find the current through the capacitor as a function of time.

S

R

A

R

V

V

R

M

I



Similarly, applying Kirchhoff’s Second Law in Loop MNSTM, we have V = I1 R +

Q + IR …(2) C

  Eliminating I from equations (1) and (2), we get V = 3 I1 R +

2Q C 2Q C







⇒

3 I1 R = V -



⇒

I1 =



⇒

1 ⎛ 2Q ⎞ dQ = ⎜V ⎟ dt 3 R ⎝ C ⎠



⇒

1 ⎛ 2Q ⎞ ⎜⎝ V ⎟ 3R C ⎠

dQ dt = 2Q 3 R VC



⇒

t

dQ dt = 2Q 3R 0 V0 C

∫

∫

This equation gives

Solution





I1 =

dQ dt

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 100

C

   V = ( I - I1 ) R + IR V = 2IR - I1 R …(1) ⇒

(a) Find the charge Q on the capacitor at time t . (b) Find the current in AB at time t . What is its limiting value as t → ∞ ?

Let at any time t charge on capacitor C be Q and currents are as shown. Since, charge Q will increase with time t . Therefore,

Q

(a) Applying Kirchhoff’s Second Law in Loop MNABM

C

B

S

T

B

R

Q

R

R

I – I1

Illustration 71

In the circuit shown in figure, the battery is an ideal one, with emf V . The capacitor is initially uncharged. The switch S is closed at time t = 0 .

A I1

S

N

Q=

(

CV 1 - e -2t/3 RC 2

)

dQ V -2t/3 RC = e dt 3 R From equation (1), we get (b) I1 =



V + I1 R I= = 2R

V+

V -2t/3 RC e 3 2R

9/20/2019 11:12:26 AM

Chapter 3: Electric Current and Circuits 3.101

So, current through AB is given by



I 2 = I - I1 =

V+

V -2t/3 RC e V -2t/3 RC 3 e 2R 3R

V -2t/3 RC V e 2R 6 R V I2 = as t → ∞ 2R I2 =

Problem Solving Technique(s) In the above ILLUSTRATION, please note that it is advisable to take the current in the branch containing the capacitor as I1 because then, directly we get dQ I1 = otherwise we had to take dt dQ ( I - I1 ) = dt

that every insulator has some conductivity (or very lightly conducting), on account of which some current flows through the capacitor which is connected to battery as in Figure (a). Similarly when the capacitor is charged and left over, then the charge does not sustain (at its value) over a longer duration of time and starts discharging. This situation can be thought of as being equivalent to the case of discharging of a capacitor in an RC series circuit. In both the cases discussed above, we calculate the resistance of the dielectric using the Ohm’s Law, according to which, we get

R=

l { s = specific conductance} sA R=

I

  (c)  Non-Ideal Practical Situation

Here l = d (the separation between the plates of the capacitor) ⇒ R=

d sA

Thus, the leakage current in the circuit shown in ­figure is

C I=0



V

I=

  (a)  Ideal Situation

Another ideal case where no current flows through the capacitor is when a capacitor is charged and the charge on the capacitor is left over for a longer duration of time, as shown. I=0 q0   (b)  Ideal Situation

However, in both the cases discussed, we get some small amount of current flowing through the c­ apacitor. This non zero current (of the order of microampere) is called the leakage current. This is due to the fact

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 101

I V

LEAKAGE CURRENT THROUGH A CAPACITOR In ideal situation, when the space between the capacitor is filled with a dielectric (insulator) then no current flows through it when it is connected to a battery as shown in figure below

d σA

V R q0 At t = 0

q At t = t

  (d)  Non-Ideal Practical Situation

Similarly, if the capacitor is given a charge q0 at time t = 0 , and left over, then after time t the charge that will remain on it is q , given by

q = q0 e - t t 

{discharging of a capacitor}

⎛ Ke0 A ⎞ ⎛ d ⎞ Here, t = RC = ⎜ ⎝ d ⎟⎠ ⎜⎝ s A ⎟⎠ ⇒ t =

Ke0 s

9/20/2019 11:12:33 AM

3.102  JEE Advanced Physics: Electrostatics and Current Electricity

Remark(s) (a) Dielectric leakage occurs in a capacitor as the result of LEAKAGE CURRENT through the dielectric. Normally it is assumed that the dielectric will effectively prevent the flow of current through the capacitor. Although the resistance of the dielectric is extremely high, a minute amount of current does flow. Ordinarily this current is so small that for all practical purposes it is ignored. However, if the leakage through the dielectric is abnormally high, there will be a rapid loss of charge and an overheating of the capacitor. (b) The power loss of a capacitor is determined by loss in the dielectric. If the loss is negligible and the capacitor returns the total charge to the ­circuit, it is considered to be a perfect capacitor with a power loss of zero. (c) The dielectric used inside the capacitor is not a perfect insulator resulting in a very small current flowing or “leaking” through the dielectric when applied to a constant supply voltage. This small current flow (in the region of micro amperes) is called the Leakage Current. This leakage current is a result of electrons physically making their way through the dielectric medium, around its edges or across the leads. The leakage current of a capacitor is sometimes called the insulation resistance. Illustration 72

A leaky parallel plane capacitor is filled completely with a material having dielectric constant k = 5 and electrical conductivity s = 7.4 × 10 -12 W -1 m -1 . If the charge on the plane at instant t = 0 is q = 8.85 mC, then calculate the leakage current at the instant t = 12 s .



R=

ρd d = A sA 

{ where ρ = resistivity & s = conductivity }

and C =

Keo A d

The capacitive time constant of the circuit is

t = CR =

Keo s

Substituting the values, we have

t=

5 × 8.86 × 10 -12

= 5.98 s 7.4 × 10 -12 Charge at any time decrease exponentially as

q = qo e - t/t

where qo = 8.85 × 10 -6 C is the charge at time t = 0 Therefore, discharging (leakage) current at time t is given by ⎛ dq ⎞ q I = ⎜ - ⎟ = o e - t/t ⎝ dt ⎠ t

So, the current at t = 12 s is

( 8.85 × 10 ) e I= -6



-12/5.98

5.98

Illustration 73

A digital voltmeter of internal resistance r is used to measure the voltage across a capacitor after the switch in Figure is closed. Because the meter has finite resistance, part of the current supplied by the battery passes through the meter. Voltmeter R

Solution

The problem deals with discharging of CR circuit, because between the plates of the capacitor, there is capacitor as well as resistance. C

R

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 102

= 0.198 × 10 -6 A = 0.198 m A

r C

S

E

(a) Apply Kirchhoff’s rules to this circuit, and use dq the fact that IC = to show that this leads to dt the differential equation

9/20/2019 11:12:40 AM

Chapter 3: Electric Current and Circuits 3.103

dq q r + = E dt C R +r rR where Req = R+r (b) Show that the solution to this differential equation is t ⎞ ⎛ CEr ⎝ Req C ⎠ q= 1- e R+r and that the voltage across the capacitor as a function of time is Req

t ⎞ ⎛ Er ⎝ Req C ⎠ VC = 1- e R+r (c) If the capacitor is fully charged, and the switch is then opened, how does the voltage across the capacitor behave in this case?

Solution

(a) Let I represent the current in the battery and IC the current charging the capacitor. Then I - IC is



the current in the voltmeter. The loop rule applied q to the inner loop is + E - IR - = 0. The loop rule C for the outer perimeter is E - IR - ( I - IC ) r = 0 . dq dq With IC = , this becomes E - IR - Ir + r = 0. dt dt Between the two loop equations we eliminate q E I= by substitution to obtain R RC q ⎞ dq ⎛E E - ( R + r )⎜ + r=0 ⎝ R RC ⎟⎠ dt dq ⎛ R+r ⎞ ⎛ R+r⎞ ⇒ E - ⎜ E+⎜ q+r =0 ⎟ ⎟ ⎝ R ⎠ ⎝ RC ⎠ dt



dq E R + r R+r ⎛ ErC ⎞ = q=⎜q⎟ dt R RrC RrC ⎝ R+r⎠ q



⇒

∫ 0

t

dq R+r dt =q - ErC RrC 0 (R+ r)

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 103

∫

q

=0

R+r t RrC

t 0

q - ErC ⎞ (R+ r) ⎟ ⎛ R+r⎞ t = -⎜ ⎝ RrC ⎟⎠ -ErC ⎟ ⎟ (R+ r) ⎠



⎛ ⎜ ⇒ log e ⎜ ⎜ ⎝



⇒ q -



t ⎞ ⎛ r Req C ⎠ where Req = Rr CE ⎝ 1 - e ⇒ q = R+r R+r

ErC ErC ⎡⎛ R + r ⎞ ⎤ =exp ⎢ ⎜ ⎟ t⎥ R+r R+r ⎣ ⎝ RrC ⎠ ⎦

The voltage across the capacitor is q Er ( -t R C = 1 - e eq ) C R+r (c) As t → ∞ the capacitor voltage approaches Er Er (1 - 0 ) = . If the switch is then opened, R+r R+r the capacitor discharges through the v ­ oltmeter. Its voltage decays exponentially according to

VC =

t

Er - rC e . R+r Illustration 74

The capacitor C1 in figure initially carries a charge q0 . When the switches S1 and S2 are shut, capacitor C1 is connected in series to a resistor R and a second capacitor C2 , which initially does not carry any charge. S1 +q0 V –q0

q ⎛ Rr ⎞ dq ⎛ r ⎞ ⇒ - ⎜ E+ +⎜ =0 ⎟ ⎝ R + r ⎟⎠ C ⎝ R + r ⎠ dt

This is the required differential equation ( b) From above, we get



ErC ⎞ ⎛ ⇒ log e ⎜ q ⎟ ⎝ R+r⎠

R

C1 C2 S2

(a) Find the charges deposited on the capacitors and the current through R as a function of time. (b) What is the heat lost in the resistor after a long time of closing the switch? Solution

(a) Suppose at a moment t the charge deposited on C1 is q ( t ) . Then,

9/20/2019 11:12:54 AM

3.104  JEE Advanced Physics: Electrostatics and Current Electricity





VC1 =

q(t ) q - q(t ) and VC2 = 0 C1 C2

dq VR = IR where I = - dt

q2 = q0 - q ( t ) = q0

R C1 VC2

C2





dq ⎛ 1 q0 1 ⎞ ⇒ R dt + ⎜ C + C ⎟ q = C ⎝ 1 2 ⎠ 2



Put



dq q0 q ⇒ R dt = C - C 2



dq 1 ⎛ C ⎞ ⇒ R dt = C ⎜ C q0 - q ⎟ ⎝ 2 ⎠ q



⇒

∫

q0



q02 2 ( C1 + C2 )

Since, Heat Loss = - ΔU = Ui - U f

dq q q0 + = dt C C2



)

q02 2C1

U f = U (∞) =

1 1 1 = + , we get C C1 C2



t

Electrostatic energy at t → ∞ is final energy given by

q ( q0 - q ) dq = IR = - R C1 C2 dt



(

C 1 - e RC C1

So, we observe that the charge on q1 decays and that on q2 grows exponentially. (b) Electrostatic energy at t = 0 is U (0) =

Applying KVL, we get

R



Charge on C2 ,

I q(t) VC1

t

dq q0 - RC ⇒ I ( t ) = - dt = RC e 1

q02 C2 ( ) ( ) ⇒ ΔU = U 0 U ∞ = 2C1 ( C1 + C2 )

Illustration 75

Two large conducting plates are arranged to form a parallel-plate capacitor of capacitance C . The plates are given charges Q1 and Q2 respectively. The plates are then connected to the terminals of a battery of emf E and internal resistance r at t = 0 . Find the charges on the four surfaces as function of time.

t

dq dt =C RC qq0 0 C2

∫

C ⎛ ⎞ qq0 ⎜ ⎟ C2 t ⎟= ⇒ log e ⎜ C RC ⎜ q0 q0 ⎟ C2 ⎠ ⎝

E

r

t



C C ⎞ - RC ⎛ ⇒ q - C q0 = q0 ⎜ 1 - C ⎟ e ⎝ 2 2 ⎠



t ⎡⎛ C ⎞ - RC C ⎤ ( ) + ⇒  q = q t = q 1 e ⎢ ⎥ where 0 ⎜ C2 ⎟⎠ C2 ⎥⎦ ⎢⎣ ⎝

C=

C1C2 C1 + C2

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 104

Solution

When the battery is connected, the charges on the inner surfaces will always be equal and opposite. Since the electric field inside the plates will be zero, the cancellation of electric field must be accomplished by the charges on the outer surfaces. For this, they have to be of same magnitude and nature. It thus follows

9/20/2019 11:13:08 AM

Chapter 3: Electric Current and Circuits

that the charges on the outer surfaces will always be Q + Q2 what they initially were, i.e., 1 . 2 Let q be the charges at time t on the inner surfaces. Applying KVL, q - Ir + E = 0 C dq dt = q r EC

-

⇒

q

∫

Q1 -Q2 2



dq = CE - q

Q1 – Q2 2 Q1 + Q2 2 t

dt

∫ rC

–

0

- log e ( CE - q ) Q1 -Q2 = q



2

⇒

⇒

⇒

CE - q ⎡ log e ⎢ ⎛ Q - Q2 ⎞ ⎢ CE - ⎜ 1 ⎟⎠ ⎝ 2 ⎣ CE - q = ⎛ Q - Q2 ⎞ CE - ⎜ 1 ⎟⎠ ⎝ 2

Q1 + Q2 2

Q1 – Q2 2

t rC

t ⎤ ⎥ = - rC ⎥ ⎦

q

⇒

Q - Q2 ⎞ ⎛ CE - q = ⎜ CE - 1 ⎟⎠ ⎝ 2

-

t rC

) + ⎛⎜⎝ Q -2 Q ⎞⎟⎠ e 1

2

-

t rC

observation We can interpret this result as a superposition of discharging and charging of a capacitor. Also note that in steady state, the charge on the inner surfaces will be CE and -CE respectively, regardless Q - Q2 of the initial value. In particular if 1 (initial 2C potential difference between the plates) is equal to E, there is no charge flow through the battery. If Q1 - Q2 > E , an extra charge is given to the battery 2C Q - Q2 and if 1 < E, the battery supplies the charge to 2C the capacitor. This all is shown in the graph shown in figure, wherein potential difference across the plates has been plotted as function of time.

–q

V

t e rC

E

(

q = CE 1 - e

3.105

Q1 – Q2 2C

r

Q1 – Q2 >E 2C Q1 – Q2 0. Also, plot the variation of current with time.

17. The switch S is closed at t = 0. The capacitor C is uncharged but C0 has a charge Q0 = 2 mC at t = 0. If R = 100 W, C = 2 mF, C0 = 2 mF, E = 4 V. Calculate I(t) in the circuit. S1

1 2 20 V 40 V

R

C1 500 Ω

C2 S2

0.5 μ F

15. The circuit shown in figure is closed at time t = 0. Calculate the total amount of heat generated in R2 during the time in which the capacitor is charged to a voltage of 20 V. E = 40 V

18. A time varying voltage is applied to the clamps A and B such that voltage across the capacitor plates is as shown in the figure. Plot the time dependence of voltage across the terminals of the resistance C and D. V A

C C

R1 = 40 Ω

K

C = 10–4 F R2 = 20 Ω

16. A battery consists of 2 identical cells, each of e.m.f. E, connected in series. The battery is used to charge a capacitor, with capacitance C, through a resistor R. (a) by connecting it across the terminals of the entire battery and (b) by connecting it first across a single cell, then across two cells until it is again fully charged. Determine, in each case, the amount of energy transferred from the battery, which is not stored in the capacitor. What has happened to this energy?

03_Physics for JEE Mains and Advanced - 2_Part 2.indd 107

t

B

R

D

19. A circuit consists of a source of a constant emf E a resistance R and a capacitor with capacitance C connected series. The internal resistance of the source is negligible. At a moment t = 0, the capacitance of the capacitor is abruptly decreased h-fold. Find the current flowing through the circuit as a function of time t. 20. The gap between the plates of a parallel plate capacitor is filled with glass of resistivity ρ = 1011 Wm. The capacitance of the capacitor C = 4 nF. Find the leakage current of the capacitor when a voltage V = 2 kV is applied across it. (Dielectric constant of glass K = 6).

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3.108  JEE Advanced Physics: Electrostatics and Current Electricity

Solved Problems Problem 1

A battery has an open circuit potential difference of 6 V between its terminals. When a load resistance of 60 W is connected across the battery, the total power dissipated by the battery is 0.4 W. What should be the load resistance R, so that maximum power will be dissipated in R. Calculate this power. What is the total power supplied by the battery when such a load is connected? Solution

When the circuit is open, V = E ⇒ E=6V Let r be the internal resistance of the battery. Then the power supplied by the battery in this case is,

Out of this 0.6 W, half of the power is dissipated in R and half in r. Therefore, maximum power dissipated in R would be

PMAX =

0.6 = 0.3 W 2

Problem 2

A conductor has a temperature independent resistance R and a total heat capacity C. At the moment t = 0 it is connected to a dc voltage V. Find the time dependence of the conductor’s temperature T assuming the thermal power dissipated into surrounding space to vary as q = k ( T − T0 ) where k is a constant, T0 is the environmental temperature (equal to conductor’s temperature at the initial moment). Solution

R

Here energy is being generated in the resistance at a V2 . Of which part of energy is being lost in R the environment and the rest is utilized in raising the temperature of conductor. So, applying the Law of Conservation of Energy, we get

rate of r

E



P=

E2 R+r

⎛ Energy ⎞ ⎛ Energy lost ⎞ ⎛ Energy used ⎞ ⎜ supplied ⎟ ⎜ ⎟ ⎜ in raising the ⎟ in the ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ by the dc ⎟ = ⎜ environment ⎟ + ⎜ temperature of ⎟ ⎜ source per ⎟ ⎜ ⎟ ⎜ the conductor ⎟ per ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ unit time ⎟⎠ ⎜⎝ unit time ⎟⎠ ⎜⎝ perr unit time ⎟⎠

Substituting the values, we have

0.4 =

( 6 )2 60 + r

⇒ r = 30 W According to Maximum Power Transfer Theorem, maximum power is dissipated in the circuit when, net external resistance is equal to net internal resistance. So, we have

R=r

Hence,

2 ⎛ dT ⎞ V = − k ( T − T0 ) ⇒ C⎜ ⎟ ⎝ dt ⎠ R

⇒

⇒ R = 30 W Further, total power supplied by the battery under this condition is,

PTotal

2

(6) E = = = 0.6 W R + r 30 + 30 2

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 108

V2 ⎛ dT ⎞ = k ( T − T0 ) + C ⎜ ⎝ dt ⎟⎠ R

dT V2 − k ( T − T0 ) R

=

dt C

Integrating the above expression, we get T



∫V

T0

dT 2

R

− k ( T − T0 )

t

=

dt

∫C 0

9/20/2019 11:17:25 AM

Chapter 3: Electric Current and Circuits 3.109

Solving this equation, we get

T = T0 +

(

kt

− V2 1− e C kR

2V

)

5Ω C1

A 1Ω

Problem 3

A battery of emf 2 V and negligible internal resistance is connected across a uniform wire of length 10 m and resistance 30 W. The appropriate terminals of a cell of emf 1.5 V and internal resistance 1 W is connected to one end of the wire, and the other terminal of the cell is connected through a sensitive galvanometer to a slider on the wire. (a) What length of the wire will be required to produce zero deflection of the galvanometer? How will the balancing change, (b) when a coil of resistance 5 W is placed in series with the battery. (c) When the cell of 1.5 V is shunted with 5 W resistor?

B

G

1.5 V

Now, the potential gradient across AB is 12 ⎛ ΔV ⎞ = voltm −1    ⎜ ⎝ Δl ⎟⎠ AB 70 Since, VAC1 = 1.5 V = E ⎛ 12 ⎞ ⇒   ⎜ ⎟ ( AC1 ) = 1.5 ⎝ 70 ⎠ ⇒   AC1 = 8.75 m (c) VAC2 = Voltage drop across 1.5 V battery 2V

Solution

(a) Potential gradient across wire AB is

ΔV 2 = = 0.2 Vm −1 Δl 10

I 5Ω AB = 10 m RAB = 30 Ω

C 1Ω

B

G

1.5 V

Now, VAC = 1.5 V ⎛ ΔV ⎞ ( ⇒  ⎜ AC ) = 1.5 ⎝ Δl ⎟⎠ ⇒   ( 0.2 )( AC ) = 1.5 = E ⇒   AC = 7.5 m 12 ⎛ RAB ⎞ ⎛ 30 ⎞ (b) VAB = ⎜ ×2= ⎜ V ⎟⎠ × 2 = ⎟ ⎝ R + 5 30 + 5 7 ⎝ AB ⎠

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 109

B

1.5 V 1 Ω

2V

A

C2

A

⎛ 5 ⎞( ) ⇒   ( 0.2 ) ( AC2 ) = ⎜ 1.5 ⎝ 5 + 1 ⎟⎠ ⇒   AC2 = 6.25 m Problem 4

A 10 m long potentiometer wire has resistance 10 W and is connected to an accumulator of 2 V and internal resistance zero. Two resistance boxes B1 and B2 are connected in series with the accumulator. A cell of emf 1.018 V and a shunted galvanometer are connected in parallel across the box B1 . Reading of galvanometer is zero. (a) For what values of resistances R1 and R2 both having integral values will be required from the boxes B1 and B2 to maintain a potential gradient of 10 −3 Vm −1 in the potentiometer wire.

9/20/2019 11:17:32 AM

3.110  JEE Advanced Physics: Electrostatics and Current Electricity

(b) Determine the length of potentiometer wire that balances the thermo emf of iron-copper couple at 300 °C which develops 1.7 × 10 −5 volt °C −1. Solution

According to the statement, we are provided with the following data E1 = 2 V, E2 = 1.018 V, AB = 10 m and RAB = 10 Ω (a) Potential gradient across potentiometer wire AB = 10 −3 Vm −1 VAB 10 −2 = = 10 −3 A …(1) RAB 10 E2 E1

G

B1

B2

Also, I =

difference

across

⇒   10

1.018 = 1018 W I

E1 R1 + R2 + RAB

−3

2 = 1018 + R2 + 10

⇒   R2 = 972 W (b) V = ( 1.7 × 10 −5 ) ( 300 ) = 5.1 × 10 −3 V ⇒  l=

5.1 × 10 −3 V = 5.1 m = Potential gradient 10 −3

Problem 5

It is desired to send a current of 8 A through a circuit whose resistance is 5 W. Find the least number of cells which must be used for this purpose and how should they be connected. The emf of each cell is 2 V and the internal resistance is 0.5 W.

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 110

mnE NE nE = = nr mR + nr mR + nr R+ m 2N 0.5 N 5m + m

⇒ 20 m2 + 2 N = Nm …(1) For least value of N, we have



⇒   IR1 = 1.018 V ⇒   R1 =

I=

dN =0 dm

Differentiating equation (1), we have

B

Further, potential B1 = E2 = 1.018 V





I A

Let N = total number of cells required to be grouped in m rows, each row carrying n cells = mn Since, we know that for n cells in series connected in m rows in parallel, the current in the external circuit is given by

⇒ 8=

∴   VAB = ( 10 −3 ) ( 10 ) = 10 −2 V    I =

Solution

40 m + 2

dN dN =m +N dm dm

dN N − 40 m = =0 dm 2−m ⇒ N = 40 m …(2) ⇒

Since N = mn …(3) Solving equations (1), (2) and (3), we get

n = 40 and m = 4

⇒ N = mn = 160 Problem 6

A resistance coil of resistance r connected to an external battery, is placed inside an adiabatic cylinder fitted with a frictionless piston of mass m and same area A. Initially cylinder contains one mole of ideal gas He. A current I flows through the coil such that temperature of gas varies as T = T0 + at + bt 2 , keeping pressure constant with time t . Atmosphere pressure above piston is P0. Find P0

m He

I r

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Chapter 3: Electric Current and Circuits 3.111

(a) Current I flowing through the coil as function of time and (b) Speed of piston as function of time.

Within what range of actual values of R will the measured values be correct to within 5% if the measurement is made using the circuit shown in figure.

Solution

Solution

Heat produced by coil inside the cylinder in time dt is

Let Rm = measured value, R = actual value,



dQ = I 2 r dt …(1)

I = current measured by the ammeter. (a) When using circuit (a), I R R = ΔV = 20000 ( I − I R )

(a) From First Law of Thermodynamics    ΔQ = ΔU + ΔW …(2) ⇒   I 2 rdt = Cv dT + PdV ⇒   I 2 r = Cv

I R = current through the resistor R

dT dT dT +R = Cp dt dt dt

⎤ ⎡ I ⇒   R = 20000 ⎢ − 1 ⎥ I ⎣ R ⎦ A

mg

But since

⇒   R = 20000

5R ( 2bt + a )  2r

5R ⎞ ⎛ ⎜⎝ Cp = ⎟ 2 ⎠

(b) PV = RT ⇒   PdV = RdT 

(pressure is constant)

⇒  v=

ΔV Rm

and

IR =

ΔV , we have R

( R − Rm ) …(1) Rm

⎛ R − Rm ⎞ When R > Rm , we require ⎜ ≤ 0.05 ⎝ R ⎟⎠ ⇒   Rm ≥ R ( 1 − 0.05 ) and from (1) we get (b) When using circuit (b), we have

dx R ⎛ dT ⎞ = ⎜ ⎟ dt PA ⎝ dt ⎠

R ( 2bt + a )  PA

I R R = ΔV − I R ( 0.5 W )

( where P0 A + mg = PA )

But since I R =

ΔV and Rm = ( 0.5 + R ) …(2) Rm

When Rm > R, we require

Problem 7

The value of a resistor R is to be determined using the ammeter-voltmeter setup shown in figure. The ammeter has a resistance of 0.5 W, and the voltmeter has a resistance of 20000 W. A

Circuit (b)

    R ≤ 1050 W .

⇒   PAdx = RdT ⇒  Velocity v =

I=

I R = I R Rm

dT = ( a + 2bt ) dt

⇒  I=

V

Circuit (a)

As T = T0 + at + bt 2 ⇒ 

R

A

V

P0 A PA

R

R

V Circuit (a)

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 111

R

A V

Circuit (b)

( Rm − R ) ≤ 0.05 R

From (2) we find R ≥ 10 W Problem 8

2 ΔV The switch in figure closes when ΔVc > and 3 ΔV opens when ΔVc < The voltmeter reads a volt3 age as plotted in figure. What is the period T of the waveform in terms of R1 , R2 and C ?

9/20/2019 11:18:05 AM

3.112  JEE Advanced Physics: Electrostatics and Current Electricity ΔVc(t) R1 ΔV R2

Voltage controlled switch

ΔV

2ΔV/3 ΔV/3

V ΔVc

C

t

T ΔVc(t)

Let ΔVC ( t ) =

ΔV

t2

2ΔV/3

⇒

ΔV/3 t

T

− 2 2 R +R C ΔV=ΔV − ΔVe ( 1 2 ) 3 3 t2

−

1 R +R C ⇒ e ( 1 2) = 2

Solution

Start at the point when the voltage has just reached 2 ΔV and the switch has just closed. The voltage is 3 2 ΔV and is decaying towards 0 V with a time con3 stant R2 C t



2 ΔV , in further time t2 , then 3

⎤ − ⎡2 ΔVC ( t ) = ⎢ ΔV ⎥ e R2C ⎣3 ⎦

⇒ t2 = ( R1 + R2 ) C log e 2 ⇒ T = t1 + t2 = ( R1 + 2R2 ) C log e 2 Problem 9

Figure shows a circuit in steady state. If charge on the capacitor is 1000 μ C , find (a) The battery current (b) The resistance R1 , R2 , R3. The current in 50 W resistor and R2 is given to be 5 A.

R1

5 μF Voltage controlled switch

Let, ΔVC ( t ) =

R2 C

ΔV A

V ΔVc

1 ΔV , at time t1 , then 3

F

t1



1 ⎡2 ⎤ − ΔV= ⎢ ΔV ⎥ e R2C 3 ⎣3 ⎦

⇒ e

−

t1 R2C

=

1   ⇒   t1 = R2 C log e 2 2

t

⎡2 ⎤ − R +R C ΔVC ( t ) = ΔV − ⎢ ΔV ⎥ e ( 1 2 ) ⎣3 ⎦

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 112

1

R2

5 A(given)

50 Ω 5 A(given)

5Ω

R1

C

D

310 V

E

Solution

1 After the switch opens, the voltage is ΔV , increas3 ing toward ΔV with time constant ( R1 + R2 ) C , so

B

0Ω

(a) The potential difference across 5 μ C capacitor, is

V=

Q ⎛ 1000 ⎞ =⎜ ⎟ V = 200 V C ⎝ 5 ⎠

The 10 W resistor and capacitor are in parallel arrangement and therefore current I through 10 W resistor is given by

9/20/2019 11:18:18 AM

Chapter 3: Electric Current and Circuits 3.113



   I =

Solution

V 200 = = 20 A R 10

(a) The distribution of current in the circuit is shown in the figure. Apply KVL to loop abefa , we get

(b) From KCL at junction A we get

I = 25 A

5 μF 20

A

I = 25 A

B 15 A

A



Ω 10 R2 5 A(given) C 10 A 50 Ω 5Ω

  − I 3 R2 +

R1

310 V

E

Now we apply KVL in the loop ABDA ,    − ( 10 )( 20 ) − 5R2 + ( 50 )( 5 ) = 0



   − ( 50 )( 5 ) − ( 5 )( 10 ) + 310 − 25R1 = 0

  I1 =

2

f

10 W = 0.4 W 25

dq1 dt

3

   − ( 20 )( 10 ) − 15R3 + ( 10 )( 5 ) + ( 50 )( 5 ) = 0 100 = 6.66 W 15

Problem 10

In the circuit shown, initially the switches are open and the capacitors are uncharged. Switches S1 and S2 are closed simultaneously at t = 0. R2 = 6 Ω S2

R1 = 3 Ω E = 18 V

C2

(a) Obtain the expression for current through switch S2 as function of time. (b) The switch S2 is opened after a long time interval. Find the heat dissipated in resistors and the charge flowing through S1.

2

e +q2 –q2

18 V

c

d I2 + I3 h

The current through capacitor C2 is given by dq   I 2 = 2 dt From equations ( 2 ) and ( 3 ) , we get, at any instant, q q   1 + 2 = E …(4) C1 C2 From equation (1), we get

  I 2 − I1 =

q1 − I3 R1C1

⇒ 

q q d q2 − q1 ) = 1 − 2 …(5) ( dt C1 R1 C2 R2

⇒ 

q ⎞⎤ d ⎡ ⎛ q2 − C1 ⎜ E − 2 ⎟ ⎥ = ⎢ dt ⎣ C2 ⎠ ⎦ ⎝

4 μF S1

1

R1 g

Apply KVL in loop ABCDA ,

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 113

q1 = E …(3) C1

a +q1 –q1 b I3 R2 C1 I1 – I3 I1 I +I –I I C2

Apply KVL in loop ADCEFA , we get

2 μF

  I 3 R2 +

The current through capacitor C1 is given by

⇒   R2 = 10 W

C1

q2 = 0 …(2) C2

Apply KVL to loop achga, we get

5 A(given) F

⇒   R3 =

q1 + ( I 2 + I 3 − I1 ) R1 = 0 …(1) C1

Apply KVL to loop bcdeb, we get R3



⇒   R1 =

  −

q ⎞⎤ q 1 ⎡ ⎛ C1 ⎜ E − 2 ⎟ ⎥ − 2 ⎢ C1 R1 ⎣ ⎝ C2 ⎠ ⎦ C2 R2 C ⎞ dq 1 ⎞ E q2 ⎛ 1 ⎛ ⇒   ⎜ 1+ 1 ⎟ 2 = − + C2 ⎠ dt R1 C2 ⎜⎝ R2 R1 ⎟⎠ ⎝

9/20/2019 11:18:30 AM

3.114  JEE Advanced Physics: Electrostatics and Current Electricity



⇒

  ⎛ C1 + C2 ⎞ ⎛ dq2 ⎞ = ⎜⎝ C ⎟⎠ ⎜⎝ dt ⎟⎠ 2

⇒   q1 = C1E −

1 ⎞ ⎛ 1 ⎜⎝ R + R ⎟⎠ EC2 ⎤ 1 2 ⎡ − q2 ⎥ ⎢ ⎛ 1 1 ⎞ C2 ⎢ R1 ⎜ ⎥ + ⎟ ⎥⎦ ⎣⎢ ⎝ R1 R2 ⎠ ⇒ 

dq2 = dt q2

⇒ 

∫ 0

t

∫ 0

dt RR ( C1 + C2 ) ⎛⎜ 1 2 ⎞⎟ ⎝ R1 + R2 ⎠ q2

⎛ EC2 R2 ⎞ ⇒   − log e ⎜ − q2 ⎟ ⎝ R1 + R2 ⎠

0

Now, I1 =

dq1 ⎛ C E ⎞ ⎛ R2 ⎞ − RC = −⎜ 1 ⎟ ⎜ e ⎝ RC ⎠ ⎝ R1 + R2 ⎟⎠ dt

t

⎛ R2 ⎞ − t C1E ⎜ ⎟ e RC R R + ⎛ R1 R2 ⎞ ⎝ 1 2 ⎠ ⎜ ⎟ ( C1 + C2 ) R + R ⎝ 1 2 ⎠

⇒   I1 = −

− C 1E e RC R 1 ( C1 + C2 )

Similarly, EC2 R2 dq I 2 = 2 = dt R1 + R2

(

t

)

− 1 e RC ⎛ R R ⎞ ( C1 + C2 ) ⎜ 1 2 ⎟ ⎝ R1 + R2 ⎠ t

t ⎛ R1 R2 ⎞ ( C1 + C2 ) ⎜ ⎝ R1 + R2 ⎟⎠

− EC2 e RC R1 ( C1 + C2 )

Let us first now substitute the values of all given in equations (6), (7), (8) to get

(

t ⎛ EC2 R2 ⎞ ⎡ ⎜⎝ R + R ⎟⎠ exp ⎢ − ⎛ RR ⎞ 1 2 ⎢ ( C1 + C2 ) ⎜ 1 2 ⎟ ⎝ R1 + R2 ⎠ ⎣⎢ t

)

⎤ ⎥ ⎥ ⎥⎦

⎛ ⇒   q2 = ⎜ ⎝

− EC2 R2 ⎞ 1 − e RC …(6) ⎟ R1 + R2 ⎠

where R =

R1 R2 and C = C1 + C2 …(7) R1 + R2

Substituting the value of q2 from above in equation (4), we get    q1 = C1E −

R1 R2 and C = C1 + C2 R1 + R2

⇒   I2 =

EC2 R2 − q2 = R1 + R2

(

(

t

− C1 ⎛ EC2 R2 ⎞ 1 − e RC ⎜ ⎟ C2 ⎝ R1 + R2 ⎠

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 114

⎤ ⎥ …(8) ⎦

where R =

⇒   I1 = −

⎛ EC2 R2 ⎞ − q2 ⎜ R1 + R2 ⎟ ⇒   log e ⎜ ⎟= ⎜ EC2 R2 ⎟ ⎜⎝ R + R ⎟⎠ 1 2

⇒ 

)

t

=

t RR ( C1 + C2 ) ⎛⎜ 1 2 ⎞⎟ ⎝ R1 + R2 ⎠

−

t

t ⎡ R2 ( 1 − e )− RC ⇒   q1 = C1E ⎢ 1 − R1 + R2 ⎣

1 ⎡ EC2 R2 ⎤ − q2 ⎥ ⎢ RR ⎦ ( C1 + C2 ) ⎛⎜ 1 2 ⎞⎟ ⎣ R1 + R2 ⎝ R1 + R2 ⎠

dq2 = ⎛ EC2 R2 ⎞ ⎜⎝ R + R − q2 ⎟⎠ 1 2

(

− C1ER2 1 − e RC R1 + R2

)

t

− ( 18 ) ( 4 ) ( 6 ) 1 − e ( 2 )( 6 )    q2 = (3 + 6)

(

⇒   q2 = 48 1 − e

−

t 12

)

)

( in μ C ) …(9)

Similarly    q1 = ( 2 ) ( 18 ) −

(

(

t − ( 2 ) ( 18 )( 6 ) 1 − e 12 (3 + 6)

⇒   q1 = 36 − 24 1 − e ⇒   q1 = 12 + 24 e

(

−

⇒   q1 = 12 1 + 2e

−

t 12

)

)

t 12 −

t 12

)

( in μ C ) …(10)

9/20/2019 11:18:44 AM

Chapter 3: Electric Current and Circuits 3.115

Please note that as per equation (4), we must q q have 1 + 2 = E (at any instant) and as per the C1 C2 relation the above two equations satisfy it at all the instants of time ranging from t = 0 to t → ∞ . Finally, from (9) and (10), we get t

   I1 =

− dq1 = −2e 12  dt

(in A)

Charge on C2 is

  Q2 = ( 12 )( 4 ) = 48 μ C

After a long time, when switch S2 is open and switch S1 still remains closed then no current is drawn from battery in steady state. Both capacitors are in parallel arrangement with battery, so as to given equivalent capacitance of ( 2 + 4 ) = 6 μF 36 μ C

t

− dq    I 2 = 2 = 4 e 12  (in A) dt The current through the switch S2 is given by

   I = I1 − I 3 = −

⇒  I=− ⇒ 

12

(

q1 + I2 R1C1

t − 1 + 2e 12

( 3 )( 2 )

) + 48 ( 1 − e )

) + 4e I = −2 ( 1 + 2e −

t 12

−

−

t 12

b 2A

c

C1

R2 = 6 Ω 2A 2A R1 = 3 Ω C2 d 2A e 48 μ C 2A

f 2A g

18 V

Total resistance of circuit is   R = 9 W So, current is given by 18 =2A 9 Potential difference across C1 is

I ′ =



V f − Ve = I ′R1 = 6 V

Charge on C1 is

R1

72 μ C 18 V

So, Qtotal = ( Ceq ) ( V ) = ( 6 )( 18 )

h

Q4 μF = ( 4 μF ) ( 18 V ) = 72 μ C Hence, the charge flowing through S1 is    ΔQ = Q f − Qi ⇒   ΔQ = ( 36 + 72 ) − ( 12 + 48 ) ⇒   ΔQ = 48 μ C Total heat dissipated in the resistors is ⎛ Work done by ⎞ ⎜ battery when ⎟ ⎜ ⎟ ⎜ charge 48 μ C ⎟ ⎛ Initial ⎞ ⎜ ⎛ Final ⎞ ΔH = ⎜ + flows through ⎟ − ⎜ ⎟ ⎝ Energy ⎟⎠ ⎝ Energy ⎟⎠ ⎜ ⎜ battery after ⎟ ⎜ switch S ⎟ 2 ⎜ ⎟ ⎜⎝ is opened ⎟⎠ ⇒   ΔH =

1 ⎛ Q12 ⎞ 1 ⎛ Q22 ⎞ + + E ( ΔQ ) − 2 ⎜⎝ C1 ⎟⎠ 2 ⎜⎝ C2 ⎟⎠ 1 ( C1 + C2 ) V 2 2

Q1 = ( 6 ) ( 2 ) = 12 μ C

Potential difference across C2 is

4 μF

Hence Q2 μF = ( 2 μF ) ( 18 V ) = 36 μ C and

(b) In the steady state, no current will pass through the capacitors. The current and charges on capacitor will be as shown. a

2 μF

⇒   Qtotal = 108 μ C

t 12

⇒   I = −2 A

12 μ C

Vb − Vc = ( 6 ) ( 2 ) = 12 V

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 115

R2

⇒   ΔH = 136 μ J

9/20/2019 11:18:58 AM

3.116  JEE Advanced Physics: Electrostatics and Current Electricity Problem 11

Consider a parallel plate capacitor of capacitance C with partially conducting medium between its plates having a resistance RC. If this capacitor is connected to a battery of emf E and a resistor R as shown. Find the C

CE − q RC Substituting for q , we get ⇒  I=

(

− E⎛ 1    I = ⎜ 1 − 1− e ⎝ R 1+ α

t ( 1+ α ) RC

)⎞⎟⎠

Problem 12 R

A capacitor initially given a charge Q0 is connected across a resistor R at t = 0. The separation d between the plates changes according to the relation

E

(a) charge on the capacitor as a function of time. (b) current through ‘ R‘ as a function of time. Solution

The equivalent circuit can be drawn as shown here.





d=

d0

(1 + t )

( 0 ≤ t < 1)



A small bulb is connected across the plates of the capacitor which lights when potential difference across the plates of the capacitor reaches V0. Find the

C I1

I

R

R

RC

(a) variation of charge with time (b) time when the bulb will light? E

Solution

Applying Kirchhoff’s Law for the two loops,

( I − I1 ) RC + IR = E …(1)



q = RC ( I − I1 ) …(2) C

(a) Eliminating I between ( 1 ) and ( 2 ) , we get   

q ⎛ R = 1+ C ⎜⎝ RC

For capacitor C , ⇒ 

⎞ ⎟⎠ + I1 R = E

(

− EC (b) Solving, we get q = 1− e 1+ α

Adding (1) and (2), we get Q    + IR = E C

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 116



  C0 =

ε0 A d0

and at time t is C =

ε0 A ε0 A ( 1 + t ) = C0 ( 1 + t ) = d d0

Using Kirchhoff’s Law, we get q − IR = 0 C q dq ⇒  +R =0 C0 ( 1 + t ) dt

  

dq = I1 dt

dq CE − q ( 1 + α ) =  dt RC

(a) Capacitance at t = 0 is given by

R ⎫ ⎧ ⎨∵ α = ⎬ RC ⎭ ⎩ t ( 1+ α ) RC

)

⇒ 

dq 1 dt =− ( q RC0 1 + t )

⇒   log e q

q Q0

=−

1 log e ( 1 + t ) RC 0

t 0

1 ⎛ q ⎞ =− ⇒   log e ⎜ log e ( 1 + t ) ⎟ RC0 ⎝ Q0 ⎠

9/20/2019 11:19:09 AM

Chapter 3: Electric Current and Circuits 3.117 1 ⎛ q ⎞ − ( ) RC0 ⇒   log e ⎜ = log 1 + t e ⎝ Q0 ⎟⎠

⇒   q = Q0 ( 1 + t )

−

⇒ q=−

1 RC0

−

From equation (2), substituting the value of q in equation (1), we have

1

q Q ( 1 + t ) RC0 (b) V = = 0 C 0 (1 + t ) C ⇒  V=

1 Q0 ( 1 + t )− RC0 +1 C0

⎛VC ⎞ ⇒   t = 1− ⎜ 0 0 ⎟ ⎝ Q0 ⎠

⎛ RC0 ⎞ −⎜ ⎝ RC0 + 1 ⎠⎟

Problem 13

In the following RC circuit, the capacitor is in the steady state. The initial separation of the capacitor plates is x0 . If at t = 0, the separation between the plates starts changing so that a constant current flows through R, find the velocity of the moving plates as a function of time. The plate area is A.

x2 + IR ⎛ dx ⎞ ε0 A ⎜ ⎟ ⎝ dt ⎠



E = −I

⇒

2 dx ⎛ I ⎞⎛ x ⎞ = v = −⎜ ⎜ ⎟ …(3) dt ⎝ ε 0 A ⎟⎠ ⎝ E − IR ⎠

⇒ −

C

Ix dx ⎛ ⎞   ⎜ where = velocity ⎟ …(2) ⎝ ⎠ dt ⎛ dx ⎞ ⎜⎝ ⎟⎠ dt

dx x2

⎛ I ⎞ ⎛ dt ⎞ =⎜ ⎟ …(4) ⎜ ⎝ ε 0 A ⎟⎠ ⎝ E − IR ⎠

Integrating the above expression w.r.t. time, we get

1 1 ⎛ I ⎞ − =⎜ t …(5) x x0 ⎝ ε 0 A ( E − IR ) ⎟⎠

From equations (3) and (5), we get

R

v=

1

I

ε 0 A ( IR − E ) ⎡ ⎛ ⎢⎜ ⎢⎜ ⎢⎣ ⎝

I ⎞ ε0 A ⎟ 1 ⎟t+ E − IR ⎠ x0

⎤ ⎥ ⎥ ⎥⎦

2

Problem 14 E

Solution

Let q be the instantaneous charge on the capacitor when a steady current I flows through the circuit. Applying KVL on the circuit, we have

S

At the instant separation between the plates of the ε A capacitor is x , its capacity C = 0 x Differentiating equation ( 1 ) with respect to time, we get

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 117

q0

E

qx + IR …(1) ε0 A

q ⎛ dx ⎞ Ix 0= +0 ⎜⎝ ⎟⎠ + ε 0 A dt ε0 A

C0

R

q E = + IR C

⇒ E=



The switch S is closed at t = 0 . The capacitor C is uncharged but C0 has a charge q0 at t = 0. Calculate the current I ( t ) in the circuit.

{

dq ∵I = dt

}

C

Solution

Let q0 and q be the instantaneous charges on C0 and C respectively. Applying KVL to the circuit, we have

q0 q + + IR = E …(1) C0 C

Differentiating this equation, we get

1 dq0 1 dq dI + +R =0 C0 dt C dt dt

9/20/2019 11:19:23 AM

3.118  JEE Advanced Physics: Electrostatics and Current Electricity

{

dI ⎛ 1 1⎞ ⇒ I⎜ + ⎟ = −R  C C dt ⎝ 0 ⎠ ⇒

where I =

dq0 dq = dt dt

}

C0 C ⎫ ⎧ ⎬ ⎨ where Ceq = C + C0 ⎭ ⎩

dI dt =−  I RCeq

Integrating this expression, we have I(t )



∫ I0

q − IR = 0 …(1) C dq q dq ⇒ ∵I= −R = 0 dt C dt



{

t

q

dI dt =− I RCeq

⇒ log e I

Applying Kirchhoff’s Law to the loop ABCDA, at time t, when charge on capacitor is q, we have

∫

⇒

0

I(t ) I0

⇒ I ( t ) = I0 e

=− −

Q0

t RCeq

t RCeq

t

dt

∫ RC 0

⎛ q ⎞ ⇒ log e ⎜ = ⎝ Q0 ⎟⎠

…(2)

where I 0 is the initial current. Further, I 0 R +

∫

dq = q

t

dt

∫ ( R + αt ) C 0

0

t

1 ⎛ q ⎞ ⇒ log e ⎜ = log e ( R0 + α t ) ⎟ ⎝ Q0 ⎠ α C 0 1 ⎛ q ⎞ ⎛ R + αt ⎞ ⇒ log e ⎜ log e ⎜ 0 = ⎟ ⎝ Q0 ⎠ α C ⎝ R0 ⎟⎠

q0 =E C0

1

q0 ⎞ ⎛ ⎜⎝ E − C ⎟⎠ 0 ⇒ I0 = …(3) R

q ⎛ R + α t ⎞ αC ⇒ =⎜ 0 Q0 ⎝ R0 ⎟⎠

Substituting I 0 from equation (3) into equation (2), we get

⎛ R + α t ⎞ αC ⇒ q = Q0 ⎜ 0 ⎝ R0 ⎟⎠



q 1⎡ I (t ) = ⎢E − 0 C0 R⎣

⎤ ⎥e ⎦

−

t RCeq

q D

I

q CR 1

Plates of a parallel plate capacitor initially charged to Q0 are connected to a variable resistance R whose resistance is given by R = R0 + α t . At time t = 0, R = R0 . Find the current through the capacitor as function of time. A

1

Since I =

Problem 15

B

}

C

Solution

Q ⎛ R + α t ⎞ αC ⇒ I= 0 ⎜ 0 RC ⎝ R0 ⎟⎠ 1

Q0 ⎛ R0 + α t ⎞ α C ⇒ I= C ( R0 + α t ) ⎜⎝ R0 ⎟⎠ Problem 16

The wire AB of a meter bridge changes linearly from radius r to 2r from left end to right end. Where should the free end of the galvanometer be connected on AB so that the deflection in the galvanometer is zero? 4Ω A

4Ω B

The resistance varies according to the relation given by

R = R0 + α t

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 118

9/20/2019 11:19:38 AM

Chapter 3: Electric Current and Circuits 3.119 Solution

Let the galvanometer be connected at a point x = x1 from end A where x = 0 .

A

Let R1 = resistance of left part, i.e., AX1 and

B

R2 = resistance of right part, i.e., X1B Length = 100 cm = 1 m Consider an element of thickness dx at a distance x from end A and of radius rx .

R1

R1

R2

R2

A

R1

R2

R1

R2

R2

Solution

Voltage across AB = V Voltage across A ′B ′ =

B

Voltage across R2 =

dx

R1

x

V 2

V 2

R1

A I V

I/2 A′

R1

I/2 R2

V/2

100 cm B

r ⎞ ⎛ Thus, rx = ⎜ r + x ⎟ = r ( 1 + x ) ⎝ 1 ⎠ Resistance of this element will be, dRx = 1



R1 = R2 =

π rx2

∫

ρ dx ρ = 2 2 2 πr π (1 + x ) r

∫

ρ dx ρ ⎡ 1 1 ⎤ = 2⎢ − 2 2 (1 + x ) r π r ⎣ 1 + x1 1 + 1 ⎥⎦

0



ρ dx

1

1 ⎤ ⎡ ⎢1 − 1+ x ⎥ ⎣ 1 ⎦

For null point of zero deflection,

R1 4 = R2 4

⇒ 1−

1 1 1 = − 1 + x1 1 + x1 1 + 1

⇒ x1 =

1 m = 33.33 cm 3

Problem 17

Consider an infinite ladder of network shown. A voltage is applied between points A and B. If the voltR age is halved after each section, find the ratio 1 . R2 Suggest a method to terminate it after a few sections without introducing much error in its attenuation.

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 119

B′

Now from Kirchhoff’s Law it is obvious that voltage V⎞ V ⎛ across must be ⎜ V − ⎟ = ⎝ 2⎠ 2 Now when the voltage is halved then the current is also halved. So, current in R2 is half of that in R1 ⎛ ⇒ IR1 = ⎜ ⎝ ⇒

I⎞ ⎟ R2 2⎠

R1 1 = R2 2

Now the attenuation produced by circuit on termination by a resistance will not be affected if equivalent resistance R becomes independent of number of sections in the circuit. This is only possible if the terminating resistance R0 is itself equal to equivalent resistance as shown. The equivalent resistance of R0 and R2 is

R′ =

R0 R2 R0 + R2

R1 is in series with it, so equivalent resistance between A and B is



Req = R1 + R ′ = R1 +

R0 R2 R0 + R2

9/20/2019 11:19:51 AM

3.120  JEE Advanced Physics: Electrostatics and Current Electricity A

R1

R0

I

R2

(Q – q)

R0

R

+ 1 –

2

+ q –

B

According to Kirchhoff’s Loop Law, we have

According to proposition we have

Req = R0



R0 = R1 +

   R0 R2 R0 + R2

⇒   IR =

Solving for R0 , we get

R0 =

q Q−q − IR − = 0 C C

⇒  R

4R ⎞ R1 ⎛ 1+ 1+ 2 ⎟ ⎜ 2 ⎝ R1 ⎠

⇒ 

Hence, the circuit may be terminated after a few sections if resistance R0 is connected in parallel as shown in figure.

∫ 0

Problem 18

In a circuit shown in figure the capacitance of each capacitor is equal to C and the resistance equal to R. One of the capacitance was connected to a voltage V0 by closing switch S1. Now at t = 0, the switch S1 is opened and S2 is closed. Calculate R

V0

C S1

C S2

(a) the current I in the circuit as a function of time t (b) the amount of heat generated provided the dependence of current I ( t ) on time t is known. Solution

(a) Initial charge on capacitor ‘1’ is Q = CV0

dq Q − 2q = dt C

dq dt = Q − 2q RC q

⇒ 

Q − 2q C

dq = Q − 2q

dt

∫ RC 0

⎛ log e ( Q − 2q ) ⎞ ⇒  ⎜ ⎟⎠ ⎝ −2

q

= 0

t RC

⇒   log e ( Q − 2q ) − log e Q = −

2t RC

⎛ ( Q − 2q ) ⎞ 2t ⇒   log e ⎜ ⎟⎠ = − RC ⎝ Q ⇒   1− ⇒ 

2q = e −2t RC Q

2q = 1 − e −2t RC Q

⇒  q=

Q [ 1 − e −2t RC ] 2

Since, Q = CV0 ⇒  q=

CV0 [ 1 − e −2t RC ] 2

Let charge q flow in the current in time t. dq Then current I = dt After time t , charge on capacitor ‘1’ = ( Q − q )

Current I =

Charge on capacitor 2 is q

⇒  I=

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 120

t

dq CV0 = 2 dt

⎡ ⎛ 2 ⎞ −2t RC ⎤ ⎢ 0 + ⎜⎝ RC ⎟⎠ e ⎥ ⎣ ⎦

V0 −2t RC e R

9/20/2019 11:20:06 AM

Chapter 3: Electric Current and Circuits 3.121

(b) Heat produced is given by ∞

   H =

∞

⎛ V0

∫ I Rdt = ∫ ⎜⎝ R e 2

0

⎞ ⎟⎠ Rdt



0

V2 ⇒  H= 0 R

⇒  H=−

−2t RC

Let I1 be the instantaneous current in 4 W , I 2 ­current in 6 W and I 3 current in 3 W , then

2

∞

∫e

−4 t RC

0

V 2 ⎛ e −4t RC dt = 0 ⎜ R − 4 ⎜ ⎝ RC

⎞ ⎟ ⎟ ⎠

∞

0

Problem 19

A two way switch S is used in the circuit as shown in figure. First the capacitor is charged by putting the switch in position 1. Now in switch is thrown to position 2. Calculate the heat generated in each resistor.

1 2

10 Ω

R1 4Ω

6Ω R3 R2 3Ω

Solution

U=

I1 2 1 2 : I1 = 1 : : = 3 : 1 : 2 3 3 3 3 This ratio remains constant throughout. Therefore ratio of heat energy dissipated in R1 : R2 and R3 is

I1 : I 2 : I 3 ≡ I1 :



H1 : H 2 : H 3 = ( 3 ) × 4 : ( 1 ) × 6 : ( 2 ) × 3

2

2

2

⇒ H1 : H 2 : H 3 = 6 : 1 : 2 H1 = 6 K , H 2 = K and H 3 = 2K Also, H1 + H 2 + H 3 = 180 J ⇒ 6 K + K + 2K = 180 J 180 = 20 9 Heat generated in R1 = 4 W is ⇒ K=

Initially the switch S is in position 1. Therefore battery of 60 V is in the circuit. The capacitor begins to collect charge. When capacitor is fully charged, the steady state is reached. In steady state there is no current in R = 10 W resistance. Energy stored by capacitor,

I1 2I and so I 3 = 1 3 3 The ratio of currents ⇒ I2 =

If K is ratio constant, then

0.1 F

S

⇒ I 3 = 2I 2 Also, I 2 + 2I 2 = I1

CV02 −∞ ( e − e −0 ) = 1 CV02 4 4

60 V + –

I 2 + I 3 = I1 and 6 I 2 = 3 I 3

1 1 2 CV 2 = 0.1 × ( 60 ) = 180 J 2 2

When switch S is thrown to position 2, the capacitor starts discharging through three resistances 4 W and 6 W and 3 W in parallel.

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 121



H1 = 6 K = 6 × 20 = 120 J

Heat generated in R2 = 6 W is

H 2 = K = 20 J

Heat generated in R3 = 3 W is

H 3 = 2K = 2 × 20 = 40 J

During discharging of capacitor, no current flows in R = 10 W , so heat generated in 10 W is zero.

9/20/2019 11:20:23 AM

3.122  JEE Advanced Physics: Electrostatics and Current Electricity

Practice Exercises Single Correct Choice Type Questions This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

A conductor of area of cross-section A having charge carriers, each having a charge q is subjected to a potential V . The number density of charge carriers in the conductor is n and the charge carriers along with their random motion are moving witha velocity v . A current I flows in the conductor. If j is the current density, then  (A) j = nqV , in the direction of current flow.  (B) j = nqv , in the direction opposite to current flow.  (C) j = nqV , in the direction perpendicular to cur-

t = ητ (B) t = τ log e η (A) 1⎞ ⎛ t = τ ( log e η − 1 ) (D) t = τ log e ⎜ 1 − ⎟ (C) ⎝ η⎠ 5.

A metal rod of radius a is concentric with a metal cylindrical shell of radius b and length l . The space between rod and cylinder is tightly packed with a high resistance material of resistivity ρ . A battery having a terminal voltage V is connected across the combination as shown. Neglect resistance of rod and cylinder. If I is the total current in the circuit then,

rent flow.  (D) j = nqv , in the direction of current flow. 2.

a

An electron in a hydrogen atom is considered to be e2 in a cirrevolving around a proton with a velocity  2 . If I is the equivalent curcular orbit of radius me 2 rent, then

l

b

I

r

V

I= (A)

me 5 me 5 (B) I = 2π  3 4π  3

(A) I=

lV 2π lV (B) I= ρ ρ ( ln b − ln a )

(C) I=

2me 5 me 5 (D) I= 3 π π 3

(C) I=

lV 4π lV (D) I= ρ ( ln b − ln a ) 4πρ ( ln b − ln a )

3.

A milliammeter of range 10 mA and resistance 9 W is joined in a circuit as shown. The meter gives full scale deflection for current I when A and B are used as its terminals, i.e. current enters at A and leaves at B ( C is left isolated). The value of I is 9 Ω, 10 mA

0.1 Ω

A

N identical light bulbs, each designed to draw P power from a certain voltage supply, are joined in series across that supply. The total power which they will draw is

NP (B) P (A) P P (C) (D) N N2 7.

0.9 Ω

B

6.

C

A battery E is connected in series to an ammeter and a voltmeter such that the respective readings are A and V . Now a resistance is joined in parallel with the voltmeter. Then

(A) 100 mA (B) 900 mA

A increases, V decreases (A)

(C) 1 A (D) 1.1 A

(B) A decreases, V increases

4.

The charge on a capacitor decreases η times in time t, when it discharges through a circuit with a time constant τ .

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 122

(C) A and V both decrease (D) A and V both increase

9/20/2019 11:20:40 AM

Chapter 3: Electric Current and Circuits 3.123 8.

A conductor of area of cross section A having charge carriers, each having a charge q is subjected to a potential V . The number density of charge carriers in the conductor is n and the charge carriers (along with their random motion) are moving with a velocity v . If σ is the conductivity of the conductor and τ is the average relaxation time, then

τ= (A)

V A

C

B A

m mσ τ= 2 (B) nq2σ nq



2mσ 1 mσ τ= τ= (C) (D) 2 2 nq2 nq



A long conductor of charge q , with charge density λ is moving with a velocity 2v parallel to its own axis. The conventional current due to motion of conductor is I . Then q (A) I = (B) I = 2λ v t

9.

(C) I = λ v (D) I = 3λv 10. An ideal battery is connected to a capacitor through a voltmeter. The reading V of the voltmeter is plotted against time t . The V -t graph for this is given by



(A) the readings of both the ammeter and the voltmeter remain constant (B) the readings of both the ammeter and the voltmeter increase (C) the reading of the ammeter remains constant but that of the voltmeter increases (D) the reading of the ammeter remains constant but that of the voltmeter decreases

13. Two resistances R1 and R2 are made of different materials. The temperature coefficient of the material of R1 is α and that of material of R2 is − β . The resistance of the series combination of R1 and R2 will not R change with temperature if 1 equals R2

α α+β (A) (B) β α −β

V (B)

(A) V

12. In the given circuit, as the sliding contact C is moved from A to B

α 2 + β2 β (D) (C) 2αβ α t

t

V (C)

V (D)

t

14. Sixteen resistors each of resistance 16 W are connected in the circuit as shown. The net resistance between AB is

t

11. In the circuit shown each battery is 5V and has an internal resistance of 0.2 W . The reading of the voltmeter is V . Then V equals

A

B

1 W (B) 2W (A) (C) 3 W (D) 4W

V

5 V (B) 10 V (A) (C) 15 V

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 123

(D) ZERO

15. When an unknown resistance is connected across a series combination of two identical batteries, each of 1.5 V, the current through the resistor is 1.0 A. When it is connected across a parallel combination of the same batteries, the current through it is 0.6 A. The internal resistance of each battery is

9/20/2019 11:20:54 AM

3.124  JEE Advanced Physics: Electrostatics and Current Electricity 1 1 (A) W (B) W 5 4

6Ω 3Ω

1 1 (C) W (D) W 3 2 16. In the circuit shown in figure, with steady current, the potential drop across the capacitor must be V

7Ω

A

B

4Ω

4.5 W (B) 12 W (A) 5.4 W (D) 20 W (C)

V 2V (C) (D) 3 3



1 W (C) 4.5 W (D) 4.5

19. The ratio of currents as measured by ammeter in two cases (when the key is open and when the key is closed) is R

2R K

2R

R

A

9 10 (A) (B) 11 8 8 (C) 9

(D) None of these

20. The equivalent resistance between A and B (of the ­circuit shown) is

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 124

20 W 3

40 10 40 20 (C) V , W (D) V, W 3 3 3 3 22. In the network shown, the equivalent resistance between A and B is 3Ω

A

B

Ω

2.0 W (A) 0.5 W (B)

(B) 20 V,

6



10 W 3

Ω

18. A cell has an emf of 1.5 V. When short circuited, it gives a current of 3 A. The internal resistance of the cell is

(A) 10 V,

8

17. A galvanometer with 50 divisions on the scale has a resistance of 25 W. A current of 2 × 10 −4 A gives a deflection of one scale division. The additional series resistance required to convert it into a voltmeter reading up to 25 V is (A) 1200 W (B) 1225 W (C) 2475 W (D) 2500 W



Ω

V (A) V (B) 2

4

2R

2V

Ω

C

21. When a resistor of 20 W is connected across a battery, the current is 0.5 A. When a resistor of 10 W is connected across the same battery, the current is 0.8 A. The emf and internal resistance of the battery are

2

R

V

6Ω

4 3 (A) W (B) W 4 3 24 17 W (C) W (D) 17 24 23. A voltmeter of range 1 V has a resistance of 1000 W. To extend the range to 10 V, the additional series resistance required is 1000 (A) W 9

(C) 9000 W

(B) 1000 W (D) 10000 W

24. A primary cell has an emf of 1.5 V. When a 5 W resistor is connected across it, the current is 0.2 A. The internal resistance of the cell is (A) 0.5 W (B) 1.25 W (C) 2.5 W (D) 3.0 W 25. Three resistors of 2 W , 3 W and 5 W are connected in parallel across a battery of 10 V and of negligible internal resistance. The potential difference across the 3 W resistor is

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Chapter 3: Electric Current and Circuits 3.125

(A) 2 V (C) 5 V

(B) 3 V (D) 10 V

26. A potentiometer wire of length 100 cm has a resistance of 10 W. It is connected in series with a resistance and an accumulator of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a 40 cm length of the potentiometer wire. The value of the external resistance is (A) 395 W (B) 790 W (C) 405 W (D) 810 W 27. In the circuit diagram shown in figure, a fuse bulb can cause all other bulbs to go out. Identify the bulb A

B D

C E

(A) B (B) C (C) A (D) D or E 28. Two coils when connected in series have an equivalent resistance of 18 W and when connected in parallel have a resistance of 4 W. The possible values of resistances for the coils is/are R1 = 4 W; R2 = 14 W (A)

E 4Ω

6Ω V



(A) zero

(B)

E 10

E E (C) (D) 5 2 32.

N identical cells, each of emf E and internal resistance r , are joined in series to form a closed circuit. The potential difference across any one cell is ΔV . Then E (A) ΔV = ΔV = E (B) N ⎛ N − 1⎞ ΔV = ⎜ E (D) ΔV = 0 (C) ⎝ N ⎟⎠ 33. A non-conducting ring of radius R has two positive point charges lying diametrically opposite to each other, each of magnitude Q . The ring rotates with an angular velocity ω . If I is the equivalent current then, 2Qω I= I = Qω (B) (A) π Qω I= (C) (D) I = ZERO π

(B) R1 = 2 W, R2 = 16 W (C) R1 = 12 W; R2 = 6 W (D) R1 = 8 W, R2 = 10 W 29. If a wire is stretched to make it 0.1% longer, the percentage change in its resistance would be (A) ZERO (B) 0.1% (C) 0.2% (D) 0.4% 30. In the given circuit the steady state current through the 2 W resistor is 2Ω

34. Two cells A and B of emf 1.3 V and 1.5 V respectively are arranged as shown in figure. The voltmeter reads 1.45 V . The voltmeter is assumed to be ideal. Then A E1 r1 V E2 r2

3Ω 0.2 μ F

4Ω

6Ω

B 4Ω

(A) r1 = 2r2 (B) r1 = 3 r2 (C) r2 = 2r1 (D) r2 = 3 r1

6V



(A) 0.6 A (C) 1.2 A

2.8 Ω

(B) 0.9 A (D) 1.5 A

31. In the circuit shown, the voltmeter has a large resistance. The emf of the cell is E. The reading of the voltmeter is

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 125

35. A straight conductor of uniform cross-section carries a dI  time varying current which varies at the rate = I . If dt s is the specific charge that is carried by each charge carrier of the conductor and l is the length of the conductor then the total force experienced by all the charge carriers per unit length of the conductor due to their drift velocities only is

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3.126  JEE Advanced Physics: Electrostatics and Current Electricity I  (B) (A) F = Is F= 2 Is F= (C)

I 2II (D) F= s s

40. N identical cells, each of emf E0 and internal resistance r0 , are joined in series. Out of these N cells, n cells 1 are wrongly connected. Assuming n < N , E and r 2 to be the respective net emf and internal resistance of the resulting battery, then

36. A milliammeter of range 10 mA has a coil of resistance 1 W . To use it as an ammeter of range 1 A, the required shunt must have a resistance of

(A) E = ( N − n ) E0, r = Nr0

1 (A) W, connected in series 101

E = ( N − 2n ) E0 , r = ( N − 2n ) r0 (C)

1 (B) W, connected in parallel 100 1 (C) W, connected in parallel 99 1 (D) W, connected in series 9 37. To use the milliammeter of PROBLEM 36 as a voltmeter of range 10 V , a resistance R is placed in series with it. The value of R is 9 W, connected in parallel (A) 99 W, connected in series (B) 999 W, connected in series (C) 1000 W, connected in parallel (D) 38.

N identical cells, each of emf E and internal resistance r , are joined in series to form a closed circuit. It is observed that mistakenly one cell has been joined with its polarity reversed. The potential difference across each cell, except the wrongly connected one is ΔV . Then

ΔV = (A)

2E ⎛ N − 1⎞ (B) ΔV = ⎜ E ⎝ N ⎟⎠ N

⎛ N − 2⎞ ⎛ 2N ⎞ ΔV = ⎜ E (D) ΔV = ⎜ E (C) ⎝ N ⎟⎠ ⎝ N − 2 ⎟⎠ 39. A rod of length L has charge distributed on it such that linear charge density λ = kx , k is a constant and x is the distance of a point from the end of the rod where no charge exists. The rod has a mass m and the mass distribution is uniform. The rod is now rotated about an axis passing through the end of the rod where no charge exists and this axis is perpendicular to the length of the rod. If I is the equivalent current then, kL2ω I= (B) I = π kLω (A) π (C) I=

kL2ω (D) I = zero 4π

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 126

E = ( N − n ) E0, r = ( N − n ) r0 (B) E = ( N − 2n ) E0 , r = Nr0 (D) 41. A block of metal is made in the cuboid form with all edges of unequal length. The shortest length is onethird the longest one. If Rmax and Rmin are the maximum and minimum resistance between parallel faces then, Rmax Rmax = 4 (B) =9 (A) Rmin Rmin R (C) max = 3 Rmin

(D) Data Insufficient

42. In the circuit, the potential difference across the capacitor is 10 V . Each resistance is of 3 W. The cell is ideal. The emf of the cell is R R

R R

C = 3 μF R

R

E

(A) 14 V (B) 16 V (C) 18 V (D) 24 V 43. Three copper wires have their lengths in the ratio 5 : 3 : 1 and their masses are in the ratio 1 : 3 : 5 . Their electrical resistance will be in the ratio 5 : 3 : 1 (B) 1: 3 : 5 (A) (C) 125 : 15 : 1 (D) 1 : 15 : 125 44. It is observed that a milliammeter of range 10 mA gives full-scale deflection for a current of 100 mA, when a shunt of 0.1 W is connected in parallel to it. The coil of the milliammeter has a resistance of 0.11 W (B) 1.1 W (A) 1 W (D) 0.9 W (C) 45. The external resistance of a circuit is η times higher than the internal resistance of the source. The ratio of the potential difference across the terminals of the source to its e.m.f. is

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Chapter 3: Electric Current and Circuits 3.127 1 (A) (B) η η

1 from the resis2 tances of the resistors in the previous elements. The equivalent resistance between A and B shown in ­figure is ­element differs by a factor of k =

1+η η (C) (D) η 1+η 46. Three voltmeters A , B and C , connected as shown, 3 have resistances R , R and 3R respectively. When 2 some potential difference is applied between X and Y , the respective voltmeter readings are VA , VB and VC . Then A

R1

kR1

R2

kR2

Y

VA ≠ VB = VC (B) VB ≠ VA = VC (A)

k 3R2

k 4R1 k 4R2

∞

B

( R − R2 ) + R12 + R22 + 6R1R2 (C) 1 2

VA = VB ≠ VC (D) VA = VB = VC (C) 47. The terminal potential difference of a cell is 2.2 V (when circuit is open) and reduces to 1.8 V (when the cell is connected to a resistance R = 5 W ). The internal resistance of cell ( r ) is



10 9 (A) W (B) W 9 10

1 ⎛ Q2 ⎞ (A) ⎜ ⎟ 2⎝ C ⎠

11 5 (C) W (D) W 9 9

(D) None of these

50. A capacitor of capacitance C has charge Q . It is connected to an identical capacitor through a resistance. The heat produced in the resistance is

1 ⎛ Q2 ⎞ (B) ⎜ ⎟ 4⎝ C ⎠

48. The current flowing through a running electric heater, I , is plotted against the time t . Taking into account the variation of resistance with temperature, which of the following curve best represents the variation of I with t ? (A) I

k 2R2

k 3R1

( R − R2 ) + 6R1R2 (B) 1 2

C



k 2R1

R1 − R2 (A) 2

B X

A

I (B)

1 ⎛ Q2 ⎞ (C) ⎜ ⎟ 8⎝ C ⎠

(D) dependent on the value of the resistance.

51. A and B are two points on a uniform ring of resistance R. The ∠ACB = θ , where C is the centre of the ring. The equivalent resistance between A and B is R θ ⎞ ⎛ R⎜ 1 − (A) 2 ( 2π − θ )θ (B) ⎟ ⎝ 2π ⎠ 4π

t



(C) I

t

(D) I

t

52. A copper wire of length l and radius r is nickelplated till its final radius is 2r . If the resistivities of the copper and nickel are ρC and ρN , then the equivalent resistance of wire is t

49. The circuit diagram shown consists of a large number of elements (each element has two resistors R1 and R2). The resistances of the resistors in each subsequent

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 127

⎛ 2π − θ ⎞ ⎛ θ ⎞ (C) R⎜ R⎜ (D) ⎝ 4π ⎟⎠ ⎝ 2π ⎟⎠

ρC l ρN l (B) (A) πr2 4π r 2 ρ ⎞ l l ⎛ (D) ρC + N ⎟ (C) 2⎜ ⎝ 3 ⎠ ⎡ ⎤ π r 1 3 πr2 ⎢ + ⎥ ⎣ ρC ρN ⎦

9/20/2019 11:22:04 AM

3.128  JEE Advanced Physics: Electrostatics and Current Electricity 53. A battery of e.m.f. E, internal resistance r is applied in the circuit in a closed loop. The current I in the circuit goes from the positive terminal of the battery to the negative terminal inside it. The potential drop across the battery is V = E − Ir (B) V = E + Ir (A) (C) V = + Ir (D) V = − Ir

10 (A) W , 5 W 3

A



1 (D) 3

(C) 0.577

56. In the adjoining circuit E = 5 V, r = 1 W, R1 = 1 W, R2 = 4 W , C = 3 μF . Then the numerical value of charge on each plate of capacitors is C

R1

R1

15 W 2

1Ω

1Ω

1Ω

2Ω

2Ω

∞

B



(A) 1 W (C) 3 W

(B) 2 W (D) 4 W

59. In PROBLEM 58, the current through the 2 W resistor nearest to the battery is (A) 1.5 A (B) 2.0 A (C) 3.0 A (D) 4.0 A 60. Three conductors draw respectively currents of 1 A , 2 A and 4 A when connected in turn across a battery. If they are connected in series across the same battery, The current drawn will be 2 3 A (A) A (B) 7 7

C

4 5 (C) A (D) A 7 7

C

61. Two resistances r1 and r2 ( r1 < r2 ) are connected in parallel. The equivalent resistance R is

R2 C

(D) 5 W,

2Ω

6V

(A) 3.2 × 10 5 ± 5% (B) 1.0 × 106 ± 10%

55. Two wires of the same material and the same length are connected to the two gaps of a metre bridge. The balancing length measured from left is 25 cm. The ratio radius of the wire is the left gap to that in the right gap is (A) 1.73 (B) 3

(C) 10 W, 15 W

58. An infinite ladder network is constructed with 1 W and 2 W resistors as shown. The effective resistance between A and B is

54. Carbon resistors used in electronic circuits are marked for their resistance values and tolerance by a colour scheme. A given resistor has colour scheme brown, black, green and gold. Its value in W is

(C) 1.0 × 106 ± 5% (D) 1.0 × 10 3 ± 5%

(B) 20 W, 30 W

(A) R > r1 + r2 (B) r2 < R < r1 + r2 r1 < R < r2 (D) R < r1 (C)

E, r

3 μC (B) 6 μC (A) (C) 12 μC (D) 24 μC 57. In a metre bridge, the gaps are closed by two resistances P and Q and the balance point is obtained at 40 cm. When Q is shunted by a resistance of 10 W, the balance point shifts to 50 cm. The values of P and Q are P

Q

62. In the circuit shown, the capacitor having capacitance C is connected to two voltmeters A and B . Voltmeter A being ideal, having infinite resistance, while voltmeter B has a resistance R . The capacitor is charged and then the switch S is closed. Both A and B show equal readings at time t . Then this is possible for A B C S

G 40 cm

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 128

(A) t → ∞ (B) t = RC (C) t = RClog e ( 2 )

(D) all the t values

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Chapter 3: Electric Current and Circuits 3.129 63. The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K its resistance is 1 W. The resistance of the wire will be 2 W at (A) 1154 K (B) 1100 K (C) 1400 K (D) 1127 K 64. Two resistors of 500 W and 300 W are connected in series with a battery of emf 20 V. A voltmeter of resistance 500 W is used to measure the potential difference across the 500 W resistor. The error in the measurement is (A) 1.4 V (B) 2.4 V (C) 3.4 V (D) 4.4 V 65. In the given bridge the value of X for which the potential difference between the points B and D will be zero, is

1

5

Ω

Ω

5

1

Ω

B

Ω

C

2

Ω

A

Ω

2

5

Ω

X D

E0 A

(A) 5 W (C) 15 W

(B) 10 W (D) 20 W

66. There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joint connections is R0 . The net resistance of the whole grid between the points A and B as shown is A

B

R0 R0 (B) (A) 2

N E1 r1

E2

r2

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 129

G

3 (C) m 2

(D) None of these

68. 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. 69. A nonconducting ring of radius R has charge Q distributed unevenly over it. If it rotates with an angular velocity ω , the equivalent current will be (B) Qω

(A) ZERO

Qω Qω (C) (D) 2π 2π R 70. Two wires of the same metal has the same length but their cross-sections are in the ratio 3 : 1 . They are joined in series. The resistance of the thicker wire is 10 W . The total resistance of the combination will be 40 W 40 W (B) (A) 3 5 100 W (C) W (D) 2 71. The switch shown in the figure is closed at t = 0. The charge on the capacitor as a function of time is given by

R R0 (C) 0 (D) 3 4 67. A battery of emf E0 = 12 V is connected across a 4 m long uniform wire having resistance 4 Wm −1. The cell of small emfs E1 = 2 V and E2 = 4 V having internal resistance 2 W and 6 W respectively are connected as shown in the figure. If galvanometer shows no deflection at the point N , the distance of point N from the point A is equal to

B

5 4 m (A) m (B) 3 3



R=4Ω

C k R

R

R

V

( ) (B) ) 3CV ( 1 − e ) (D) ) (C) CV ( 1 − e CV ( 1 − e (A) CV 1 − e

−

t RC

−

3t RC

−

−

t RC

t 3 RC

9/20/2019 11:22:33 AM

3.130  JEE Advanced Physics: Electrostatics and Current Electricity 72. Two electric bulbs A and B having power ratings PAand PB , ( < PA ) respectively are designed for the same voltage. They are joined in series across a supply of voltage V . Then A and B will draw the same power. (A) A will draw more power than B . (B) B will draw more power than A . (C) (D) the ratio of powers drawn by them will depend on V . 73. In the circuit diagram, the current through the battery immediately after the switch S is closed is E

S

76. A battery is formed by connecting N identical cells together. On short circuiting the terminals of the battery a current I flows in the circuit. To maximise the value of I , N (A)  two rows of cells each are connected in 2 parallel.

(B) N rows of N cells each are connected in parallel, given that N is an integer. (C) all the cells are connected in series. (D) all the cells are connected in parallel.

77. In the circuit shown in figure, the steady state voltage drop across the capacitor is VC . Then R3

C1 R1

C2

R3



E E (D) (C) RR R1 + R2 R1 + 2 3 R2 + R3 74. In the circuit shown, the conductor PQ is of negligible resistance. Then R1

+ –

R2 P

E1

Q

(A) VC =

VR1 VR2 (B) VC = R2 + R3 R1 + R3

(C) VC =

VR1 VR2 (D) VC = R1 + R2 R1 + R2

78. In a balanced Wheatstone bridge shown, if the positions of the battery E and the galvanometer G are interchanged, then G will show zero deflection R1

+

C

R1

E (B) R1

(A) zero

R2

V

R2

–

R2 G

E2

R4

E − E2 E1 + E2 ≠ 1 R1 + R2 R1 − R2



(A) current will flow through PQ if



(B) current will flow through PQ if E1 ≠ E2



(C) no current will flow through PQ



(D) current will flow through PQ if

E1 E2 ≠ R1 R2

75. A fuse wire of radius r , resistivity ρ , length l has a current I passing through it. If ε is the emissivity of the fuse wire and T is the excess safe temperature above the surroundings then

R3

C

R1 R2 = . R3 R4



(A) only if



(B) only if R1 = R3 and R2 = R4 .



(C) only if all the resistances are equal. (D) in all cases.

79. A cell develops the same power across two resistors r1 and r2 when connected separately. If r is the internal resistance of the cell then

T= (A)

I 2ρ I 2ρ T= 2 3 (B) 2 3 π εr 2π ε r

r= (A)

1 r = r1r2 r1r2 (B) 2

T= (C)

2I 2 ρ 4I 2ρ T = (D) π 2ε r 3 π 2ε r 3

(C) r=

1 r = r1 + r2 ( r1 + r2 ) (D) 2

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 130

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Chapter 3: Electric Current and Circuits 3.131 80. A uniform conductor has its two free ends joined to a cell of emf E and some non zero internal resistance. Consider a point P right at the middle of the conductor. Let us now move away from the midpoint of the conductor in the direction of the current and then return to P . During this process we study the variation of the potential V at every point on the path and then plot it against the distance covered ( x ) . The curve that best represents the variation of V with x is v (A)

v (B)

(A) 1 V (B) 2V (C) 3 V (D) 4V 84. The capacitive growth current and the capacitive decay current in series RC circuit are I g and I d respectively. If I 0 is the value of growth current as t → ∞ then at any instant of time (A) I g = I d (B) I g = I0 Id = −I0 (C) I g = − I d (D)

E

E

x

x

85. In the circuit shown, the switch S is closed initially. The capacitor C is uncharged but C0 has a charge q0 at t = 0. The capacitive current immediately after switch S is closed is IC . Then

v (D)

v (C)

R

q0

S

C0

E E2 ) , when placed in series produce null deflection at a distance of 204 cm in a potentiometer. When placed in opposition, they produce null deflection at a distance of 36 cm. If E2 = 1.4 V , E1 is

(A) 14 V (C) 4.2 V

(B) 10 V (D) 2 V

97. The given four terminal network is part of a larger circuit. The points A, B, C are at the same potential. The potential difference between any one of A, B or C and D is 40 V. The potential difference between A and O is C R R

B

O

D

R

R A



(A) 10 V (C) 18 V

(B) 15 V (D) 20 V

98. For a cell, the graph between the potential difference (V) across the terminals of the cell and the current (I) drawn from the cell is shown in the figure. The e.m.f. and the internal resistance of the cell are V (volts)

1.5 1.0 0.5 1

B



(A) 5 W (C) 14 W

(B) 7 W (D) 35 W

100. A galvanometer can be converted into a voltmeter or an ammeter by using either of the two resistances R1 and R2 ( R1  R2 )

(A) R1 in series with the galvanometer for voltmeter and R2 in parallel for ammeter



(B)  R1 in parallel with the galvanometer for voltmeter and R2 in series for ammeter



(C)  R2 in series with the galvanometer for voltmeter and R1 in parallel for ammeter



(D) R2 in parallel with the galvanometer for voltmeter and R1 in series for ammeter

101. All the edges of a block with parallel faces are unequal. Its longest edge is thrice its shortest edge. The ratio of the maximum to minimum resistance between parallel faces is (A) 3 (B) 6 (C) 9 (D) indeterminate unless the length of the third edge is specified 102. A 2.0 V potentiometer is used to determine the internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 75 cm. When a resistor of 10 W is connected across the cell, the balance point shifts to 60 cm. The internal resistance of the cell is (A) 1.5 W (B) 2.5 W (C) 3.5 W (D) 4.5 W 103. A galvanometer of resistance 50 W is connected to a battery of 3 V along with resistance of 2950 W in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the above series resistance should be

2.0

0

99. Seven resistors each of value 5 Ω are connected as shown. The equivalent resistance between the points A and B is

2

3

4

5

I (amperes)

(A) 2 V, 0.5 W (B) 2 V, 0.4 W (C) >2 V, 0.5 W (D) >2 V, 0.4 W

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 133

4450 W (B) 5050 W (A) (C) 5550 W (D) 6050 W 104. The effective resistance between points P and Q of the electrical circuit shown in figure is

9/20/2019 11:23:39 AM

3.134  JEE Advanced Physics: Electrostatics and Current Electricity 2R

108. Twelve wires, each of resistance R, are connected to form a cube as shown in the figure. The effective resistance between A and B is

2R 2R

r P

r

Q

2R 2R

A 2R

2Rr 8R ( R + r ) (A) (B) R+r 3R + r

B

5R 2r + 4 R (D) + 2r (C) 2

R R (A) (B) 6 3

105. Five cells, each of e.m.f. E and internal resistance r , are connected in series. If by mistake, one of the cells is connected wrongly, the equivalent e.m.f. and internal resistance of the combination are

5R 5R (C) (D) 3 6

5E , 5r (B) 3E , 5r (A) 5E, 3 r (D) 3E, 3 r (C) 106. The e.m.f. and the internal resistance of a source which is equivalent to two batteries which are connected in parallel having e.m.f.’s E1 and E2 and internal resistances r1 and r2 respectively are E and r . E r + E2 r2 E= 11 (A) ; r = r1 + r2 ( r1 + r2 )

A 12 V

3R

5R 11

110. The deflection in a galvanometer falls from 50 divisions to 20 divisions when a 12 W shunt is applied. The galvanometer resistance is (A) 18 W (B) 24 W (C) 30 W (D) 36 W

E r + E2 r2 rr E= 11 (D) ; r= 12 r1 + r2 ( r1 + r2 ) 107. Twelve identical resistors each of value 1 W are connected as shown. Net resistance between C and D is R C G B F H D

7 4 R = W (B) R= W (A) 6 3 3 (C) R = 1 W (D) R= W 4

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 134

2R

(C) 6 W (D) 8W

E r + E2 r1 rr E= 1 2 (C) ; r= 12 r1 + r2 ( r1 + r2 )

A

R

(A) 2 W (B) 4W

E r + E2 r1 E= 1 2 (B) ; r = r1 + r2 ( r1 + r2 )

E

109. In the circuit shown the reading of ammeter is 2 A . The ammeter has negligible resistance. The value of R equals.

111. A straight conductor of uniform cross-section carries a current I . If s be the specific charge of an electron. The momentum of all the free electrons per unit length of the conductor, due to their drift velocities only, is I Is (B) (A) s 2

I ⎛ I⎞ (C) (D) ⎜⎝ ⎟⎠ s s 112. The circuit shown here is used to compare the e.m.f.’s of the cells E1 and E2 ( E1 > E2 ) . When the galvanometer is connected to E1 , the null point is at C. When the galvanometer is connected to E2 , the null point will be

9/20/2019 11:23:57 AM

Chapter 3: Electric Current and Circuits 3.135 9 Ω, 10 mA C

A

B

E1

0.1 Ω

G

E2

A



(A) to the left of C (C) at C itself

(B) to the right of C (D) no where on AB

113. In the given circuit, the ammeter A , assumed to have negligible resistance, reads 0.1 A. The value of R is R

12 V



1 A (D) 1.1 A (C) 117. In the given three section network, the equivalent resistance between the points A and B is A Req

50 Ω

30 Ω

(A) 6 W (C) 16 W

B

(B) 8 W (D) 20 W

R

R

R

A

100 Ω 6Ω

2Ω

25 Ω

30 Ω

30 Ω

3Ω

15 Ω

20 Ω

22.26 W (D) 12.13 W (C) 118. In the network, the current flowing through the resistance 2R as shown in the figure is E

B

R

9Ω

42.64 W (B) 32.42 W (A)

114. In the network shown below, the equivalent resistance between A and B is R

C

B

100 mA (B) 900 mA (A)

20 Ω

A

0.9 Ω

R R (A) (B) 2

4R

R

2R

2R (D) 4R (C)

E E (A) (B) R 7R

115. Five identical resistors, each of value 1100 W, are connected to a 220 V battery as shown. The reading of the ideal ammeter A is

2E 2E (C) (D) 7R R

220 V

119. The numerical value of charge on either plate of capacitor C shown in figure C

R2

A

1 2 A (B) A (A) 5 5 3 4 (C) A (D) A 5 5 116. A milliammeter of range 10 mA and resistance 9 W is joined in a circuit as shown. The metre gives fullscale deflection for current I when A and B are used as its terminals. The value of I is

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 135

R1 E

r

CER1 (A) CE (B) R1 + r CER CER1 (C) 2 (D) R2 + r R2 + r

9/20/2019 11:24:09 AM

3.136  JEE Advanced Physics: Electrostatics and Current Electricity 120. 24 identical cells, each of internal resistance 0.5 W, are arranged in a parallel combination of n rows, each row containing m cells in series. The combination is connected across a resistor of 3 W. In order to send maximum current through the resistor, we should have m = 12 , n = 2 (B) m=8, n=3 (A) (C) m = 2 , n = 12 (D) m=3, n=8 121. Switch S is closed at time t = 0. Which one of the following statements is correct? E1

E 2 r2

r1

S R



increases if

connected with reverse polarities, the balancing length obtained is 200 cm . Ratio of emf of cells is (A) 3 : 2 (B) 3 :1 4 : 1 (D) (C) 4:3 124. The charge flowing through a resistor R varies as Q ( t ) = α t − βt 2 . The total heat produced in R is

α 3R α 3R (A) (B) 2β β α 3R α 3R (C) (D) 3β 6β 125. In the circuit shown in figure switch S is thrown to position 1 at t = 0 . When the current in the resistor is 1 A , it is shifted to position 2. The total heat generated in the circuit after shifting to position 2 is

(A)  Current in the resistance E1r2 < E2 ( R + r1 )

R



(B)  Current in the resistance E1r2 > E2 ( R + r1 )

R

increases if

1



(C)  Current in the resistance E1r2 > E2 ( R + r1 )

R

decreases if

2



(D)  Current in the resistance E1r2 = E2 ( R + r1 )

R

decreases if

10 V

i(A) 10

2.5 2



t(s)

(A) The initial potential difference across the capacitor is 100 V 1 F (B) The capacitance of the capacitor is 10 ln 2

2 μF 5V



122. The figure shows a graph of the current in a discharging circuit of a capacitor through a resistor of resistance 10 W .

5Ω

S

(B) 625 μ J

(A) zero

(C) 100 μ J

(D) None of these

126. In the metre bridge arrangement, an unknown resistance R1 is connected in series with or resistance of 10 W . When the combination is connected to left gap of the metre bridge and R2 is connected to right gap, balance point from left end is at 50 cm . When 10 W resistor is removed, the balance point shifts to 40 cm. The value of R1 is (A) 20 W (B) 10 W (C) 60 W (D) 40 W 127. All resistances shown in circuit are 2 W each. The current in the resistance between D and E is B

A D

C

E

(C) The total heat produced in the circuit will be 500 J ln 2 (D) All of the above

(A) 5 A (B) 2.5 A

123. Two cells of emf E1 and E2 are to be compared in

1 A (D) (C) 7.5 A



a potentiometer ( E1 > E2 ) . When the cells are used in series correctly, the balancing length obtained is 400  cm. When they are used in series but E2 is

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 136

10 V F

G

H

128. Each resistor shown in figure is an infinite network of resistance 1 W . The effective resistance between points A and B is

9/20/2019 11:24:30 AM

Chapter 3: Electric Current and Circuits 3.137 1Ω

1Ω

1Ω 1Ω

1Ω

1Ω

1Ω

1Ω B 1Ω

1Ω A

1Ω 1Ω

G A

1Ω

(A) less than 1 W 1W (B)

1Ω

X

B

J E

(C) more than 1 W but less than 3 W

(A) 0.25 W (B) 0.8 W

(D) 3W

(C) 0.2 W (D) 0.16 W

129. Two wires A and B made of same material and having their lengths in the ratio 6 : 1 are connected in series. The potential difference across the wires are 3 V and 2 V respectively. If rA and rB are the radii r of A and B respectively, then B is rA

133. A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is 3 μF

S

1 1 (A) (B) 4 2

(C) 1

3Ω

(D) 2

130. A leaky parallel capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity σ = 7.4 × 10 −12 W −1m −1 . Charge on the plate at instant t = 0 is q = 8.885 μC . Then, time constant of leaky capacitor is (A) 3 s (B) 4s (C) 5 s (D) 6s 131. The given figure represents an arrangement of potentiometer for the calculation of internal resistance ( r ) of the unknown battery ( E ) . The balance length is 70.0 cm with the key opened and 60.0 cm with the key closed. R is 132.40 W . The internal resistance ( r ) of the unknown cell will be E0 A

E

B r

R

G

Y

6Ω

9V



(A) zero

(B) 54 μC

(C) 27 μC (D) 81 μC 134. Consider a capacitor – charging circuit. Let Q1 be the charge given to the capacitor in time interval of 20 ms and Q2 be the charge given in the next time interval of 20 ms . Let 10 μC charge be deposited in a time interval t1 and the next 10 μC charge is deposited in the next time interval t2 . Then, (A) Q1 > Q2 , t1 > t2 (B) Q1 > Q2 , t1 < t2 Q1 < Q2 , t1 > t2 (D) Q1 < Q2 , t1 < t2 (C) 135. The temperature coefficient of resistance of a conductor varies as α ( T ) = 3T 2 + 2T . If R0 is resistance at T = 0 and R be resistance at T then (A) R = R0 ( 6T + 2 ) R = 2R0 ( 3 + 2T ) (B)

K

22.1 W (B) (A) 113.5 W (C) 154.5 W (D) 10 W 132. In the metre bridge, the balancing length AJ = l = 20 cm The unknown resistance X is equal to

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 137

6 μF

X

(C) R = R0 ( 1 + T 2 + T 3 ) R = R0 ( 1 − T + T 2 + T 3 ) (D) 136. A source of emf E = 10 V and having negligible internal resistance is connected to a variable resistance. The resistance varies as shown in figure. The total charge that has passed through the resistor R during the time interval from t1 to t2 is

9/20/2019 11:24:56 AM

3.138  JEE Advanced Physics: Electrostatics and Current Electricity R

(A) 0.5 W (B) 1W (C) 2 W (D) 4W

40 Ω

140. The equivalent resistance between the points A and B for the circuit shown in figure is

20 Ω t

9Ω

t1 = 10 s t2 = 30 s

40 log e 4 (B) 30 log e 3 (A) B

137. A capacitor C is connected to two equal resistances as shown in the figure. Consider the following statements.

E



C

R

(1) At the time of charging of capacitor time constant of the circuit is 2CR (2) At the time of discharging of the capacitor the time constant of the circuit is CR (3) At the time of discharging of the capacitor the time constant of the circuit is 2CR (4) At the time of charging of capacitor the time constant of the circuit is CR (A) Statements (1) and (2) only are correct (B) Statements (2) and (3) only are correct (C) Statements (3) and (4) only are correct (D) Statements (1) and (3) only are correct.

jτ 2Iτ (A) (B) nq nqA 2jτ Iτ (C) (D) nq 2nqA

141. The potential difference VA − VB between points A and B for the circuit segment shown in figure at the given instant is 9 μC 3Ω A 6V 2Ω 3A

139. A charged capacitor is allowed to discharge through a resistor by closing the key at the instant t = 0 . At the instant t = ( ln 4 ) μs , the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to

K A

(C) 6 V (D) −6 V 142. For the circuit shown in the figure, find the charge stored on capacitor in steady state.

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 138

E

R

E

C

E0

Rn

RC RC (A) E (B) ( E − E0 ) R + R0 R0

(C) zero

(D)

RC ( E − E0 ) R + R0

143. A cell of internal resistance r, when applied across an external resistance R draws a current I . The current drawn in the circuit is maximum when R = r (B) Rr (A) R  r (D) R = 2r (C) 144. In the circuit shown, switch S is closed at t = 0 . Let i1 and i2 be the current at any finite time t , then the ratio

2Ω

B

1 μF

(A) 12 V (B) −12 V

(All symbols carry their usual meaning).

C = 0.5 F

A

3 W (B) 6W (A)

138. The mean free path of charge carriers in a current carrying conductor is



3Ω

9 W (D) 12 W (C)

R

S

9Ω

9Ω

20 log e 2 (D) 10 log e 2 (C)

i1 i2

9/20/2019 11:25:12 AM

Chapter 3: Electric Current and Circuits 3.139

i1

3C

2R

C

R

shown. If the potential difference between points a and b is zero, the resistance R in W is 3 V, 1 Ω a

i2

(A) (B) (C) (D)

is constant increases with time decreases with time first increases and then decreases

145. 24 cells each of internal resistance 1 W are arranged such that n cells in series are connected in m rows in parallel across a load resistance of 10 W . The power consumption across the load is maximum for n = 24 , m = 1 (B) n = 12 , m = 2 (A) (C) n = 8 , m = 3 (D) n=6, m=4 146. For the circuit arrangement shown in figure, in the steady state condition charge on the capacitor is 12 V 2 μF

15 V, 2 Ω

S

V



b

2Ω 4Ω

R



(A) 5 (C) 3

150. Six identical resistors each of value 1 W are joined to form a hexagon. Further six more resistors of the same value are connected between the six vertices and the centre of the hexagon. The equivalent resistance between any two vertices lying opposite to each other is 4 5 W W (B) (A) 4 5 3 4 W (D) W (C) 5 3 151. When the switch is closed, the initial current through the 1 W resistor is

6Ω

(A) 12 μC (B) 14 μC

(B) 7 (D) 1

1Ω 2 μF

6Ω

12 V

(C) 20 μC (D) 18 μC

3Ω S

147. A uniform wire of resistance 4 W is bent into a circle of radius r . A specimen of the same wire is connected along the diameter of the circle. What is the equivalent resistance across the ends of this wire?

2 A (B) 4A (A)

4 3 W (A) W (B) (4 +π ) (3 + π )

152. In the circuit shown in figure, the capacitor is charged with a cell of 5 V . If the switch is closed at t = 0 , then at t = 12 s , charge on the capacitor is

2 1 W (C) W (D) (2 + π ) (1 + π ) 148. 9 identical light bulbs, each designed to extract power P from a certain voltage supply are connected in series across that supply. The total power drawn by them is P 9P (B) (A)

(C) 3 A (D) 6A

2 μF

3 MΩ

S 2

( 0.37 ) 10 μC (B) ( 0.37 ) 10 μC (A)

P P (D) (C) 9 81

( 0.63 ) 10 μC (D) ( 0.63 ) 10 μC (C)

149. Two batteries one of the emf 3 V , internal resistance 1 W and the other of emf 15 V , internal resistance 2 W are connected in series with a resistance R as

153. A wire has a non-uniform cross-section as shown in figure. A steady current flows through it. The drift speed of electrons at points P and Q is vP and vQ .

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 139

2

9/20/2019 11:25:31 AM

3.140  JEE Advanced Physics: Electrostatics and Current Electricity 158. For the circuit shown the equivalent resistance between A and C is P

Q

r

B

r

r

(A) vP = vQ (B) vP < vQ (C) vP > vQ

(D) Data Insufficient

154. A moving coil galvanometer is converted into an ammeter reading up to 0.03 A by connecting a shunt r of resistance . What is the maximum current which 4 can be sent through this galvanometer, if no shunt is used. (Here, r = resistance of galvanometer )

A

155. Kirchhoff’s junction law and loop law are manifestations of conservation of (A) charge only. (B) energy only. (C) charge and energy respectively. (D) energy and charge respectively. 156. The potential difference between points A and B , in a section of a circuit shown, is 1A 2Ω 2Ω

2Ω 3A

A

2A

2Ω 1Ω

3V

2V

r

r

r

C

r

r D

12 13 (A) r (B) r 11 11 14 15 r (C) r (D) 11 11 159. The current in 1 W resistance and charge stored in the capacitor are 1Ω

0.004 A (B) 0.005 A (A) (C) 0.006 A (D) 0.008 A

r

2Ω 2V 4Ω 3Ω

5V

2 μF

6V

(A) 4 A , 6 μC (B) 7 A , 12 μC (C) 4 A , 12 μC (D) 7 A , 6 μC 160. The space between two concentric shells of radii a and b ( a < b ) is filled with a medium of resistivity ρ . The resistance offered by the medium to the flow of current when a source is connected across the two spheres is

ρ ρ (A) (B) 4π a 4π b

1Ω B

(A) 5 volt (B) 1 volt (C) 10 volt (D) 17 volt 157. In the network shown in figure, each resistance is R. The equivalent resistance between points A and B is

ρ ρ ⎛ 1 1⎞ (D) (C) ⎜ − ⎟ 4π ( a − b ) 4π ⎝ a b ⎠ 161. A capacitor C1 is charged to a potential V and connected to another capacitor in series with a resistor R as shown. It is observed that heat H1 is dissipated across resistance R , till the circuit reaches steady state. Same process is repeated using resistance of 2R . If H 2 is heat dissipated in this case, then C1 R

A

B

C2

19 20 R (A) R (B) 11 20

H2 H2 = 1 (B) =4 (A) H1 H1

8 R R (D) (C) 15 2

H2 1 H2 =2 (C) = (D) H1 4 H1

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 140

9/20/2019 11:25:49 AM

Chapter 3: Electric Current and Circuits 3.141 162. In the circuit shown, when keys K1 and K2 both are closed, the ammeter reads I 0 . But when K1 is open I and K 2 is closed, the ammeter reads 0 . Assuming 2 that ammeter resistance is much less than R2 , the values of r and R1 in W are K1

(A) 10 W (B) 15 W 10 (C) W (D) 7.5 W 3 167. The equivalent resistance between the points A and B is ( R is the resistance of each side of smaller square)

K2 100 Ω

A

R

R2 = 100 Ω

R1

R

A E, r



(A) 25, 50 (C) 0, 100

(B) 25, 100 (D) 0, 50

163. In a series RC circuit if we assume that the key K is opened as soon as the steady state is reached in the circuit as a result of which the charge had build up. Due to this a pulse is obtained at a regular interval t0. (A) t0 → ∞ (B) t0 = 5 RC t0 = 10 RC (D) t0 = 15 RC (C) 164. Potentiometer wire of length 1 m is connected in series with 490 W resistance and 2 V battery. If 0.2 mVcm −1 is the potential gradient, then resistance of the potentiometer wire is approximately 4.9 W (B) 7.9 W (A) (C) 5.9 W (D) 6.9 W 165. In a series RC circuit a steady state charge of 10 μC is established in time of 10 ms . If 1 ms is the time constant of the circuit and 3 μC is the charge at any instant for the growth part then the decay charge in the circuit is

B

3R R (B) (A) 2 R (C) 2R (D) 2 168. A capacitor is charged from a cell with the help of a resistor. The circuit has a time constant τ . The capacitor collects 10% of the steady charge at time t given by ⎛ 10 ⎞ τ ln ( 1.1 ) (B) τ ln ⎜ ⎟ (A) ⎝ 9 ⎠

τ ln ( 0.9 ) (D) τ ln ( 0.1 ) (C) 169. Two cells, two resistors and two capacitors are connected as shown in figure, The charge on 2 μF capacitor is 3 μF

18 V 1Ω

5Ω

15 V 2Ω

4Ω

qd = 3 μC (B) qd = 7 μC (A) qd = −3 μC (D) qd = −7 μC (C) 166. A capacitor of capacitance 2 μF is charged to a potential difference of 5 V . Now the charging battery is disconnected and the capacitor is connected in parallel to a resistor of 5 W and another unknown resistor of resistance R as shown in figure. If the total heat produced in 5 W resistance is 10 μ J , then the unknown resistance R is equal to

2 μF

(A) 30 μC (B) 20 μC (C) 25 μC (D) 48 μC 170. Consider an infinite ladder network shown in figure. A voltage V is applied between the points A and B. This applied value of voltage is halved after each section.

2 μF A 5Ω R

03_Physics for JEE Mains and Advanced - 2_Part 3.indd 141

R1 R2

R1 R2

R1 R2

R1 R2

R1 R2

B

9/20/2019 11:26:09 AM

3.142  JEE Advanced Physics: Electrostatics and Current Electricity R1 R1 1 = = 1 (B) (A) R2 R2 2 R1 R1 (C) = 2 (D) =3 R2 R2 171. In order to increase the resistance of a given wire of uniform cross-section to four times its value, a fraction of its length is stretched uniformly till the full 3 times the original length of the wire becomes 2 length. What is the value of this fraction? 1 1 (A) (B) 4 8 1 1 (C) (D) 16 6



(A) 200 (C) 10

1Ω

E

D

C

1Ω

CE CE (A) = 1 (B) =2 ED ED CE 1 CE (C) = = 2 (D) ED ED 2 174. In the circuit shown in the figure ammeter and voltmeter are ideal. If E = 4 V , R = 9 Ω and r = 1 Ω , then readings of ammeter and voltmeter are V R

R

R

E, r

V

10 V

B

1Ω

1Ω

172. In the given circuit, the voltmeter records 5 volt . The resistance of the voltmeter in Ω is

100 Ω

1Ω

A

A 50 Ω

K

(B) 100 (D) 50

173. ABCD is a square where each side is a uniform wire of resistance 1 Ω . A point E lies on CD such that if a uniform wire of resistance 1 Ω is connected across AE and constant potential difference is applied across A and C then B and E are equipotential.

(A) 1 A , 3 V (B) 2A, 3V (C) 3 A , 4 V (D) 4 A, 4 V 175. A capacitor of capacitance C has a charge Q0 . This capacitor is now connected to another identical capacitor through a resistance R . The heat produced in the resistance is Q02 Q02 (B) (A) C 2C Q02 Q02 (D) (C) 4C 8C

Multiple Correct Choice Type Questions This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 1.

A metal rod of radius a is concentric with a metal cylindrical shell of radius b and length l . The space between rod and cylinder is tightly packed with a high resistance material of resistivity ρ . A battery having a terminal voltage V is connected across the combination as shown. Neglect resistance of rod and cylinder. If I is the total current in the circuit then, J is the current density and E is the electric field, then

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 142

b

a

I +

l

r

V

9/20/2019 11:17:52 AM

Chapter 3: Electric Current and Circuits 3.143

(A) J=

I ρI (B) E= 2π rl 2π rl

(C) J=

I (D) E = ZERO πr2

2.

In the network shown, the capacitor C is initially uncharged. The time constant of the circuit is t and the charge on C at time t after the switch S is closed is q . Then R3 R1

R2

C S

V

(A) t = CR1

the battery is ideal and has an emf E = 2 V. The resistance of the coil of the galvanometer G is 1 Ω . The current that flows through the galvanometer G is I, the potential difference across the capacitors C1 and C2 , in steady state is DV1 and DV2 respectively. Then 1 I= A (A) I = 0 (B) 5 (C) DV1 = 1.6 V (D) DV2 = 2.4 V 5.

For a metallic conductor (A) the number density of charge carriers is 10 28 m -3 .



(B) the number density of charge carriers is 10 26 m -3 .



(C) the drift velocity of electrons is 1 cms -1 .



(D) the drift velocity of electrons is 10 cms -1 .

6.

In the network shown, points A , B and C are at potentials VA = 70 V , VB = 0 , VC = 10 V and VD respectively. Then

RR ⎞ ⎛ (B) t = C ⎜ R1 + 2 3 ⎟ R2 + R3 ⎠ ⎝ (C) q=

CVR2 ( 1 - e -t t ) R2 + R3

B (0 V) A (70 V)

CVR1 ( (D) q= 1 - e -t t ) R1 + R2 3.

Two batteries of having emfs E1 and E2 and internal resistances r1 and r2 respectively are joined as shown. VA and VB are the potentials at A and B respectively.



(A) The potential difference across one battery will be greater than its emf.

(B) VA - VB = 4.



7.

( E1r2 + E2r1 ) . ( r1 + r2 )

(C) The potential difference across both the battery will be equal. (D) One battery will continuously supply energy to the other. In the circuit shown, C2 = 10 μ F

8Ω

2Ω

G

10 Ω

C1 = 8 μ F 4V

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 143

30 Ω

(A) VD = 40 V

E2, r2 B

D

C (10 V)

E1, r1

A

20 Ω

10 Ω

(B) The currents in the sections AD, DB, DC are in the ratio 3 : 2 : 1 . (C) The currents in the sections AD, DB, DC are in the ratio 1 : 2 : 3 . (D) The network draws a total power of 200 W . A long round conductor of cross-sectional area A is made of a material whose resistivity depends on the radial distance r from the axis of the conductor as α ρ = 2 , α is a constant. The total resistance per unit r length of the conductor is R and the electric field strength in the conductor due to which a current I flows in it is E . Then

R= (A)

2πα 4πα (B) R= 2 A2 A

(C) E=

2πα I A2

8.

(D ) E =

4πα I A2

A voltmeter and an ammeter having resistances RV and RA respectively are connected in series to an ideal battery of emf E . The voltmeter reading is V, the ammeter reading is I , the potential difference across the ammeter is DVA and the total resistance of the

9/20/2019 11:18:14 AM

3.144  JEE Advanced Physics: Electrostatics and Current Electricity circuit is R. Which of the following statement(s) seem to be correct? V (A) V < E (B) RV = I DVA = E - V (D) RA + RV = (C) 9.

E I

In the circuit shown E, F , G and H are cells of e.m.f. 2 V , 1 V , 3 V and 1 V respectively and their internal resistances are 2 Ω , 1 Ω , 3 Ω and 1 Ω respectively. F D

G

VD - VB = (A) VD - VB = (B)

B

2Ω C

H

2 V 13

2 V 13

VG = (C)

21 V = Potential difference across G . 13

VH = (D)

19 V = Potential difference across H . 13

10. A uniform wire of resistance R is shaped into a regular, n even sided polygon. The equivalent resistance between any two corners can have

(A) the maximum value

R 4



(B) the maximum value

R n



⎛ n - 1⎞ (C) the minimum value R ⎜ 2 ⎟ ⎝ n ⎠



(D) the minimum value

R n

11. A voltmeter and an ammeter are joined in series to an ideal battery such that the respective readings shown by them are V and A . A resistance equal to the resistance of the ammeter is now joined in parallel to the ammeter. Then, V will not change. (A) (B) V will increase slightly. (C) A will become exactly half of its initial value.

(D) A will become slightly more than half of its initial value.

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 144

C

2R

2C

R S

E



E

A

12. In the circuit shown in the figure, switch S is closed at time t = 0 . Select the correct statements.

(A) Rate of increase of charge is same in both the capacitors (B) Ratio of charge stored in capacitors C and 2C at any time t would be 1 : 2



(C) Time constants of both the capacitors are equal



(D) Steady state charges on capacitors C and 2C are in the ratio of 1 : 2

13. A number of resistors R1 , R2 , R3 , ........... are connected in parallel such that Rp is the equivalent resistance of parallel combination. R2 is the resistance of least magnitude. A current I is flowing in the circuit due to a potential V applied across the circuit. I1 , I 2 , I 3 , ........... are the currents in resistors R1 , R2 , R3 , ........... respectively and V1 , V2 , V3 ..... are potentials across R1 , R2 , R3 ,........... respectively. V1 = V2 = V3 = ........ (A) ⎛ Rp ⎞ ⎛ Rp ⎞ ⎛ Rp ⎞ (B) I1 = ⎜ I ; I2 = ⎜ I ; I3 = ⎜ I ;..…… ⎝ R1 ⎟⎠ ⎝ R2 ⎟⎠ ⎝ R3 ⎟⎠ (C) Rp > R2 (D) Rp < R2 14. In the circuit shown, the points A and B are connected across a battery of emf E . The equivalent resistance between A and B is RAB , the potential difference and the current across the 5 Ω resistor are DV and I respectively. 2Ω A

3Ω

5Ω

4Ω

B

6Ω

(A) I=

5E (B) DV = 0 18

(C) RAB =

10 18 Ω (D) RAB = Ω 3 5

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Chapter 3: Electric Current and Circuits 3.145 15. In the circuit shown, the battery is ideal, with emf E = 15 V and it sends a current I in the circuit. All resistors are identical and each resistor has resistance R = 3 Ω . The potential difference across the capacitor in steady state is VC , then 1

R

2

C = 3 μF

R 7 6

R

8

E = 15 V

R

3

4 5

(C) VC = 12 V (D) I=3A 16. A number of resistors R1 , R2 , R3 , ........... are connected in series such that Rs is the equivalent resistance of series combination. A current I is flowing in the circuit due to a potential V applied across the circuit. V1 , V2 , V3 , .......... are potentials across R1 , R2 , R3 , ........... respectively. (A) Same current I will flow through each resistor.

(B) V1 + V2 + V3 + ....... = V ⎛R ⎞ ⎛R ⎞ ⎛R ⎞ (C) V1 = ⎜ 1 ⎟ V ; V2 = ⎜ 2 ⎟ V ; V3 = ⎜ 3 ⎟ V ; ...... ⎝ Rs ⎠ ⎝ Rs ⎠ ⎝ Rs ⎠

(D) Data Insufficient

17. A vacuum diode consists of plane parallel electrodes separated by a distance d and each having an area A . On applying a potential V to the anode with respect to the cathode a current I flows through the diode. Assume that the electrons are emitted with zero velocity and they do not change the field between the electrodes. The electron velocity is v and the charge density is ρ at any point between the electrodes at a distance x from the cathode . If I is the equivalent current, m is the mass of each charge carrier, then (A) v=

2eVx md (B) ρ=I md 2eVxA 2

(C) v=

eVx 2md (D) ρ=I 2md eVxA 2

18. In the circuit, the battery connected is ideal. A voltmeter of resistance 600 Ω is connected one by one across R1 and R2 , giving readings of V1 and V2 respectively.

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 145

E = 120 V

V2 = 40 V (B) V1 = 80 V (A) R

(A) I = 1 A (B) VC = 9 V



R1 = 600 Ω R2 = 300 Ω

(C) V2 = 30 V (D) V1 = 60 V 19. In household electric circuits, the electrical appliance and the usage of the switch with the appliance are governed by the following facts. (A) All electric appliances drawing power are joined in parallel. (B) A switch may be either in series or in parallel with the appliance which it controls. (C) If a switch is in parallel with an appliance, it will draw power when the switch is in the ‘off’ position (open). (D) If a switch is in parallel with an appliance, the fuse will blow (burn out) when the switch is put ‘on’ (closed). 20. A piece of germanium (material used for making semiconductors) and a piece of copper (material used for making conducting wires) are cooled from room temperature to 85 K . R1 and R2 be the resistances of the pieces respectively and α 1 and α 2 are the temperature coefficients of resistance of materials respectively. (A) R1 decreases and R2 increases (B) R1 increases and R2 decreases (C) α1 < 0 , α 2 > 0 (D) α1 > 0 , α 2 > 0 21. In the circuit shown, the cell has emf E = 10 V and internal resistance = 1 Ω . 3Ω E = 10 V r=1Ω

2Ω

8Ω

2Ω

2Ω

8Ω 2Ω

4Ω

2Ω



(A) The current through the 3 Ω resistor is 1 A.



(B) The current through the 3 Ω resistor is 0.5 A.



(C) The current through the 4 Ω resistor is 0.5 A.



(D) The current through the 4 Ω resistor is 0.25 A.

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3.146  JEE Advanced Physics: Electrostatics and Current Electricity 22. Two identical fuses are rated at 10 A. When joined in series the combination acts as a fuse of rating I1 and when joined in parallel the combination acts as a fuse of rating I 2 . Then (A) I1 = 10 A (B) I1 = 20 A I 2 = 20 A (D) I2 = 5 A (C)

(C) I=

3 I 0 , from D to C 5

(D) I=0 26. When the switch K is open, the equivalent resistance between A and B is 20 Ω . Which of the following statement(s) is/are correct? 20 Ω

23. The ammeters, A1 and A2 , each of resistance 5 Ω are connected as shown. An ideal cell of emf 10 V is applied between A and B , then 3Ω

C1

5Ω B

2Ω

A2 D2



(A) the current drawn from the cell is 1 A



(B) the reading of A1 is 1 A



(C) the reading of A2 is 1 A



(D) for C1 joined to C2 and D1 joined to D2 , the ammeter readings will become equal

(B) No current flows through K when it is closed



(C) The powers dissipated in R and in the 5 Ω resistor are always equal (D) The powers dissipated in the two 20 Ω resistors are unequal

27. Three voltmeters V1 , V2 and V3 , all having different resistances, are connected between two points A and B , such that V1 and V2 are in series which in turn fall to be in parallel with V3 . When some potential difference is applied across A and B , the respective readings of the voltmeters are V1 , V2 and V3 . Which of the following mathematical relations support the argument given here? V1

20 Ω

C

5Ω

A

B I0 D

(A) VC = VD (B) RAB = 8 Ω

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 146

20 Ω

B V3

V1 = V2 (B) V1 ≠ V2 (A) V1 + V2 = V3 (D) V1 + V2 > V3 (C) 28. A battery of emf E , internal resistance r drives a current I through an external resistance R . The battery supplies a power P and produces per second a heat H across the external circuit and a heat h across the internal circuit of the battery. Then E2 P = EI (B) P= (A) R+r (C) H=

5Ω

V2

A

qσ (D) I0 = - 0 ε0k

25. A current I 0 enters the network at A and leaves at B, when some potential difference is maintained across the points A and B . If RAB is the equivalent resistance between A and B , VC and VD be the potentials at the points C and D respectively and I be the current flowing through the branch CD , then which of the following is correct?

20 Ω



qσ ε0k (B) I= 0 σ ε0k

(C) I = I0e

D

(A) R = 80 Ω



24. A leaky parallel plate capacitor is filled completely with a material having dielectric constant k , electrical conductivity σ . The charge on the plate initially is q0 and the leakage current at time t is I . The time constant of the circuit is t . Then

⎛ tσ ⎞ ⎜⎝ - ε k ⎟⎠ 0

B

K

4Ω

D1

t= (A)

R

C2

A1

A 1Ω

A

C

E2 r EIR h= (D) R+r ( R + r )2

29. Two circuits (as shown in figure) are called ‘Circuit A ’ and ‘Circuit B’. The equivalent resistance of ‘Circuit A’ is x and that of ‘Circuit B’ is y between 1 and 2.

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Chapter 3: Electric Current and Circuits 3.147 2R

1

2R

R

2R

R

2R

R

∞

R

2 Circuit A 1

2R R

2R

2R

R

2R

R



(A) The battery absorbs energy at the rate of EI. (B) The battery stores chemical energy at the rate of



(C)  The potential difference across the battery is E + Ir . (D) Some heat is produced in the battery.

∞

R

2 Circuit B

y > x (B) y = ( 3 + 1)R (A)

( EI - I 2r ) .

33. A milliammeter of range 10 mA and resistance 9 Ω is joined in a circuit as shown. The milliammeter gives full scale deflection for current I when A and B are used as its terminals. It gives a full scale deflection again for a current I ′ when A and C are used as terminals. Then 9 Ω, 10 mA

xy = 2R2 (D) x - y = 2R (C) 30. Three ammeters A1 , A2 and A3 having resistances R1 , R2 and R3 respectively are connected between two points X and Y , such that A1 and A2 are in series which in turn fall to be in parallel with A3 . When some potential difference is applied across the terminals X and Y , the respective readings of the ammeters are I1 , I 2 and I 3 . Which of the following mathematical relations support the argument given here? A1

A2

X

Y

I1 = I 2 (B) I1R1 + I 2 R2 = I 3 R3 (A) I 2 R3 I1 R3 = (D) = (C) I 3 R1 I 3 R1 + R2 31. A circuit has an equivalent resistance R0 . A voltmeter of resistance Rv is applied across the circuit to measure the potential drop across R0 . The new equivalent resistance of the circuit is RR R0 ( for R0  Rv ) (B) 0 v (A) R0 + Rv (D) Data Insufficient

32. A current I is being driven through a battery of emf E and internal resistance r , as shown. Then which of the following statement(s) best suit the above mentioned condition? I

+ E, r

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 147

A

0.9 Ω

B

C

I ′ = 111 mA (B) I ′ = 900 mA (A) I = 1 A (D) I = 1.1 A (C) 34. A cube is made of twelve identical wires each of resistance R0 . The equivalent resistance between the two points lying on the diagonal corners of cube is x and the equivalent resistance between the two opposite corners of a face of cube is y . R x 10 x-y= 0 (A) = (B) 12 y 9

A3

R0 + Rv (C)

0.1 Ω

x 10 R (D) (C) = x-y= 0 y 7 4 35. Two electric bulbs, A and B , with respective ratings 25 W , 220 V and 100 W , 220 V are connected in series across a voltage source of 220 V . The bulbs A and B draw powers PA and PB respectively. Then (A) PA = 16 W (B) PA = 4 W PB = 4 W (D) PB = 16 W (C) 36. In the circuit shown, that contains a capacitor C and a resistor R , the capacitor C is uncharged initially. When the terminal X is joined to the terminal Y for a long time, it is observed that a heat H1 is produced in the resistor. When the terminal X is joined to the terminal Z , again for a long time, it is observed that a heat H 2 is produced in the resistor. Also, energy supplied by the battery during the process of charging is H . Then

9/20/2019 11:19:52 AM

3.148  JEE Advanced Physics: Electrostatics and Current Electricity C



R

Y Z

(A) H1 = H 2 =

X

(B) a potential drop of 4.2 V appears across the resistance of 2.8 Ω. (C)  a potential drop of 1.8 V appears across the capacitor C. (D) a potential of 4.2 V appears across the capacitor C.

40. In the given circuit the point A is 9 V higher than point B. A

H 2

B 6V 1Ω

H1 + H 2 = 2 H (B)

15 V 2Ω

(D) The maximum energy stored in C at any time is H1.

37. Two sources of current of equal e.m.f. are connected in series and have different internal resistances r1 and r2 . An external resistance R is connected in series with two sources of e.m.f.

(A) R = ( r1 - r2 ), for terminal potential difference to be zero across source 1.



(B)  R = r1r2 , for terminal potential difference to be zero across both sources.



(C)  R = ( r2 - r1 ), for terminal potential difference to be zero across source 2.



(D) Data insufficient to arrive at a conclusion.

38. Two heaters A and B , designed for the same voltage V have different power ratings. When connected individually across a source of voltage V , they produce H amount of heat each in times tA and tB respectively. When connected together across the same source, they produce H amount of heat in time T (when in series) and t (when in parallel). Then (A) T = 2 ( tA + tB ) (B) T = tA + tB t= (C)

tAtB t t (D) t= A B 2 ( tA + tB ) tA + tB

39. In the circuit shown, 2Ω 3Ω C

6V

2.8 Ω

(A) a current of 0.9 A flows through 2 Ω resistor when steady state is reached.

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 148

R=1Ω (A) R=7 Ω (B)

(C) Potential difference between B and D is 30 V .



(D) Potential difference between B and C is 15 V .

41. In the circuit shown in figure, the switch S is closed at t = 0. The voltage across the capacitor C at time t after the switch S is closed is V . The voltage as t → ∞ is V0. S C

R

E R

(A) V=

E( E 1 - e -3t RC ) (B) V0 = 3 3

V= (C)

E( E 1 - e -2t RC ) (D) V0 = 2 2

42. Two heaters designed for the same voltage V have different power ratings. When connected individually across a source of voltage V , they produce H amount of heat each in time t1 and t2 respectively. When used together across the same source, they produce H amount of heat in time t

(A) If they are in series, t = t1 + t2



(B) If they are in series, t = 2 ( t1 + t2 )



(C) If they are in parallel, t =



(D) If they are in parallel, t =

4Ω

0.2 μ F



24 V 1Ω

R

H1 + H 2 = H (C)

D

C

t1t2

( t1 + t2 ) t1t2 2 ( t1 + t2 )

43. The charge, Q flowing in a conductor varies with time in accordance with the relation Q = at - bt 2 . If I be the current at an instant of time, then I

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Chapter 3: Electric Current and Circuits 3.149

(A) decreases linearly with time. (B) reaches a maximum and then decreases. 1a . (C) falls to zero after time t = 2b (D) charges at a rate -2b .

44. In the circuit shown in figure, the charge on each capacitor is Q and the potential difference between the points B and D is V .

⎛ 10 ⎞ ⎛ 10 ⎞ t2 - t1 = t log e ⎜ ⎟ (B) t1 + t2 = 2t log e ⎜ ⎟ (A) ⎝ 3 ⎠ ⎝ 9 ⎠ (C) t1 + t2 = 2t log e ( 3 ) (D) t2 - t1 = 2t log e ( 3 ) 48. A battery of e.m.f. E and internal resistance r is connected to a variable resistor R as shown. Which one of the following is true? E

3Ω D 6Ω A

r R

C

1 μF B 2 μF 12 V, 1 Ω



Q = 7.2 mC (B) Q = 7.2 mC (A) (C) V = 3.6 V (D) V = -3.6 V 45. In the circuit shown, when the switch is closed, the capacitor charges with a time constant t1 and when the switch is opened (after the capacitor has been charged) then the capacitor discharges with a time constant t2 . Then C





(A) Potential difference across the terminals of the battery is maximum where R = r (B) Power delivered to resistor is maximum when R=r (C) Current in the circuit is maximum when R = r (D) Current in the circuit is maximum when R >> r

49. In the circuit shown in figure, reading of ammeter will R

S1

A

R

E

R r

E

r

S2

R + _ B

t1 (A) = 2 (B) t1 + t2 = 3 RC t2 t1 (C) = 1 (D) t1 + t2 = 2RC t2 46. A capacitor is charged and then made to discharge through a resistance. The time constant is t . In time t1 the potential difference across the capacitor decreases by 10% and in time t2 the potential difference across the capacitor falls to 10% of its initial value. Then ⎛ 9⎞ ⎛ 10 ⎞ t1 = t log e ⎜ ⎟ (B) t1 = t log e ⎜ ⎟ (A) ⎝ 8⎠ ⎝ 9 ⎠ t2 = 2.303 t (D) t2 = 0.693 t (C) 47. A capacitor charges from a cell through a resistance. The time constant is t . The capacitor collects 10% of its final charge in a time t1 and after a time t2 , the charge on the capacitor becomes 10% less than its final charge. Then

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 149



(A) increase if S1 is closed



(B) decrease if S1 is closed



(C) increase if S2 is closed



(D) decrease if S2 is closed

50. When a resistance of 9.5 Ω is connected across a battery, the voltage across the resistance is 11.4 V. If the resistance connected across the same battery is 11.5 Ω, the voltage across the resistance is 11.5 V (A) The emf of the battery is 12.0 V (B) The internal resistance of the battery is 0.5 Ω (C) The internal resistance of the battery is 11.45 Ω (D) The emf of the battery is 11.5 V 51. In the given circuit is steady state

5Ω

C2

C1

2 μF

5 μF

14 V

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3.150  JEE Advanced Physics: Electrostatics and Current Electricity

(A) the potential difference across C1 is 10 V



(B) the potential difference across C2 is 10 V (C) the charge on C1 is 20 mC (D) the charge on C2 is 8 mC

54. In the potentiometer experiment shown in figure, the null point length is l . Choose the correct options given below. E1 l

52. A capacitor of 2 F (practically not possible to have a capacity of 2 F ) is charged by a battery of 6 V . The battery is removed and circuit is made as shown. Switch is closed at time t = 0 . Choose the correct options.

E2

2F

1Ω

S 2Ω





(A) At time t = 0 current in the circuit is 2 A



(B) At time t = ( 6 ln 2 ) second, potential difference across capacitor is 3 V (C) At time t = ( 6 ln 2 ) second, potential difference across 1 Ω resistance is 1 V. (D) At time t = ( 6 ln 2 ) second, potential difference across 2 Ω resistance is 2 V. 53. In the circuit shown in figure, the charge on each capacitor is Q and the potential difference between the points B and D is V . 3Ω D 6Ω A

G S

6V – +



J

C

1 μF B 2 μF 12 V, 1 Ω

Q = 7.2 mC (B) Q = 7.2 mC (A)



(A) If jockey J is shifted towards right, l will increase



(B) If value of E1 is increased, l is decreased



(C) If value of E2 is increased, l is increased



(D) If switch S is closed, l will decrease

55. In the circuit shown, A and B are equal resistances. When S is close, the capacitor C charges from the cell of emf ε and reaches a steady state. C B + –

S A

ε



(A) During charging, more heat is produced in A than in B (B) In steady state, heat is produced at the same rate in A and B 1 (C) In the steady state, energy stored in C is Cε 2 4 (D) In the steady state energy stored in C is

1 2 Cε 8

V = 3.6 V (D) V = -3.6 V (C)

Reasoning Based Questions This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A)  Bubble (B)  Bubble (C)  Bubble (D) 

If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE. If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 150

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Chapter 3: Electric Current and Circuits 3.151 1.

Statement-1: In a wire of non-uniform cross-section, the current is same everywhere. Statement-2: The current in a wire is due to the drift of electrons along the wire. 2.

Statement-1: A conductor carrying electric current becomes electrically charged. Statement-2: A conductor carrying electric current contains same number of positive and negative charges and is electrically neutral.

Statement-2: Finite resistance of voltmeter changes current flowing through the resistance across which potential difference is to be measured. 13. Statement-1: In a simple battery circuit, the point at the lowest potential is the positive terminal of the battery. E

I

3.

Statement-1: Electric field outside the conducting wire which carries a constant current is zero. Statement-2: Net charge on conducting wire is zero.

R

4.

Statement-1: Direction of electronic current can not be from negative potential to positive potential. Statement-2: Direction of current is opposite to flow of electrons.

Statement-2: The current flows towards the point of lowest potential of the battery from the outside in the same manner as it flows in a circuit from positive to the negative terminal.

5.

14. Statement-1: If potential difference between two points is zero and resistance between those points is zero, current may flow between the points. Statement-2: Kirchhoff’s 1st Law is based on conservation of charge.

Statement-1: Insulators do not allow flow of current through them. Statement-2: Insulators have no free charge carriers. 6.

Statement-1: If there is current in a wire potential drop has to be there. Statement-2: If potential drop is zero, the resistance may be zero. 7.

Statement-1: Constant potential difference is applied across a conductor. If temperature of conductor is increased, the drift speed of electrons will decrease. Statement-2: Resistivity increases with increase in temperature. 8.

Statement-1: When a conductor is stretched to double its length, its resistance will get doubled. Statement-2: Resistance is directly proportional to the length of a conductor. 9.

Statement-1: The drift velocity of electrons in a metallic wire decreases when the temperature of the wire is increased. Statement-2: On increasing temperature, conductivity of metallic wire decreases. 10. Statement-1: Voltmeter always reads the e.m.f. of a cell if it is connected across the terminals of a cell. Statement-2: Terminal potential of a cell is given by V = E - Ir .

11. Statement-1: A voltmeter is an inherently inaccurate instrument. Statement-2: A voltmeter is always connected in parallel in a circuit. 12. Statement-1: Potential measured by a voltmeter across a wire is always less than actual potential difference across it.

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 151

15. Statement-1: Internal resistance of the battery is connected in parallel to it in an electrical circuit. Statement-2: Heat generated in battery is due to internal resistance. 16. Statement-1: Two bulbs of 25 W and 100 W rated at 200 V are connected in series across 200 V supply. Ratio of powers of both the bulb in series is 2 : 1. Statement-2: In series, current in both bulbs is same, therefore power depends on resistance of bulb. 17. Statement-1: Consider the two situations shown in the figure. Potential difference between points A and B , in Case-1 is more as compared to Case-2. E

A I

B r

Case I

E

A I

B r

Case II

Statement-2: In Case-1 VA - VB = E + Ir

In Case-2 VA - VB = E - Ir

18. Statement-1: Since all the current coming to our house returns to power house (Since current travels in a closed loop). So there is no need to pay the electricity bill. Statement-2: The electricity bill is paid for the electrical energy used and not for the current. 19. Statement-1: When current through a bulb is increased by 2% power increases by 4% .

9/20/2019 11:20:44 AM

3.152  JEE Advanced Physics: Electrostatics and Current Electricity Statement-2: Current passing through the bulb is inversely proportional to its resistance.

Statement-2: Resistance of a metal increases with increase in temperature.

20. Statement-1: The e.m.f. of the driver cell in the potentiometer experiment should be greater than the e.m.f. of the cell to be determined. Statement-2: The fall of potential across the potentiometer wire should not be less than e.m.f. of the cell to be determined.

24. Statement-1: For zero value of R in circuit, power transfer in external resistance will be maximum.

21. Statement-1: When a cell is charged by connecting its positive electrode with positive terminal of the charger battery then potential difference across the electrodes of cell will be smaller than the EMF of cell. Statement-2: Potential difference across electrodes in a cell providing electric current I is E − Ir where E is the EMF and r internal resistance. 22. Statement-1: In metre bridge experiment, a high resistance is always connected in series with a galvanometer. Statement-2: As resistance increases, current through the circuit increases. 23. Statement-1: In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance

R R1 = 5 Ω E = 20 V r = 10 Ω

Statement-2: Since R1 < r in the given circuit. So, power transfer in external resistance will be maximum when R = 0. 25. Statement-1: The switch S shown in the figure is closed at t = 0, Initial current flowing through battery E is . R+r R

C

E

r

S

Statement-2: Initially capacitor was uncharged, so resistance offered by capacitor at t = 0 is zero.

Linked Comprehension Type Questions This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options)

Comprehension 1

1.

In a certain particle accelerator, electrons emerge with an energy of 40 MeV. The electrons emerge not in a steady stream but rather in pulses at the rate of 250 pulses/s. This corresponds to a time interval between pulses of 4 ms. Each pulse has a duration of 200 ns and the electrons in the pulse constitute a current of 250 mA. The current is zero between pulses. Based on the information supplied, answer the following questions. 2 × 10–7 s

4 ms

(A) 2.12 × 1011 (B) 21.2 × 1010 (C) 3.13 × 1011 (D) 31.3 × 1010 2.

The average current per pulse delivered by the accelerator is 0.005% of peak current (A) (B) 0.05% of peak current (C) 0.001% of peak current (D) 0.01% of peak current

I

t(s)

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 152

How many electrons are delivered by the accelerator per pulse?

3.

The peak power delivered by the electron beam is (A) 1 MW  (B)  10 MW  (C)  100 MW  (D)  100 kW

4.

The average power (instead of the peak power) delivered by the electron beam is (A) 500 MW  (B)  50 MW  (C)  25 W  (D)  500 W



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Chapter 3: Electric Current and Circuits 3.153

Comprehension 2 A galvanometer (coil resistance 99 Ω ) is converted into an ammeter using a shunt of 1 Ω and connected as shown in figure (a). The ammeter reads 3 A. The same galvanometer is converted into a voltmeter by connecting a resistance of 101 Ω in series. This voltmeter is connected as shown in figure (b). 12 V

12 V

r

r

10. The potential difference across the points AB is VAB. Then VAB = 4 V (B) VAB = 8 V (A) VAB = 6 V (D) VAB = 0 (C)

Comprehension 4 Based on the circuit diagram shown, answer the following questions. 6 μF

A

2Ω V

2Ω Figure (a)

Figure (b)

4 times the full scale reading. 5 Based on the facts and figures provided answer the following questions.

V = 18 V

S

Its reading is found to be

5.

The internal resistance r of the cell is

3Ω

b

6Ω

a

3 μF

11. The potential of point a with respect to point b in ­figure when switch S is open is

(A) 10.1 Ω (B) 1Ω

(A) Vab = 9 V (B) Vab = 18 V

(C) 1.01 Ω (D) 11.1 Ω

Vab = −9 V (D) Vab = −18 V (C)

6.

The range of the ammeter and voltmeter respectively is (A) 5 A, 9 V (B) 50 A, 9 V (C) 5 A, 9.95 V (D) 5.95 A, 9 V

12. The final potential of point b with respect to ground when switch S is closed is (A) 18 V (B) 9 V (C) 6 V (D) 3 V

The full scale deflection current in the galvanometer is given by (A) 5 A (B) 0.05 A (C) 0.5 A (D) 50 A

13. The change in the value of charge on 3 μF capacitor is Δq and on 6 μF capacitor is Δq′ , when the switch S is closed . Then

7.

Δq = Δq′ = −36 μC (A)

Comprehension 3

Δq = 36 μC, Δq′ = −36 μC (B)

Based on the circuit shown, answer the following questions. Assume that all the resistances have value in ohm.

Δq = Δq′ = −18 μC (C)

A I

8 16

3

5

2

I1

Comprehension 5

16

Based on the circuit diagram shown answer the following questions.

1

24 V

Δq′ = 18 μC , Δq = −18 μC (D)

2A

B

8.

The current I in the circuit is given by (A) 2 A (B) 4 A (C) 6 A (D) 8 A

9.

The current I1 is given by

1 (A) 1 A (B) A 2 1 1 (C) A (D) A 4 8

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 153

R E2 +

E1 + 4Ω

3Ω

3A

6Ω 5A

14. The current in the 3 Ω resistor is (A) 2 A (B) 4 A (C) 8 A (D) 10 A

9/24/2019 11:26:10 AM

3.154  JEE Advanced Physics: Electrostatics and Current Electricity 15. The respective E1 and E2 values, in volt, are given by (A) 34, 56 (B) 36, 54 (C) 54, 36 (D) 56, 34 16. The resistance R , in ohm is

20. In steady state, the potential difference across the capacitor is given by (A) zero (B) 10 V (C) 20 V (D) 25 V

(A) 3 Ω (B) 6Ω

21. The capacitor current as a function of time, when the battery is disconnected is given by

(C) 7.5 Ω (D) 9Ω

3t ⎞ ⎛ 100 exp ⎜ − mC (A) ⎝ 1000 ⎟⎠

Comprehension 6 A perfect insulator is filled between the plates of a capacitor. For practical purposes this insulator is assumed to be perfect when its resistance is taken to be very large or infinite. Due to this, if a capacitor is connected to a battery, no current flows through it in steady state. However, an insulator must have some finite resistance which can be calcuρl lated by using the relation R = . So, due to the finite value A of the resistance a non zero current must pass through the resistance. This small non zero current is called leakage current. Consider a leaky parallel plate capacitor is filled completely with material having dielectric constant K = 18 and its resistivity is ( 4π × 10 3 ) Ωm . The charge on the plate at instant t = 0 is 2 μC . Based on the above facts and the data provided answer the following questions. 17. Time constant of circuit is (A) 3 μs (B) 2 μs

t ⎞ ⎛ 100 exp ⎜ − μC (B) ⎝ 1000 ⎟⎠ 3t ⎞ ⎛ (C) 200 exp ⎜ − μC ⎝ 1000 ⎟⎠ t ⎞ ⎛ (D) 200 exp ⎜ − mC ⎝ 1000 ⎟⎠ 22. The time taken by the capacitor to discharge until the potential difference across it is 1 V (A) 17.66 ms (B) 1.766 ms (C) 1766 ms (D) 1 ms

Comprehension 8 In the circuit shown, the switch S has been open for a long time. It is then suddenly closed. The time constant of the circuit before the switch is closed is τ 1 and that after the switch is closed τ 2 . Assuming the switch to be closed just when t = 0 , give answers to the following questions. 50 kΩ

(C) 9 μs (D) 1 μs 18. Charge on the capacitor at t = 2 μs is (A) 0.54 μC (B) 0.37 μC

10 V

(C) 1 μC (D) 0.74 μC 19. Leakage current at t = 4 μs is (A) 0.48 μA (B) 0.36 μA (C) 0.24 μA (D) 0.13 μA

Comprehension 7 Consider the circuit shown here. Based on the circuit and the basic knowledge of the Kirchhoff’s Laws, answer the following questions.

10 μ F

S 100 kΩ

23. The value of τ 1 is (A) 0.5 s (C) 1.5 s

(B) 1 s (D) 2 s

24. The value of τ 2 is (A) 0.5 s (C) 1.5 s

(B) 1 s (D) 2 s

25. The current in the switch as a function of time is (A) 100 ( 2 + e − t ) mA (B) 100 ( 2 − e − t ) mA (C) 100 ( 2 + e − t ) μA (D) 100 ( 2 − e − t ) μA

10 Ω

40 Ω

36 V 10 μ F 80 Ω

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 154

20 Ω

Comprehension 9 A certain circuit consists of a variable capacitor, a resistor R and an emf source of emf E as shown in figure. Initially the capacitance of the capacitor is 10 μF which is abruptly changed to 5 μF at t = 0 . If E = 10 V and R = 50 Ω , then answer the following questions

9/24/2019 11:26:18 AM

Chapter 3: Electric Current and Circuits 3.155 31. The total heat dissipated in the circuit during the discharging process of C1 is

R

q2 q02 (A) 0 2 × C (B) 2C 2C1 E

26. The current in the circuit just after the capacitance has been changed is 1 2 A (A) A (B) 5 5 3 4 (C) A (D) A 5 5

q02C2 q02 (D) (C) 2 2C1C2 2C1

Comprehension 11 In the circuit shown in Figure, the capacitors are initially uncharged. Based on the facts & figure provided, answer the following questions.

27. The current in the circuit as function of time is

10 Ω

(A) 0.2 exp ( −2000t ) (B) 0.2 exp ( −4000t ) 10 Ω

0.1 exp ( −2000t ) (D) 0.1 exp ( −4000t ) (C) 28. The charge on the capacitor as function of time is (A) 50 ( 1 + e −2000t ) (B) 5 ( 1 + e −4000t )

10 Ω

(C) 50 ( 1 + e −4000t ) (D) 5 ( 1 + e −2000t )

10 Ω 10 Ω

10 μ F

10 Ω

10 Ω

10 Ω

5 μF 10 Ω

10 Ω

10 Ω

29. The heat produced in the resistor as function of time is (A) 25 ( 1 − e −8000t ) mJ (B) 250 ( 1 − e −4000t ) μ J (C) 250 ( 1 − e −8000t ) μ J (D) 25 ( 1 − e −4000t ) mJ

Comprehension 10 The capacitor C1 in the figure shown initially carries a charge q0 . When the switches S1 and S2 are closed, capacitor C1 is connected in series to a resistor R and a second capacitor C2 , which is initially uncharged. S1 C1

q0

30. The charge flown through wires as a function of time t is t RC

+

t ⎤ − q0C ⎡ C q0 (B) × ⎣⎢ 1 − e RC ⎦⎥ C1 C2

t

C1C2 C1 + C2

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 155



(A) 2 A (C) 6 A

(B) 4 A (D) 8 A

33. The battery current after a long time is 33 73 (A) A (B) A 70 30 (D) None of these

q5 = q10 = (A)

2500 μC 33

(B) q5 = q10 =

5000 μC 33

t

− C − CR q0 e (C) (D) q0 e RC C1

where, C =

S

34. The charges on the 10 μF and 5 μF capacitors after the steady state is reached are q10 and q5 respectively. Then

S2

−

50 V

32. The initial value of the battery current when the switch S is closed is

70 (C) A 33

R

C2

q0 e (A)

10 Ω

(C) q5 =

2500 5000 μC , q10 = μC 33 33

(D) q5 =

5000 2500 μC , q10 = μC 33 33

9/24/2019 11:26:28 AM

3.156  JEE Advanced Physics: Electrostatics and Current Electricity

Comprehension 12

36. What is battery current a long time after both switches are closed? (A) 0.08 A (B) 0.04 A (C) 0.02 A (D) 0.01 A

In the circuit shown the capacitors are initially uncharged. The switch S2 is closed and then switch S1 is closed. Based on the facts & figure provided, answer the following questions. 100 Ω B 50 Ω

A

C1 = 10 μ F

H

D

S2

S1 12 V

C

37. What is final voltage across C1 ? (A) 2 V (B) 4 V (C) 6 V (D) 8 V

C2 = 50 μ F

150 Ω

G

38. What is final voltage across C2 ? (A) 2 V (B) 4 V (C) 6 V (D) 8 V

F

39. Switch S2 is opened again after a long time. The current, (in A) in the 150 Ω resistor as a function of time is

E

⎛ 400t ⎞ ⎛ 200t ⎞ 0.02 exp ⎜ − −0.04 exp ⎜ (A) ⎟ (B) ⎝ ⎝ 3 ⎟⎠ 3 ⎠

35. What is the battery current immediately after S1 is closed? (A) 1.2 A (B) 12 A (C) 0.12 A (D) None of these

⎛ 400t ⎞ ⎛ 400t ⎞ (C) −0.04 exp ⎜ − −0.02 exp ⎜ − ⎟ (D) ⎟ ⎝ ⎝ 3 ⎠ 3 ⎠

Matrix Match/Column Match Type Questions Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of bubbles will look like the following: A B C D

1.

p p p p p

q q q q q

COLUMN-I contains some of the laws while COLUMN-II contains the corresponding expression. Match the Laws in COLUMN-I with the expressions in COLUMN-II. COLUMN-I

COLUMN-II

(A)  Ohm’s Law

(p) ∑V = 0 in a closed mesh

(B) Kirchhoff’s Junction Law

V (q) R = under same I physical conditions

r

s

t

r r r r

s s s s

t t t t

2.

2

(C) Kirchhoff’s Loop Law

(r) W = I RT joule, heat produced in a resistance

(D)  Joule’s Law

(s) ∑I = 0 at any junction

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 156

3.

Three wires of same material are connected in parallel to a battery. The length ratio of wires is 1 : 2 : 3 and the ratio of their area of cross-section is 2 : 4 : 1 . Match the ratio of quantities in COLUMN-I with their respective answers in COLUMN-II COLUMN-I

COLUMN-II

(A) Resistance

(p)  1 : 1 : 1

(B) Current

(q)  6 : 6 : 1

(C) Power

(r)  1 : 6 : 6

(D) Conductivity

(s)  1 : 1 : 6

Consider a series RC circuit. Match the quantities in COLUMN-I with the quantities in COLUMN-II

9/24/2019 11:26:31 AM

Chapter 3: Electric Current and Circuits 3.157 C

R

COLUMN-I

COLUMN-II

(A) Charging current at time t = 0

1 (p)  CV 2 2

(B) Discharging current at t = 0

(q) Maximum

(C) While charging energy stored

(r) Capacitor becomes short circuit

(D) While charging energy dissipated as heat

(s)  Exponential Law

6.

1Ω

b

c

e

+2 V

1 Ω +6 V 2Ω +4 V d

5.

7.

COLUMN-I

COLUMN-II

(A)  Current in wire ae

(p)  1 A

(B)  Current in wire be

(q)  2 A

(C)  Current in wire ce

(r)  0.5 A

(D)  Current in wire de

(s)  None of these

COLUMN-I (A) For maximum power transfer to load if cells are connected in series

(s)  Eeq = 2E, req = 2r

(B) For maximum power transfer to load if cells are connected in parallel

(q) 

Current i is flowing through a wire of non-uniform cross-section as shown. Match the following two columns. 2

i

COLUMN-I

COLUMN-II

(A)  Current density

(p)  is more at 1

(B)  Electric field

(q)  is more at 2

(C) Resistance per unit length

(r) is same at both sections 1 and 2

(D) Potential difference per unit length

(s)  data is insufficient

With reference to the circuit diagram shown match the following

R=3Ω 6V

COLUMN-II (p) 

r 2

A

Consider two identical cells each of emf E and internal resistance r connected to a load resistance R

2

E 4r E2 2r

(Continued)

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 157

(r)  Eeq = E, req =

i

2Ω a

(C) For series combination of cells

1

For the circuit shown in figure, match the two columns. +4 V

COLUMN-II

(D) For parallel connection of cells

V

4.

COLUMN-I

3 μF

X=2Ω

D

B

S=7Ω

Y=4Ω C

COLUMN-I

COLUMN-II

(A) Potential difference, in volt, across A and D

(p) zero

(B) Potential difference, in volt, across capacitor

(q) 

9 5

(C) Value of Y, in ohm, for which no energy is stored across capacitor

(r) 

1 5

(Continued)

9/20/2019 11:21:46 AM

3.158  JEE Advanced Physics: Electrostatics and Current Electricity

COLUMN-I

COLUMN-II

COLUMN-I

(D) Steady state current, in ampere, in the branch containing the capacitor is

(s) 14

(D) In parallel connection, VB2 (s)  2 the ratio of power VA consumed in A and B

14 (t)  3 8.

Match the consumption of power given in COLUMN-II to the corresponding appliance(s) given in COLUMN-I. COLUMN-I

COLUMN-II

(A) 100 W, 220 V bulb connected across 220 V supply

(p)  300 W

(B) 200 W, 220 V bulb connected across 220 V supply

(q)  66.7 W

9.

(A) In series connection, the ratio of potential difference across A and B

COLUMN-II

(A) 

(p) 

COLUMN-II (p) 

(q)  r

b

(B) 

a

(C) 

b

a

b

RB (r)  RA

r 2

(s) 

5 r 8

11. In the circuit shown in figure, match the following two columns A

B

4 V, 1 Ω

(C) In parallel connection, the ratio of current in A and B

(r) 

a

RA RB

(q) 

r 4

b

(D) 

(B) In series connection, the ratio of power consumed by A and B

VA2 VB2

1 V, 1 Ω 1Ω

(Continued)

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 158

COLUMN-I

(s)  200 W

Two bulbs A and B consume same power P when operated at voltage VA and VB respectively. Bulbs can be connected with a dc voltage source. Match the quantities for respective combinations in COLUMN-I to their appropriate answers in COLUMN-II. COLUMN-I

10. COLUMN-I represents a combination of resistors of equal identical resistors of resistance r while COLUMN-II represents the equivalent resistance across (a) and (b). Match identities in COLUMN-I to their corresponding match(es) in COLUMN-II.

a

(C) Above 100 W and 200 W (r)  100 W bulbs connected in series across a 220 V supply (D) Above 100 W, 200 W bulbs connected in parallel across a 220 V supply

COLUMN-II

COLUMN-I

COLUMN-II (In SI Units)

(A) Potential difference across battery A

(p) zero

(Continued)

9/20/2019 11:21:51 AM

Chapter 3: Electric Current and Circuits 3.159

COLUMN-I

COLUMN-II (In SI Units)

(B) Potential difference across battery B

(q) 1

(C) Net power supplied/ consumed by A

(r) 2

(D) Net power supplied/ consumed by B

(s) 3

COLUMN-I

COLUMN-II

(C)  Rac (D)  Rch, when points c and h are shorted

(r) 

5R 6

(s) 

7R 12

13. With reference to the circuit diagram shown match the following

12. For the given network if Rij is the equivalent resistance between the points i , j of the circuit. Then match the resistances in COLUMN-I with their respective equivalent values in COLUMN-II

2V

2 μF

e g

b

c

e

f

h

g

a

d

4 μF

24 V a

c

b

d

h f

COLUMN-I

COLUMN-II

(A)  Rab

(p) zero

(B)  Rbc

(q) 

3R 4

COLUMN-I

COLUMN-II

(A)  vd - vc (in volt)

(p)  -8

(B)  va - vb (in volt)

(q)  -4

(C) work done by 12 V cell is (in mJ)

(r)  -192

(D) work done by 24 V cell is (in mJ)

(s) 384

(Continued)

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s). 1.

2.

An aluminium wire having a cross-sectional area of 4 × 10 -6 m 2 carries a current of 5 A. Find the drift speed of the electrons in the wire, in mms -1. The density of aluminium is 2.7 gcm -3. Assume that one conduction electron is supplied by each atom. A carbon wire and a Nichrome wire are connected in series, so that the same current exists in both wires. If the combination has a resistance of 10 kΩ at 0 °C ,

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 159

what is the resistance of each wire, in ohm, at 0 °C, so that the resistance of the combination does not change with temperature? 3.

A high-voltage transmission line with a diameter of 2 cm and a length of 200 km carries a steady current of 1000 A. If the conductor is copper wire with a free charge density of 8.49 × 10 28 electron / m 3, how long does it take one electron to travel the full length of the line? Give your answer in years.

9/20/2019 11:21:56 AM

3.160  JEE Advanced Physics: Electrostatics and Current Electricity 4.

5.

The potential difference across the filament of a lamp is maintained at a constant level while equilibrium temperature is being reached. It is observed that the steady-state current in the lamp is only one tenth of the current drawn by the lamp when it is first turned on. If the temperature coefficient of resistivity for -1 the lamp at 20 °C is 0.00450 ( °C ) , and if the resistance increases linearly with increasing temperature, what is the final operating temperature, in °C, of the filament? Taking R = 1 kΩ and E = 250 V in figure, determine the direction and magnitude of the current, in mA, in the horizontal wire between a and e. R

b + –

E

2R

c

4R

d + –

3R

a

6.

e

In the circuit of figure, determine the current, in ampere, in each resistor and the voltage across the 200 Ω resistor. 40 V 200 Ω

7.

20 Ω

1Ω 1Ω

3Ω

(b) the maximum charge, in mC, on the capacitor after the switch is closed.



(c) find the current in the resistor 10 s after the switch is closed, in mA to the nearest integer.

9.

A capacitor in an RC circuit is charged to 60% of its maximum value in 0.9 s. What is the time constant of the circuit? (Express your result in second to the nearest integer).

10. A 10 mF capacitor is charged by a 10 V battery through a resistance R. The capacitor reaches a potential difference of 4 V in a time 3 s after charging begins. Find R, in kΩ. 11. When two unknown resistors are connected in series with a battery, the battery delivers 225 W and carries a total current of 5 A. For the same total current, 50 W is delivered when the resistors are connected in parallel. Determine the values of the two resistors (in ohm). 12. Three 60 W, 120 V light bulbs are connected across a 120 V power source, as shown in figure. Find R1 120 V

70 Ω

For the network shown in figure, show that the resisx ⎞ ⎛ tance Rab = ⎜ 1 + ⎟ Ω. Find x ⎝ 17 ⎠ 1Ω

(a)  the time constant of the circuit, in second.



80 V

360 V

80 Ω

a

8.

2E



b

5Ω



(a) the total power delivered to the three bulbs (in watt) and (b) the voltage across each (in volt).  Assume that the resistance of each bulb is constant. 13. Switch S has been closed for a long time, and the electric circuit shown in figure carries a constant current. Take C1 = 3 mF, C2 = 6 mF , R1 = 4 kΩ, and R2 = 7 kΩ. The power delivered to R2 is 2.4 W.

Consider a series RC circuit (see figure) for which R = 1 MΩ, C = 5 mF and E = 30 V. Find

C1

R1 S

R = 1 × 106 kΩ R2 C

R3

R2

C2

5 μF

E

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 160

S



(a) Find the charge on C1 , in mC.



(b) Now the switch is left open for a long time. Find the change in the value of charge on C2 , in mC.

9/20/2019 11:22:10 AM

Chapter 3: Electric Current and Circuits 3.161 14. In figure, suppose the switch has been closed for a time sufficiently long for the capacitor to become fully charged. Find 12 kΩ S 9V





R2 = 15 kΩ

10 μ F 3 kΩ

(a) the steady-state current in each resistor, in mA. (b) the charge Q on the capacitor, in mC. (c) the switch is now opened at t = 0. Write an equation for the current I through R2 as a function of time. (d) the time interval required for the charge on the capacitor to fall to one-fifth its initial value, in millisecond.

15. A 75 W tungsten light bulb has a resistance of 190 Ω when switched on and 15 Ω when switched off. Estimate the temperature, in kelvin of the filament when the bulb is switched on. Given α = 4.5 × 10 -3 °C -1 16. Two wires of different materials x and y have resis-1 -1 tances per unit lengths, 100 Ω ( km ) , 50 Ω ( km ) -1 and temperature coefficients 0.0025 ( °C ) , -1 0.00075 ( °C ) respectively. It is desired to make a coil having 1000 Ω resistance and a temperature coef-1 ficient of 0.001( °C ) by using suitable lengths of the two wires in series. Calculate their respective lengths in metre. 17. A block of metal is heated directly by dissipating power in the internal resistance of block. Because of temperature rise, the resistance increases exponentially with time and is given by R ( t ) = 0.5e 2t , where t is in second. The block is connected across a 110 V source and dissipates 7644 J heat energy over a certain period of time. Calculate this period of time (in millisecond). 18. An electric kettle has two coil. When one coil is switched on, it takes 15 minutes to boil water and when the second coil is switched on it takes double the time for the same job to be done. How long, in minutes will it take to boil water when both the coils are used in

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 161



(a) series? (b) parallel?

19. A wire of length 1 m and radius 10 -3 m is carrying a heavy current and is assumed to radiate as a black body. At equilibrium its temperature is 900 K while that of the surroundings is 300 K. The resistivity of the material of the wire at 300 K is π 2 × 10 -8 Ωm and its temperature coefficient of resistance is 7.8 × 10 -3 per °C. Find the current, in ampere in the wire. Stefan’s constant, σ = 5.68 × 10 -8 Wm -2K -4 20. With only switch S1 closed in the circuit shown, 3 μF

20 V

12 Ω S1

V

6Ω

S3

S2 2 μF

V

5Ω 20 Ω





(a) what is the steady state reading of the voltmeter (in volt)? (b) what is the charge on the 3 mF capacitor (in mC)? (c) how much power does the 12 V battery supply in the steady state after all the switches are closed (in watt)? (d) what is the charge on 2 mF capacitor (in mC)?

21. 200 identical electrical bulbs, each having resistance 300 Ω, are connected in parallel to a current source of emf 200 V and internal resistance 0.5 Ω.

(a) What is the power expended, in watt on each bulb? (b) The fractional change in power expended on each bulb when one bulb burns out is f. Find 10000 f .

22. In which branch of the circuit shown in figure, an 11  V battery be inserted so that it dissipates minimum power. What will be the current, in ampere, through the 2 Ω resistance for this position of the battery?

2Ω

4Ω

6Ω

9/20/2019 11:22:22 AM

3.162  JEE Advanced Physics: Electrostatics and Current Electricity 23. Figure shows part of a circuit. Calculate the power, in ohm, dissipated in 3 Ω resistance. What is the potential difference VC - VB ? 1Ω C

12 V

D

2Ω

5A

3V

E

4Ω 2A 6A

3Ω

ble internal resistance. Two resistances R1 and R2 are 4000 Ω and 6000 Ω respectively. Find the reading of the voltmeters V1 and V2 , in volt, when E

B

6Ω

V1

V2 S

24. An ammeter and a voltmeter are connected in series to a battery with an emf E = 6 V . When a certain resistance is connected in parallel with the voltmeter, the reading of the latter decreases two times, whereas the reading of the ammeter increase the same number of times. Find the voltmeter reading, in volt, after the connection of the resistance. 25. A voltmeter of resistance R1 and an ammeter of resistance R2 are connected in series across a battery of negligible internal resistance. When a resistance R is connected in parallel to voltmeter, reading of ammeter increases three times while that of voltmeter reduces to one third. Find R1 and R2 when R = 3 Ω. 26. The emf of a storage battery is 90 V before charging and 100 V after charging. When charging began the current was 10 A. What is the current, in ampere, at the end of charging if the internal resistance of the storage battery during the whole process of charging may be taken as constant and equal to 2 Ω? 27. The circuit shown in figure is made of a homogeneous wire of uniform cross-section. 1234 is a square. Q The ratio 12 of the amounts of heat liberated per Q34 unit time in conductor 1-2 and 3-4 is found to be 11x + y 2x , where x and y are non zero positive integers. Find the values of x and y . 3

R2

R1



(a) the switch S is open and



(b) the switch S is closed.

29. It is required to send a current of 8 A through a circuit whose resistance is 5 Ω. What is the least number of cells which must be used for this purpose and how should they be connected? The emf of each cell is 2 V and the internal resistance is 0.5 Ω. 30. What will be the change in the resistance, in ohm, of a circuit between A and F consisting of five identical conductors each of resistance 2 Ω if two similar conductors added as shown by the dashed line in figure. B

A

D

F

C

E

31. A hemispherical network of radius a is made by using a conducting wire of resistance per unit length ⎛ 64 ⎞ 1 λ=⎜ . Find the equivalent resistance, in ohm, ⎝ 2 + π ⎟⎠ a across OP . P

4 B O

A

C

D 1

2

28. In the circuit shown in figure, V1 and V2 are two voltmeters having resistances 6000 Ω and 4000 Ω respectively. E.M.F. of the battery is 250 V, having negligi-

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 162

32. Find the velocity of charge leading to 1 A current which flows in a copper conductor of cross-section 1 cm 2 and length 10 km. Free electron density of copper is n = 8.5 × 10 28 per m 3 . How long will it take the electric charge to travel from one end of the conductor to the other. Give your answer in years.

9/20/2019 11:22:34 AM

Chapter 3: Electric Current and Circuits 3.163 33. Compute the value of battery current I, in ampere, shown in figure. All resistances are in ohm.

series with an ammeter and two cells identical with the others. The current 3 A when the cells and battery aid each other and 2 A when the cells and battery oppose each other. How many cells in the battery are wrongly connected?

6

I 4 4

12 V

8 2

34. Calculate the potential, in volt, of points A , B , C and D shown in Figure 1. What would be the new potential values, in volt, if polarity of 6 V battery is reversed as shown in Figure 2? All resistances are in Ω . A

B

D

C

1

2

12 V

3

0V

B

D

C 2

12 V

37. The gap between the plates of a parallel plate capacitor is filled up with an inhomogeneous poorly conducting medium whose conductivity varies linearly in a direc-

6V

G

1

be ∗ × 10 -13 ampere, where ∗ is not readable. Find the value of ∗ .

tion perpendicular to the plates from σ = 10 -12 Ω -1m -1 to 2 × 10 -12 Ω -1m -1 . Each plate has an area 230 cm 2 and the separation between the plates is d = 2 mm . Find the current, in nanoampere, flowing through the capacitor due to a voltage V = 300 V .

Figure 1 A

36. A parallel plate capacitor has plate of area 10 cm 2 separated by a distance of 1 mm. It is filled with the dielectric mica the resistivity of mica is 1 × 10 3 Ωm , the leakage current through the capacitor is found to

3

6V

G 0V

38. An electric bulb rated for 500 W at 100 V is used in a circuit having a 200 V supply. Calculate the resistance R that one must be put in series with the bulb, so that the bulb delivers 500 W. Give your answer in ohm. 39. A fuse of lead wire has an area of cross-sectional 0.2 mm 2 . On short circuiting, the current in the fuse wire reaches 30 A. How long after the short-circuiting, will fuse begin to melt? Give your answer in millisecond. -1

 For lead, specific heat 0.032 calg -1 ( °C ) , melting point is 327 °C , density is 11.34 gcm -3 and resistiv-

Figure 2

ity is 22 × 10 -6 Ω cm . The initial temperature of wire is

35. Twelve cells each having the same e.m.f. are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in

20°C . Neglect heat losses.

ARCHIVE: JEE MAIN 1. [Online April 2019] A 200 Ω resistor has a certain colour code. If one replaces the red colour by green in the code, the new resistance will be 400 Ω (B) 500 Ω (A) (C) 300 Ω (D) 100 Ω 2. [Online April 2019]  For the circuit shown, with R1 = 1.0 Ω, R2 = 2.0 Ω, E1 = 2 V and E2 = E3 = 4 V, the potential difference between the points a and b is approximately (in V )

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 163

R1

R1

a

E3

R2

E1

E2 R1



(A) 2.7 (C) 2.3

R1

b

(B) 3.7 (D) 3.3

9/20/2019 11:22:45 AM

3.164  JEE Advanced Physics: Electrostatics and Current Electricity 3. [Online April 2019] A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when R=r R = 1000 r (B) (A) R = 2r (D) (C) R = 0.001r 4. [Online April 2019]  In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between A and B . The resistance per unit length of the potentiometer wire is r = 0.01 Ωcm -1 . If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be 1.5 V, 1.5 V 0.5 Ω 0.5 Ω

6Ω

72 V

4Ω

10 μ F

10 Ω

(A) 200 mC (B) 60 mC (C) 10 mC (D) 2 mC 7. [Online April 2019] A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is ( E is mid-point of arm CD ) A

B

V A

J 50 cm D

E

C

3 R (A) R (B) 4

1Ω B

100 cm

(A) 0.75 V (B) 0.50 V (C) 0.20 V (D) 0.25 V 5. [Online April 2019] In the figure shown, what is the current (in Ampere) drawn from the battery? You are given

2Ω

R = 15 Ω, R2 = 10 Ω, R3 = 20 Ω, R4 = 5 Ω, R5 = 25 Ω, 1 R6 = 30 Ω , E = 15 V R1

R3 R2

E

R6

R4

R5

13 9 (A) (B) 32 24 20 7 (C) (D) 3 18 6. [Online April 2019] Determine the charge on the capacitor in the following circuit

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 164

1 7 R (C) R (D) 16 64 8. [Online April 2019] A moving coil galvanometer has resistance 50 Ω and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a 5 kΩ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to 10 V (B) 20 V (A) (C) 15 V (D) 40 V 9. [Online April 2019] A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be 5 5 Ω (A) Ω (B) 3 2 12 7 Ω (C) Ω (D) 2 5 1 0. [Online April 2019]  The resistance of a galvanometer is 50 Ω and the maximum current which can be passed through it is 0.002  A. What resistance must be connected to it in order to convert it into an ammeter of range 0 - 0.5 A?

9/20/2019 11:23:03 AM

Chapter 3: Electric Current and Circuits 3.165 (A) 0.2 Ω (B) 0.002 Ω (C) 0.5 Ω (D) 0.02 Ω 1 1. [Online April 2019] In a conductor, if the number of conduction electrons per unit volume is 8.5 × 10 28 m -3 and mean free time is 25 fs (femto second), its approximate resistivity is

( me = 9.1 × 10-31 kg )

(A) 0.5 Ω (B) 0Ω (C) 1.5 Ω (D) 1Ω 1 5. [Online April 2019] In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure.

(C) 10 -7 Ωm (D) 10 -8 Ωm

G

1 2. [Online April 2019] In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that

E

S. No.

1 T2

R(T ) = (A)

R0 (B) R ( T ) = R0 e T2

(C) R ( T ) = R0 e

-

T02 T2

T2

T2

T02

(D) R ( T ) = R0 e T0

2

1 3. [Online April 2019] A moving coil galvanometer allows a full scale current of 10 -4 A. A series resistance of 2 MΩ is required to convert the above galvanometer into a voltmeter of range 0 - 5 V. Therefore, the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0 - 10 mA is (A) 200 Ω (B) 500 Ω (C) 100 Ω (D) 10 Ω 1 4. [Online April 2019]  In the given circuit, an ideal voltmeter connected across the 10 Ω resistance reads 2 V . The internal resistance r , of each cell is 15 Ω

10 Ω

1.5 V, 1.5 V r Ω, r Ω

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 165

2Ω

Unknown resistance

l

lnR(T)

-

X

R Resistance box

10 -5 Ωm (B) 10 -6 Ωm (A)



K

l(cm)

R(Ω)

1.

1000

60

2.

100

13

3.

10

1.5

4.

1

1.0

Which of the readings is inconsistent? (A) 1 (B) 2 (C) 3 (D) 4

1 6. [Online April 2019] A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10 -8 Ωm ) of radius of cross-section 5 mm . Find the mobility of the charges if their drift velocity is 1.1 × 10 -3 ms -1 . (A) 1.3 m 2 V -1s -1 (B) 1.8 m 2 V -1s -1 (C) 1.5 m 2 V -1s -1 (D) 1.0 m 2 V -1s -1 1 7. [Online April 2019] Space between two concentric conducting spheres of radii a and b ( b > a ) is filled with a medium of resistivity ρ . The resistance between the two spheres will be

ρ ⎛ 1 1⎞ ρ ⎛ 1 1⎞ (A) ⎜ - ⎟ (B) ⎜ - ⎟ 4π ⎝ a b ⎠ 2π ⎝ a b ⎠ ρ ⎛ 1 1⎞ ρ ⎛ 1 1⎞ (C) ⎜ + ⎟ (D) ⎜ + ⎟ 2π ⎝ a b ⎠ 4π ⎝ a b ⎠ 1 8. [Online April 2019] The resistive network shown below is connected to a D.C. source of 16 V . The power consumed by the network is 4 W . The value of R is

9/20/2019 11:23:19 AM

3.166  JEE Advanced Physics: Electrostatics and Current Electricity 4R

6R

R

4R

V

R

1.5 V

12 R V0 E = 16 V

If V0 is almost zero, identify the correct statement

1Ω (A) 8 Ω (B) (C) 16 Ω (D) 6Ω



1 9. [Online April 2019] A galvanometer of resistance 100 Ω has 50 divisions on its scale and has sensitivity of 20 mA/division . It is to be converted to a voltmeter with three ranges, of 0 - 2 V, 0 -10 V and 0 - 20 V. The appropriate circuit to do so is



(A) G

R1

R2

2V

(B) G

R1

(C) G

R1

2V

R3



(A) The emf of the battery is 1.5 V and its internal resistance is 1.5 Ω (B) The emf of the battery is 1.5 V and the value of R is 1.5 Ω (C) The value of the resistance R is 1.5 Ω (D)  The potential difference across the battery is 1.5 V when it sends a current of 1000 mA

2 1. [Online April 2019] A moving coil galvanometer, having a resistance G, produces full scale deflection when a current I g flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I 0 ( I 0 > I g ) by connecting a shunt resistance RA to it and (ii) into a voltmeter

R1 = 1900 Ω R2 = 9900 Ω R3 = 19900 Ω

of range 0 to V ( V = GI 0 ) by connecting a series resistance RV to it. Then,

(A) RA RV = G 2 and

20 V R3

10 V R2

R1

R1 = 1900 Ω R2 = 8000 Ω R3 = 10000 Ω 20 V

10 V R2

2V

(D) G

10 V R2

2V

R3

R1 = 1900 Ω R2 = 9900 Ω R3 = 1900 Ω 20 V

R3

10 V

R1 = 2000 Ω R2 = 8000 Ω R3 = 10000 Ω 20 V

2 0. [Online April 2019] To verify Ohm’s law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained V

RA ⎛ I g ⎞ = RV ⎜⎝ I 0 - I g ⎟⎠

RA RV = G 2 and (C)

2

Ig RA = RV ( I 0 - I g )

2 ⎛ ( I0 - I g ) ⎞ Ig ⎞ RA ⎛ = (D) RA RV = G 2 ⎜ and ⎟ Ig RV ⎜⎝ ( I 0 - I g ) ⎟⎠ ⎝ ⎠

2 2. [Online January 2019] A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is 0.5% (B) 2.0% (A) (C) 2.5% (D) 1.0% 2 3. [Online January 2019] A resistance is shown in the figure. Its value and tolerance are given respectively by Orange

Ammeter R

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 166

2

⎛ Ig ⎞ ⎛ I0 - I g ⎞ R RA RV = G 2 ⎜ (B) and A = ⎜ ⎟ RV ⎝ I g ⎟⎠ ⎝ I0 - I g ⎠

Red Internal resistance

I

1000 mA

Violet

Silver

9/20/2019 11:23:34 AM

Chapter 3: Electric Current and Circuits 3.167 (A) 27 kΩ , 20% (B) 270 Ω , 5% (C) 27 kΩ , 10% (D) 270 Ω , 10% 2 4. [Online January 2019] When the switch S, in the circuit shown, is closed, then the value of current i will be 20 V i1 A

i

10 V



V=0

2 A (B) (A) 5A 4 A (D) (C) 3A 2 5. [Online January 2019] Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm 2 , is v . If the electron density in copper is 9 × 10 28 m -3 the value of v in mms -1 is close to (Take charge of electron to be = 1.6 × 10 -19 C ) (A) 0.02 (C) 3

(B) 0.2 (D) 2

2 6. [Online January 2019] A carbon resistance has a following colour code. What is the value of the resistance?

GOY

Golden

(A) 6.4 MΩ ± 5% (B) 5.3 MΩ ± 5% (C) 64 kΩ ± 10% (D) 530 kΩ ± 5% 2 7. [Online January 2019] In the given circuit the internal resistance of the 18 V cells is negligible. If R1 = 400 Ω , R3 = 100 Ω and R4 = 500 Ω and the reading of an ideal voltmeter across R4 is 5 V, then the value of R2 will be R3 R1

20 Ω R2

20 Ω

B

4Ω 2Ω S



R1

i2 10 V

C 2Ω

2 8. [Online January 2019] In the given circuit the cells have zero internal resistance. The currents (in amperes) passing through resistance R1 and R2 respectively, are

10 V

(A) 1, 2 (C) 0.5, 0

(B) 0, 1 (D) 2, 2

2 9. [Online January 2019] A 2 W carbon resistor is colour coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is 20 mA (B) 0.4 mA (A) (C) 100 mA (D) 63 mA 3 0. [Online January 2019] A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is 4 Ω (B) 12 Ω (A) 2Ω (C) 8 Ω (D)

6Ω

6Ω

A

B

6Ω

3 1. [Online January 2019] A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ε and interε nal resistance r. A cell C having emf and internal 2 resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is

R4

D ε, r

R2 J

A 18 V

(A) 230 Ω (B) 450 Ω (C) 550 Ω (D) 300 Ω

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 167

Cε , 3r 2

B

G

9/20/2019 11:23:54 AM

3.168  JEE Advanced Physics: Electrostatics and Current Electricity 11 11 (A) L (B) L 12 24 5 13 (C) L (D) L 12 24 3 2. [Online January 2019] The Wheatstone bridge shown in figure here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80 Ω and 40 Ω respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3 , would be R1

R2

3 5. [Online January 2019]  Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be 60 W (B) 30 W (A) (C) 120 W (D) 240 W 3 6. [Online January 2019] In a Wheatstone bridge (see figure), resistances P and Q are approximately equal. When R = 400 Ω , the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405 Ω. The value of X is close to B

G R3

P

R4 A



(A) Brown, Blue, Black (C) Grey, Black, Brown

V using the standard formula R = , where V and I I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is V A

R

C

K2 R

(B) Red, Green, Brown (D) Brown, Blue, Brown

3 3. [Online January 2019] The actual value of resistance R, shown in the figure is 30 Ω. This is measured in an experiment as shown

Q

G

X D K1

(A) 404.5 Ω (B) 401.5 Ω (C) 402.5 Ω (D) 403.5 Ω 3 7. [Online January 2019] The resistance of the metre bridge AB in given figure is 4 Ω. With a cell of emf E = 0.5 V and rheostat resistance Rh = 2 Ω the null point is obtained at some point J . When the cell is replaced by another one of emf E = E2 the same null point J is found for Rh = 6 Ω . The emf E2 is E

(A) 570 Ω (B) 600 Ω (C) 350 Ω (D) 35 Ω 3 4. [Online January 2019] A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W . Dissipated power when an ideal power supply of 11 V is connected across it is (A) 11 × 10

-5

W (B) 11 × 10 W

(C) 11 × 10

-3

W (D) 11 × 10 -4 W

5

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 168

A

B

J

6V

Rh

(A) 0.6 V (B) 0.5 V (C) 0.3 V (D) 0.4 V

9/20/2019 11:24:12 AM

Chapter 3: Electric Current and Circuits 3.169 3 8. [Online January 2019]  A galvanometer having a resistance of 20 Ω and 30 divisions on both sides has figure of merit 0.005 ampere/division . The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt , is: (A) 100 Ω (B) 125 Ω (C) 80 Ω (D) 120 Ω 3 9. [Online January 2019] In the circuit shown, the potential difference between A and B is: M

A

5Ω

1Ω

1V

1Ω

2V

10 Ω

D N

C

B

3V

1Ω

R2

K2

R1 = 220 Ω

G

K1

(A) 22 Ω (B) 25 Ω (C) 5 Ω (D) 12 Ω 4 2. [Online January 2019] In a meter bridge, the wire of length 1 m has a nondR uniform cross-section such that, the variation of dl dR 1 its resistance R with length l is ∝ . Two equal dl l resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP ?

(A) 6 V (B) 3V (C) 2 V (D) 1V 4 0. [Online January 2019] In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A . If a 10 Ω resistor is connected in series with R1 , the null point shifts by 10 cm . The resistance that should be connected in parallel with ( R1 + 10 ) Ω such that the null point shifts back to its initial position is: R1

R2 G

A

B

(A) 60 Ω (B) 30 Ω (C) 40 Ω (D) 20 Ω 4 1. [Online January 2019] The galvanometer deflection, when key K1 is closed but K 2 is open, equal θ 0 (see figure). On closing K 2 also and adjusting R2 to 5 Ω , the deflection in galva-

θ0 . The resistance of the galvanom5 eter is, then given by (neglect the internal resistance of battery) nometer becomes

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 169

R′

R

G P A

l

1–l

B

(A) 0.2 m (B) 0.35 m (C) 0.25 m (D) 0.3 m 4 3. [Online January 2019] An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is 480 Ω (B) 490 Ω (A) (C) 495 Ω (D) 395 Ω 4 4. [Online January 2019]  Two electric bulbs, rated at ( 25 W, 220 V ) and ( 100 W , 220 V ) , are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then (A) P1 = 9 W , P2 = 16 W

(B) P1 = 4 W , P2 = 16 W P1 = 16 W , P2 = 9 W (D) P1 = 16 W , P2 = 4 W (C)

9/20/2019 11:24:36 AM

3.170  JEE Advanced Physics: Electrostatics and Current Electricity 4 5. [Online January 2019]  A galvanometer, whose resistance is 50 Ω , has 25 divisions in it. When a current of 4 × 10 -4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of 6250 Ω (B) 250 Ω (A)

X B

A

P

I6

Q

I3

I5 S I 4

I2

I1

R

1.1 A , 0.4 A , 0.4 A (A) 1.1 A , -0.4 A , 0.4 A (B) (C) 0.4 A , 1.1 A , 0.4 A (D) -0.4 A , 0.4 A , 1.1 A 47. [2018] Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 Ω . The internal resistances of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between (A) 11.6 V and 11.7 V (B) 11.5 V and 11.6 V (C) 11.4 V and 11.5 V (D) 11.7 V and 11.8 V 48. [2018] On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm . The resistance of their series combinations is 1 kΩ . How much was the resistance on the left slot before the interchange? 990 Ω (B) 505 Ω (A) (C) 550 Ω (D) 910 Ω 49. [Online 2018] In a meter bridge, as shown in the figure, it is given that resistance Y = 12.5 Ω and that the balance is obtained at a distance 39.5 cm from end A (by Jockey J). After interchanging the resistances X and Y , a new balance point is found at a distance l2 from end A . What are the values of X and l2 ?

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 170

G

I1 39.5

J

(100 – I1) Wire

C

Meter scale

(C) 200 Ω (D) 6200 Ω 4 6. [Online January 2019] In the given circuit diagram, the currents, I1 = -0.3 A, I 4 = 0.8 A and I 5 = 0.4 A, are flowing as shown. The currents I 2 , I 3 and I 6 , respectively, are:

Y

Battery

Key

(A) 19.15 Ω and 60.5 cm

8.16 Ω and 60.5 cm (B) (C) 8.16 Ω and 39.5 cm (D) 19.15 Ω and 39.5 cm

50. [Online 2018]  In the given circuit all resistances are of value R ohm each. The equivalent resistance between A and B is

A

B

5R 2R (B) (A) 2 5R 3R (D) (C) 3 51. [Online 2018] A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be (A) Increased 8 times (B) Unchanged (C) Doubled (D) Halved 52. [Online 2018] A copper rod of cross-sectional area A carries a uniform current I through it. At temperature T , if the volume charge density of the rod is ρ , how long will the charges take to travel a distance d ?

ρdA ρdA (A) (B) I IT 2ρdA 2ρdA (C) (D) I IT

9/20/2019 11:24:59 AM

Chapter 3: Electric Current and Circuits 3.171 53. [Online 2018]  In the following circuit, the switch S is closed at t = 0 . The charge on the capacitor C1 as a function of

56. [2017] In the following circuit the current in each resistance is

CC ⎞ ⎛ time will be given by ⎜ Ceq = 1 2 ⎟ C1 + C2 ⎠ ⎝ C1

2V

2V

1Ω

C2

1Ω

2V S

R

E



t ⎞⎤ ⎡ ⎛ (A) CeqE ⎢ 1 - exp ⎜⎝ RCeq ⎟⎠ ⎥ ⎢⎣ ⎥⎦ ⎡ ⎛ tR ⎞ ⎤ (B) C1E ⎢ 1 - exp ⎜ - ⎟ ⎥ ⎝ C1 ⎠ ⎦ ⎣ t ⎞ ⎛ CeqE exp (C) ⎜⎝ RCeq ⎟⎠ t ⎞⎤ ⎡ ⎛ C2E ⎢ 1 - exp ⎜ (D) ⎟⎥ RC ⎝ 2⎠⎦ ⎣ 54. [Online 2018] A heating element has a resistance of 100 Ω at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500 °C more than room temperature. What is the temperature coefficient of resistance of the heating element? (C) 0.5 × 10

-1

-4

°C (D) 5 × 10 °C

r r1

C



r2

r1 CE (A) CE (B) ( r2 + r )

CE (C)

r2

( r + r2 )

r1 CE (D) ( r1 + r )

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 171

(B) 0.25 A (D) 0 A

58. [Online 2017] A 9 V battery with internal resistance of 0.5 Ω is connected across an infinite network as shown in the figure. All ammeters A1 , A2 , A3 and voltmeter V are ideal. Choose correct statement. A1 9V

E

2V

57. [2017] Which of the following statement is false? (A)  Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude. (B) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed. (C) A rheostat can be used as a potential divider. (D)  Kirchhoff’s second law represents energy conservation.

-1

55. [2017] In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be

1Ω

2V

(A) 1 A (C) 0.5 A

1 × 10 -4 °C -1 (B) 2 × 10 -4 °C -1 (A) -4

2V

1Ω

A3

A2

V

4Ω

0.5 Ω 1Ω

1Ω

1Ω

1Ω



(A) Reading of V is 9 V



(B) Reading of A1 is 2 A



(C) Reading of V is 7 V



(D) Reading of A1 is 18 A

4Ω

4Ω ∞

1Ω

59. [Online 2017] A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1 A when two way key K 3 is open. The balance point is at a length l1 cm from P when two way key K 3 is plugged in between 2 and 1, while the balance points is at a length l2 cm from P when key K 3

9/20/2019 11:25:14 AM

3.172  JEE Advanced Physics: Electrostatics and Current Electricity is plugged in between 3 and 1. The ratio of two resisR tances 1 , is found to be R2 Rh2

E2

A

R2 R1

3 K3

1

G

P1 > P3 > P2 (D) P1 > P2 > P3 (C) 62. [Online 2017] A uniform wire of length l and radius r has a resisr tance of 100 Ω . It is recast into a wire of radius . The 2 resistance of new wire will be 400 Ω (B) 100 Ω (A)

K2

2

P3 > P2 > P1 (B) P2 > P1 > P3 (A)

(C) 200 Ω (D) 1600 Ω Q

P K1

Rh1

E1

l l1 (A) 2 (B) l2 - l1 l2 - l1 l l1 (C) 1 (D) l1 + l2 l1 - l2 60. [Online 2017] In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P = 4 Ω and the neutral point N is at 60 cm from A . Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A . The value of unknown resistance R is P

63. [2016]  The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by (A) Linear decrease for Cu, linear decrease for Si. (B) Linear increase for Cu, linear increase for Si. (C) Linear increase for Cu, exponential increase for Si. (D) Linear increase for Cu, exponential decrease for Si. 64. [Online 2016] In the circuit shown, the resistance r is a variable resistance. If for r = fR, the heat generation in r is maximum then the value of f is R

R

r

Q G

A

E

Rh

1 1 (A) (B) 2

B

N

1 3 (C) (D) 4 4

K

20 33 (A) Ω (B) Ω 3 5

65. [Online 2016] The resistance of an electrical toaster has a temperature

(C) 6 Ω (D) 7Ω

dependence given by R ( T ) = R0 ⎡⎣ 1 + α ( T - T0 ) ⎤⎦ in its

61. [Online 2017] The figure shows three circuits I, II and III which are connected to a 3 V battery. If the powers dissipated by the configurations I, II and III are P1, P2 and P3 respectively, then

range of operation. At T0 = 300 K , R = 100 Ω and at

1Ω 1Ω

1Ω

1Ω

1Ω

1Ω

1Ω 3V 1Ω

I

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 172

1Ω 1Ω 1Ω II

1Ω 3V

T = 500 K, R = 120 Ω. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s . The total work done ( in kJ ) in raising the temperature is

1Ω

⎛ 6⎞ ⎛ 3⎞ (A) 40 ln ⎜ ⎟ (B) 20 ln ⎜ ⎟ ⎝ 2⎠ ⎝ 5⎠

1Ω

⎛ 60 ln ⎜ (C) ⎝

1Ω

3V III

6⎞ ⎛ 15 ⎞ 40 ln ⎜ ⎟ ⎟⎠ (D) ⎝ 13 ⎠ 5

66. [2015] When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is

9/20/2019 11:25:32 AM

Chapter 3: Electric Current and Circuits 3.173 2.5 × 10 -4 ms -1. If the electron density in the wire is 8 × 10 28 m -3 , the resistivity of the material is close to

(A) 1.6 × 10-8 Ωm (B) 1.6 × 10 -7 Ωm

70. [Online 2015] In the electric network shown, when no current flows through the 4 Ω resistor in the arm EB , the potential difference between the points A and D will be

(C) 1.6 × 10 -6 Ωm (D) 1.6 × 10 -5 Ωm 67. [2015] In the circuit shown, the current in the 1 Ω resistor is 6V

9V



Q 3Ω

68. [Online 2015] Suppose the drift velocity vd in a material varied with the applied electric field E as Vd ∝ E . Then V -I graph for a wire made of such a material is best given by V (B)

I

I

(C) V

V (D)

I

V



(A) 11.9 V (C) 13.1 V

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 173

(B) 12.5 V (D) 24.5 V

(A) 3 V (C) 5 V

B

3V

C

(B) 4 V (D) 6 V

73. [2012] The figure shows an experimental plot for discharging of a capacitor in an R-C circuit. The time constant t of this circuit lies between 25 20 15 10 5 0 50 100 150 200 250 300 Time t in seconds

1Ω

0.6 Ω

9V

72. [2013] The supply voltage to a room is 120 V . The resistance of the lead wires is 6 Ω . A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb? (A) zero volt (B) 2.9 volt (C) 13.3 volt (D) 10.04 volt

69. [Online 2015] A 10 V battery with internal resistance 1 Ω and a 15 V battery with internal resistance 0.6 Ω are connected in parallel to a voltmeter (shown in figure). The reading in the voltmeter will be close to

15 V

R

71. [2014] In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be (A) 8 A (B) 10 A (C) 12 A (D) 14 A

I

10 V

D

4Ω

A

(A) 1.3 A , from P to Q (B) 0A (C) 0.13 A , from Q to P (D) 0.13 A , from P to Q

(A) V

E

4V

Potential difference V in volts

3Ω

2Ω

2Ω

P 2Ω 1Ω

2V

F



(A) 0 and 50 sec (C) 100 sec and 150 sec

(B) 50 sec and 100 sec (D) 150 sec and 200 sec

74. [2012]  Two electric bulbs marked 25 W - 220 V and 100 W - 220 V are connected in series to a 440 V supply. Which of the bulbs will fuse?

9/20/2019 11:25:45 AM

3.174  JEE Advanced Physics: Electrostatics and Current Electricity

(A) 100 W (C) neither

(B) 25 W (D) Both

75. [2011] If a wire is stretched to make it 0.1% longer, its resistance will (A) increase by 0.05% (B) increase by 0.2% (C) decrease by 0.2% (D) decrease by 0.05% 76. [2010] Two conductors have the same resistance at 0 °C but their temperature coefficients of resistance are α 1 and α 2 . The respective temperature coefficients of their series and parallel combinations are nearly

α1 + α 2 α1 + α 2 , (A) 2 2 α1 + α 2 , (C)

α1 + α 2 2

(B)

77. [2009] This question contains Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement 1: The temperature dependence of resistance is usually given as R = R0 ( 1 + αΔt ) . The resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from 27 °C to 227 °C . This implies that α = 2.5 × 10 −3 °C −1 . Statement 2: R = R0 ( 1 + αΔt ) is valid only when the change in the temperature ΔT is small and ΔR = ( R − R0 )  R0 .

α1 + α 2 , α1 + α 2 2



α 1α 2 α1 + α 2



(D) α 1 + α 2 ,



(A) Statement 1 is true, Statement 2 is false. (B) Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation of Statement 1. (C) Statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation of Statement 1. (D) Statement 1 is false, Statement 2 is true.

ARCHIVE: JEE advanced In this section each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. [JEE (Advanced) 2016] An infinite line charge of uniform electric charge density λ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ . The electrical conduction in the material follows Ohm’s law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j ( t ) at any point in the material?

m

m

AI Fe

j(t) (B)

(A) j(t)

2. [JEE (Advanced) 2015] In an aluminium ( Al ) bar of square cross section, a square hole is drilled and is filled with iron ( Fe ) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 × 10 −8 Ωm and 10 × 10 −7 Ωm , respectively. The electrical resistance between the two faces P and Q of the composite bar is

50

Single Correct Choice Type Problems

2 mm

7 mm

(0, 0)

t

t

1875 2475 (C) μΩ (D) μΩ 49 132

j(t) (D)

j(t) (C)

(0, 0)

(0, 0)

t

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 174

(0, 0)

2475 1875 (A) μΩ (B) μΩ 64 64

t

3. [JEE (Advanced) 2014] During an experiment with a meter bridge, the galvanometer shows a null point when the jockey is pressed at 40 cm using a standard resistance of 90 Ω , as show in the scale used in the meter bridge is 1 mm . The unknown resistance is

9/24/2019 11:59:05 AM

Chapter 3: Electric Current and Circuits 3.175 (C) R1 G1

90 Ω

R

G2

RT

R2

40 cm

(A) 60 ± 0.15 Ω (B) 135 ± 0.56 Ω

V

(C) 60 ± 0.25 Ω (D) 135 ± 0.23 Ω

(D) R2

4. [IIT-JEE 2011] A meter bridge is set-up as shown in figure, to determine an unknown resistance X using a standard 10 Ω resistor. The galvanometer shows null point when t­apping key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B . The determined value of X is 10 Ω

X

A

G1

G2

RT

R1 V

6. [IIT-JEE 2010] Consider a thin square sheet of side L and thickness t , made of a material of resistivity ρ . The resistance between two opposite faces, shown by the shaded areas in the figure is

B

(A) 10.2 Ω (B) 10.6 Ω 11.1 Ω (C) 10.8 Ω (D) 5. [IIT-JEE 2010] To verify Ohm’s law, a student is provided with a test resistor RT , a high resistance R1 , a small resistance R2 , two identical galvanometers G1 and G2 and a variable voltage source V . The correct circuit to carry out the experiment is 10 Ω B

Heater R 100 V

(A) G1 R2

G2

RT

R1

(B) G1 G2

RT

L



(A) (B) (C) (D)

directly proportional to L directly proportional to t independent of L independent of t

7. [IIT-JEE 2010] Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100 W , 60 W and 40 W bulbs have filament resistances R100 , R60 and R40 , respectively, the relation between these resistances is 1 1 1 + R100 = R40 + R60 (B) (A) = R100 R40 R60 1 1 1 > R100 > R60 > R40 (D) > (C) R100 R60 R40

V

R1

C

t

R2

8. [IIT-JEE 2008] Figure shows three resistor configurations R1 , R2 and R3 connected to 3 V battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3 , respectively, then

V

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 175

9/20/2019 11:26:15 AM

3.176  JEE Advanced Physics: Electrostatics and Current Electricity

1Ω

1Ω

1Ω 1Ω

1Ω

3V

1Ω

1Ω

1Ω

1Ω 3V

11. [IIT-JEE 2007] A circuit is connected as shown in the figure with the switch S open. When the switch is closed, the total amount of charge that flows from Y to X is 3 μF

1Ω

1Ω

S

( (

R2

R1

6 μF

X

Y

3Ω

1Ω

6Ω

9V 1Ω

1Ω

3V



(B) 54 mC

(A) zero

(C) 27 mC (D) 81 mC

1Ω R3

P1 > P2 > P3 (B) P1 > P3 > P2 (A)

12. [IIT-JEE 2006] Two bars of radius r and 2r are kept in contact as shown. An electric current I is passed through the bars. Which one of the following is correct?

P2 > P1 > P3 (D) P3 > P2 > P1 (C) 9. [IIT-JEE 2007] A resistance of 2 Ω is connected across one gap of a metre bridge (the length of the wire is 100 cm) and an unknown resistance, greater than 2 Ω , is connected across the other gap. When these resistances are interchanged, the balance point shifts by 20 cm. Neglecting any corrections, the unknown resistance is 4Ω 3 Ω (B) (A) (C) 5 Ω (D) 6Ω 10. [IIT-JEE 2007] A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric d constant K = 2 . The level of liquid is initially. 3 Suppose the liquid level decreases at a constant speed V , the time constant as a function of time is

l/2

l/2

2r

r

I

C A



B

(A) Heat produced in bar BC is 4 times the heat produced in bar AB (B) Electric field in both halves is equal (C)  Current density across AB is double that of across BC (D) Potential difference across AB is 4 times that of across BC

13. [IIT-JEE 2006] Find the time constant for the given RC circuits in ­correct order V

V

R1 R2

C2 C1

R1

C1

C d

d/3

R

C2

R2

   V

( 15d + 9Vt ) ε 0 R 6ε R (A) 0 (B) 5d + 3Vt 2d 2 - 3 dVt - 9V 2t 2 ( 15d - 9Vt ) ε 0 R 6ε R (D) (C) 0 5d - 3Vt 2d 2 + 3 dVt - 9V 2t 2

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 176



R1

C1

R2

C2

R1 = 1 Ω , R2 = 2 Ω , C1 = 4 mF , C2 = 2 mF

9/20/2019 11:26:24 AM

Chapter 3: Electric Current and Circuits 3.177



(A) 18, 4,

8 9

(B) 18,

8 ,4 9



(C) 4, 18,

8 9

(D) 4,

8 , 18 9

(A) P and Q (B) Q and R (C) P and R (D) any two points

14. [IIT-JEE 2005] A moving coil galvanometer of resistance 100 Ω is used as an ammeter using a resistance 0.1 Ω . The maximum deflection current in the galvanometer is 100 mA . Find the minimum current in the circuit, so that the ammeter shown maximum deflection (A) 100.1 mA (B) 1000.1 mA (C) 10.01 mA (D) 1.01 mA 15. [IIT-JEE 2005] A rigid container with thermally insulated walls contains a coil of resistance 100 Ω, carrying current 1 A. Change in internal energy after 5 minute will be (A) zero (B) 10 kJ (C) 20 kJ (D) 30 kJ 16. [IIT-JEE 2005] Find out the value of current through 2 Ω resistance for the given circuit. 5Ω

10 V

10 Ω

20 V

19. [IIT-JEE 2004] For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between B

C

A

B1

C1

B and C (B) (A) C and D A and D (D) B1 and C1 (C) 20. [IIT-JEE 2004] A capacitor is charged using an external battery with a resistance x in series. The dashed line shows the variation of log e I with respect to time. If the resistance is changed to 2x , the new graph will be loge I

2Ω



(A) 5 A (C) zero

D

S

(B) 2 A (D) 4 A

R

Q

17. [IIT-JEE 2005]

P

t

A 4 mF capacitor, a resistance of 2.5 MΩ is in series with 12 V battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor: ⎡⎣ Given log e ( 2 ) = 0.693 ⎤⎦

P (B) (A) Q



21. [IIT-JEE 2003] Express which of the following set-up can be used to verify Ohm’s Law?

(A) 13.86 s (C) 7 s

(B) 6.93 s (D) 14 s

18. [IIT-JEE 2004]  Six equal resistances are connected between points P, Q and R as shown in the figure. Then, the net resistance will be maximum between

R (D) (C) S

(A)

A V

P

(B) A Q

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 177

V

R

9/20/2019 11:26:31 AM

3.178  JEE Advanced Physics: Electrostatics and Current Electricity (C)

A

V

(D)

2Rr (A) R+r

(B)

8R ( R + r ) 3R + r

(C) 2r + 4 R

(D)

5R + 2r 2

25. [IIT-JEE 2002] A 100 W bulb B1 , and two 60 W bulbs B2 and B3 , are connected to a 250 V source, as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1 , B2 and B3 respectively. Then

V

A

22. [IIT-JEE 2003] In the shown arrangement of the experiment of the meter bridge if AC corresponding to null deflection of galvanometer is x , what would be its value if the radius of the wire AB is doubled?

B1

B2 B3

250 V R2

R1

W1 > W2 = W3 (B) W1 > W2 > W3 (A)

G A

x

W1 < W2 = W3 (D) W1 < W2 < W3 (C) B

C

x (A) x (B) 4 (C) 4x (D) 2x 23. [IIT-JEE 2003] The three resistances of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipation I

i

II

26. [IIT-JEE 2001] In the given circuit, it is observed that the current I is independent of the value of the resistance R6 . Then, the resistance values must satisfy R5 I

R1

R6

R2

R3 R4

i

R1R2 R5 = R3 R4 R6 (A)

III

i

IV

1 1 1 1 = + (B) + R5 R6 R1 + R2 R3 + R4

i

(C) R1R4 = R2 R3 R1R3 = R2 R4 (D)



(A) III < II < IV < I (C) I < IV < III < II

(B) II < III < IV < I (D) I < III < II < IV

24. [IIT-JEE 2002] The effective resistance between points P and Q of the electrical circuit shown in figure is 2R

P

r

2R 2R

r

2R 2R

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 178

Q

27. [IIT-JEE 2001] A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by DT in time t. A number N of similar cells is now connected in series with a wire of the same material and cross section but of length 2L. The temperature of the wire is raised by the same amount DT in the same time t. The value of N is (A) 4 (B) 6 (C) 8 (D) 9

2R

9/20/2019 11:26:40 AM

Chapter 3: Electric Current and Circuits 3.179 28. [IIT-JEE 2001] In the circuit shown in figure, with steady current, the potential drop across the capacitor must be V

R

V

C

2V

2R

32. [IIT-JEE 1997]  A parallel combination of 0.1 M Ω resistor and a 10 mF capacitor is connected across a 1.5 V source of negligible resistance. The time required for the capacitor to get charged upto 0.75 V is approximately (in second) ∞ (B) log e 2 (A) log10 2 (C)

V (A) V (B) 2 V 2V (C) (D) 3 3

33. [IIT-JEE 1995] A battery of internal resistance 4 Ω is connected to the network of resistances as shown. In order that the maximum power can be delivered to the network, the value of R in Ω should be

29. [IIT-JEE 1999] In the circuit shown P ≠ R, the reading of galvanometer is same with switch S open or closed. Then P

(D) zero

R

E 4Ω

Q

R

R R

6R R

4R

S R

4 2 (A) (B) 9

G

(A) IR = IG (B) IP = IG I R = I P (D) (C) IQ = IR 30. [IIT-JEE 1998] In the circuit shown in figure, the current through 3Ω A

2Ω B 2Ω

I 8Ω

9V

8Ω

4Ω

8 (C) (D) 18 3 34. [IIT-JEE 1993] Read the following statements carefully : Y: The resistivity of semiconductor decreases with increase of temperature. Z: In a conducting solid, the rate of collisions between free electrons and ions increases with increase of temperature Select the correct statement(s) from the following : Y is true but Z is false (A)

2Ω

D 2Ω C 2Ω

(B) Y is false but Z is true

(C) Both Y and Z are true



(A) the 3 Ω resistor is 0.5 A.



(B) the 3 Ω resistor is 0.25 A.

(D) Y is true and Z is the correct reason for Y



(C) the 4 Ω resistor is 0.5 A.

35. [IIT-JEE 1988]



(D) the 4 Ω resistor is 0.25 A.

 A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of

31. [IIT-JEE 1997] A steady current flows in a metallic conductor of nonuniform cross-section. The quantity/quantities constant along the length of the conductor is/are (A) current, electric field and drift speed (B) drift speed only (C) current and drift speed (D) current only

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 179



(A) each of them increases. (B) each of them decreases. (C)  copper increases and that of decreases. (D)  copper decreases and that of increases.

germanium germanium

9/20/2019 11:26:47 AM

3.180  JEE Advanced Physics: Electrostatics and Current Electricity 36. [IIT-JEE 1983] The current i in the circuit (see figure) is

(B) The key S1 is kept closed for long time such that capacitors are fully charged. Now key S2 is closed, at this time, the instantaneous current across 30 Ω resistor (between points P and Q) will be 0.2 A (round off to 1st decimal place).

i 30 Ω

30 Ω

2V

30 Ω

1 1 (A) A (B) A 45 15 1 1 (C) A (D) A 10 5 37. [IIT-JEE 1981] In the circuit shown in figure the heat produced in the 5 Ω resistor due to the current flowing through it is 10 calories per second. 4Ω

6Ω

5Ω



The heat generated in the 4 Ω resistor is

1 cals -1 (B) 2 cals -1 (A) (C) 3 cals -1 (D) 4 cals -1

Multiple Correct Choice Type Problems (In this section each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct)

(C) At time t = 0, the key S1 is closed, the instantaneous current in the closed circuit will be 25 mA.



(D) If key S1 is kept closed for long time such that capacitors are fully charged, the voltage difference between point P and Q will be 10 V. 2. [JEE (Advanced) 2019] Two identical moving coil galvanometers have 10 Ω resistance and full scale deflection at 2 mA current. One of them is converted into a voltmeter of 100 mV full scale reading and the other into an Ammeter of 1 mA full scale current using appropriate resistors. These are then used to measure the voltage and current in the Ohm’s law experiment with R = 1000 Ω resistor by using an ideal cell. Which of the following statement(s) is/are correct? R will be (A)  The measured value of 978 Ω < R < 982 Ω (B)  The resistance of the ammeter will be 0.02 Ω (round off to 2nd decimal place) (C) If the ideal cell is replaced by a cell having internal resistance of 5 Ω then the measured value of R will be more than 1000 Ω (D) The resistance of the voltmeter will be 100 kΩ 3. [JEE (Advanced) 2016] In the circuit shown below, the key is pressed at time t = 0. Which of the following statement(s) is (are) true?

1. [JEE (Advanced) 2019] In the circuit shown, initially there is no charge on capacitors and keys S1 and S2 are open. The val-

40 μ F V

ues of the capacitors are C1 = 10 mF , C2 = 30 mF and C3 = C4 = 80 mF . Which of the following statement(s)

A

is/are correct? S1 C1 70 Ω

P

C4 5 V 30 Ω S1

C2 30 Ω

C3 10 V Q



03_Physics for JEE Mains and Advanced - 2_Part 4.indd 180

50 kΩ

Key



100 Ω

(A) If key S1 is kept closed for long time such that capacitors are fully charged, the voltage across the capacitor C1 will be 4 V .

25 kΩ



– + 20 μ F

5V

(A) The voltmeter display -5 V as soon as the key is pressed and displays +5 V after a long time. (B) The voltmeter will display 0 V at time t = ln 2 seconds. 1 (C) The current in the ammeter becomes of the inie tial value after 1 second. (D) The current in the ammeter becomes zero after a long time.

9/20/2019 11:26:58 AM

Chapter 3: Electric Current and Circuits 3.181 4. [JEE (Advanced) 2016] An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true? (A) The temperature distribution over the filament is uniform. (B) The resistance over small sections of the filament decreases with time. (C) The filament emits more light at higher band of frequencies before it breaks up. (D)  The filament consumes less electrical power towards the end of the life of the bulb. 5. [JEE (Advanced) 2016] Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance R of the galvanometers RC < , which of the following 2 statement(s) about any one of the galvanometers is(are) true? (A) The maximum voltage range is obtained when all the components are connected in series. (B) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer. (C) The maximum current range is obtained when all the components are connected in parallel. (D) The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors. 6. [IIT-JEE 2012] For the resistance network shown in the figure, choose the correct option(s) P I2 2 Ω

S

2Ω

2Ω 1Ω

I1

4Ω

Q

1Ω 4Ω

T

4Ω

12 V



(A) The current through PQ is zero

(B) I1 = 3 A

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 181



(C) The potential at S is less than that at Q

(D) I2 = 2 A 7.

[IIT-JEE 2009] For the circuit shown in the figure I

24 V



2 kΩ

R1

6 kΩ

R2

RL

1.5 kΩ

(A) the current I through the battery is 7.5 mA (B) the potential difference across RL is 18 V (C) ratio of powers dissipated in R1 and R2 is 3 (D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9

8. [IIT-JEE 1999] When a potential difference is applied across, the current passing through (A) an insulator at 0 K is zero (B) a semiconductor at 0 K is zero (C) a metal at 0 K is finite (D) a p -n diode at 300 K is finite, if it is reverse biased 9.

[IIT-JEE 1991]

A microammeter has a resistance of 100 Ω and full scale range of 50 mA . It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s) (A) 50 V range with 10 kΩ resistance in series (B) 10 V range with 200 kΩ resistance in series (C) 5 mA range with 1 Ω resistance in parallel (D) 10 mA range with 1 Ω resistance in parallel 10. [IIT-JEE 1989] Capacitor C1 of capacitance 1 mF and capacitor C2 of capacitance 2 mF are separately charged fully by a common battery. The two capacitors are then separated allowed to discharge through equal resistors at time t = 0 (A) the current in each of the two discharging circuits is zero at t = 0 (B) the currents in the two discharging circuits at t = 0 are equal but not zero (C) the currents in the two discharging circuits at t = 0 are unequal (D) the capacitors C1 and C2 lose their 50% charge in same duration

9/20/2019 11:27:02 AM

3.182  JEE Advanced Physics: Electrostatics and Current Electricity

Integer/Numerical Answer Type Questions In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data given in the question(s). 1. [JEE (Advanced) 2015] In the following circuit, the current through the resistor R ( = 2 Ω ) is I amperes. The value of I is 1Ω

R=2Ω

are connected in parallel across R, the rate is J 2 . If J1 = 2.25 J 2 then the value of R in Ω is 5. [IIT-JEE 2010] At time t = 0 , a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have no charge initially, at what time (in second) does the voltage across them become 4 V? [Take: ln 5 = 1.6 , ln 3 = 1.1 ] 2 MΩ

8Ω 6Ω 6.5 V 12 Ω

2Ω

2Ω

4Ω

A

B 2 MΩ

10 Ω

4Ω

2 μF

2 μF

Assertion and Reasoning Type Problems 2. [IIT-JEE 2011] Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volt is 6V

1Ω

A

B

3V

2Ω

3. [JEE (Advanced) 2014] A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 Ω resistance, it can be converted into a voltmeter of range 0 - 30 V . If con2n nected to a Ω resistance, it becomes an ammeter 249 of range 0 -1.5 A . The value of n is

For the following Assertion-Reason type questions given below, choose the correct option: (A) Both statement I & II are correct and statement II is correct explanation of statement I. (B) Both statement I & II are correct and statement II is not correct explanation of statement I. (C) Statement I is true and statement II is false. (D) Statement II is correct and statement I is false. 1. [IIT-JEE 2008]  Statement-1: In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance Statement-2: Resistance of a metal increases with increase in temperature.

4. [IIT-JEE 2010] When two identical batteries of internal resistance 1 Ω each are connected in series across a resistor R, the rate of heat produced in R is J1 . When the same ­batteries

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 182

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Chapter 3: Electric Current and Circuits 3.183

Answer Keys—Test Your Concepts and Practice Exercises Test Your Concepts-I (Based on Current Definition) 1. (b)  1.05 mA 2. (a)  17 A

4. 1 + α 0 ( T1 - T2 ) m ⎡ ⎤ 5.  ρ2 = ρ1 ⎢ 1 + ( T2 - T1 ) ⎥ ρ 1 ⎣ ⎦ 6. Percentage change in Length = 0.0017%

     (b)  85 kAm -2

    Percentage change in Area = 0.0034%

3. 0.265 C

    Percentage change in Resistance = 0.39%

⎧ ⎛ q1 q2 ⎞ ⎪ - v ⎜⎝ a + b ⎟⎠ , ⎪ ⎪ q ( ) 4. I t = ⎨ -v 2 , b ⎪ ⎪ 0, ⎪ ⎩

t
250 ms

)A

18.  VCD

     (c)  8 Ω 7. B 8. 

Vx V0 Rx , V = 0 for R  R0 x⎞ L ⎛ RL + R0 x ⎜ 1 - ⎟ ⎝ L⎠

Test Your Concepts-XI (Based on RC Circuit)

t ηt

19. 

E ( η - 1 ) e CR R

20.  1.5 mA

1. (a)  ( 0.632 ) Ι 0t

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 186

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Chapter 3: Electric Current and Circuits 3.187

Single Correct Choice Type Questions   1. D

  2. A

  3. C

  4. B

  5. B

  6. C

  7. A

  8. B

  9. B

 10. C

 11. D

 12. C

 13. D

 14. C

 15. C

 16. C

 17. C

 18. A

 19. C

 20. A

 21. D

 22. A

 23. C

 24. C

 25. D

 26. B

 27. C

 28. C

 29. C

 30. B

 31. C

 32. D

 33. C

 34. B

 35. C

 36. C

 37. C

 38. A

 39. C

 40. D

 41. B

 42. A

 43. C

 44. D

 45. D

 46. D

 47. A

 48. D

 49. C

 50. B

 51. A

 52. C

 53. B

 54. C

 55. A

 56. B

 57. A

 58. B

 59. A

 60. C

 61. D

 62. D

 63. B

 64. C

 65. C

 66. C

 67. D

 68. D

 69. C

 70. A

 71. C

 72. C

 73. B

 74. C

 75. B

 76. D

 77. D

 78. D

 79. B

 80. D

 81. D

 82. C

 83. B

 84. C

 85. C

 86. A

 87. B

 88. D

 89. D

 90. D

 91. B

 92. B

 93. B

 94. A

 95. B

 96. D

 97. A

 98. B

 99. B

100.  A

101.  C

102.  B

103.  A

104.  A

105.  B

106.  C

107.  D

108.  D

109.  C

110.  A

111.  B

112.  A

113.  B

114.  A

115.  C

116.  C

117.  C

118.  C

119.  B

120.  A

121.  B

122.  D

123.  B

124.  D

125.  C

126.  A

127.  B

128.  C

129.  B

130.  D

131.  A

132.  A

133.  C

134.  B

135.  C

136.  D

137.  C

138.  B

139.  C

140.  B

141.  A

142.  D

143.  A

144.  B

145.  B

146.  D

147.  A

148.  C

149.  C

150.  B

151.  B

152.  B

153.  C

154.  C

155.  C

156.  D

157.  D

158.  D

159.  B

160.  D

161.  A

162.  D

163.  C

164.  A

165.  B

166.  C

167.  B

168.  B

169.  A

170.  B

171.  B

172.  B

173.  D

174.  A

175.  C

Multiple Correct Choice Type Questions 1.  A, B

2.  B, C

3.  A, B, C, D

4.  B, D

5.  A, C

6.  A, B, D

7.  A, C

8.  A, B, C, D

9.  A, C

10.  A, C

11.  B, D

12.  B, C, D

13.  A, B, D

14.  B, D

15.  C, D

16.  A, B, C

17.  A, B

18.  C, D

19.  A, C, D

20.  B, C

21.  A, D

22.  A, C

23.  B, C, D

24.  A, C, D

25.  A, B, C

26.  A, B, C, D

27.  B, C

28.  A, B, C, D

29.  A, B, C, D

30.  A, B, D

31.  A, B

32.  A, B, C, D

33.  A, C

34.  A, B

35.  A, C

36.  A, C, D

37.  A, C

38.  B, D

39.  A, B, D

40.  B, C

41.  C, D

42.  A, C

43.  A, C, D

44.  A, D

45.  A, B

46.  B, C

47.  B, D

48.  B, C

49.  A, C

50.  A, B

51.  B, C

52.  A, B, C, D

53.  A, D

54.  A, B, C, D

55.  A, B, D

Reasoning Based Questions 1.  B

2.  D

3.  A

4.  D

5.  A

6.  D

7.  B

8.  D

9.  B

10.  D

11.  B

12.  A

13.  D

14.  B

15.  D

16.  D

17.  A

18.  D

19.  B

20.  A

21.  D

22.  C

23.  D

24.  C

25.  A

Linked Comprehension Type Questions 1.  C

2.  A

3.  B

4.  D

5.  C

6.  C

7.  B

8.  B

9.  C

10.  A

11.  B

12.  C

13.  A

14.  C

15.  B

16.  D

17.  B

18.  D

19.  D

20.  C

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 187

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3.188  JEE Advanced Physics: Electrostatics and Current Electricity 21.  C

22.  B

23.  C

24.  B

25.  C

26.  A

27.  B

28.  C

29.  C

31.  A

32.  B

33.  C

34.  C

35.  C

36.  B

37.  D

38.  C

39.  C

30.  B

Matrix Match/Column Match Type Questions 1.  A → (q)

B → (s)

C → (p)

D → (r)

2.  A → (s)

B → (q)

C → (q)

D → (p)

3.  A → (q, r, s)

B → (q, s)

C → (p, s)

D → (p, s)

4.  A → (q)

B → (s)

C → (q)

D → (s)

5.  A → (q)

B → (q)

C → (s)

D → (r)

6.  A → (p)

B → (p)

C → (p)

D → (p)

7.  A → (q)

B → (r)

C → (t)

D → (p)

8.  A → (r)

B → (s)

C → (q)

D → (p)

9.  A → (p, q)

B → (p, q)

C → (r, s)

D → (r, s)

10.  A → (s)

B → (r)

C → (p)

D → (q)

11.  A → (s)

B → (r)

C → (s)

D → (r)

12.  A → (s)

B → (q)

C → (r)

D → (p)

13.  A → (q)

B → (p)

C → (r)

D → (s)

Integer/Numerical Answer Type Questions 1.  130

2.  5560, 4440

6.  1 in 200 Ω, 4 in 70 Ω, 3 in 80 Ω, 8 in 20 Ω 10.  587

11.  6, 3

3.  27

4.  2020

5.  50

7.  10

8.  (a) 5 (b) 150 (c) 4

9.  1

12.  (a) 240, 360, 40 (b) 80, 40, 40

13.  (a) 222 (b) 444

14.  (a) 333 (b) 50 (c) 278 (d) 290

15.  2885

17.  500

18.  (a) 45 (b) 10

19.  36

20.  (a) zero (b) 36 (c) 2 (d) 30

21.  (a) 75 (b) 25

22.  1

23.  20

24.  2

25.  24, 8

26.  5

27.  1, 6

28.  (a) 100 (b) 125

29.  160

30.  4

31.  8

32.  431

33.  6

34.  (a) 12, 9, 3, - 6 (b) 12, 11, 9, 6

36.  6

37.  5

38.  20

16.  6250, 7500

35.  1

39.  95

ARCHIVE: JEE MAIN 1.  B

2.  D

3.  B

4.  D

5.  B

6.  A

7.  D

8.  B

9.  A

10.  A

11.  D

12.  B

13.  (*)

14.  A

15.  D

16.  D

17.  A

18.  A

19.  A

20.  A

21.  A

22.  D

23.  C

24.  B

25.  A

26.  D

27.  D

28.  C

29.  A

30.  A

31.  D

32.  D

33.  A

34.  A

35.  D

36.  C

37.  C

38.  C

39.  C

40.  A

41.  A

42.  C

43.  D

44.  D

45.  C

46.  A

47.  B

48.  C

49.  B

50.  A

51.  A

52.  A

53.  A

54.  B

55.  C

56.  D

57.  B

58.  B

59.  B

60.  A

61.  B

62.  D

63.  D

64.  A

65.  C

66.  D

67.  C

68.  C

69.  C

70.  C

71.  C

72.  D

73.  C

74.  B

75.  B

76.  A

77.  A

(*) None of the options given in exam is correct

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 188

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Chapter 3: Electric Current and Circuits 3.189

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems 1.  C

2.  B

3.  C

4.  B

5.  C

6.  C

7.  D

8.  C

9.  A

10.  A

11.  C

12.  A

13.  B

14.  A

15.  D

16.  C

17.  A

18.  A

19.  C

20.  B

21.  A

22.  A

23.  A

24.  A

25.  D

26.  C

27.  B

28.  C

29.  A

30.  D

31.  D

32.  D

33.  B

34.  C

35.  D

36.  C

37.  B

Multiple Correct Choice Type Problems 1.  A, C

2.  A, B

3.  A, B, C, D

4.  C, D

6.  A, B, C, D

7.  A, D

8.  A, B, D

9.  B, C

5.  B, C 10.  B

Integer/Numerical Answer Type Questions 1.  1

2.  5

3.  5

4.  4

5.  2

Assertion and Reasoning Type Problems 1.  D

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 189

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This page is intentionally left blank

03_Physics for JEE Mains and Advanced - 2_Part 4.indd 190

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Hints and explanations

01_Ch 1_Hints and Explanation_P1.indd 1

9/20/2019 11:30:23 AM

This page is intentionally left blank

01_Ch 1_Hints and Explanation_P1.indd 2

9/20/2019 11:30:23 AM

Chapter 1: Electrostatics

1.

Fair =

q1q2 4πε 0 r 2

 Let us now replace the slab of thickness t with an equivalent amount of air such that two charges placed in air at separation x will experience the same force if the separation between them completely has a slab of dielectric constant K between them. In such a case

q1q2 q1q2 = 2 4πε 0 x 4πε 0 Kt 2



⇒ Equivalent separation in air = x = t K



So, now the effective separation between the charges is

r ′ = r − t + x = r − t + t K

⇒ r′ = r − t + t K

q1q2 So, F ′ = 4πε 0 r ′ 2

⇒ F′ =



⇒ q = ne =



⇒ q=

Now, one piece of Cu has positive charge and the other has a negative charge, so both will attract each other with a force of F =



⇒ F = 2 × 1016 N

dq

dq = λ dx

q1q2

2





⇒

Q

⇒

F′ 4 4 = = Fair ( 1 + K )2 9



⇒ K=4

2.

4.

63.5 g of Cu contains 6 × 10 23

6 × 10 63.5

23

Cu atoms



1 g of Cu contains



⎤ ⎡ 6 × 10 23 × 10 ⎥ Cu atoms 10 g of Cu contains ⎢ ⎣ 63.5 ⎦



Since for every 1000 atoms, an electron is transferred,

So, n =

23

6 × 10 × 10 63.5 × 1000

01_Ch 1_Hints and Explanation_P1.indd 3

Cu atoms

∫

x

dx

L

dq =

O



K =2

O

⎛ x2 ⎞ ⇒ dq = λ 0 ⎜ ⎟ dx ⎝ L⎠ 





2

⎝ 127 ⎠ ≅ 2.0 × 1016 N 2 ⎛ 1 ⎞ ⎜⎝ ⎟ 100 ⎠

⇒ F=



⇒

( 9 × 109 ) ⎛⎜ 1920 ⎞⎟



4 q1q2 1 ⎡ ⎤ ⇒ F′ = ⎢ 2 2 ⎥ 4πε 0 ( 1 + K ) r ⎣ ⎦



1 q2 4πε 0 r 2

Consider an element of length dx, carrying a charge dq at a distance x from O . Then, by definition 

r Now, as per the problem we have t = 2 q1q2 1 ⇒ F′ = 2 4πε 0 ⎛ r r ⎞ K⎟ ⎜⎝ r − + ⎠ 2 2

⇒ 1+ K = 3

1920 C 127

3.

4πε 0 ( r − t + t K )



6 × 10 24 × 1.6 × 10 −19 C 63500

CHAPTER 1

Test Your Concepts-I (Based on Coulomb’s Law)

λ0 x 2 dx L

∫ O

λ ⎛ L3 ⎞ ⇒ Q= 0⎜ ⎟ L ⎝ 3⎠ λ 0 L2 3 Each ball is in equilibrium under the following three forces: ⇒ Q=



(i) tension ( T ) 



(ii) electric force ( F )



(iii) weight ( W )



So, Lami’s theorem can be applied.

T cos θ = W …(1) T sin θ = F …(2) T ′ cos θ = W ′ …(3) T ′ sin θ = F ′ …(4)

9/20/2019 11:30:36 AM

H.4  JEE Advanced Physics: Electrostatics and Current Electricity

θ θ

θ θ

T

F

F′ W

W′ In vacuum



T′

In liquid

So, from (1), (2), (3) and (4), we get



W W′ = F F′

F where W ′ = W − U = V ρ g − Vσ g and F ′ = K Vρ g



⇒



⇒ K=

5.

Between Two Electrons:

F

Fe =

=

V ρ g − Vσ g

ρ ρ −σ

F K

1 e2 m2 and Fg = G 2e 2 4πε 0 r r

Fe e2 1 = 2 Fg me 4πε 0G



⇒



F ⎛ 1.6 × 10 −19 ⎞ 9 × 109 ⇒ e =⎜ ⎟ 31 − Fg ⎝ 9.1 × 10 ⎠ 6.67 × 10 −11



⇒



Between Two Protons:

2

F e2 1 ⇒ e = 2 Fg mp 4πε 0G



⇒

Fe ⎛ 1.6 × 10 −19 ⎞ 9 × 109 =⎜ −27 ⎟ ⎝ ⎠ Fg 1.67 × 10 6.67 × 10 −11

F ⇒ e ≅ 10 36 Fg

q Let the specific charge be for which both the forces m become equal, then

q = 4πε 0G m



⇒

q = 8.6 × 10 −11 Ckg −1 m

6.

  F = F21 =



q1q2     3 ( r − r0 ) 4πε 0 r − r0   where r − r0 = 6iˆ − 8 ˆj



  ⇒ r − r0 = 10 m

 50 × 10 −6 × 2 × 10 −6 × 9 × 109 So, F = 6iˆ − 8 ˆj ( 10 )3  ⇒ F = 9 × 10 −4 6iˆ − 8 ˆj N

(

(

1 q2 Gm2 = 2 4πε 0 r 2 r

01_Ch 1_Hints and Explanation_P1.indd 4

)

)



 2 ⇒ F = 9 × 10 −4 6 2 + ( −8 )



 ⇒ F = 9 × 10 −3 N

7.

  Here, rP = iˆ + 2 ˆj − 4 kˆ and rQ = 4iˆ + 6 ˆj − 16 kˆ



  ⇒ rP − rQ = −3iˆ − 4 ˆj + 12 kˆ



  2 2 2 ⇒ rP − rQ = ( 3 ) + ( −4 ) + ( 12 ) = 13 m

 Since F =

q1q2 1 4πε 0 rP − rQ

P

2



⇒

Fe ≅ 4 × 10 42 Fe

Gmp2 1 e2 Fe = and F = g 4πε 0 r 2 r2



r

rP



3



( rP − rQ )

F Q

rQ rQ + r = rP





 17 × 10 −12 × 9 × 109 −3iˆ − 4 ˆj + 12kˆ ⇒ F= ( 13 )3



 17 × 9 × 10 −3 −3iˆ − 4 ˆj + 12kˆ ⇒ F= 169 × 13

(

(

)



 170 × 90 × 10 −5 ⇒ F= −3iˆ − 4 ˆj + 12kˆ 169 × 13  ⇒ F ≅ ( 1 )( 7 ) ( 10 −5 ) −3iˆ − 4 ˆj + 12kˆ



 ⇒ F ≅ −21iˆ − 28 ˆj + 84 kˆ × 10 −5 N



(

(

(

)

)

)

)

9/20/2019 11:30:52 AM

Hints and Explanations H.5

Since λ =

Q 2π R

and the tension developed in the ring when a charge is placed at its centre is given by T =

q0 λ q0Q = (we have done this already) 4πε 0 R 8π 2ε 0 R2



Further, by Laws of Elasticity, we know that ⎛T⎞ ⎜⎝ ⎟⎠ Stress = A Y = Strain ⎛ ΔR ⎞ ⎜⎝ ⎟ R ⎠ TR ⇒ Y= AΔR

⇒ ΔR =

⇒



q0Q TR = AY 8π 2ε 0 RAY

d2 y 2 g y=0 +  dt 2

 Comparing with the standard equation of SHM according to which we have  y + ω 2y = 0 where we get ω =



2g 

11. When the bead is displaced through x , then the separation between the two charges becomes x 2 + L2 and at that instant let the Coulomb force of attraction between the two charges be F. We observe that the component of force F cos θ restores the charge q to its mean position O.



m X

–q

Putting q0 = 10 −8 C , Q = π C , R = 0.1 m , A = 10 −6 m 2, we get ΔR = 2.25 mm

9.

DO YOURSELF (same kind of illustration done before) 33 × 10 −9 C

+Q

10. In equilibrium position, gravitational force is balanced by Coulomb repulsive force

j i

+ L2

+Q Finally

Initially



–q

F F sin θ x2



m

θ

l

l

Y = 2 × 10 −11 Nm −2

x F cos θ

O

From Newton’s Second Law,

− F cos θ = ma

Qq mg = 4πε 0 2

where the negative sign indicates that the component of force has a restoring nature and restores the charge −q to its mean position.

If charge q is displaced in positive y − direction, such that y   , from Newton’s Second Law,



⎛ ⎞⎛ ⎞ Qq x ⇒ −⎜ ⎟⎜ ⎟ = ma 2 2 2 2 ⎝ 4πε 0 ( x + L ) ⎠ ⎝ x + L ⎠



⎛ ⎛ Qq ⎞ ⎜ x ⇒ a = −⎜ ⎟ 3 ⎜ ⎝ 4πε 0 m ⎠ ⎝ ( x 2 + L2 ) 2



Qq 4πε 0 (  + y )

2

− mg = ma

Qq ⎡ 1 ⎤ − mg = ma 2 ⎥ 4πε 0 2 ⎢ ⎛ y ⎞ ⎢ ⎜ 1+ ⎟ ⎥ ⎢⎣ ⎝  ⎠ ⎥⎦ Y F m q

l



X

l

mg Q

Equilibrium position

m q

Q



2y ⎞ ⎛ ⇒ mg ⎜ 1 − ⎟ − mg = ma ⎝  ⎠



2 gy ⇒ a=− 

01_Ch 1_Hints and Explanation_P1.indd 5

CHAPTER 1

8.

⎞ ⎟ ⎟ ⎠

For a small linear displacement, x  L , we can ignore x 2 in comparison to L2 , so we get ⎛ Qq ⎞ a  − ⎜ ⎟x ⎝ 4πε 0 mL3 ⎠

⎛ Qq ⎞ ⇒ x + ⎜ ⎟x=0 ⎝ 4πε 0 mL3 ⎠

Comparing with the standard equation of SHM, which is given by x + ω 2 x = 0 , we get ω =



Qq 2π = = 2πν T 4πε 0 mL3

⇒ T = 2π

4πε 0 mL3 Qq

9/20/2019 11:31:12 AM

H.6  JEE Advanced Physics: Electrostatics and Current Electricity 12. Since the spring is metallic so the charge distributes Q resides equally on both the blocks. Hence a charge 2 on each block, due to which the blocks will repel each other with a force given by

14.

q1 + q2 = 50 × 10 −6



⇒ q1 + q2 = 5 × 10 −5 …(1)

Also 1 = 9 × 109

q1q2

( 2 )2

⎛ Q⎞⎛ Q⎞ 1 ⎜⎝ 2 ⎠⎟ ⎝⎜ 2 ⎠⎟ = k ( L − L0 ) F = 4πε 0 L2



⇒ q1q2 =





⇒ q1q2 = 4.44 × 10 −10 …(2)



Put value of q2 from (2) in (1), we get

Solving for Q

Q = 2L 4πε 0 k ( L − L0 ) 13. Let q1 and q2 be the initial charges on the spheres, then according to Coulomb’s Law 0.108 = 9 × 109

q1q2

( 0.5 )2

⇒ q1q2 = 3 × 10 −12 …(1)

After connection, the charge flows from one sphere to another till both acquire the same potential (or in this case equal charge). So, final charge on both is q f =

q1 − q2 2

(as initially they have opposite nature)

Now, again according to Coulomb’s Law, the repulsive force is ⎛ q1 − q2 ⎞ ⎜ ⎟ 2 ⎠ 9⎝ 0.036 = 9 × 10 2 ( 0.5 )

2

0.036 × 0.25 × 4 9 × 109



⇒ q1 − q2 =



⇒ q1 − q2 = 2 × 10 −6 …(2)



Solving (1) and (2), we get

q1 − q12

3 × 10 q1

−12

= 2 × 10 −6

−6

− 2 × 10 q1 − 3 × 10

−12

=0



⇒



⇒ q1 =

2 × 10 −6 ± 4 × 10 −12 + 12 × 10 −12 2



⇒ q1 =

2 × 10 −6 ± 4 × 10 −6 2



⇒ q1 = 3 × 10 −6 C and q2 = −1 × 10 −6 C



⇒ q1 = 3 μC and q 2 = −1 μC

01_Ch 1_Hints and Explanation_P1.indd 6

4 9 × 109

4.44 × 10 −10 = 5 × 10 −5 q1

q1 +

⇒ q12 − 5 × 10 −5 q1 + 4.44 × 10 −10 = 0



Solving the quadratic, we get

q1 = 3.8 × 10 −5 C = 38 μC q1 = 1.2 × 10 −5 C = 12 μC OR So, we conclude that charge on one sphere is 38 μC and that on the other is 12 μC . 1 e2 4πε 0 r 2

15.

Fe = F =



⇒ F=



⇒ F = 8.2 × 10 −8 N

9 × 109 × ( 1.6 × 10 −19 )

2

( 0.53 × 10 −10 )2

This electrostatic force between the proton and electron, provides the necessary centripetal force to electron to revolve in a circle of radius 0.53 Å.

⇒ mrω 2 = F



⇒ ac = rω 2 =



⇒ ac = 9 × 10 22 ms −2

F 8.2 × 10 −8 = m 9.1 × 10 −31

Further rω 2 = 9 × 10 22 9 × 10 22 0.53 × 10 −10



⇒ ω2 =



⇒ ω = 4.1 × 1016 rads −1

Also, T =

2π ω

⇒ T = 1.5 × 10 −16 s

9/20/2019 11:31:35 AM

Hints and Explanations H.7

Fg =

Gme mp

…(1)

2

r where me = mass of electron = 9.1 × 10 −31 kg and mp = mass of proton = 1.67 × 10 −27 kg Fe =

1 e2 …(2) 4πε 0 r 2

⇒ F8 ≅ 3.6 × 10 −47 N and Fe ≅ 8.2 × 10 −8 N

So, we observe the gravitational force to be negligible in comparison to the electrostatic force. (But this result is not to be generalised to the solar system). So, this result is true in case of subatomic particles and we can also conclude here that gravitational forces are long range forces having significant values for massive bodies (like the sun-earth system). 17.

q2 N sin 60 = mg and N cos 60 = 4πε 0 R2 N 60°

q2 4πε0R2

N cos 60



mg



⇒ tan 60° =

⇒ q2 =

mg ⎛ q2 ⎞ ⎜ 4πε R2 ⎟ ⎝ ⎠ 0

4πε 0 mgR2 3



⎛ 4πε 0 mgR2 ⎞ ⇒ q=⎜ ⎟⎠ ⎝ 3

18.

q2 = μmg 4πε 0 r 2



⇒

9 × 109 × ( 10

)

−7 2

( 10 × 10 −2 )2

1 2

⎛ 5 ⎞( ) = μ⎜ 10 ⎝ 1000 ⎟⎠



⇒ μ = 0.18

19.

F=



For this to be minimum, we have



1 q(Q − q ) 4πε 0 r2

Q ⇒ q= 2 q Q 2 = =1 ⇒ Q −q Q −Q 2

01_Ch 1_Hints and Explanation_P1.indd 7

1. Let us redraw the diagram to locate the coordinates of all the points. Considering origin at 7, we get Now, according to Principle of Superposition, net force on 1 is given by         F1 = F12 + F13 + F14 + F15 + F16 + F17 + F18 y

4

dF =0 dq

1

z

5 (a, 0, a) 6 (a, 0, 0) 7 (0, 0, 0) 8 (0, 0, a)

x

6

7



1 (a, a, a) 2 (a, a, 0) 3 (0, a, 0) 4 (0, a, a)

2

3



N sin 60



Test Your Concepts-II (Based on Principle of Superposition)

5

8

CHAPTER 1

16.

From Coulomb’s Law in vector form, we have

 Fij =

qi q j 1  4πε 0 ri − rj

 So, F12 =

1 Q2 ( ˆ ) ⎛ Q2 ⎞ ˆ ak = ⎜ ⎟k 4πε 0 a 3 ⎝ 4πε 0 a 2 ⎠

 F13 =

1 4πε 0

 F13 =

Q2 ⎡ 1 ( ˆ ˆ ) ⎤ i+k ⎥ 4πε 0 a 2 ⎢⎣ 2 2 ⎦

 F14 =

1 Q2 ( ˆ ) ⎛ Q2 ⎞ ˆ ai = ⎜ ⎟i 4πε 0 a 3 ⎝ 4πε 0 a 2 ⎠

 F15 =

⎛ Q2 ⎞ ˆ 1 Q2 ˆ = aj ⎜ ⎟j 4πε 0 a 3 ⎝ 4πε 0 a 2 ⎠

 F16 =

1 4πε 0

 F16 =

Q2 ⎡ 1 ˆ ˆ ⎤ j+k ⎥ 4πε 0 a 2 ⎢⎣ 2 2 ⎦

 F17 =

1 4πε 0

 F17 =

Q2 ⎡ 1 ˆ ˆ ˆ ⎤ i+ j+k ⎥ 4πε 0 a 2 ⎢⎣ 3 3 ⎦

 F18 =

1 4πε 0



3

Q2

(

2a )

3



( ri − rj )

( aiˆ + akˆ )

( )

Q2

(

2a )

3

( ajˆ + akˆ ) (

Q2

(

3a )

3

)

( aiˆ + ajˆ + akˆ ) (

Q2

(

2a )

3

)

( aiˆ + ajˆ )

9/20/2019 11:31:51 AM

H.8  JEE Advanced Physics: Electrostatics and Current Electricity  F18 =  ⇒ F1 =



3.

Q2 ⎡ 1 ˆ ˆ ⎤ i+j ⎥ 4πε 0 a 2 ⎢⎣ 2 2 ⎦

(

)

Fnet = 2 F cos θ

1 Q2 ⎡ ˆ ˆ ˆ 2iˆ + 2 ˆj + 2kˆ + i+ j+k + 2 2 4πε 0 a 2 ⎢⎣

(

)

(

)

1

(

)

⎤ iˆ + ˆj + kˆ ⎥ 3 3 ⎦





 ⇒ F1 =



 ⇒ F1 =



The force on 1 is equally inclined to x , y and z -axis.

2.

The force due to one charge on the charge at P is shown here. When all the forces are taken into account, then the sine components cancel each other. So,



1 1 ⎤ ˆ ˆ ˆ Q2 ⎡ 1+ + i+ j+k 2 3 3 ⎥⎦ 4πε 0 a 2 ⎢⎣

(

)

P

l

D l 2 l

F

C

x

h

l

O

A

B

⇒ Fnet = 4

x

O

l 2

Q(1)

h

⎛ 2 ⎞ 4πε 0 ⎜ h 2 + ⎟ ⎝ 2 ⎠

2

h2 +

2 2

⎧ ⎪∵ cos θ = ⎨ ⎪ ⎩ ⇒ Fnet =

4Qh ⎛ 2 ⎞ 4πε 0 ⎜ h 2 + ⎟ ⎝ 2⎠

32

⎫ ⎪  ⎬ h2 + 2 ⎪⎭ h

2

(UPWARDS)

 For equilibrium of m , this Fnet must balance the weight of the particle mg (downwards).

4Qh

⇒

3

⇒ Q=

mgπε 0 ⎛ 2  2 ⎞ ⎜ h + ⎟⎠ 2 h ⎝

01_Ch 1_Hints and Explanation_P1.indd 8

2Qq0 r

32 4πε 0 ( r 2 + a 2 )

θ

2F cos θ F

θ

F sin θ

F sin θ

r2 + a2

y x

r2 + a2

r

a

a

Fy will be MAXIMUM, when



d ( Fy ) = 0 dr Qq0 d ⎡ r ⎤ =0 2πε 0 dr ⎢⎣ ( r 2 + a 2 )3 2 ⎥⎦



⇒



−5 2 −3 2 ⎡ 3 ⎤ 2r ⎥ + ( r 2 + a 2 ) ⇒ r ⎢ − ( r 2 + a2 ) =0 ⎣ 2 ⎦



⇒ −



⇒ 3r 2 = r 2 + a2



⇒ 2r 2 = a 2



⇒ r=±

3r 2 +1= 0 r + a2 2

a 2 ⎛ a ⎞ ⎜⎝ ⎟ Qq0 2⎠ = 32 2πε 0 ⎛ a 2 ⎞ + a2 ⎟ ⎜⎝ ⎠ 2



⇒ Fmax



⇒ Fmax =

Qq0 a ( 2 ) 2πε 0 3 3 a 3



⇒ Fmax =

Qq0 3 3πε 0 a 2

= mg

⎛ 2 ⎞ 2 4πε 0 ⎜ h 2 + ⎟ ⎝ 2⎠

⇒ Fy =



y

θ

l2 2

h2







This force is directed radially outwards when q0 is positive in nature.

Fnet = Fy = 4 F cos θ

Qq0 ⎡ ⎤ cos θ ⇒ Fnet = Fy = 2 ⎢ 2 2)⎥ ( ⎣ 4πε 0 r + a ⎦

F

F sin θ

A



3Q 2 ⎛ 1 1 ⎞ + ⎜ 1+ ⎟ MAGNITUDE 2 3 3⎠ 4πε 0 a 2 ⎝

F cos θ

x

From the figure drawn, we see that on resolution of the two identical forces, F sin θ and F sin θ cancel. So, net force equals

32

9/20/2019 11:32:10 AM

Hints and Explanations H.9 Z

q2 1 ( 2q ) = 4πε 0 ( 2R )2 4πε 0 R2 2

F=

F sin θ

Fnet = F 2 + F 2 + 2 F 2 cos 60

P F sin θ

2 r = z 2+ a 2

θ

2q

F cos θ F cos θ

F

R

+q z

F

R

a 2 a

+q

F



⇒ Fnet = 3 F = 3



⇒ Fnet =

q2 4πε 0 R2

Assume, no charge at 6,       F0 = F01 + F02 + F03 + F04 + F05

Qq ⎛ 1 ⇒ FR = −4 ⎜ 4πε 0 ⎛ 2 L2 ⎞ ⎜ ⎜⎝ z + ⎟⎠ ⎝ 2



⇒ FR = −

a 4



 ⇒ F0 =

6.

(a) zero



2

q  4πε 0 a 2

a

–q a

7.

6

2

4Qqz 3

3⎛ Qq ⎞ FR = mz = −4 ( 2 ) 2 ⎜ z 3⎟ 4 πε ⎝ 0a ⎠

5

{from 0 to 3}

qQ 4πε 0 a 2

Let us first visualise the situation given in the problem and draw it here. Now let us consider two diagonally opposite charges having charge +q each. Each of the charge will exert an attractive force F on −Q such that the horizontal component of this force F cancels, whereas vertical components add. Similarly for other two diagonally opposite charges the horizontal components cancel and the vertical components still stay so as to make the resultant force as

01_Ch 1_Hints and Explanation_P1.indd 9

⎞ ⎟ L ⎟ z2 + 2 ⎠ z

For z  L , we may neglect z 2 term in the denominator so as to get

(b) Assume, a to be the distance between q and Q , then

F =

⎞⎛ ⎟⎜ ⎟⎜ ⎠⎝

⎛ L2 ⎞ 2 4πε 0 ⎜ z 2 + ⎟ ⎝ 2⎠

1 0

+q



5.

2

a

O

FR = −4 F cos θ

3 q2 4πε 0 R2

3

+q

a

R

CHAPTER 1

4.



3 ⎞ ⎛ 4 ( 2 ) 2 Qq ⎟ ⎜ z+ z=0 ⇒  ⎜ 4πε 0 ma 3 ⎟ ⎠ ⎝

Comparing with the standard equation of SHM given by  z + ω 2 z = 0 , we get 3

ω =

8.

4 ( 2 ) 2 Qq 4πε 0 ma 3

=

2π = 2πν T

⎛ Qq ⎞ ⇒ T = 2π 2 2 ⎜ ⎟ ⎝ πε 0 ma 3 ⎠ If we place one more charge q at the 24th vertex, the total force on the central charge will add up to zero. So according to the principle of superposition we have

F =

qQ 4πε 0 a 2

9/20/2019 11:32:22 AM

H.10  JEE Advanced Physics: Electrostatics and Current Electricity

9.

FCB = FCA = F =

q2 4πε 0 a 2

FC2 = F 2 + F 2 + 2 F 2 cos 120

⇒

2

2

= 2F − F = F

(a = 3 m) a

FCB = F

–q

a

FCA = F



C

q2  4πε 0 a 2

{parallel to AB}



⇒ FC = 10 N ( parallel to AB )

3.

Force on +2 μC = qE = ( 2 × 10 −6 )( 10 5 ) = 0.2 N   (due North)



Force on −5 μC = ( 5 × 10 −6 )( 10 5 ) = 0.5 N  (due South)

4.

DO YOURSELF

2

5.

q 4πε 0 a 2

1.

E=

2.

Let us assume that the small length the  is not cut from  loop. In that case, at the centre, Ecomplete loop = 0    =0 ⇒ Edue to removed + Edue to the



  ⇒ Edue to the remaining = Edue to removed



⎛q ⎞ ⇒ dτ = x ⎜ dx ⎟ E ⎝L ⎠



qE ⇒ τ = dτ = x dx L



L

∫

⇒ τ=

∫ 0

qEL 2

τ total =

portion of length x (  R )

qEL qEL + ( qE ) L + = ( 2qE ) L 2 2



L ⎛ L⎞ ⇒ τ = 2qEL = ( mg ) + ( mg ) L + mg ⎜ ⎟ ⎝ 2⎠ 2

x



⇒ τ = 2qEL = 2mgL

θ/2 θ/2



⇒ E=

6.

Force on the particle at the centre of the ring is zero.

x

R R

Now, the removed portion is actually an arc having ⎛ Q ⎞ charge q = ⎜ x and subtending an angle θ at the ⎝ 2π R ⎟⎠ centre. In that case, we know

01_Ch 1_Hints and Explanation_P1.indd 10

Consider the arm AB . Let us consider an element of length dx charge dq at a distance x from XY . Then torque on this element due to the field E is

remaining wire

wire

R

6q 4πε 0 2

dτ = x ( dqE )

Test Your Concepts-III (Based on Electric Field)

portion

Qx 8π 2ε 0 R3

⇒ E



9 × 109 × ( 100 × 10 −6 ) ⇒ FC = 9

x  1 , so R



B

FC

⇒ FC = F =

Q ⎛ x ⎞ sin ⎜ ⎝ 2R ⎟⎠ 4π 2ε 0 R2

(done already)

x ⎛ x ⎞ sin ⎜  ⎝ 2R ⎟⎠ 2R

A

a

+q

⇒ E=

Since

2

+q

Q ⎛θ⎞ sin ⎜ ⎟  ⎝ 2⎠ 4π 2ε 0 R2



⎛ 1⎞ ⇒ FC2 = 2 F 2 + 2 F 2 ⎜ − ⎟ ⎝ 2⎠ FC2

E =

mg q

Let −Q be now displaced through x (  R ) along axis of ring. Then a restoring force F will act on −Q , where F = −QEdue to ring at axis

9/20/2019 11:32:39 AM

Hints and Explanations H.11 A force of same magnitude but in opposite direction acts on a corresponding element in the region of negative charge.

Qλ ( 2π R ) x

⇒ F=−

4πε 0 ( R2 + x

)

2 32

Since x  R

{

Qλ ( 2π R ) x  4πε 0 R3

∵ ( R2 + x 2 )



⇒ F=−



Qλ ( 2π R ) x ⇒ mx = − 4πε 0 R3



⎛ Qλ ⎞ x=0 ⇒ x + ⎜ ⎝ 2ε 0 mR2 ⎟⎠

32

≅ R3

}

7.

λQ 2ε 0 mR2

⇒ τ = 2λ R E0



Equation for pure rolling motion is

Στ = Ια



(b) Since the electric field is zero at a distance a from the point 2, therefore Q Q    1 2 − 22 = 0 ( + a) a Q1 ⎛  + a ⎞ =⎜ ⎟ Q2 ⎝ a ⎠

2

(c) For all points x > a ,

9.

8.

1

=b −1

Consider an infinitesimal element subtending an angle dθ at the centre and at angle θ as shown in figure. E0

(

4πε 0 R2 + x02

)3 2

2

R + x02 

{towards the centre}

(b) The motion of the bead is periodic between ± x0



m (c) 

2Qqx0 d2x =− dt 2 4πε 0 R2 + x02

(

)

32

with

dx = 0 at x = ±x0 dt

d2x ⎛ 2Qq ⎞ m 2 = mx = − ⎜ x (d) dt ⎝ 4πε 0 R3 ⎟⎠ ⎛ 2Qq ⎞ x=0 ⇒   x + ⎜ ⎝ 4πε 0 mR3 ⎟⎠ ⇒   x + ω 2 x = 0 where ω =



mg

01_Ch 1_Hints and Explanation_P1.indd 11

2Qqx0

x0

)

2Qq 4πε 0 mR3 and T = 2π 3 2Qq 4πε 0 mR

   x = x0 cos ( ωt ) and v = − x0ω sin ( ωt )

θ

dF = λ Rdθ E0

Qq

(

4πε 0 R2 + x02

such that

dF dθ N



(a) Fnet = F = 2

and x = x0 at t = 0

dE 2Q1 2Q2 + =0=− dx (  + x )3 x 3 13

f = λ RE0 along positive x-axis.



Q E= − 22    2 ( + x ) a

⎛ Q1 ⎞ ⎜⎝ Q ⎟⎠ 2

Solving equations (1), (2) and (3), we get

⇒   F =

Q1

Thus, x =

⇒ 2λ R2E0 − fR = mR2α …(1)

and a = Rα …(3)

(a) The charge Q2 is negative and the charge Q1 is positive.

For max. E ,

2

where f = ma …(2)





∫

⇒ τ = 2λ R2E0 sin θ dθ





2ε 0 mR2 ⇒ T = 2π λQ

⇒  

π 2

0

2 x + ω x = 0



Net torque due to both types of charges is

dτ = ( λ Rdθ E0 ) ( 2R sin θ )



which can be compared to the standard equation of SHM i.e.,

where ω =



CHAPTER 1



(e) The velocity of bead will vanish for first time at T t= 2

f

⇒   t =

4π 3ε 0 mR3 2Qq

9/20/2019 11:33:01 AM

H.12  JEE Advanced Physics: Electrostatics and Current Electricity 10. Let us first calculate the effective dipole moment of the arrangement.



    Enet = E2 {As E1 = − E2 }

y

1 E

+ dθ

x

z

– θ

θ θ

θ

dθ

E3

E2 C R

2



dp = (Either charge) (Separation between the charges)



⎡⎛ q ⎞ ⎤ ⇒ dp = ⎢ ⎜ ⎟ ( R dθ ) ⎥ 2R sin θ ⎣⎝ π R ⎠ ⎦



⇒ dp =

2qR sin θ dθ π

+ + + +

+

z

⇒ p=

E

12. From figure we can write for equilibrium of ball T sin θ = qE

x

T cos θ = mg

+ +

π



2qR 2qR ⎡ sin θ dθ = − ⎢ cos θ π π ⎣

∫ 0



2qR ( cos π − cos 0 ) ⇒ p=− π



⇒ p=

π 0

Ι pE

mRπ mR 2 = 2π T = 2π 4 qE ⎛ 4q R ⎞ ⎜⎝ ⎟⎠ E π 11. Here we can clearly see in the figure  E1 (field due to wire 1) cancels out with  E3 (field due to wire 3) and  E2 (field due to wire 2)     Thus net field is Enet = E1 + E2 + E3

01_Ch 1_Hints and Explanation_P1.indd 12

θ

⎤ ⎥ ⎦

4 qR π

Since, T = 2π

3

⎛ πR ⎞ ⎛π⎞ 2K ⎜ λ sin ⎜ ⎟  ⎝ 2 ⎟⎠ ⎝ 4⎠ Enet = π 2 R 2  λ Enet = 2 2πε 0 R

y +

E1

T

l

θ

T qE d

⇒ tan θ =



⇒



⇒ E=

qE mg

d 2

qE

T sin θ mg

mg



T cos θ

 −d

2

=

qE mg

mgd q 2 − d2

13. Different forces acting on bob are shown in figure From the figure T sin α = qE sin β …(1) T cos α + qE cos β = mg T cos α = mg − qE cos β …(2)

9/20/2019 11:33:14 AM

Hints and Explanations H.13 From equations (1) and (2), we have

qE sin β tan α = mg − qE cos β

2qEd m = qE m

⇒ t=

Thus time period of oscillations of block is



qE sin β ⎞ ⎛ …(3) ⇒ α = tan −1 ⎜ ⎝ mg − qE cos β ⎟⎠





The net acceleration on the bob due to mg and qE is

T = 4t = 2

2

⎛ qE ⎞ ⎛ qE ⎞ + 2g ⎜ cos ( 180 − β ) a = g 2 + ⎜ ⎝ m ⎟⎠ ⎝ m ⎟⎠ ⎛ qE ⎞ ⎛ qE ⎞ − 2g ⎜ cos β ⇒ a = g2 + ⎜ ⎝ m ⎟⎠ ⎝ m ⎟⎠

15. Electric Field due to 1 is

qE cos β O qE

T cos α qE

E

β T α

T

O α



qE sin β

 a

T = 2π

 2

⎛ qE ⎞ ⎛ qE ⎞ g2 + ⎜ cos β − 2g ⎜ ⎝ m ⎟⎠ ⎝ m ⎟⎠

14. Here acceleration of block is a =

qE m

Time taken by block to reach wall

t =

T sin α

Time period of pendulum can be given as

d =

iˆ

λ ( cos 45° + cos 45° ) iˆ 2πε 0 1 ⎛ 2 2λ ⎞ ˆ ⎜ ⎟i 4πε 0 ⎝  ⎠



⇒ E1 =



Electric Field due to 2 is E2 =

1 ⎛ 4 2λ ⎞ ˆ ⎜− ⎟i  ⎠ 4πε 0 ⎝



Electric Field due to 3 is E3 =

1 ⎛ 6 2λ ⎞ ˆ ⎜ ⎟j 4πε 0 ⎝  ⎠



Electric Field due to 4 is E4 =



 ⇒ Etotal



 1 ⎛ 2 2λ 4 2λ ⎞ ˆ 1 ⎛ 6 2λ 8 2λ ⎞ ˆ ⇒ E= − + ⎜ ⎟i+ ⎜ ⎟j  ⎠  ⎠ 4πε 0 ⎝  4πε 0 ⎝ 



 λ 2 ⇒ E= −2iˆ + 14 ˆj 4πε 0

mg

T = 2π



E1 =

β α

mg β

2md qE

Since the restoring force is independent of x , (the displacement from mean position), hence this is not a simple harmonic motion but periodic in nature.

2



2m qE



1 ⎛ qE ⎞ 2 ⎜ ⎟t 2⎝ m ⎠

1 ⎛ 8 2λ ⎞ ˆ ⎜ ⎟j 4πε 0 ⎝  ⎠     = E = E1 + E2 + E3 + E4

(

)

16. The electric field strength due to the three rods AB , BC and CA are as shown in figure

2md qE

y A

Velocity at the time of impact is

v = 2 ad

CHAPTER 1



⇒ v=

λ

2qEd m

C

When the block rebounds, time taken by block to come to rest is given by 0 =

2qEd ⎛ qE ⎞ −⎜ t ⎝ m ⎟⎠ m

01_Ch 1_Hints and Explanation_P1.indd 13

λ

EAC E BC B

60° –λ

. x

30° EAB C

9/20/2019 11:33:29 AM

H.14  JEE Advanced Physics: Electrostatics and Current Electricity  EAC = iˆ

 EAB = iˆ

 EBC = iˆ

1 −2λ ( 2 sin 60° ) cos θ iˆ + sin θ ˆj 4πε 0 ⎛  ⎞ ⎜⎝ ⎟ 3⎠

)

1 2λ ( 2 sin 60° ) cos θ iˆ − sin θ ˆj 4πε 0 ⎛  ⎞ ⎜⎝ ⎟ 3⎠

)

(

(



1 2λ ( 2 sin 60° ) ˆj 4πε 0 ⎛  ⎞ ⎜⎝ ⎟ 3⎠

    Enet = EAC + EAB + EBC  ⎛ 3λ ⎞ ˆ j Enet = − ⎜ ⎝ πε 0 ⎟⎠

∑ F = −T sinθ + qA = 0 …(1) and ∑ F = +T cosθ + qB − mg = 0 …(2) x

mv 2 2qR

Similarly for the y-motion, we have

   v 2 = 0 + 2 ay R ⇒   ay = + ⇒   Ey =

v 2 qEy = 2R m

mv 2 2qR

The magnitude of the field E is given by

E = Ex2 + Ey2 ⇒ E =

17. Again, Newton’s Second Law

⇒   Ex = −

mv 2 2qR

at an angle of 135° in the counterclockwise direction from the x-axis. 19. From the free-body diagram shown,

y



(a) Substituting T =

15°

qA , in equation (2), we get sin θ

qA cos θ + qB = mg sin θ ⇒ q =

T =

mgA

( A cos θ + B sin θ )

18. (a) Each ion moves in a quarter circle. The electric force causes the centripetal acceleration.

∑ F = ma



∑F = 0 y



⇒ T cos ( 15° ) = 1.96 × 10 −2 N



⇒ T = 2.03 × 10 −2 N .



Again for the equilibrium, we have



∑F = 0 x



⇒ qE = T sin ( 15° )



⇒ q=



⇒ q=

vx2 = ux2 + 2 ax ( x f − xi )



⇒ q = 5.25 μC

⇒ 0 = v 2 + 2 a x R

20. One

qE = E =

mv 2 R 2

mv qR

(b) For the x-motion, we have

⇒ ax = −

01_Ch 1_Hints and Explanation_P1.indd 14

v 2 Fx qEx = = 2R m m

qE

Fg = 0.0196 N

mg ( A cot θ + B )

(b) Substituting this value in equation (1), we get

T q

T sin ( 15° ) E

( 2.03 × 10 −2 N ) sin ( 15° ) 1.00 × 10 3 NC −1

= 5.25 × 10 −6 C

of the charges creates at P a field 1 ⎛ Q⎞ 1 E= at an angle θ to the x-axis as ⎜⎝ ⎟⎠ 2 4πε 0 n R + x 2 shown.

9/20/2019 11:33:49 AM

Hints and Explanations H.15 π 2

R x

Ex

P

θ

1 4πε 0

⇒ E=

1 4πε 0



1 ⎞ ⎡ ⎛ Q⎞⎛ ⎤ cos θ ⎥ iˆ ⎢ n ⎜⎝ n ⎟⎠ ⎜⎝ 2 2⎟ ⎠ R +x ⎣ ⎦ Qx

( R2 +

3 x2 2

)

∫

E = dE =

E =

1 4πε 0

1 4πε 0

∞

∫

x0

∫

0

 If rod were infinite in both direction. The answer λ would become Ex = 0 , Ey = 2πε 0b

1.

θ

λ 0 x0 dx λ 0 x0 = 4πε 0 x3

∞

∫x

−3

p

dx

x0

2.

∞

⎞ λ0 ⎟= 8 πε x0 ⎠ 0 x0

qi

22. From figure, electric field due to small length element dx at observation point at distance r .

λ dx 4πε 0 r 2 dE

xi

qi xi

yi

qi yi

zi

qi zi

2q

0

0

a

2qa

a

2qa

q

0

0

–a

–qa

a

qa

–q

0

0

0

0

–a

qa

z

dEy

dEx

p

r x

y λ

dx

and dEy = dE cos θ and dEx = dE sin θ Put x = b tan θ

1

px = py =

dx = b sec θ dθ b = b sec θ cos θ

⎡ λ b sec 2 θ dθ ⎤ dEy = ⎢ cos θ 2 2 ⎥ ⎣ 4πε 0b sec θ ⎦

∑q x = 0 i

i

i

∑ q y = 2qa − qa i

i

i

2

01_Ch 1_Hints and Explanation_P1.indd 15

4

θ

θ

b

Also r =

θ 2 p

directed towards the negative x axis.

dE =

pnet = 2 p cos

λ dx x2

λ x ⎛ 1 E = 0 0 ⎜ − 2 4πε 0 ⎝ 2x

∫

Test Your Concepts-IV (Based on Dipoles and Dipole Moment)

iˆ

This result is identical to the electric field due to a uniformly charged ring at a point lying at its axis. 21.

λ λ sin θ dθ = to left. 4πε 0b 4πε 0 b

Similarly Ex =

CHAPTER 1

The total field is iˆ

∫

π 2

When all the charges produce field, for n > 1 , the components perpendicular to the x-axis add to zero.

E =

λ λ cos θ dθ = directed upward 4πε 0b 4πε 0b 0

E





Then Ey =



⇒

py = qa

pz =

∑ q z = 2qa + qa + qa i

i

i



⇒

pz = 4 qa

9/20/2019 11:34:05 AM

H.16  JEE Advanced Physics: Electrostatics and Current Electricity  Since, p = px i + py j + pz k

⇒

 p = qa j + 4 k 



⇒

 p = 17 qa

3.

(



)

{in yoz plane}

⎛ 1⎞ at an angle tan −1 ⎜ ⎟ with the z axis. ⎝ 4⎠

Thus torque experienced by dipole has magnitude

⎛ σ ⎞ sin θ τ = p ⎜ ⎝ 4ε 0 ⎟⎠

θ

θ

x Z

τ

For small angular displacement,

⎛ σ ⎞ θ = Ια τ = − ⎜ ⎝ 4ε 0 ⎟⎠

⎛ ⇒ α = −⎜ ⎝

T = 2π

md 2 4 + md 2 4 ( λ Ld ) E

T = 2π

md 2λ LE

(b) Work done required to turn the dipole from initial angle θ i to θ f is

  (i)  For θ i = 0 , θ f =

p

E X



Ι pE

W = pE ( cos θ i − cos θ f   

Y

τ

E p θ

T = 2π



Y

The time period of oscillation of dipole is

   W = λ LdE (ii)  For θ i = 0 , θ f = π ,    W = λ LdE ( cos 0 − cos π )    W = 2λ LdE

σ ⎞ θ 4πε 0Ι ⎟⎠

5.

(a) We consider two differential elements on the rod as shown in figure +λ Rdθ

α = −ω 2θ we obtain ω =

T = 2π

R sin θ

2π σ = 4πε 0Ι T

R sin θ

4πε 0Ι σ

dθ θ dθ θ

–λ Rdθ



Potential energy of dipole is U = − pE cos θ



⎛ σ ⎞ ⇒ U = −p ⎜ cos θ ⎝ 4πε 0 ⎟⎠

4.

(a) The figure shows a side view of the system, it can be represented as an electric dipole of dipole moment  p = ( λ L ) d E +λ d

p –λ

Slide view

01_Ch 1_Hints and Explanation_P1.indd 16

π , 2

   W = ( λ Ld ) E [ 1 − 0 ]

 On comparing the above expression with standard expression for SHM



)

 These elements constitute an electric dipole whose dipole moment is  dp = ( λ Rdθ )( 2R sin θ )   

Net dipole moment of the system is

    p =

2θ 0

∫ ( λRdθ ) 2R sinθ 0

 p = 2λ R2 ( 1 − cos 2θ 0 )        p = 4 λ R2 sin 2 θ 0 Time period of oscillation of a dipole in a uniform magnetic field is T = 2π   

Ι pE

9/20/2019 11:34:18 AM

Hints and Explanations H.17



(b) Work done required to rotate the dipole through 180° is W = pE(cos θ i − cos θ f ) = pE(cos 0 − cos 180°)

(

)

W = 2 pE = 2 4 λ R2 sin 2 θ 0 E = 8 λ R2E sin 2 θ 0 6.

(a) In this case the resultant force on the dipole is zero as shown in figure (a) +q

pλ 2πε 0 r 2   λp ⇒   F = −  2πε 0 r 2    F=−





–q

λ 2πε 0 r 2 + a 2

and force on the charges is

  F1 = F2 = F =   

–q

qE

{force is attractive}

(c) The magnitude of electric field at the position of two charges is

qE

p

(a)

qλ 2πε 0 r 2 + a 2 F sin θ

(b) The electric field at the position of negative charge ( −q ) ,

r2

λ 2πε 0 ( r − a )

E1 =   

d⎛ λ ⎞ dr ⎜⎝ 2πε 0 r ⎟⎠

  E1 = E2 =    +q



   F=p

Thread



The electric field at the position of positive charge λ E2 =    2πε 0 ( r + a )

CHAPTER 1

mR2 m =π 2 2 λ sin 2 θ 0E 4 λ R sin θ 0E

   T = 2π

r2

θ θ

+

+

a2

a

E1 θ E1 = F F cos θ

r E2 = F

F sin θ a

a2

θ F cos θ

E2 (c)

r qE1

qE2

p –q a

a

+q

   FR = 2 F sin θ    FR =

(b)



Resultant force on the dipole is

   F = qE1 − qE2 F=   

qλ ⎡ 1 1 ⎤ + 2πε 0 ⎢⎣ r − a r + a ⎥⎦

qλ 2a F=    2πε 0 ( r 2 − a 2 ) for r  a , F 

( 2aq ) λ = 2πε 0 r

2

pλ 2πε 0 r 2

 Alternatively, the electric force on an electric dipole in a non-uniform field is given by    F=p

01_Ch 1_Hints and Explanation_P1.indd 17

dE dr

 As shown in figure (c), resultant force on the dipole is

2qλ 2

2πε 0 r + a

a 2

2

r + a2

For r  a and 2aq = p ,    FR 

pλ 2πε 0 r 2



Resultant force is parallel to dipole moment   λp FR =    2πε 0 r 2 7.

Consider two differential charge elements at A and B as shown in figure. Dipole moment of this pair is

dp = ⎡⎣ ( λ 0 cos θ ) Rdθ ⎤⎦ 2R cos θ

⇒ dp = 2λ 0 R2 cos 2 θ dθ

9/20/2019 11:34:32 AM

H.18  JEE Advanced Physics: Electrostatics and Current Electricity dA

A

B

dθ

dθ

θ

θ

E

θ

q

θ x dx

C R

L



+

p = 2λ 0 R2

The electric field due to the point charge on the strip is q E = 4πε 0 ( R2 + x 2 )

π 2

∫ cos θdθ = π R λ 2

−

8.



So, the Dipole moment of the charge distribution is 2

If dϕ is the electric flux through the strip, then

0

π 2

dϕ = E dA cos θ

(a) Applying Energy Conservation Principle, Increase in ⎛ ⎜ kinetic energy of ⎜⎝ the dipole

decrease in ⎞ ⎛ ⎟ = ⎜ electrostatic potential ⎟⎠ ⎜⎝ energy of the dipole

⎞ ⎟ ⎟⎠

Kinetic energy of dipole at distance d from origin = Ui − U f ⎛ 1 q ˆ⎞   ⇒   KE = 0 − − p.E = p.E = piˆ . ⎜ i ⎝ 4πε o d 2 ⎟⎠

(

iˆ

⇒   KE =

)

( )

qp 4πε o d 2



(b) Electric field at origin due to the dipole,  1 2p ˆ   i Eaxis ↑↑ p E = 3 4πε o d

(

dϕ =



2π qR2 ⎡ dx ⎤ 4πε 0 ⎢⎣ ( x 2 + R2 )3 2 ⎥⎦

Total flux through the lateral surface of cylinder is given by integrating the above result for the complete lateral surface. So

∫

ϕ = dϕ =

)

∴  Force on charge q   pq ˆ i F = qE = 2πε o d 3

⇒ dϕ =

q R 1 ( 2π Rdx ) ⎛⎜ 2 2 ⎞⎟ 4πε 0 ( x 2 + R2 ) ⎝ x +R ⎠

2

qR 2ε 0

+

L 2

∫ (x

−

L 2

q ⎛ L ⎞ ε 0 ⎜⎝ L2 + 4 R2 ⎟⎠

3.

Consider an infinitesimal area element on the surface of the circle. E(2π xdx)

ϕtotal = ( ϕdue to + q ) + ( ϕdue to − q )

dA

Since, we have seen already that q q ⎛  ⎞ ( 1 − cos α ) = 1− ϕ = 2 2 ⎟ 2ε 0 2ε 0 ⎜⎝ R + ⎠ q ⎛  ⎞ 1− ⎟ ε 0 ⎜⎝ R2 +  2 ⎠ (Do not expect the answer to come to be zero as one charge is +q and other −q . If both had been +q or −q then it would have been zero) ⇒ ϕtotal = 2ϕ =

2.

For this we consider an infinitesimal strip of width dx on the surface of cylinder as shown. The area of this strip is dA = 2π Rdx

01_Ch 1_Hints and Explanation_P1.indd 18

32

⇒ ϕ=





+ R2 )



Test Your Concepts-V (Based on Flux) 1.

dx 2

r

dr

y

Ex

x

Ey



Let this element have radius r and thickness dr . Then  dA = 2π rdrjˆ

Since,  E = Ex iˆ + Ey − ˆj

( )

where Ex = Ey =   Since, dϕ = E ⋅ dA

λ 4πε 0 r

9/20/2019 11:34:45 AM

Hints and Explanations H.19

⎛ λ ⎞ ⇒ dϕ = −2π rdr ⎜ ⎝ 4πε 0 r ⎟⎠

Now A1 = A2

R



⇒ ϕ=

λ dr 2ε 0

⎛ π R2 ⎞ A1 = ⎜ sin θ ⎝ 2 ⎟⎠

λR 2ε 0



∫ 0



⇒ ϕ=

Perpendicular component of area A⊥ = A1 + A2 + A3

and A3 = 2 HR cos θ

Total ⊥ component of area is

q =ϕ= − flux through the mouth ε0

4.

ϕlateral surface



⇒ ϕ=

q q ⎛ h ⎞ 1− − 2 2 ⎟ ε 0 2ε 0 ⎜⎝ h +R ⎠



⇒ ϕ=

q ⎛ 1+ 2ε 0 ⎜⎝



⎛ 3⎞ q ⇒ ϕ = ⎜ 1+ ⎟ ⎝ 2 ⎠ 2ε 0

⎛ π R2 ⎞ sin θ × 2 + 2 HR cos θ A⊥ = ⎜ ⎝ 2 ⎟⎠

CHAPTER 1



A⊥ = 2RH cos θ + π R2 sin θ

⎞ 2 2 ⎟ h +R ⎠ h

Thus flux passing through cylinder is

ϕ = EA⊥ ϕ = E ( 2RH cos θ + π R2 sin θ ) 7.

R h

In general, we have

E = ayiˆ + bzjˆ + cxkˆ ϕ



In the xy plane, z = 0 and so, E = ayiˆ + cxkˆ z

y=0

5.

The cross sectional area perpendicular to the electric field is

A = ( 2R ) H Thus electric flux crossing the cylinder is ϕ = EA

x=l

dA = hdx



By definition of flux, we have   ˆ Φ E = E ⋅ dA = ayiˆ + cxkˆ ⋅ kdA

⇒ ϕ = E ( 2RH ) = 2 HRE

∫(

∫





H

⇒ Φ E = ch

∫

xdx = ch

x=0

8. Looking at the cylinder from − x-axis it appears as shown in figure. A2

A3 θ

)

x2 2

 x=0

=

ch 2 2

(a) Φ E = EA cos θ

Φ E = ( 3.5 × 10 3 ) ( 0.35 × 0.7 ) cos 0° Φ E = 858 Nm 2 C −1

θ = 90° (b) ⇒   Φ E = 0 Φ E = ( 3.5 × 10 3 ) ( 0.35 × 0.7 ) cos 60° (c)

A1

01_Ch 1_Hints and Explanation_P1.indd 19

y

x

2R

6.

y=h

x=0

Φ E = 428.75 Nm 2 C −1

9/20/2019 11:35:03 AM

H.20  JEE Advanced Physics: Electrostatics and Current Electricity Φ E = EA cos θ = (2 × 10 4 NC −1 )(18 m 2 )cos 60° = 180 kN m 2 C −1  1 0. Φ E = EA cos θ 9.

2



Since A = π r 2 = π ( 0.4 ) = 0.504 m 2



⇒ 5.04 × 10 5 = E ( 0.504 ) cos 0°



⇒ E = 106 NC −1 = 1 MNC −1

13. The flux entering the closed surface equals the flux exiting the surface. The flux entering the left side of the cone is given by   Φ E = E ⋅ dA = ERh

∫

This is the same flux that exits the right side of the cone. Here we must observe that for a uniform field only the cross sectional area matters, note the shape. 14. (a) With δ being very small, all points on the hemisphere are nearly at a distance R from the charge, so the field everywhere on the curved surface is

11. (a) A1 = ( 10 cm )( 30 cm ) A1 = 300 cm 2 = 0.03 m 2

1 Q radially outward (normal to the surface). 4πε 0 R2 Therefore, the flux is this field strength times the area of half a sphere or hemisphere

ϕ1 = EA1 cos θ ϕ1 = ( 7.8 × 10 4 ) ( 0.03 ) cos 180° ϕ1 = −2.34 kNm 2 C −1 30 cm

10 cm

∫

⇒   Φ curved = E ⋅ dA = Elocal Ahemisphere l E

60°

⎛ 1 Q ⎞⎛ 1 2⎞ ⇒   Φ curved = ⎜ ⎜ 4π R ⎟⎠ ⎝ 4πε 0 R2 ⎟⎠ ⎝ 2 ⇒   Φ curved =

A1

1 Q Q ( 2π ) = + 4πε 0 2ε 0

A

(b) ϕ2 = EA cos θ = ( 7.8 × 10 4 ) ( A ) cos 60°

δ →0

Q R

⎛ 10 cm ⎞ Now, A = ( 30 cm ) (  ) = ( 30 cm ) ⎜ ⎟ ⎝ cos 60° ⎠

⇒   A cos 60 = 30 × 10 ⇒   ϕ2 = 7.8 × 10 4 × 30 × 10 = +2.34 kN m 2 C −1





Φ curved + Φ flat = 0

(c) The bottom and the two triangular sides all lie parallel to E , so Φ E = 0 for each of these. Thus,

Φ E, total = −2.34 kN m 2 C −1 + 2.34 kN m 2 C −1 +0 + 0 + 0 = 0  12. Φ E = EA cos θ through the base

Φ E = ( 52 )( 36 ) cos 180° = −1.87 kNm 2 C −1

 Note the same number of electric field lines pass through the base as are passing through the surface of the pyramid excluding the base.

6m



6m

For the inclined surfaces, Φ E = +1.87 kNm 2 C −1

01_Ch 1_Hints and Explanation_P1.indd 20

(b)  The closed surface encloses zero charge, so according to Gauss’s Law we have

⇒ Φ flat = − Φ curved = −

Q 2ε 0

Test Your Concepts-VI (Based on Gauss’s Law) 1.

Net charge enclosed by the sphere is qenc = 0 . Therefore, according to Gauss’s Law net flux passing through the sphere is zero.

2.

(a) According to Gauss’s Law, q q ϕtotal = in = ε0 ε0

(b) The cube is a symmetrical body with 6 faces and the point charge is at its centre, so electric flux linked with each face will be,

9/20/2019 11:35:17 AM

Hints and Explanations H.21 q ϕtotal = 6 6ε 0



⎛α⎞ ⇒ dϕ = ⎜ ⎟ dA ⎝ R⎠

We can also do this by another method discussed below :

y n

Here the total solid angle subtended by cube surface at the point charge q is 4π . As q is at centre of cube, we can say the each face of cube subtend equal solid angle at the centre, thus solid angle subtended by each face at point charge is Ωface =   

P (x, y, z) dA x

O

4π steradian 6

z

Thus electric flux through each face is

 Here, we note that dϕ is independent of the coordinates x , y and z . Therefore, total flux passing through the sphere q

∫

ϕ = dϕ =

⎛ q ⎞    ϕ=⎜ Ω ⎝ 4πε 0 ⎟⎠

α ⎛α⎞ dA = ⎜ ⎟ ( 4π R2 ) ⎝ R⎠ R

∫



⇒ ϕ = 4πα R



From Gauss’s Law, ϕ =

qenc ε0

⇒   ϕface =

q ⎛ 4π ⎞ ⎜ ⎟ 4πε 0 ⎝ 6 ⎠



⇒ ( 4πα R ) =

⇒   ϕface =

q 6ε 0



⇒ qenc = 4πε 0α R

4.

At time t , length of rod inside the cube is  = vt



So, charge enclosed qenc ( t ) = λ  = λ vt

If charge is not at the centre, the answer of part (a) will remain same while that of part (b) will change. 3.

At any point P ( x , y , z ) on the sphere a unit vector perpendicular to the sphere directed radially outwards is,

nˆ =

x 2

2

x +y +z

2

iˆ +

y 2

2

x +y +z

2

2

x + y 2 + z2

λa ε0

kˆ

λv –ε 0

λv ε0

a v

0

2a v

t

}



⇒ Flux = ϕ1 ( t ) =

Let us find the electric flux passing through a small area dA at point P on the sphere, then



So, flux increases as a function of time When the rod enters the cube completely, flux is

 ⎧⎪ α y2 α x2 dϕ = E ⋅ nˆ dA = ⎨ + 2 2 R x2 + y 2 ⎪⎩ R x + y

ϕmax =



⇒ nˆ =

{∵ x

qenc ε0

ϕ

ˆj +  z

 xˆ yˆ z ˆ i+ j+ k R R R

(

01_Ch 1_Hints and Explanation_P1.indd 21

)

(

2

+ y 2 + z 2 = R2

)

⎫⎪ ⎬ dA ⎪⎭

CHAPTER 1

ϕeach face =

1 λ vt {By Gauss’s Law} ⎡ qenc ( t ) ⎤⎦ = ε0 ε0 ⎣

1 ( λa ) ε0

9/20/2019 11:35:31 AM

H.22  JEE Advanced Physics: Electrostatics and Current Electricity and again when the rod starts leaving the cube it starts decreasing as λ ϕ2 ( t ) = ( a − vt ) ε0 and ultimately falls to zero as soon as rod is out of the cube 5.

Draw a Gaussian cylinder of length  and radius s. For this surface, Gauss’s Law states  1 E ⋅ d A = Qenc ε0

∫

8.

The charge density is determined by Q =

ρ =

(a) The flux is that created by the enclosed charge within radius r .

ϕE =

Q . Note that the answers to parts (a) and (b) ε0 agree at r = a .



∫

 Now, symmetry dictates that E must point radially outward, so for the curved portion of the Gaussian cylinder we have      E ⋅ d A = E dA = E dA = E 2π x  while the two ends contribute nothing (here E is per pendicular to d A ). Thus,

∫

∫

6.

(c) ϕE Q ε0

∫

 1 2 π k x 3 E 2π s = ε0 3  1 kx 2 ⇒ E =E= 3ε 0 (a) One half of the total flux created by the charge q goes through the plane. Thus,

q 1 1⎛ q ⎞ ϕE , plane = ϕE , total = ⎜ ⎟ = 2 ⎝ ε 0 ⎠ 2ε 0 2 (b) The square looks like an infinite plane to a charge very close to the surface. Hence, q ϕE , square ≈ ϕE , plane = 2ε 0

0

9.

(c) The plane and the square behave in the same manner for the charge.

7.

The total charge enclosed is Qenc = Q + ( −6 q ) = Q − 6 q. Q − 6q , ε0 of which one-sixth goes through each face so , flux through each face is given by

The total outward flux from the cube is ϕE =

( ϕE )one face =

Q − 6q 6ε 0

For, the cube not to be associated with any flux, we Q must have q = 6

01_Ch 1_Hints and Explanation_P1.indd 22

r

a

CASE-1

 When R ≤ d , the sphere encloses no charge and q ϕE = enc = 0 ε0 CASE-2  When R > d , the length of line falling within the sphere is 2 R2 − d 2 . So, according to Gauss’s Law ϕE =





qenc 4π r 3 ρ ⎛ 4π r 3 ⎞ ⎛ 3Q ⎞ Qr 3 = =⎜ ⎜ ⎟= ε0 3ε 0 ⎝ 3ε 0 ⎟⎠ ⎝ 4π a 3 ⎠ ε 0 a 3

ϕE = (b) 

S

The enclosed charge is 2 Qenc = ρ dV = π kx 3 3

3Q 4π a 3

4 3 πa ρ 3





qenc 2λ R2 − d 2 = ε0 ε0

) = qenc

10.

∫ E ⋅ dA = E ( 4π r



(a) For r > R , we have

2

ε0

R

∫

qenc = ρ0 r 2 ( 4π r 2 ) dr = 0

⇒ Eoutside =

4πρ0 R5 5

ρ0 R 5 5ε 0 r 2

(b) For r < R , we have r

∫

qenc = ρ0 r 2 ( 4π r 2 ) dr = 0

⇒ Einside =

4πρ0 r 5 5

ρ0 r 3 5ε 0

9/20/2019 11:35:44 AM

Hints and Explanations H.23 11. Consider the Gaussian surface, a Gaussian pill box as shown y

13. (a) A point mass m creates a gravitational acceleraGm  tion g = − 2 rˆ at a distance r . r

x

A x

z

(a) For x >

d 2

dq = ρdV = ρ Adx = Aρ0 x 2 dx Since





1

∫ E ⋅ dA = ε ∫ dq 0

d 2

ρ A 2 1 ⎛ ρ A ⎞ ⎛ d3 ⎞ ⇒   EA = 0 x dx = ⎜ 0 ⎟ ⎜ ⎟ 3 ⎝ ε0 ⎠ ⎝ 8 ⎠ ε0

∫



The flux of this field through a sphere is Gm      g ⋅ dA = − 2 ( 4π r 2 ) = −4π Gm r

∫

Since the flux is independent of r , so we can visualize the field as unbroken field lines. The same flux would go through any other closed surface around the mass. If there are several or no masses inside a closed surface, each creates field to make its own contribution to the net flux according to   menc    g ⋅ dA = −4π G

∑

∫

CHAPTER 1

Gaussian surface



(The direction is radially outward from the center for positive ρ0 and radially inward for negative ρ0 )



(b) Take a spherical Gaussian surface of radius r . The field is inward so   ϕ = g ⋅ dA = g 4π r 2 cos 180° = − g 4π r 2   

(

∫

)

(

)

⇒   ϕ = − g ( 4π r 2 )

0

⇒   E =

3

ρ0 d 24ε 0

menc

⎡ ρ0 d ˆ d for x > i ⎢ 24 2 ε 0 ⇒   E = ⎢ ⎢ ρ d3 d ⎢ − 0 iˆ for x < − 2 ⎢⎣ 24ε 0 3



d d (b) For − < x < 2 2

  1 E ⋅ dA = ε0

∫

∫

x

dq =

ρ0 A 2 ρ Ax 3 x dx = 0 ε0 3ε 0

∫ 0

⎡ ρ0 x ˆ for x > 0 i ⎢ 3ε 0 ⇒   E = ⎢ ⎢ ρ x3 ⎢ − 0 iˆ for x < 0 ⎢⎣ 3ε 0

∫

  q 1 E ⋅ dA = enc = ε0 ε0

r

∫ 0

4πρ0 E ( 4π r 2 ) =

4 3 πr ρ 3 So, according to Gauss’s Law, we have



     g ⋅ dA = −4π G

∫

∑m

enc

⎛4 ⎞ ⇒   − g ( 4π r 2 ) = −4π G ⎜ π r 3 ρ ⎟ ⎝3 ⎠ ⇒   g =

4 π Grρ 3 ME ⎛ GME ⎞ ⇒ g=⎜ r , inwards 4 3 ⎝ RE3 ⎟⎠ π RE 3

But ρ =

r

14. (a) Consider a cylindrical shaped Gaussian surface, also called Gaussian Pill box, perpendicular to the yz plane with one end in the yz plane and the other end containing the point x.

∫



ρ ⇒ E = 0 = constant in magnitude 2 ε0

0

rdr =

4πρ0 r 2 2ε 0

⇒

01_Ch 1_Hints and Explanation_P1.indd 23

Now, we have

ρ0 4π r 2 dr r



ε0



r

menc =   

3

12.

RE

E

9/20/2019 11:35:58 AM

H.24  JEE Advanced Physics: Electrostatics and Current Electricity y v(x) Q 4πε0a

x

x

z

V(y) = (b)



According to Gauss’s Law, we have   q    E ⋅ dA = enc ε0

∫

By symmetry, the electric field is zero in the yz plane and is perpendicular to dA over the wall of the gaussian cylinder. Therefore, the only contribution to the integral is over the end cap containing the point x:   q    E ⋅ dA = enc ε0

∫

⇒   EA =

1.5

1 0.5 0 –4 –3 –2 –1 0 1 2 3 4 x/a

Gaussian surface A

2.5 2

1 Q1 1 Q2 + 4πε 0 r1 4πε 0 r2 −Q ⎞ 1 ⎛ +Q + y + a ⎟⎠ 4πε 0 ⎜⎝ y − a

⇒   V ( y ) = ⇒   V ( y ) =

⇒  

ρ ( Ax ) ε0

1 Q ⎛ 1 − ⎜ y y 4πε 0 a −1 +1 ⎜⎝ a a

V(y) ⎛ Q ⎞ ⎜⎝ 4πε a ⎟⎠ 0

=

1 ⎛ 1 − ⎜ y y −1 +1 ⎜⎝ a a

–5 –4 –3 –2 –1

F ( −e ) E ⎛ ρe ⎞ a= = = −⎜ x (b) m m ⎝ mε 0 ⎟⎠

1 2 3 4 5 –2 –4 –6 –8 –10

The acceleration of the electron is of the form a = −ω 2 x with ω =

ρe 2π = mε 0 T

Thus, the motion is simple harmonic with period T = 2π

mε 0 ρe

Test Your Concepts-VII (Based on Electrostatic Potential and Energy) (a) V ( x ) =

⇒  

2Q 4πε 0

Q ⎛ 2 ⎞ = ⎜ ⎟ 2 2 2 a πε 4 0 x +a ⎜ ⎛⎜ x ⎞⎟ + 1 ⎟ ⎜⎝ ⎝ a ⎠ ⎟⎠

V(x) = ⎛ Q ⎞ ⎜⎝ 4πε a ⎟⎠ 0

01_Ch 1_Hints and Explanation_P1.indd 24

y/a

2.

In this problem,

r41 = r43 = r32 = r21 = 1 m 2 2 and r42 = r31 = ( 1 ) + ( 1 ) = 2 m

15. DO YOURSELF (Already Discussed)  ρa E = 2ε 0

1.

⎞ ⎟ ⎟⎠

10 8 6 4 2

v(y) Q 4πε0a

so that at distance x from the mid-line of the slab, ρx E= ε0

⎞ ⎟ ⎟⎠

2 2

⎛ x⎞ ⎜⎝ ⎟⎠ + 1 a



⇒ U=

1 4πε 0

⎡ q1q2 q2 q3 q3 q4 q4 q1 ⎢ r + r + r + r + 23 34 14 ⎣ 12

q2 q4 q1q3 ⎤ + r24 r13 ⎥⎦  Substituting the proper values with sign in above equation, we get ⎡ ( 1 )( 2 ) ( 2 ) ( −3 ) + + U = ( 9 × 109 )( 10 −6 )( 10 −6 ) ⎢ 1 ⎣ 1 

( −3 ) ( 4 ) 1

+

( 1 )( 4 ) 1

+

( 2 )( 4 ) 2

+

( 1 ) ( −3 ) ⎤ 2

⎥ ⎦

5 ⎤ ⎡ U = ( 9 × 10 −3 ) ⎢ −12 + 2 ⎥⎦ ⎣ U = −7.62 × 10 −2 J

9/20/2019 11:36:09 AM

Hints and Explanations H.25

ΔV = −14 V and



Q = − N A e = − ( 6.02 × 10 23 ) ( 1.6 × 10 −19 ) = −9.63 × 10 4 C



W Q

⇒ W = QΔV = ( −9.63 × 10 4 C ) ( −14 JC −1 )



⇒ W = 1.35 × 10 J = 1.35 MJ

4.

The net electric potential at origin is,



1 4πε 0

⇒   m



3⎞ −6 ⎟ × 10 3⎠

5.

(a) Let us consider V = 0 at 0. Then at other points

   V = − Ex and U e = QV = −Q ( Ex ) Between the extreme points of the motion, we have    ( K + U s + U e )i = ( K + U s + U e ) f

The period of the motion is then 2π m = 2π k ω

1 2 −QExmax ⇒   0 + 0 + 0 − μ k mgxmax = 0 + kxmax 2    xmax =

E

2 ( QE − μ k mg ) k

The electric potential at point P , at distance r from circumference is

V = x=0

Q 2πε 0 r

2 2 where r = ( 3 ) + ( 4 ) = 5 m and

1 2 ⇒   0 + 0 + 0 = 0 + kxmax − QExmax 2



k m

(d) ( K + U s + U e )i + ΔEmech = ( K + U s + Ue ) f

6.

⇒   xmax

d2X k + X=0 dt 2 m

T=   

⇒ V = 9 × 10 3 V

Q

d2X = − kX dt 2

with ω =



k

d2X d2X = 2 dt 2 dt

This is the equation for simple harmonic motion

Substituting the values, we have

⎛1 2 V = ( 9 × 10 ) ⎜ − + ⎝1 2

QE , k

 + ω 2 X = 0 ⇒   X

⎡ q1 q2 q3 ⎤ ⎢r +r +r ⎥ 2 3 ⎦ ⎣ 1

9

⇒  

⇒  

6

V =

d2x QE ⎞ ⎛ = −k ⎜ x − ⎟ ⎝ k ⎠ dt 2

Let X = x −

3.

Since ΔV =

⇒   m

Q = 10 μC = 10 −5 C

2QE = k



(b) At equilibrium,

⇒ V=

( 9.0 × 109 )( 10 −5 ) ( 5.0 )

∑

   Fx = − Fs + Fe = 0



So the equilibrium position is at x =

(c) The block’s equation of motion is



∑

01_Ch 1_Hints and Explanation_P1.indd 25

d2x Fx = − kx + QE = m 2 dt

= 1.8 × 10 4 V

z P

⇒   kx = QE

CHAPTER 1

The negative sign of U implies that positive work has to be done by electrostatic forces in assembling these charges at respective positions from infinity.

QE k

r

4m y

3m

x Q

9/20/2019 11:36:26 AM

H.26  JEE Advanced Physics: Electrostatics and Current Electricity

7.

For the entire motion, y f − yi = uy t +



⇒ 0 − 0 = uy t +



⇒ ay = −

Since,

1 2 ay t 2

1 2 ay t 2



2uy

∑ F = ma y

of y

⇒ − mg − qE = −



 ⎞ ⎞ m ⎛ 2uy m ⎛ 2uy − g ⎟ OR E = − ⎜ − g ⎟ ˆj ⇒ E= ⎜ ⎝ ⎠ ⎠ ⎝ q t q t



For the upward flight, we have

t

vy2 = uy2 + 2 ay ( y f − yi )

⎛ 2u y ⎞ ⇒ 0 = uy2 + 2 ⎜ − ⎟ ( ymax − 0 ) ⎝ t ⎠



⇒ ymax =

1 uy t 4 ymax

Edy = +

0

⎞ m ⎛ 2uy − g⎟ y ⎜⎝ ⎠ q t

⇒ ΔV =



⇒ ΔV =

8.

Initially, the potential energy of Q is



ymax 0

⎞⎛ 1 m ⎛ 2uy ⎞ − g ⎟ ⎜ uy t ⎟ ⎜ ⎠ ⎠⎝ 4 q⎝ t



2 ⎛ 2 ( 20.1 ) ⎞⎡1 ⎤ − 9.8 ⎟ ⎢ ( 20.1 ) ( 4.1 ) ⎥ ≅ 40 kV −6 ⎜ ⎝ ⎠⎣4 4.1 ⎦ 5 × 10

U1 =

x2 ⎞ 1 2Qq ⎛ ⎜⎝ 1 − 2 ⎟⎠ 4πε 0 a a

1 ⎡ Qq Qq ⎤ + a ⎥⎦ 4πε 0 ⎢⎣ a

x2 ⎞ 1 2Qq ⎛ ⎜⎝ 1 + 2 ⎟⎠ 4πε 0 a a −1 ⎧⎪ ⎛ ⎫⎪ x2 ⎞ x2 ⎨∵ ⎜ 1 − 2 ⎟ ≅ 1 + 2 for x 2  a 2 ⎬ a ⎠ a ⎩⎪ ⎝ ⎭⎪



So, the change in electrostatic potential energy of the system, 1 2Qq x 2 dU = U 2 − U1 = 4πε 0 a a 2 1 2Qqx 2 4πε 0 a 3



⇒ dU =

9.

Arbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length,

V = −Ed and U e = − λ LEd ( K + U )i = ( K + U ) f

⇒ 0+0=



⇒ v=



⇒ v=

When Q is displaced by a small amount x to the right, the potential energy of Q becomes

2λ Ed μ 2( 40 × 10 −6 Cm −1 )(100 NC −1 )(2 m)

(

1 x=0

q –a

q Q

1 μLv 2 − λ LEd 2

0.1 kgm −1

)

= 0.4 ms −1

10. (a) The work that must be done on q3 by an external force is equal to the difference of potential energy U when the charge is at x = 2 m and the potential energy when it is at infinity.

1 2Qq ⇒ U1 = 4πε 0 a

x

a

–1 μ C +1 μ C 2

+1 μ C 3

x=1m

x=2m

x

∴  Wext = U f − U i = U f − U ∞ V3 = V13 + V23 =   

01_Ch 1_Hints and Explanation_P1.indd 26

−1

x2 as x  a a2

U 2 =

2muy



∫

⇒ U2 =

Qq ⎤ 2Qqa 1 ⎡ Qq 1 + = 4πε 0 ⎢⎣ ( a + x ) ( a − x ) ⎥⎦ 4πε 0 ( a 2 − x 2 )

Expanding binomially and neglecting higher powers

t

Since ΔV = −

U 2 =

q1 q2 + 4πε 0 r13 4πε 0 r23

9/20/2019 11:36:44 AM

Hints and Explanations H.27

1 ⎡ q3 q2 q3 q1 ⎤ + r13 ⎥⎦ 4πε 0 ⎢⎣ r23 Substituting the values, we have

⇒   W =

⎡ ( 1 )( 1 ) ( 1 ) ( −1 ) ⎤ + ⇒   W = ( 9 × 109 )( 10 −12 ) ⎢ ( 2 ) ⎥⎦ ⎣ (1) ⇒   W = 4.5 × 10

−3

J



(b) The total potential energy of the three charges is given by, 1 ⎛ q3 q2 q3 q1 q2 q1 ⎞ U= + +    4πε 0 ⎜⎝ r23 r13 r12 ⎟⎠ ⇒   U = ( 9 × 109 ) 

12. (a) E =

Q 4πε 0 r 2

V =

Q 4πε 0 r

⇒   r =

V 3000 V = =6m E 500 Vm −1

(b) V = −3000 V ⇒   −3000 =

⎡ ( 1 )( 1 ) ( 1 ) ( −1 ) ( 1 ) ( −1 ) ⎤ ( −12 ) ⎢ ( 1 ) + ( 2 ) + ( 1 ) ⎥ 10 ⎣ ⎦



⎛ a⎞ This implies that a charge +Q ⎜⎝ ⎟⎠ has been transb ferred to earth leaving negative charge on A.

⇒   U = −4.5 × 10 −3 J  Since U < 0 , the system of three charges has lower potential energy then 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.

⇒   Q = −



P





⎛ Potential ⎞ ⎛ potential ⎞ ⎛ potential ⎞ ⎜ due to ⎟ = ⎜ due to ⎟ + ⎜ due to ⎟ ⎜ ⎝ inner sphere ⎟⎠ ⎜⎝ charge on A ⎟⎠ ⎜⎝ charge on B ⎟⎠ VA =

q′ q′ Q + q′ = VEarth = 0 − + 4πε 0 a 4πε 0b 4πε 0b

⎛ ⇒ q′ = −Q ⎜ ⎝

01_Ch 1_Hints and Explanation_P1.indd 27

a⎞ ⎟ b⎠

–3q

Case I

q 3q − =0 4πε 0 r 4πε 0 ( 1 + r )

CASE-2: +q

P r

VP =    v

1m

Solving this, we get r = 0.5 m

Q + q1

a

+q r

The charge q′ on A induces − q′ on inner surface of B and + q′ on outer surface of B , in equilibrium the charge distribution is as shown in figure.

A

3000 × 6 = −2 μC 9 × 109

CASE-1:

   VP =

B

Q 4πε 0 ( 6 )

13. Let P be the point on the axis either to the left or to the right of charge +q at a distance r where potential is zero. Hence,

11. When an object is connected to earth (grounded), its potential is reduced to zero. Let q′ be the charge on A after it is earthed.

q1 –q

CHAPTER 1

Now W∞→ 3 = q3V3

–3q 1– r Case II

q 3q − = 0, 4πε 0 r 4πε 0 ( 1 − r )

which gives r = 0.25 m Thus, the potential will be zero at point P on the axis which is either 0.5 m to the left or 0.25 m to the right of charge +q. 14. The original electrical potential energy is q2 ⎛ q ⎞ U e = qV = q ⎜ = ⎟ ⎝ 4πε 0 d ⎠ 4πε 0 d In the final configuration we have mechanical equilibrium. The spring and electrostatic forces on each charge are

9/20/2019 11:36:59 AM

H.28  JEE Advanced Physics: Electrostatics and Current Electricity

− k ( 2d ) +

⇒ k=

q

q2 4πε 0 ( 3 d )

=0

2

2

4πε 0 ( 18 d 3 )



⇒ V=

1 [ log e 3 + λπ + log e 3 ] 4πε 0



⇒ V=

λ ( π + 2 log e 3 ) 4πε 0

18. For an element of area which is a ring of radius r and dq 1 q q 4 q 1 2 1 width dr , dV = ( 2 d )2 + q kx + qV = = 2 πε 4 3 0 + x2 r 4πε 0 ( 3 d ) 9 4πε 0 d 2 2 4πε 0 ( 18 d ) dq = σ dA = σ 0 r ( 2π rdr ) and q2 q 1 2 1 4 q2 ( 2 d )2 + q kx + qV = = R 2 2 4πε 0 ( 18 d 3 ) 4πε 0 ( 3 d ) 9 4πε 0 d r 2 dr ⎛ 2π ⎞ V = σ 0 ⎜⎝ 4πε ⎟⎠ The missing energy must have become internal energy, 0 r 2 + x2 0 as the system is isolated x ⎛ ⎞⎤ ⎛ πσ 0 ⎞ ⎡ 2 2 2 q2 4 ⎛ q2 ⎞ ⇒ V =⎜ ⎟⎠ ⎢ R R + x + x log e ⎜ = ⎜ So, + ΔEint 2 2 ⎟⎥ ⎟ 4 πε ⎝ 0 ⎝ R + R + x ⎠ ⎥⎦ 4πε 0 d 9 ⎝ 4πε 0 d ⎠ ⎢⎣ In the final configuration the total potential energy is

2

2

∫



⇒ ΔEint =

15. ΔV = V2 R

5 ⎛ q2 ⎞ 9 ⎜⎝ 4πε 0 d ⎟⎠

1 − V0 = 4πε 0

ΔV =

19. (i) Potential at x = 0 will be

1 Q − 2 2 ( πε 4 0 R R + 2R ) Q

V=

⇒   V =

1 Q⎛ 1 Q ⎞ − 1 ⎟ = −0.553 ⎜ ⎝ ⎠ 4πε 0 R 4πε 0 R 5

E0

⇒ ΔU = −Wi → f = − ⎡⎣ q0E0 ( 2 a − a ) ⎤⎦

⇒   E =



⇒ ΔU = − q0E0 a



 Here work done by electrostatic force is positive. Hence the potential energy must be decreasing.





all charge

dq r

3R ⎡R ⎤ 1 ⎢ λ dx λ ds λ dx ⎥ ⇒ V= + + x R x ⎥ 4πε 0 ⎢ semicircle R ⎣ 3R ⎦

∫

⇒ V=

∫

R

1 1 λ log e ( x ) + πε 4πε 0 4 0 3R

01_Ch 1_Hints and Explanation_P1.indd 28

x=4

q x=8

This is an infinite G.P. sum of ∞ terms of G.P. is a Sn = 1− r



∫

q

2q q ⎡ 1 ⎤ q ⎡ 1 ⎤ = = 4πε 0 ⎢ 1 − 1 ⎥ 4πε 0 ⎢⎣ 1 2 ⎥⎦ 4πε 0 ⎢ ⎥ 2⎦ ⎣ Now electric field q q 1 ⎡ q ⎤ E= + + + ...∞ ⎥    4πε 0 ⎢⎣ ( 1 )2 ( 2 )2 ( 4 )2 ⎦

Since, Wi → f = −ΔU

1 4πε 0

q

⇒   V =

q0

17. V =

1 1 ⎡ ⎤ ⎢⎣ 1 + 2 + 4 + ....∞ ⎥⎦

x=0 x=1x=2

{along positive y-direction} q0E0

q 4πε 0 q

16. Electrostatic force F on the test charge, F = q0E0 

1 ⎛q q q q ⎞ ⎜ + + + + ...∞ ⎟⎠ 4πε 0 ⎝ 1 2 4 8

⎡λ ⎢ π R + λ log e x ⎣R

q ⎡ 1 ⎤ q ⎡ 1 ⎤ = ⎢ ⎥ 4πε 0 ⎣ 1 − 1 4 ⎦ 4πε 0 ⎢⎣ 3 4 ⎥⎦

⇒   E =

q 4q = 4πε 0 × 3 3πε 0

(ii) When consecutive charges are negative then

V=    3R R

⎤ ⎥ ⎦

1 1 1 ⎡ ⎤ ⎢⎣ 1 + 4 + 16 + 64 + ...∞ ⎥⎦ {This is also an infinite G.P.}

⇒   E =



∫

q 4πε 0

⇒   V =

q 1 ⎡q q q q q ⎤ − + − + − + ...∞ ⎥ 4πε 0 ⎢⎣ 1 2 4 8 16 32 ⎦ 1 ⎡q q q ⎤ ⎡q q q ⎤ + + + ...∞ ⎥ − ⎢ + + + ...∞ ⎥ 4πε 0 ⎢⎣ 1 4 16 ⎦ ⎣ 2 8 32 ⎦

9/20/2019 11:37:18 AM

Hints and Explanations H.29 q ⎧⎡ 1 1 ⎤⎫ ⎤ 1⎡ − ⎪ ⎪ 1⎞ ⎤⎥ 2⎢⎡ 1⎞ ⎤ ⎥⎬ 4πε 0 ⎨ ⎢ ⎡ ⎛ ⎛ ⎢ ⎢1 − ⎜ ⎟ ⎥ ⎥ ⎢ ⎢1 − ⎜ ⎟ ⎥ ⎥ ⎪ ⎢⎣ ⎣ ⎝ 4 ⎠ ⎦ ⎥⎦ ⎝ 4 ⎠ ⎦ ⎥⎦ ⎪⎭ ⎣⎢ ⎣ ⎩

q ⎡4 2⎤ 1 ⎛ 2q ⎞ ⇒   V = − = ⎜ ⎟ 4πε 0 ⎢⎣ 3 3 ⎥⎦ 4πε 0 ⎝ 3 ⎠

Electric field at x = 0 is q q q q 1 ⎡ q ⎤ − + − + ...∞ ⎥ + 4πε 0 ⎢⎣ ( 1 )2 ( 2 )2 ( 4 )2 ( 8 )2 ( 16 )2 ⎦  q q q q 1 ⎡ q ⎤ E= − + − + ...∞ ⎥ + 4πε 0 ⎢⎣ ( 1 )2 ( 2 )2 ( 4 )2 ( 8 )2 ( 16 )2 ⎦

   E=



⇒   E =

q 4πε 0

⇒   E =

⇒   E =

 1 ⎡1 1 ⎤ ⎢⎣ 4 + 64 + 1024 + ... ⎥⎦

q ⎧ 1 ⎫ 1⎡ 1 ⎤ ⎪− ⎪ 1 4πε 0 ⎨ ⎡ ⎤ ⎬ 4 ⎢1− 1 ⎥ 1 − ⎢ ⎥ ⎪⎩ ⎢⎣ 16 ⎦ ⎣ 16 ⎥⎦ ⎪⎭ q ⎡ 16 1 16 ⎤ q ⎡ 4q ⎤ − = 4πε 0 ⎢⎣ 15 4 15 ⎥⎦ 4πε 0 ⎢⎣ 5 ⎥⎦

K ( 1 × 10 −6 ) K ( 2 × 10 −6 ) + 3R 3R

⇒ v=

7Qq 4πε 0 m

22. For potential energy of the system of charges, total number of charge pairs will be 8 C2 or 28 of these 28 pairs, 12 unlike charges are at a separation ’a ’, 12 like charges are at separation 2 a and 4 unlike charges are at separation 3 a. Therefore, the potential energy of the system 1 4πε o



⎛ 1 q2 ⎞ ⇒ U = −5.824 ⎜ ⎟ ⎝ 4πε o a ⎠



The binding energy of this system is therefore,

⎛ 1 q2 ⎞ U = 5.824 ⎜ ⎟ ⎝ 4πε o a ⎠ So, work done by external forces in disassembling, this system of charges is

1 q 4πε 0 a

V =

R = 1 m 1 μC

2 μC

R 2R

3R

⎡ ( 12 ) ( q ) ( − q ) ( 12 ) ( q )( q ) ( 4 ) ( q ) ( − q ) ⎤ + + ⎢ ⎥ a 2a 3a ⎢⎣ ⎥⎦

23. Let q be the charge on the bubble, then

3 R = 3

P



⎛ 1 q2 ⎞ W = 5.824 ⎜ ⎟ ⎝ 4πε o a ⎠

20. Potential at point P 9000 =

⇒

U= 

1 1 ⎡ ⎤ ⎢⎣ 1 + 16 + 256 + ... ⎥⎦ −



qQ ⎡ 7 ⎤ 1 mv 2 = 2 4πε 0 ⎢⎣ 2 ⎥⎦





⇒ q = 4πε 0 aV

After collapsing, let the radius of droplet becomes R. Then equating the volume, we get

(

)

4 3 πR 3

4π a 2 t = 21. We can here use Law of Conservation of Energy, according to which ( U + K )at  + ( U + K )at A 4

Q

–q

O(0, 0)

A(2l, 0)

x

− qQ − qQ 1 1 2 + m( 0 ) + mv 2 = ( )  4πε 0 2 2 ⎛ ⎞ 2 4πε 0 ⎜ ⎟ ⎝ 4⎠



⇒



qQ ⎡ 4 1 ⎤ 1 ⇒ mv 2 = − 2 4πε 0 ⎢⎣  2 ⎥⎦

01_Ch 1_Hints and Explanation_P1.indd 29

CHAPTER 1

⇒   V =

(

)

1/3



⇒ R = 3 a 2t



Now, potential of new droplet will be V ′ =



Substituting the values, we have

V ’ =



1 q 4πε 0 R

1 4πε 0 aV 4πε 0 3 a 2t 1/3

(

⎛ a⎞ ⇒ V’= V⎜ ⎟ ⎝ 3t ⎠

)

1/3

24. There will be two points where electric potential will be zero. One point lies between the charges and other point lies in region outside the charges.

9/20/2019 11:37:36 AM

H.30  JEE Advanced Physics: Electrostatics and Current Electricity



For inside point, 2 μC

–4 μ C

P x

VP =

A

6–x

1 ⎛ 2 × 10 −6 −4 × 10 −6 ⎞ + ⎜ ⎟ =0 x 4πε 0 ⎝ 6−x ⎠

8

2 4 = x 6−x



⇒



⇒ 12 − 2x = 4 x



⇒ x=2m For outside point

0

3. –4 μ C

x

V ( x , y , z ) =

2 4 = x 6+x



⇒ 12 + 2x = 4 x



⇒ x=6m

Test Your Concepts-VIII (Based on Equipotential Surfaces) B

C

B

∫

A

∫

A



The surface on which V ( x , y , z ) = 0



is given by



From (1), we get 4 ( x + R ) + 4 y 2 + 4 z 2 = x 2 + y 2 + z 2



⎛8 ⎞ ⎛4 ⎞ ⇒ x 2 + y 2 + z 2 + ⎜ R ⎟ x + ⎜ R2 ⎟ = 0 …(2) ⎝3 ⎠ ⎝3 ⎠

2



y



)∫

(

dy − E cos 90ο

−0.3



⇒ VB − VA = ( 325 )( 0.8 ) = +260V

2.

(a) EA > EB since E =

EB = − (b)

0.4

)∫

dx

−0.2

ΔV Δ

(6 − 2) V ΔV =− = 200 NC −1 downwards Δ 0.02 m

01_Ch 1_Hints and Explanation_P1.indd 30

( −2z0 ) z + ( x02 + y02 + z02 − r 2 ) = 0 …(3)

8 4 −2x0 = R, −2 y0 = 0 , −2 z0 = 0 , x02 + y02 + z02 − r 2 = R2 3 3 0.5

(

2

Comparing equations (2) and (3), it is seen that the equipotential surface for which V = 0 is indeed a sphere and that

E

⇒ VB − VA = − E cos 180ο

2

⇒ x 2 + y 2 + z 2 + ( −2x0 ) x + ( −2 y0 ) y +



B

x

A

2

( x − x0 ) + ( y − y0 ) + ( z − z0 ) − r 2 = 0

C

C

1 ⎛ 1 2⎞ q − = 0   ⇒   2r1 = r2 4πε 0 ⎜⎝ r1 r2 ⎟⎠

The general equation for a sphere of radius r centered at ( x0 , y0 , z0 ) is

      VB − VA = − E ⋅ d  = − E ⋅ d  − E ⋅ d 

∫

1 (q) 1 ( −2q ) + …(1) 4πε 0 r1 4πε 0 r2

2 where r1 = ( x + R ) + y 2 + z 2 and r2 = x 2 + y 2 + z 2

1 ⎛ 2 × 10 −6 −4 × 10 −6 ⎞ VP = + ⎜ ⎟ =0 x 4πε 0 ⎝ 6+x ⎠ ⇒

4

For the given charge distribution,

6m



2

6

B

2 μC

P

1.

(c)  The figure is shown here with the field lines sketched.



4 ⎛ 16 ⇒ x0 = − R, y0 = z0 = 0 , and r 2 = ⎜ − ⎝ 9 3

4⎞ 2 4 2 ⎟R = R 3⎠ 9

The equipotential surface is therefore a sphere cen2 ⎛ 4 ⎞ tered at ⎜ − R, 0 , 0 ⎟ , having a radius R ⎝ 3 ⎠ 3 4.

Let us consider that the equipotential surface corresponding to this situation mentioned lies at a perpendicular distance r from all the charge densities. Then, at the equipotential surface, we have

9/20/2019 11:37:56 AM

Hints and Explanations H.31



 

λ λ ⎛r ⎞ ⎛r ⎞ ⇒ log e ⎜ 1 ⎟ + log e ⎜ 2 ⎟ − ⎝ r ⎠ 2πε 0 ⎝ r⎠ 2πε 0

 2λ ⎛r ⎞ log e ⎜ 3 ⎟ = constant ⎝ r⎠ 2πε 0

⎛ rr ⎞ ⎛r ⎞ ⇒ log e ⎜ 1 22 ⎟ − 2 log e ⎜ 3 ⎟ = constant ⎝ r⎠ ⎝ r ⎠ ⎛ r2 ⎞ ⎛ rr ⎞ ⇒ log e ⎜ 1 22 ⎟ + log e ⎜ 2 ⎟ = constant ⎝ r ⎠ ⎝ r3 ⎠ ⎧ ⎛ 1⎞⎫ ⎨∵ log e x = − log e ⎜⎝ ⎟⎠ ⎬ x ⎭ ⎩



⎛ rr ⎞ ⇒ log e ⎜ 1 22 ⎟ = constant ⎝ r3 ⎠ {∵ log e m + log e m = log e ( mn ) }



⇒



r1r2 = constant r32

Test Your Concepts-IX (Based on Relation Between Electrostatic Field and Potential)

∂ ⎛ 3 2 2 2 ⎜ V0 − E0 z + E0 a z x + y + z ∂y ⎝

⇒ Ey = 3E0 a 3 yz x 2 + y 2 + z 2

(

Ez = −

)

(



⇒ Ez = E0 + E0 a 3 2 z 2 − x 2 − y 2

3.

 ⎛ ∂V ˆ ∂V ˆ ∂V ˆ ⎞ E = −⎜ i+ j+ k ∂y ∂z ⎟⎠ ⎝ ∂x



where,





 ⇒ E = −2iˆ − 3 ˆj + kˆ



 ⇒ E

4.

  dV = − E ⋅ d 



⇒ E=

Ex2

+

A

Ey2

+

∫ B

2 2 E = ( −5 ) + ( −5 ) + 0 2 = 5 2 NC −1

2.



∂V ∂x





⎡ ⎛ 3⎞ ⇒ Ex = − ⎢ E0 a 3 z ⎜ − ⎟ x 2 + y 2 + z 2 ⎝ 2⎠ ⎣

(

(

(

⇒ Ex = 3E0 a 3 xz x 2 + y 2 +

Ey = −

∂V ∂y

01_Ch 1_Hints and Explanation_P1.indd 31

5 2

)(

5 2

∫ ) ( xiˆ − 2yjˆ + zkˆ ) ⋅ ( dx iˆ + dy ˆj + dz kˆ )

( 0, 2, 4

⇒ VA − VB = −

∫

( xdx − 2 ydy + zdz )

( 0, 2, 4 )

∂ ⎡ ⇒ Ex = − ⎢ V0 − Eo z + Eo a 3 z x 2 + y 2 + z 2 ∂x ⎢⎣



2y

( 2 , 1, 0 )

Inside the sphere, Ex = Ey = Ez = 0 .

Outside, Ex = −

5 2

( 2 , 1, 0 )

dV = −

Ez2

−

∂V ∂ = ( 2x + 3 y − z ) = 2 ∂x ∂x

∂V ∂ = ( 2x + 3 y − z ) = −1 ∂z ∂z

0, −2 )

−

)

)

∂V ∂z



( 1,

)

−



(

  ⎛ ∂V ˆ ∂V ˆ ∂V ˆ ⎞ Since E = −∇V = − ⎜ i+ j+ k ∂y ∂z ⎟⎠ ⎝ ∂x  ⇒ E = ( −5 + 6 xy ) iˆ + ( 3 x 2 − 2z 2 ) ˆj − 4 yz kˆ = −5iˆ − 5 ˆj

x2 + y 2 + z2

⎛ 3⎞ ⇒ Ey = − E0 a 3 z ⎜ − ⎟ x 2 + y 2 + z 2 ⎝ 2⎠

V = 5x − 3 x 2 y + 2 yz 2

iˆ

⎞ ⎟ ⎠



∂V ∂ = ( 2x + 3 y − z ) = 3 ∂y ∂y



3 2

⇒ Ey = −



1.

−

(



CHAPTER 1

ΣV = constant

5 − z2 2

)

)

−

5 2

)

−

3 2

⎤

( 2x ) ⎥ ⎦

⎤ ⎥ ⎥⎦



( 2 , 1, 0 ) ⎤ ⎡⎛ 2 x z2 ⎞ ⎥ − y2 + ⎟ ⇒ VAB = − ⎢ ⎜ ⎢⎝ 2 ⎥ 2⎠ ( 0, 2, 4 ) ⎦ ⎣



⇒ VAB = 3 V

5.

  (a) dV = − E ⋅ d  A

∫

dV = − B

( 0, 0, 0 )

∫ (

( yiˆ + xjˆ ) ⋅ ( dx iˆ + dy ˆj + dz kˆ )

1, 1, 1 )

9/20/2019 11:38:16 AM

H.32  JEE Advanced Physics: Electrostatics and Current Electricity y

( 0, 0, 0 )

∫

⇒   VA − VB = −

( ydx + xdy )

( 1, 1, 1 )

( 0, 0, 0 )

⇒   VAB = −

∫

( 1, 1, 1 )

x

d ( xy )  { as y dx + xdy = d ( xy ) }

( 0, 0, 0 ) ⎤ =1V ⇒   VAB = − ⎡ ( xy ) ( 1, 1, 1 ) ⎦⎥ ⎣⎢

(b)

  (b) dV = − E ⋅ d  ⇒  

iˆ

∫

7.

( 0, 0, 0 )

A

dV = −

∫

(3 x yiˆ + x j ) ⋅ (dx iˆ + dy ˆj + dz kˆ ) 3ˆ

2

( 1, 1, 1 )

B

( 0, 0, 0 )

∫

⇒   VA − VB = −

( 3x 2 ydx + x 3 dy )

( 1, 1, 1 ) ( 0, 0, 0 )

⇒   VA − VB = −

∫ (

(

3

d x y

(

(

Shape of the equipotential surface i.e. for V = constant

(

( 0, 0, 0 ) ⎤

Since in both the cases, the line integral of the field is an exact differential and hence both the fields are conservative in nature.

(

(a) Given, φ = a x 2 − y 2

)

)

Since, V = a x 2 + y 2 + bz 2

⎡ =1V ⇒   VAB = − ⎢ x 3 y ( 1, 1, 1 ) ⎥⎦ ⎣

6.

)

 Hence E = 2 a 2 x 2 + y 2 + b 2 z 2

1, 1, 1 )

( )

)

  Since, E = −∇V  ⇒ E = − 2 axiˆ + 2 ayjˆ + 2bzkˆ



)

(

Given , V = a x 2 + y 2 + bz 2

)



⇒



⇒

2

2

y x z2 + + =1 ⎛V⎞ ⎛V⎞ ⎛V⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ a a b x2 ⎛ V⎞ ⎜⎝ ⎟ a⎠

2

+

y2 ⎛ V⎞ ⎜⎝ ⎟ a⎠

2

+

z2 ⎛ V⎞ ⎜⎝ ⎟ b ⎠

2

=1

 which is an ellipsoid of revolution with semi-axis V V V . , , a a b

y

8.

Given, again   dV = − E ⋅ d  x

(a)

  So, E = −∇φ = −2 a xiˆ − yjˆ

(

)

The sought shape of field lines is as shown in the figure. (b) Since φ = axy   So, E = −∇φ = − ayiˆ − axjˆ

The sought shape of field lines is as shown in the figure.

01_Ch 1_Hints and Explanation_P1.indd 32

(

) ( dxiˆ + dyjˆ + dxkˆ )



⇒ dV = − ayiˆ + ( ax + bz ) ˆj + bykˆ



⇒ dV = − a ( ydx + axdy ) + b ( zdy + ydz )



⇒ dV = − ⎡⎣ ad ( xy ) + bd ( yz ) ⎤⎦



Integrating we get,

V = − ( axy + byz ) + C

Test Your Concepts-X (Based on Motion of Charged Particles in Electric Field) 1.

(a) ΔV = Ed = ( 6 × 10 3 Vm −1 ) ( 0.01 m ) = 60 V



(b)

1 mv 2f = qΔV 2

9/20/2019 11:38:31 AM

Hints and Explanations H.33 1( 9.1 × 10 −31 ) v 2f = ( 1.6 × 10 −19 ) ( 60 ) 2

⇒   v f = 4.6 × 106 ms −1 2.

(a) Energy of the proton-field system is conserved as the proton moves from higher to lower potential, which can be defined for this problem as moving from 120 V down to 0 V .

K i + U i + ΔEmech = K f + U f    ⇒   0 + qV + 0 = ⇒   ( 1.6 × 10

−19

1 mvp2 + 0 2

) ( 120 ) = 1 ( 1.67 × 10 −27 ) vp2 2

1 ⇒   0 + ( 40 × 10 −9 C)(3 × 10 3 V) = (2 × 10 −13 kg )v 2 + 0 2 ⇒   v =

2 ( 1.2 × 10 −4 J ) = 3.46 × 10 4 ms −1 2 × 10 −13 kg

4.

Using conservation of energy for the alpha particlenucleus system,



we have K f + U f = K i + U i .

where U i =

1 qα qgold 4πε 0 ri

Since ri → ∞

⇒ Ui = 0

⇒   vp = 1.52 × 10 5 ms −1

also K f = 0 because, at turning point, we have v f = 0





so we get, U f = K i



⇒



⇒ rmin =



⇒ rmin =



⇒ rmin = 2.73 × 10 −14 m = 27.3 fm

5.

Each charge moves off on its diagonal line. All charges have equal speeds.

(b) The electron will gain speed in moving the other way, from Vi = 0 to V f = 120 V . So,

   K i + U i + ΔEmech = K f + U f ⇒   0 + 0 + 0 = ⇒   0 = ⇒  

1 mve2 + qV 2

1 mve2 + ( − e )V 2

1( 9.1 × 10 −31 ) ve2 = ( 1.6 × 10 −19 ) ( 120 ) 2

⇒   ve = 6.5 × 106 ms −1 3.

(a) U =

10 −18 ⎡ ( 20 )( 10 ) ( 10 ) ( −20 ) ( 20 ) ( −20 ) ⎤ + + ⎥⎦ 0.08 4πε 0 ⎢⎣ 0.04 0.04

⎡ 200 200 400 ⎤ ⇒   U = 9 × 10 −9 × 100 ⎢ − − 4 8 ⎥⎦ ⎣ 4



⎛ 400 ⎞ ⇒   U = 9 × 10 −9 × 100 × ⎜ − ⎟ ⎝ 8 ⎠



⇒   U = −4.5 × 10 −5 J

(b) The three fixed charges create this potential at the location where the fourth is released:



   V = V1 + V2 + V3 ⎛ 20 × 10 −9 10 × 10 −9 20 × 10 −9 ⎞ ⇒   V = (9 × 109 ) ⎜ + − ⎟ ⎝ 0.05 0.03 0.05 ⎠ ⇒   V = 3 × 10 3 V Energy of the system of four charged objects is conserved as the fourth charge flies away ⎞ ⎛1 ⎞ ⎛1 2 2    ⎜⎝ mv + qV ⎟⎠ = ⎜⎝ mv + qV ⎟⎠ 2 2 i f

01_Ch 1_Hints and Explanation_P1.indd 33

6.

CHAPTER 1

⇒  

1 qα qgold 1 = mα vα2 4πε 0 rmin 2 1 2qα qgold 4πε 0 mα vα2 2 ( 9 × 109 ) ( 2 ) ( 79 ) ( 1.6 × 10 −19 )

2

( 6.67 × 10 −27 ) ( 2 × 107 )2

∑( K + U ) = ∑( K + U ) i

f

2q 2 ⎞ 1 ⎛ 4 q 2 2q 2 ⎞ 1 ⎛ 4 q2 ⎛1 2⎞ + + ⎜⎝ ⎜⎝ ⎟ ⎟⎠ = 4 ⎜⎝ mv ⎟⎠ + 4πε 0 L 4πε 0 2L 2 2L ⎠ 2 2L 2q 2 ⎞ 1 ⎛ 4 q 2 2q 2 ⎞ 1 ⎛ 4 q2 ⎛1 2⎞ 0+ + + ⎜⎝ ⎜⎝ ⎟ ⎟⎠ = 4 ⎜⎝ mv ⎟⎠ + 4πε 0 L 4πε 0 2L 2 2L ⎠ 2 2L  1 ⎛ 1 ⎞ q2 = 2mv 2 ⇒ ⎜⎝ 2 + ⎟ 4πε 0 2⎠ L ⇒ 0+

1 ⎞ 1 q2 ⎛ ⇒ v = ⎜ 1+ ⎟ ⎝ 8 ⎠ 4πε 0 mL The potential created by the ring at the electron’s starting point is

Vi =

1 4πε 0

Q xi2

+a

2

=

1 ( 2πλ a ) 4πε 0 x 2 + a 2 i

9/20/2019 11:38:51 AM

H.34  JEE Advanced Physics: Electrostatics and Current Electricity



1 ( 2πλ ) 4πε 0 From Law of Conservation of Energy, we have while at the center, it is V f =

R

Q

( U + K )i = ( U + K ) f

Uniformly charged ring

1 ⇒ − eVi + 0 = − eV f + meV f2 2 v 2f

a 2e 1 4π eλ = ( Vf − Vi ) = 4πε m ⎛⎜ 1 − 2 2 ⎞⎟ me 0 e ⎝ xi + a ⎠



⇒



4π ( 1.6 × 10 −19 ) ( 9 × 109 ) ( 1 × 10 −7 ) ⇒ v 2f = × 9.1 × 10 −31 0.2 ⎛ ⎞ 1− ⎜ 2 2 ⎟ ( ) ( ) ⎝ 0.1 + 0.2 ⎠

 7

−1



⇒ v f = 1.45 × 10 ms

7.

Initially, since no external horizontal forces act on the set of four balls, so the center of mass of the system stays fixed at the initial location of the center of the square. As the charged balls 4 and 3 swing out and away from each other, balls 1 and 2 move down with equal y-components of velocity. The maximum kinetic energy point is illustrated. For system energy to be conserved,

1 q2 1 q2 1 1 1 1 = + mv 2 + mv 2 + mv 2 + mv 2  4πε 0  4πε 0 3 2 2 2 2 v

v + 4

v



8.

l

CM

1 2q 2 = 2mv 2 ⇒ 4πε 0 3 ⇒ v=

⇒ vf =

9.

By Law of Conservation of Energy,

( U + K )(

1 q2 4πε 0 3 m

1 1 Q2 Mv 2f + 0 = 0 + 2 4πε 0 R

01_Ch 1_Hints and Explanation_P1.indd 34

= ( U + K )(

final for A + B ) system

q2 q2 1 +0+0= + mu2 + K A 4πε 0 ( 2 −  ) 2 4πε 0 2

⇒ mg +



q2 1 1 (1 − 2 ) ⇒ K A = mg − mu2 + 2 4πε 0 2

10. Let rmin be the distance of closest approach between the two charges. Then, from Conservation of Angular Momentum, we get Q +

r

90°

v0

+q

l

v

mv0 = mvr

⇒ v0 = vr …(1)



From Law of Conservation of Energy, 1 1 1 Qq ⋅ …(2) mv02 = mv 2 + 2 2 4πε 0 r Solving equations (1) and (2), we get 2

rmin =

Qq Qq ⎞ ⎛ + ⎜ + 2 4πε 0 mv02 ⎝ 4πε 0 mv02 ⎟⎠

11. Since E =

V at =  



⇒ m



⇒

dv ⎛ at ⎞ =⎜ ⎟e dt ⎝  ⎠

v

From Law of Conservation of Energy,

⇒

initial for A + B ) system



v

K f + U f = K i + U i

+ 3

Q2 2πε 0 MR





2

1

x

v

∫ 0

t

dv =

ae t dt m

∫ 0



⎛ ae ⎞ 2 ⇒ v=⎜ t …(1) ⎝ 2m ⎟⎠



⇒

dx ⎛ ae ⎞ =⎜ ⎟t dt ⎝ 2m ⎠

9/20/2019 11:39:06 AM

Hints and Explanations H.35





⇒ x

0

=

ae ⎛ t 3 ⎞ ⎜ ⎟ 2m ⎝ 3 ⎠

Test Your Concepts-XI (Based on Conductors)

t 0



⎛ ae ⎞ 3 ⇒ =⎜ t …(2) ⎝ 6 m ⎟⎠



Substituting value of t from (2) in (1), we get

1.

The charge divides equally between the identical Q spheres, with charge on each. Then they repel like 2 point charges at their centers so, we have

⎛ Q⎞⎛ Q⎞ Q2 1 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 1 F = = 2 4πε 0 ( L + R + R ) 4πε 0 4 ( L + 2R )2

12. Using Law of Conservation of Angular Momentum about +Q , we get

F =

2

9 × 109 × ( 60 × 10 −6 ) 4 × ( 2.01 )

v1 r2

–q, m

Q +

r1

2.

– q, m

Since E =

v2

CHAPTER 1

2 ⎛ ae ⎞ ⎛ 6 m ⎞ 3 v = ⎜ ⎟ ⎜ ⎝ 2m ⎠ ⎝ ae ⎟⎠

2

2

≅2N

σ 2ε 0 E+

E+

E+

E–

E–

E–

(I)

(II)

(III)

mv1r1 = mv2 r2

⇒ v1r1 = v2 r2 …(1)



From Law of Conservation of Energy,

( U + K )initial = ( U + K )final

1 1 Qq 1 1 Qq mv12 − = mv22 − …(2) 2 4πε 0 r1 2 4πε 0 r2 On solving equations (1) and (2), we get ⇒

v1 =

Qqr2 and v2 = 2πε 0 mr1 ( r1 + r2 )

Qqr1 2πε 0 mr2 ( r1 + r2 )

13. Since, no external force acts on the system of protons, so by Law of Conservation of Linear Momentum, we get v0

Rest

m

m

v

m

v

rmin

mv0 = 2mv v0 …(1) 2 At the closest distance of approach, both the protons have same velocity, so By Law of Conservation of Energy, we get



⇒ v=

e2 1 ⎡1 ⎤ …(2) mv02 = 2 ⎢ mv 2 ⎥ + 2 ⎣2 ⎦ 4πε 0 rmin

On solving equations (1) and (2) for rmin , we get

rmin =

e2 πε 0 mv02

01_Ch 1_Hints and Explanation_P1.indd 35



Let us designate left to right as positive direction. Then 1 EI = − (σ A + σB ) 2ε 0 EII =

1 (σ A − σB ) 2ε 0

EIII =

1 (σ A + σB ) 2ε 0

For σ A = +σ and σ B = −σ , we get EI = EIII = 0 and EII =

σ ε0

3.

Consider as a Gaussian surface a box with horizontal area A , lying between 500 m and 600 m elevation.   q E ⋅ dA = ε0

∫

( +120 NC −1 ) A + ( −100 NC −1 ) A = ρA ( 100 m )



⇒



⇒ ρ=

ε0

( 20 NC −1 ) ( 8.85 × 10 −12 C2Nm −2 )

ρ = 1.77 × 10

100 m −12

Cm −3 = 1.77 pCm −3

The charge has to be positive so that it produces the net outward flux of electric field.

9/20/2019 11:39:21 AM

H.36  JEE Advanced Physics: Electrostatics and Current Electricity 4.

The electric field on the surface of a conductor varies inversely with the radius of curvature of the surface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa. The local charge density and the electric field intensity are related by

E =

σ ε0



⇒ σ = Eε 0



(a) At the point on the surface where the radius of curvature is the greatest, we have

8.

σ = ε 0Emin = (8.85 × 10

−12

2

−2

4

−1

C Nm )(5.6 × 10 NC )

(a)





∫ E ⋅ dA = E ( 4π r

2

) = qenc ε0

⎛4 ⎞ For r < a , qenc = ρ ⎜ π r 3 ⎟ ⎝3 ⎠

⇒   σ = 496 nCm −2

Insulator

(b) At the point on the surface where the radius of curvature is the smallest, we have

a

σ = ε 0Emax = (8.85 × 10 −12 C 2 Nm −2 )(11.2 × 10 4 NC −1 ) ⇒   σ = 992 nCm 5.

(a) The charge +q at the center induces charge −q on the inner surface of the conductor, where its surface density is given by

⇒   E =

Q+q 4π b 2

6.

Charge on the outer surface of the sphere is −Q



Charge on the inner surface of the shell is +Q



Charge on the outer surface of the shell is +2Q

ρr 3ε 0

For a < r < b and c < r , qenc = Q ⇒   E =

(b)  The outer surface carries charge Q + q with density

   σb =

c

−2

−q σa =    4π a 2

Conductor

b

Q 4π r 2ε 0

For b ≤ r ≤ c , E = 0 (since E = 0 inside a conductor).

(b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.

Therefore, q1 + Q = 0 q1 −Q = 4π b 2 4π b 2  Let q2 = induced charge on the outside surface of the hollow sphere. since the hollow sphere is uncharged, we require ⇒   σ 1 =

Using Gauss’s Law we evaluate the electric field in each region, keeping in mind that the electric field is zero everywhere within conducting materials. Then Inside the sphere and within the material of the shell, we have E = 0

q1 + q2 = 0   

Between the sphere and shell the field is directed radi-

⇒   σ 2 =

ally inwards given by E =

Q 4πε 0 r 2

9.

q1 Q = 4π c 2 4π c 2

(a) qenc = +3Q − Q = +2Q

Outside the shell the field is directed radially outwards 2Q given by E = 4πε 0 r 2 7.

An approximate sketch is given at the right. Note that the electric field lines should be perpendicular to the conductor both inside and outside.

01_Ch 1_Hints and Explanation_P1.indd 36

–Q 3Q a

c

Gaussian surface

b

9/20/2019 11:39:35 AM

Hints and Explanations H.37

(b) The charge distribution is spherically symmetric and qenc > 0. Thus, the field is directed radially outward.

1 ⎛ qenc ⎞ 2Q E= for r ≥ c.    ⎜ ⎟= 4πε 0 ⎝ r 2 ⎠ 4πε 0 r 2



(j)  Since the total charge on the conducting shell is qnet = qouter + qinner = −Q, we have qouter = −Q − qinner = −Q − ( −3Q ) = +2Q



(k) This is shown in the figure. E

(c)  Since all points within this region are located inside conducting material, E = 0 for b < r < c .

∫

⇒   qenc = ε 0ϕE = 0

a

–Q Gaussian surface

3Q a

c

Test Your Concepts-XII (Based on Charge Distribution) 1.

qenc = +3Q (e) –Q

a

1 (Q – q1)

c

b

a≤r q1 Again electric field is zero between q1 and q2 , therefore, both charges must be of same nature. Since electric field due to q1 is towards right and that of q2 is towards left therefore q1 and q2 are positive charges. Hence, the correct answer is (A).

109. Both A and B have same radius and same potential. Hence QA = QB 

II

A

d qQ - q2 = 0 dq

Q - 2q = 0



–σ

⎧ ⎫ Q for both ⎬ ⎨∵V = 4 πε r 0 ⎩ ⎭

Hence, the correct answer is (C).

01_Ch 1_Hints and Explanation_P2.indd 56

σ σ σ and is uniform. + = 2ε o 2ε o ε o



Hence, the correct answer is (B).

112.

V=



⇒ V=



⇒ V=

q ⎛ 1 1 1 ⎞ ⎜⎝ 1 + + + + .... ⎟⎠ 4πε 0 2 4 8

E =

q ⎛ 1 ⎞ 4πε 0 ⎜ 1 - 1 ⎟ ⎜⎝ ⎟ 2⎠ 1 ( 2q ) 4πε 0 q ⎛ 1 1 ⎞ + .... ⎟ ⎜ 1+ + ⎠ 4πε 0 ⎝ 4 16 q 1 1⎞ 4πε 0 ⎛ ⎜⎝ 1 - ⎟⎠ 4



⇒ E=



⇒ E=



Hence, the correct answer is (D).

1 ⎛ 4q ⎞ ⎜ ⎟ 4πε 0 ⎝ 3 ⎠

9/20/2019 11:35:07 AM

Hints and Explanations H.57

V=



⇒ V=



⇒ V=

Vouter =

q ⎛ 1 ⎞ 4πε 0 ⎜ 1 + 1 ⎟ ⎜⎝ ⎟ 2⎠ 1 ⎛ 2q ⎞ ⎜ ⎟ 4πε 0 ⎝ 3 ⎠

E = E+ - Eq ⎛ q ⎛1 1 1 ⎞ ⎞ + ... ⎟ + + ... ⎟ ⎜ 1+ ⎠ 4πε 0 ⎜⎝ 4 64 ⎠ 4πε 0 ⎝ 16

Q2 Q1 + 4πε 0b 4πε 0b



⇒ Vinner > Vouter as b > a



Hence, the correct answer is (B).

117. Electric field at O depends only on the induced charge on the outer surface of the spherical conductor. Shifting of q within the cavity does not change distribution of induced charge over the outer surface. Hence, the correct answer is (D).   118. E = -∇V



⇒ E=



1 ⎤ ⎡ ⎥ q ⎢ 1 - 4 ⎥ ⇒ E= ⎢ 4πε 0 ⎢ 1 - 1 1 - 1 ⎥ 16 16 ⎦ ⎣



⇒ E=

q ⎛ 16 4 ⎞ - ⎟ ⎜ 4πε 0 ⎝ 15 15 ⎠

∂U = -16 xy - 8 + 6 z ∂y



⇒ E=

q ⎛ 12 ⎞ ⎜ ⎟ 4πε 0 ⎝ 15 ⎠

∂U = 6 y ∂z



⇒ E=

q ⎛ 4⎞ ⎜ ⎟ 4πε 0 ⎝ 5 ⎠





Hence, the correct answer is (C).



∂U = 6 - 8 y 2 - 8 x ∂x

114. Net electric field between the plates is E = q q 2 Since σ = = A 2A ⇒ E=



2 q/2

–q/2

σ ε0

∂U = -8 ∂y



Hence, the correct answer is (D).

119. If angle made by the electric field due to the dipole with the radius vector is ϕ then

Q2 b a



q/2

Hence, the correct answer is (A).

115.

∂U = 6 ∂x

iˆ

1



At (0, 0, 0)

∂U = 0 ∂z  ⇒ E ( 0 , 0 , 0 ) = -6iˆ + 8 ˆj  ⇒ E = 10 NC -1   ⇒ F =q E  ⇒ F = 20 N

q 2ε 0 A q/2

 ⎛ ∂U  ∂U  ∂U ⎞ ⇒ E = - ⎜ i +j +k ∂y ∂z ⎟⎠ ⎝ ∂x

CHAPTER 1

q ⎛ 1 1 1 ⎞ ⎜ 1 - + - + .... ⎟⎠ 4πε 0 ⎝ 2 4 8

113.

Vinner =

Q1 Q2 + 4πε 0 a 4πε 0b

01_Ch 1_Hints and Explanation_P2.indd 57

ϕ

p

Q1

tan ϕ =

E

θ

1 tan q 2

9/20/2019 11:35:23 AM

H.58  JEE Advanced Physics: Electrostatics and Current Electricity

In this equation

ϕ =

π -q 2



⎛π ⎞ 1 ⇒ tan ⎜ - q ⎟ = tan q ⎠ 2 ⎝2



⇒ cot q =



⇒ q = tan -1 ( 2 )



Hence, the correct answer is (D).

120.



1 tan q 2

A

B

C

FCB

FCD FCA

Fnet

Fnet

⎛ 1 + 2 2 ⎞ q2 =⎜ ⎟ ⎝ 2 ⎠ 4πε 0 a 2



Hence, the correct answer is (C).

121.

VB - V0 = - Ex dx = -



Hence, the correct answer is (A).

Since earthing of B will make potential of B zero but potential difference will not change because potential difference between two concentric spheres depends only on the charge on the inner sphere. Hence Difference

=



⇒ VA′ - VB′ = VA - VB



⇒ VA′ - 0 = VA - VB



⇒ VA′ = VA - VB



Hence, the correct answer is (C). qQ 4πε0a2

125.   

Q

q

45°

Initial

Potential

Q2 4πε 0(2a2) qQ 4πε 0 a2

400 ( 0.03 ) 2

∫

VA - VB = - Ey dy = -

400 ( 0.02 ) 2

where V0 is potential at origin. So, 400 ( 0.03 - 0.02 ) VA - VB = 2 4 = 2.8 V 2



⇒ VA - VB =



Hence, the correct answer is (D).

122.

q r + qR = Q …(1)

q r2 4π r 2σ r = = 2 2 qR 4π R σ R

q r2 ⇒ r = 2 …(2) qR R



From (1) & (2) Qr 2 QR2 and qR = 2 2 2 R +r R + r2

01_Ch 1_Hints and Explanation_P2.indd 58

⇒ V=

 Final Potential Difference

2q 2 q2 = + 2 2 4πε 0 a 4πε 0 ( 2 a )

∫

Q ⎛ r+R ⎞ ⎜ ⎟ 4πε 0 ⎝ r 2 + R2 ⎠



124. Initially potential difference between the spheres is VA - VB

D

qr =

qr qR + 4πε 0 r 4πε 0 R

So, V =

q

Q



For net force on Q to be zero

qQ qQ Q2 + cos 45 + cos 90 = 0 2 2 2) ( 4πε 0 a 4 πε 4πε 0 2 a 0a

⇒ Q = -2 2 q



Hence, the correct answer is (D).

126. Method 1  Let origin be at A, then co-ordinates of points A, B ⎛ 3 ⎞ and C are ( 0 , 0 ) , ( , 0 ) and ⎜ , respectively. ⎝ 2 2 ⎟⎠ C

A

q

–2q

q

B

9/20/2019 11:35:37 AM

Hints and Explanations H.59

∫ S

⇒ px = 0

py = q ( 0 ) + q ( 0 ) + ( -2q )

⇒ py = - 3 q



⇒ p = px2 + py2



⇒ p = 3 q



Method 2 (Shortcut)



3 2

qenc …(1) ε0

A( 4π r 2 dr ) = 4π A ( rdr ) r So, total charge enclosed by Gaussian-surface is

dq = ρdV =

r1

⎛ r2 ⎞ qenc = dq = 4π A rdr = 4π A ⎜ 1 ⎟ ⎝ 2⎠

∫

because Either Distance of p = ⎛ charge ⎞ ⎛ separation ⎞ ⎝ ⎠⎝ ⎠ Hence, the correct answer is (C).



⇒ L = rp sin q



⇒ L = x0 ( mv ) sin ( 90° ) = x0 mv



Now, at time t, the velocity is given by

0



⎛ r2 ⎞ ⇒ qenc = 4π A rdr = 4π A ⎜ 1 ⎟ = 2π Ar12 ⎝ 2⎠



From Equation (1), we get

∫ 0

(

)

E 4π r12 =

qE0 ⎫ ⎧ ⎨∵ a = ⎬ m ⎭ ⎩

⎛ qE ⎞ v = at = ⎜ 0 ⎟ t  ⎝ m ⎠

∫

r1

128. Angular momentum is given by    L = r × p



)

Now, let us calculate qenc . The sphere can be regarded as consisting of a large number of spherical shells. Consider a shell of inner and outer radii r and r + dr . Its volume will be dV = 4π r 2 dr. Charge dq on this shell will be

⎛ 3 ⎞ p = ( 2q ) ⎜ ⎝ 2 ⎟⎠



(

⇒ E 4π r12 =

⎛ qE ⎞ ⇒ L = x0 m ⎜ 0 ⎟ t = ( qE0 x0 ) t ⎝ m ⎠

Since the speed of the particle is increasing with time, therefore angular momentum is increasing with time. Hence, the correct answer is (C).

2π Ar12 ε0

A 2ε 0



⇒ E=



Hence, the correct answer is (D).

130. Direction of electric field with the axis of dipole is given by ⎛ tan q ⎞ q + tan -1 ⎜ ⎝ 2 ⎟⎠

Here given that E is along y-axis, so

⎛ tan q ⎞ π q + tan -1 ⎜ = ⎝ 2 ⎟⎠ 2 E

129.

P is any inside point at distance r1 from O . Consider a spherical surface of radius r1 as Gaussian surface. So,   q E ⋅ dA = enc ε0

β

P

∫

–q

r1

P

By symmetry, E at all points on the surface is same   and angle between E and dA is zero everywhere. So,

01_Ch 1_Hints and Explanation_P2.indd 59

θ

θ

S

O

CHAPTER 1



  q E.dA = EA = enc ε0

 2

px = q ( 0 ) + q (  ) + ( -2q )

+q

tan q = cot q 2



⇒



⇒ tan q = 2 cot q



⇒ tan 2 q = 2



⇒ tan q = 2 Hence, the correct answer is (B).



9/20/2019 11:35:55 AM

H.60  JEE Advanced Physics: Electrostatics and Current Electricity 131.

q1

q2

r

q1 + q2 2

Initially

Initially, F =

r q1 + q2 2 2

Finally

q1q2 4πε 0 r 2

After contact i.e., finally, we have

( q + q )2 F ′ = 1 2

k( 40)( 40) r2



So, new force F ′ =



⇒



⇒ F′ =



Hence, the correct answer is (C).

F ′ 40 × 40 16 = = 9 F 10 × 90 16 F 9

134. (due to –4q and –3q) E

4πε 0 r 2

E (due to +4q and 3q)

Since, F ′ = 4.5 F

O

( q1 + q2 )2 = 4.5q1q2



⇒



⇒ 2q12 + 2q22 + 4 q1q2 = 9q1q2



⇒ 2q12 + 2q22 - 5q1q2 = 0



q2 ⎛q ⎞ ⇒ 12 + 2 - 5 ⎜ 1 ⎟ = 0 q2 ⎝ q2 ⎠



⇒ 2x 2 - 5x + 2 = 0 , where x =



⇒ 2x 2 - 4 x - x + 2 = 0



⇒ 2x ( x - 2 ) - 1 ( x - 2 ) = 0



⇒ ( 2x - 1 ) ( x - 2 ) = 0



1 ⇒ x = or x = 2 2



⇒ q1 = 2q2



Hence, the correct answer is (A).

r2

r2

E (due to +6q and +5q)

 Since all fields are equal in magnitude and are inclined to each other at 120°, so Enet = 0

Hence, the correct answer is (D).

135. q1 q2

132. Due to induction, charges will appear on the surface of sphere as shown in figure. Electric field lines never enter any conducting surface and are perpendicular at the surface.

+q

–q

r

q2  F= and when 25% of charge of one is trans4πε 0 r 2 ferred to the other, then we have, q–

q 4

–q +

r

q 4

⎛ 3q ⎞ ⎛ 3q ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 9 4 4 F F ′ = = 2 16 4πε 0 r

Hence, the correct answer is (B).

136. We can consider all the charge inside the sphere to be concentrated at the centre of sphere, consider an elementary shell of radius x and thickness dx , then dx x

r



Hence, the correct answer is (C).

k(10)(90) 1 33. F = r2



Charge on each, when contact is made is given by

q1′ = q2′ =

01_Ch 1_Hints and Explanation_P2.indd 60

E =

10 + ( -90) = -40 μC 2

∫

∫ 4π x dx ( α x ) 2

dq

1 = 4πε 0 r 2

0

4πε 0 r

2

=

αr2 4ε 0

Hence, the correct answer is (B).

137. If 20 th charge is also placed, then   Fnet = 0 (by symmetry)      ⇒ F0 , 1 + F0 , 2 + F0 , 3 + ... + F0 , 20 = 0

9/20/2019 11:36:12 AM

Hints and Explanations H.61    ⇒ F0 , 1 + F0 , 2 + ... + F0 ,



Hence, the correct answer is (B).

138.

 = - F0 ,

19

20

= F0 ,

20

=

Qq 4πε 0 a 2

2qE k



⇒ x=



Hence, the correct answer is (A).

143.   



2q, m

2

F = 2qE E F = qE

2q, 2m

1



⇒ VA =



1 ⎛ 2qEt ⎞ m⎜ ⎝ m ⎟⎠ KEA = 2 ⇒ 2 KEB 1 2qEt ⎞ ( 2m ) ⎛⎜ ⎟ ⎝ 2m ⎠ 2

2qE t and m

qE



l 2

l 2

E

  ( E1+ 3 ) y =

2λ  4πε 0 2



⇒ qE = mg



⇒ neE = mg -3

mg 1.6 × 10 × 9.8 = eE 1.6 × 10 -19 × 109



⇒ n=



⇒ n = 9.8 × 107



Hence, the correct answer is (A).

141.

F=



When allowed to touch, we have





⇒ Enet =



Hence, the correct answer is (A).

144.

λ ( 2 5 - 1) 2 5πε 0



F′ =

F 8 Hence, the correct answer is (C).

142. Since work done by electric force is equal to the gain in elastic potential energy of the spring, so 1 kx 2 = qEx 2

01_Ch 1_Hints and Explanation_P2.indd 61

O

So, Fnet = 2 F -

F 2

45° q0

F=

⎛ Q⎞⎛ Q⎞ K⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠

⇒ F′ =

2λ ⎛ 2 1 ⎞ + ⎜2⎟ 4πε 0 ⎝ 5 5⎠

⇒ Enet = ( E1+ 3 )y + ( E2 )y =

F

r2

4λ ⎛ 1 ⎞ ⎜1⎟ 4πε 0 ⎝ 5⎠

2λ ⎛ 1 ⎞ λ 2 sin α = ⎜ ⎟ 4πε 0 4πε 0  ⎝ 5 ⎠



K (Q)(2Q) r2

F ’ =

( cos 0° - cos q ) =

Similarly, we have

E2 = ( E2 )y = mg

E1

E2

So, on resolution, their x components will cancel. However their y - components will add up to give

Hence, the correct answer is (A).

140.

θ

  By Symmetry, E1 = E3

2



θ

E3

Since, VA = uA + aAt

3

α α

1 2 KEA 2 mAvA = 1 KEB mBvB2 2

5l 2

5l 2

CHAPTER 1



qq0 4πε0 a2

F 2



1 ⎞ qq0 ⎛ ⇒ Fnet = ⎜ 2 - ⎟ ⎝ 2 ⎠ 4πε 0 a 2



Hence, the correct answer is (C).

145.

mg = qE



⇒ mg = ( Ne ) E 

{∵ q = Ne }

9/20/2019 11:36:26 AM

H.62  JEE Advanced Physics: Electrostatics and Current Electricity

⇒



148.

4 3 π r × ρ × g = NeE 3

E θ



4π r 3 ρ g ⇒ N= 3 eE



Number of moles of the metal in sphere is n =



⇒ n=

T1

4 π r 3ρ 3 A

Total number of electrons =



So, required fraction =



Hence, the correct answer is (C).

4 π r 3 ρN A × Z = N0 3 A

gA N = N 0 eEZN A

 

q, m

T2 sin α

mg Figure 2



⇒ tan q =

mv 2 …(2) r

v2 …(3) rg

When an external field is applied such that the pendulum makes an angle q to the opposite side, then α = -q . From Figure 2, we get

° 30 2R

r⊥

T2 cos α = mg …(4) p

f

T2 sin α - qE =

qE





For system to be in equilibrium, τ = 0



Let us take the torque about the point P, so



qE [ 2R sin ( 30° ) ] = ⎡⎣ mg sin ( 30° ) ⎤⎦ R mg 2q



⇒ E=



Hence, the correct answer is (B).

147. Assume a charge q to be placed at the eight corner too. Then by symmetry, we have   E = 0     ⇒ E01 + E02 + ... + E08 = 0     ⇒ E01 + E02 + ... + E07 = E08 =

q = E0 = 3πε 0 2



⇒ Ecentre



Hence, the correct answer is (B).

01_Ch 1_Hints and Explanation_P2.indd 62

qE

T1 cos q = mg …(1) T1 sin q =

30°

in

T2

When no field is applied as in Figure 1. Then, we have,

qE



s mg



q, m

Figure 1

4 π r ρ × NA 3 A



146.

T1 sin θ

mg

So, number of metal atoms =

α = –θ

θ

m A

3



T2 cos α

T1 cos θ

q ⎛ 3 ⎞ 4πε 0 ⎜ ⎝ 2 ⎟⎠

⇒ T2 sin α =



⇒ -



⇒ qE = -



⇒ E=-



mv 2 + qE …(5) r

mv 2 + qE ⇒ tan α = - tan q = r mg v2 ⇒ - = rg

2

mv 2 r

mv 2 + qE r mg

mv 2 mv 2 = + qE r r 2mv 2 r

2mv 2  qr (negative sign implies opposite direction) 2mv 2 qr



⇒ E =E=



Hence, the correct answer is (C).

9/20/2019 11:36:41 AM

Hints and Explanations H.63



F = 10 × 10 -3 × 10 N = 10 -1 2

q Since F = 4πε 0 r 2



⇒ q = 4πε 0 Fr 2



⇒ q=



Hence, the correct answer is (D).

150.

E=



0.1 × 0.36 = 2 × 10 -6 C 9 × 109



⇒ T=



Hence, the correct answer is (D).

2

q2 + mω 2 16πε 0 2

Eeq

eρ a 3 mε 0

Acceleration =

ρea 3 mε 0

–q

2r

1 ⎛ ρea ⎞ 2 t = 2r 2 ⎜⎝ 3 mε 0 ⎟⎠



P +q

p

π 4 Hence, the correct answer is (B). ⇒ ϕ = tan -1 ( 0.5 ) +

154.



6 2 rε 0 m ρea



⇒ t=



Hence, the correct answer is (A).

151.

⎛ σ2 ⎞ F=⎜ A ⎝ 2ε 0 ⎟⎠

T QE mg

 Since a soap bubble has 2 free surfaces, so F = 2T ( 2π R )

σ2 ( 2 ) πR 2ε 0



⇒ 2 [ T ( 2π R ) ] =



⇒ T=



Hence, the correct answer is (A).

σ 2R 8ε 0

152.  

01_Ch 1_Hints and Explanation_P2.indd 63

Eaxial

⎛ tan 45 ⎞ ⇒ ϕ = tan -1 ⎜ +q ⎝ 2 ⎟⎠



T - Fe = mω 2

θ

O

θ

⎛ tan q ⎞ Angle ϕ = tan -1 ⎜ +q ⎝ 2 ⎟⎠

45° a

q, m

α

r

P



4πε 0 ( 2 )

E

⇒ F = eE =

Fe

+ mω 2

⇒ T=

153. Angle of resultant with x-axis will be α + q , where 1 tan α = tan q 2

ρa 3ε 0

Since s =

q2



CHAPTER 1

149.

T

l

ω

Since ( T + QE ) - mg =

mv 2 L mv 2 L



⇒ 10 mg + QE - mg =



⇒ 9mg +



QE ⎞ ⎛ ⇒ v = L ⎜ 9g + ⎟ ⎝ m ⎠



Hence, the correct answer is (C).

155.

F=



⇒ log e F = log e

QE mv 2 = m L

q1q2 4πε 0 r 2 q1q2 - 2 log e r 4πε 0

9/20/2019 11:36:55 AM

H.64  JEE Advanced Physics: Electrostatics and Current Electricity

⇒ y = c + mx

159.



⇒ m = -2





Hence, the correct answer is (A). p

(0, 0, d)

q

(a, 0, 0) p –q

x-axis

4πε 0 ( r 2 +  2 ) 2

where p = q ( 2d )



⇒ E=  ⇒ E= iˆ

1 4πε 0 1 4πε 0

)



Magnitude of torque is

τ = 8 2 × 10 -5 Nm Hence, the correct answer is (C).

160.

Σq = ϕnet = -8 × 10 3 + 4 × 10 3 = -4 × 10 3 ε0



⇒ ∑ qenc = -4000ε 0 coulomb



Hence, the correct answer is (D).

3

( a2 + d2 ) 2 2qd

q

( - kˆ ) 3

( a2 + d2 ) 2

∫

  q E ⋅ dS = enc , where qenc is the charge ε0

enclosed by the Gaussian-surface which, in the present case, is the surface of given sphere. As shown, length AB of the line lies inside the sphere.

ϕGaussian =

q ε0



q 1 ⇒ ϕ vessel = ϕGaussian = 2 2ε 0



Hence, the correct answer is (A).

162.

B

m

u = 2 ms–1

O

y-direction

5m

R

A

Final position R

( y < R)



In ΔOO ’ A R2 = y 2 + ( O ’ A )

2

Along y-direction Sy = uy t +

⇒ O ’ A = R2 - y 2

and AB = 2 R2 - y 2





Charge on length AB is qenc = 2 R2 - y 2 × λ



So, electric flux is ϕ =

∫ S

  2λ R 2 - y 2 E ⋅ dA = ε0

Hence, the correct answer is (C).

01_Ch 1_Hints and Explanation_P2.indd 64

x-direction E = 2 NC–1

O′ y



)

q × 2d

S



(

 ⇒ τ = ( 4 × 10 -6 ) ⎣⎡ iˆ - ˆj + kˆ × 20iˆ ⎤⎦ = 8 × 10 -5 kˆ + ˆj

161.

Hence, the correct answer is (C).

158. Electric flux

(





E

(0, 0, –d)

Since the point p ( a , 0 0) lies at the equitorial line of p dipole, so we have E = 3



 Here, 2 a = ( 2 - 1 ) iˆ + ( -1 - 0 ) ˆj + ( 5 - 4 ) kˆ = iˆ - ˆj + kˆ  E = 0.20iˆ Vcm -1 = 20iˆ Vm -1 iˆ

z-axis

157.   



     τ = p × E = q ( 2a ) × E

⇒ 5=0+

t =

1 2 ay t …(1) 2

1 × g × t2 2

10 =1s g

Along x-direction Sx = uxt +

1 2 axt …(2) 2

9/20/2019 11:37:11 AM

Hints and Explanations H.65

ϕ =

-9

1 2 × 10 × 2 × ×1 2 2 × 10 -9



⇒ R = 2 ×1+



⇒ R= 2+1= 3 m



Hence, the correct answer is (D).

163.

Earc



⇒ Etotal



Hence, the correct answer is (D).

B(q)

E

2E

E E E E

A(–q)

–q O

Hence, the correct answer is (A).

165.

O

cos q =

⎛ ⎜⎝

r 2



⇒ q = 60°



Length of ring in sphere is

2π r  = r × 2q = 3 So, charge enclosed in the sphere is qenc =

Q 2π r Q × = 2π r 3 3

01_Ch 1_Hints and Explanation_P2.indd 65

Enet

x

⎛ Q⎞ q ⎜ ⎟ dx ⎝ L⎠ 4πε 0 x 2

Qq 4πε 0 L

L 2

dx

∫x

r-

2

=-

L 2

Qq ⎛ 1 1 ⎞ 4πε 0 L ⎜ ⎛ L⎞ ⎛ L⎞ ⎟ ⎜⎝ r + ⎜⎝ ⎟⎠ r - ⎜⎝ ⎟⎠ ⎟⎠ 2 2

Qq Qq = ⎛ 2 L2 ⎞ πε 0 ( 4 r 2 - L2 ) 4πε 0 ⎜ r - ⎟ ⎝ 4⎠

⇒ F=



Hence, the correct answer is (B).

167. Effective distance in vacuum is given by

A

A

Q1Q2

(

4πε 0 t1 k1 + t2 k2 - t1 - t2 Q1Q2



⇒ F=



Hence, the correct answer is (D).

168.

4πε 0 ⎡⎣



(

)

k1 - 1 t1 +

σ

(

)

2

)

k2 - 1 t2 ⎤⎦

2

σ

E

C

r⎞ ⎟ 2⎠ = 1 r 2

dx



So, F =

2E

D(q)



2E

O

E

q

⇒ F=

q

r = k1 t1 + k2 t2 - ( t1 + t2 )

–q

q

r

r+



164. Since magnitude of each charges are same and situated at equal distance from centre O, so all charges will produce same magnitude of electric field at centre.

C(q)

dF =

6Q 2λ ⎛ 1 ⎞ λ , where λ = ⎜ ⎟= 2π R 4πε 0 R ⎝ 2 ⎠ 4πε 0 R

3Q 6λ = , where λ = 4πε 0 R πR

E E

Q



L

 Field due to a single arc subtending an angle of 60°



Q 3ε 0

Hence, the correct answer is (C).

166.

2λ ⎛q⎞ = sin ⎜ ⎟ ⎝ 2⎠ 4πε 0 R

at centre is E1 =

Hence, flux by Gauss Law is

CHAPTER 1

1 ⎛ qE ⎞ 2 ⇒ R= 2×t+ ⎜ ⎟t 2⎝ m ⎠



E

Electric field due to infinite sheet is independent of distance and hence electric field due to both sheets will be equal and opposite. Hence, the correct answer is (D). 170.

f ≥ Fe , where Fe =

q2 4πε 0 r 2

Since f = μ N = μmg

⇒ μmg ≥

q2 4πε 0 r 2

9/20/2019 11:37:24 AM

H.66  JEE Advanced Physics: Electrostatics and Current Electricity 176. q is inversely proportional to L Hence, the correct answer is (A).

q2 4πε 0 r 2 mg



⇒ μ≥



Hence, the correct answer is (B).

177.

E

    θ

T T cos θ θ T sin θ

171. Using Gauss Theorem, we have E =

1 Q for r ≤ R and 4πε 0 r 2

E = 0 for r ≥ R So, correct graph is (A) Hence, the correct answer is (A).

Fe = qE

mg



For equilibrium

T cos q = mg and

172.

( -10 °C ice )

( 0 °C water )

T sin q = qE



k1 q2   q1 r1 = 25 cm

k = 80 q2 q1 2 r2 = 5 cm



⇒ T = (mg )2 + (qE)2



⇒ T = (80 × 10 × 10 -6 )2 + (2 × 10 -8 × 2 × 10 4 )2



⇒ T = 64 × 10 -8 + 16 × 10 -8



⇒ T = 80 × 10 -8 = 8.8 × 10 -4 N

Given Fice = Fwater 1 q1q2 1 q1q2 = 2 4πε 0 k1r1 4πε 0 k2 r22



⇒



⇒ k1r12 = k2 r22



Hence, the correct answer is (C).



⎛r ⎞ ⎛ 5 ⎞ ⇒ k1 = k2 ⎜ 2 ⎟ = 80 × ⎜ ⎟ = 3.2 ⎝ 25 ⎠ ⎝ r1 ⎠

178.

ϕ=



Hence, the correct answer is (B).



Hence, the correct answer is (D).

173.

ϕnet = E0 ( x +  )  - E0 x = E0



Since, ∑ q = ( ϕnet ) ε 0



⇒ qenc = ε 0 E0 3

179. Since, positive charge diverts in the direction of field, so field must be upwards So, particle (2) deflects down, (3) up and particle (4) deflects down Hence, the correct answer is (A).  180. A = 100 kˆ   ⇒ ϕ = E ⋅ A = 300 units

2

2

2

2

3

E = E0 xi

x



Hence, the correct answer is (B).

181.

EP =

x+l



Hence, the correct answer is (B).

175. Since ϕ A + ϕC + ϕB =

q ε0

and ϕ A = ϕC

q ⇒ ϕA + ϕA + ϕ = ε0



1⎛ q ⎞ ⇒ ϕA = ⎜ -ϕ⎟ 2 ⎝ ε0 ⎠



Hence, the correct answer is (D).

01_Ch 1_Hints and Explanation_P2.indd 66

qenclosed λ  1 C/cm × 100 cm 100 C = = = ε0 ε0 ε0 ε0

q 4πε 0 r 2

EQ =

q 4πε 0 ( 2r )

2

+

3q 4πε 0 ( 2r )

2



⇒ EP : EQ = 1 : 1



Hence, the correct answer is (C).

=

q 4πε 0 r 2

182. For downward motion with constant velocity, we have q = 2e and m = 3.2 × 10 -17 kg  Since motion is downwards, so viscous force is upwards.

9/20/2019 11:37:42 AM

Hints and Explanations H.67

mg

6πηrv = mg …(1)

4T Q2 = r 16π 2 r 4 × 2ε 0



⇒



⇒ Q = 8π r 2rT ε 0



Hence, the correct answer is (B).

186.



σ

E

For upwards motion with same constant speed, we have 6πηrv + mg = qE …(2) qE

θ

T T cos θ θ q T sin θ

E

qE

mg

T sin q = qE …(1)

Fv

T cos q = mg …(2)

mg



Substitute (1) in (2), we get

qE = 2mg 2mg 2 × 3.2 × 10 -17 × 10 = = 2 × 10 3 Vm -1 q 2 × 1.6 × 10 -19

qE qσ = mg 2ε 0 mg



⇒ tan q =



⇒ σ=



Hence, the correct answer is (B).

2ε 0 mg tan q q



⇒ E=



Hence, the correct answer is (A).

183.

ϕcube =



q 4π q 1 = ⇒ ϕface = ϕcube = 6 6ε 0 24πε 0



  F1 =



Hence, the correct answer is (A).



Hence, the correct answer is (C).

tK

r

187.

q ε0

q1

184. The electric field due to Q at any point of square plate will be along the plane of square, therefore angle between electric field and area vector of plate is 90° and hence ϕ = 0 Hence, the correct answer is (D).

CHAPTER 1

Fv

q2

q1

med

q2

Kq1q2 qq q1q2 = 1 2 2 and F2 = 4πε 0 r 4πε 0 (r - t + t k )2 r2

Multiple Correct Choice Type Questions 1.

T cos q + qEy = mg ...(1)



T sin q = qEx ...(2)

4T greater than outside r pressure in bubble. This excess pressure is provided by charge on bubble. So, we have

185. Inside pressure must be

T

qEy θ

T cos θ qEx

T sin θ

Pout r

4T σ 2 = r 2ε 0

Q Since σ = 4π r 2

01_Ch 1_Hints and Explanation_P2.indd 67

mg Pin



⇒   T ( 0.6 ) = 3 × 10 5 q



⇒   q =



⇒   q = 2 × 10 -6 T

6 × 10 -5 T 30

9/20/2019 11:37:57 AM

H.68  JEE Advanced Physics: Electrostatics and Current Electricity



⇒

T ( 0.8 ) + ( 2 × 10 -6 T ) ( 5 × 10 5 ) =



⇒

T ( 0.8 ) + T =



⇒

T=



⇒

T = 5.55 × 10 -3 N

1 ( 10 ) 1000

10 1000

mv 2 ⎛ λ ⎞ = qE = q ⎜ r ⎝ 2πε 0 r ⎟⎠



⇒ v=



Hence, (A), (B) and (C) are correct.

14.

E = E0t

qλ 2πε 0 m

1

( 100 )( 1.8 )



⇒

q = 11.1 × 10



⇒

q  11 nC

-9

C

Hence, (A) and (C) are correct.     2. EA is along OA and EB is along OB, where   OA = iˆ + 2 ˆj + 3 kˆ and OB = iˆ + ˆj - kˆ   Since, we observe that OA ⋅ OB = 0   ⇒ EA ⊥ EB (dot product of perpendicular vectors is

zero)  1 Also, E ∝ 2 r   and OC = 2 OB   ⇒ EB = 4 EC Charge density at A < charge density at B. Since field inside cavity is zero, hence Potential at A = potential at B = a constant value. By Gauss Theorem,

⎛ Total electric ⎞ 1 ⎛ charge ⎞ ⎛ q ⎞ ⎜ ⎟⎠ = ⎜⎝ enclosed ⎟⎠ = ⎜ ⎟ flux ⎝ ε0 ⎝ ε0 ⎠

Hence, (C) and (D) are correct.

7.

Field lines must enter a surface or leave a surface normally.



Hence, option (D) is correct.

8.

t=



⇒ t=

qE qV 2d when a = = a m md 2md 2 qV

( 2 d )2 t =2 = T d2



⇒



⇒ t = 2T



Hence, option (D) is correct.

01_Ch 1_Hints and Explanation_P2.indd 68

dv = E0 st   ⇒   dv = E0 st dt dt

⇒



⇒ v = E0 s t dt =

t

∫ 0

⇒



⇒ dx = ⇒

1 E0 st 2 2

dx 1 = E0 st 2 dt 2



∫ 0

5.

qE qE0t = = E0 st m m





Hence, (A) and (C) are correct.

2d = qV md

Since a =

x





10.

1 E0 st 2 dt 2 t

1 dx = E0 s t 2 dt 2

∫ 0

1 E0 st 3 6



⇒ x=



Hence, (B) and (D) are correct.

16.

1 mv 2 = QV 2



⇒



⇒ p = 2mQV



2QV v m  ⇒ t= = a Q⎛V⎞ ⎜ ⎟ m⎝ d ⎠



⇒ t=



Hence, (B) and (D) are correct.

18.

a=

p2 = QV 2m

{∵ v = 0 + at }

2md 2 QV

qE = sE m

Since, L = ut

⇒ t=

L u

9/20/2019 11:38:18 AM

Hints and Explanations H.69 1 2 at 2



L2 1 ⇒ Δ = ( sE ) 2 2 u



⇒ Δ=

EsL2 2u2

at = tan q = vx u

Est EsL = 2 ⇒ tan q = u u



Hence, (A), (B), (C) and (D) are correct. 1 ⎛ qA qB ⎞ + ⎜ ⎟ = 2V …(1) 4πε 0 ⎝ R 2R ⎠

and,

1 ⎛ qA qB ⎞ 3 + ⎜ ⎟ = V …(2) 4πε 0 ⎝ 2R 2R ⎠ 2

Solving equations (1) and (2), we get



⇒ Q1 + Q3 = -Q2 …(2)



Solving equations (1) and (2), we get

Q1 = -

vy

20.



qA 1 = qB 2



Hence, (A), (B) and (C) are correct.

24.

a=



⇒ a=

qα qβ x m m



⇒ x =

qα qβ x m m

Let

{charge on A remains same}

q′A = -1 qB′



⇒



Also after earthing,

VA - VB =



qA ⎛ 1 qA 1 ⎞ ⎜ ⎟= 4πε 0 ⎝ R 2R ⎠ 8πε 0 R



⇒ x = -



⇒ -



 + qβ X = 0 ⇒ X m



 + ω 2 X = 0 ⇒ X





Hence, (A) and (D) are correct.

V 2

22. Potential of innermost shell is zero, so Q Q Q 1 + 2 + 3 = 0 r 2r 3 r

⇒ 6Q1 + 3Q2 + 2Q3 = 0 …(1)

 Similarly, potential of the outermost shell is also zero. So,

⇒

Q1 Q2 Q3 + + =0 3r 3r 3r

01_Ch 1_Hints and Explanation_P2.indd 69

m  X qβ

m  X=X qβ

At the mean position, no force acts on the particle

α β Equation (1) can be written as, ⇒ F = 0 at x = x0 =

v ⋅

V 2

Since B is earthed, so VB = 0 and hence VA =

qβ  x = X m

⇒ -



1 qA V = 4πε 0 2R 2

⇒ VA - VB =

qα qβ x=X m m

So, the motion of the particle is oscillatory and we α observe that a = 0 at x = β

Substituting qB = 2qA in equation (1), we get

F qE q = = ( α - β x ) …(1) m m m



When B is earthed, potential of B becomes zero. So, qB′ = - q′A = - qA 

Q2 Q3 Q 3 = 3 and 3 = , 4 Q1 Q2 4

CHAPTER 1

Also, Δ =

dv q = ( α - βx ) dx m

v



⇒

∫

v dv =

0



⇒ v=

q m

x

∫ ( α - βx ) dx 0

2qx ⎛ β ⎞ ⎜ α - x ⎟⎠ 2 m ⎝

v = 0 at x = 0 and x =

2α β

 So, the particle reverses its direction of motion at 2α x= (extreme position) and hence the particle will β 2α oscillate between x = 0 to x = with mean position β

9/20/2019 11:38:39 AM

H.70  JEE Advanced Physics: Electrostatics and Current Electricity α α . So, amplitude of particle is with maxiβ β mum acceleration of particle at extreme positions (at qα 2α x = 0 and x = and amax = {from equation (1)}. β m

at x =



Hence, (A), (B), (C) and (D) are correct.

force eE = mass m

28. Retardation a =

⇒ a=

1.6 × 10 -19 × 1000 = 1.78 × 1014 ms -2 9 × 10 -31

Since, v 2 - v02 = 2 as

26. The electron will experience a downward acceleration a , given by



⇒ 0 2 - ( 5 × 106 ) = 2 × ( -1.78 × 1014 ) s



⇒ s = 0.07 m

eE 1.6 × 10 -19 × 2000 a = = m 9 × 10 -31

Now, v = v0 + at



eE ⇒ a= = 3.6 × 1014 ms -2 m

 So, we can consider the electron to be a projectile launched with some initial velocity between the plates under influence of downward acceleration a. From the point of launch let us consider the rightward horizontal as the x-axis and upward vertical as y-axis, let us calculate the height to which the electron will rise and its horizontal range. Considering upward vertical motion, we have 0 = 6 × 106 sin 45°t +

1 ( × -3.5 × 1014 ) t 2 2



⇒ t = 2.4 × 10 -8 s



Considering motion along x-axis, we get



⇒ t = 2.8 × 10 -8 s = 0.03 μs



Loss of energy = work done W = Fd = ( eE ) d



⇒ Fraction f =



eEd 2eEd = 1 2 mv 2 mv 2 ⇒ %age Fraction is

T 2 The vertical height is given by

(

)(

) 12 ( 3.5 × 1014 ) ( 1.2 × 10 )

y = 6 × 106 sin 45° 1.2 × 10 -8 -



⇒ y = 0.0509 - 0.0252



⇒ y = 0.0257 = 2.57 cm

-8 2

So, the electron will not touch the upper plate and will strike the lower plate at the other edge. Hence, (B) and (C) are correct. Q2 1 mv 2 = 2 4πε 0 d Q2 2πε 0 mv 2



⇒ d=



Hence, (A), (B) and (D) are correct.

01_Ch 1_Hints and Explanation_P2.indd 70

9 × 10 -31 × ( 5 × 106 )

2

= 11%

Hence, (B), (C) and (D) are correct.

⇒ x = 0.102 m = 10.2 cm



2 × 1.6 × 10 -19 × 1000 × 0.8 × 10 -2

f =

V =

t =

27.

⇒ 0 = 5 × 106 - ( 1.78 × 1014 ) t

30. The potential due to a pair of rings each of radius R and carrying charges +q and -q at a distance x from the mid point along the axis is given by

The vertical displacement in half the total time i.e., at maximum height, we have







x ( range ) = 6 × 106 cos 45° × 2.4 × 10 -8

2

qx 32 4πε 0 ( R2 + x 2 )

Now, consider a pair of rings of radius z and thickness dz. If dq be the infinitesimal charge on each ring, then dq = 2π zσ dz and if dV be the potential due to this infinitesimal pair of elements at P

⇒ dV =



⇒ dV =



⇒ V=



⇒ V=

( 2π zσ dz ) x

4πε 0 ( z 2 + x 2 )

32

zdz σ x 2ε 0 ( z 2 + x 2 )3 2

σ x 2ε 0

∞

∫ (z R

zdz 2

+ x2 )

32

σ x

2ε 0 ( R 2 + x 2 )

Since, E = -

12

∂V ∂x

9/20/2019 11:39:00 AM

Hints and Explanations H.71 38.  

σ R 2



⇒ E=-



Hence, (A) and (B) are correct.

2ε 0 ( R 2 + x 2 )

Q1

32

Q2

31.

E ≠ 0, V = 0

E = 0, V = 0

Number of lines on Q1 is greater and number of lines is directly proportional to magnitude of charge. So, Q1 > Q2 Electric field will be zero to the right of Q2 as it has small magnitude and opposite sign to that of Q1 .

E = 0, V ≠ 0



E ≠ 0, V ≠ 0

Hence, (A), (B), (C) and (D) are correct.

32. Equating the forces in the vertical and horizontal directions T cos q = mg and T sin q = F = QE P θ

T T

T cos θ

QE 2 × 10 -8 × 2 × 10 4 Dividing, tan q = = mg 71 × 10 -6 × 9.8



⇒ q = 30°

40. Using Work Energy Theorem, we get 1 mv 2 = qE (  -  cos ( 60° ) ) 2 qE m



⇒ v=



So, OPTION (B) is correct. At point B, we have

QE

mg



Hence, (A) and (D) are correct.

T - qE =

T sin θ

1 ⇒ tan q = 0.57 ≅ 3



mv 2  2

m ⎛ qE ⎞ = 2qE  ⎜⎝ m ⎟⎠



⇒ T = qE +



So, OPTION (D) is also correct. Hence, (B) and (D) are correct.

41. The distribution of charges is shown in figure. We see that the inner surface of B and C possess the same charge -Q . Also if x is charge on outer surface of C, then x

71 × 10 -6 × 9.8 = 803 μN cos ( 30° ) Hence, (C) and (D) are correct.

Now, T =

34. The given graph is of charged conducting sphere of radius. The entire charge distributes on the surface of the sphere. Hence, (A), (B), (C) and (D) are correct. 36. Torque about Q of charge -q is zero, so angular momentum charge -q is constant, but distance between charges is changing, so force is charging, so speed and velocity are changing. Hence, option (A) is correct.

01_Ch 1_Hints and Explanation_P2.indd 71

CHAPTER 1

From the diagram, it can be observed that Q1 is positive, Q2 is negative.

+Q –Q

–Q Q

B

C

A c

VC = 0 Q ⎛ -Q + Q ⎞ -Q + x + =0 + 4πε 0c ⎜⎝ 4πε 0c ⎟⎠ 4πε 0c



⇒



⇒ x=0 Hence, (A) and (D) are correct.



9/20/2019 11:39:12 AM

H.72  JEE Advanced Physics: Electrostatics and Current Electricity

42.

 E=

E = 104 NC–1

Q2 ⎛ a ˆ ˆ⎞ ⎛ a ˆ ˆ⎞ - i + aj ⎟ ⎜⎝ i + aj ⎟⎠ + 3⎜ ⎝ 2 ⎠ 2 ⎛ ⎞ ⎛ ⎞ 5 5 4πε 0 ⎜ 2 4πε 0 ⎜ a ⎟ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ Q1

3

u = 10 ms–1

H

Since, x component is zero, so R

Q1 = Q2

Hence, (A) and (C) are correct.

43. Now electric field outside the sphere is due to charge on outer surface of sphere as E due Q and -Q (induced charge) gets cancelled. Also E due to conducting sphere for from it will be zero Hence, (B) and (D) are correct.

Fi ∝ q1q2

⇒ Fi ∝ Q 2



After touching, new charges on each ball is Q1 + Q2 2Q = =Q 2 2

Fnew ∝ Q

⇒ R=



Hence, (A), (B) and (C) are correct.



Hence, (A), (B) and (C) are correct.

51.    Einduced

2



If q1 ≠ q2 , then 2

q 2 q 2 2q q ⎛ q +q ⎞ =⎜ 1 2⎟ = 1 + 2 + 1 2 ⎝ 2 ⎠ 4 4 4



⇒ Fnew > Fi



Hence, (A) and (B) are correct.

46. As we are moving away from P towards sheet S spacing between electric lines of force is increasing. P

Inside the conducting sphere, we have Enet = 0

Since, Enet = Einduced - Edue to q ⇒ Edue to q =



Hence, (A) and (D) are correct.

52. At a distance h above the sheet E = Esheet + Eslab ⇒ E=

-σ ρD ρD - σ + = 2ε 0 2ε 0 2ε 0

At a distance h below the top surface of slab, we have

R S

⇒ ER < EQ

 Also, in the direction of electric field potential decreases.

⇒ VR < VQ



Hence, (A) and (C) are correct.

47. Time of flight ( t ) =

H=

2u 2 × 10 = = 2 sec. g 10

u2 10 × 10 = =5m 2 g 2 × 10

01_Ch 1_Hints and Explanation_P2.indd 72

kq = Einduced d2





Q



q

Eq

⇒ Fi = Fnew

Fnew

1 ⎛ 10 -3 × 10 4 × 2 × 2 ⎞ ⎜ ⎟⎠ = 10 m 2⎝ 2



50. Charge is distributed over the surface of conductor in such a way that net field due to the charges inside and outside the conductor is zero inside. Field only due to the charge q inside the conductor is non-zero

44. If q1 = q2 = Q , then



1 ⎛ qE ⎞ 2 R = 0 + ⎜ ⎟t 2⎝ m ⎠

Eslab =

ρ ( D - 2h ) 2ε 0 σ ρ(D - 2 h) + 2ε 0 2ε 0



⇒ E = Esheet + Eslab =



⇒ E=



At a distance h below the bottom surface of the slab

E =

σ + ρ(D - 2 h) 2ε 0 -σ ρD ρD - σ + = 2ε 0 2ε 0 2ε 0

Hence, (A), (B) and (D) are correct.

9/20/2019 11:39:28 AM

Hints and Explanations H.73

B

Q

A



–Q

3m

Q E = , when a ≤ r ≤ b 4πε 0 r 2 1 ⎛ Q Q⎞ V = ⎜ - ⎟ , where a ≤ r ≤ b 4πε 0 ⎝ r b ⎠

Potential of B is VB = 0



Potential of A is VA =



So, potential difference is VA - VB =

1 ⎛ Q Q⎞ ⎜ - ⎟ 4πε 0 ⎝ a b ⎠ Q ⎛ 1 1⎞ ⎜ - ⎟ 4πε 0 ⎝ a b ⎠

⇒ -

kq1 kq2 + =0 x 3+x



⇒ +

q2 16 q2 = x 3+x



⇒ 3 + x = 16 x



⇒ 3 = 15x



⇒ x=



For a point P lying between the charges Vat P = 0

1 = 0.2 m 5 3–x

r

⎛4 ⎞ ρ ⎜ πr3 ⎟ ⎝3 ⎠ = ε0

ρr 3ε 0



⇒ E=



Slope of E-r graph (inside) i.e. for r < R is



⇒ Slope ∝ ρ

ρ . 3ε 0

⇒



⇒ 16 x = 3 - x



3 meter 17 Hence, (B) and (C) are correct.

58.

qExmax =



⇒ xmax =



At Equilibrium, we have

⇒ x=

1 2 kxmax 2 2qE k

qE = kxeq

E

–q2

P

kq1 kq = 2 3-x x





R

qenc ε0

x

q1



E ( 4π r 2 ) =

P



Hence, (A), (C) and (D) are correct.

qenc

–q2

Vat P = 0

P

54.

x

q1

a

O



For a point P lying outside the two charges

CHAPTER 1

53.

⇒ xeq =

qE k

Amplitude of oscillation is equal to the displacement of block from the mean position, So, amplitude is

O



R

3m q1

Since

2qE qE qE = k k k

Hence, (A), (B) and (C) are correct.

59. When E0 makes an angle of 45° with x-axis, then

1m –q2

A =

Hence, (B) and (C) are correct.

56.   



r

y

E=0

kq2 kq1 = 16 12

(0, b)

B

E0

⇒ q1 = 16 q2

Now, potential can be zero both inside and outside the charges on the line joining then

01_Ch 1_Hints and Explanation_P2.indd 73

45° A

(a, 0)

x

9/20/2019 11:39:47 AM

H.74  JEE Advanced Physics: Electrostatics and Current Electricity

Reasoning Based Questions

 E iˆ E ˆj E = 0 + 0 2 2

1.

  Now, VA - VB = - E ⋅ BA



(



E ⎞ ⎛ E ⇒ VA - VB = - ⎜ 0 iˆ + 0 ˆj ⎟ ⋅ + aiˆ - bjˆ ⎝ 2 2 ⎠



⇒ VA - VB = -

)

E0 a E0b E0 (b - a) + = 2 2 2

2.

The rate of decrease of electric field is different in the 1 two cases. In case of a point charge, it decreases as 2 r but in the case of electric dipole it decreases more rap1 idly, as E = 2 . r



Hence, the correct answer is (D).

4.

V0 =

⇒ VA - VB = 0

If a < b

⇒ VA - VB > 0



⇒ VA > VB

If a > b

⇒ VA - VB < 0



⇒ VA < VB



Hence, (B), (C) and (D) are correct.

60.



θ

σ 2ε 0

Since, T0 cos q = mg T0 sin q = qE ⇒ tan q =

q ⎛ σ ⎞ mg ⎝⎜ 2ε 0 ⎟⎠

( mg )



⇒ T0 =



⇒ T0 > mg

2

+ ( qE )

Hence, (A) and (D) are correct.

61. Near positive charge net potential is positive and near negative charge potential is negative. Q1 is positive & Q2 is negative, at 1 potential is zero and this point is near to Q2 . So, magnitude of Q1 is more than magnitude of Q2 . Also, slope of V - r graph gives E , which is zero at 3. Hence, (A), (C) and (D) are correct.

01_Ch 1_Hints and Explanation_P2.indd 74

–

∑ Induced charge ⎞⎟ ⎠

r

– + +

+

–

E=0

+

–

O

r + +

–

–

– +

+

q

1 q +0 4πε 0 r



⇒ V0 =



Hence, potential is constant Hence, the correct answer is (C).

6.

Electric field due to line charge E =



Hence, the correct answer is (D).

7.

Electron being a negative charge, will move from lower to higher potential. Hence, the correct answer is (D).

2

i.e. the effective value of g is increased, hence time period T of oscillation decreases.

–

qE mg



⎛ 1 ⎜q + 4πε 0 ⎝ r

T0 T0 cos θ

T0 sin θ

E =

Force of interaction between two charge is independent of presence of other charge. Hence, the correct answer is (B).

3.

If a = b

Coulombic attraction exists even when one body is charged and the other is uncharged. Hence, the correct answer is (C).



λ ( sin α + sin β ) 4πε 0 r

10. Since field is zero for a charged conductor on the surface and inside it also, so the surface of a charged conductor is always equipotential. Also, ΔV = 0 for equi  potential surface. Hence E ⋅ d  = 0

∫



  ⇒ E ⊥ d . Hence both are true and Statement-2 is the correct explanation to Statement-1. Hence, the correct answer is (A).

12. Path of charged particle is parabolic. Hence, the correct answer is (C).

9/20/2019 11:40:03 AM

Hints and Explanations H.75

14. See theory part in detail to get the answer. Hence, the correct answer is (B). 15. Statement-1 is also practical experience based; so it is true. Statement-2 is also true but is not the correct explanation of Statement-1. Correct explanation is “there is increase in normal reaction when the object is pushed and there is decrease in normal reaction when object is pulled”. Hence, the correct answer is (B). q0 4πε 0 r 2 but electric field due to induced charge is towards q0 so that right and will have same magnitude 4πε 0 r 2 electric field inside the sphere is zero.

16. Electric field due to q0 is towards left and is

⎛ q 1 ⎜ = V0 = ⎜ 4πε 0 r 4πε 0 ⎝

17.   

⎞ ⎟ ⎟⎠

1 q  4πε 0 r



Hence, the correct answer is (A).

{

∵

∫ dq

inside

=0

}

19. In a hollow spherical shield, the charge is present only on its surface but charge is zero at every point inside the hollow sphere. Hence, the metallic shield in form of hollow shell may be built to block an electric field. Hence, the correct answer is (A). 20. Apply the concept of electric image. Hence, the correct answer is (A).

Linked Comprehension Type Questions 1.

The magnitude of the force is given by 1 e2 4πε 0 r 2

Now we can substitute our numerical values and find that the magnitude of the force between the proton and the electron in the hydrogen atom is Fe =

Hence, the correct answer is (D).

R

⇒ V0 =

r



induced



Fe = q0

∫ dq

CHAPTER 1

13. Inside the conductor, charged atoms are also present. Hence, the correct answer is (B).

( 9.0 × 109 Nm 2C -2 ) ( 1.6 × 10 -19 C )2 ( 5.3 × 10 -11 m )2

Fe = 8.2 × 10 -8 N q

O



At point O    Enet = Ein + Eq  Enet = 0

2.

The magnitude of the electric field due to the proton is given by 1 q ( 90 × 109 Nm 2 C -2 ) ( 1.6 × 10 -19 C ) = 4πε 0 r 2 ( 0.5 × 10 -10 m )2

E = 5.76 × 1011 NC -1

⇒ Ein + Eq



1 q ≠0 ⇒ Eq = 4πε 0 r 2



Hence, the correct answer is (D).

18.  

01_Ch 1_Hints and Explanation_P2.indd 75

Hence, the correct answer is (C).

E =



O



q



Hence, the correct answer is (B).

3.

The mass of the electron is me = 9.1 × 10 -31 kg and the mass of the proton is mp = 1.7 × 10 -27 kg . Thus, the ratio of the magnitudes of the electric and gravitational force is given by

⎛ 1 e2 ⎞ 1 2 ⎜⎝ 4πε r 2 ⎟⎠ 4πε e 0 0 = Ratio = ⎛ mp me ⎞ Gmp me ⎜⎝ G 2 ⎟⎠ r

9/20/2019 11:40:13 AM

H.76  JEE Advanced Physics: Electrostatics and Current Electricity

( 9 × 109 ) ( 1.6 × 10 -19 )2 Ratio = ( 6.67 × 10 -11 ) ( 1.7 × 10 -27 ) ( 9.1 × 10 -31 ) Ratio = 2.2 × 10 39 which is independent of r, the distance between the proton and the electron.

Hence, the correct answer is (C).

⎛4 ⎞ ⎝3 ⎠ where the oil drop is assumed to be a sphere of radius 4π r 3 r with volume V = 3

4. M = ρoilV = ρoil ⎜ π r 3 ⎟



⎛4 ⎞ ⇒ M = ρoil ⎜ π r 3 ⎟ ⎝3 ⎠ ⇒

3 ⎛ 4π ⎞ ( 1.64 × 10 -6 m ) M = 8.51 × 10 2 kgm -3 ⎜ ⎝ 3 ⎟⎠

(

)



⇒ M = 1.57 × 10 -14 kg



Hence, the correct answer is (A).

5.

The oil drop will be in static equilibrium when the gravitational force exactly balances the electrical force   ⇒ Fg + Fe = 0



Since the gravitational force points downward so, the electric force on the oil must be acting upwards. So, we get   0 = mg + qE

⇒ mg = - qE

( 1.57 × 10

-14

)

kg ( 9.80 ms 5

1.92 × 10 NC

-2

 Since the electron has charge e = 1.6 × 10 -19 C , the charge of the oil drop in units of e is



q 8.02 × 10 -19 C = =5 e 1.6 × 10 -19 C

Hence, the correct answer is (D).

Since the electron has a negative charge, q = - e , the force on the electron is    Fe = qE = - eE = ( - e ) ( - E ) ˆj = eE ˆj

6.

01_Ch 1_Hints and Explanation_P2.indd 76

The acceleration of the electron is  qE ˆ eE ˆ  qE =j= j a = m m m and its direction is upward.  The time of passage for the electron is given by t1 = . v0 The time t1 is not affected by the acceleration because v0 , the horizontal component of the velocity which determines the time is not affected by the field. Hence, the correct answer is (B).  8. The electron has an initial horizontal velocity, v0 = v0 iˆ . Since the acceleration of the electron is in the +y direction, only the y-component of the velocity changes. The velocity at a later time t1 is given by  Since v = vx iˆ + vy ˆj = v0 iˆ + ay t1 ˆj  ⎛ eE ⎞ v = v0 iˆ + ⎜ ⎟ t1 ˆj ⎝ m⎠  ⎛ eE ⎞ ⇒ v = v0 iˆ + ⎜ ⎟ ˆj ⎝ m⎠ Hence, the correct answer is (C). ⇒

9.

From the figure (in the question), we see that the elec tron travels a horizontal distance  in the time t1 = v0 and then emerges from the plates with a vertical displacement y1 . Then

)

-1

q = -8.03 × 10 -19 C

N =

7.



With the electrical field pointing downwards, we conclude that the charge on the oil drop must be negative. mg q = Ey q = -

 where the electric field is written as E = - Ejˆ , with Fy > 0 . The force on the electron is upward. Note that the motion of the electron is analogous to the motion of a mass that is thrown horizontally in a constant gravitational field. The mass follows a parabolic trajectory downward. Since the electron is negatively charged, the constant force on the electron is upward and the electron will be deflected upwards on a parabolic path. Hence, the correct answer is (D).

y1 =

2

eE 2 1 2 1 ⎛ eE ⎞ ⎛  ⎞ ay t1 = ⎜ ⎟ ⎜ ⎟ = 2 2 ⎝ m ⎠ ⎝ v0 ⎠ 2mv02

Hence, the correct answer is (C).

10. When the electron leaves the plates at time t1 , the electron makes an angle q with the horizontal given by ⎛ eE ⎞ ⎛  ⎞ ⎜⎝ ⎟⎠ ⎜ ⎟ m ⎝ v0 ⎠ eE tan q = = = vx v0 mv02 vy



Hence, the correct answer is (C).

11. After the electron leaves the plate, there is no longer any force on the electron so it travels in a straight path. The deflection y 2 is

9/20/2019 11:40:28 AM

Hints and Explanations H.77 eEL mv02





and the total deflection becomes

eE 2 eEL eE ⎛  ⎞ + = y = y1 + y 2 = ⎜ + L ⎟⎠ 2mv02 mv02 mv02 ⎝ 2

Hence, the correct answer is (B).

36 × 10 -9 Cm -2 σ 1 2. Since, E = = = 4.07 kNC -1 ε 0 8.85 × 10 -12 C2 Nm -2 and V = Ed So, the potential difference between the plates is

⇒ v0 =



Hence, the correct answer is (B).

18.

v=

v0 2

( U + K )initial = ( U + K )final Here U initial → 0 and K initial = qq0

1 mv 2 2 and K final = 0



where as U final =



0 + 2qq0 1 ⎛ v2 ⎞ ⇒ 0 + m⎜ 0 ⎟ = 2 ⎝ 4 ⎠ 4πε x 2 + a 2 0

V = Ed = ( 4.07 × 10 3 NC -1 ) ( 0.12 m ) V = 488 V

qq0 πε 0 ma



4πε 0 x 2 + a 2



Hence, the correct answer is (D).

13.

1 mv 2 = qV 2

qq0 1 ⎛ qq0 ⎞ m ⎜ = ⎟ 2 ⎝ 4πε 0 ma ⎠ 2πε x 2 + a 2 0



⇒ KE = ( 1.6 × 10 -19 ) ( 488 ) = 7.81 × 10 -17 J



⇒



Hence, the correct answer is (B).



⇒ x = 15 a



Hence, the correct answer is (D).

19.

0+0= -



⇒



⇒ v = v0 =



Hence, the correct answer is (C).

14. Since

1 mv 2 = qV 2 2qV m



⇒ v=



⇒ v = 306 ms -1



Hence, the correct answer is (C).

15. Since v 2 - u2 = 2 a ( Δx )

( 3.06 × 105 )2 = 2a ( 0.12 )



⇒



⇒ a = 3.9 × 1011 ms -2



Hence, the correct answer is (A).



(

) ( 3.90 × 1011 ms-2 )

Hence, the correct answer is (A).

2qq0 1 + 0 = mv02 + 0 ⇒ 4πε 0 a 2

(0, 0)

q0 q

01_Ch 1_Hints and Explanation_P2.indd 77

qq0 1 mv 2 = 2 2πε 0 a qq0 πε 0 ma

GPEi + EPEi = GPE f + KE f + EPE f



x

Q1Q2 m v2 Q Q = m1 gh1 + 1 + 1 2 4πε 0 h 2 4πε 0 h1

On solving for v , we get

v = 2 g ( h - h1 )

q v0

2qq0 1 + mv 2 4πε 0 a 2

m1 gh +

17. For q0 , we have ( U + K )at 0 = ( U + K )at ∞

x 2 + a2 = 4 a

20. The forces of interaction, the force of gravity and Coulomb force are conservative, so we can apply the Law of Conservation of Energy. We have

16. Since, we know that F = ma -27 kg ⇒ F = 1.67 × 10 -16 F = 6.51 × 10 N

CHAPTER 1

y 2 = L tan q =



Assuming that h1  h , equation (1) reduces to

v = 2 gh

2Q1Q2 ( h - h1 ) …(1) 4πε 0 mhh1

2Q1Q2 Q1Q2 = 2 gh …(2) 4πε 0 m1h1 2πε 0 m1h1

Hence, the correct answer is (A).

9/20/2019 11:40:43 AM

H.78  JEE Advanced Physics: Electrostatics and Current Electricity 21. Substituting v = 0 in (2), and replacing h1 by h2, we get



4πε 0 gm1hh2 Q2 = Q1





QQ i.e., 1 22 = m1 g 4πε 0 h3 ⇒ h3 =

Q1Q2 4πε 0 m1 g

If the object is displaced upward, the force of gravity will restore it to equilibrium position and if displacement is downward. Coulombic force will be restoring. Thus the motion of the object will be periodic and oscillatory. Motion will be simple harmonic for very small displacement from equilibrium position. Hence, the correct answer is (C). 23. Let the charges on A and C be qA and qC respectively . From conservation of charge, we have qA + qC = 0 Hence qA = - qC Since A and C are connected by a conducting wire, so they have same potential. So, potential of A and C is given by VA =

1 qA 1 Q 1 qC + + 4πε 0 R 4πε 0 2R 4πε 0 4 R

1 qA 1 Q 1 qC + + VC = 4πε 0 4 R 4πε 0 4 R 4πε 0 4 R 4R

qc

R 2R

A B C



Equalising the potentials of A and C , we get

qA Q qC qA Q qC + + = + + R 2R 4 R 4 R 4 R 4 R

01_Ch 1_Hints and Explanation_P2.indd 78

Q 3

Hence, the correct answer is (D).

24.

VA =



Hence, the correct answer is (C).

25.

VB =

Q 1 ⎡ Q Q Q⎤ - + + = 4πε 0 R ⎢⎣ 3 2 12 ⎥⎦ 16πε 0 R

1 qA 1 Q 1 qC + + 4πε 0 2R 4πε 0 2R 4πε 0 4 R Q⎤ 1 ⎡ Q 5Q - +Q+ ⎥ = 8πε 0 R ⎣⎢ 3 6 ⎦ 48πε 0 R

VB =

Hence, the correct answer is (D).

26.

C λ ⎡C 1 ⎤ [ α ] = ⎡⎢ ⎤⎥ = ⎢ ⎛⎜ ⎞⎟ ⎥ = ⎡⎢ 2 ⎤⎥ = M 0 L-2TA ⎣ x ⎦ ⎣ m⎝ m⎠ ⎦ ⎣ m ⎦



Hence, the correct answer is (B).

27.

V=

1 4πε 0

V =

∫

dq 1 = r 4πε 0

∫

L

x dx λ dx 1 α = d+x r 4πε 0

∫ 0

α L⎞ ⎤ 1 ⎡ ⎛ α ⎢ L - d log e ⎜ 1 + ⎟ ⎥ = [ L - log e 5 ] ⎝ ⎠ d ⎦ 4πε 0 4πε 0 ⎣



Hence, the correct answer is (A).

28.

V=

1

∫ 4πε

0

dq 1 = r 4πε 0

Consider z =

Then x =

α V = 4πε 0 V =

∫

α x dx ⎛L ⎞ b2 + ⎜ - x ⎟ ⎝2 ⎠

2

L -x 2

L - z , and dx = - dz 2

VA = VC

Q 3





Q qA

qA = -



Hence, qC =

Hence, the correct answer is (B).

22. The object m1 will be in equilibrium where gravitational and Coulombic force balance,



⇒ 4 qA + 2Q = qA + Q

⇒ V=-

∫

αL 8πε 0

⎛L ⎞ ⎜⎝ - z ⎟⎠ ( - dz ) 2 b2 + z2

∫

dz 2

b +z

(

2

+

α 4πε 0

αL log e z + z 2 + b 2 8πε 0

∫ )+

zdz b2 + z2

α z2 + b2 4πε 0

9/20/2019 11:40:54 AM

Hints and Explanations H.79 L

2

α ⎛L ⎞ 2 ⎜ - x ⎟⎠ + b 4pε 0 ⎝ 2





2 ⎡L L ⎢ - L + ⎛⎜ ⎞⎟ + b 2 ⎝ 2⎠ αL 2 ⇒ V=log e ⎢⎢ 2 8pε 0 ⎢ L + ⎛⎜ L ⎞⎟ + b 2 ⎝ 2⎠ ⎢⎣ 2

⎡ ⎛ ⎢ b2 + ⎜ ⎝ αL ⇒ V=log e ⎢⎢ 8pε 0 ⎢ b 2 + ⎛⎜ ⎝ ⎢⎣ But at b =

2

L2 ⎞ L ⎤ ⎟ - ⎥ 4⎠ 2⎥ ⎥ L2 ⎞ L ⎥ ⎟⎠ + ⎥ 4 2⎦

3L , we get 2

⎡ ⎛ L⎞ ⎢ ⎜⎝ ⎟⎠ αL V = log e ⎢ 2 8pε 0 ⎢ ⎛⎜ 3 L ⎞⎟ ⎢⎣ ⎝ 2 ⎠

⇒ Ex = -



In cd

V = constant = 5 V

⇒ E=0



In de



⇒ V = 5x - 5



⇒ Ex = -



Hence, the correct answer is (D).

⎤ ⎥ αL ⎛ 1⎞ log e ⎜ ⎟ ⎥=⎝ 3⎠ 8 pε ⎥ 0 ⎥⎦

V + 10 = 25 ( x + 3 )

⇒ V = 25x + 65



So, for V = 0 , we get





Hence, the correct answer is (C).

65 13 =m 25 5

Hence, the correct answer is (C).

34. Since, work done by electric forces on a charge particle is the gain in K.E. E

αL log e ( 3 ) 8pε 0

⇒ V=

dV = -5 V dx

33. In the region ab ,

x = -



dV 10 = Vm -1 dx 3

V - 10 = 5 ( x - 3 )

⎤ α ⎡ ⎛L L ⎢ ⎜ - L ⎞⎟ + b 2 - ⎛⎜ ⎞⎟ + b 2 ⎥ ⎝ ⎠ ⎝ ⎠ ⎥⎦ 4pε 0 ⎢⎣ 2 2





0

⎤ ⎥ ⎥+ ⎥ ⎥ ⎥⎦

2



L



CHAPTER 1



2 ⎤ ⎡ L αL ⎛ ⎞ ⎛L ⎞ ⇒ V=log e ⎢ ⎜ - x ⎟ + ⎜ - x ⎟ + b 2 ⎥ + ⎠ ⎝2 ⎠ ⎥⎦ 0 ⎢⎣ ⎝ 2 8pε 0

q = 3 × 10–9 C

29. The correct answer is (B). 30. The correct answer is (A).



Work done = DK.E.

31. The correct answer is (D).



Thus work done = 4.5 × 10 -5 J



Hence, the correct answer is (B).

32. The correct answer is (C). Combined solution to 29, 30, 31, 32 In ab V + 10 = 25 ( x + 3 )

⇒ V = 25x + 65 …(1)



⇒ Ex = -



⇒ Ex = 25



In bc

dV = -25 Vm -1 dx

V - 15 =

10 (x + 2) 3

10 25 ⇒ V = - x+ 3 3

01_Ch 1_Hints and Explanation_P3.indd 79

35. Work done is also equal to qE

⇒ 4.5 × 10 -5 = 3 × 10 -9 × 5 × 10 -2 × E



⇒ E = 3 × 10 5 NC -1



Hence, the correct answer is (A).

36. By energy conservation q ( V1 - V2 ) =

1 mv 2 2

1 mv 2 2 ( V1 - V2 ) = q ( V1 - V2 ) =

4.5 × 10 -5 3 × 10 -9

9/20/2019 11:32:50 AM

H.80  JEE Advanced Physics: Electrostatics and Current Electricity 39. Let us review the formula of uniformly (volume) charged solid sphere according to which we have ρr E= 3ε 0

( V1 - V2 ) = 1.5 × 10 4 V

Hence, the correct answer is (D).

37. Electric field at r = R is E =

Q 4pε 0 R2

E

ρ (r)

d

a

R

r



Q = Total charge within the nucleus = Ze



1 Ze ⇒ E= 4pε 0 R2



Hence the electric field is independent of a. Hence, the correct answer is (A).

38.



⇒ E ∝ r , ρ should be constant throughout the ­volume of nucleus.



This will be possible only when a = R .



Hence, the correct answer is (C). R 2

40.

3

4 ⎛ R⎞ pα R3 Q1 = ρ ( r ) 4p r 2 dr = α p ⎜ ⎟ = ⎝ ⎠ 3 2 6

∫ 0

R

∫

2

Q = ρr 4p r dr for a = 0 , and the figure for the distri-

∫

Q2 = ρ ( r ) 4p r 2 dr R 2

bution looks like as shown ρ

R



d

R

r



d ρ Slope = = r R R-r



⇒ Q=

∫ 0

⇒ Q=

⎞ ⎛ 4p d 4p d ⎛ R 4 R 4 ⎞ ⎜ R r 2 dr - r 3 dr ⎟ = ⎜ ⎟= 4 ⎠ R ⎜ R ⎝ 3 ⎟⎠ ⎝ 0 0

⇒

p dR3 3



Q=

∫

R

∫

p dR3 ⇒ Q = Ze =  3

4 ⎛ R3 ⎞ ⇒ Q2 = 2α p ⎜ R3 ⎟ 3 ⎝ 8 ⎠ 7αp R3 15αp R3 11 = pα R3 3 8 24 So, total charge is ⇒ Q2 =

3 Ze p R3 Hence, the correct answer is (B).

{∵ Q = Ze }



1 11 ⇒ Q = pα R3 + pα R3 6 24



⇒ Q=



⇒



Hence, the correct answer is (A).

15 pα R3 24

Q1 4 = Q 15

41. Since F = ma = qE where q = - e and E =

⎛ - eα ⎞ ⇒ a=⎜ r ⎝ 3ε 0 m ⎟⎠



⎛ eα ⎞ r+⎜ r=0 ⇒  ⎝ 3ε 0 m ⎟⎠

⇒ d=

01_Ch 1_Hints and Explanation_P3.indd 80

∫

Q = Q1 + Q2

d ( R - r ) 4p r 2 dr R R





d ⇒ ρr = ( R - r ) R R

r⎞ ⎛ ⇒ Q2 = 2α ⎜ 1 - ⎟ 4p r 2 dr ⎝ R⎠ R 2

ρr



r

a=R

ρr αr = 3ε 0 3ε 0

9/20/2019 11:33:00 AM

Hints and Explanations H.81 eα 3ε 0 m

⇒ ω=



⇒ T = 2p



⇒ α=



Substituting the value of α , we get

3ε 0 m 15 , where Q = pα R3 eα 24

8 5Qp R3



⇒ E=-

2α r ⎛ 3r ⎞ α R3 + ⎜1⎟ 4R ⎠ 96ε 0 r 2 3ε 0 ⎝



Hence, the correct answer is (D).

Hence, the correct answer is (D).

44.

∫

  Q E ⋅ dA = 1 + ε0

⇒ E ( 4p r 2 ) =

⇒ E ( 4p r 2 ) =

r

2α ⎛ r⎞ 2 ⎜⎝ 1 - ⎟⎠ 4p r dr R 0

∫ε R 2

pα R3 + 6ε 0

r

∫ R 2

2α ⎛ r⎞ 2 ⎜ 1 - ⎟⎠ 4p r dr R ε0 ⎝ r

pα R 8pα ⎛ 2 r ⎞ + ⎜ r - ⎟⎠ dr 6ε 0 ε0 ⎝ R 3

3

∫





⇒ E ( 4p r 2 ) =



⇒ E ( 4p r 2 ) =



pα R3 8pα ⎡ 3 ⎛ 1 r ⎞ R3 ⎛ 1 1 ⎞ ⎤ + ⇒ E ( 4p r 2 ) = ⎟+ ⎜- + ⎟⎥ ⎢r ⎜ ε 0 ⎣ ⎝ 3 4R ⎠ 8 ⎝ 3 8 ⎠ ⎦ 6ε 0



⇒ E ( 4p r 2 ) =

pα R3 8pα ⎡ 3 ⎛ 1 r ⎞ -5R3 ⎤ + ⎟+ ⎢r ⎜ ⎥ 6ε 0 ε 0 ⎣ ⎝ 3 4 R ⎠ 8 ( 24 ) ⎦



⇒ E ( 4p r 2 ) =

pα R3 8pα ⎛ 1 r ⎞ 3 5pα R3 + ⎜ ⎟r 6ε 0 ε 0 ⎝ 3 4R ⎠ 24ε 0



⇒ E ( 4p r 2 ) =

pα R3 ⎛ ⎜16ε 0 ⎝

3

r R 2

⎤ r ⎥ 4R R ⎥ 2 ⎦ 4 r

pα R3 8pα ⎡ 1 ⎛ 3 R3 ⎞ 1 ⎛ 4 R 4 ⎞ ⎤ + ⎢ ⎜r ⎜r ⎟⎥ ⎟16 ⎠ ⎦ 6ε 0 ε0 ⎣ 3 ⎝ 8 ⎠ 4R ⎝ pα R3 8pα ⎡ r 3 r 4 R3 R3 ⎤ + + ⎢ ⎥ 6ε 0 ε 0 ⎣ 3 4 R 24 64 ⎦

⇒

pα R E ( 4p r 2 ) = -

3

⇒

pα R E ( 4p r 2 ) = -

3

qb

Now E outside the sphere is due to ( qa + qb ) because E due to qa , qb and -qa and -qb on the inner surface of cavity gets cancelled. So, we have E =

qa + qb 4pε 0 r02

Hence, the correct answer is (C).

45. Due to presence of charge on outside of sphere nothing will happen to the field inside cavity. So, E will remain same. Hence, the correct answer is (C). 46. Correct answer is (B). Since ultraviolet light ejects electrons from zinc so the ball becomes positively charged. This is known as photoelectric effect. Hence, the correct answer is (B). 47. Let n be the number of electrons lost, so that the ball acquires a positive charge given by q = ne .  So, ball experiences a force qE along horizontal direction.

The situation is shown in following figure. For equilibrium,

ΣFx = 0 θ

5 ⎞ 8pα ⎛ 1 r ⎞ 3 ⎟⎠ + ⎜⎝ ⎟r 4 ε 0 3 4R ⎠

8pα ⎛ 1 r ⎞ 3 + ⎜⎝ ⎟r 24ε 0 ε 0 3 4R ⎠

01_Ch 1_Hints and Explanation_P3.indd 81

qa + qb

qa

pα R 8pα ⎡ r 3 ⎢ + ⇒ E ( 4p r 2 ) = 6ε 0 ε0 ⎢ 3 ⎣







R 2



2α ⎛ 3r ⎞ α R3 + ⎜⎝ 1 ⎟r 2 3 4 R⎠ ε 96ε 0 r 0

Hence, the correct answer is (B).

42. Since,



⇒ E=-

43. In cavity there is no electric field, so no force acts on charge qa .

15pε 0 mR3 T = 2p 8Qe



CHAPTER 1



8pα ⎛ 3r ⎞ 3 + ⎜⎝ 1 ⎟r 24ε 0 3ε 0 4R ⎠

y

T θ qE

x

mg



⇒ qE - T sin θ = 0 …(1)

ΣFy = 0

9/20/2019 11:33:19 AM

H.82  JEE Advanced Physics: Electrostatics and Current Electricity



⇒ T cos θ - mg = 0 …(2) From (1) and (2), we get



qE = tan θ mg

mg tan θ E Now by Charge Quantisation, are have q = ne . So ⇒ q=

mg tan θ ne = E mg tan θ ⇒ n= eE

⎛ θ ⎞ qE ⇒ tan ⎜ ⎟ = ⎝ 2 ⎠ mg



⎛ qE ⎞ ⇒ θ = 2 tan -1 ⎜ ⎝ mg ⎟⎠



Hence, the correct answer is (C).

∫

θ

l

l cos θ

l sin θ l – l cos θ

B

r⎞ ⎛ ⇒ Q = 4pρ0 ⎜ 1 - ⎟ r 2 dr ⎝ R⎠

∫ 0



⎛ R3 R 4 ⎞ ⎛ R3 ⎞ ⇒ Q = 4pρ0 ⎜ = 4pρ0 ⎜ ⎟ ⎝ 12 ⎟⎠ ⎝ 3 4R ⎠

 To find the electric field outside the ball, the total charge Q is considered to be concentrated at the centre of ball. Hence, electric field at a distance r from the centre of ball is

v=0

Circular path



E



v=0

By Modified Work-Energy Theorem, we have

Between the two positions A and B , total change in KE is zero i.e. DK = 0 ⇒ Wext = DU ⇒ qE (  sin θ ) = mg ( 1 - cos θ )

⎛θ⎞ Since 1 - cos θ = 2 sin 2 ⎜ ⎟ ⎝ 2⎠

⎛θ⎞ ⇒ mg × 2 sin 2 ⎜ ⎟ = qE sin θ ⎝ 2⎠

⎛θ⎞ ⎛θ⎞ Since sin θ = 2 sin ⎜ ⎟ cos ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

⎡ ⎛θ⎞ ⎛θ⎞ ⎛θ⎞⎤ ⇒ mg × 2 × sin 2 ⎜ ⎟ = qE ⎢ 2 sin ⎜ ⎟ cos ⎜ ⎟ ⎥ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎦ ⎣

01_Ch 1_Hints and Explanation_P3.indd 82

⎛ R3 ⎞ 1 1 Q 1 pρ = 4 ⎜⎝ ⎟× 0 4pε r 2 4pε 12 ⎠ r 2

ρ0 R 3 12ε r 2 Hence, the correct answer is (B). ⇒ E=

50. For E to be maximum (which is possible inside), we have

dE =0 dr

where E = Einside =

Also T acts along radial direction, so WTension = 0

∫

R



E =

Wext = DU + DK





Since Q = ρdV = ρ ( 4p r 2 dr )

48. As soon as a charge ( q ) appears on the ball it starts experiencing a horizontal force ( qE ) due to presence of electric field. Due to the constraint of string,it will start moving on a circular path in a vertical plane. At maximum deflection of the string, velocity of the ball reduces to zero as shown in figure.



⎛θ⎞ ⎛θ⎞ ⇒ mg sin ⎜ ⎟ = qE cos ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

49. For calculating the electric field outside the ball, we should calculate first the total charge present in the ball.

Hence, the correct answer is (C).

A





⇒

Qinside ρ0 ⎛ r r 2 ⎞ = ⎜ ⎟ ε ⎝ 3 4R ⎠ 4pε r 2

ρ0 ⎛ 1 2 r ⎞ ⎜ ⎟ =0 ε ⎝ 3 4R ⎠



2R 3 Hence, the correct answer is (C).

51.

Em =





⇒ rm =

ρ0 ⎛ rm rm2 ⎞ ρ0 ⎛ 2R 4 R2 ⎞ ⎜ ⎟ ⎜ ⎟= ε ⎝ 3 4 R ⎠ ε ⎝ 9 36 R ⎠

ρ0 R 9ε Hence, the correct answer is (A). ⇒ Em =

9/20/2019 11:33:38 AM

Hints and Explanations H.83 52. Let the speed of charges A and B be vA and vA when the separation between them is  . Then by Conservation of Linear Momentum, we have

So, the direction of electric field at P should be in x-y plane and normal to y-axis . Hence direction of electric field is along positive x direction.

- mvA + ( 2m ) vB = 0





⇒ vA = 2vB



By Law of Conservation of Energy, we have

U i + K i = U f + K f

57. Consider two small elements of ring having charge +dq and -dq as shown in figure. The pair constitutes a dipole of dipole moment dp given by y

2



⇒

2q 2q 1 1 +0+0= + mvA2 + ( 2m ) vB2 4pε 0 4pε 0 ( 2 ) 2 2



⇒

2q 2 ⎛ 1 1 1⎞ mvA2 + 2mvB2 = ⎜ 1 - ⎟⎠ 2 2 4pε 0 ⎝ 2



q2 1 1 ⎛v ⎞ ⇒ mvA2 + ( 2m ) ⎜ A ⎟ = ⎝ 2 ⎠ 4pε 0 2 2



Solving, we get the speed of charge A as

–dq

θ

vA =

Hence, the correct answer is (D).

53. The work done by electrostatic force on charge A, from work energy theorem, in the given duration equals the change in kinetic energy of A . So, q2 1 W = DK A = mvA2 = 2 6pε 0

Hence, the correct answer is (B).

dθ

+λ

2

q2 3pε 0 m

θ

CHAPTER 1

2

Hence, the correct answer is (A).

x –λ

+dq

dp = ( dq ) ( 2R ) Since dq = λ d = λ ( Rdθ )

⇒ dp = λ ( Rdθ ) 2R

By symmetry the resultant dipole moment is along negative x-direction .

∫

∫

So, px = dpx = dp cos θ p 2



∫ cosθdθ

⇒ px = 2 λ R 2

54. The net work done by electrostatic force on system of two charged particle is equal to decrease in electrostatic potential energy of the system. So,

-

p 2

y

We = U i - U f ⇒ We =



q2 ⇒ We = 4pε 0



dp cos θ θ

2q 2 ⎛ 1⎞ ⎜ 1 - ⎟⎠ 4pε 0 ⎝ 2



Hence, the correct answer is (C).

55. By symmetry, potential at any point at y-axis is zero.

R⎞ ⎛ Hence potential at P ⎜ 0 , ⎟ is zero. ⎝ 2⎠



Hence, the correct answer is (B).

56. Since all charge lies in x -y plane, hence direction of electric field at point P should be in x -y plane. Also y-axis is an equipotential (zero potential) line. Hence direction of electric field at all points on y-axis should be normal to y-axis .

01_Ch 1_Hints and Explanation_P3.indd 83

dp sin θ

dp



⇒ px = 2λ R2 sin θ



⇒ px = 4 λ R 2

∫

x

p 2 p 2

∫

Similarly py = dpy = dp sin θ p 2



⇒ p y = 2λ R 2

∫ sinθdθ = 0

-



p 2

Hence, the correct answer is (C).

9/20/2019 11:33:55 AM

H.84  JEE Advanced Physics: Electrostatics and Current Electricity 58.

+q –q

R

a b

σ R =

q 4p R2

σ a =

-q 4p a 2

q

q σ b = 4p b 2

Hence, the correct answer is (B).

59.

Vcentre =



Hence, the correct answer is (D).

1 ⎛ q q q⎞ ⎜ + - ⎟ 4pε 0 ⎝ b R a ⎠

60. On earthing the outer surface, we have Vouter surface = 0 If x be the charge on outer surface, then v =

1 (q - q + x) = 0 4pε 0 b



⇒ x=0



⇒ Vcentre =



q ⎛ 1 1⎞ ⎜ - ⎟ 4pε 0 ⎝ R a ⎠

Hence, the correct answer is (A).

C → (p, q) D → (p, r, s, t) When Q = + q is displaced along +x axis, then before being displaced, force on Q is zero. However after being displaced along x-axis (say towards B ), then the repulsive force due to B is stronger on Q than the repulsive force due to A . So, Q has a tendency to go back to O (i.e. the mean position) and hence equilibrium is stable i.e. U is minimum. On the similar approach, the remaining answers can be given. 4. A → (q, r, s) B → (p) C → (p, q, r, s) D → (p, q, r, s)  Electric field is zero inside a charged conducting sphere or a charged conducting shell or a charged non-conducting shell. However for all of these, electric field outside them is inversely proportional to r 2 i.e, 1 Eoutside ∝ 2 . For the charged non-conducting sphere r ρr Einside = , Ecentre = 0 and E is maximum at the 3ε 0 surface. 5. A → (q, r) B → (s) C → (t) D → (p) E1 =

Matrix Match/Column Match Type Questions 1. A → (p, q, r, s) B → (p, q, s) C → (p, q, s) D → (p, q, r, s, t)  Electrostatic force is conservative in nature, obeys Newton’s Third Law (i.e. action-reaction force), depends upon the nature of medium between the interacting particles, obeys Principle of Superposition, attractive for unlike charges and repulsive for like charges. So, A → (p, q, r, s) Similarly, gravitational force is conservative in nature, obeys Newton’s Third Law (i.e. action-reaction force), does not depend upon the nature of medium between the interacting particles, obeys Principle of Superposition and is always attractive in nature. So, B → (p, q, s), C → (p , q, s) and D → (p, q, r, s, t) 2. A → (p, r, s, t) B → (p, q)

01_Ch 1_Hints and Explanation_P3.indd 84



λ Rx 3

2ε 0 ( R2 + x 2 ) 2

E2 =

x σ ⎛ ⎞ 12 2 ⎟ 2ε 0 ⎜⎝ R +x ⎠

E3 =

λ 2pε 0 x

E4 =

σ 2ε 0

6. A → (s, t) B → (p, q, r) C → (q, r) D → (p) E = 0 inside the body of the conductor. So E = 0 in the region between inner and outer surface of shell (both small as well as big). So, A → (s, t). For the space between the shells and outside the shells, E ≠ 0 . So, B → (p, q, r). Charge is 3Q on outersurface of inner shell and -3Q on inner surface of outershell. So, magnitude of charge on these surfaces is 3Q . So, C → (q, r).

9/20/2019 11:34:08 AM

Hints and Explanations H.85

7. A → (r, s) B → (p, r, s) C → (q) D → (t) For dipole, Vcentre = 0 and Vaxial = 0. So, A → (r, s)  E is parallel to p (i.e. dipole moment) and Eequitorial  axial    is anti-parallel to p . So, at C , E is along p and at D,   E is opposite to p . So, B → (p, r, s) and C → (q). For r  a , Eaxial = 2 ( Eequitorial ) . Hence EC > ED . So D → (t)

8. A → (q, s) B → (p) C → (t) D → (r) Since E and V are field and potential at centre of square having charges placed at the corners, so For (p), E = 0 , V °≠ 0 For (q), E ≠ 0 , V ≠ 0 For (r), E ≠ 0 , V = 0 For (s), E ≠ 0 , V ≠ 0 For (t) E = 0 , V = 0 9. A → (s, t) B → (q) C → (p) D → (r, s)

1 r3 For a uniformly charged ring of radius R , field at a Q point P on axis (at a distance r  R ) is E ≈ 4pε 0 r 2 1 ⇒ E∝ 2 r

For a small dipole, E ∝

For line charge, charged cylinder or charged conduct1 ing cylinder E ∝ (outside them) r σ ,  For an infinite charged conducting plate, E = 2ε 0 independent of r So, E ∝

1 r0

10. A → (q, t) B → (s) C → (p) D → (r)

⎛ 1 -3 2⎞ Jm is the unit of energy density ⎜⎝ = 2 ε 0E ⎟⎠ and hence electrostatic pressure ( = Pe ) . So, A → (q, t)

01_Ch 1_Hints and Explanation_P3.indd 85



V will have SI unit of metre. So, B → (s) E

Since U =

QQ Q1Q2 , so ε 0 = 1 2 . Hence C 2 J -1m -1 is 4pε 0 r 4pε 0Ur

the unit of ε 0 . So, C → (p) The unit of capacitance is farad. Since energy stored 1 ( U ) in the capacitor is U = CV 2 , so farad represents 2 U . So, D → (r) V2 11. A → (r) B → (s) C → (p) D → (p) Between the plates, E = constant . So, D → (p) When a particle of charge -q , mass m moves between the plates from -V to V , then K increases, U decreases, however T ( = U + K ) remains constant. So, A → (r), B → (s), (C) → (p)

CHAPTER 1

Charge on outer surface of bigger (or outer) shell is 8Q . So, D → (p)

12. A → (s) B → (r) C → (q, t) D → (p)

[ ε 0 ] = M -1L-3T 4 A2 , So, A → (s) [ ϕE ] = ML3T -3 A -1 So, B → (r) σ ⎤ -3 -1 ⎥ = MLT A ⎣ 2ε 0 ⎦

[ E ] = ⎡⎢

So, C → (q, t)

[ V ] = ML2T -3 A -1 So, D → (p) 13. A → (p) B → (q, s) C → (p) D → (q, r) There will be no field inside the body of the conductor, so E = 0 for a < r < b and c < r < d . So, C → (p). Also, inside shell A , E = 0 . So, A → (p)  In the region b < r < c , E is directed radially out2q q = wards and E = 2 4pε 0 r 2pε 0 r 2

So, B → (q, s).

9/20/2019 11:34:30 AM

H.86  JEE Advanced Physics: Electrostatics and Current Electricity

2q –2q

(B) Electrostatic potential energy U=

B

a

A

b

 Finally, in the region outside B i.e. for r > d , E is directed radially inwards (because charge on outer 2q q = surface of B is -2q ). Hence E = 2 4pε 0 r 2pε 0 r 2 So, D → (q, r) 14. A → (r, s) B → (r, s) C → (p, s) D → (p, s)

When S is open then in Region II, E =



⇒ E≠0



In Region III, E =



⇒ E≠0



q Potential is in both regions. 4pε 0 r

q 4pε 0 r 2

q 4pε 0 r 2

⇒ V≠0  When S is closed, then all charge goes from inner sphere to outer sphere. O q

U=

So, in Region I and II

E=0 But in Region I and II, V ≠ 0 15. A → (p, s) B → (q, s) C → (q, s) D → (s) (A) Electrostatic potential energy Q2 1 ( -Q ) = 4pε 0 2 a 8pε 0 a 2

  

01_Ch 1_Hints and Explanation_P3.indd 86

⎞ 3 Q2 = ⎟ 20 pε 0 a ⎛ 5a ⎞ 2⎜ ⎟ ⎟ ⎝ 2 ⎠⎠

1 3Q 2 3 Q2 = 4pε 0 5 a 20 pε 0 a

(D) Electrostatic potential energy U=

1 ⎛ 3Q 2 ( -Q 2 ) ( -Q ) × ( -Q ) ⎞ 27Q 2 + + = ⎜⎝ ⎟ ⎠ 80pε 0 a 4pε 0 5 a 2 ( 2a ) 2a

16. A → (p, r, s, t) B → (p, r, s, t) C → (p, q, r, s) D → (q) For arrangement (A), potential at any point on z-axis is zero, y-component of electric field at any point on z-axis is non-zero, U system is negative, z-component of electric field at any point on z-axis is zero. So A → (p, r, s, t). Similarly B → (p, r, s, t), C → (p , q, r, s) and D → (q) 17. A → (q) B → (r) C → (s) D → (p) Q 8pε 0 R2



For hemispherical shell, E at centre is E =



So, A → (q)



For hemispherical sphere, E =



So, B → (r) For ring at a point near its axis ( x  R ) , we get E=

U=

( -Q )2

(C) Electrostatic potential energy

d c



1 ⎛ ( -Q ) × ( -Q ) + ⎜ 5a 4pε 0 ⎜ ⎝ 2

3Q 8pε 0 R2

3x 2 ⎞ λx ⎛ 1⎟ 2⎜ 2ε 0 R ⎝ 2R 2 ⎠

So, C → (s) For semi-infinite wire, E at the given point is E =

λ 2 2pε 0 x

So, D → (p)

18. A → (s) B → (r) C → (p) D → (q)

9/20/2019 11:34:40 AM

Hints and Explanations H.87 ∂V ∂V . Also, tangent to elecand Ey = ∂x ∂y tric lines of force will give direction of electric field. So,

Since Ex = -

dy Ey = dx Ex 2

⇒   log y = log x + c



⇒ 



Straight line equation passing through origin

y =c x ⇒   y = cx

2

(A) V=x -y  ⇒   E = -2xiˆ + 2 yjˆ dy Ey - y ⇒  = = dx E x

Electric lines of force

CHAPTER 1





x



⇒   log y = - log x + c ⇒   xy = c (Rectangular Hyperbola) (D) V=

y

x

Corresponding curve is drawn for electric lines of force. V = xy (B)  E = - yiˆ - xjˆ    dy x = ⇒  dx y





x y

 1 x    E = - iˆ + 2 ˆj y y x dy y 2 x    = =dx - 1 y y    - ydy = xdx ⇒   x 2 + y 2 = constant (circle) y

x

⇒   y 2 = x2 + c y

x



Corresponding curve for field lines is shown 2

2

V=x +y (C)  ⇒   E = -2xiˆ - 2 yjˆ Tangent to electric line of force will give direction of electric field. So

 

⇒ 

dy Ey y = = dx Ex x dy dx = y x

01_Ch 1_Hints and Explanation_P3.indd 87

Corresponding curve is given.

19. A → (r) B → (q) C → (p) D → (s) For S1 , imagine the charge q to be enclosed inside a q cube with S1 as one of its face, then ϕcube = . Since ε0 S1 is only one sixth of the total surface area of cube

⇒ ϕS1 =

q 1⎛ q ⎞ = . 6 ⎜⎝ ε 0 ⎟⎠ 6ε 0

So, A → (r) For an sheet having charge q placed near it, we have q ( 1 - cos θ ) . ϕ= 2ε 0

9/20/2019 11:34:57 AM

H.88  JEE Advanced Physics: Electrostatics and Current Electricity



Since sheet is infinite, so θ →



⇒ ϕS2 =



So, B → (q)

q . 2ε 0



⇒ ϕS3 =

a 2

a + 15 a

2

=

1 4

q ⎛ 1 ⎞ 3q . ⎜ 1 - ⎟⎠ = 2ε 0 ⎝ 4 8ε 0

So, C → (p)

Similarly ϕS4 =

1



⇒ R = n3r

q 1⎛ q ⎞ = . 4 ⎜⎝ ε 0 ⎟⎠ 4ε 0

So, D → (s)



⇒ Ebig =



So, A → (p) 2



So, B → (r) 1

Since Cbig = 4pε 0 R = n 3 Csmall .



q1 R1

r

So, C → (p)

For, r < R1 E=0+0=0 1 ⎛ q1 q2 ⎞ + = constant 4pε 0 ⎜⎝ R1 R2 ⎟⎠

R1 < r < R2

V= and

q1 +0 4pε 0 r 2 1 ⎛ q1 q2 ⎞ + 4pε 0 ⎜⎝ r R2 ⎟⎠

21. A → (p) B → (r) C → (p) D → (p) n drops coalesce, then radius ( R ) of bigger drop having charge nq is

  

⎛4 ⎞ 4 n ⎜ p r 3 ⎟ = p R3 ⎝3 ⎠ 3

⇒ R3 = nr 3

01_Ch 1_Hints and Explanation_P3.indd 88

nq = n 3 σ small . 4p R2

So, D → (p)

22. A → (q) B → (r) C → (p) D → (s) For (A), we have

E=

nq = n 3 Vsmall 4pε 0 R

1

R2



2

= n 3 Esmall

4pε 0n 3 r 2

Since σ big =

q2

V= and

1

nq

Similarly Vbig =



20. A → (s) B → (r) C → (p) D → (q)



nq 4pε 0 R2

Now Ebig =

q ( 1 - cos θ ) where cos θ = 2ε 0

Since ϕS3 =

p 2



Einside = ( rb ) So A → (q) For (B), we have

Einside = ( rb )

Q 4pε 0 r 2

So, B → (r) For (C), we have

Einside = 0 ( rb )

So, C → (p)

Q 4pε 0 r 2

9/20/2019 11:35:11 AM

Hints and Explanations H.89 For (D), we have -Q Einside = 2 4 pε ( rb ) So, D → (s). Please note that, while answering all these we must keep in mind that the field inside the body of the conductor is zero.

σ - Edisc 2ε 0

6.

Enet =



⇒ Enet =



⇒ Enet =



⇒ K = 9 × 109 × 1.6 × 10 -19 ×



⇒ x = 12

Integer/Numerical Answer Type Questions

7.

2 2 From figure AC = ( AO ) + ( OC )

1.

AC = 3 2 + 4 2 = 5 m

33

Q 2. Since λ = 2p R and the tension developed in the ring when a charge is placed at its centre is given by T=



Further, by Laws of Elasticity, we know that

TR ADR



⇒ Y=



q0Q TR = ⇒ DR = AY 8p 2ε 0 RAY

Substituting, q0 = 10-8 C, Q = p C, R = 0.1 m, A = 10-6 m2 and Y = 2 × 10-11 Nm-2, we get DR = 225 × 10-5 m ⇒ x = 225 2t t 3. t = 1 2 = 12 ms t1 - t2

–q



⇒

1 2 ay t 2

+q

( q ) × ( -q )



P.E. at C is UC = 2 × 9 × 109 ×



⇒ UC = -2 × 9 × 109 ×



⇒ UC = -9 J



K.E. at C is 4 J



Total energy at C is EC = P.E. + K.E.



⇒ EC = -9 + 4 = -5 J



Potential energy at D is U D , given by

AC

( 5 × 10 -5 )2 5



-45 J r K.E. at D = 0



So, total energy at D is ED = 0 +



-31

2 × 9.1 × 10 × 16 × 10 1.6 × 10 -19 × 4 × 200

01_Ch 1_Hints and Explanation_P3.indd 89

B



1 1 ( 1.6 × 10 -19 ) ( E ) ⎛ 2 ⎞ = ⎜ ⎟ 200 2 9.1 × 10 -31 ⎝ 4 × 108 ⎠

⇒ E=

C

2

y = uy t +

⇒ y=

–q

O

D

-2 × 9 × 109 ( 5 × 10 -5 ) r where AD = BD = r (say)

2



+q

U D =

1 2 ay t , where 2 eE  and t = uy = 0 , ay = m u



A

4 × 10 -8 9

Similarly, BC = 5 m

⎛T⎞ ⎜⎝ ⎟⎠ Stress Y= = A Strain ⎛ DR ⎞ ⎜⎝ ⎟ R ⎠

5.

σz

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

2ε 0 z 2 + a 2

q0 λ q0Q = (we have done this already) 4pε 0 R 8p 2ε 0 R2



4.

z σ σ ⎛ ⎞ 1⎟ 2 2 ⎠ 2ε 0 2ε 0 ⎝⎜ z +a

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

CHAPTER 1



16

2

= 2275 NC -1

⇒ UD =

-45 -45 = J r r By Law of Conservation of Energy,

UC + KC = U D + K D -45 = -5 r



⇒



⇒ r=9m

9/20/2019 11:35:32 AM

H.90  JEE Advanced Physics: Electrostatics and Current Electricity

2 2 Now OD = ( AD ) - ( AO )



⇒ OD = 92 - 3 2 = 72 m



⇒ x = 72

8.

Done already (Illustration 61)

vmin =





39 ⇒ x= ≅1 40

9.

⎡ Qq ⎛ ⎞ Since, θ = 2 sin -1 ⎢ ⎜ ⎢ ⎝ 32pε mg 2 ⎟⎠ 0 ⎣



⎤ ⎥ ⎥ ⎦ (As already done in Illustration 6)



⎛ 1⎞ ⇒ θ = 2 sin -1 ⎜ ⎟ ⎝ 2⎠



⎛p⎞ ⇒ θ = 2⎜ ⎟ ⎝ 6⎠

10.

q2 mg = 4pε 0 r 2



⇒ T ′ = 2.6 s



For 20 oscillation, time taken is t = 20 × 2.6



⇒ t = 52 s

12. Let us first the force on a-q charge placed at a distance x from centre of ring along its axis. Figure shows the respective situation. In this case force on particle P is

Q + + +

+ + R + + + +

+ + +

FP = -

9 × 109 × ( 1.6 × 10 -19 ) × 10 = r2

9 × 10 × ( 1.6 × 10 1.7 × 10 -26 9



⇒ r2 =



⇒ r = 0.116 m



⇒ r = 11.6 cm ≅ 12 cm

2

)

-19 2

Finally T ′ = 2p

qE q = 4 × 10–6 C mg

geff = g –

qE m

+

+

q Qx 4pε 0 ( x 2 + R2 )3 2

v 2 = 2 × ( 1.76 × 1011 ) × 9 × 109 × . So

1 qQx 4pε 0 R3 Now, acceleration of particle is

F = -

a =

L …(2) qE ⎞ ⎛ ⎜⎝ g ⎟⎠ m

P –q

For small x , x  R , we can neglect x , compared to



1 ⎞ ⎛ 1 …(1) 11. Initially 9 × 109 ⋅ q2 ⎜ ⎝ 0.40 0.50 ⎟⎠

F x

+

+

⇒ 1.7 × 10 -27

01_Ch 1_Hints and Explanation_P3.indd 90

qE 4 × 10 -6 × 2.5 × 10 4 T = 1= 1T′ mg 40 × 10 -3 × 10

+ + + +

p ⇒ θ = 60° 3

⇒ θ=

⇒

FP = - qE

1⎤ ⎡ -1 ⎢ ⎛ 4pε 0 ⎞ 3 ⎥ ⇒ θ = 2 sin ⎜ ⎢ ⎝ 32pε 0 ⎟⎠ ⎥ ⎣ ⎦



1 3

Qq = 4pε 0 mg 2



qE m g

g-

T = T′

39 ⎛ Qq ⎞ 40 ⎜⎝ 2pε 0 m ⎟⎠



Dividing equation (1) and (2), we get

F 1 qQ =x 4pε 0 mR3 m

qQ ⎞ 6 × 10 -8 × 0.15 × 0.30 ⎛ x + ⎜ x=0 ( 0.30 - 0.15 ) ⎝ 4pε 0 mR3 ⎟⎠



⇒



which is a standard SHM equation. So

ω =

Qq 4pε 0 mR3 2p 4pε 0 mR3 = 2p Qq ω



⇒ T=



⇒ T = 2p



⇒ T=

3

0.9 × 10 -3 × ( 1 ) 9 × 109 × 10 -5 × 10 -6

p s  ⇒   k = 5 5

9/20/2019 11:35:52 AM

Hints and Explanations H.91

N

sin

3

T1 mg O Reference level

mg mg cos 30°

By definition, function is



mg sin 30° - μ N - qE cos 30° = ma

T + qE =

⇒ ma = mg sin 30° - μmg cos 30° -

  a = g sin 30° - μ g cos 30° -

μqE qE cos 60° cos 30° m m

⎛ 3 ⎞ ( 0.2 )( 0.01 )( 100 ) ⎛ 1⎞ ⎟⎜⎝ ⎟⎠ - ( 0.2 )( 10 ) ⎜⎝ 2 2 ⎠ 1 1

⎜⎝ ⎟ 2 ⎠

3 2



⇒ a = 5 - 3 - 0.1 -



⇒ a = 2.3 ms -2



Now, distance travelled in time t is 1 2 at 2

⇒ t=

2×2  a



⇒ t=

4 2.3



⇒ t = 1.32 s



⇒ t = 1320 × 10 -3 s



⇒ t = 1320 ms



1⎞ ⎟2⎠

( 0.01 )( 100 ) ⎛ 3 ⎞



s = 0 +

01_Ch 1_Hints and Explanation_P3.indd 91

v

at position A we have mv 2 + mg 

Since T = 15mg

μ qE cos 60° - qE cos 30°  Thus acceleration of the particle down the incline is

⎛ ⇒ a = 10 ⎜ ⎝

A

1 1 mu2 + 2mg - 2qE = mv 2 …(1) 2 2

If a is the acceleration of the particle down the incline, then



T2 qE

Here by using work energy theorem between position B and A , we have

f = μ N = μ mg cos 30° + μ qE cos 60°



E

mg

From figure,

N = mg cos 30° + qE cos 60°

B

u

μN qE 30° qE sin 30°

q

30°



qE

0°

30°

mg

° 30

os

c qE

14. The corresponding situation is shown in figure

CHAPTER 1

13. The different forces on the particle are shown in figure

1 ⎧ ⎫ = 2⎬ ⎨∵ s = sin 30 ⎩ ⎭



⇒ qE = mv 2 - 14 mg 

{∵ at the lowest point, T = 15 mg}

From equation (1), we have

qE + 14 mg = mu2 + 4 mg - 4 qE

⇒ u2 =

5qE 10 mg + m m



⇒ u=

5qE + 10 g m



⇒ u=

5 ( qE + 2mg ) m



⇒ u=

5 ⎛ 6 mg ⎞ + 2mg ⎟ = ⎜ ⎠ m⎝ 5



⇒ u = 4 g = 4 ( 10 )( 3.6 ) = ( 4 ) ( 6 ) = 24 ms -1

5 ⎛ 16 mg ⎞ ⎜ ⎟ m⎝ 5 ⎠

15. Let q be the initial charge on each of spheres A and B , then given FAB = 2.0 × 10 -5 N

So from Coulomb’s Law



1 q2 = 2.0 × 10 -5 …(1) 4pε 0 r 2

When the uncharged sphere C is touched with A , the charge on A and C is distributed equally i.e.,

9/20/2019 11:36:12 AM

H.92  JEE Advanced Physics: Electrostatics and Current Electricity A q/2

r/2

q/2

B

C

q

Y P(3, 1, 1) r1

r

0+q q = 2 2 The arrangement is shown, where C is mid point of A and B qA = qC =



AB r ⇒ AC = BC = = 2 2



The force on C due to A

 FCA =

⎛ q⎞⎛ q⎞ 1 qB qC 1 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ = 2 4pε 0 rBC 4pε 0 ⎛ r ⎞ 2 ⎜⎝ ⎟⎠ 2

1 q2 from A → C 4pε 0 r 2



⇒ FCA =



The force on C due to B is

 FCB



⇒ FCB =

q 1 qA qC 1 2(q) = = 2 4pε 0 rBC 4pε 0 ⎛ r ⎞ 2 ⎜⎝ ⎟⎠ 2 1 2q 2 from B → C 4pε 0 r 2



By Principle of Superposition, net force on C is    FC = FCA + FCB

⎛ 1 2q 2 1 q2 ⎞ ⇒ FC = ⎜ from B to C 4pε 0 r 2 ⎟⎠ ⎝ 4pε 0 r 2



1 q2 ⇒ F= from B to C 4pε 0 r 2



Using (1), we get  FC = 2.0 × 10 -5 N from B to C

⇒ F = 20 μN

16. 2 17.

T = 7.72 × 10

-2

N  77 milli newton

m = 7.96 g  8 g 18. The electric field strength at a point having position  vector r relative to charge Q is given by   1 r (vector form) E = 4pε 0 r 3

01_Ch 1_Hints and Explanation_P3.indd 92

r2 Q

O P

Q1

A(2, 0, 0)

X



The position vector of P relative to origin is  r1 = 3iˆ + ˆj + kˆ m  modulus of r1 is

(

)

2 2 2 r1 = ( 3 ) + ( 1 ) + ( 1 ) = 11 m

The position vector of P relative to point A ( 2, 0 , 0 ) is given by  r1 = 3iˆ + ˆj + kˆ - ( 2iˆ ) = iˆ + ˆj + kˆ  modulus of r2

(

)

2 2 2 r2 = ( 1 ) + ( 1 ) + ( 1 ) = 3 m



The electric field at P due to charge Q , at O is

(



)

1 Q1 3iˆ + ˆj + kˆ …(1) 4pε 0 ( 11 )3 2

 E1 =

The electric field at P due to charge Q , at A is

 E2 =

(

)

1 Q iˆ + ˆj + kˆ …(2) 4pε 0 ( 3 )3 2

∴ Net field at P ,    E = E1 + E2



 E =

)

(

)

X -component of electric field at P is

Ex =

(

Q iˆ + ˆj + kˆ ⎤ 1 ⎡ Q1 3iˆ + ˆj + kˆ ⎢ ⎥ …(3) + 32 4pε 0 ⎢⎣ ( 11 ) ( 3 )3 2 ⎥⎦ 1 ⎡ Q1 ( 3 ) Q ( 1 ) ⎤ + =0 4pε 0 ⎢⎣ ( 11 )3 2 ( 3 )3 2 ⎥⎦

This gives

⎛ 3⎞ Q = - ⎜ ⎟ ⎝ 11 ⎠

32

35 2

( 3Q1 ) = -

⇒ Q=-



⇒ Q = -3 × 11 × 10 -9 C



⇒ Q = -33 nC



⇒ Q = 33 nC

32

11

× 11 ×

113 2



33 2

3 5 2 × 10 -9 113 2

C

× 10 -9

9/20/2019 11:36:37 AM

Hints and Explanations H.93



0.3 = 0.95

    W = V ( ρair - ρHe ) g …(1)

 

 where V = volume of balloon, ρair and ρHe are densities of air and helium respectively

⇒   q2 =

(ii)  Electric repulsive force acting horizontally given by

q2 1 q2 9 = × × …(2) 9 10 4pε 0 r 2 ( 0.6 )2  (iii) Tension in string T along AC for balloon A

    Fe =



For equilibrium of balloon A say

2

( 1 ) - ( 0.3 )

2

=

0.3 …(6) 0.95

Using (2), (5) and (6), we get from (3)

    W = V ρair g - V ρHe g



0.3

⇒   tan θ =

 

9 × 109 ×

q2

( 0.6 )2

24.5 × 10 -3

( 0.6 )2 × ( 24.5 × 10 -3 ) 9 × 109

⇒   q = 55 μC

(b) From equations (1) and (5), we get mg = V ( ρair - ρHe ) g 2 So, volume of balloon, is given by

 

Fe = T sin θ  

V =

m 2 ( ρair - ρHe )



⇒  V=

5 × 10 -3 2 ( 1.29 - 0.2 )



⇒  V=

5 × 10 -3 = 2.293 × 10 -3 m 3 2 × 1.09



⇒   V = 2293 cc

Dividing (2) by (1)

  tan θ =

Fe …(3) W W 0.6 m A

D θ

T cos θ

T 2T cos θ T θ θ

B

Fe

1m

Fe

W

C mg



For equilibrium of weight C

 

2T cos θ = mg

∴  T cos θ =

mg …(4) 2

20. Let q1 and q2 be the respective charges distributed over two concentric spheres of radii r and R such that =

   

W=

 

W = 24.5 × 10 -3 N …(5)

1 × 5 × 10 -3 × 9.8 2

From figure, tan θ =

01_Ch 1_Hints and Explanation_P3.indd 93

AD CD

Q ⎡1 1 ⎤ …(1) 4pε 0 ⎢⎣ R ( d - R ) ⎥⎦

As surface densities are given to be equal, therefore

σ 1 = σ 2

⇒

q1 q = 2 …(2) 4p r 2 4p R2



⇒

q1 r 2 = q2 R2



⇒

q1 r2 +1= 2 +1 q2 R



⇒

q1 + q2 r 2 + R2 = q2 R2



Using (1), we get

From (3) and (4) mg W = T cos θ = 2

0.3 0.95

⇒   q = 55 × 10 -6 C

  W = T cos θ



×

CHAPTER 1

19. (a) The forces acting on each balloon are shown in figure and are as under (i) The effective weight of balloon W acting vertically upward



Q r 2 + R2 = q2 R2

9/20/2019 11:36:55 AM

H.94  JEE Advanced Physics: Electrostatics and Current Electricity



⎡ R2 ⎤ This gives q2 = ⎢ 2 ⎥Q ⎣ r + R2 ⎦

Therefore q1 = Q - q2

⎛ R2 ⎞ ⎛ r2 ⎞ ⇒ q1 = Q - ⎜ 2 Q=⎜ 2 Q ⎟ 2 ⎝ r +R ⎠ ⎝ r + R2 ⎟⎠

⇒ 0+



⇒ n = 30

22. Potential at the centre of a loop is the sum of potentials due to charge of its own and that due to other loop.

The potential V1 at common centre due to charge q1 is given by V1 =

1 ( - q1 ) q2 1 1 = m1v12 + m2v22 4pε 0 d 2 2





Potential at the centre C1 of loop A

1 q1 4pε 0 r

V1 =

q⎞ 1 ⎛ q + 4pε 0 ⎜⎝ R1 r1 ⎟⎠ 1 1 ⎡ 1 ⎤ q + 2 2 ⎥ 4pε 0 ⎢⎢ R1 R2 + d ⎥⎦ ⎣



⇒ V1 =

1 ⎛ r2 ⎞ Q ⎜ ⎟ 4pε 0 ⎝ r 2 + R2 ⎠ r



⇒ V1 =



⇒ V1 =

1 Qr 2 4pε 0 r + R2



1 ⎡ 1 ⎤ ⇒ V1 = 9 × 109 × 10 -6 ⎢ + 2 2 ⎥ R1 ( 0.09 ) + ( 0.12 ) ⎦ ⎣



The potential V2 at common centre due to charge q2 is



1 ⎤ ⎡ 1 ⇒ V1 = 9 × 10 3 ⎢ + ⎣ 0.05 0.15 ⎥⎦



⇒ V1 = 9 × 10 3 ×



The potential at the centre C2 of loop B

1 q2 1 ⎛ R2 ⎞ Q V2 = = ⎜ ⎟ 4pε 0 R 4pε 0 ⎝ r 2 + R2 ⎠ R 2e ( DV ′ ) = m

-19

2 × 1.602 × 10 × 30 × 10 1.672 × 10 -27



⇒ v=



∴ Net potential at common centre,

2

V = V1 + V2

V2 =

80 = 2.4 × 10 5 V 3

q⎤ 1 ⎡ q + ⎥ ⎢ 4pε 0 ⎣ R2 r2 ⎦



⇒ V=

1 Q [r + R] 2 ( 4pε 0 r + R2 )



⇒ V2 =



⇒ V=

1 Q( R + r ) 4pε 0 R2 + r 2





⇒ V = 9 × 109 ×

1 ⎡ 1 ⎤ ⇒ V2 = 9 × 109 × 10 -6 ⎢ + 2 2 ⎥ R1 ( 0.05 ) + ( 0.12 ) ⎦ ⎣



⇒ V=9V



1 ⎤ ⎡ 1 ⇒ V2 = 9 × 10 3 ⎢ + ⎣ 0.09 0.13 ⎥⎦



⎡1 1 ⎤ ⇒ V2 = 9 × 10 3 × 100 ⎢ + ⎥ ⎣ 9 13 ⎦



⎡1 1 ⎤ ⇒ V2 = 9 × 10 5 ⎢ + ⎥ ⎣ 9 13 ⎦



⎡ 13 + 9 ⎤ ⇒ V2 = 9 × 10 5 ⎢ ⎣ 117 ⎥⎦



⇒ V2 = 9 × 10 5 ×



⇒ V2 = 1.692 × 10 5 V



Potential difference, DV is given by

5 9 × 10 -9 × × 100 100 45

21. The horizontal force on dust particle is F = qE If v is speed of particle in air, then by Stoke’s Law,

Viscous force = 6pηrv



For uniform speed v , we have

qE = 6pηrv If n is the number of excess electrons, then q = ne

⇒ neE = 6pηrv 6pηrv ⇒ n= eE Substituting given values

n =

-5

-7

6 × 3.14 × 1.6 × 10 × 5 × 10 × 0.02 1.6 × 10 -19 × 6.28 × 10 5

01_Ch 1_Hints and Explanation_P3.indd 94

1 1 ⎡ 1 ⎤ q⎢ + 2 2 ⎥ 4pε 0 ⎢ R1 R + d ⎥⎦ 2 ⎣

22 117

DV = V1 - V2

⇒ DV = 2.4 × 10 5 - 1.692 × 10 5

9/20/2019 11:37:20 AM

Hints and Explanations H.95 ⇒ DV = 7.08 × 10 4 volt



⇒ v = 1.54 × 107 ms -1

⇒ DV = 7080 V



⇒ v = 154 × 10 5 ms -1

23. The potential differences between spherical conductors



⇒ x = 154

1 ⎛ q q⎞ Vab = ⎜ - ⎟ 4pε 0 ⎝ a b ⎠

24. Initial potential energy of system

b = 4 cm = 4 × 10 -2 m

1 ⎛ 1 ⎞ ⇒ 3000 = 9 × 109 q ⎜ -2 ⎟ ⎝ 10 4 × 10 -2 ⎠



⎡ ( 4 - 1) ⎤ ⇒ 3000 = 9 × 109 q ⎢ ⎥ ⎣ 4 × 10 -2 ⎦



⇒ q=



The potential at point distant r1 = 3 cm from centre

3000 × 4 × 10 -2 4 = × 10 -8 C 9 9 × 109 × 3

V1 =

1 ⎛ q q⎞ 4pε 0 ⎜⎝ r1 b ⎟⎠

The potential at point distant r2 = 2 cm from centre, 1 ⎛ q q⎞ 4pε 0 ⎜⎝ r2 b ⎟⎠

V2 =

The gain in kinetic energy of electron

K = e ( V2 - V1 ) 1 4pε 0

⎡ q q q q⎤ ⎢r -b - r + b⎥ 1 ⎣ 2 ⎦



⇒ K=e



1 ⎡1 1⎤ eq ⇒ K= 4pε 0 ⎢⎣ r2 r1 ⎥⎦



9

⇒ K = 9 × 10 × 1.6 × 10

1 ⎡ q1q2 q2 q3 q3 q1 ⎤ + + r23 r13 ⎥⎦ 4pε 0 ⎢⎣ r12

U1 =

Here a = 1 cm = 10 -2 m ,



⎡ 2× 3 3 ×1 2×1⎤ + + ⇒ U1 = q × 109 ⎢ 1 ⎥⎦ 1 ⎣ 1



⇒ U1 = 9 × 109 × 11



⇒ U1 = 99 × 109 J



Final potential energy of system

U 2 =

1 ⎡ q1q2 q2 q3 q3 q1 ⎤ + + r23 r13 4pε 0 ⎢⎣ r12 ′ ′ ′ ⎥⎦



⎡ 2× 3 3 ×1 2×1⎤ + + ⇒ U 2 = 9 × 109 ⎢ 0.5 ⎥⎦ 0.5 ⎣ 0.5



⇒ U 2 = 9 × 109 × 22 = 198 × 109 J



So, work done W = U 2 - U1



⇒ W = 198 × 109 - 99 × 109



⇒ W = 99 × 109 J



⇒ x = 99

25. Initial electrostatic energy of system of three charges, U i =

1 ⎛ q1q2 q2 q3 q3 q1 ⎞ + + 4pε 0 ⎜⎝ r12 r23 r13 ⎟⎠ A

-19



CHAPTER 1



4 × × 10 -8 × 9 1 1 ⎡ ⎤ ⎢⎣ 2 × 10 -2 - 3 × 10 -2 ⎥⎦

0.1 C

32 × 10 -17 J 3



⇒ K=



⇒



⇒ v2 =



2 ⎛ ⎞ 32 ⇒ v =⎜ × × 10 -17 ⎝ 9 × 10 -31 ⎟⎠ 3



⇒ v2 =

1 32 mv 2 = × 10 -17 J 2 3 2 32 × × 10 -17 J m 3

2

64 × 1014 27

01_Ch 1_Hints and Explanation_P3.indd 95

B 0.1 C

A′

C 0.1 C

Here q1 = q2 = q3 = q (say) = 0.1 C r12 = r23 = r13 = r (say) = 1 m ⎛ 1 q2 ⎞ U i = 3 ⎜ ⎟ ⎝ 4pε 0 r ⎠

9/20/2019 11:37:42 AM

H.96  JEE Advanced Physics: Electrostatics and Current Electricity



⇒ U i = 3 × 9 × 109 ×

( 0.1 )2

ay =

1



8

⇒ U i = 2.7 × 10 J

When charge A is moved to new position A′ which is mid point of side BC , then final electrostatic energy, U f =

⇒ Uf =

1 ⎛ q1q2 q2 q3 q3 q1 ⎞ + + 4pε 0 ⎜⎝ r12 r23 r13 ⎟⎠ 1 4pε 0

⎡ q ⎤ q q + + ⎢( ⎥ ) ( ) [ ] BC A′C ⎦ ⎣ A′B

But A′B = A′C =

2

2

2

⎡ ( 0.1 ) ( 0.1 ) ( 0.1 ) ⎤ + + U f = 9 × 109 ⎢ ⎥ 1 0.5 ⎦ ⎣ 0.5

⇒ U f = 4.5 × 108 J



Work done

2

2

⇒ W = 4.5 × 108 - 2.7 × 108



⇒ W = 1.8 × 108 J



Work done Since, by definition, Power = Time W ⇒ P= t W 1.8 × 108 = second P 10 3



⇒ t=



⇒ t = 1.8 × 10 5 s



⇒ t=



⇒ t = 50 h

1.8 × 10 5 h 3600

26. The electron on entering the electric field, will experience a force F = eE opposite to direction of electric field. On account of this force, it will be deflected in transverse direction. The path of electron is shown in figure.

E= and

x

l

Assuming initial velocity along X-axis, the transverse deflection will be along Y-direction,

01_Ch 1_Hints and Explanation_P3.indd 96

⎫ ⎧ ⎨∵ t = ⎬ u⎭ ⎩

1 eE ⎛  2 ⎞ ⎟ 2 2 m ⎜⎝ uMAX ⎠ d 2

V d

d  e ⎛ V ⎞ 2 = ⎜ ⎟ 2 2 2 m ⎝ d ⎠ uMAX



⇒



2 ⇒ uMAX =

e V ⎛ 2 ⎞ ⎜ ⎟ md⎝ d⎠



⇒ uMAX =

 eV d m



Substituting given values

uMAX =

⎛ 1.6 × 10 -19 × 300 ⎞ 10 -1 ⎟⎠ -2 ⎜ ⎝ 2 × 10 9 × 10 -31



⇒ uMAX = 3.65 × 107 ms -1



⇒ uMAX = 365 × 10 5 ms -1



⇒ x = 365

27. (a) For equilibrium of oil drop between the plates the weight ( mg ) of drop must be balanced by upward electric force qE where E is electric field strength and q the charge on drop. qE = mg      ⇒   q =

y d

2

1 ⎛ eE ⎞ ⎛  ⎞ ⎜ ⎟⎜ ⎟  2⎝ m ⎠ ⎝ u⎠

Here y MIN =

W = U f - U i

⇒ y=

1 2 ay t 2

So, y MIN =

r = 0.5 m , so we get 2 2

∴ Initial velocity along Y-direction is zero and initial velocity along X-direction is u (say) therefore transverse deflection

y =

F eE = = constant m m

mg E

where E = ⇒   E =

V 1.5 kV = d 1.5 × 10 -2 m

1.5 × 10 3 Vm -1 = 10 5 Vm -1 1.5 × 10 -2

and m = 4.9 × 10 -15 kg

9/20/2019 11:38:05 AM

Hints and Explanations H.97 At the position of maximum elongation of spring, the sphere moving downward comes to rest. Let maximum elongation of spring be x , then by Law of Conservation of Energy

4.9 × 10 -15 × 9.8 10 5

⇒   q = 4.8 × 10 -19 C  If n is the number of electrons attached to oil drop and e is elementary charge, then by charge quantisation q = ne      ⇒   n =

   F = mg + qE = 2mg {since mg = qE} Initial Acceleration of drop,

F 2mg = 2 g downward    a = = m q ⇒   a = 2 × 10 = 20 ms -2 downward



⇒

1 ( q )( -q ) 1 q ( -q ) + mg x0 + 0 = + x0 4pε 0 4pε 0 ( x0 - x )

1 mg ( x0 - x ) + K ( x 2 ) 2

q 4.8 × 10 -19 = =3 e 1.6 × 10 -19

(b) If the polarity of plates is reversed, then both weight of drop and electric force act downward therefore net downward force,



( U + K )initial = ( U + K )final

(c) If the downward force F is balanced by upward viscous force, then drop attains terminal velocity vT given by Stoke’s Law

   F = 6pηrvT ⇒   2mg = 6pηrvT ⇒   vT =

2mg mg = 6pηr 3pηr

⇒   vT =

4.9 × 10 -15 × 10 3 × 3.14 × 1.8 × 10 -5 × 5 × 10 -6

Total energy in initial position = Total energy in final position

⇒

1 2⎛ 1 1 ⎞ 1 2 q = Kx - mgx …(1) 4pε 0 ⎜⎝ x0 - x x0 ⎟⎠ 2

Given x = 0.1 m , x0 = 0.5 m ,     K = 10 4 Nm -1 , g = 10 ms -2 q2 ⎛ 1 1 ⎞ 1( 4 ) 2 ⎜ ⎟ = 10 ( 0.1 ) - ( 5 )( 10 )( 0.1 ) 4pε 0 ⎝ 0.4 0.5 ⎠ 2



⇒



⇒ 4.5 × 109 q2 = 45



⇒ q2 =



⇒ q = 10 4 C = 100 μC

45 = 108 4.5 × 109

29. Electric field due to a straight conductor,

⇒   vT = 5.8 × 10 -5 ms -1

E =

1 2λ 4pε 0 r 1 2λ q 4pε 0 r



∴ Force, F = qE =

⇒   vT = 58 μms -1



⇒ F = 9 × 109 ⋅

28. When the thread is burnt, the positively charged sphere moves downward due to two forces (i)  Weight mg , acting vertically downward



⇒ F = 3.45 × 10 -4 N



⇒ F = 345 × 10 -6 N = 345 μN



Since, by definition, we have

⇒   vT = 58 × 10 -6 ms -1

(ii) Coulomb force of attraction downward

1 qq , acting 4pε 0 x 2

x

01_Ch 1_Hints and Explanation_P3.indd 97

Final position

2 × 6 × 10 4 × 1.6 × 10 -19 0.5

W = q ( V2 - V1 ) 1 ⎛r ⎞ 2λ log e ⎜ 2 ⎟ 4pε 0 ⎝ r1 ⎠



⇒ W = q⋅



⇒ W=



⎛ 0.5 ⎞ ⇒ W = 9 × 109 × ( 2 × 6 × 10 4 × 1.6 × 10 -19 ) × log e ⎜ ⎝ 0.2 ⎟⎠

x0 Initial position

CHAPTER 1

⇒   q =

1 r 2λ q ) log e 2 ( 4pε 0 r1

9/20/2019 11:38:29 AM

H.98  JEE Advanced Physics: Electrostatics and Current Electricity ⇒ W = 1.58 × 10 -4 J



⇒ W ≅ 16 mJ

32. Considering the equilibrium of ball P.



⇒ v = 8 × 106 ms -1



So, the part not readable is 8.

60° B

60°

Q

2

Q 50 cm



(i) A large solid sphere of charge density ( ρ )



(ii) A small solid sphere of charge density ( - ρ )

The electric potential due to whole solid sphere of radius R at a distance r from centre

( 3R2 - r 2 ) 1 q 4pε 0 2R 3

mg



⇒ T=

mg

( 0.866 × 10 -3 ) × 10 mg = cos 30 0.866

⇒ T = 10 -2 N From equation (1)

T sin 30° =



⎛ 4 3 ⎞( 2 p R ρ ⎟ 3R - r 2 ) ⎠ 1 ⎜⎝ 3 ⇒ V1 = 4pε 0 2R 3



Solving we get,



The electric potential at external point at a distance

q = 373 nC

q = 3.726 × 10 -7 C

2

r ′ = x + y = 5 cm

1 q2 4pε 0 r 2

1 9 × 109 × q2 = 2 ( 0.5 )2

⇒ 10 -2 ×

q ⎛ a ⎞ 1 2 2 ⎟ 2ε 0 ⎜⎝ R +a ⎠

33.

ϕE =



⇒ ϕE =



⇒ ϕE =

106 ⎛ 4⎞ ⎜ 1 - ⎟⎠ 2 ⎝ 5



⇒ ϕE =

106 1 × = 10 5 NC -1 2 5

where R = 5 cm = 5 × 10 -2 m,



⇒ ϕE = 100 kNC -1

r = 4 cm = 4 × 10 -2 m,

34. By Law of Conservation of Energy



4 p r′3 ( -ρ ) 1 q′ 1 3 ⇒ V2 = = ⋅ r′ 4pε 0 r ′ 4pε 0 So, net potential

V = V1 + V2

⇒ V=

1 ⎡2 4 ⎤ p Rρ ( 3 R2 - r 2 ) - p r 2 ρ ⎥ 4pε 0 ⎢⎣ 3 3 ⎦

r ′ = 5 cm = 5 × 10 ρ =

-2

m,

10 -6 p

Substituting given values, V = 35.16 V

So, V ≅ 35 V

01_Ch 1_Hints and Explanation_P3.indd 98

Fe

T cos ( 30° ) = mg …(2)



2

T sin 30 Q Fe

T sin ( 30° ) = F = Fe …(1)

31. The electric potential at P is due to two spheres

V1 =

T

°



A

T cos 30

T

T

1 1 ⎛ 1 1⎞ ⇒ mv 2 = e q⎜ - ⎟ 2 4pε 0 ⎝ a b ⎠ 2e ⎡ 1 ⎛ ab ⎞ ⎤ ⇒ v = ⎢ ⎜q ⎟ m ⎣ 4pε 0 ⎝ b - a ⎠ ⎥⎦

T 30



30°

30. Gain in KE = e ( V2 - V1 )

T

30°



8.85 × 10 -6

2 ( 8.85 × 10

-12

{as derived earlier}

40 ⎞ ⎛ ⎜⎝ 1 - ⎟⎠ ) 50

( U + K )initial = ( U + K )final

⇒

Qq Qq 1 1 2 mv 2 + = m( 0 ) + 2 4pε 0 r1 2 4pε 0 r2



⇒

Qq ⎛ 1 1 ⎞ 1 mv 2 = 2 4pε 0 ⎜⎝ r2 r1 ⎟⎠

9/20/2019 11:38:49 AM

Hints and Explanations H.99

⇒

(

)

(



⇒

100 1 = - 10 r2 9



⇒

1 190 = m r2 9



⇒ r2 =



⇒ r2 = 4.7 × 10 -2 m



⇒ r2 = 47 × 10 -3 m



⇒ r2 = 47 mm

1

3

  F2 = 0   ⇒ F2 x iˆ + F2 y ˆj = 0

23

copper atoms.

6 × 10 23 × 10 63.5 × 1000

Magnitude of charge q is given by 6 × 10 23 × 10 × 1.6 × 10 -19 C 63.5 × 1000

1920 C 127



⇒ q=



Separation between pieces is r = 1 cm = 10 -2 m

One piece of copper has positive charge and the other negative charge, so force of attraction between the pieces is 1 q1q2 F = 4pε 0 r 2



⎛ 1920 ⎞ ⎛ 1920 ⎞ × 9 × 109 ⎜ ⎝ 127 ⎟⎠ ⎜⎝ 127 ⎟⎠ N ⇒ F= ( 10 -2 )2



⇒ F = 2.057 × 106 N

So, x ≅ 2 36. Let us consider the equilibrium of charge Q placed at 2. Then for equilibrium

01_Ch 1_Hints and Explanation_P3.indd 99

y

q

x

6 × 10 23 × 10 63.5

q = ne =

F25 F24 F21

2

4

Since only one electron is transferred for every 1000 atoms, therefore the number of electrons transferred, n =

a

5



⇒

F2 x = 0



⇒

F21 + F24 cos 45 + F25 cos 45 + F23 cos 90 = 0



⇒



⇒

Q+



⇒

Q ( 2 2 + 1 ) + 4q = 0



⇒

q=-

Q ( 2 2 + 1) 4

Since Q =

8 (1 - 2 2 ) 7

Therefore number of copper atoms in 10 g copper is

N =



F23

⎞ ⎟ ⎟⎠

9 m 190

35. 63.5 g copper contains N A = 6 × 10

)

1 1 ⎛ 1 2 2 × 10 -6 ( 1 ) = ( 10 -9 )( 10 -8 ) 9 × 109 ⎜ r2 ⎛ 10 ⎞ 2 ⎜⎝ ⎟ ⎜⎝ 100 ⎠

Qq 1 1 Q2 Q2 + + +0=0 2 2 2 2 2 4pε 0 a a ⎛ ⎞ ( ) 4pε 0 2 a 4pε 0 ⎜ ⎝ 2 ⎟⎠

CHAPTER 1



Q + 2q = 0 2 2



1⎡8 ⎤ ⇒ q = - ⎢ ( 1 - 2 2 )( 1 + 2 2 ) ⎥ 4⎣7 ⎦



1⎛ 8⎞ ⇒ q = - ⎜ ⎟ ( -7 ) = 2 C 4⎝ 7⎠

37. Beta particles (i.e., electrons) emitted in t seconds N = 5 × 1010 t ∴  The deficiency of electrons from a conductor results in positive charge As 40% beta-particles escape from the source, positive charge gained by source (metal sphere),



q =

40 40 Ne = × 5 × 1010 et 100 100



⇒ q = 2 × 1010 t × 1.6 × 10 -19 = 3.2 × 10 -9 t C



Due to this charge the potential acquired by sphere

V =

1 q 4pε 0 R

9/20/2019 11:39:06 AM

H.100  JEE Advanced Physics: Electrostatics and Current Electricity

Substituting given values 3.2 × 10 t 10 -2

2 = 9 × 109 ×

2 × 10 -2 9 × 109 × 3.2 × 10 -9



⇒ t=



⇒ t = 7 × 10 -4 s = 700 μs





2

2

q mv = …(1)  4pε 0 2

At the topmost point T2 = 0

mg -

⇒ Q = 2pε 0 R DV



⇒ Q=



Potential at the surface of first sphere,

2 × 100 9 × 109

 V1 = Potential due to charge on 1 + Potential due to charge on 2

38. For the ball just to complete the circle, the tension must vanish at the topmost point, i.e., T2 = 0 From Newton’s Second Law, T2 + mg -



-9

⇒ V1 =



Potential at the surface of second sphere,

 V2 = Potential due to charge on 1 + Potential due to charge on 2

q2 mv 2 …(2) = 2  4pε 0

By Law of Conservation of Energy

⎛ Energy at the ⎞ ⎛ Energy at the ⎞ ⎜ = ⎝ lowest point L ⎟⎠ ⎜⎝ highest point H ⎟⎠

Q Q Q ⎡1 1 ⎤ = ⎢ ( ) ( 4pε 0 R 4pε 0 d - R 4pε 0 ⎣ R d - R ) ⎥⎦



⇒ V2 =

+Q Q Q ⎡ 1 1⎤ = 4pε 0 ( d - R ) 4pε 0 R 4pε 0 ⎢⎣ ( d - R ) R ⎥⎦

Potential difference between the surfaces of the two spheres is DV ′ given by DV ′ = V1 - V2

⇒ DV ′ =

200 ⎡ 1 1 ⎤ 200 ⎡ 1 1 ⎤ - -2 ⎢ - ⎥ -2 ⎢ ⎥ 10 40 40 10 ⎣ ⎦ ⎣ ⎦ 10 10



⇒

1 1 mu2 = mv 2 + mg ( 2 ) …(3) 2 2



⇒ DV ′ = 30 × 10 2 V



⇒ v 2 = u2 - 4 g …(4)



Kinetic energy gained by proton = + eDV ′ =



From equation (2), v 2 = g -

q2 …(5) 4pε 0 m



⇒ v=

2e ( DV ′ ) m



Equating (4) and (5), u = 5 g -



⇒ v=

2 × 1.602 × 10 -19 × 30 × 10 2 1.672 × 10 -27



⎛ 275 ⎞ ⇒ u=⎜ ⎝ 8 ⎟⎠



⇒ v = 7.58 × 10 5 ms -1



⇒ v = 758 × 10 3 ms -1 = 758 kms -1



12

⇒ u = 586 cms

q2 4pε 0 m

= 5.86 ms -1

-1

1 mv 2 2

40. Using the concept of Work-Energy Theorem,

39. Since the spheres are initially neutral, so when a charge Q is transferred from one sphere to the other, the charges on the sphere become +Q and -Q respectively. The potential difference between them is +Q

v l

60°

qE

mg E

O

–Q

l

R

R

C 60°

u Sphere 1

+Q d = 50 cm

Q Q ⎞ ⎛ DV = - 4pε 0 R ⎜⎝ 4pε 0 R ⎟⎠

01_Ch 1_Hints and Explanation_P3.indd 100

Sphere 2



we get

1 m ( v 2 - u2 ) = - mg ( 1 + sin 60° ) + qE cos 60° 2

Substituting the values, we get

u2 - v 2 = 32.32 …(1)

9/20/2019 11:39:25 AM

Hints and Explanations H.101



Since, tension in the string at C is zero. ⇒

y E (or a)

2

mv = mg sin 60° - qE cos 60° 



⇒ v = 3.66 …(2)



From equations (1) and (2), we get

0 = m1v1iˆ + m2v2 ( - iˆ ) ⇒ v2 =

m1v1 m2



By Law of Conservation of Energy



0+

1 ( - q1 ) q2 1 1 1 ( - q1 ) q2 = m1v12 + m2v22 + 4pε 0 d 2 2 4pε 0 r1 + r2

⇒  

1 q1q2 1 q1q2 1 1 m12v12 = m1v12 + 4pε 0 r1 + r2 4pε 0 d 2 2 m2

⇒   v1 = v1 =

2m2 q1q2 ⎛ 1 1 1⎞ 4pε 0 m1 ( m1 + m2 ) ⎜⎝ r1 + r2 d ⎟⎠

Substituting values, we get 2 ( 0.7 ) ( 9 × 109 ) ( 2 × 10 -6 ) ( 3 × 10 -6 ) ⎛ 1 1⎞ - ⎟ ⎜⎝ ( 0.1 )( 0.8 ) 8 × 10 -3 1 ⎠

⇒   v1 = 10.8 ms -1 ⇒   v2 =

m1v1 0.1 kg ( 10.8 ms -1 ) = 0.7 kg m2

30°

O

u = 6 ms -1 41. (a) Consider the two spheres as a system By Law of Conservation of Momentum





42.

(b) If the spheres are conductors/metal, then electrons will move around them such that the centres of the excess charges can be placed inside the spheres without any loss in energy. Then just before they touch, the effective distance between charges will be less than r1 + r2 and the spheres will really be moving faster than that calculated in (a). u = 20 ms -1 , α = 45°

ax =

qE = 40 ms -2 m

ay = g = -10 ms -2

01_Ch 1_Hints and Explanation_P3.indd 101

x R

Since at maximum height vy = 0 , so

vy = uy + ay t

⇒ 0 = u sin α + ( - g ) T



⇒ T=

2uy ay

2u sin α 2 × 20 1 = × =2 2 s g 10 2

=

In time t = T , the projectile touches the ground at a distance R from the launch point. So, Range is given by R = uxT +

1 ax T 2 2



⎛ 20 ⎞ ( ⎛ ⇒ R=⎜ 2 2)+⎜ ⎝ ⎝ 2 ⎟⎠



⇒ R = 40 + 160



⇒ R = 200 m

43.

ax =

qE m



⇒

dvx q = ( 5 - 2x ) dt m

2 1⎞ ⎟⎠ ( 40 ) ( 2 2 ) 2

At time t v

⇒   v2 = 1.55 ms -1 v1 = 1080 cms -1 and v2 = 155 cms -1 So, 

g

u

2

CHAPTER 1



E = (5 – 2x) × 106

dv q = ( 5 - 2x ) dx m



⇒ v



⇒



⇒ vx2 =



⇒

vx2 q ( = 5x - x 2 ) 2 m

dx = dt

2q ( 5x - x 2 ) m 2q ( 5x - x 2 ) m

For x to be maximum

⇒ 5x - x 2 = 0



⇒ x=5m

dx =0 dt

9/20/2019 11:39:47 AM

H.102  JEE Advanced Physics: Electrostatics and Current Electricity 44. The situation is visualised in the figure



10 cm + + + + + + + + + +

vyfinal = vyinitial + at

m u q



⇒ E=

E



y = 1 cm – – – – – – – – – –

⎛ eE ⎞ ⎜⎝ ⎟⎠ tan θ = = mu vx u vy

300 = 15000 Vm -1 2 100

Since the particle does not come out hence its maximum deflection in vertical direction has to be y = 1 cm = 10 y =

-2



m 2

1 2 1 qE ⎛  ⎞ at = ⎜ ⎟  2 2 m ⎝ u⎠

⎛ eE ⎞ ⎛  ⎞ ⇒ vy = ⎜ ⎟ ⎜ ⎟ ⎝ m ⎠⎝ u⎠

If θ be the angle of deviation, then

V d

Since, E =

 u

From (1), we get t =

qE ⎫ ⎧ and t = ⎬ ⎨ As a = m u ⎩ ⎭

1 ⎛ qE ⎞ 2 ⇒ u = ⎜ x 2 ⎝ my ⎟⎠ 2

eE ⎡ ⎤ ⎛ 1 ⎞ = tan -1 ⎜ ⇒ θ = tan -1 ⎢ = 30° eE ⎞ ⎥ ⎝ 3 ⎟⎠ ⎛ ⎢ m⎜ 3 ⎥ ⎟ m ⎠⎦ ⎣ ⎝

46. Lets consider V = 0 at point P. Then the potential at the original position of the charge is Vi = - Ex = - EL cos θ

1 ( 1.6 × 10 ) ( 15000 ) ⎛ 1 ⎞ 8 ⎜ ⎟ = 10 2 ( 12 × 10 -24 )( 10 -2 ) ⎝ 10 ⎠ 2

-19

⎛ eE ⎞ ⇒ θ = tan -1 ⎜ ⎝ mu2 ⎟⎠

At the final point A , V f = - EL . Since the table is frictionless, by Law of Conservation of Energy



⇒ u2 =



⇒ u = 10 4 ms -1

( K + U )i = ( K + U ) f



⇒ u = 10 kms -1



⇒ 0 - qEL cos θ =



⇒ v=



⇒ v=



⇒ v = 0.3 ms -1



⇒ v = 30 cms -1

45. Along the x-direction speed of electron remains uniform; because of no component of field along horizontal. So, vx = u and x = ut

⇒  = ut …(1) +

+

+

+

+

+

+

y

u –

–

θ

–

– l

–

–

–

2qEL ( 1 - cos θ ) m 2 ( 2 × 10 -6 C )( 300 NC -1 ) ( 1.5 m ) ( 1 - cos 60° ) 0.01 kg

47. Total charge x

∫

r=R

Q = ρdV =

Along y direction uy = 0 Acceleration along y-direction of motion of electron is given by eE ay = a = m

∫

Q′ = ρdV =

⎛ eE ⎞ ⇒ vy = 0 + ⎜ ⎟ t ⎝ m⎠

01_Ch 1_Hints and Explanation_P3.indd 102

a

2

a+ 3

)

r=0

R 2

∫

( Kr a ) ( 4p r 2dr ) =

r=0

4p k ⎛ R ⎞ ⎜ ⎟ a+3⎝ 2 ⎠

a+ 3

According to question





4p k

∫ ( Kr )( 4p r dr ) = a + 3 ( R

r=

Since, vy = uy + ay t

1 mv 2 - qEL 2

Q′ 1 1⎛ 1 Q ⎞ = 4pε 0 ⎛ R ⎞ 2 8 ⎜⎝ 4pε 0 R2 ⎟⎠ ⎜⎝ ⎟⎠ 2

Putting the value of Q and Q′ get

a = 2

9/20/2019 11:40:06 AM

Hints and Explanations H.103

ARCHIVE: JEE MAIN T cos θ = mg



T sin θ = qE

1

θ

qE

T

θ

θ

mg



⇒ tan θ =

qE mg



⇒ tan θ =

5 × 10 -6 × 2000 1 = 2 2 × 10 -3 × 10

2.



⎛ ⎞ x2 ⇒ V1 - V2 = ⎜⎝ - A - Bx ⎟⎠ 2 -5



⎞ ⎛ A ⎞ ⎛A ⇒ V1 - V2 = ⎜ - - B ⎟ + ⎜ 25 + B ( -5 ) ⎟ ⎠ ⎝ 2 ⎠ ⎝ 2



⇒ V1 - V2 = 12 A - 6B = 240 - 60 = 180 V



Hence, the correct answer is (A).

4.

Since τ = Ια q d 2 θ

1⎞ ⇒ tan ⎜ ⎟ ⎝ 2⎠ Hence, the correct answer is (D). -1 ⎛

For Case-1 Electric field between spherical surface

 E =

Q 4pε 0 r 2

E



For Case-2 Electric field between the surfaces remains unchanged. +Q –Q –3Q



⇒

v1

1

v2

-5

∫ dV = ∫ - ( Ax + B ) dx

01_Ch 1_Hints and Explanation_P3.indd 103

⎛ md 2 ⎞ ⎛ d 2θ ⎞ d 2θ = 2 ⎜⎝ ⎟⎜ ⎟ 4 ⎠ ⎝ dt 2 ⎠ dt 2

⇒ - ( qE ) dθ = Ι



⇒



⇒ ω=



Hence, the correct answer is (A).

5.

E=

d 2θ 2qEθ md dt 2 2qE md

λ 2pε 0 r VP





λ

∫ dV = ∫ - 2pε r dr 0

VG

⇒ VP - VG =

Since, So, potential difference between them also remains unchanged. Hence, the correct answer is (B).   3. Since dV = - Edr = - ( Ax + B ) dx

–q



+Q –Q +Q

d 2

qE

qE

CHAPTER 1

1.

λ ⎛ r⎞ n⎜ ⎟ 2pε 0 ⎝ r0 ⎠

1 mv 2 = q ( VP - VG ) 2

qλ 1 ⎛ r⎞ mv 2 = n ⎜ ⎟ 2 2pε 0 ⎝ r0 ⎠



⇒



⎡ ⎛ r ⎞ ⎤2 ⇒ v ∝ ⎢ n ⎜ ⎟ ⎥ ⎣ ⎝ r0 ⎠ ⎦



Hence, the correct answer is (A).

1

9/20/2019 11:40:19 AM

H.104  JEE Advanced Physics: Electrostatics and Current Electricity

6.



U=

Qq Qq ⎞ 1 ⎛ q2 - + d⎞ ⎟ 4pε 0 ⎜⎜ d ⎛ d ⎞ ⎛ ⎜⎝ + D ⎟⎠ ⎜⎝ D - ⎟⎠ ⎟ ⎝ 2 2 ⎠

⇒ U=

1 ⎛ q2 qQd ⎞ ⎜ - - 2 ⎟⎠ 4pε 0 ⎝ d D

y

+q

2 ⎡ 2 ⎛ q2 ⎞ ⎤ 2 ⎢ ⎥ m ⎢⎣ 15 ⎜⎝ 4pε 0 a ⎟⎠ ⎥⎦

⇒ v0 =



Hence, the correct answer is (D).

x

D

B

+q

Since, U1 =

–q

2qD 3

4pε 0 ( D2 + 4 d 2 ) 2

-

mB = 4 μ g = 4 × 10–9 kg

1 mm

A (Fixed)

d

B′

q1q2 qq    U f = 1 2 4pε 0 ri 4pε 0 r2

By Law of Conservation of Energy, we have

( U + K )i = ( U + K ) f

2qD 3

4pε 0 ( D2 + d 2 ) 2



(along -ive x direction)  ⇒ E =

4 ⎛ 1 ⎞ ⎛ q2 ⎞ ⎜ ⎟ 15 ⎜⎝ 4pε 0 ⎟⎠ ⎝ ma ⎠



D >> d

d



2 ⇒ v0 =

1 μC

d

 E =



9.

–q d

⇒

1

Hence, the correct answer is (A).

7.

Q ⎛Q Q⎞ 1 mv02 = - ⎟ ⎜ 2 4pε 0 ⎝ 3 a 5 a ⎠



2qD ⎡ 1 1 3 ⎢ 3 3 4pε 0 D 2 2 ⎢⎛ 2d ⎞ ⎞ 2 ⎛ d ⎞ ⎞2 ⎛ ⎛ ⎢ ⎜ 1+ ⎜ ⎜⎝ 1 + ⎜⎝ ⎟⎠ ⎟⎠ ⎝ D ⎟⎠ ⎟⎠ ⎢⎣ ⎝ D

⎤ ⎥ ⎥ ⎥ ⎥⎦

2q ⎛ 3 4d2 3 d2 ⎞ 1-1+ ⎟ 2⎜ 2 2 D 2 D2 ⎠ 4pε 0D ⎝

q1q2 qq 1 = 1 2 + mv 2 4pε 0 r1 4pε 0 r2 2

v 2 =

2q1q2 ⎛ 1 1 ⎞ 4pε 0 m ⎜⎝ r1 r2 ⎟⎠ 1⎞ 2 × 9 × 109 × 10 -12 ⎛ 1 - ⎟ = 4 × 109 -9 -3 ⎜ ⎝ 9⎠ 4 × 10 × 10



2 ⇒ v =



⇒ v = 40 × 10 4 ms -1 = 6.32 × 10 4 ms -1 Hence, the correct answer is (D). T = 2p



 ⇒ E =



 9qd 2 ⇒ E = 4pε 0 D 4

10.



Hence, the correct answer is (C).

⎛ gE ⎞ 2 where, geff = g + ⎝⎜ ⎟ m⎠

8.

+Q, R

Potential at any point of the charged ring

VP =

Q 4pε 0 R2 + x 2

If v0 is the minimum velocity for the point charge to reach the centre, then 1 mv02 = Q ( VC - VP ) 2

01_Ch 1_Hints and Explanation_P3.indd 104

2

L



⇒ T = 2p



Hence, the correct answer is (C).

Q



L geff

⎛ qE ⎞ g2 + ⎜ ⎝ m ⎟⎠

2

V = 0 and   p 3 cos 2 θ + 1 E = 4pε 0 r 3 11.

p and r = d , we get 2   -p E = 4pε 0 d 3 For θ =



Hence, the correct answer is (C).

9/20/2019 11:40:34 AM

Hints and Explanations H.105

13. According to Gauss Law, we have E ( 4p a

) = qenclosed

2

…(1)

ε0

Qq

FQ =



⎛ d⎞ 4pε 0 ⎜ ⎟ ⎝ 2⎠

+

2

Q2 =0 4pε 0 d 2

⇒ 4Qq + Q 2 = 0



Q 4 Hence, the correct answer is (C).

15.

E( x ) =



⇒ q=-

P

Qx

CHAPTER 1

12. Since net charge on the dipole is zero, so the inside surface shall have non-zero and non-uniform charge distribution. Net field outside the region would be same as that would have been for point charge at surface. Hence, the correct answer is (C).

3

4pε 0 ( R2 + x 2 ) 2 Q R x

a



∫

⇒ qenc = ( kr ) × 4p r dr 2

0

⎛ a4 ⎞ = 4p k ⎜ ⎟ = 2Q …(2) ⎝ 4⎠



⇒ qenc



From (1), we get

E ( 4p a 2 ) =





4p ka 4 4ε 0

For maximum,

dE =0 dx

3 1 ⎡ 2 ⎤ 2 )2 ( ( x ) 3 ( R 2 + x 2 ) 2 ( 2x ) ⎥ + R x ⎢ Q 2 ⎢ ⎥=0 ⇒ 4pε 0 ⎢⎣ ⎥⎦ ( R 2 + x 2 )3 1

ka 2 ⇒ E= 4ε 0

( R2 + x 2 ) 2 Q ( 2 2 x + R - 3x 2 ) = 0 3 4pε 0 ( R2 + x 2 )



⇒

Since both charges do not experience any force, so ( Fnet )A = 0



⇒ x=

R 2





⇒ h=

R 2



Hence, the correct answer is (A).



⇒ ka 2 Q2 ×Q = ⇒ 4ε 0 4pε 0 × 4 a 2

Since, 2Q =

4p kR2 4

16.

(0, 3) 3

2 Q2 ⎛ 2Q ⎞ ⎛ a Q ⎞ = ⎜ ⎟ ⎜ ⎟ 4 ⎝ p R ⎠ ⎝ 4ε 0 ⎠ 4pε 0 × 4 a 2

14.

P

2Q ⇒ k= p R4

q1(0, 3)

⇒ 8 a4 = R4 ⇒ a=

R 1 84

=8

q2(4, 0)

1 -

1 4R

4



Hence, the correct answer is (D). +Q

q

+Q

0

d/2

d

01_Ch 1_Hints and Explanation_P3.indd 105

By Principle of Superposition, we have    E = E1 + E2 where,  E1 =

1 10 × 10 -6 ⎛ iˆ + 3 ˆj ⎞ × ⎜⎝ ⎟ 4pε 0 10 10 ⎠

9/20/2019 11:40:48 AM

H.106  JEE Advanced Physics: Electrostatics and Current Electricity

( -25 × 10 -6 ) ⎛ 4iˆ + 3 ˆj ⎞ 1 × ⎜⎝ ⎟ 5 ⎠ 4pε 0 25

 E2 =



Hence, the correct answer is (A).



(

(

(

Q=

⎛ A

∫ 4p r ⎜⎝ r 2

2

2r e a

⇒ Q=

-2

-

2R a

0

VA =

2R e a

Q 2p aA



⇒ Q=e



⇒ e



⇒ R=



Hence, the correct answer is (A).

18.

V∝

1 d2



⇒

4 qa 2qa = 2 ( R - x )2 x



x ⇒ (R - x) = 2



⇒ x=



Hence, the correct answer is (B).

19.

Q = 4pσ ( a 2 + b 2 + c 2 )



⇒ V=



2R a

⇒ V=

=

= 1-

(

= 2p aA 1 -

1 Q ⎞ ⎛ ⎜⎝ 1 ⎟ 2p aA ⎠

1 a ⎛ ⎞ ln ⎜ Q ⎟ 2 ⎜⎝ 1 ⎟ 2p aA ⎠

)

Hence, the correct answer is (A).

01_Ch 1_Hints and Explanation_P3.indd 106

1 ⎛ 2Q ⎞ Q+ ⎟ 4pε 0 ⎜⎝ 2 5⎠

So, potential energy is given by

U =

(4, –2)

q Q 1 ⎛Q Q ⎞ + + + ⎜ ⎟ 2 2 2 4pε 0 ⎝ 2 2 2 + 42 ⎠ 2 +4



1 ⎞ Q2 ⎛ ⎜⎝ 1 + ⎟ 4pε 0 5⎠



Hence, the correct answer is (D).

21.

F=

p Q , for electric dipole 4pε 0 y 3

and F ′ =



p ⎛ y⎞ 4pε 0 ⎜ ⎟ ⎝ 3⎠

3

Q=

27 pQ 4pε 0 y 3

⇒ F ′ = 27 F Hence, the correct answer is (C).

22. For spherical shell V =

q for r ≤ r0 4pε 0 r0

q for r > r0 4pε 0 r V q 4πε0 r0

2R

Q ⎛ a+b+c ⎞ ⎜ ⎟ 4pε 0 ⎝ a 2 + b 2 + c 2 ⎠

Q

⇒ VA =

O

2 + 1)

σ ⎛ 4p a 2 4p b 2 4p c 2 ⎞ + + ⎜ ⎟ 4pε 0 ⎝ a b c ⎠

(4, 2) x



V= and

(

Q

(0, –2) Q

⎞ ⎟⎠ dr

-2 r R 4p Aa a e

Q

A (0, 0) Q

)

0



(0, 2)

)

)

R

17.

y

⎡ iˆ + 3 ˆj ⎛ 4iˆ + 3 ˆj ⎞ ⎤  9 -6 -⎜⇒ E = 9 × 10 × 10 ⎢ ⎟⎥ ⎝ 10 5 ⎠⎦ ⎣  9 × 10 3 - iˆ + 3 ˆj + 8iˆ - 6 ˆj ⇒ E= 10  ⇒ E = 9 × 10 2 +7 iˆ - 3 ˆj  2 -1 ⇒ E = 63iˆ - 27 ˆj × 10 Vm



20.

r = r0



Hence, the correct answer is (A).

23.

0=U =



1 ⎞ ⎛ ⇒ -q = Q ⎜ 1 + ⎟ ⎝ 2⎠



r

1 ⎛ q2 Qq Qq ⎞ + + ⎜ ⎟ 4pε 0 ⎝ a a 2a ⎠

-q 2 2 +1 Hence, the correct answer is (A). ⇒ Q=

9/20/2019 11:41:06 AM

Hints and Explanations H.107 24.

  U = -p ⋅ E

28. The potential of the shell B,



⇒ U = - pE cos ( 45° )

VB =



⇒ U = -10 -29 × 10 3 ×



⇒ U = -7 × 10 -27 J



Hence, the correct answer is (C).

25. By Law of Conservation of Energy, we get





⇒ VB =

4p ⎛ σ × a 2 σ × b 2 σ × c 2 ⎞ + ⎜ ⎟ b c ⎠ 4pε 0 ⎝ b



⇒ vB =

⎞ σ ⎛ a2 - b 2 + c⎟ ⎜ ⎠ ε0 ⎝ b

1 Q2 Q2 = + mv 2 4pε 0 R0 4pε 0 R 2

⇒ v=

C B

2Q ⎛ 1 1⎞ - ⎟ ⎜ 4pε 0 m ⎝ R0 R ⎠

b

–2q l , 3 l 2 2

(0, 0) +q

l (l, 0) l

+q

x

⎛ ⇒ px = ( q ) ( 0 ) + ( q ) (  ) + ( -2q ) ⎜ ⎝



⇒ px = 0

⎞ ⎟ 2⎠

Similarly, py = q1y1 + q2 y 2 + q3 y 3

⎛ 3 ⎞ ⇒ py = ( q ) ( 0 ) + ( q ) ( 0 ) + ( -2q ) ⎝⎜ ⎟ 2 ⎠



⇒ py = - 3 q



 ⇒ p = px iˆ + py ˆj



 ⇒ p = - 3 qˆj



Hence, the correct answer is (C).

27. Since E =

Q σ = ε 0 Aε 0



⇒ Q = ε 0 AE = ( 8.85 × 10 -12 ) ( 1 ) ( 100 )



⇒ Q = 8.85 × 10 -10 C



Hence, the correct answer is (A).

01_Ch 1_Hints and Explanation_P3.indd 107



Hence, the correct answer is (B).

Q 6ε 0



So, flux through each face ϕ =



Hence, the correct answer is (B).

30. Charge density on given solid ball varies as

Since px = q1x1 + q2 x2 + q3 x3

c

29. Charged particle can be considered at the centre of a cube of side a, and the given surface represents its one side.

y

l

+σ

+σ

a

Hence v increases and attains a finite value after a large time. Hence, the correct answer is (D). 26.

–σ

A

2

CHAPTER 1

1 2

1 ⎛ qA qB qC ⎞ + + 4pε 0 ⎜⎝ rb rb rc ⎟⎠

r⎞ ⎛ ρ = ρ0 ⎜ 1 - ⎟ , for 0 ≤ r ≤ R ⎝ R⎠

Electric field outside the ball is given by

E =

1 q …(1) 4pε 0 r 2

Now, dq = ρdV = ρ ( 4p r 2 ) dr

R

r⎞ ⎛ ⇒ q = dq = ρ0 ⎜ 1 - ⎟ ( 4p r 2 ) dr ⎝ R⎠

∫

∫ 0

R



⎛ R3 R3 ⎞ ⎡ r3 1 r4 ⎤ ⇒ q = ( 4pρ0 ) ⎢ - × ⎥ = 4pρ0 ⎜ ⎟ ⎝ 3 4 ⎠ ⎣ 3 R 4 ⎦0



⎛ R3 ⎞ ⇒ q = 4pρ0 ⎜ …(2) ⎝ 12 ⎟⎠



From (1) and (2), E =



Hence, the correct answer is (A).

ρ0 R 3 12ε 0 r 2

9/20/2019 11:41:24 AM

H.108  JEE Advanced Physics: Electrostatics and Current Electricity

31. Initially force between spheres A and B, F =

1 q2 4pε 0 r 2

When A and C are touched, charge on both will be q . Again C is touched with B the charge on B is 2 given by q + q 3q qB = 2 = 2 4 q

q r



 T2 = p cos θ iˆ + p sin θ ˆj × 3E1 ˆj

(

iˆ

)



⇒ -τ kˆ = 3 pE1 cos θ kˆ …(2)



From (1) and (2), we get

pE sin θ = 3 pE cos θ

⇒ tan θ = 3



⇒ θ = 60°



Hence, the correct answer is (C).

35.

ϕ=

Required force between spheres A and B is given by

qenclosed ε0

For S1 , ϕ1 =

2q ε0

For S2 , ϕ2 =

3 q - q 2q = ε0 ε0

32. Equilibrium position will shift to a point where resultant force is zero. So,

For S3 , ϕ3 =

q + q 2q = ε0 ε0

kxeq = qE

For S4 , ϕ 4 =

8 q - 6 q 2q = ε0 ε0

q 3q 3 ⎛ 1 q2 ⎞ 3 1 qA qB 1 2× 4   = = ⎜ F′ = = F 2 2 8 ⎝ 4pε 0 r 2 ⎟⎠ 8 4pε 0 r 4pε 0 r

Hence, the correct answer is (A).

qE k



⇒ xeq =



Total energy of the system is

So, ϕ1 = ϕ2 = ϕ3 = ϕ 4 =

1 1 q 2E2 E = mω 2 A 2 + 2 2 k

2q ε0

Hence, the correct answer is (A).

36. Applying Gauss’s theorem for radius r, we get   1 E ⋅ dA = ( Q + q ) ε0

∫

Hence, the correct answer is (D).  33. p = p cos θ iˆ + p sin θ ˆj  E1 = Eiˆ

iˆ



⇒ E ( 4p r 2 ) =

1 ( Q + q ) …(1) ε0 P

y r

a

Q p

θ

x



   ⇒ T1 = p × E1  ⇒ T1 = p cos θ iˆ + p sin θ ˆj × E ( iˆ ) iˆ

(

01_Ch 1_Hints and Explanation_P3.indd 108

r

q =

)

⇒ τ kˆ = pE sin θ ( - kˆ ) …(1)  E2 = 3E1 ˆj

dx

where q is charge enclosed between x = a and x = r .

z



b

x

∫ a

⇒

r

r

⎛ x2 ⎞ A 4p x 2 dx = 4p A xdx = 4p A ⎜ ⎟ ⎝ 2 ⎠a x

∫

q = 2p A ( r 2 - a 2 )

a

Substituting the value of q in equation (1), we get

E ( 4p r 2 ) =

1 ⎡ 2 2 ⎣ Q + 2p A ( r - a ) ⎤⎦ ε0

9/20/2019 11:41:44 AM

Hints and Explanations H.109 1 ⎛Q 2p Aa 2 ⎞ ⎜⎝ 2 + 2p A ⎟ 4pε 0 r r2 ⎠



⇒ E=



E will be constant if it is independent of r

Vcentre =

V 3V0 2

2



⇒



⇒ A=



Hence, the correct answer is (A).

V0

Q 2p a 2

3V0 4 V0 4 R1 = 0 R2 = R R R3 = 4R R4 = 4R 3 2

⎧⎪ 30 - 5 z 2 for z ≤ 1 m 37. Since, V ( z ) = ⎨ ⎪⎩ 35 - 10 z for z ≥ 1 m and E ( z ) = -

dV = 10 z for z ≤ 1 m and dz

Now, V =

E ( z ) = 10 for z ≥ 1 m So, the source is an infinite non-conducting thick plate of thickness 2 m .

⇒ E=

q q 2E since ρ0 = ε0 = 2 Aε 0 At t

2 × 10 ε 0 = 10ε 0 2 Hence, the correct answer is (B). ⇒ ρ0 =

⎛ DV ⎞ 38. Since, E = - ⎜ ⎝ Dr ⎟⎠ V0

V0 + ΔV V0 + 2ΔV

r0 Δr

Δr

where, DV and Dr are same for any pair of surfaces and hence, E = constant Since, electric field inside the spherical charge distribution is given by E =

39.

3V0 , so R1 = 0 …(1) 2

ρ r = constant 3ε 0

(

Q 5V0 3 R2 - R22 = 4 8pε 0 R3

r

)

R2 5 = 3 - 22 2 R



⇒



⇒ R2 =



Similarly, at R3 , V =



⇒ R3 =



Again, at R4 , V =



⇒ R4 = 4 R …(4)



So, we observe from (1), (2), (3) and (4), that

R …(2) 2 3V0 Q = 4 4pε 0 R3

4R …(3) 3 V0 Q = 4 4pε 0 R4

R1 = 0 , R4 > 2R and R4 - R3 = 4 R

4R 8R = > R2 3 3

Hence, the correct answer is (C) and (D).

40. Electric field lines must originate from the positive charge and end at the negative charge. Also they must exit and enter a surface normally, must be smooth and continuous.

⇒ ρr = constant

41. Electric field due to complete disc ( R = 2 a ) ,

1 r Hence, the correct answer is (C).

E1 =

⇒ ρ(r ) ∝

⎡ Q ( 2 2) ⎢ 8pε R3 3 R - r , for r < R 0 V=⎢ ⎢ Q , for r ≥ R ⎢ ⎣ 4pε 0 r

So, Vsurface

Q = V0 = , 4pε 0 R

01_Ch 1_Hints and Explanation_P3.indd 109

CHAPTER 1

Q 2p Aa = r2 r2



x h σ ⎛ σ ⎛ ⎞ ⎞ = 1⎜⎝ 1 2 2 ⎟ 2 2 ⎟ 2ε 0 ⎜⎝ 2 ε ⎠ 0 4a + h ⎠ R +x h ⎞ σ ⎛ ⎜ 1 - ⎟⎠  2ε 0 ⎝ 2a



⇒ E1 =



Electric field due to disc of radius R = a is

E2 =

σ ⎛ ⎜12ε 0 ⎝

{∵ h  a }

h⎞ ⎟ a⎠

9/20/2019 11:42:04 AM

H.110  JEE Advanced Physics: Electrostatics and Current Electricity

Hence, electric field due to given annular disc is

σh ⇒ E = E1 - E2 = 4ε 0 a σ 4 aε 0



⇒ C=



Hence, the correct answer is (C).



 ⎛ 4 × 10 3 ⎞ ˆ ⎛ 4 × 10 3 ⎞ ˆ i =⎜ i ⇒ Enet = ⎜ ⎝ 4 × 0.04 ⎟⎠ ⎝ 4 × ( 0.2 )2 ⎟⎠  ⇒ Enet = ( 25 × 10 3 NC -1 ) iˆ iˆ

Hence, the correct answer is (B).

  44. Since DV = - E ⋅ dr

∫

42.

2

+Q σ1



σ2

  ⇒ V2 - V1 = - E ⋅ dr

∫

–Q



1

On outer surface there will be no charge

So Q2 = σ 2 = 0 on inner surface total charge will be zero, however, charge distribution will be there, so Q1 = 0 and σ 1 ≠ 0

 where, dr = dxiˆ + dyjˆ and  E = 25iˆ + 30 ˆj NC -1

(

)

( 2, 2 )



⇒ DV = V - 0 = -

Hence, the correct answer is (C).

( 25iˆ + 30 ˆj ) ⋅ ( dxiˆ + dyjˆ ) ∫ ( ) 0,0

43. Due to quarter ring electric field intensity is

2 ⎞ ⎛2 ⇒ V = - ⎜ 25dx + 30 dy ⎟ ⎟⎠ ⎜⎝ 0 0

p So, due to each quarter section, θ = , so field inten2 sity is



⇒ V = - 25 ( x )0 + 30 ( y )0



⇒ V = - ( 25 × 2 + 30 × 2 ) V = -110 V = -110 J/C

y ++ – – – – + +



Hence, the correct answer is (D).

λ ⎛θ⎞ sin ⎜ ⎟ ⎝ 2⎠ 2pε 0 R

E

45.

Enet E

 2 2λ ˆ 2λ ˆ ⇒ Enet = 2E ˆi= i= i 4pε 0 R 4pε 0 R

2

)

∫

VA - VO = - Ex dx 2



and p R = L = 20 cm  ⎛ 4 × 10 3 ε 0 ⎞ ˆ ⎛ 4 × 10 3 ⎞ ˆ ⇒ Enet = ⎜ ⎟i ⎟ i = ⎜⎝ 4 L2 ⎠ ⎝ 4p 2ε 0 R2 ⎠

∫

⇒ VA - VO = -30 x 2 dx 0



⎛ x3 ⇒ VA - VO = -30 ⎜ ⎝ 3



⇒ VA - VO = -80 V



Hence, the correct answer is (C).

46.  

 2 ( 2Q ) ˆ ⎛ 4Q ⎞ ˆ i =⎜ 2 i ⇒ Enet = 4pε 0p R2 ⎝ 4p ε 0 R2 ⎟⎠

01_Ch 1_Hints and Explanation_P3.indd 110

2

O

F

2⎞

⎟ = -10 ( 8 )

0⎠

F y

θ

q

where Q = 10 3 ε 0C



(

x

Q 2Q = Since λ = ⎛ pR ⎞ pR ⎜⎝ ⎟ 2 ⎠

∫

A

2λ λ ⎛p⎞ sin ⎜ ⎟ = E = ⎝ 4 ⎠ 4pε 0 R 2pε 0 R

∫

––

–––

++ + + +

++

––

+

E =

a

q

0

a

Fnet = 2 F cos θ

⇒ Fnet =

2 × kq2

(

2 a2 + y 2

×

y

) ( a2 + y 2 )1 2

9/20/2019 11:42:21 AM

Hints and Explanations H.111 O

kq2 y (for y  a ) a3

⇒ F∝y



Hence, the correct answer is (A).

∫

V=



kQ ⇒ V= L



⇒ V=

L

∫ O

A

F

T sin θ B

x mg 2

dL kQ = ln 2 L+x L

T sin θ = F =

Q ln 2 4pε 0 L



⇒

q2 x =  2 4pε 0 x 2 mg



⇒

q2 x ∝ 2 2 x

When a charge q is taken from the centre to the surface, the change in potential energy is.



⇒ q2 ∝ x 3

⎛ R2 ρ 1 R2 ρ ⎞ 1 R 2 ρq DU = ( VC - VS ) q = ⎜ q= ⎟ 6 ε0 ⎝ 2ε 0 3 ε 0 ⎠



⇒ q ∝ x2

48. Potential at the centre of the sphere, VC =



R2 ρ 2ε 0

Potential at the surface of the sphere, VS =

1 R2 ρ 3 ε0

Statement 1 is false. Statement 2 is true. Hence, the correct answer is (B).

49. For uniformly charged sphere E

E =

r

1 Qr (For r < R ) 4pε 0 R3

E =

1 Q (For r = R ) 4pε 0 R2

E =

1 Q (For r > R ) 4pε 0 r 2

x ⎫ ⎧ ⎨∵ tan θ = ⎬ 2 ⎭ ⎩

when charge begins to leak from both the spheres at a constant rate, then 1



dq 3 2 dx ∝ x dt 2 dt

⇒ v∝x

-

1 2

⎧ dq ⎫ ⎨∵ = constant ⎬ ⎩ dt ⎭



Method-II  At any instant, loss in electrostatic potential energy is equal to gain in kinetic energy. So,

q2 ⎛1 ⎞ = 2 ⎜ mν 2 ⎟ ⎝2 ⎠ 4pε 0 x -

1 2



⇒ ν ∝x



Hence, the correct answer is (A).

51.

ϕ = ar 2 + b



dϕ = -2 ar …(1) dr   q According to Gauss’s theorem, E ⋅ dA = inside ε0 Electric field, E = -

The variation of E with distance r from the centre is as shown in figure. Hence, the correct answer is (B).



50. Method-I



⇒ ( -2 ar ) 4p r 2 =

 Figure shows equilibrium positions of the two spheres, so T cos θ = mg and



⇒ qinside = -8ε 0 ap r 3

01_Ch 1_Hints and Explanation_P3.indd 111

mg

3



R

x

F

q2 1 2 4pε 0 x mg

⇒ tan θ =

Hence, the correct answer is (D).

C

1 q2 4pε 0 x 2





l T

T

θ

kdq 1 where k = L+x 4pε 0

47.

l θ θ

T cos θ



CHAPTER 1

So, F ≈

∫

qinside  ε0

{using (1)}

9/20/2019 11:42:39 AM

H.112  JEE Advanced Physics: Electrostatics and Current Electricity

Charge density inside the ball is



According to given problem, we have θ ′ = θ

qinside 4 3 pr 3



From equations (3) and (4), we get K =

-8ε 0 ap r 3 4 3 pr 3



⇒ K=



Hence, the correct answer is (D).

ρinside =



⇒ ρinside =



⇒ ρinside = -6 aε 0



Hence, the correct answer is (D).

1.6

=

=2

53. Consider a thin spherical shell of radius x and thickness dx as shown in the figure. Volume of the shell,

52.

dV = 4p x 2 dx

30°

θ

ρ

( ρ - σ ) ( 1.6 - 0.8 )

1 σ⎞ ⎛ ⎜⎝ 1 - ρ ⎟⎠

dx θ

T

T

F

F mg

Gaussian surface

x

dx

O

r

R

mg



Initially, the forces acting on each ball are (i) Tension T (ii) Weight mg

Let us draw a Gaussian surface of radius r ( r < R ) as shown in the figure above.



(iii) Electrostatic force of repulsion F For its equilibrium along vertical

Total charge enclosed ( Qenc ) inside the Gaussian surface is

T cos θ = mg …(1)

and along horizontal

r

∫

⇒

When the balls are suspended in a liquid of density σ and dielectric constant K, the electrostatic force will become ( 1/K ) times, i.e. F ′ = ( F/K ), whereas weight mg ′ = mg - upthrust

⇒ mg ′ = mg - Vσ g 



σ⎞ ⎛ ⇒ mg ′ = mg ⎜ 1 - ⎟  ρ⎠ ⎝



For equilibrium of balls, we have

tan θ ′ =

{∵ Upthrust = Vσ g } ⎧ m⎫ ⎨∵V = ⎬ ρ⎭ ⎩

F′ F …(4) = mg ′ Kmg ( 1 - ( σ / ρ ) )

01_Ch 1_Hints and Explanation_P3.indd 112

0

r

Dividing equation (2) by (1), we get

F tan θ = …(3) mg

∫

0

T sin θ = F …(2)

r

⎛5 x⎞ Qenc = ρdV = ρ0 ⎜ - ⎟ 4p x 2 dx ⎝ 4 R⎠ ⎛5 x3 ⎞ Qenc = 4pρ0 ⎜ x 2 - ⎟ dx ⎝4 R⎠

∫ 0

r



⎛ 5 ⎛ 5 x4 ⎞ r4 ⎞ ⇒ Qenc = 4pρ0 ⎜ x 3 = 4pρ0 ⎜ r 3 ⎟ ⎟ ⎝ 12 ⎝ 12 4R ⎠ 0 4R ⎠



⇒ Qenc =



According to Gauss’s Law, E ( 4p r 2 ) =



⇒ E ( 4p r 2 ) =



⇒ E=



Hence, the correct answer is (C).

⎛ 5 3 r4 ⎞ 4pρ0 ⎛ 5 3 r 4 ⎞ ⎜⎝ r - ⎟⎠ = pρ0 ⎜⎝ r - ⎟⎠ R R 4 3 3 Qenc ε0

pρ0 ⎛ 5 3 r 4 ⎞ ⎜ r - ⎟⎠ ε0 ⎝ 3 R

pρ0 r 3 ⎛ 5 r ⎞ ρ0 r ⎛ 5 r ⎞ ⎜ - ⎟ ⎜ - ⎟= 4p r 2ε 0 ⎝ 3 R ⎠ 4ε 0 ⎝ 3 R ⎠

9/20/2019 11:42:55 AM

Hints and Explanations H.113 qQ 4πε 0a2

q pr

Consider a small element AB of length d subtending an angle dθ at the centre O as shown in the figure. j

d

A

dE cos θ θ

dE

dl

θ

i

O dE sin θ

d ⎫ ⎧ ⎨∵ dθ = ⎬ r ⎭ ⎩

⇒ dq = λ ( rdθ ) 

the electric field at the centre O due to the charge element is 1 dq λ rdθ = dE = 4pε 0 r 2 4pε 0 r 2 Resolving dE into two rectangular components, we get by symmetry,

∫

dEcos θ = 0

iˆ

( )

∫ 0

p

λ rdθ

∫ 4pε r 0

0

2

p

 qr sin θ dθ ˆ ⇒ E=j 4p 2ε 0 r 3 iˆ

∫ 0

Work done in moving 100 negative charges i.e. electrons from the positive to the negative potential i.e. from P to Q is W = ( 100 e - ) ( VQ - VP ) = ( -100 × 1.6 × 10 -19 ) ( -14 V )

⇒ W = 2.24 × 10 -16 J



Hence, the correct answer is (D). Q r, p R4

57. If the charge density, ρ =

The electric field at the point P distant r1 from the centre, according to Gauss’s theorem is

( )

sin θ - ˆj

r1 R

q ⎫ ⎧ ⎬ ⎨∵ λ = pr ⎭ ⎩

iˆ

(

)

qenc ε0

(

)

1 ρdV  ε0

p

E 4p r12 =

0



p  q sin θ dθ ˆ q ⇒ E=j = - 2 2 ( - cos θ ) ˆj 2 2 4p ε 0 r 4p ε 0 r 0

⇒

–4 V Q

P

Q p



q

So, the net electric field at O is

 E = dE sin θ - ˆj =



qQ 4πε0 a2

Hence, the correct answer is (A).

56. +10 V

Charge dq on the element is dq = λ d

45°

Q

B

θ r

Q

q

Q2 4πε 0 (2a2)

CHAPTER 1

54. Linear charge density, λ =

∫

 q ⎛ ⎞ E = - ⎜ 2 2 ⎟ ˆj ⎝ 2p ε 0 r ⎠

Hence, the correct answer is (D).



⇒ E 4p r12 =

(

⇒ E 4p r1

55. For net force on Q to be zero qQ qQ Q2 + cos 45 + cos 90 = 0 2 4pε 0 a 4pε 0 a 2 4pε 0 ( 2 a 2 )

⇒ Q = -2 2q



Q ⇒ = -2 2 q

01_Ch 1_Hints and Explanation_P3.indd 113

2

)

1 = ε0

∫ r1

Qr

∫ pR

4

{

∫

∵ qenc = ρdV

}

( 4p r 2dr )

0

2

Qr1 4pε 0 R 4



⇒ E=



Hence, the correct answer is (C).

58. Work done = Potential difference × charge

⇒ W = ( VB - VA ) × q ,

9/20/2019 11:43:12 AM

H.114  JEE Advanced Physics: Electrostatics and Current Electricity VA and VB only depend on the initial and final posi  tions and not on the path. Electrostatic force is a conservative force. if the loop is completed, VA - VA = 0 . No network is done as the initial and final potentials are the same. Both the statements are true but Statement-2 is not the reason for Statement-1.

A



B

Hence, the correct answer is (C).

ARCHIVE: JEE advanced Single Correct Choice Type Problems 1.

For uniformly distributed charged shell, the surface charge density ( σ ) is

σ =

Q 4p R2

So, charge of small area dA = α 4p R2 is

dq = σ dA = αQ

R/2 C

B

A

 Since potential at the surface before removing the charge dq is V0 , so V0 =

⇒ DEA = -



Hence, field at C increases by



R (dq)

αV0 R αV0 So, at A, magnitude of electric field decreases by R αQ E = Similarly, ( EC )initial = 0 and ( C )final 4pε 0 R2 αQ αV0 = ⇒ DEC = R 4pε 0 R2

Q 4pε 0 R

αV0 R Hence, the correct answer is (D).

2.

The sphere with cavity can be assumed as a complete sphere with positive charge of radius R1 + another complete sphere with negative charge and radius R2 .  E+ → E due to total positive charge  E- → E due to total negative charge. E = E+ + E

If we calculate it at P , then E- comes out to be zero.

When dq is removed, then potential at the centre is

So, E = E+

Vcentre = V0 - V( dq )

and E+ =

Q αQ = V0 ( 1 - α ) 4pε 0 R 4pε 0 R



⇒ Vcentre =



Similarly, potential at the point B (after removing dq) is

VB = V0 -



⇒

αQ = V0 ( 1 - 2α ) ⎛ R⎞ 4pε 0 ⎜ ⎟ ⎝ 2⎠

Vcentre VC 1-α = = VB VB 1 - 2α

Here, q is total positive charge on whole sphere.

1 ⎛ Q αQ ⎞ αV0 = Eshell R 4pε 0 ⎜⎝ ( 2R )2 ( R )2 ⎟⎠

01_Ch 1_Hints and Explanation_P3.indd 114

It is in the direction of OP or a.

Now, inside the cavity electric field comes out to be uniform at any point. This is a standard result. Hence, the correct answer is (D). 3.

According to principle of superposition, we get the electric field at A as EA =

1 q ( OP ) , in the direction of OP . 4pε 0 R13



At the shown position, net force on both charges is zero. Hence they are in equilibrium. But equilibrium of +q is stable equilibrium. So, it will start oscillations when displaced from this position. These small oscillations are simple harmonic in nature. While equilibrium of -q is unstable. So, it continues to move in the direction of its displacement. Hence, the correct answer is (C).

9/20/2019 11:43:28 AM

kQ 1 , where k = 2 4pε 0 R



⇒ X=

mgd q

2



⇒ X=

2kQ ⇒ E2 = 2 R

( 1.67 × 10 -27 ) ( 9.8 ) ( 10 -2 ) ( 1.6 × 10 -19 )



⇒ X ≈ 1 × 10 -9 V

7.

Hence, the correct answer is (C).   Electric flux, ϕ = E ⋅ A



⇒ ϕ = EA cos θ

E1 =

4.

E2 =

E3 =

k ( 2Q ) R

k ( 4Q ) R

( 2 R )3

kQ 2R 2



⇒ E3 =



Hence, the correct answer is (C).

5.

For inside points ( r ≤ R )

E = 0 ⇒ V = constant =

  Here, θ is the angle between E and A  In this question θ = 45° , because A is perpendicular to surface E = E0 and

1 q 4pε 0 R

A = ( 2 a ) ( a ) = 2 a 2

For outside points ( r ≥ R )

1 q 1 E = or E ∝ 2 2 4pε 0 r r V= and



⇒ ϕ = ( E0 ) ( 2 a 2 ) cos ( 45° ) = E0 a 2



Hence, the correct answer is (C).

CHAPTER 1

Hints and Explanations H.115

8.

1 q 1 or V ∝ 4pε 0 r r

F

F

On the surface ( r = R )

V = ⇒ E=



1 q 4pε 0 R



1 q σ = 2 ε0 4pε 0 R

⎛ σ2 ⎞ Fele = ⎜ p R2 ⎝ 2ε 0 ⎟⎠

where, σ =

q = surface charge density 4p R2

Electrostatics repulsive force :

F = Fele =

σ 2p R2 2ε 0

corresponding to above equations the correct graphs are shown in option (D). Hence, the correct answer is (D).



Hence, the correct answer is (A).

9.

In equilibrium, mg = qE

6.



In absence of electric field, mg = 6pηrv



⇒ qE = 6pηrv

qE

45°

qE mg

Net force

m = mg

qE 4 p Rr 3 d = 3 g 3



Net force is at 45° from vertical.



⇒ qE = mg



qX = mg  ⇒ d

01_Ch 1_Hints and Explanation_P3.indd 115

qE 4 ⎛ qE ⎞ d= p ⎜ ⎟ g 3 ⎝ 6pηv ⎠ X⎫ ⎧ ⎨∵ E = ⎬ d⎭ ⎩

After substituting value we get,

q = 8 × 10 -19 C

Hence, the correct answer is (D).

9/20/2019 11:43:48 AM

H.116  JEE Advanced Physics: Electrostatics and Current Electricity

10.



Q1 = σ ( 4p R2 ) = 4p R2σ –(Q2 + Q1) Q2 + Q1

(Q3 + Q2 + Q1)

–Q1 Q1 3R

⎛ 2q ⎞ ⎛ q ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 4 × 2q 2 1 1 3 3 F = = 2 4pε 0 ( 2R sin 60° ) 4pε 0 9 × 4 × 3 R2

⇒ F=

2q 2 (attractive) 9 × 3 × 4pε 0 R2



⇒ F=

1 q2 54 pε 0 R2



Potential at O is

R 2R

A B C

⎛ q q 2q ⎞ + 1 ⎜⎝ 3 3 3 ⎟⎠ V = =0 R 4pε 0

2

Q2 + Q1 = 16p R σ

⇒ Q2 = 16p R2σ - Q1 = 12p R2σ 2

Q3 + Q2 + Q1 = 36p R σ

Force between B and C

2

2



Hence, the correct answer is (C).

13.

A ≡ ( - a, 0, 0 ) , B ≡ ( 0, a, 0 )

2

⇒ Q3 = 36p R σ - 16p R σ = 20p R σ



⇒ Q :Q :Q = 1: 3 : 5 1 2 3



Hence, the correct answer is (B).

y

B

11. From Gauss’s Law, we have flux equal to

–Q

⎛ 6C ⎞ ⎛ 8C ⎞ - 7C + ⎜ ⎝ 2 ⎟⎠ ΣQenc ⎜⎝ 4 ⎟⎠ 2C = ϕ = = ε0 ε0 ε0

Hence, the correct answer is (A).

q 1 2. Net electric field due to both charges , will get can3 ⎛ 2q ⎞ celled. Electric field due to ⎜ - ⎟ will be directed in ⎝ 3 ⎠ –ve axis q/3 B C –2q/3 q/3

E 60°

O

A



⇒ E=

q 6pε 0 R2

⎛ 1 ⎜ ⎜ P.E. of system = 4pε 0 ⎝ P.E. of system ≠ 0

01_Ch 1_Hints and Explanation_P3.indd 116

z



Q

Since point charge is moved from A to B and

VA = VB = 0

⇒ W=0



Hence, the correct answer is (C).

14. Charge will be induced in the conducting sphere, but net charge on it will be zero. Hence, the correct answer is (D). 15. Inside the cavity, field at any point is uniform and non-zero. Hence, the correct answer is (B). 16. There will be an electric field between two cylinders (using Gauss theorem). This electric field will produce a potential difference. Hence, the correct answer is (A).

⎛ 2q ⎞ 1 ⎜⎝ 3 ⎟⎠ E = 4pε 0 R2

x

A

17. All the three plates will produce electric field at P along negative z-axis. Hence , q ⎛ 2q ⎞ q ⎛ 2q ⎞ ⎞ ⎛ q⎞ ⎜⎝ ⎟⎠ ⎜⎝ - ⎟⎠ ⎜- ⎟ ⎟ 3 + 3 3 + 3⎝ 3 ⎠ ⎟ 2R 2R sin 60° 2R cos 60° ⎠ 2

 ⎡ σ 2σ 2σ ˆ σ ⎤ ˆ + + k EP = ⎢ ⎥ -k = 2 2 2 ε ε ε εo o o ⎦ ⎣ o

( )



The correct answer is (B). Hence, the correct answer is (B).

9/20/2019 11:44:01 AM

Hints and Explanations H.117

Don’t confuse with the electric flux which is zero (net) passing over the Gaussian surface as the net charge enclosing the surface is zero. The correct answer is (C). Hence, the correct answer is (C). 19. According to option (D), the electric field due to P and S and due to Q and T add to zero. While due to U and R will be added up. Hence, the correct answer is (D).

Hence, the correct answer is (D).

20. Electric field is zero everywhere inside a metal or conductor i.e. field lines do not enter a metal. Also, field lines must enter or leave the conductor surface at right angles because it is an equipotential surface. Hence, the correct answer is (C). 21.

Ui =



⇒ DU = U f - U i

2KQqx 2 a3 For x  a , we have

⇒ DU =

DU ∝ x 2 Hence, the correct answer is (B). 23. Electrostatic lines of force never form a closed loop. Therefore, options (B) and (D) are wrong, Electrostatic lines of force emanate from positive charge and terminate on negative charge, therefore, option (A) is also wrong. Hence, the correct answer is (C). 24. Potential decreases in the direction of electric field. Dotted lines are equipotential lines. So VA = VC and VA > VB y

2Qq 4pε 0 a

Uf = and

CHAPTER 1

18. At any point over the spherical Gaussian surface, net electric field is the vector sum of electric fields due to + q1 , - q1 and q2 .

C

Qq ⎡ 1 1 ⎤ + 4pε 0 ⎢⎣ a + x a - x ⎥⎦ q x = –a

Q

Q

E



q

Hence, the correct answer is (B).

25. Net electrostatic energy of the configuration will be

x=a x=x Final position

U =

DU = U f - U i

2Qqx 2 for x  a ⇒ DU = 4pε 0 a 3



⇒ DU ∝ x 2



Hence, the correct answer is (B).

22. q

Q

x = –a

x=0

q

q

x = +a x = –a

Initial position

Q

q

x=x

x=a

Final position

1 4pε 0

⎡ q2 Qq Qq ⎤ + ⎢ + ⎥=0 a ⎦ 2a ⎣ a

-2q 2+ 2



⇒ Q=



Hence, the correct answer is (B).

26.

V=

q ⎛ 1 1 1 1 1 ⎞ ⎜ 1 - + - + - ......... ⎟⎠ 4pε 0 x0 ⎝ 2 3 4 5 6



⇒

V=



Hence, the correct answer is (D).

27.

s=

q log e 2 4pε 0 x0

1 2 at 2

U i =

1 ⎛ 2Qq q2 ⎞ + ⎟ ⎜ 4pε 0 ⎝ a 2a ⎠

where a =

and Uf =

Qq ⎡ 1 q2 1 ⎤ + + 4pε 0 ⎢⎣ a + x a - x ⎦⎥ 4pε 0 ( 2 a )



01_Ch 1_Hints and Explanation_P3.indd 117

x

B

q

x = +a x=0 Initial position

q x = –a

A

⇒

qE m

1 ⎛ qE ⎞ 2 1 ⎛ qE ⎞ 2 t1 = t2 2 ⎜⎝ me ⎟⎠ 2 ⎜⎝ mp ⎟⎠

9/20/2019 11:44:13 AM

H.118  JEE Advanced Physics: Electrostatics and Current Electricity 1



⎛ mp ⎞ 2 t ⇒ 2 =⎜ t1 ⎝ me ⎟⎠



Hence, the correct answer is (B). =0

28.

-

∫

q ( Q1 - Q2 ) ( 2 - 1 )



⇒

W=



Hence, the correct answer is (B).

4 2pε 0 a

32.  

+Q

–Q

  E ⋅ d is the line integral of electric field which is

Q a

 =∞

b

also called electrostatic potential. Since, charge is distributed non-uniformly, so E ≠ 0 at the centre. =0



⇒

-

∫

  E ⋅ d =

 =∞



V = Vsurface - Vouter

q =2 4pε 0 R

sphere



shell

Q Q 4pε 0 a 4pε 0b

⇒ V=

Hence, the correct answer is (A).

29. Field lines always enter or leave a surface normal to it. Hence, the correct answer is (D).  30. The electrical field E at all points on the x-axis will not have the same direction. Eq E

x –q (d, 0)

For - d ≤ x ≤ d, electric field is along positive x-axis, while for all other points it is along negative x-axis.  The electric field E at all points on the y-axis will be parallel to the x-axis (i.e. along iˆ ) (OPTION (C)). The electrical potential at the origin due to both the charges is zero, hence, no work is done in bringing a test charge from infinity to the origin. Dipole moment is directed from the -q charge to the +q charge (i.e. along - iˆ direction).

Hence, the correct answer is (C).

31.

Q1 Q2 V1 = + 4pε 0 a 4pε 0 2 a



V2 =

Q1 Q2 + 4pε 0 2 a 4pε 0 a



⇒

W1→ 2 = q ( V2 - V1 )



⇒

W1→ 2

q ( Q1 - Q2 ) ⎛ 1 ⎞ = ⎜⎝ 1 ⎟ 4pε 0 a 2⎠

01_Ch 1_Hints and Explanation_P3.indd 118

b

Potential difference = V ′ = Vsurface - Vouter sphere

E–q

q (–d, 0)

a



y

–3Q + Q = –2Q Q

–Q

shell



-2Q ⎞ -2Q ⎞ ⎛ Q ⎛ Q -⎜ + ⇒ V′ = ⎜ + ⎟ ⎝ 4pε 0 a 4pε 0b ⎠ ⎝ 4pε 0b 4pε 0b ⎟⎠



⇒ V′ =



Hence, the correct answer is (A).

Q ⎛ 1 1⎞ ⎜ - ⎟ =V 4pε 0 ⎝ a b ⎠

33. For equilibrium of Q l/2





Qq ⎛ 4pε 0 ⎜ ⎝

⎞ ⎟ 2⎠

2

l/2 q

Q

+

Q

Q2 =0 4pε 0  2

⇒ 4q + Q = 0

Q 4 Here, we are asked to calculate equilibrium for system. In this system q is already in equilibrium. So, we are calculating equilibrium condition for Q. Hence, the correct answer is (B).

⇒ q=-

34. By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will

9/20/2019 11:44:25 AM

Hints and Explanations H.119

F = 2



i.e., F =

- qQ 1 × 4pε 0 ( a 2 + x 2 )

1 4pε 0

Since h > 2R and r = sin α =

x 1

( a2 + x 2 ) 2



2 qQ x 3

( a2 + x 2 ) 2

As the restoring force F is not linear, motion will be oscillatory (with amplitude 2a ) but not simple harmonic. Hence, the correct answer is (D). 35. The potential at the centre of a hollow sphere is equal to the potential at its surface. Hence, the correct answer is (B). 36. From conservation of mechanical energy  Decrease in kinetic energy = increase in potential energy

⇒ cos α =

3R , so we get 5

r 3 = R 5 4 5

⎡ Q ( ⎤ 1 - cos α ) ⎥ …(1) Since ϕ = 2 ⎢ 2 ε ⎣ 0 ⎦

⇒ ϕ=

Q 5ε 0

Please note that a factor of 2 is used in equation (1), because we have two surfaces i.e., AB and CD of the shell that are enclosed by the cylinder. For OPTION (B) h
2R and r > R

cm.

r

Hence, the correct answer is (C).

R

Multiple Correct Choice Type Problems 1.

h

For OPTION (A) r A Q

R

B

α R

D

01_Ch 1_Hints and Explanation_P3.indd 119

C

h > 2R



⇒ Qenclosed = Q



⇒ ϕ=



For OPTION (D)

Q ε0

h > 2R and r =

4R i.e., r < R 5

9/20/2019 11:44:35 AM

H.120  JEE Advanced Physics: Electrostatics and Current Electricity Since Eeq = E0

r A

B

α

D



 ⇒ EB = Eeq - E0 = 0



Hence, (A) and (C) are correct.

3.

(a) Ω = 2p ( 1 - cos θ ) , where θ = 45°

C

+Q 45° 45°

R

Qenclosed = Q [ 1 - cos α ]

R

r 3 Since, sin α = = R 5 3 5



⇒ cos α =



2Q ( 1 - cos α ) ⇒ ϕ= 2ε 0



3⎞ Q⎛ ⇒ ϕ = ⎜1- ⎟ ε0 ⎝ 5⎠



⇒ ϕ=



Hence, (A), (B) and (C) are correct.

2.

 p E0 is the external field in the direction of p = 0 iˆ + ˆj 2

2Q 5ε 0

(

)

E0 B

A

Eeq

⇒  ϕ = -

Ω Q × 4p ε 0

⇒  ϕ = -

2p ( 1 - cos θ ) Q 4p ε0

⇒  ϕ = -

Q ⎛ 1 ⎞ ⎜1⎟ 2ε 0 ⎝ 2⎠

The negative sign signifies that the flux is due to field lines entering a surface. (b) The component of the electric field perpendicular to the flat surface will decrease so we move away from the centre as the distance increases (magnitude of electric field decreases) as well as the angle between the normal and electric field will increase. Hence, the component of the electric field normal to the flat surface is not constant.

P

A′

 So, ϕ through the curved and flat surface will be less than

E0 = Eeq (in magnitude)

but both in opposite direction (as shown)



⇒



⎛ p0 ⎞ 3 ⇒ R=⎜ ⎝ 4pε 0E0 ⎟⎠



p0 = E0 4pε 0 R3

Q . ε0

(d) Since, the circumference is equidistant from Q it will be equipotential V =

1

Also, A lies at axial line of dipole, so       2p + E0 = 2E0 + E0 = 3E0 EA = 3 4pε 0 R

Q . ε0

45°

B′

For an equipotential surface of radius r , the point B is the point at the equitorial line of the dipole. Since, on an equipotential surface, no tangential field exists, so we have

01_Ch 1_Hints and Explanation_P3.indd 120

(c) Total flux ϕ due to charge Q is

Q 4pε 0 2R

.

Hence, (A) and (D) are correct. 4.

Q λ σ = = 2 2 pε 2 ε0 r 4pε 0 r0 0 0



⇒ Q = 2pσ r02

9/20/2019 11:44:44 AM

Hints and Explanations H.121

(A) is incorrect, r0 =

λ pσ

⎛r ⎞ (B) is incorrect, E1 ⎜ 0 ⎟ = 4E1 ( r0 ) ⎝ 2⎠ 1 As E1 ∝ 2 r



1 ⎛r ⎞ ⇒ E2 ⎜ 0 ⎟ = 2E2 ( r0 ) as E2 ∝ ⎝ 2⎠ r



(C) is correct

⇒



Hence, (B) and (D) are correct.

7.

(A) +q



Hence, option (C) is correct.

5.  

P –ρ

ρ

C2 R2

C1 R1

F

E

O

A +2q

⎛r ⎞ E3 ⎜ 0 ⎟ = E3 ( r0 ) = E2 ( r0 ) ⎝ 2⎠ as E3 ∝ r 0

ρ1 32 =( ρ1 must be negative) ρ2 25



EB EE

2K EA ED

EC EF

+q B

–q

C

D ≡O –2q

120°

4K ≡ O

2K 4K

= 6K 2K –q EB = EE = EC = EF = K and EA = ED = 2K

 Resultant of 2K and 2K (at 120° ) is also 2K towards 4K . Therefore, net electric field is 6K . (B) V0 =

1 ⎡ qA qB qC qD qE qF ⎤ + + + + + ⎥ L L L L L ⎦ 4pε 0 ⎢⎣ L

V0 =

1 ( qA + ... + qF ) = 0 4pε 0 L

CHAPTER 1



Because qA + qB + qC + qD + qE + qF = 0

For electrostatic field,

ρ ( -ρ ) C P C1P + EP = E1 + E2 = 1 3ε 0 3ε 0



(C) Only line PR , potential is same ( = 0 ) .



Hence, (A), (B) and (C) are correct.

8.

OPTION (A) is correct due to symmetry. OPTION (B) is wrong again due to symmetry.



ρ ⇒ EP = ( C1P + PC2 ) 3ε 0



ρ C1C2 ⇒ EP = 3ε 0



Hence, (C) and (D) are correct.

OPTION (C) is correct because as per Gauss’s theorem, net electric flux passing through any closed surface q = in ε0

6.

At point P

Here, qin = 3 q - q - q = q C1

Q 2R



P C2 2R

If resultant electric field is zero, then



KQ1 KQ2 = R 4R2 8R3

ρ1 =4 ρ2



⇒



At point Q If resultant electric field is zero then

KQ1 KQ2 + =0 4 R2 25R2

01_Ch 1_Hints and Explanation_P3.indd 121

So, net electric flux =

q ε0

OPTION (D) is incorrect because there is no symmetry in two given planes. Hence, (A) and (C) are correct. 9.



If charges are of opposite signs, then the two fields are along the same direction. So, they cannot be zero. Hence, the charges should be of same sign. Therefore, OPTION (C) is correct.

⎛ Work done by ⎞ ⎛ Change in electrostatic ⎞ = Further, ⎜ potential energy ⎟⎠ ⎝ external force ⎟⎠ ⎜⎝

⇒ WA→ B = q ( DV ) = ( +1 ) ( VB - VA )



⇒ WA→ B = VB - VA

9/20/2019 11:44:53 AM

H.122  JEE Advanced Physics: Electrostatics and Current Electricity So, OPTION (D) is also correct. Hence the correct OPTIONS are (C) and (D). Hence, (C) and (D) are correct.

10. Inside a conducting shell electric field is always zero. Therefore, OPTION (A) is correct. When the two are connected, their potentials become the same. ⇒ VA = VB



⇒

⎧ 1 Q⎫ ⎨∵ V = ⎬ 4pε 0 R ⎭ ⎩

QA QB =  RA RB

Since, RA > RB

⇒ QA > QB



So, OPTION (B) is correct.



Potential is also equal to, V =

Since VA = VB

⇒ σ A RA = σ B RB



σ R ⇒ A = B α B RA



⇒ σ A < σB



So, OPTION (C) is correct.



Electric field on surface, E =



⇒ E ∝σ



Hence, option (A) is correct.

13. The given graph is of charged conducting sphere of radius. The entire charge distributes on the surface of the sphere.

Hence, (A), (B), (C) and (D) are correct.

14. Charge density at A < charge density at B. Since field inside cavity is zero, hence Potential at A = potential at B = a constant value. By Gauss Theorem, ⎛ Total electric ⎞ 1 ⎛ charge ⎞ ⎛ q ⎞ ⎜ ⎟⎠ = ⎜⎝ enclosed ⎟⎠ = ⎜ ⎟ flux ⎝ ε0 ⎝ ε0 ⎠

σR ε0



Hence, (C) and (D) are correct.

15. Inside the sphere i.e. for r < R E =

1 Q r 4pε o R3

⇒ E∝r

i.e., E at centre is Ecentre = E

σ ε0

= 0 (r = 0)

r=R

=

1 Q 4pε o R2

E

Since, σ A < σ B ⇒ EA < EB So, OPTION (D) is also correct. Hence the correct OPTIONS are (A), (B), (C) and (D). Hence, (A), (B), (C) and (D) are correct.

E=

1 Q 4πε0 R2

E



r =0

and E at surface is Esurface = E

O

11.   

Q1 Q2

r



12. Net torque on (-q) about a point (say P) lying over +Q is zero. Therefore, angular momentum of ( - q ) about point P should remain constant.

∝



1 r2

r=R

r

Outside the sphere, i.e. for r > R

E =

E∝

⇒ E∝

1 Q 4pε o r 2 1 r2

From the diagram, it can be observed that Q1 is positive, Q2 is negative.

Thus, variation of electric field ( E ) with distance ( r ) from the centre will be as follows.

Number of lines on Q1 is greater and number of lines is directly proportional to magnitude of charge.



So, Q1 > Q2 Electric field will be zero to the right of Q2 as it has small magnitude and opposite sign to that of Q1 .

Hence, (A) and (C) are correct.

16. Let Q be the charge on the ring, the negative charge -q is released from point P ( 0 , 0 , z0 ) . The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be

Hence, (A) and (D) are correct.

01_Ch 1_Hints and Explanation_P3.indd 122

9/20/2019 11:45:02 AM

Hints and Explanations H.123 2. x Q

E –q

O

P (0, 0, z0)

R

z

Fe

1 E = 4pε 0

Qz0

V0 r k



⇒ q=



Inside the cylinder, electric field is

E = ⎡⎣ V0 - ( -V0 ) ⎤⎦ h

3

( R2 + z02 ) 2

E = 0 at centre of the ring because z0 = 0

Force on charge at P will be towards centre as shown, and its magnitude is 1 Fe = qE = 4pε 0

As the balls keep on carrying charge from one plate to another, current will keep on flowing even in steady state. When at bottom plate, if all balls attain charge q, then kq = V0 r

Qqz0

3

( R2 + z02 ) 2

…(1)

Similarly, when it crosses the origin, the force is again towards centre O . Thus, the motion of the particle is periodic for all values of z0 lying between 0 and ∞ .

(



Secondly, if z0  R , then R2 + z02



⇒ Fe ≈

1 Qq z0  4pε 0 R3

3

) 2 ≈ R3 {From equation (1)}

i.e. the restoring force Fe ∝ - z0 . So, the motion of the particle will be simple harmonic. (The negative sign implies that the force is towards the mean position).



⇒ E = 2V0 h



⇒ Acceleration of each ball, a =



⇒ Time taken by balls to reach other plate is

t =

2h = a

2 hmk 1 = 2 hrV02 V0

mk r



If there are n balls, then



Average current, iav =



⇒ iav ∝ V02



Hence, the correct answer is (B).

3.

Electric field at r = R is

E =

qE ⎛ 2 hr ⎞ 2 = V0 m ⎜⎝ k m ⎟⎠

CHAPTER 1

y

nq Vr r = n × 0 × V0 t k mk

Q 4pε 0 R2 ρ (r)

Reasoning Based Questions 1.

Statement-1 is also practical experience based; so it is true. Statement-2 is also true but is not the correct explanation of Statement-1. Correct explanation is “there is increase in normal reaction when the object is pushed and there is decrease in normal reaction when object is pulled”. Hence, the correct answer is (B).

Linked Comprehension Type Questions 1.



After hitting the top plate, the balls will get negatively charged and will now get attracted to the bottom plate which is positively charged. The motion of the balls will be periodic but not SHM. Hence, the correct answer is (A).

01_Ch 1_Hints and Explanation_P3.indd 123

d

a

R

r



Q = Total charge within the nucleus = Ze



⇒ E=



Hence the electric field is independent of a. Hence, the correct answer is (A).

4.

Q = ρr 4p r 2 dr for a = 0 , and the figure for the distri-

1 Ze 4pε 0 R2

∫

bution looks like as shown

9/20/2019 11:45:10 AM

H.124  JEE Advanced Physics: Electrostatics and Current Electricity ρ

P

d ρr

30

° 30°

O

R

r

A

d ρ Slope = = r R R-r

⇒ ρr =



⇒ Q=

d (R - r) R R

∫ 0



d ( R - r ) 4p r 2 dr R

R ⎞ ⎛ R 4p d 4p d ⎛ R 4 R 4 ⎞ ⎜ R r 2 dr - r 3 dr ⎟ = ⇒ Q= ⎜ ⎟ R ⎜ R ⎝ 3 4 ⎠ ⎟⎠ ⎝ 0 0

∫

Q=

∫

p dR3 3



⇒ Q = Ze =



⇒ d=

p dR3  3

{∵ Q = Ze }

3 Ze p R3



Hence, the correct answer is (B).

5.

Let us review the formula of uniformly (volume) charged solid sphere according to which we have ρr E= 3ε 0

B



⎛ AM ⎞ ⇒ ∠AOM = tan -1 ⎜ ⎝ OM ⎟⎠



⎛ 1 ⎞ ⇒ ∠AOM = tan -1 ⎜ = 30° ⎝ 3 ⎟⎠



Electric flux passing from the whole cylinder qin λ L = ε0 ε0

ϕ1 =

 Electric flux passing through ABCD plane surface (shown only AB ) = Electric flux passing through cylindrical surface ANB . So ⎛ 60° ⎞ ϕ = ⎜ ϕ ⎝ 360° ⎟⎠ ( 1 )

λL 6ε 0



⇒ ϕ=



⇒ n=6

2.

Volume of cylinder per unit length (  = 1 ) is

V = p R2 = ( p R2 )

E

M

N a

So, charge per unit length is

⎛ Volume per ⎞ ⎛ Volume charge ⎞ 2 λ = ⎜ × ⎟⎠ = ( p R ρ ) density ⎝ unit length ⎟⎠ ⎜⎝

a=R

r



Now at P , Etotal = Eremainder + Ecavity



⇒ ER = ET - EC



⇒ E ∝ r , ρ should be constant throughout the volume of nucleus.



This will be possible only when a = R.

T = Total portion and



Hence, the correct answer is (C).

C = cavity portion

Integer/Numerical Answer Type Questions 1.

ANBP is cross-section of a cylinder of length L . The line charge passes through the centre O and perpendicular to paper.

AM =

a 3a , MO = 2 2

01_Ch 1_Hints and Explanation_P3.indd 124

Where R = Remainder portion



⇒ ER =

1 Q λ 2pε 0 ( 2R ) 4pε 0 ( 2R )2

where Q is charge on sphere given by Q =

3

p R3 ρ 4 ⎛ R⎞ p⎜ ⎟ ρ = 3 ⎝ 2⎠ 6

9/20/2019 11:45:19 AM

Hints and Explanations H.125 Substituting the values, we get

ER

( p R2 ρ ) = 4pε 0 R

ρ⎞ ⎛ p R3 ⎟ 1 ⎜⎝ 6⎠ 4pε 0 4 R2

23 ρR 23 ρR = 96ε 0 ( 16 )( 6 ) ε 0

⇒ ER =



⇒ k=6

3.

F1 = Net electrostatic force on any one charge due to rest of three charges. So



⎛ q2 ⎞ 3 1 ⎞ ⎞ 3 ⎛ q2 ⎞ 3 ⎛ 1 ⎛ ⇒ a=⎜ ⎜⎝ 2 + ⎟⎠ ⎟ ⎜ ⎟ = k ⎜ ⎟ ⎝ γ ⎠ ⎝ 4pε 0 2 ⎠ ⎝ γ ⎠



⇒ N=3

4.

Total charge

∫

r=R

F1 45°

∫

Q′ = ρdV =

F2

C

45° F1



F2 = Force due to surface tension = γ a



If we see the equilibrium of line BC , then

2 F1 cos ( 45° ) = F2

⇒



⇒

2 F1 = F2



a

2

a+ 3

)

r=0

R 2

∫

( Kr a ) ( 4p r 2dr ) =

r=0

4p k ⎛ R ⎞ ⎜ ⎟ a+3⎝ 2 ⎠

a+ 3

According to question





4p k

∫ ( Kr )( 4p r dr ) = a + 3 ( R

r=

F1

1

1 ⎞⎞3 ⎛ 1 ⎛ where, k = ⎜ ⎜2+ ⎟ ⎝ 4pε 0 ⎝ 2 ⎠ ⎟⎠

Q = ρdV = B

D

1

1

1 q2 ⎛ 1⎞ F1 = ⎜ 2 + ⎟⎠ 4pε 0 a 2 ⎝ 2 A

⇒ a3 =

1



F1

1 ⎛ 1 ⎞ q2 ⎜⎝ 2 + ⎟ 4pε 0 2⎠ γ



CHAPTER 1



Q′ 1 1⎛ 1 Q ⎞ = 4pε 0 ⎛ R ⎞ 2 8 ⎜⎝ 4pε 0 R2 ⎟⎠ ⎜⎝ ⎟⎠ 2

Putting the value of Q and Q′ get

a = 2

1 q2 ⎛ 1 ⎞ ⎜2+ ⎟ =γa 4pε 0 a 2 ⎝ 2⎠

01_Ch 1_Hints and Explanation_P3.indd 125

9/20/2019 11:45:25 AM

Chapter 2: Capacitance and Applications

Test Your Concepts-I (Based on General Capacitance)

⇒   380 × 10 −9 = 50 × 10 −12 × V1

1.

⇒   V1 =

The common potential,

V =

C1V1 + C2V2 C1 + C2



Here C1 = 20 mF = 20 × 10 −6 F , V1 = 500 V C2 = 10 mF = 10 × 10 −6 F , V2 = 200 V 20 × 10 −6 × 500 + 10 × 106 × 200 20 × 10 −6 + 10 × 10 −6



⇒ V=



⇒ V = 400 V

2.

The energy stored in the condenser is given by

U =

−6

F , V = 200 V

1 2 × 100 × 10 −6 × ( 200 ) = 2 J 2 When discharged through a resistor ( 2 Ω ) , the whole energy is dissipated as heat. So, heat produced is given by U =

Q = U = 2 J 3.

When the dial is at 0° , the capacitance of the capacitor is given by

C1 = 50 pF = 50 × 10 −12 F

When dial is at 180° , the capacitance is given by

C2 = 950 pF = 950 × 10 −12 F The potential difference across capacitor C2 , is given by V2 = 400 V

Charge on capacitor C2 is

q = C2V2 = 950 × 10 −12 × 400

⇒ q = 380 × 10 −9 C



(a) When battery is disconnected the charge remains the same and so, q = constant. Let V1 be the potential difference across capacitor when dial reads 0° . Then

   q = C1V1

02_Ch 2_Hints and Explanation_P1.indd 126

(b) Work required to turn the dial from 180° to 0° is W = Gain in energy of capacitor

⇒   W =

2 q2 q2 q2 ⎛ 1 1 ⎞ q ( C2 − C1 ) − = ⎜ − = ⎟ 2C1 2C2 2 ⎝ C1 C2 ⎠ 2C1C2

⇒   W =

( 380 × 10 −9 )2 ( 950 − 50 ) × 10 −12 2 × 50 × 10 −12 × 950 × 10 −12

⇒   W = 1.368 × 10 −3 J

1 CV 2 2

where C = 100 mF = 100 × 10

380 × 10 −9 = 7600 V 50 × 10 −12

4.

Initial energy stored in the capacitors

U1 =

⇒ U1 =

1 1 C1V12 + C2V22 2 2 1 1 × 0.1 × 10 2 + C2 × 0 = 5 J 2 2

When charging battery is removed, the charge remains constant. This charge is collected by first capacitor ( q1 = C1V1 ) then redistributed equally in such a way that their potentials are equal i.e., V1′ = V2′ q1′ q′ = 2 C1 C2



⇒



Now since, q1′ = q2′ {given}



⇒ C1 = C2 = C ( say )



∴ Common potential

V =

q1 + q2 CV1 + 0 V1 = = =5V C1 + C2 C+C 2

So, final energy stored is

U 2 =

1 ( C1 + C2 )V 2 2 1 ( 2C )V 2 = CV 2 = 0.1 × ( 5 )2 = 2.5 J 2



⇒ U2 =



⇒

5.

(a) Let the charge on sphere of radius R1 be q1 ( = q ) and that on the sphere of radius R2 be q2 ( = Q − q ) , then the total energy is

U 2 2.5 1 = = U1 5 2

9/20/2019 11:33:25 AM

Hints and Explanations H.127

Test Your Concepts-II (Based on Series and Parallel Combination of Capacitors)

1 q12 1 q22 + 2 C1 2 C2

q2 (Q − q ) + 8πε 0 R1 8πε 0 R2

2

⇒   U =

For U to be minimum we must have

1 1 2 2 2 C ( DV ) + C ( DV ) = C ( DV ) 2 2

1.

(a) U =



C (b) The altered capacitor has capacitance C ’ = . The 2 total charge is the same as before.

  

dU =0 dq

So, qinitial = qfinal

⇒  

2(Q − q ) 2q ( −1 ) = 0 + 8πε 0 R1 8πε 0 R2

⇒   C ( DV ) + C ( DV ) = C ( DV ′ ) +

⇒  

q Q−q = R2 R1

⇒   DV ′ =

4 DV 3 2

1 ⎛ 4 DV ⎞ 1 ⎛ C ⎞ ⎛ 4 DV ⎞ C⎜ ⎟⎠ + ⎜⎝ ⎟⎠ ⎜⎝ ⎟ ⎝ 2 3 2 2 3 ⎠

⎛ R1 ⎞ ⇒   q = ⎜ Q = q1 ⎝ R1 + R2 ⎟⎠

(c) U’=

⎛ R2 ⎞ and Q − q = ⎜ Q = q2 ⎝ R1 + R2 ⎟⎠

⇒   U ′ = 4C

(b) V1 =

1 q1 1 R1Q = 4πε 0 R1 4πε 0 R1 ( R1 + R2 )

⇒   V1 =   V2 =





⇒   V2 =

6.

R2Q 1 q2 1 = 4πε 0 R2 4πε 0 R 2 ( R1 + R2 )



Q 1 4πε 0 R1 + R2

⇒ C =

( C + C0 )V1 = C0V0

⇒ V1 =

C0V0 C + C0

C0V1 = ( C + C0 )V2 2

⎛ C0 ⎞ ⎛ C0 ⎞ ⎛ C0 ⎞ ⇒  V2 = ⎜ V0 = ⎜ V0 ⎝ C + C0 ⎟⎠ ⎜⎝ C + C0 ⎟⎠ ⎝ C + C0 ⎟⎠ and so on 10



⎛ C0 ⎞ ⇒ V10 = V = ⎜ V0 ⎝ C + C0 ⎟⎠



1 ⎡ ⎤ ⎛ V0 ⎞ 10 ⎢ ⇒ C= ⎜ − 1 ⎥ C0 ⎟ ⎢⎣ ⎝ V ⎠ ⎥⎦

02_Ch 2_Hints and Explanation_P1.indd 127

d d 1 1 1 = + = + C C1 C2 ε 0 A ε 0 A

ε0 A 2d When the plates are moved in the given manner, net capacity C’ of the system is given by   



Let V2 be the potential of C0 after second charging then



3 (d) The extra energy comes from work put into the system by the agent pulling the capacitor plates apart.

2. (a) Capacitors connected in series. Initially, capacitance C is

Let V1 be the potential of C0 after first charging, then

2

( DV )2

Q 1 4πε 0 R1 + R2

⇒   V1 − V2 = 0

C ( DV ′ ) 2

CHAPTER 2

   U = U1 + U 2 =

1 d − Dd d + Dd = + ε0 A ε0 A C’

{

∵ C1′ =

ε0 A ε A and C2′ = 0 d − Dd d + Dd

}

ε0 A 2d Net capacitance remains same as before.

⇒   C ’ =

(b)  Two identical capacitors C1 and C2 connected in parallel. Initially the net capacity of the system is    C = C1 + C2 = 2

ε0 A d

 where, A is the area of the plates, and d the distance between the plates, for both the capacitors.  When the plates are brought closer by Dd in ε0 A one capacitor, its capacity increases to ; d − Dd whereas the capacity of the second capacitor,

9/20/2019 11:33:45 AM

H.128  JEE Advanced Physics: Electrostatics and Current Electricity whose plates are moved away by Dd, decreases to ε0 A . Net capacity of the system becomes d + Dd 1 ⎞ 2ε 0 Ad ⎛ 1 + C ’ = ε 0 A ⎜ = ⎝ d − Dd d + Dd ⎟⎠ d 2 − ( Dd )2 ⇒ C ’ =

2ε 0 A ⎛ Dd 2 ⎞ d−⎜ ⎝ d ⎟⎠

We find that C ’ > C and hence net capacitance increases. 3.

The initial charge on the larger capacitor is

Q = C DV



An additional charge q is pushed through the 50 V battery, giving the smaller capacitor charge q and the larger charge 150 + q ( in mC ) .

C1

C2

⇒ 9C2 = C1 + 2 2C2 +2 3

2 μF



⇒ 9C2 =



⇒ 25C2 = 6



6 ⇒ C2 = = 0.24 mF 25

100 V

and C1 = 0.16 mF 6.

⇒ Q = ( 10 mF ) ( 15 V ) = 150 mC

⇒ C2 ( 90 ) = C1 ( 10 ) + ( 2 ) ( 10 )



1 1 CPV 2 = ( C1 + C2 )V 2 2 2 1 2 ⇒ ( C1 + C2 ) ( 2 ) = 0.1 2

U=

⇒

( C1 + C2 ) = 5 × 10 −2 …(1)

Similarly

1 CSV 2 = 1.6 × 10 −2 2

1 ⎛ C1C2 ⎞ ( 2 )2 = 1.6 × 10 −2 2 ⎜⎝ C1 + C2 ⎟⎠



⇒ 50 V=

q 150 mC + q + 10 mF 5 mF



⇒



⇒ 500 mC = 2q + 150 mC + q



⇒ C1C2 = 4 × 10 −4 …(2)



⇒ q = 117 mC



Solving equations (1) and (2), we get



Across the 5 mF capacitor DV =



Across the 10 mF capacitor

DV =

4.

150 mC + 117 mC = 26.7 V 10 mF

Uf =

⇒   Wext

7.

In the figure, C1 =

C2 =

ε0 A 2d

and  C3 =

ε0 A d

⎛ q2 ⎞ q2 =⎜ x2 2C f ⎝ 2ε 0 A ⎟⎠

q2 = W = U f − Ui = ( x2 − x1 ) 2ε 0 A

(b) Ui =

1 1⎛ ε A⎞ CiV 2 = ⎜ 0 ⎟ V 2 2 2 ⎝ x1 ⎠



Uf =

Initially C1 ( 60 ) = C2 ( 40 )

3C1 = 2C2 Now, q2 = q1 + qacross 2 mF

ε0 A 3d

A

C1

3d

B

C2

2d

C3

d

Now, C2 is short circuited so that C1 and C3 are in series.

1 1⎛ ε A⎞ C fV2 = ⎜ 0 ⎟ V2 2 2 ⎝ x2 ⎠

⇒   Wexternal = U f − U i =

02_Ch 2_Hints and Explanation_P1.indd 128

C1 = 40 mF and C2 = 10 mF



q2 ⎛ q2 ⎞ (a) U i = = x1 and 2Ci ⎜⎝ 2ε 0 A ⎟⎠



5.

q 117 mC = = 23.3 V C 5 mF

1⎞ 1 ⎛ 1 ε 0 AV 2 ⎜ − ⎟ x x 2 ⎝ 2 1⎠



⇒ C AB =

C1C3 C1 + C3



⇒ C AB =

ε0 A 4d

8.

In series the charge on each capacitor is same. q Therefore, potential difference given by V = across C the capacitors are in inverse ratio of their capacitors. Hence

9/20/2019 11:34:09 AM

Hints and Explanations H.129 q = 130 V and C1

V2 =

q = 100 V C2

This is the required expression for the net capacitance and is independent of d and hence of the position of the central H-shaped part.



⇒

C2 V1 130 = = C1 V2 100



⇒

C1 10 = …(1) C2 13

and so the smaller capacitance is C1 . As capacitance of a capacitor is proportional to the dielectric constant ( K ) , so

New capacitance C1′ of the first capacitor is

C1′ =

5 C1  2

 New capacitance of second capacitor C2′ does not change, so C2′ = C2

dC =

ε 0 dA ε ( adx ) ε ( adx ) = 0 = 0 ( EF ) d + x tan θ d + xθ {since for small θ , tan θ = θ }



⎧ C1′ K1′ 5 ⎫ = = ⎬ ⎨∵ ⎩ C1 K1 2 ⎭

5 C1′ 2 C1 5 10 25 ⇒ =  = = × 2 13 13 C2′ C2

10. The given capacitor may be supposed to be formed of a large number of differential capacitors each connected in parallel. Consider one such capacitor of width dx at a distance x from O . The area of each plate of this small capacitor, dA ( = adx ) . Separation between these plates, EF = d + x tan θ If dC be the capacitance of this small capacitor, then



⇒ dC =



⇒ dC =

xθ ⎞ ε 0 ( adx ) ε ( adx ) ⎛ = 0 ⎜⎝ 1 + ⎟ xθ ⎞ d d ⎠ ⎛ d⎜ 1 + ⎟ ⎝ d ⎠

ε 0 ( adx ) ⎛ θ ⎞ ⎜⎝ 1 − x ⎟⎠ {using Binomial Theorem} d d a

{using (1)}

d

V1′ C2′ 13 …(2) = = V2′ C1′ 25

O

dx F dx

Also, V1′ + V2′ = 230 V …(3)

9.



x a

Solving (2) and (3) , we get

V1′ = 78.7 V and V2′ = 151.3 V . Let d be the separation between the plates of first capacitor, the separation between the plates of second capacitor will be [ a − ( b + d ) ] . The capacitances of these capacitors are

ε A ε0 A C1 = 0 and C2 = d a − (b + d) Since the capacitors are in series, so the equivalent capacitance of the combination C is given by

1 1 1 1 1 = + = + ε0 A C C1 C2 ⎛ ε 0 A ⎞ ⎛ ⎞ ⎝⎜ d ⎟⎠ ⎜⎝ [ a − ( b + d ) ] ⎟⎠



1 1 a−b [d + a − b − d] = = ⇒ C ( ε0 A ) ε0 A



⇒ C=

ε0 A a−b

02_Ch 2_Hints and Explanation_P1.indd 129

E

θ

If V1′ and V2′ are the respective potential differences, then

−1

CHAPTER 2

V1 =

The net capacitance of capacitor is obtained by integrating the expression with respect to x between 0 to a i.e., a

a

ε a ⎛ θ θ x2 ⎤ ⎞ ε a⎡ C = 0 ⎜ dx − xdx ⎟ = 0 ⎢ x − ⎥ ⎠ d ⎝ d d ⎣ d 2 ⎦0

∫ 0



⇒ C=

aθ 2 ⎤ ε 0 a 2 ⎡ ε0a ⎡ aθ ⎤ 1− ⎢a − ⎥= ⎢ d ⎣ 2d ⎦ d ⎣ 2d ⎥⎦

11. (a) T  he equivalent of Arrangement 1 is shown. Let C be capacitance of each capacitor. x 2 3 P

Q

y

z

2 1

4 3

The net capacitance of y and z connected in series is

9/20/2019 11:34:26 AM

H.130  JEE Advanced Physics: Electrostatics and Current Electricity

   C yz =

 Capacitors C1 and C23 are in series; so the equivalent capacitance C of the entire combination is

C 2

 Now C yz and capacitor x are connected in parallel. Therefore net capacitance between P ­ and Q ,    CPQ

C 3 3⎛ ε A⎞ =C+ = C= ⎜ 0 ⎟ 2 2 2⎝ d ⎠

(b) The equivalent of Arrangement 2 is shown. x

2

1 z

P

2

y

 Now Cxy and z are connected in series. Therefore net capacitance of arrangement between P and Q is C + Cxy

=

C ( 2C ) 2 2⎛ ε A⎞ = C= ⎜ 0 ⎟ C + 2C 3 3⎝ d ⎠

12. The capacitance is maximum when the plates of one group are parallel to the plates of other group, so that the effective area of each plate is A. Since the alternate plates are connected together the potential difference across any two consecutive plates is same. The n plates of given arrangement form ( n − 1 ) capacitors connected in parallel. Therefore the maximum capacitance of radio capacitor is

ε A ε A C = 0 + 0 + ... ( n − 1 ) times d d



(b) The total charge that flows from the battery is

 Q = CeqV = ( 4 × 10 −6 ) ( 12 ) = 4.8 × 10 −5 C  Capacitors C1 and C23 are in series, hence they carry the same charge Q = 4.8 × 10 −5 C .

ε A ⇒ C = ( n − 1) 0 d

C23 = C2 + C3 = 4 + 8 ⇒ C23 = 12 mF C1

C23

C1

02_Ch 2_Hints and Explanation_P1.indd 130

The voltage across the combination C23 is Q 4.8 × 10 −5 = =4V C23 12 × 10 −6

C2

C23

C1 +Q –Q

C3

+Q –Q

+ –

+ –

C2 + – Q2 Q2 Q3 Q3 + – C3

C23 is parallel combination of C2 and C3 and  hence they have the same potential  V2 = V3 = 4 V

The charges on C2 and C3 are

 Q2 = C2V2 = ( 4 × 10 −6 ) ( 4 ) = 1.6 × 10 −5 C  Q3 = C3V3 = ( 8 × 10 −6 ) ( 4 ) = 3.2 × 10 −5 C

So we have V1 = 8 V , Q1 = 48 mC

 V2 = 4 V , Q2 = 16 mC  V3 = 4 V , Q3 = 32 mC Also, we note that Q2 + Q3 = Q1 = Q

14. (a) T  he two condensers in the middle are connected together and hence act as a single conductor. Thus effectively there are three plates which form two capacitors in parallel. Capacitance of each capacitor,  C =

a b

Q 4.8 × 10 −5 = =8V C1 6 × 10 −6

 V23 =



13. (a) C  apacitors C2 and C3 are in parallel, so their equivalent capacitance is

The voltage across C1 ,

 V1 =

3

   Cxy = 2C

CCxy

 Ceq = 4 mF

4Q

The net capacitance of x and y connected in parallel is

   CPQ =

1 1 1 1 1 3 + = + =   = Ceq C1 C23 6 12 12

ε0 A d

So, net capacitance of combination is

 CPQ = 2C =

2ε 0 A d

9/20/2019 11:34:45 AM

Hints and Explanations H.131 (b) The four plates are alternatively connected and form three capacitors in parallel. So, the net capacitance of combination is given by

   CPQ = 3C =

3ε 0 A d

ε 0 A1 = d1

ε 0π R12 d1 2

1 ⎛ ⎞ ⎛ 0.1 ⎞ π ⎜⎝ 9⎟ ⎠ ⎜⎝ 2 ⎟⎠ 1 ⇒ C1 = 36π × 10 −3 = F 2 × 10 288 × 108

The capacitance of capacitor ( C , D ) is

C2 =

ε 0 A ε 0π R22 = d2 d2 2

1 ⎛ ⎞ ⎛ 0.12 ⎞ π ⎜⎝ 9⎟ ⎠ ⎜⎝ 2 ⎟⎠ 1 × π 36 10 ⇒ C2 = = F 3 × 10 −3 300 × 108  The capacitors C1 and C2 are in series. Therefore, equivalent capacitance C is given by ⇒

   U =

15. (a) The capacitance of capacitor ( A , B ) is C1 =

Therefore, the energy density (u) should also be constant.

1 1 1 = + = 288 × 108 + 300 × 108 C C1 C2 1 = 588 × 108 C

q2 1 ε 0 E2 = 2 2 A 2ε 0

So, total stored energy, U = ( u ) (total volume)

⎛ q2 ⎞ q2 U = ⎜ ⎟ ( A.d ) = ⎛ Aε ⎞ 2 ⎝ 2A ε0 ⎠ 0 2⎜ ⎝ d ⎟⎠ ⇒ U =

q2  2C

E =

1 q 4πε 0 r 2





1 1 2 CV 2 = × ( 17 × 10 −12 ) × ( 120 ) 2 2

1 ⎛ 1 q⎞ ε0 2 ⎜⎝ 4πε 0 r 2 ⎟⎠

}

q

2

Energy stored in a volume dV = ( 4π r 2 ) dr is

  

dU = u dV

⇒ U = 1.224 × 10 −7 J

r dr

16. (a) E  lectric field is uniform between the plates of the capacitor. The magnitude of this field is E =

Aε 0 d

R

1 u ( r ) = ε 0E2 2 ⇒ u ( r ) =

U =

∵C=

(b) In case of a spherical conductor (of radius R ) the excess charge resides on the outer surface of the conductor. The field inside the conductor is zero. It extends from surface to infinity. And since the potential energy is stored in the field only, it will be stored in the region extending from surface to infinity. Also the field is non-uniform, the energy density u is also non-uniform. Hence, the total energy can be calculated using the concept of integration. Electric field at a distance r from the centre is,

1 = 17 × 10 −12 F = 17 pF ⇒ C = 588 × 108 (b) Energy stored by capacitors

{

CHAPTER 2



q σ = ε 0 Aε 0 + + + + E=0 + + + + +

E = εσ

0

02_Ch 2_Hints and Explanation_P1.indd 131

– – – – – E=0 – – – –

r =∞

⇒ Total energy stored is, U =

∫ dU

r=R



Substituting the values and integrating, we get

   U =

q2 2 ( 4πε 0 R )

⇒ U =

q2  2C

{∵ C = 4πε 0 R }

9/20/2019 11:35:00 AM

H.132  JEE Advanced Physics: Electrostatics and Current Electricity

Test Your Concepts-III (Based on Dielectrics and Breakdown) 1.

Arrangement A is equivalent to a combination of two d capacitors each of area A, separation and connected 2 in series. So,



1 1 1 (d 2) + d 2 = + = CS C1 C2 ε 0 K1A ε 0 K 2 A



1 ε A⎡ K K ⎤ = 0 ⎢ 1 2 ⎥ CS d ⎣ K1 + K 2 ⎦

where KS =

Please note that this situation is not the same as the situation where the battery is removed from the circuit before inserting the dielectric. C 3. (a) Net capacitance without inserting the slab is, 2 ⎛ C⎞ ⇒ q = ⎜ ⎟ V ⎝ 2⎠

K 1K 2 K1 + K 2

Arrangement B is equivalent to combination of two A capacitors each of area , separation d and con2 nected in parallel. So,

where K P = Now 2.

⇒

When the dielectric is inserted at constant ­voltage, then

⇒

KC

q

q

q

q

V

Net capacitance after inserting the slab is

Q DV0

1 2 C0 ( DV0 ) 2

⎛ K ⎞ ⇒ q′ = CV ⎜ ⎝ K + 1 ⎟⎠

   E′ =

ε 0 A Q0 = d DV0

U= and

C

Electric field

CS 4 × 2 × 3 24 = = CP ( 2 + 3 )2 25

Since, U 0 =

C

⎛ K ⎞ C ′ = ⎜ C ⎝ K + 1 ⎟⎠

K1 + K 2 2

   C = κ C0 =

C

V



CS 2 K 1K 2 2 4 K 1K 2 = × = CP K1 + K 2 ( K1 + K 2 ) ( K1 + K 2 )2

(a) C0 =

Since, electric field

⎛V⎞ ⎜⎝ ⎟⎠ V Potential Difference    E= = 2 = d 2d separation between the plates

⎛ A⎞ ⎛ A⎞ ε 0 K1 ⎜ ⎟ ε 0 K 2 ⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠ + C p = C3 + C4 = d d

ε A ⎡ K + K2 ⎤ C p = 0 ⎢ 1 ⎥⎦ 2 d ⎣

Q =κ Q0

⇒

(

κ C0 DV02 1 2 C ( DV ) = 2 2

)

U =κ U0

Potential Difference Separation between the plates

⎛ 1 ⎞ V⎜ ⎝ K + 1 ⎟⎠ V    E′ = ( K + 1)d d

Electric field decreases by a factor

⎛ V⎞ ⎜⎝ ⎟ E K +1 2d ⎠    = = V E′ ⎛ 2 ⎞ ⎜⎝ ( ⎟ K + 1)d ⎠

(b) Charge that flows through the battery is q′ − q

   Charge Flowing =

ε0 A

CV ( K − 1 ) 2( K + 1)

C0  t t 1− + d kd

{

ε0 A d

}

The extra energy comes from (part of the) electrical work done by the battery in separating the extra charge.

4.

C=

Q0 = C0 DV0 (b)



where t = 2 ( 0.1 mm ) = 0.2 mm , d = 4 mm and K = 3

and Q = C DV0 = κ C0 DV0



Putting values, we get C0 = 44.25 pF

02_Ch 2_Hints and Explanation_P1.indd 132

t d−t+ k

=

∵ C0 =

9/20/2019 11:35:16 AM

Hints and Explanations H.133 44.25 0.2 0.2 + 1− 4 ( 3 )( 4 )

44.25 pF 1 1 1− + 20 60



⇒ C=



44.25 × 60 2655 ⇒ C= = 60 − 3 + 1 58



⇒ C = 45.77%

⇒ Percentage increase in the value of capacitance is 3.4%

5.

(a) C =

ε0 A d

C= (b)

ε0 A d−t

The system of three parallel plates form two parallel plate capacitors connected in series. The capacitance of first capacitor



C2 =

ε 0 A 2ε 0 A = d2 d

(a) S  ince both are connected to the same potential, hence they are in parallel. So, ⎛ ε A⎞ ⎛ ε A⎞ = K1 ⎜ 0 ⎟ + K 2 ⎜ 0 ⎟ ⎝ d1 ⎠ ⎝ d2 ⎠

   Cnet

K ⎞ ⎛K ⇒ Cnet = ε 0 A ⎜ 1 + 2 ⎟ ⎝ d1 d2 ⎠ Q A

K1ε 0 A d1

and capacitance of second capacitor K 2ε 0 A d2

The net capacitance C is given by



Since, t → 0

(b) σ=

1 ⎛ ε 0 K1V 2 ⎞ 2 ⎜⎝ d12 ⎟⎠

C1 =



6.

⇒ U1 =



DC ⎛ 45.77 − 44.25 ⎞ × 100 = ⎜ ⎟⎠ × 100 = 3.4% ⎝ 44.25 C0

C= (c)

⎛ V2 ⎞ 1 ( K1ε 0 ) ⎜ 2 ⎟ 2 ⎝ d1 ⎠

7.

So, %age change is

⇒ U1 =

1 1 1 1 1 = + = + C C1 C2 ⎛ K1ε 0 A ⎞ ⎛ K 2ε 0 A ⎞ ⎜⎝ d ⎟⎠ ⎜⎝ d ⎟⎠ 1 2 1 1 ⎡ d1 d2 ⎤ 1 d1K 2 + d2 K1 = + = C ε 0 A ⎢⎣ K1 K 2 ⎥⎦ ε 0 A K 1K 2



⇒



⇒ C=

8.

The capacitances C2 and C3 are in series, each having A d and separation . area 2 2

ε 0 K 1K 2 A d1K 2 + d2 K1

⎛ A⎞ ε ⎜ 0 2⎟ ⎛ ε A⎞ = K2 ⎜ 0 ⎟ C2 = K 2 ⎜ ⎝ d ⎠ d ⎟ ⎟⎠ ⎜⎝ 2 ⎞ ⎟ ⎛ ε0 A ⎞ ⎟ ⎟ = K 3 ⎜⎝ d ⎠ ⎟ ⎠

⇒ σ =

CV A



⇒ σ =

ε 0 AV ⎛ K1 K 2 ⎞ + A ⎜⎝ d1 d2 ⎟⎠

⎛ ⎛ A⎞ ⎜ ε 0 ⎜⎝ 2 ⎟⎠ ⇒ C3 = K 3 ⎜ d ⎜ ⎝ 2



Equivalent capacitance of C2 and C3 is

K ⎞ ⎛K ⇒ σ = ε 0V ⎜ 1 + 2 ⎟ ⎝ d1 d2 ⎠ U1 = (c)

2

⎛ ε A⎞ K2K 3 ⎜ 0 ⎟ ⎝ d ⎠ C 2C 3 ⎛ K2K3 ⎞ ε 0 A = C ′ = = C2 + C3 ⎛ ε 0 A ⎞ ⎜⎝ K 2 + K 3 ⎟⎠ d ( K 2 + K 3 ) ⎜⎝ d ⎟⎠

1 2 ε E1 2

Since ε = K1ε 0 and E1 =    U1 =

02_Ch 2_Hints and Explanation_P1.indd 133

V 1 ( K1ε 0 ) ⎛⎜ ⎞⎟ 2 ⎝ d1 ⎠

CHAPTER 2

Now, C =

V1 , so d 2

⎛ ⎛ A⎞ ⎞ ⎜ ε 0 ⎜⎝ 2 ⎟⎠ ⎟ ⎛ ε0 A ⎞ Also C1 = K1 ⎜ ⎟⎠ = K1 ⎜⎝ ⎟ ⎝ d 2d ⎠

9/20/2019 11:35:33 AM

H.134  JEE Advanced Physics: Electrostatics and Current Electricity P

d



+Q

Final energy of both capacitors

C2, K2

d/2

U f =

C3, K3

d/2



⇒ Uf =



⇒

C1, K1

l

–Q

l

1 1 C ′V 2 + C ′V ′ 2 2 2 2

V 1 1 5 ( 3C )V 2 + ( 3C ) ⎛⎜ ⎞⎟ = CV 2 ⎝ 3⎠ 2 2 3

Ui CV 2 3 = = 5 Uf 5 2 CV 3

From figure it is clear that C1 is in parallel with C ′ (combination of C2 and C3 ). Hence net capacitance between P and Q .

10. (a) C =

⎛ ε A ⎞ ⎛ K2K3 ⎞ ε 0 A C = C1 + C ′ = K1 ⎜ 0 ⎟ + ⎜ ⎝ 2d ⎠ ⎝ K 2 + K 3 ⎟⎠ d

ε0 [ (  − x )  + κ x ] = ε 0 [  2 +  x ( κ − 1 ) ] d d

(b) U=

2 1 1 ⎛ ε ( DV ) ⎞ 2 2 C ( DV ) = ⎜ 0 ⎟⎠ [  + x (κ − 1 ) ] 2 2⎝ d

K2K3 ⎞ ε 0 A ⎛ K1 + d ⎜⎝ 2 ( K 2 + K 3 ) ⎟⎠



⇒ C=



Substituting given values

C =

8.8 × 10 −12 × ( 1 × 10 −4 ) ⎛ 4 6 × 2 ⎞ ⎜⎝ + ⎟ 2 6+ 2⎠ 2 × 10 −3

2



⎛ dU ⎞ ˆ ε 0 ( DV ) (  κ − 1 ) to the left (out F = −⎜ i= (c)  ⎝ dx ⎟⎠ 2d of the capacitor)

F= (d)

( 2000 )2 ( 8.85 × 10 −12 ) ( 0.0500 ) ( 4.50 − 1 ) 2 ( 2.00 × 10 −3 )



⎛ 7⎞ ⇒ C = 0.44 × 10 −12 ⎜ ⎟ ⎝ 2⎠



⇒ C = 1.54 × 10 −12 F = 1.54 pF

11. The vertical orientation sets up two capacitors in ­parallel, with equivalent capacitance

9.

With the switch S closed, the potential difference across capacitors A and B is same. So,

⎛ A⎞ ⎛ A⎞ ε 0 ⎜ ⎟ κε 0 ⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠ ⎛ κ + 1 ⎞ ε0 A CP = + =⎜ ⎝ 2 ⎟⎠ d d d

V =

QA QB = C C

The initial charges on the capacitors are given by

QA = QB = CV When dielectric is introduced, the new capacitance of either capacitor. C ′ = KC = 3C Now, when the switch S is opened, let the potential difference across capacitor A be V volt and that across the capacitor B be V ′ volt. When dielectric is introduced with the switch open (i.e., battery disconnected) then, the charge on capacitor B remains unchanged, so QB = CV = C ′V ′ C V V= volt C′ 3



⇒ V’=



Initial energy of both capacitors

U i =

F = 1.55 × 10 −3 N

where A is the area of either plate and d is the separation of the plates. The horizontal orientation produces two capacitors in series. If f is the fraction of the horizontal capacitor filled with dielectric, the equivalent capacitance is

fd 1 (1 − f )d = ⎡ f + κ (1 − f ) ⎤ d = + ⎢ ⎥ Cs κε 0 A ε0 A κ ⎣ ⎦ ε0 A



κ ⎡ ⎤ ε0 A ⇒ Cs = ⎢ ⎥ ⎣ f + κ (1 − f ) ⎦ d



For having C p = Cs we get



⇒ (κ + 1 ) ⎡⎣ f + κ ( 1 − f ) ⎤⎦ = 2κ

κ +1 κ = 2 f + κ (1 − f

)

For κ = 2 , we get 3 ⎡⎣ 2 − ( 1 ) f ⎤⎦ = 4 2 ⇒ f = 3

1 1 CV 2 + CV 2 = CV 2 2 2

02_Ch 2_Hints and Explanation_P1.indd 134

9/20/2019 11:35:51 AM

Hints and Explanations H.135

Test Your Concepts-IV (Based on Capacitor Circuits, Kirchhoff's Laws, Charge Flown and Generation of Heat)



1.



q2

C q1

A

2C

2C

2C q3

q2

B

q1 = 15 2 ⇒ q1 = 10 mC

Solving these three equations, we have

q 2 3 q1 = q , q2 = q and q3 = − 5 5 5 Now, let Ceq be the equivalent capacitance between A and B . Then, q q q = 1+ 2 VA − VB = Ceq C 2C q 2q 3q 7q = + = Ceq 5C 10C 10C



⇒



⇒ Ceq =

2.

VA −

10 C 7

q1 q + E − 2 − VB = 0 C1 C2

q1 q + 10 − 2 = 0 1 2 q2 ⇒ q1 + = 15 …(1) 2 Also, we observe that − q1 + q2 = 0 5 −



⇒ q1 = q2 …(2)

02_Ch 2_Hints and Explanation_P1.indd 135

B

and across the second is q V2 = 2 = 5 V C2 3.

Here, we observe that q1 , q2 , q3 and q4 are in mC (microcoulomb) and any battery supplies same magnitude of charge from its both terminals. 3 μF

Applying Second Law, we have q q q − 1 − 3 + 2 = 0 C 2C 2C



C2





q2 + q3 − q1 = 0 …(3)

E

So, voltage across the first capacitor is q V1 = 1 = 10 V C1

q1

Plates inside the dotted line form an isolated system. Hence,

C1



C

⇒ q2 − q3 − 2q1 = 0 …(2)

A

+q2 –q2

q2 = 10 mC

q1 + q2 = q …(1)



+q1 –q1

q1 +

5V

q1 q2

5V

6 μF q3

2 μF q4

CHAPTER 2

The given circuit forms an unbalanced Wheatstone bridge. Let us suppose that the point A is connected to the positive terminal of a hypothetical battery and B to the negative terminal of the same battery. Hence, a total charge q is stored in the capacitors. Seeing the symmetry about the input and output points, we can say that charges will be distributed as shown.

From (1) and (2), we get

4 μF

10 V

 Select the sign of the charges on the capacitors ­arbitrarily, we get − q1 − q2 + q3 + q4 = 0 …(1)

Applying Loop Law in three loops, one by one, we get

5 −

q1 q2 + = 0 …(2) 3 6

10 − and 5−

q2 q3 − = 0 …(3) 6 2

q3 q4 + = 0 …(4) 2 4



Solving these four equations, we get 40 mC q2 = q3 = 10 mC and q1 = q4 = 3 4.

The charge distribution is shown in figure. The circuit is equivalent to two capacitors connected in series with a battery in between them. Let the charge on C1 is q . The same charge will also appear on C2 . The reason is as follows : Whatever charge appears on plate of C1 (say- q ) then equal but opposite charge goes to C2 through E1 . The potential drops across C1 and C2 q q will be and respectively. Which plate would C1 C2 be at higher potential will be decided by relative strength of E1 and E2. If we move along the complete loop, the total potential difference must be zero. Let E2 > E1 .

9/20/2019 11:36:11 AM

H.136  JEE Advanced Physics: Electrostatics and Current Electricity C1

E2

–q +q



C2 –q +q

C34 =



Ceq = C12 + C34

q q + E1 + − E2 = 0 C1 C2

⇒ q=



Charge Q divides into charge Q12 in C1 or C2 and Q34 in C3 or C4 . We have

C1 + C2 E2 − E1 C1C2 C1 + C2

Q12 = C12V and Q34 = C34V

It is obvious from the figure that the addition of a ­battery in between two capacitors is equivalent to charging the potential applied to the two capacitors in series.



5.



Charge distribution is shown in figure. Consider the loop NMLKN. Applying − DV = 0 , we have E1

K P

E2 E3

N

− E3 +

1

C1

Solving these equations, we get

E C − E1C2 − E1C3 + E3C3 …(3) q1 + q2 = 2 2 C3 C2 + +1 C1 C1



⇒ V1 =

{∵ V0 = 0 }

E1 ( C2 + C3 ) − E2C2 − E3C3 C1 + C2 + C3

Similarly, V2 = V3 =

C1

E2 ( C1 + C3 ) − E1C1 − E3C3 C1 + C2 + C3

E3 ( C1 + C2 ) − E1C1 − E2C2 C1 + C2 + C3

6.

Capacitors C1 and C2 are in series; their effective capacitance is CC C12 = 1 2 C1 + C2

02_Ch 2_Hints and Explanation_P1.indd 136

V3 =

Q34 C4 = V C3 C3 + C 4

Potential of A and B are

C4 ⎞ ⎛ VB = V − V3 = ⎜ 1 − V and C3 + C4 ⎟⎠ ⎝

Similarly, for loop OLKPO , we get

Now V1 − V0 = V1

Potential drop at capacitor C3

C2 ⎞ ⎛ VA = V − V1 = ⎜ 1 − V C1 + C2 ⎟⎠ ⎝

q1 ⎛ q1 + q2 ⎞ + + E1 = 0 …(1) C3 ⎜⎝ C1 ⎟⎠

(q + q ) =− 1 2 

Potential across capacitor C1 is Q C2 V V1 = 12 = C1 C1 + C2



L –(q1 + q2) (q + q ) 1 2 C2 2 +q2 –q2 O C3 3 +q1 –q1 M

q ⎛ q +q ⎞ ⎜ 1 2 ⎟ + E1 − E2 + 2 = 0 …(2) C2 ⎝ C1 ⎠

total charge on the capacitors is

Q = CeqV

( E2 − E1 ) C1C2

q =

C3C 4 C3 + C 4

C12 and C34 are in parallel; equivalent capacitance of  the circuit is

E1

Thus,

Effective capacitance of C3 and C4 is

 Thus, potential differences between A and B are equal to

C2 ⎞ C1C4 − C2C3 ⎛ C4 VA − VB = ⎜ − V= V ⎟ C + C C + C C ⎝ 3 ( 1 + C2 ) ( C3 + C 4 ) 4 1 2⎠



This will be zero, if C1C4 − C2C3 = 0



⇒



This is the condition for a balanced Wheatstone bridge.

7.

Let the charges on capacitors C1 , C2 , and C3 be Q1 , Q2 and Q3 . Potential differences are therefore

C1 C3 = C2 C 4

Q1 Q2 Q , and 3 respectively. Look at the inner plates C1 C2 C3 of the three capacitors connected to point A . From conservation of charge, we have Q1 − Q2 + ( ±Q3 ) = 0 …(1) C1 –Q1 Q1 C3

V1

C2

A

–Q2 Q2 Q3

B

V2

9/20/2019 11:36:31 AM

Hints and Explanations H.137

−V1 + V2 − and

Q1 Q3 ∓ = 0 …(2) C1 C3

Q2 Q3 ∓ = 0 …(3) C2 C3

In a closed loop the net potential drop must be zero, from KVL . Therefore, for loop 1, loop 2 and loop 3, we have For Loop 1

For Loop 2



q1 q2 q3 − − =0 C C C q2 q4 q5 − + =0 C C C

For Loop 3 q3 q5 q6 − + = 0 …(3) C C C

If Q3 is positive, then potential falls as we go from A ⎛Q ⎞ to B . Hence ∓ Q3 in (1) implies ∓ ⎜ 3 ⎟ in (2) and (3). ⎝ C3 ⎠ From (2) and (3), we get



Q ⎞ Q ⎞ ⎛ ⎛ Q1 = ⎜ V1 + 3 ⎟ C1 and Q2 = ⎜ V2 ∓ 3 ⎟ C2 C3 ⎠ C3 ⎠ ⎝ ⎝

q3 + q5 − q2 = 0 …(4)



Substituting for Q1 and Q2 in (1),we get



V1C1 − V2C2 ± Q3 ( C1 + C2 + C3 ) C3

=0

⇒



Thus potential difference between A and B

8.



On solving, we get

q1 = q2 = q5 = q6 = Therefore, C AB = 2C

V C − V1C1 = 2 2 C1 + C2 + C3

Connect a to b

Let the potential difference across the battery terminals be V and the charge of the battery Q. To find the capacitance of the battery means to find the capacitance of a capacitor which would have the same charge Q on its plates as the battery at voltage V. Hence, q1

A q2

1

q3

2

q5

3

q6



Q …(1) V

where from conservation of charge we must have

Q = q1 + q2 + q3 = q4 + q5 + q6 …(2) V = V4 = and

q4 C

02_Ch 2_Hints and Explanation_P1.indd 137

b

C

Let the charges in three capacitors be as shown in figure B

2 μF

A

10 V

4 μF

C

D

q2

q1 6 μF

B

C AB =

Balanced wheatstone bridge

a

9.

V q4

q4 and q3 = 0 2

 Method 2. Connect the connection of point a to b and note the indicated Wheatstone bridge. Now simplify the circuit to obtain the desired result.

Q3 V C − V1C1 = (±) 2 2 C3 C1 + C2 + C3



VAB

The conductor that connects the second, third and fifth capacitors is electrically neutral. Hence,

CHAPTER 2

Because all the inner three plates were neutral initially, therefore, whatever charge is induced on one is obviously drawn from others. Q3, the charge on the upper plate of C3 , may be positive or negative depending upon the value of ( Q1 − Q2 ) . As we move along the closed loops in the sense shown, we have

q3 F

20 V

E

Charge supplied by 10 V battery is q1 and that from 20 V battery is q2 . Then, q1 + q2 = q3

9/20/2019 11:36:46 AM

H.138  JEE Advanced Physics: Electrostatics and Current Electricity This relation can also be obtained by using the Junction Law. The charges on the three plates which are in contact add to zero. Because these plates taken together form an isolated system which can’t receive charges from the batteries. Thus,

Wbattery = V1 DQ = V1V2C …(3)

q3 − q1 − q2 = 0





⇒ q3 = q1 + q2 …(1)

Applying Kirchhoff’s Second Law to loops BCFAB and CDEFC , we get −

q1 q3 − + 10 = 0 2 6

is transferred to the right plate of the capacitor. Work done by battery V1 in the process of charge transfer is given by

A part of this work changes the energy of the capacitor

DUC =

Q 2f 2C

q2 q and − 20 + 3 = 0 4 6

Qi2 2C

1 1 2 DUC = V12C − ( V2 − V1 ) C 2 2

(

)

1 2V1V2 − V22 C 2 and the remaining part is lost as Joule heating. Hence the heat generated H is given by DUC =

⇒ q3 + 3 q1 = 60 …(2)

−



⇒ 3 q2 + 2q3 = 240 …(3)

1 DH = Wbattery − DUC = V22C 2



Solving equations (1), (2) and(3), we have

11. When the switch is open, then Ceq = 4 mF

10 140 mC , q2 = mC and q3 = 50 mC q1 = 3 3 10 μ C 3

140 μC 3

50 μ C

10 V

20 V

10. Initially, when the switch is closed on position 1, the capacitor C is connected in series with batteries V1 and V2 . From KVL we have

Qi − V2 + V1 = 0 C



⇒ q = 800 mC

This charge divides equally amongst the upper branch and lower branch. So, V1 − Vb =

400 V 3

V1 − Va =

400 200 V= V 6 3



⇒ Va − Vb =

Q f = −V1C …(2) The net charge flow through the circuit is

DQ = Q f − Qi = ⎡⎣ −V1 − ( V2 − V1 ) ⎤⎦ C = −V2C We can say that a net positive charge equal to V2C is pulled by the battery of emf V1 from the left plate of the capacitor, which flows through battery V1 and

02_Ch 2_Hints and Explanation_P1.indd 138

200 V 3

When the switch is closed, then Va − Vb = 0 The charges on the capacitors before the switch S is closed are shown.



+400 μ C –400 μ C +400 μ C –400 μ C

⇒ Qi = ( V2 − V1 ) C …(1)

depending upon the sign of ( V2 − V1 ) , charge Qi on the left plate may be positive (if V2 > V1 ), or negative ( if V2 < V1 ) . However, charge on the right plate would be equal and opposite. When the switch is moved to position 2, the left plate earlier having charge +Qi , will have charge

So, q = CeqV

C1 3 μ F

6 μF

C2

200 V

0V C3 6 μ F

3 μF

C4

+400 μ C –400 μ C +400 μ C –400 μ C



Now, when the switch is closed, then Ceq ′ =



⇒ q′ = Ceq ′ V = 900 mC

q mF 2

This 900 mC is shared amongst 3 mF and 6 mF in direct ratio of their capacitances. So ⎛ 3 ⎞ 900 = 300 mC and q3 mF = ⎜ ⎝ 3 + 6 ⎟⎠

9/20/2019 11:37:05 AM

Hints and Explanations H.139 ⎛ 6 ⎞ q6 mF = ⎜ 900 = 600 mC ⎝ 3 + 6 ⎟⎠

2.

C1 = 3 μ F 100 μ C 200 μ C 6 μ F = C2 V2 = 0 V S 300 μ C

Let q′ charge comes on the inner sphere. Then,

200 μ C

100 μ C 3 μ F = C4

+600 μ C –600 μ C +300 μ C –300 μ C

Initial Charge

Final Charge

Charge Flown |Dq|

b = 4 × 10 −2 m

⇒

1 ⎡ q′ q ⎤ + =0 4πε 0 ⎢⎣ a b ⎥⎦

Now, Vinner − Vouter =

Dq1 = 100 mC

C2

400 mC

600 mC

Dq2 = 200 mC

⇒ 0 − Vouter = −10 −6 × 109 × 10 2 ×

C3

400 mC

600 mC

Dq3 = 200 mC

⇒ Vouter = 2.5 × 10 4 V

C4

400 mC

300 mC

Dq4 = 100 mC



b

1 4

(b) Charges appearing on different surfaces are as shown in figure. Hence the charge retained on the outer surface of the outer sphere is +10 −6 C or +1 mC . +10–6 C +10–6 C –10–6 C

Let us assume a positive charge +q on A and a negative charge −q on B . Then their potentials are 1 q 1 q 1 q 1 q − − and VB = 4πε 0 R 4πε 0 d 4πε 0 d 4πε 0 R A R

B

+q

R

–q

d

The potential difference is

V = VA − VB =

1 2q 1 2q − 4πε 0 R 4πε 0 d

q ⎛ d−R⎞ 1 ⎡ 1 1⎤ − q = ⎜ ⎟ 4πε 0 ⎢⎣ R d ⎥⎦ 2πε 0 ⎝ Rd ⎠

Since d  R q ⇒ V= 2πε 0 R

Since, C =

a

q′ ⎛ 1 1 ⎞ ⎜ − ⎟ 4πε 0 ⎝ a b ⎠

300 mC

⇒ V=2

q′

a ⇒ q′ = − ⋅ q = −10 −6 C b

400 mC

VA =

and

q

C1

1.



a = 2 × 10 −2 m , (a) Given, −6 q = 2 × 10 C

Vinner = 0

Test Your Concepts-V (Based on Spherical and Cylindrical Capacitors)



⇒ C = 2 pF

+300 μ C –300 μ C +600 μ C –600 μ C

C3 = 6 μ F



⇒ C = 2 × 10 −12 F

CHAPTER 2

V1 = 200 V



q V



⇒ C = 2πε 0 R



⇒ C=

2 9 × 9 × 109 1000

02_Ch 2_Hints and Explanation_P1.indd 139

3.

Let q , v and c be the charge, potential and capacitance of individual (small) drop. Also let Q , V , C be the corresponding quantities for bigger drop. As total charge is conserved, therefore the charge of bigger drop Q = N × charge on small drop



⇒ Q = Nq



⇒

Q = N …(1) q

Since the capacitance of a spherical drop is proportional to its radius ( C = 4πε 0 r ) , therefore if R and r are radii of big drop and small drop respectively, then

C R = …(2) c r

Now, since total mass of the drops is conserved, so, we have Mass of bigger drop = N × Mass of small drop

9/20/2019 11:37:28 AM

H.140  JEE Advanced Physics: Electrostatics and Current Electricity

4 3 ⎛4 ⎞ π R ρ = N ⎜ π r 3 ρ ⎟ { ρ is density of mercury} ⎝3 ⎠ 3



⇒



⇒ R = N1 3r



⇒

C R = = N 1 3 …(3) c r 1 N3

C = c



⇒



Again, by definition

V =

⇒

V Q c ⎛ Q⎞⎛ c ⎞ N = ⋅ = = N 2 3 …(4) ⎜ ⎟= v C q ⎜⎝ q ⎟⎠ ⎝ C ⎠ N 1 3



⇒

V = N3 v



q2 Q2 and u = 2C 2c

U = N2 × u U = u



⇒



Equation (3) gives

1 ⎡ − q2 q1 + q2 ⎤ + = 0 …(3) b ⎥⎦ 4πε 0 ⎢⎣ a − q2 q1 + q2 − q2 Q + =0 + = 0 i.e., a b a b q2 a = + …(4) Q b



⇒



Using this equation (1) gives

1 − N 3

=



a⎞ b− a ⎛ ⇒ q1 = Q ⎜ 1 − ⎟ = Q ⎝ b⎠ b

Potential difference between A and B

Vab = VB − VA = VB − 0 = VB =

5 N3



5 N3

4.

Let Q be the charge given to outer sphere B . This charge will be partly on outer surface and partly on inner surface. Let q1 be charge on outer surface and q2 on inner surface. Then

q1 + q2 = Q …(1) + +

+ +

–

– b

–

a –

–

+ + +

+ + +q2 – + q1 –q2 + A– B – + +

+

+

+

The charge induced on inner sphere = −q2 The potential of B ,



1 q1 4πε 0 b

∴ Capacitance

C =

⇒

VB =

As sphere A is earthed VA = 0

Q b …(5) i.e., = q1 b − a





1 ⎛ − q2 q1 + q2 ⎞ + ⎜ ⎟ b ⎠ 4πε 0 ⎝ a





Q2 2 U 2C ⎛ Q ⎞ ⎛ c ⎞ ⇒ = 2 =⎜ ⎟ ⎜ ⎟ u q ⎝ q ⎠ ⎝ C⎠ 2c ⇒

VA =

a q1 + Q = Q b

2



The potential of A,



q Q and v = C c

Also, U =



Q = VAB

Q Q = 4πε 0b 1 q1 q1 4πε 0 b

Using (5), we get

2 ⎛ b ⎞ 4πε 0b C = 4πε 0b ⎜  = ⎟ ⎝ b− a⎠ b−a

5.

{done already}

Charge on inner sphere is given by

a − q2 = − Q b

⇒ − q2 = −

8 cm × 30 nC = −24 nC 10 cm b

–q2 a

q1 q2

Q

1 q1 1 q1 + q2 − q2 …(2) = b 4πε 0 b 4πε 0

02_Ch 2_Hints and Explanation_P1.indd 140

9/20/2019 11:37:47 AM

Hints and Explanations H.141 Charge on inner surface of outer sphere is



FOR OPTION (B)

q2 = +24 nC

d = 10 −5 m

The charge Q is distributed into q1 and q2 so that the electrostatic potential of inner sphere is zero.



⇒

E=

V d

6.



⇒

E=

20 10 −5



⇒

E = 2 × 106 Vm −1 < 3 × 106 Vm −1



The given arrangement is a parallel combination of two capacitors. The capacity of a cylindrical capacitor is given by

C =







2πε 0 L ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠

Hence OPTION (B) is right to give as an answer. Hence, the correct answer is (B). 4.

⎛ L⎞ 2πε 0 K1 ⎜ ⎟ ⎝ 2⎠ ⇒ C1 = ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠

10 μ F 1 μ F 10 μ F 11

⎛ L⎞ 2πε 0 K 2 ⎜ ⎟ ⎝ 2⎠ ⇒ C2 = ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠ ⇒ Ceq = C1 + C2 =

πL ( K1 + K 2 ) ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠

Single Correct Choice Type Questions 1.



When the switch is closed, the inner plates of the two capacitors get connected whereas the outer plates still are not connected and hence the circuit is not complete. Hence, the correct answer is (A).

3.

C = kε 0



⇒

1.77 × 10 −6 = 200 ( 8.85 × 10 −12 )

⇒

A = 10 3 d



A d A d



Hence, the correct answer is (A).

5.

The circuit show is a balanced Wheat Stone Bridge. Hence, the correct answer is (D).

6.

The given system is equivalent to a spherical capacitor of inner radius b and outer radius c . So, the capaci⎛ bc ⎞ tance of the system will be 4πε 0 ⎜ . ⎝ c − b ⎟⎠



Hence, the correct answer is (D).

7.

C = 4πε 0 R



⇒ C = 711 mF



⇒ C ≅ 1000 mF



⇒ C ≅ 10 −3 F = 1 mF



Hence, the correct answer is (B).

8.

Q2 = −Q 3 

{By induction}

Further, Q2 2Q2 Q2 − ( −Q2 ) = = 2C 2C C

This ratio is satisfied by both OPTIONS (A) and (B). So to arrive at a conclusion we take help of the break through strength. FOR OPTION (A)

DV =

⇒

d = 10 −6 m



Hence, the correct answer is (C).

DV =

Q2 − Q3 2C



⇒

E=

V d

10. On connecting both with a thin wire, the common potential V is given by



⇒

E=

20 10 −6

( C1 + C2 )V = C1V1 + C2V2



⇒

E = 2 × 107 Vm −1 > 3 × 106 Vm −1



which is impossible

02_Ch 2_Hints and Explanation_P1.indd 141

CHAPTER 2





⇒

V=

C1V1 + C2V2 C1 + C2

9/20/2019 11:38:00 AM

H.142  JEE Advanced Physics: Electrostatics and Current Electricity



Total initial energy = U i =

1 1 C1V12 + C2V22 2 2

1 ( C1 + C2 )V 2 2 Loss = Total Initial Energy – Total Final Energy



⇒

ab = na b−a



⇒

b =n b−a



⇒ b = nb − na



⇒ na = ( n − 1 ) b



Total final energy = U f =



⇒ Loss = U i − U f



⇒ Loss =

1 ⎛ C1C2 ⎞ ( V1 − V2 )2 2 ⎜⎝ C1 + C2 ⎟⎠



⇒



Hence, the correct answer is (C).



Hence, the correct answer is (B).

12.

C′ =

17.

1 1 1 1 = + + + ........... C 2 4 8



⇒



⇒ C = 1 Hence, the correct answer is (D).

C 1−

1 2

C ⎫ ⎧ ⎪∵ C ′ = ⎪ t⎬ ⎨ − 1 ⎪⎩ d ⎪⎭



C ′ = 2C Hence, the correct answer is (B). 14. The capacitance of the capacitor increases initially and then decreases and hence the positive charge on plate A first increases and then decreases. Due to this the current in the outer circuit first flows from B to A and then from A to B . Hence, the correct answer is (C). 15.

CS = 2 pF

12 1 = C 1−1 2

18. Since, the charges on the facing surfaces are −Q and Q (used Gauss’s Law to find them) So, E =

q = CSV

b n = a n−1

Q Aε 0

⇒ VP − VA = +

⇒



when connected in parallel 2Q

CP = C1 + C2

⇒

CP = 9 pF



⇒

V=

q CP



⇒

V=

10 kV 9

V = 1111 V

+Q

–Q

q = 10 mC



2Q E A



⇒



Hence, the correct answer is (D).

16.

C1 = 4πε 0 a

P

B

Qd Q Q = = ε A 2 Aε 0 ⎛ ⎞ 2C0 2⎜ 0 ⎟ ⎝ d ⎠



⇒ VP − VA =



Hence, the correct answer is (B).

19. For series combination CS =

⎛ ab ⎞ and C2 = 4πε 0 ⎜ ⎝ b − a ⎟⎠

Ed 2

Given that C2 = nC1

a C1

a b

C1C2 C1 + C2

2 mF 3 When connected in series the maximum charge that can flow through the combination equals the lower value of charge accommodated by the first capacitor i.e., 6000 mC

⇒

CS =



∴

Q1 = 6000 mC and Q2 = 8000 mC

C2

02_Ch 2_Hints and Explanation_P1.indd 142

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Hints and Explanations H.143 22. k1 in series with half of k3 and hence equivalent dieleckk tric constant is 1 3 k1 + k3 k2 in series with half of k3 and hence equivalent  k2 k3 dielectric constant is and then both in parallel k2 + k3

⇒

VS =



⇒

VS =



⇒

VS = 9 kV



Hence, the correct answer is (C).

6000 mC 2 mC 3

20. Let C be the capacitance of the capacitor without slab. Then after removing the slab i.e., finally the net capaciC tance is 2

to give k =

Hence, the correct answer is (D).

24. Charges on the capacitors are

CE q1 = q2 = 2

q + –

q + –

2 pF

2 pF

⎛ K ⎞ ⇒ Cnet = ⎜ C ⎝ K + 1 ⎟⎠



⎛ KCE ⎞ ⇒ Q1 = Q2 = ⎜ ⎝ K + 1 ⎟⎠ q2 K + 1 = Q2 2K



⇒



Hence, the correct answer is (A). 1 C0E2 …(1) 2 On inserting a dielectric new capacitance is Ui =

C = kC0 …(2)

and new potential difference is { V = Ed }

 E′ = f e′d where f e is the electrostatic field. But on 1 inserting the dielectric, the new field f e′ becomes k times the original field.

⇒

E′ =

E …(3) k

E 1 ( kC0 ) ⎛⎜⎝ ⎞⎟⎠ 2 k



⇒

Uf =



⇒

⎛1 ⎞1 U f = ⎜ C0 E 2 ⎟ ⎝2 ⎠k



V2 = 20 V



⇒ q2 = ( 20 )( 3 ) = 60 pC



⇒ q1 = q2 = q  (say)

The situation is similar to the two capacitors in series which are first charged with a battery of emf 50 V and then disconnected. So, when S3 is closed, V1 = 30 V and V2 = 20 V Other way of looking at the thing is, when S1 and S3 both are closed, due to attraction with opposite charge, no flow of charge takes place through S3 . Therefore, potential difference across capacitor plates remains unchanged or V1 = 30 V and V2 = 20 V . Hence, the correct answer is (D). C

26.

C C

Q/2

C/2

Q

C

2

1 ⎛1 ⎞ ⇒ DU = C0E2 ⎜ − 1 ⎟ ⎝k ⎠ 2 Hence, the correct answer is (D).

02_Ch 2_Hints and Explanation_P1.indd 143

3Q/2

3C/2

E

E

1 So U f = CE′ 2 2



V1 = 30 V 50 V

f So f e′ = e k

q = 60 pC + –

q1 = ( 30 ) ( 2 ) = 60 pC





q = 60 pC + –

Before the slab is removed

C1 = C and C2 = KC

21.

k1k3 k k + 2 3 k1 + k3 k2 + k3

CHAPTER 2

Q1 CS



E

{∵ Q = CE }



Hence, the correct answer is (B).

27.

W=



{

(

)

}

1 1 C ( 100 − 25 )  ∵W = C V22 − V12 2 2 75 ⇒ W = C …(1) 2



⇒ W′ =

1 C ( 225 − 100 ) 2



⇒ W′ =

125 C …(2) 2

9/20/2019 11:38:37 AM

H.144  JEE Advanced Physics: Electrostatics and Current Electricity

From (1) and (2)

5 W ′ = W 3

Hence, the correct answer is (C).

28.

⎛ t d−t⎞ V = 4πσ ⎜ + k2 ⎠⎟ ⎝ k1



⇒



Hence, the correct answer is (B).

29.

Vcombined =



⇒

V=

4π Q ⎛ t d − t ⎞ + k2 ⎟⎠ A ⎜⎝ k1

Vcombined =

400  133 V 3

Loss =



⇒



Hence, the correct answer is (C).

1 ⎛ C1C2 ⎞ ( V1 − V2 )2 2 ⎜⎝ C1 + C2 ⎟⎠ 1⎛ C⎞ 1 2 2 ⎜ ⎟ ( V1 − V2 ) = C ( V1 − V2 ) 2⎝ 2 ⎠ 4

⇒

2

q = kx 2ε 0 A 2

( CV ) 4d ⎞ ⎛ = k⎜ d − ⎟  ⎝ 2ε 0 A 5 ⎠ ⎛ ⎜ ⎜⎝

{∵ q = CV }

2

ε0 A ⎞ 2 V ⎛ 4d ⎞ ⎟ ⎜⎝ ⎟⎠ ⎟ 5 ⎠ = 0.2kd 2ε 0 A

X

3.2 2



⇒

X  1.6 m F



Hence, the correct answer is (C). d 2

ε0 A 4⎛ ε A⎞ = ⎜ 0 ⎟ ⎝ d ⎠ d d 3 d− + 2 2K 1 4 = ⇒ d⎛ 1 ⎞ 3d ⎜ 1 + ⎟⎠ K 2⎝ ⇒

1 3 = K 2



⇒ 1+



⇒ K=2



Hence, the correct answer is (A).

3C 2 Total potential across the series combination of 2C, C and 2C is 90 V. If Q is the total charge then C AB =

Q = C AB ( 90 )

⇒



Charge in SR branch = Q1 = 90C



Charge in SPQR branch = Q2 = 45C



⇒

VMN = Potential across C



⇒

VMN =

Q = 135C

Q2 = 45 V C





⇒



⇒ k=



Hence, the correct answer is (C).

32.

X (1) +1= X X +1

Hence, the correct answer is (C).  35. The spring force FS acting on plate a is given by  FS = − kx iˆ  Similarly, the electrostatic force Fe due to the electric field created by plate b is

ε 0 AV 2 4ε 0 AV 2 ≈ 0.256 d 3 d3

 Q2 ˆ ⎛ σ ⎞ˆ Fe = QEiˆ = Q ⎜ i= i ⎟ 2 Aε 0 ⎝ 2ε 0 ⎠

1

A

2



⇒

X = X −1



⇒

X2 − X − 1 = 0

CAB = X



⇒

X=

1+ 5 2

B

02_Ch 2_Hints and Explanation_P1.indd 144

⇒

34.

31. During electrostatic equilibrium. Electrostatic attraction between the plates = Spring force







1+ 2

30.

⇒

1 + 2.236 2

t =

( 1 ) ( 200 ) + 2 ( 100 )

Hence, the correct answer is (C).



X=





Loss =

⇒

33. Let d be the thickness of the slab, then

C1V1 + C2V2 C1 + C2

Vcombined =



1

X

where A is the area of the plate. Notice that charges on plate a cannot exert a force on itself, as required by Newton’s Third Law. Thus, only the electric field due to plate b is considered. At equilibrium the two forces cancel and we have

9/20/2019 11:38:56 AM

Hints and Explanations H.145 V = 250 V 2 1 U i = CV 2 2



which gives

x =



Q2 2kAε 0

U f =

Hence, the correct answer is (C).

36. The charge distribution on the surfaces of the shells are given. As per the given condition, the surface charge densities of the outer surfaces are equal. So,

⇒

Q1 Q + Q2 Q + Q2 + Q3 = 1 = 1 2 2 2 4π R 4π ( 2R ) 4π ( 3 R ) Q1 Q2 Q3 = = 1 3 5 Q3 Q1

D

Q2

B

A

R

C

R

3R 2R

3R

After charge distribution q A = Q1 qD = –(Q2 + Q1) qB = –Q1 qE = Q3 + Q1 + Q2 qC = Q2 + Q1

Hence, the correct answer is (B).

37.

C

C

C

2

1 CV 2 4 1 ⇒ Change = CV 2 4



1( 2 100 × 10 −6 ) ( 500 ) 4



⇒ Change =



⇒ DU = 6.25 J



Hence, the correct answer is (B).

39.

C=



⇒ C=



⇒ C = 1.8 × 10 −10 F

ε0 A d 1 π ( 0.08 ) 4π × 9 × 109 10 −3

2

Since Q = CV

⇒ Q = 1.8 × 10 −8 C



Hence, the correct answer is (B).

40. Total Initial Charge Qi = ( 2C )( 2V ) − CV = 3CV

Total Final Charge

Q f = 2CV ′ + CV ′

C C

P

1 1 ⎛V⎞ C ′V ′ 2 = ( 2C ) ⎜ ⎟ ⎝ 2⎠ 2 2

U f =

E

2R

Initially



⇒ V′ =

C

Q

where V ′ is common potential. By Law of Conservation of Charge Qi = Q f

C C C 3 μF 6 μF 9 μF6 μF 3 μF Ceq = 0.9 μ F



⇒ 3CV = 3CV ′



⇒ V′ = V



Hence, the correct answer is (C).



So, final energy of combination is

38.

q V= C

U f =

Since C ∝

1 d



So, as d is halved, C becomes double



⇒ V′ =



⇒

q C′

V C′ = =2 V′ C

02_Ch 2_Hints and Explanation_P1.indd 145

CHAPTER 2

⎛ Q ⎞ kx = Q ⎜ ⎝ 2 Aε 0 ⎟⎠

1 3 ( C + 2C )V 2 = CV 2 2 2



Hence, the correct answer is (B).

41.

⎛ ab ⎞ C = 4πε 0 ⎜ ⎝ b − a ⎟⎠

C =

⇒ C=

( 0.5 )( 0.6 ) 1 0.1 9 × 109 0.3 9 × 108

9/20/2019 11:39:12 AM

H.146  JEE Advanced Physics: Electrostatics and Current Electricity 49. Total initial energy is



⇒ C ′ = kC



⇒ C ′ = 2 × 10 −9 F



Hence, the correct answer is (B).

42. The 3 mF capacitor has energy 600 m J 1( 3 × 10 −6 )V02 2



⇒ 600 × 10 −6 =



⇒ V0 = 20 V



⇒ VAB = 3V0 = 60 V

This 60 V has to divide in between 2 mF and 6 mF in the inverse ratio of their capacitances i.e., 3 : 1 .

⎛1 ⎞ Ei = 2 ⎜ CV 2 ⎟ = CV 2 ⎝2 ⎠ Initial charge across A is CV Initial charge across B is CV On, opening the switch when a dielectric with k = 3 is introduced in both A and B, then C A = 3C and CB = 3C  The new potential across A is still V. Let V′ be new potential across B. So, new charge across B is ( 3C )V ′. By Law of Conservation of Charge

⎛ 3 ⎞ 60 V = 45 V So, V2 mF = ⎜ ⎝ 3 + 1 ⎟⎠

CV = ( 3C )V ′





Hence, the correct answer is (C).

V 3 So, total final energy is ⇒ V′ =

45. The arrangement is a balanced Wheatstone Bridge. So, no charge will exist across the branch PQ. Hence



Cnet = 10 m F

E f =



Hence, the correct answer is (D).

47. On connecting by a wire, let V be the common potential. Then, by Law of Conservation of Charge Qinitial = Q final

1 V 1 ( 3C )V 2 + ( 3C ) ⎛⎜ ⎞⎟ ⎝ 3⎠ 2 2



⇒ Ef =

3 1⎞ ⎛ CV 2 ⎜ 1 + ⎟ ⎝ 2 9⎠



⇒ Ef =

5 CV 2 3

Ei 3 = Ef 5



⇒ C1V1 + C2V2 = C1V + C2V



⇒



⇒ V=

C1V1 + C2V2 C1 + C2



Hence, the correct answer is (B).

If Ei be total initial energy and E f be the total final energy, then

50.

Cs =

Loss = Ei − E f



⇒ Cs = 2.4 m F



⇒ Loss =

1 1 1 C1V12 + C2V22 − ( C1 + C2 )V 2 2 2 2



⇒ Loss =

1 C1C2 ( V1 − V2 )2 2 C1 + C2

We get above result by putting value of V and then solving. Hence, the correct answer is (C).

4×6 mF 4+6

q = CsV = 2.4 × 500 × 10 −6 q = 1200 mC

Hence, the correct answer is (C).

52.

Q1 = C0V0

Q2 = ( kC0 )V0

C C C + + + ... 2 4 8



⇒

Q1 1 = Q2 k

⎛ 1 ⎞ ⎜ 1⎟ ⎜⎝ 1 − ⎟⎠ 2



⇒

40 1 = 100 k



⇒ k = 2.5



⇒ CP = 2C = 2 m F



Hence, the correct answer is (B).



Hence, the correct answer is (C).

48.

CP = C +



⇒ CP = C

02_Ch 2_Hints and Explanation_P1.indd 146

2

9/20/2019 11:39:34 AM

53. It is a balanced Wheatstone Bridge. 5

3

A

B 5

3

5×3 5×3 + 5+3 5+3

⇒ C AB =



⇒ C AB = 2 ×



⇒ C AB = 3.75 m F



Hence, the correct answer is (A).

15 8

54. Initial charge on first capacitor is CV = Q1 Initial charge on second capacitor is 2CV = Q2

Final capacitance of first capacitor is KC If V’ is the common potential then V ′ =

Q1 + Q2 C1′ + C2

⇒ V′ =

CV + 2CV KC + 2C



⇒ V′ =

3V 2+K



Hence, the correct answer is (A).

55.

Ui =

=

C2 and C3 in parallel



⇒ V2 = V3 . Also,

V = V1 + V2 = V1 + V3 Q2 + Q3 = Q1

Hence, the correct answer is (C).

1



⇒ C1 > C2



Hence, the correct answer is (C).

Qinitial = 5 × 12 = 60 mC

3 2



⇒ ( C + 5 ) 3 = 60



⇒ C = 15 m F



Hence, the correct answer is (B).

Hence, the correct answer is (B).

56. By Law of Conservation of Charge C1V = C1V ′ + C2V ′ ⎛ C1 ⎞ ⇒ V′ = ⎜ V ⎝ C1 + C2 ⎟⎠



Hence, the correct answer is (A).

57.

1 U i = U = C1V 2 2

2

⎛ C1 + C2 ⎞ V ′ = U ⎜⎝ C1 ⎟⎠ V 2

02_Ch 2_Hints and Explanation_P1.indd 147

62. Potential difference across the branch de is 6 V. Net capacitance of de branch is 2.1 m F .

So, q = CV

1 1 C1V ′ 2 + C2V ′ 2 2 2

Uf

Qfinal = ( C + 5 ) 3 mC





⇒

59.

q2 q2 = 2C f 2 ( 2C0 )





Hence, the correct answer is (A).

Qinitial = Qfinal

⇒

U f =



61. By Law of Conservation of Charge



Ui

⎛ C1 ⎞ U ⇒ Uf = ⎜ ⎝ C1 + C2 ⎟⎠

2

q2 q2 = 2Ci 2 ( 3C0 )

Uf



60. When we move from left to right the voltage increases and this increase is just within the capacitors and remains constant in the conducting wires. Also for 1 series combination V ∝ and according to graph C V >V



U f =

⎡ C + C2 ⎛ C1 ⎞ 2 ⎤ ⇒ Uf = ⎢ 1 ⎜ ⎟ ⎥U ⎢⎣ C1 ⎝ C1 + C2 ⎠ ⎥⎦

58. While drawing the dielectric plate outside, the capacitance decreases till the entire plate comes out and then becomes constant. So, V increases and then becomes constant. Hence, the correct answer is (B).







CHAPTER 2

Hints and Explanations H.147



⇒ q = 2.1 × 6 mC



⇒ q = 12.6 mC



Potential across 3 m F capacitance is

V =

12.6 = 4.2 volt 3

 Potential across 2 and 5 combination in parallel is 6 − 4.2 = 1.8 V

9/20/2019 11:39:53 AM

H.148  JEE Advanced Physics: Electrostatics and Current Electricity So, q′ = ( 1.8 )( 5 ) = 9 mC

64.

C

A

4

Hence, the correct answer is (C).

68.

V0 =

B

Next equivalent circuit diagram is C

1

A 8 3

8

8 9

B

8 3 C 32 9

32 C =1 ⇒ 9 32 +C 9 32 mF 23



⇒ C=



Hence, the correct answer is (A).

65.

1 1 1 = 1 + + + ... Cs 2 4



1 1 = =2 Cs 1 − 1 2



⇒ Cs = 0.5 m F



Hence, the correct answer is (C).

66.

C=

ε0 A d−t+

t K

02_Ch 2_Hints and Explanation_P1.indd 148



⇒

V C0 = V0 C



⇒

C 500 20 = = C0 75 3



By definition

B

20 3



⇒ k=



Hence, the correct answer is (C).

69.

3⎞ σ σ ⎛ d = ⎜ d + 2.4 − 3 + ⎟ k⎠ ε0 ε0 ⎝



⇒ d = d − 3 + 2.4 +



⇒ 3 − 2.4 =



⇒ 0.6 =



⇒ k=5



Hence, the correct answer is (B).

C

A

q C

C = kC0 B

A

q C0

V =

4

4





1

8



⇒ C=

Hence, the correct answer is (B).

63. Since all are in series. Hence, the correct answer is (D).

ε0 A 1⎞ ⎛ d − t⎜ 1 − ⎟ ⎝ K⎠



3 k

3 k

3 k

70. A

ε1

d1



⇒



⇒

ε2

A

d2

1 1 1 = + Cs C1 C2 1 1 = A ε ε A Cs 1 + 2 d1 d2 1 1 ⎛ d1 d2 ⎞ = + Cs A ⎜⎝ ε1 ε 2 ⎟⎠

9/20/2019 11:40:05 AM

Hints and Explanations H.149 On partial insertion of dielectric, the net capacitance of capacitor is

⇒ Cs =



Hence, the correct answer is (A).

71.

A

C

C

C

C

C

C



B

q2  2C dU Since, F = − , so we get dx

 The figure shows two independent balanced Wheatstone Bridges connected in parallel each having a capacitance C. So,



Cnet = C AB = 2C



72.

Hence, the correct answer is (B). A

B C

C

The circuit contains such two combinations in series. Hence Ctotal =

( 3C )( 3C )

3C = 2



⇒ Ctotal



Hence, the correct answer is (B).

73. (A), (B) and (C) are different forms of wheatstone bridge. Hence, the correct answer is (D). 74. When disconnected from the source, q = constant. When immersed in dielectric, we have C ′ = KC . So, we get q q V = = V ′ = C ′ KC K Hence E′ = Vd =

Vd Vd E = = K K K

Hence, the correct answer is (A).

75. Let area of plate be A, then A = bL, where b is width of plate

{

q2 d ε 0b ( L + ( K − 1)x ) 2C 2 dx d ε 0b ( K − 1 ) dx q2 d ⇒ dU = − 2 2 ⎛ ε b ⎞ [ L + ( K − 1)x ] 2⎜ 0 ⎟ ⎝ d ⎠



⇒ F=−



⇒ F=

But

3C + 3C

q2 ( −1 ) C −2 dC 2 dx

⇒ dU = −

C

C AB = 3C

q2 ( K − 1 ) ⎛ ε ( bL ) ⎞ ⎛ ⎛ K − 1⎞ ⎞ 2⎜ 0 ⎟ x⎟ ⎟ L ⎜ 1 + ⎜⎝ ⎝ d ⎠ ⎝ L ⎠ ⎠

2

ε 0 ( bL ) = C0 d q2 ( K − 1 )



⇒ F=



Hence, the correct answer is (C).

⎛ ⎛ K − 1⎞ ⎞ 2C0 L ⎜ 1 + ⎜ x ⎝ L ⎟⎠ ⎟⎠ ⎝

2

, attractive

76. Potential across the first capacitor is V, equal to emf of the battery. Thus no charge will further flow through the circuit. Hence, the correct answer is (D). 77. As soon as the switch is closed, for the charge to flow in the circuit (through S ), there must exist a potential difference in the circuit. Here, we observe, that for the closed loop abcd , we have q q q DV = − + + = 0 2 3 6 2 μF + –

a

b

q d

q + – 6 μF

02_Ch 2_Hints and Explanation_P1.indd 149

}

q2 ( K − 1 ) d dU = dx 2ε 0b [ L + ( K − 1 ) x ]2

L b

{∵ q = constant }

Now U =

dU =

C

C

ε 0b ( L − x ) Kε 0bx + d d ε 0b [ L + ( K − 1)x ] ⇒ C= d

C =

CHAPTER 2

Aε1ε 2 ε 2 d1 + ε1d2



q + – S

c

3 μF

9/20/2019 11:40:18 AM

H.150  JEE Advanced Physics: Electrostatics and Current Electricity

Hence, no charge will flow through S, when closed. Hence, the correct answer is (A).

⇒



⇒ 1+



⇒ K=2



Hence, the correct answer is (A).

81.

F=



⇒ F ∝ q2

78. Equivalent capacitance of the circuit ⎛ ε A⎞ C = 2 ⎜ 0 ⎟ ⎝ d ⎠ x

y Equivalent circuit

x

y d

V

1 CV 2 2



Energy stored, U =



⇒ U=



Hence, the correct answer is (B).

1 ⎛ 2 ( ε 0 A ) ⎞ 2 ε 0 AV 2 ⎜ ⎟V = 2⎝ d ⎠ d

79. Starting from positive plate, at higher potential (say V0 ) and going to negative plate at lower potential, we have potential after covering a σ . (a) distance d in air, is V1 = V0 − E0 d where E0 = ε0 (b) further d in conductor, is still V1 , because field inside conductor is zero and hence potential stays constant. σ . (c) further d in air is V2 = V1 − E0 d where E0 = ε0 (d) further d in dielectric is V3 = V2 − Ed where σ E= ( < E0 ) Kε 0 σ (e) last d in air is V4 = V3 − E0 d where E0 = . ε0 In all the cases except (b), potential decreases linearly with field and in (d) the field has lesser value than cases (a), (c) and (e). So OPTION (C) is correct. Hence, the correct answer is (C). 80.



4 ⎛ ε0 A ⎞ ε0 A d , where t = ⎜ ⎟= 2 3⎝ d ⎠ d−t+ t k ⇒

4 ⎛ ε0 A ⎞ ε0 A ⎜ ⎟= 3⎝ d ⎠ d− d + d 2 2k

02_Ch 2_Hints and Explanation_P1.indd 150

4 1 = 1⎞ 3d d ⎛ ⎜ 1 + ⎟⎠ 2⎝ K



1 3 = K 2

q2 2ε 0 A

 Since capacitor stays connected to the battery, so V = constant. Hence when separation is halved, C becomes twice. So, q ( = CV ) becomes twice. Hence F becomes four times. Hence, the correct answer is (C). 82. This circuit is a balanced Wheatstone Bridge with one capacitor connected in parallel to the bridge. So, the equivalent circuit is shown here. The net capacitance of the circuit between the points A and B is 2C . C A

C B

C C

C

C



Hence, the correct answer is (D).

83. Since both free terminals are earthed, hence they being at same potential can be connected to give the following equivalent circuit. Now, all three being in parallel between A and B, will given net capacitance as 10 mF . 4 μF A

4 μF

B 4 μF

4 μF

A

4 μF

B

4 μF

2 μF



Hence, the correct answer is (D).

84. Charge attained by the plates of capacitor is q0 = CV0 = ( 10 mF ) ( 50 V ) = 500 mC

9/20/2019 11:40:33 AM

Hints and Explanations H.151 –500 μ C

q (700 – q)

B Initially

A



–q

P

–(500 – q)

B Finally

A

Now, potential across

V2 =

15C is 4

q 150C = = 40 V 15C ⎛ 15C ⎞ ⎜⎝ ⎟ 4 4 ⎠

So, potential across 3C and combination of 3C and C is also 40 V . 3C

Now, when a charge 200 mC is given to the positive plate of capacitor, then the net positive charge becomes 700 mC , while charge on negative plate is stil −500 mC . Let a charge q appear on inner surface of A , then surface of B facing A gets a charge −q . Such that a charge ( 700 − q ) and − ( 500 − q ) appears on outer surfaces of A and B respectively. Consider a point P inside plate B . Then net field at P , inside the conductor is zero. So

This 40 V is distributed between C and 3C (in series), in the inverse ratio of their capacitance i.e., 3 : 1 . Hence

EP = 0

VAB = 10 V

700 − q 500 − q q q + − + =0 2ε 0 A 2 ε 0 A 2ε 0 A 2ε 0 A





⇒



⇒ q = 600 mC



So, potential difference between the plates is

DV = 85.

3C

B

Hence, the correct answer is (A).

86. The equivalent circuit is shown here for convenience 10 V

q 600 mC = = 60 V C 10 mF

4 μF

Hence, the correct answer is (B). Cnet

A

C

CHAPTER 2

500 μ C

 Potential across each capacitor is divided)

15C = 19 190 V

2 μF

190 V



⇒ q = CV = ( 4 mF ) ( 5 V ) = 20 mC



Hence, the correct answer is (B).

5V

(equally

87. Using Kirchhoffs’ Law, we get 3C

C

A C

3C

C

B 3C

3C 4

190 V

8 − B

q1 q −2− 3 −3=0 C1 C3

Since, C1 = C3 = 2 mF

⇒ q1 + q2 = 6 …(1)

Also 8 − V1 C



⇒ q = CnetV



⎛ 15C ⎞ ( ⇒ q=⎜ 190 ) = 150C ⎝ 19 ⎟⎠

02_Ch 2_Hints and Explanation_P1.indd 151

V2

15C 4

q1 q2 + −2=0 C1 C2

Since C1 = C2 = 2 mF

⇒ q1 − q2 = 12 …(2)



Also, we observe that

− q1 − q2 + q3 = 0

⇒ q1 + q2 = q3 …(3)

9/20/2019 11:40:51 AM

H.152  JEE Advanced Physics: Electrostatics and Current Electricity

Solving equations (1), (2) and (3) we get

So, we have

q1 = 6 mC , q2 = −6 mC and q3 = 0

Hence, the correct answer is (C).

88. Initial charge on plates of capacitor will be q0 ( say ) . Then q0 = CV q0

–q0

q0 + Q – q

q

B

A

Initially



CV + Q − q = −CV + q

⇒ q = CV +

Q 2

Q q CV + 2 Q ⇒ DV = = =V+ C C 2C

P A

q0 + Q − q = − ( q0 − q ) , where q0 = CV

⇒ 2q = 2CV + Q

–(q0 – q)

–q



B Finally

When charge Q is given to the positive plate of capacitor, then the net positive charge becomes q0 + Q while charge on negative plate is still −q0 . Finally the charge distribution on the plates is also shown such that both inner faces have charge q and −q respectively. When a point P is considered inside plate B , then net field at P , inside the conductor, is always zero. So

3C 4 When key K was open, charge stored in the capaci3CV tors was qinitial = 4 When key K is closed, capacitor 3C becomes short circuited and hence the new charge on C is qfinal = CV 89. In the circuit Cnet =

EP = 0

1 ⇒ ⎡ ( q0 + Q − q ) + q − q + ( q0 − q ) ⎤⎦ = 0 2ε 0 A ⎣



⇒ q0 + Q + q0 − 2q = 0



⇒ 2q = 2q0 + Q

Since q0 = CV , so we get Q 2 So, potential difference between the plates is

q = CV +

Q q CV + 2 Q DV = = =V+ C C 2C

Hence, the correct answer is (C).

Problem Solving Technique(s) We can also arrive at the result by using the concept that (a) Charges on the surfaces facing each other are equal and opposite. (b) Charges on the outer surfaces are equal. (c) Capacitance of a parallel plate capacitor depends upon its shape i.e., A and d, but not on charge distribution.

02_Ch 2_Hints and Explanation_P1.indd 152

+ –

+ –

C

3C

V



So, charge flowing is

Dq = qfinal − qinitial 3CV CV = 4 4



⇒ Dq = CV −



Hence, the correct answer is (C).

1 CV 2 = 3 J 2 On connecting this capacitor to an uncharged capacitor, since charge distributes equally, hence both capacitors must be same capacitance.

90. Energy stored in capacitor is,

CV + C ( 0 ) V = C+C 2



Now common potential, V ′ =



Total energy stored in two capacitors is

U ′ =

2

1 1 1 ⎛V⎞ CV ′ 2 + CV ′ 2 = C ⎜ ⎟ = CV 2 ⎝ 2⎠ 2 2 4 3 = 1.5 J 2



⇒ U′ =



Hence, the correct answer is (A).

9/20/2019 11:41:08 AM

Hints and Explanations H.153

C1 + C2 =



Hence, the correct answer is (A).

92. Fe



T

T

m2g

m1g

For upper plate, we have

T = Fe + m2 g …(1) For m1 , we have



σ 25 = = 5000 NC −1 ε 0 5 × 10 −3

⇒ E0 =

The capacitor is disconnected from the battery but charge on it will not change so that σ has the same value. When a dielectric slab of thickness 3 mm is placed between the plates, the thickness of air between the plates will be t = 5 − 3 = 2 mm Electric field strength in air will have the same value

( 5000 NC −1 )

E =

but inside the dielectric, it will be

5000 5000 = = 500 NC −1 K 10

T = m1 g …(2)



So potential difference is V = Eair dair + Emed dmed





⇒ V = 5000 × ( 2 × 10 −3 ) + 500 × ( 3 × 10 −3 )



⇒ V = 11.5 V



Hence, the correct answer is (C).

Equating (1) and (2), we get

m1 g =

2

q + m2 g 2 Aε 0 2

q + m2 2 Aε 0 g



⇒ m1 =



Hence, the correct answer is (C).

93.

VC1 = 10 − 4 = 6 V

VC2 = 4 − 0 = 4 V

97.

ε 0 KRdθ × h b Since all small dC are in parallel, so dC =

π 6

VC1 VC2

=

θ =0

C2 C1

K = 5.5 R θ

C1 VC2 4 2 = = = C2 VC1 6 3



⇒



Hence, the correct answer is (B).

1 2 4 3

B

Ceq =

2ε 0 A d

A



b

E0 =

σ V = ε0 d

02_Ch 2_Hints and Explanation_P1.indd 153

6 KhR b

π 6

π ⎛ ε 0 KhR ⎞ ⎟ b ⎠

∫ dθ = 6 ⎜⎝



⇒ Ceq =



Hence, the correct answer is (A).

99.

Hence, the correct answer is (B).

96. The capacitor is charged by a battery of 25 V . Let the magnitude of surface charge density on each plate be σ . Before inserting the dielectric slab, electric field strength between the plates,

dθ

h

94. It is a Wheatstone bridge Hence, the correct answer is (A). 95.

∫ dC

Ceq = ∑ dC =

Since, in series, we have



CHAPTER 2

25 ⎛ C1C2 ⎞ 6 ⎜⎝ C1 + C2 ⎟⎠

91.

+ + + +

K

– q = CV1 – – – + + + +

V1

V =

q1 + q2 C1 + C2

⇒ V=

CV1 C + KC



θ =0

C

– – – –

9/20/2019 11:41:23 AM

H.154  JEE Advanced Physics: Electrostatics and Current Electricity

⇒     CV + KCV = CV1



⇒     K =



V −V ⇒     K = 1 V



Hence, the correct answer is (D).

CV1 − CV CV

Hence charge flown is 300 mC.



Hence, the correct answer is (A).

101. Since F = kx q2 = kx 2 Aε 0

100. When switch is open

+200 V

6 μF

3 μF

3 μF

6 μF

V=0

q = CeqV (for both branches) ⇒ q = 2 × 200 = 400



Hence charge on each capacitor is 400 mC



When switch is closed 3 μF

3 μF



Hence, the correct answer is (A).

102.

Ceq = C1 + C2 = 4πε 0 ( r1 + r2 )



Hence, the correct answer is (B).

104.

Loss = −DU =



C1C2 ( V1 − V2 )2 2 ( C1 + C2 )

Vc = 0

B V1

100 V

6 μF

3 μF

3 μF

6 μF

200 V

q6 = 600 mC and q3 = 300 mC Final charge on both plates is 300 mC a

5 μ F 15 μ F A

V=0

100 V

Equivalent

C2 ( 2V )2 = CV 2 4C Hence, the correct answer is (B). ⇒ −DU =

105.

6 μF

6 μF

⇒ x=



6 μF +200 V

q2 2 Aε 0 k



Since V1 = V and V2 = −V







3 μF

2000 V

V1 =

2000 × 5 = 500 20

⇒ VB − VC = 500

Since VC = 0

⇒ VB = 500



Hence, the correct answer is (D).

106. Equivalent capacitance of Arrangement-1 a A/2

600 μ C 300 μ C

A/2

K b 3 μ F, 300 μ C 6 μ F, 600 μ C



C′ b

Initial charge on both plates is zero +

–

+

–

400 μ C 400 μ C + – 400 μ F

02_Ch 2_Hints and Explanation_P1.indd 154

+ – 400 μ C

C″ =

C ′ =

C0 2

(Both C′ and C″ are in parallel)

ε0 A CK K= 0 2d 2

⇒ C1 = C ′ + C ′′ C0 C0 K C0 (1 + K ) + = 2 2 2 Equivalent capacitance of Arrangement-2 ⇒ C1 =

9/20/2019 11:41:37 AM

Hints and Explanations H.155



a

ε0 A

t d−t+ K

d

Both are same, so

t

C ε0 A 0 ( K + 1 ) = t 2 d−t+ K ε A ε0 A ⇒ 0 ( 2 + 1) = t 2d d−t+ 2

K b



⇒ Wbattery = 20 × 4 = 80 m J



Hence, the correct answer is (C).

109. Since between the plates, we have E =

2Q Q + 2 Aε 0 2 Aε 0



⇒ E=

3Q 2 Aε 0



⇒ E=

3 Q  2 Cd

{

∵ C=



3t ⇒ 3 d − = 2d 2



3t ⇒ d= 2



⇒ Ed =



Also, force on one plate due to other is



2d ⇒ t= 3

F = Q2E1



Hence, the correct answer is (A).

107.



⎛ 2Q ⎞ ⇒ F = ( −Q ) ⎜ ⎝ 2ε 0 A ⎟⎠



⇒ F=−



Energy =



⇒ U=



Hence, the correct answer is (D).

B

3 2

A

1 1

A

2

3 2

5

4

B

3 4



ε A The capacity of each capacitor is, C0 = 0 d



From figure it is clear that Ceq =



Hence, the correct answer is (C).

Q2 Q2 =− Aε 0 Cd

1 ε 0E2 Ad 2 2

1 ⎛ 3Q ⎞ 9 Q2 ε0 ⎜ Ad = ⎟ 2 ⎝ 2Cd ⎠ 8 C

110. When no capacitor, resistor or battery exist between any two points in a branch, then those two points are taken at same potential. Using this concept to solve the circuit, we get A

B

108. Before the switch S is closed, then



⇒ qi = 5 × 4 = 20 mC



When the switch S is closed, then

q f = CV

X X

5 5⎛ ε A⎞ C0 = ⎜ 0 ⎟ 3 3⎝ d ⎠

qi = CeqV

}

3Q =V 2C

5 4

ε0 A d

X Y

X

A

B

1 1 1 1 = + + Ceq C 2C C



⇒



⇒ Ceq =

Hence, the correct answer is (B).

2C 5



⇒ q f = 10 × 4 = 40 mC





⇒ Wbattery = ( Dq )Vbattery = ( q f − qi )Vbattery

112. Earthing a single point in a circuit does not change the charge or current flowing in the circuit, So,

02_Ch 2_Hints and Explanation_P1.indd 155

CHAPTER 2

C2 =

9/20/2019 11:41:55 AM

H.156  JEE Advanced Physics: Electrostatics and Current Electricity 10 V C/3

N

A

117. C

V1



10 × C 30C = = 7.5 VA − VN = C 4C C+ 3

⇒ VA − 0 = 7.5



⇒ VA = 7.5 V



Hence, the correct answer is (B).

113.

Q2 Q 2 Q 2 ⎛ 2d d ⎞ W = DU = U f − U i = − = − 2C f 2Ci 2 ⎜⎝ ε 0 A ε 0 A ⎟⎠



⇒ W=



Hence, the correct answer is (A).

Q2 d C 2V 2 CV 2 ε 0 AV 2 = = = 2 ε0 A 2C 2 2d

114. Energy density =

V 1 ε 0E2 and E = 2 d



Hence, the correct answer is (C).

115.

Q = CV = 120 mC and

Q′ = C ′V ′ = 50 × 12 = 600 mC

⇒ Charge Flown = DQ = 600 − 120 = 480 mC



Hence, the correct answer is (C).

116.

+ A

Q

Q + –

–

1 μF

10 V

2 μF

⇒ ( VA − VB ) + 10 =



⇒ Q = 10 mC



⇒ V1 m C



3Q 2 × 10 −6

Q 10 mC = = = 10 V and C 1 mF

V2 m C =

Q 10 mC = =5V C 2 mF

Hence, the correct answer is (C).

02_Ch 2_Hints and Explanation_P1.indd 156

+ + + +

– – – – Q2 = CV

Since loss in energy is given by

−DU =

C1C2 ( V1 − V2 )2 2 ( 1 + C2 )



⇒ −DU =

C2 [ V − ( −V ) ]2 2 × 2C



⇒ −DU =

C × 4V 2 = CV 2 4



Hence, the correct answer is (B).

119.

i=



⇒ q = it + a , where a = constant

dq dt

Since V =

⇒ V=

q C

it + a C

So, V is proportional to time Hence, the correct answer is (D). 120.

Ceff =

ε0 A d

Since effective capacitance between plates A and E is zero, so

B

Q Q − VB = 0 + 10 − −6 1 × 10 2 × 10 −6



– Q1 = CV – – –

U =

Applying Kirchhoff’s Voltage Law ( KVL ) between A and B , we get VA −

+ + + +



1 ε A CV 2 = 0 V 2 2 2d

Hence, the correct answer is (C).

121. Gravitational force = mg = ρπ ( b 2 − a 2 ) gh dU 1 2 dC = V dh 2 dh where h is the height of liquid. Now let us calculate C as a function of h .

Net upward force F =

1

L–H

K

h

Since, Ceq = C Air + CK =

2πε 0 ( L − h ) 2πε 0 Kh + ⎛ b⎞ ⎛ b⎞ log e ⎜ ⎟ log e ⎜ ⎟ ⎝ a⎠ ⎝ a⎠

9/20/2019 11:42:13 AM

Hints and Explanations H.157





2πε 0 ( L + ( K − 1 ) h ) ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠

⇒ C=

dC 1 2 2πε 0 ( K − 1 ) 1 = V × ⇒ F = V2 × dh 2 2 ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠



⇒ W=

KCV 2 CV 2 CV 2 ⎛ K 1⎞ − = − ⎟ ⎜ 2(1 + K ) 4 2 ⎝ 1+ K 2⎠



⇒ W=

CV 2 ⎛ 2K − 1 − K ⎞ CV 2 ⎛ K − 1 ⎞ ⎜ ⎟ ⎜ ⎟= 2 ⎝ 2(1 + K ) ⎠ 4 ⎝ K + 1⎠



Hence, the correct answer is (B).

124. Total charge = 1.25 mC + 0.75 mC = 2 mC

1 2 V × 2πε 0 ( K − 1 ) = ρπ ( b 2 − a 2 ) gh 2 ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠

Since, q1′ : q2′ = R1 : R2 = 1 : 2



⇒ h=

πε 0V 2 ( K − 1 )

⎛ b⎞ ρπ ( b 2 − a 2 ) g log e ⎜ ⎟ ⎝ a⎠

q2′ = and

2 4 × 2 = mC 3 3

125.

ε 0V 2 ( K − 1 )

⇒ h=



Hence, the correct answer is (B).

122.

U=



Work done, W = qV



qV 1 So, Loss of energy = qV − qV = 2 2



Hence, the correct answer is (D).

0

0

+q –q

1 qV 2

C

Energy loss = −DU =



⇒ −DU =



Hence, the correct answer is (A).

126.

Ui =

1 ⎛ C ⎞ 2 CV 2 ⎜ ⎟V = 2⎝ 2 ⎠ 4

Q2 2Ci Q2 2C f



⇒ U f > Ui



⇒



⇒ Ci > C f



Hence, the correct answer is (B).

C V KC

Finally, energy stored is

KCV 2 1 ⎛ C × KC ⎞ 2 U f = ⎜ V = ⎟ 2 ⎝ C + KC ⎠ 2(1 + K ) Since, W = DU = U f − U i

q2 q2 d = 4C 4 ε 0 A

As surface was deformed in such a way that charge on original surface are coming closer or moving perpendicular to electric force acting on them, so total energy of foil increases.

Initially, energy stored is

U i =

Q2 Q2 > 2C f 2Ci

127.

C1V0 1 1′

11′33′

2 2′ 3 3′ Middle plate

02_Ch 2_Hints and Explanation_P1.indd 157

1 ⎛ C 2 ⎞ 2 1 C q2 × ⎜ ⎟V = 2⎝ C +C⎠ 2 2 C2



U f =

C



Hence, the correct answer is (A).

⎛ b⎞ ρ ( b 2 − a 2 ) g log e ⎜ ⎟ ⎝ a⎠

V



1 2 × 2 = mC 3 3





123.

⇒ q1′ =



CHAPTER 2

Since F = mg , so we get

ε1 C V ε2 2 0

22′

V0

9/20/2019 11:42:28 AM

H.158  JEE Advanced Physics: Electrostatics and Current Electricity

Total charge on 2 and 2′ plate is

⎛ ε ε A ε ε A⎞ q = CeqV = ⎜ 1 0 + 2 0 ⎟ V d2 ⎠ ⎝ d1

q ε ⎞ ⎛ε ⇒ σ = = ε 0V ⎜ 1 + 2 ⎟ A ⎝ d1 d2 ⎠



Hence, the correct answer is (A).

128.

Vcommon =



C C

Hence, the correct answer is (B). b C2

KC1 x

b

l–x

Ceq = C1 + C2 =

ε 0bxK ε 0 (  − x ) b + d d

ε 0b ( b + x ( K − 1 ) ) d



⇒ Ceq =



⇒



Hence, the correct answer is (B).

dC ε 0b ( K − 1 ) dx = ε 0b ( K − 1 ) v = dt d dt d

130. Charge in the circuit flows only when potential difference across C1 is either greater or less than that across C2

q q ⇒ 1 ≠ 2 C1 C2



⇒ q1C2 ≠ q2C1



Hence, the correct answer is (D).

3

1 5C 2 5 × V = CV 2 2 3 6



So, final Energy is U f =



Charge supplied by battery after closing K 2 is 5 2 CV − CV = CV 3 3

Dq =

Energy supplied by battery is ( Dq )Vbattery



Also energy supplied by battery = U f − U i + DH



⇒



⇒ DH =



Hence, the correct answer is (C).

132.

E=



Hence, the correct answer is (A).

133.

2 5 1 CV 2 = CV 2 − CV 2 + DH 3 6 2 1 CV 2 3

V d

9 μF

6 μF V

131. When K1 is closed

A, 1 C

V

Q = CV 1 So, initial Energy is U i = CV 2 2

When K 2 is also closed

12 μ F

2, B

90 V

+Q –Q

02_Ch 2_Hints and Explanation_P1.indd 158

2

2C 5C = 3 3

Ceq = C +

129.

C

1 C



3

Equivalent circuit is shown here.

C1V C1 + C2

⇒ Vcommon =

3

1 2 V

C1V1 + C2V2 C1 × V + C2 × 0 = C1 + C2 C1 + C2





2

1

V ′ =

90 × 9 = 54 15



⇒ q = CV ′ = 6 × 54 = 324 mC



Hence, the correct answer is (C).

134.

q . If potential difference between plates is zero V then capacitance will be infinite. Hence, the correct answer is (D). C=

9/20/2019 11:42:43 AM

Hints and Explanations H.159 ε0 A

CK =



If we set b = 0 , we get

b d−b+ K

CK =

ε0 A =C d

If CK = 2C



⇒

ε0 A

= b d−b+ K 2b ⇒ K= 2b − d

2ε 0 A d



⎧ C = 2 mF ⎪ For Capacitor 2, we have ⎨ Vmax = 4 KV ⎪Q ⎩ max = 8 mC



For series combination of capacitors, we have

1× 2 2 ⎧ ⎪ Ceq = 1 + 2 = 3 mF ⎪ ⎪ Qmax = 8 mC ⎨ Qmax 8 mC ⎪ and V = = 12 KV max = ⎪ 2 Ceq m F ⎪⎩ 3

Hence, the correct answer is (B).



Since, we know that K > 0 and b ≤ d

138. All the 3C capacitors are in series, so



2b ⇒ K= and 2b − d

1 1 1 1 + = + 3C 3C 3C Ceq

2b − d > 0





d ⇒ b> 2



⇒



Hence, the correct answer is (B).

⇒ Ceq = C

Since C and C are in parallel

d CQR . The upper surface of Q will have greater charge than the lower surface. As the force of attraction between the plates of a capacitor is proportional to Q 2 so, there will be a net upward force on Q which can balance its weight. Hence, (B) and (D) are correct. The magnitude and direction of electric field in different regions are shown in figure. The direction of the electric field remains the same. Hence, OPTION (B) is correct.

02_Ch 2_Hints and Explanation_P1.indd 164

x = 4d

E

The capacitances

The capacitors are in series. So, the effective capacitance is

4.

E0

 Similarly, electric lines always flow from higher to lower potential, therefore, electric potential increases continuously as we move from x = 0 to x = 4 d . Therefore, OPTION (C) is also correct.

Multiple Correct Choice Type Questions C1 =

x = 3d

× 3 × 10 3 = 2 × 10 −2

⇒ q1 =



x=d

d



1.

E0

V0 O

8.

d

2d

3d

4d

X

Let the length, width, thickness and dielectric constant of the slab be  , b , d and K respectively. The initial capacitance of the capacitor is given by

C0 =

ε 0 K b d

When the dielectric is pulled out of the capacitor with a speed v then, at time t , the length of the slab that has been pulled out is vt . At this instant, the capacitor may be considered as a parallel combination of two capacitors, an air capacitor having area b ( vt ) and a dielectric capacitor having area b (  − vt ) .

(l – vt) vt d

9/20/2019 11:44:15 AM

Hence, the net capacitance as a function of time will be equal to ε ( vt ) b ε 0 Kb (  − vt ) + C = 0 d d ε 0b ⇒ C= ( vt + K (  − vt ) ) d ε b ⇒ C = 0 ( K  − ( K − 1 ) vt ) d It is observed that the net capacitance C varies with time t linearly and slope of this line is negative and has positive intercept on C-axis. Hence, the OPTION (D) is correct. When the battery is disconnected, the charge on the capacitor remains constant and at time t, the potenq and tial difference across the capacitor will be V = C hence the curve between C and V is a rectangular hyperbola as shown in OPTION (C). Now since we observe that C decreases with time as a result of q . which V also increases from initial value V0 = C0 So, OPTION (A) is also correct. Let the potential difference between the plates of the capacitor be V , then the energy stored in it is equal to



Hence, (A) and (D) are correct.

10. Let C0 be the capacitance initially and C be the capacε A itance finally . Then C0 = 0 d Since, Q = C0V

⇒

ε 0 AV d

Q=

Further E0 =

V E and E = 0 d K

V Kd

1 qV , where q remains constant. Hence, energy 2 U ∝ V . So, the curve between U and V will be a straight line passing through origin but since the min-



ε 0 b , d therefore, the curve must start from this minimum value of C . Hence, the OPTION (B) is also correct. Hence, (A), (B), (C) and (D) are correct.

After the introduction of slab if U f be the final energy, then

U=

imum possible non-zero value of C is C MIN =



a force of attraction on the dielectric slab. On pulling the dielectric slab out of the capacitor, a work has to be done by the external force F . But if the slab is pulled out slowly then a very small current will flow through the circuit. In that case, heat generated in the circuit will be negligible. Hence, it is not necessary that significant amount of heat is produced in the circuit. Therefore, the OPTION (C) is incorrect.   It is also observed that an anticlockwise current flows in the circuit during the process i.e., the battery is being charged which further means that the internal energy of the battery increases. Therefore, the OPTION (D) is also correct.

9.

 ince the given circuit is a series combination of two S identical dielectric capacitors, therefore, charges on these two capacitors are equal. So, when the dielectric slab of the capacitor B is pulled out, its capacitance decreases, as a result of which the equivalent capacitance of the series combination also decreases. Hence the charge on series combination begins to decrease as a result of which an anticlockwise current starts flowing through the circuit or a charge flows from a to b . Hence, the OPTION (A) is correct.   Also the capacitors are in series with each other, so the charge on two capacitors remains same or both the capacitors possess a common charge at all instants and hence, the OPTION (B) is incorrect.   As studied and discussed earlier, we know that a charged capacitor always attracts the dielectric in the space between the plates. So, capacitor B exerts

02_Ch 2_Hints and Explanation_P2.indd 165

⇒

E=

CHAPTER 2

Hints and Explanations H.165

Also, if U i is the initial energy, then U i =

U f =

1 1 ⎛V⎞ 2 CVslab = ( KC0 ) ⎜ ⎟ ⎝ K⎠ 2 2

1 C0V 2 2

2

1 C0V 2 2 K



⇒

Uf =



⇒

ΔU = U 2 − U1

1 ⎛ 1 ⎞ C0V 2 ⎜ − 1 ⎟ ⎝K ⎠ 2 Since work done = Decrease in Potential Energy ⇒ W = −ΔU

⇒



⇒



ΔU =

1 ε 0 AV 2 ⎛ 1⎞ ⎜ 1 − ⎟⎠ K 2 d ⎝  ence, (A), (C) and (D) are correct. H W=

11. Capacitance without water is

C0 =

ε 0 A ε 0 × 0.01 = = ε0 d 0.01

⇒ C0 = 8.85 × 10 −12 F

9/20/2019 11:43:31 AM

H.166  JEE Advanced Physics: Electrostatics and Current Electricity When water rises by 1 cm C =

ε 0 × 81 × 0.001 ε 0 ( 0.01 − 0.001 ) + 0.01 0.01



⇒ C = 8.1ε 0 + 0.9ε 0 = 9ε 0



⇒ C = 9 8.85 × 10 −12 F = 80 F

(

)

Let m be the mass of water that rises to a height of 100 cm, then m = ( 0.001 × 0.01 ) × 1000 = 0.01 kg By Law of Conservation of Energy, we have

2

2

q q mgh = + 2C0 2C 2

where A is the area of the plate. Notice that charges on plate a cannot exert a force on itself, as required by Newton’s Third Law. Thus, only the electric field due to plate b is considered. At equilibrium the two forces cancel and we have ⎛ Q ⎞ = QE kx = Q ⎜ ⎝ 2 Aε 0 ⎟⎠ which gives x =

Q2 QE = 2kAε 0 k

Hence, (A), (B) and (C) are correct.

16. The effective capacitance C =

q2 q2 ⇒ = + mgh C0 C

4 × 5 20 = μF 9 9

The charge on C1 is given by

Now, q = C0V

Q1 = CV =

V 2C02 V 2C02 = + mgh C0 C

20 × 9 = 20 μC 9



⇒



C ⎞ ⎛ ⇒ V 2C0 ⎜ 1 − 0 ⎟ = mgh ⎝ C ⎠

2 ⎛ C2 ⎞ Q1 = × 20 = 8 μC Q2 = ⎜ 5 ⎝ C2 + C3 ⎟⎠



ε ⎞ ⎛ ⇒ V 2ε 0 ⎜ 1 − 0 ⎟ = 0.01 × 9.8 × 0.01 9ε 0 ⎠ ⎝

The charge on C3 is given by



⇒ V=

The charge on C2 is given by

9 × 0.01 × 9.8 × 0.01 8.85 × 10 −12 × 8



⇒ V = 1110 V



Hence, (A), (C) and (D) are correct.

12. O  n connecting the conducting sphere having charge Q with a conducting shell having no charge, the entire charge must flow out to the uncharged shell. {OPTION (C)} The capacitance of the outer shell of radius 2R is C = 4πε 0 ( 2R ) . So, energy liberated is U =

Q2 Q2 = . 2C 16πε 0 R

Hence, (C) and (D) are correct.  13. The spring force FS acting on plate a is given by  FS = − kx iˆ  Similarly, the electrostatic force Fe due to the electric field created by plate b is

 Q2 ˆ ⎛ σ ⎞ˆ Fe = QEiˆ = Q ⎜ i = i 2 Aε 0 ⎝ 2ε 0 ⎟⎠

02_Ch 2_Hints and Explanation_P2.indd 166

3 ⎛ C3 ⎞ Q1 = × 20 = 12 μC Q3 = ⎜ 5 ⎝ C2 + C3 ⎟⎠ Since,

Q2 Q3 = C2 C3 C3 ⎛ C + C3 ⎞ Q2 = Q2 ⎜ 2 C2 ⎝ C2 ⎟⎠



⇒ Q1 = Q2 + Q3 = Q2 +



⎛ C2 ⎞ ⎛ 2 ⎞ ⇒ Q2 = ⎜ Q1 = ⎜ 20 = 8 μC ⎝ 2 + 3 ⎟⎠ ⎝ C2 + C3 ⎟⎠

⎛ C3 ⎞ ⎛ 3 ⎞ and Q3 = ⎜ Q1 = ⎜ 20 = 12 μC ⎝ 2 + 3 ⎟⎠ ⎝ C2 + C3 ⎟⎠

Hence, (A), (C) and (D) are correct.

17. W  hen either A or C is earthed (both not earthed together), a parallel plate capacitor is formed with B , with ±Q charges on the inner surfaces. The other plate, which is not earthed, plays no role, hence charge of amount +Q flows to the earth. When both are earthed together, A and C effectively become connected. The plates now form two capacitors in parallel, with capacitances in the ratio 1 : 2 , and hence share charge Q in the same ratio. Hence, (A), (B) and (C) are correct.

9/20/2019 11:43:42 AM

Hints and Explanations H.167 18. C  ondition for a balanced Wheat Stone Bridge. Hence, (A) and (C) are correct.



q2 2ε 0 A

Q = CV =

 Since all the capacitors are in series and hence all capacitors will have the same charge given by

Hence, (A) and (D) are correct.

21. T  he distribution of charges is shown in figure. Applying Kirchhoff’s loop rules for loops ①, ②, and ③, we get LOOP ① −

q2 q q + 3 + 1 =0 5 0.75 15



⇒ q1 − 3 q2 + 20 q3 = 0 …(1)



LOOP ②

−

q2 + q3 q q −q q − 3 + 1 3 − 3 =0 15 0.75 5 0.75

⇒ 3 q1 − q2 − 44 q3 = 0 …(2) 15 μ F + – + – q1

a

5 μF + – + – q1 – q3

e

15 μ F + – + – q1

++ – –q 0.75 μ F – – q3 0.75 μ F ++ 3 1 2 3 q2 q2 + q3 q2 + – + – + – + – + – + – b f 5 μF 5 μF 15 μ F 4 23 V



LOOP ③

q q + q3 q2 − =0 23 − 2 − 2 5 5 15

⇒ 345 = 7 q2 + q3 …(3)

Solving for q1 , q2 , q3 , we get q1 =

20 × 10 −6 × 90 = 6 × 10 −4 C=600 μC 3

19 × 345 13 × 345 345 , q2 = , q3 = 92 92 92

Potential Difference between a and b is

Q = CV The potential difference across each capacitor will be in the inverse ratio of their capacitances. So, the potential difference across C1 is 30 V, the potential difference across C2 is 20 V and the potential difference across C3 is 40 V and hence all the OPTIONS (A), (B), (C) and (D) are correct. Hence, all the options are correct. 30.

q = CV = C0V0

where C0 =

⇒

ε0 A ε AV ε AV and q = 0 0 = 0 (d + ) d d

(d + ) ε0 A ε A ⎞ ⎛ V0 = 0 V ⇒ V= V0 = ⎜ 1 + ⎟ V0 ⎝ (d + ) d d d⎠

and C =

ε0 A ε0 A C0 = = ⎞ ⎛ ⎞ d+ ⎛ d⎜ 1 + ⎟ ⎜ 1 + ⎟ ⎝ d⎠ d⎠ ⎝



Hence, (B) and (C) are correct.

31.

Cs =

2×8 = 1.6 μ F 2+8

Since Q = CsV = 1.6 × 10 −6 × 300 Q = 4.8 × 10 −4 C V1 =

4.8 × 10 −4 = 240 V 2 × 10 −6

V2 =

4.8 × 10 −4 = 60 V 8 × 10 −6

U =

Q2 Q2 + 2C1 2C2

q3 345 4 = × =5V 0.75 92 3 Potential Difference between e and f is



⇒ U=

Vef = −5 V



⇒ U = 3 × 2.4 × 10 −2 J



⇒ U = 2.4 × 10 −2 J



Hence, (A) and (D) are correct.

Vab =



Hence, (A) and (B) are correct.

27. T  he equivalent capacitance of the arrangement here is C where

20 1 1 1 1 = + + μF ⇒ C= C C1 C2 C3 3

02_Ch 2_Hints and Explanation_P2.indd 167

CHAPTER 2

19. Since, F =

The potential difference across the combination is 90 V and hence the total charge is

( 4.8 × 10 −4 )2 2

1 ⎛ ⎞ ⎜⎝ ⎟ 1.6 × 10 −6 ⎠

32. Since battery is still in connection. So, V = V0

9/20/2019 11:43:52 AM

H.168  JEE Advanced Physics: Electrostatics and Current Electricity



⇒ Q0 = C0V0 and



Q = kC0V0

⇒ Q = kQ0

Since k > 1

⇒ Q > Q0

1 Also U 0 = Q0V0 and 2 1 U = QV = kU 0  2

{∵ Q = kQ0 and V = V0 }

Hence U > U 0

Hence, (A) and (D) are correct.

33. Since same charge flows through 7 μ F and 3 μ F (as both are in series), so Q7 = Q3 = 42 μC

⇒ 7 × 6 = 3V3



⇒ V3 = 14 V

Q−q Q−q + =E KC C



⎛ K ⎞ ⇒ Q−x=⎜ CE ⎝ K + 1 ⎟⎠



⇒



1 ⎞ KCE ⎛ K − 1 ⎞ ⎛1 ⇒ q = KCE ⎜ − = ⎜ ⎟ ⎝ 2 K + 1 ⎟⎠ 2 ⎝ K + 1⎠

KCE ⎛ K ⎞ −q=⎜ CE ⎝ K + 1 ⎟⎠ 2

⎛ K − 1⎞ ⇒ q = Q⎜ ⎝ K + 1 ⎟⎠ Since K is always greater than one, so q is positive and hence we have the charge flowing from a to b . So, option (A) is correct. Final charge on B equals the final charge on A , so option (B) is incorrect. During this process, energy received by the is battery is ( Δq )Vbattery = qE . So, option (D) is correct. Hence, (A) and (D) are correct.

35. The charges stored in different capacitors before and after the switch S is closed are shown in figure(s).

So, V3.9 = 20 V

20 μ C

Charge on 12 μ F capacitor is

60 μ C

40 μ C

Q12 = Q3.9 + Q7 7×3 × 20 7 + 30

20 μ C



⇒ Q12 = 3.9 × 20 +



⇒ Q12 = 120 μC = Total charge in the circuit.

120 = 10 V 12 Voltage across battery is 14 + 6 + 10 = 30 V

34.

⇒ V12 =

30 V Before

30 V After

The amount of charge flown through the battery is Δq = q f − qi = 20 μC

Hence, (A), (C) and (D) are correct.

So, energy supplied by the battery is

–Q KC +Q A



a

+Q E

–Q B

Wbattery = ( Δq )Vbattery

KC

b

KCE 2 On removal of the dielectric from B , let a charge q flow from a to b . Then by applying Kirchhoff’s Loop Law, we have Q =

q –Q + q KC Q–q A a q b

02_Ch 2_Hints and Explanation_P2.indd 168

Energy stored in all the capacitors before closing the switch S is 1 1⎛ 4 ⎞ 2 CnetV 2 = ⎜ × 10 −6 ⎟ ( 30 ) = 0.6 mJ ⎠ 2 2⎝ 3 and after closing the switch is

U i =

U f =

(

)

1 1 2 CnetV 2 = 2 × 10 −6 ( 30 ) = 0.9 mJ 2 2

Since we know that Wbattery = ΔUC + ΔH

Q–q –Q + q

⇒ Wbattery = ( 20 × 10 −6 ) ( 30 ) = 0.6 mJ

C

So, heat generated ΔH is given by ΔH = Wbattery − ΔUC = 0.3 mJ

E q

9/20/2019 11:44:03 AM

Hints and Explanations H.169



Hence, (A), (C) and (D) are correct.

37.

CV –CV

39.

(

q1 = ( CE ) 1 − e

V

E =

Q σ = ε 0 Aε 0

−

)

t τC

)

⇒

t

Now,

dq1 CE − τ C e = τC dt t

So, E remains constant and hence option (A) is correct Work is done against attractive force by the external force Fext



Fele

Fext

So, option (B) is also correct. 1 CV 2 2  When the battery stays connected to the capacitor, then V = constant . When the separation between the ε A plates is increased, then capacitance given by C = 0 d decreases and hence U also decreases. So option (C) is also correct. Hence, (A), (B) and (C) are correct. Since, U =

A +30 –30 q

(

t τC

and q2 = ( 2CE ) 1 − e

and

38.

−

q1 1 = (at any time) q2 2 So, the ratio of charges in steady state is also 1 : 2



 When the battery is disconnected, then the charge will remain constant. Now when the plates are pulled apart, then we have

τ C1 = τ C2 = 2RC = τ C (say)

CHAPTER 2

and charge flown through the switch is 60 μC .

–30 +30 D

The common potential is given by C V − C2V2 30 − 30 = =0 V = 1 1 C1 + C2 C1 + C2

dq2 2CE − τ C e = τC dt

⎛ dq ⎞ ⎛ dq ⎞ ⇒ ⎜ 1⎟ ≠⎜ 2⎟ ⎝ dt ⎠ ⎝ dt ⎠

The option (A) is wrong. Hence, (B), (C) and (D) are correct. 40. In steady state, the potential difference across C1 is 20 V and potential difference across C2 is 0 V . Then, charge stored on C1 is 40 μC and charge stored on C2 is 0 μC . 20 V 1Ω

2Ω 20 V

C2 1Ω 40 V



Hence, (B) and (D) are correct.

41. L  et the charge flowing from the positive terminal to the negative terminal of the battery be q, then ⎛ Q⎞ q = CV = C ⎜ ⎟ = Q ⎝ C⎠ Q+q Q–q Q Q Q +q –q Q q

Q2′ = C2V = 0

Q C

So, final energy stored in the arrangement is zero. Let a charge q flow from A to D , then

Total charge on plate x is

30 − q = 0

Q + q = Q + Q = 2Q

⇒ q = 30 μC



Hence, (A) and (C) are correct.

02_Ch 2_Hints and Explanation_P2.indd 169

3Ω 60 V

120 V

Q1′ = C1V = 0



60 V 3Ω

C1

Final charge on the first capacitor is Final charge on the second capacitor is

40 V 2Ω

q

Total charge on plate y is Q − q = Q − Q = 0

9/20/2019 11:44:11 AM

H.170  JEE Advanced Physics: Electrostatics and Current Electricity The energy supplied by the cell is Q2 ⎛ Q⎞ Wbattery = ⎜ ⎟ q = = CV 2 ⎝ C⎠ C Hence, all the options are correct. 1 CV 2 2  As source between the plates is connected, so potential difference remains constant. But capacitance C becomes KC hence energy stored is increased by factor K . 42. (A) U =





V is not changed. d (C) Charge on each plate is increased by factor K hence force between them increases by a factor of K 2.

(B) Electric field

(D) Since Q = CV

Hence charge becomes KQ as C becomes KC and V remains unchanged. Hence, (A), (C) and (D) are correct.

VA =

VB =

11. T  he capacitance depends upon the geometrical parameters only. And if Q is increased then V increase. Hence, the correct answer is (C). 12. In the given case V = V0 (constant) 1 Energy stored in the capacitor = CV 2 2 C ′ = CK, so energy stored will become A times



1 ⎛ QA QB ⎞ + ⎜ ⎟ b ⎠ 4πε 0 ⎝ a

Since V =

10. The sum of charges on both the plates should be zero. Hence, the correct answer is (A).



Reasoning Based Questions 6.

Q C If sizes of two capacitors are different then potentials will also be different. Thus potential difference may exist between them although they carry same amount of positive charge. Hence, the correct answer is (D). 9.

Q = CV , so Q will become K times Kq = Kσ 0 A Hence, the correct answer is (C). ∴ surface charge density σ ′ =

13. A

1 ⎛ QA QB ⎞ + ⎜ ⎟ b ⎠ 4πε 0 ⎝ b QB QA a

B



⇒ VA − VB =

QA ⎛ 1 1 ⎞ 1 ⎛ QA QA ⎞ − ⎜ − ⎟ ⎜ ⎟= 4πε 0 ⎝ a b ⎠ 4πε 0 ⎝ a b ⎠

Hence, the correct answer is (A). 7.

Equivalent capacitance of parallel combination is

Cparallel = C1 + C2 + C3

Hence, the correct answer is (C).

8.

 hen the battery is disconnected from the capacitor, W then Q = constant

Energy =

2

2

Q Q d = 2C 2ε 0 A



⇒ Energy ∝ d



Hence, the correct answer is (D).

02_Ch 2_Hints and Explanation_P2.indd 170

B – +

A +

–

+

– +

+

–

+

– +

+

–

+

– +

+

–

V = VA1 – VB + VB

A

b

+

V′ = VA1 – VB ⇒ V′ < V



Hence, the correct answer is (C).

14.

C1 =



B

ε 0 KA d C1 ⎛ K1 ⎞ ⎛ d2 ⎞ ⎛ K1 ⎞ ⎛ d2 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 1 = = =⎜ ⎟⎜ ⎟ = C2 ⎜⎝ d1 ⎟⎠ ⎜⎝ K 2 ⎟⎠ ⎜⎝ K 2 ⎟⎠ ⎜⎝ d1 ⎟⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ 6

Hence, the correct answer is (B).

Linked Comprehension Type Questions 1.

Since,

C =



⇒ C=

ε0 A d−t+

t k

ε0 A d d d− + 3 6

9/20/2019 11:44:18 AM

Hints and Explanations H.171 6 μF

ε0 A d d− 6



⇒ C=



⇒ C=

6 ⎛ ε0 A ⎞ ⎜ ⎟ 5⎝ d ⎠



⇒ C=

6 C0 5

200 V

⎛ 600 + q ⎞ ⎛ 150 + q ⎞ 200 − ⎜ ⎟ −⎜ ⎟ = 0 ⇒   q = 100 μC ⎝ 6 ⎠ ⎝ 3 ⎠

According to question



ε0 A 3 ε0 A = t 2 d d− 2

Hence, charge on 6 μF capacitor will be 700 μC and that on the 3 μF capacitor will be 250 μC .  Hence, the correct answer is (B).

t 2 Solving we get = d 3  Hence, the correct answer is (A). 3.

For dielectric U1 =

q2 q2 q2 = = 2C1 ⎛ ε0 A ⎞ ⎛ 3ε A ⎞ 2 2⎜ 0 ⎟ ⎝ 2d ⎠ ⎜ t⎟ ⎜⎝ d − ⎟⎠ 2

6.

From all the switches 100 μC of charge will flow.



Hence, the correct answer is (D).

7.

Let d be the separation between the plates and  be the length of the silk thread, then in the final position, we have

θ

Hence, the correct answer is (B).

4.

 late b and c are joined together and they are neither P connected to any of the terminals of the battery nor to any other source of charge. So, they together form an isolated system. Hence, the correct answer is (D).

5.

100 V

sin θ =

S2

⇒ θ = 30° 



Hence, the correct answer is (A).

8.

 hen the ball again touches the plate A, the charge on W plate A, i.e., Q0 = C0V0 , redistributes on A and ball. Let the charge on plate be Q0′ and on ball be Q after redistribution such that the common potential is V . Then

50 V S3

200 V

In steady state, after closing the switch, let q charge goes from battery.

d 50 1 = =  100 2



3 μF c d

S1

02_Ch 2_Hints and Explanation_P2.indd 171

V0

d

 he charges, before the switches are closed are shown T here 6 μF a b

C1

S

q2 q2 U2 3 = U 2 =    ⇒   = 2C2 U1 2 ⎛ ε0 A ⎞ 2⎜ ⎟ ⎝ d ⎠

B

A

For conductor with air filling the complete space, we have

CHAPTER 2

Then charge on each will increase by q Applying Kirchhoff’s Voltage Law, we get

 Hence, the correct answer is (D). 2.

3 μF

{with plate A }

Q0′ + Q = C0V0

Q0′ C0 = Q C

⇒ 1+

Q0′ C = 1+ 0 Q C

9/20/2019 11:44:25 AM

H.172  JEE Advanced Physics: Electrostatics and Current Electricity



⎛ C ⎞ ⇒ Q=⎜ ( Q + Q0′ ) ⎝ C + C0 ⎟⎠



⎛ C ⎞ ⇒ Q=⎜ C0V0 ⎝ C + C0 ⎟⎠



Q CV ⇒ V= = 0 0 C C + C0



Hence, the correct answer is (B).

9.

I n the final position of the ball, the free body diagram shows that

15.

C∝r



⇒



Hence, the correct answer is (A).

C A rA 2 = = CB rB 3

16. T  he charge will be shared till both attain a common potential V (say). Then

Q2 C2V C2 3 = = = Q1 C1V C1 2

⎛V⎞ T sin θ = QE = Q ⎜ ⎟ ⎝ d⎠



⇒

Q2 3 +1= +1 Q1 2

and T cos θ = mg



⇒

Q2 + Q1 3 + 2 = 2 Q1



⇒

Q 5 = Q1 2

T cos θ T θ

V F = QE = Q d

T sin θ



mg

QV mgd



⇒ tan θ =



QV ( CV )V CV 2 ⇒ mg tan θ = = = d d d

Since V =

C0V0 C + C0 C ⎛ C0V0 ⎞ d ⎜⎝ C0 + C ⎟⎠

2



⇒ mg tan θ =



⇒



C + C mgd tan θ ⇒ V0 = 0 C0 C



C ⎞ ⎛ 50 ⎞ tan 30 ⎛ ⇒ V0 = ⎜ 1 + m ( 10 ) ⎜ ⎟ ⎝ 1000 ⎟⎠ C C0 ⎠ ⎝



C ⎞ m ⎛ ⇒ V0 = ⎜ 1 + C0 ⎟⎠ 2 3C ⎝

C02V02

( C0 + C )

2

=

mgd tan θ C

 Hence, the correct answer is (C). 10.

CA 1 = CB K



Hence, the correct answer is (B).

02_Ch 2_Hints and Explanation_P2.indd 172

2Q 5 Hence, the correct answer is (C). ⇒ Q1 =

17.

( U A )i (UA ) f



Hence, the correct answer is (C).

=

2

qi2 ⎛ 2 ⎞ 4 =⎜ ⎟ = 25 q2f ⎝ 5 ⎠

1 C1V 2 Ui C1 2 2 2 = = = = 18. 1 + + Uf C C 2 3 5 2 1 2 ( C1 + C2 )V 2 Hence, the correct answer is (B). 22. C  apacitor A is a combination of two capacitors CK and CO in parallel. Hence, C A = C K + CO =

Kε o A ε o A ε A + = ( K + 1) o d d d

Here, A = 0.02 m 2 Substituting the values, we have C A = ( 9 + 1 )

8.85 × 10 −12 ( 0.02 )

( 8.85 × 10 ) −4



⇒ C A = 2 × 10 −9 F = 2 nF



Hence, the correct answer is (B).

F

23. E  nergy stored in capacitor A, when connected with a 110 V battery is U A =

(

1 1 C AV 2 = 2 × 10 −9 2 2

)( 110 )



⇒ U A = 1.21 × 10 −5 J ≅ 12 μ J



Hence, the correct answer is (C).

2

9/20/2019 11:44:35 AM

Hints and Explanations H.173

(

qA = C AV = 2 × 10 −9

⇒ qA = 2.2 × 10

−7

)( 110 )

C

Now, this charge remains constant even after battery is disconnected. But when the slab is removed, capacitance of A will get reduced. Let the new capacitance be C ’A. So C ’A =

(

)



⇒ Cnet = 2C



Hence, the correct answer is (C).

27. I f they were initially uncharged, C1 stores the same charge as C2 and C3 together. With greater capacitance, C3 stores more charge than C2 . Then Q1 > Q2 > Q3 Hence C1 > C2 > C3 (in terms of charges)

8.85 × 10 −12 ( 0.04 ) ε o ( 2A ) = F d 8.85 × 10 −4

28. The

Energy stored in this case would be

( (

) )



2

J

⇒ U ’A = 6.05 × 10 −5 J > U A

W = U ’A − U A = ( 6.05 − 1.21 ) × 10 −5 J ⇒ W = 4.84 × 10 −5 J = 48.4 μ J



Hence, the correct answer is (A).

25. Capacitance of B when filled with dielectric is CB =

(

)

(

Hence, the correct answer is (C).

30. E  ach face of P2 carries charge, so the three-plate system is equivalent to

( 0.02 ) K ε o A ( 9 ) 8.85 × 10 = F −4 d 8.85 × 10 −12

equivalent capacitor stores the same

29. I f C3 is increased, the overall equivalent capacitance increases. More charge moves through the battery and Q increases. As ΔV1 increases, ΔV2 must decrease so Q2 decreases. Then Q3 must increase even more: Q3 and Q1 increase; Q2 decreases. Hence charge on C1 and C3 increase but that on C2 decreases. Hence, the correct answer is (B).

So, work done to remove the slab would be



( C2  C3 )

charge as C1. Since it has greater capacitance, so Q implies that it has smaller potential differΔV = C ence across it than C1. In parallel with each other, C2 and C3 have equal voltages. ΔV1 > ΔV2 = ΔV3 . Hence C1 > C2 = C3 (in terms of potentials)

⇒ C ’A = 0.4 × 10 −9 F

2 −7 1 ( qA ) 1 2.2 × 10 = U ’A = 2 C A′ 2 0.4 × 10 −9

Hence, the correct answer is (A).

)

12 V

⇒ CB = 1.8 × 10 −9 F

P1

P2

P2 P3

These two capacitors are in parallel. Therefore, net capacitance of the system is

Each capacitor by itself has capacitance

C = C ’A + CB = ( 0.4 + 1.8 ) × 10 −9 F

C =



⇒ C = 2.2 × 10 −9 F

K ε 0 A ( 8.85 × 10 −12 ) ( 7.5 × 10 −4 ) = = 5.58 pF ( 1.19 × 10 −3 ) d

So, equivalent capacitance = 5.58 + 5.58 = 11.2 pF

Charge stored in the system is q = qA = 2.2 × 10

−7

C

1 q2 qA2 = 2 C 2C

So, energy stored, U =

( (

) )

2



Hence, the correct answer is (B).

31.

Q = C ΔV + C ΔV



⇒ Q = ( 11.2 × 10 −12 ) ( 12 V ) = 134 pC



Hence, the correct answer is (C).



−7 1 2.2 × 10 ⇒ U= 2 2.2 × 10 −9



⇒ U = 1.1 × 10 −5 J = 11 μ J

C = 3 ( 5.58 pF ) = 16.7 pF



Hence, the correct answer is (B).



26. C  apacitors 2 and 3 are in parallel and give equivalent capacitance 6C, Which further is in series with capaci1 1 1 tor 1, so the battery sees capacitance = + Cnet 3C 6C

02_Ch 2_Hints and Explanation_P2.indd 173

CHAPTER 2

24. Charge stored in the capacitor

32. N  ow P3 has charge on two surfaces and in effect three capacitors are in parallel Hence, the correct answer is (D).

33. Only one face of P4 carries charge

Q = C ΔV = 5.58 × 10 −12 F ( 12 V ) = 67 pC

Hence, the correct answer is (B).

9/20/2019 11:44:48 AM

H.174  JEE Advanced Physics: Electrostatics and Current Electricity 34. L  et V1 and V2 be the voltage across 1 and 2 μF capacitors respectively. Then V1 + V2 = 300 V  In series the charge Q on each capacitor is same, therefore we have

1 ⎞ ⎛ 1 + Q = 300 ⇒ ⎜ ⎝ C1 C2 ⎟⎠

Q″1

C = 200 μC

⇒ Q = 200 × 10



⇒ V1 =

Q 200 μC = = 200 V C1 1 μF

V2 =

Q 200 μC = = 100 V C2 2 μF

Q″2

 

If V ′′ is common potential difference across the capacitors, then V ′′ =

Q1 − Q2 ( 200 μF − 200 μC ) =0 = C1 + C2 1 μF + 2 μF

Q1′′= C1V ′′ = C1 ( 0 ) = 0 Q2′′ = C2V ′′ = C2 ( 0 ) = 0

35. W  hen the plates of same polarity are connected together, the capacitors are in usual parallel combination. In parallel the potential difference across each capacitor is the same. C1

Hence, the correct answer is (A).

37. W  hen the opposite plates of SITUATION (II) are connected together, the charge will flow from one having larger quantity of charge of that of smaller quantity of charge i.e., from C2 to C1, to make common potential V ′′′ (say) given by





C2



V′

400 ⎞ 400 ⎛ Q1′′′= C1V ′′′ = ⎜ 1 μF × V⎟ = μC ⎝ ⎠ 9 9

∴ The common potential difference

V = ⇒ V=

2 × ( 200 μC ) 400 μC 400 = = 3 ( 1 μF + 2 μF ) 3 μF

⇒ Charges after sharing are

and Q2′′′= C2V ′′′ =

Q1 + Q2 2Q = C1 + C2 C1 + C2

V

If Q1′ and Q2′ are new charges on the respective capacitors, then ⎛ 400 ⎞ 400 V⎟ = μC Q1′ = C1V = ( 1 μF ) × ⎜ ⎝ 3 ⎠ 3 ⎛ 400 ⎞ 800 Q2′ = C2V = ( 2 μF ) × ⎜ V⎟ = μC ⎝ 3 ⎠ 3 Hence, the correct answer is (C).

02_Ch 2_Hints and Explanation_P2.indd 174

⎛ 800 400 ⎞ − ⎟ μC 400 3 3 ⎠ V = ( 1+2 ) μF 9

( Q2′ − Q1′ ) = ⎝⎜ V ′′′ = C1 + C2

Q′2

V″ (b)

The charges after sharing

Hence, the correct answer is (D).

Q′1

C1

C2

100 V (a)

300C1C2 300 ( 1 × 10 −6 ) ( 2 × 10 −6 ) = ⇒ Q= ( 1 × 10 −6 + 2 × 10 −6 ) C1 + C2 −6

C1 200 V C2





Charge flow

Q Q + = 300 C1 C2

Here C1 = 1 μF , C2 = 2 μF

36. W  hen the opposite plates of SITUATION (I) are connected together, the charge will flow from the capacitor of higher potential to that of lower potential until their potentials are same. The net charge is then ( Q1 − Q2 ).

800 μC 9

Hence, the correct answer is (B).

38. I nitial charges on first and second capacitors are q1 = C1V0 = 100 μC and q2 = −C2V0 = −300 μC finally, let the common potential be V , then qfinal = ( C1 + C2 )V . So, by Law of Conservation of Charge, we get ( C1 + C2 )V = C1V0 + C2 ( −V0 )

⎛ C − C2 ⎞ ⇒ V=⎜ 1 V0 ⎝ C1 + C2 ⎟⎠ ⎛ 3 − 1⎞ ⇒ V=⎜ 100 = 50 V ⎝ 3 + 1 ⎟⎠ Hence, the correct answer is (B).

9/20/2019 11:45:01 AM

Hints and Explanations H.175 39.

( q1 ) f



Hence, the correct answer is (A).

40.

( q2 ) f



Hence, the correct answer is (C).



41.

Uf =

1 ( C1 + C2 )V 2 2 1 ⇒ U f = ( 4 × 10 −6 ) ( 2500 ) 2

So, x1 =

= C2V = ( 3 μF ) ( 50 V ) = 150 μC



⇒ U f = 2 × 10 −6 × 2500



⇒ U f = 5 × 10 −3 J = 5 mJ

and U i =

1 ( C1 + C2 )V02 = 200 mJ 2



⇒ Loss = U i − U f = 15 mJ



Hence, the correct answer is (C).

42.

2k A B

k



⇒



⇒



x2 + x2 = 6 2

3 x2 =6 2 ⇒ x2 = 4 mm x2 = 2 mm 2 Hence, the correct answer is (B).

44. S  ince force between plates of the capacitor connected to the battery is given by F =

1 C2V 2 10 × 10 −3 = =5N 2 d2 2 × 10 −3



⇒ ( 2k ) x1 = 5



⇒ k A = 2k =



Hence, the correct answer is (C).

5 5 = = 2500 Nm −1 x1 2 × 10 −3

CHAPTER 2



= C1V = ( 1 μF ) ( 50 V ) = 50 μC

45. Since, we know that Ceq = S

50 V

Final energy of capacitor is 10 × 10 −3 J

When plate was absent, then

1 2 Cfinal ( 50 ) 2



⇒ 10 × 10 –3 =



⇒ Cfinal = C2 = 8 μF

Since, C =

ε0 A ε0 A ε A = = 0 …(1) 0 . 3 0 . 7 d d 1 0.4 d ⎛ ⎞ + d − t⎜ 1 − ⎟ ⎝ 1 7 K⎠

t = 0.7d

ε0 A d



C d ⇒ 1 = 2 C2 d1



⎛d ⎞ ⎛ 2⎞ ⇒ C1 = ⎜ 2 ⎟ C2 = ⎜ ⎟ ( 8 ) = 2 μF ⎝ 8⎠ ⎝ d1 ⎠

So, the capacitance of the capacitor when the switch S is not closed is 2 μF. Hence, the correct answer is (C).

K=7 d

C0 =

ε0 A = 10 μF d

So from equation (1), we get Ceq =

10 = 25 μF 0.4

Charge stored = CeqV = 25 × 10 = 250 μC 1 ( 25 )( 10 )2 = 1250 μ J 2 Hence, the correct answer is (B).

43. I f x1 and x2 be the extensions in springs connected to plates A and B respectively, then we observe that

Energy stored =

F = ( 2k ) x1 = kx2





⇒ 2x1 = x2

Also, x1 + x2 = d1 − d2 = 8 − 2 = 6 mm

02_Ch 2_Hints and Explanation_P2.indd 175

46. A  fter disconnecting the battery, when the slab is removed, then let new charge on the capacitor plate be q f . So, we have

9/20/2019 11:45:17 AM

H.176  JEE Advanced Physics: Electrostatics and Current Electricity –q q

+q

C = 10 μ F 10 V

−10 +

qf 10

q

where, C =

=0



⇒ q f = 100 μC

New energy =

⎛ 3Q ⎞ Qinside ⎜⎝ 2 ⎟⎠ 9Q V = = = Ceq ⎛ ε 0 A ⎞ 2C ⎜⎝ ⎟ 3d ⎠



( 100 )2 2 × 10

47. Since Wbattery = ΔUC + ΔH

Since electric energy density is given by

where, Wbattery = ( Δq )Vbattery

uelectric =

Since the positive plate of the capacitor is connected to the negative of the battery, so qfinal = −100 μC , whereas we observe that qinitial = 250 μC . Hence the charge flown through the battery is Δq = qfinal − qinitial = 350 μC Also, ΔUC = U final − U initial = 500 − 1250 = −750 μ J

⇒ 3500 = −750 + ΔH



⇒ ΔH = 4250 μ J Hence, the correct answer is (D).



⇒ E=

2

1 ⎛ 3Q ⎞ 9Q 2 ε0 ⎜ = ⎟ 2 ⎝ 2 Aε 0 ⎠ 8ε 0 A 2

So, the electric energy U is obtained by taking the product of the volume of the capacitor with the energy density. Hence



Method-2

U =

( Qinside )2 2Ceq 2

3Q 2 Aε 0

Potential difference between the plates is 9Q ⎛ 3Q ⎞ ΔV = E ( 3 d ) = ⎜ 3d = 2C ⎝ 2 Aε 0 ⎟⎠

⇒ uelectric =

⎛ 9Q 2 ⎞ 27Q 2 ⎛1 ⎞ U = A ( 3 d ) ⎜ ε 0E2 ⎟ = 3 Ad ⎜ ⎟ = 8C ⎝2 ⎠ ⎝ 8ε 0 A 2 ⎠

2Q Q and E2 = 2 Aε 0 2 Aε 0

⇒ E = E1 + E2



1 ε 0 E2 2

⎛1 ⎞ U = A ( 3 d ) ⎜ ε 0E2 ⎟ ⎝2 ⎠

48. Method-1



9Q 2C Hence, the correct answer is (B). ⇒ V=

49. Method-1 Since field between the plates is 3Q E = 2 Aε 0

= 500 μ J

Hence, the correct answer is (A).

E1 =

ε0 A as said in the question. d



⎛ 3Q ⎞ ⎜⎝ ⎟ 2 2 ⎠ = 27Q ⇒ U= 8C ⎛ ε A⎞ 2⎜ 0 ⎟ ⎝ 3d ⎠



Hence, the correct answer is (C).

50. Q

–2Q

2v

v

Method-2 Q

– Q 3Q – 3Q – Q 2 2 2 2 3d

Ceq =

ε0 A 3d

02_Ch 2_Hints and Explanation_P2.indd 176

m

A

–2Q

3d

2m

Q –2Q d

By Law of conservation of Energy, we have ⎛ Loss in Electrostatic ⎞ ⎛ Gain in Kinetic ⎞ ⎜ = ⎝ Energy of system ⎟⎠ ⎜⎝ Energy of system ⎟⎠

9/20/2019 11:45:32 AM

Hints and Explanations H.177

1 1 2 m ( 2v ) + ( 2m ) v 2 2 2

⇒ U initial − U final =



⇒

27Q 2 9Q 2 − = 3 mv 2 8C 8C



⇒

9Q 2 = 3 mv 2 4C



⇒ v=



9Q 1 2 3 mC Hence, the correct answer is (D).

C = C1 + C2 K ε 0Wh ε 0W ( L − h ) + d d



⇒ C=



εW ⇒ C = 0 [ ( K − 1) h + L ] d ⎡ ( K − 1) h ⎤ ⇒ K eff = ⎢ + 1⎥ L ⎣ ⎦ Hence, the correct answer is (A).

53. When the tank is one-fourth empty then it is three3L fourth full, so h = and hence 4 K +1 4 Hence, the correct answer is (D).

K eff =

54. I t is observed that potential difference across C3 is 12 V, so we have C1

C2

1 ⎛ ⎜ C3 9 = ⎜ ⎜ 1 + 1 + 1 ⎜⎝ C C ′ C 1 3 2



⇒ C1V1 = C2′ V1



⇒ C1 = C2′ =



⇒ C2 = 3C1 …(3)

Now, when the dielectric material ( K = 3 ) between the plates of C2 is removed then the new capacitance becomes C2′ , so that

02_Ch 2_Hints and Explanation_P2.indd 177

C2 3

Substituting (3) in (1) and (2), we get 1 ⎞ ⎛ ⎟ ⎜ V0 C3 …(4) 12 = ⎜ ⎟ V0 = 4 C 4 1 3 ⎟ ⎜ + + 1 ⎜⎝ 3C C ⎟⎠ 3 C1 1 3 1 ⎛ ⎜ C3 9 = ⎜ ⎜ 2 + 1 ⎜⎝ C ′ C 2 3

⎞ ⎟ V0 …(5) ⎟ V0 = 2C 3 ⎟ + 1 ⎟⎠ C1

C3 = x and dividing equation (4) by C1 equation (5), we get Now assuming



V0

⎞ ⎟ ⎟ V0 …(1) ⎟ ⎟⎠

⎞ ⎟ ⎟ V0 …(2) ⎟ ⎟⎠

It is also observed that the potential differences across C1 and C2′ are equal, so we have

C3

K

1 ⎛ ⎜ C3 12 = ⎜ ⎜ 1 + 1 + 1 ⎜⎝ C C C3 1 2

C3

V0

52. Since both the capacitors are in parallel, so we have



C′2

3Q 2 3Q 1 = 4 mC 2 3 mC

vrel = 3v =



In this situation, the new potential difference across C3 is found to be 9 V , so C1

So the relative velocity of approach of the plates is



C2 3

C2′ =

CHAPTER 2



12 2x + 1 = 9 4x + 3



⇒ 16 x + 12 = 18 x + 9



⇒ x=

3 2



⇒ x=

C3 3 = C1 2

Substituting the value of x in equation (4), we get 12 =

V0 × 3 4x + 3

9/20/2019 11:45:49 AM

H.178  JEE Advanced Physics: Electrostatics and Current Electricity V0 6+3

VC3 = VC4 =



⇒ 4=



⇒ V0 = 36 V



Hence, the correct answer is (D).





C2 6 = = 2 μF 3 3 Hence, the correct answer is (D).

56.

C3 = xC1 =



Hence, the correct answer is (C).

1 ⎛ 14CV ⎞ 2V ⎜ ⎟= 7 C ⎝ 23 ⎠ 23

Hence (B) → (p)

55. If C2 is 6 μ F, then C2 = 3C1

⇒ VC3 = VC4 =

q 7C

Energy stored across C3 is

⇒ C1 =

U 3 =

3 × 2 = 3 μF 2

1 1 ⎛ q ⎞ C3VC3 2 = ( 3C ) ⎜ ⎝ 7 C ⎟⎠ 2 2



⇒ U3 =

3q2 q2 1 ( 3C ) = 2 98C 2 49C

Matrix Match/Column Match Type Questions



⇒ U3 =

3q2 3 ⎛ 14CV ⎞ = ⎜ ⎟ 98C 98C ⎝ 23 ⎠

1. A → (q, s) B → (p) C → (q, s) D → (r)



⇒ U3 =

3 × 196CV 2 CV 2  98 × 23 × 23 88

Energy stored across C4 is U 4 = C3 = 3C

C1 = C C2 = 2C

C4 = 4C

1 1 1 1 = + + Ceq C 2C 7 C

⇒ Ceq =

14C 23

So, charge across C1 and C2 is q =



⇒ U4 =

⎛ q2 ⎞ 1 ( 4C ) ⎜ ⎝ 49C 2 ⎟⎠ 2



⇒ U4 =

4 q2 4 × 196CV 2 CV 2 =  98C 98 × 23 × 23 66

4. A → (s) B → (q) C → (r) D → (p) After reconnection, charges are redistributed as shown

1 14 + 7 + 2 23 = = Ceq 14C 14C



2

3. A → (p, r, s) B → (p, r) C → (p, r, s) D → (p, r)

Effective capacitance is given by

⇒

1 1 ⎛ q ⎞ C4VC24 = 4C ⎜ ⎟ 2 2 ⎝ 7C ⎠

Hence (D) → (r)

V



2

Hence (C) → (s)

2. A → (q) B → (p) C → (s) D → (r)



2

14CV 23

Charge will get distributed across C3 and C4 in direct ratio of their capacitances and hence charge across C3 is 3 q 3 14CV 6CV = × = 7 7 23 23 Hence, (A) → (q) Potential differences across C3 and C4 would be the same, because they are in parallel so,

Q1 = ( 2 μF )V Q2 = ( 3 μF )V Q3 = ( 6 μF )V

qC3 =

02_Ch 2_Hints and Explanation_P2.indd 178

2 μF

+Q1 3 μF –Q1

+Q2 6 μF –Q2

+Q3 –Q3

9/20/2019 11:46:03 AM

Since, initially the capacitors were in series, so, 12 μC charge appears on each of the positive plates and −12 μC charge appears on each of negative plates. So  a total of 36 μC charge on the three positive plates now redistribute as Q1 , Q2 and Q3 . Q1 + Q2 + Q3 = 36 μC ( 11 μF )V = 36 μC 36 V 11



⇒ V=



⇒ Q1 =

72 μC 11



⇒ Q2 =

108 μC 11



⇒ Q3 =

216 μC 11

V3 =

q3 − q 7V = 3C 5

V4 =

q4 − q 14V = 4C 5

6. A → (q) B → (r) C → (p) D → (s) The charges on three plates which are in contact add to zero, because these plates taken together form an isolated system which cannot receive charges from the batteries Thus, q3 − q1 − q2 = 0 …(1) 2 μF B

5. A → (r) B → (p) C → (q) D → (s)

A

q4 = 16 CV

2C

– +

q2 = 4 CV

3C + –

q

– + + –

D

– + + –

B (q2 – q) C

When circuit is closed, let charge q flows in the circuit; then applying Loop Law in ABCDA q1 − q q2 − q q3 − q q4 − q + + + =0 C 2C 3C 4C

Substituting the value of q1 , q2 , q3 and q4 we get ⎛ 24 ⎞ ( q = ⎜ CV ) ⎝ 5 ⎟⎠

V2 =

E 20 V

⇒ q3 + 3 q1 = 60 …(2) q2 q3 − =0 4 6

⇒ 3 q2 + 2q3 = 240 …(3)

Solving the above three equations, we have q1 =

10 140 μC, q2 = μC, q3 = 50 μC 3 3

Potential difference across 6 μC is

(q3 – q)

Hence V1 =

+ q – 3

q1 q3 − + 10 = 0 2 6

and 20 −

⇓ (q1 – q)

−

q3 = 9 CV

(q4 – q)

D

Applying Kirchhoff’s Law in loop ABCFA and CDEFC

+ 4C –

A

+ – q2

F

10 V – + C



+ – q1 6 μF

q1 = CV

4 μF

C

CHAPTER 2

Hints and Explanations H.179

q1 − q 19V = C 5

V6 μ C =

q3 50 μC 25 = = V 6 6 μF 3

7. A → (q) B → (p) C → (s) D → (r) 2

4

1

5

3 A

B

C

D

V

q2 − q 2V = 2C 5

02_Ch 2_Hints and Explanation_P2.indd 179

9/20/2019 11:46:16 AM

H.180  JEE Advanced Physics: Electrostatics and Current Electricity These five plates constitute four identical capacitors in ε A parallel, each of capacity 0 . Now as plate 1 is cond nected to positive terminal of battery and is a part of one capacitor only, so charge on it ⎛ ε AV ⎞ ⎛ ε AV ⎞ q1 = + ⎜ 0 = 1⎜ 0 ⎝ d ⎟⎠ ⎝ d ⎟⎠ So (A) → (q) However the plate 4 is connected to negative terminal of battery and in common to two identical capacitors in parallel. 2ε AV ⎛ ε AV ⎞ = −2 ⎜ 0 So q4 = − 0 ⎝ d ⎟⎠ d Hence (B) → (p) From circuit, it is obvious that between the plates 2 and 3, battery is connected so potential difference will ⎛ 1 ⎞ be V ⎜ = ( 2V ) ⎟ . Between 1 and 5. Plates 1 and 5 gets ⎝ 2 ⎠ connected through connecting wire, so potential difference is zero ( = 0 ( 2 V ) ).

So, when Q = constant , then F = constant.

(B)  As battery is connected to capacitor, so V is constant

when slab is inserted, C increases So, energy U = Since F =

1 CV 2 also increases 2

CV 2 2d

So, as U increases, F also increases.

(C) If area of plates of capacitor increases, then C increases. As capacitor is isolated, so Q = constant Since, V =

Q C

As C increases, V decreases. Q2 2C

Hence (C) → (s), (D) → (r)

Since, U =

8. A → (q) B → (r) C → (s) D → (p) Done Already

As C increases, U decreases.

9. A → (p, s) B → (q, r, s) C → (q, r, s) D → (p) 10. A → (p, q) B → (r, q) C → (p, q, s) D → (r, q) (A) For an isolated capacitor Q = constant So, if distance between plate decreases, then capacitance will increase. Since Q = CV

⇒ V∝

1 C

So as C increases V decreases. Q2 2C So as C increases energy decreases. Since Energy U =

Q2 Since Force F = 2 Aε 0

02_Ch 2_Hints and Explanation_P2.indd 180

Since, F =

q2 q2 = 2ε 0 A 2Cd

As U decreases i.e. C increases, F decreases. (D) Capacitor is connected with battery V = constant As d decreases, C increases Since, U =

1 CV 2 2

As C increases, U increases. Since, F =

Q2 C 2V 2 = 2 Aε 0 2 Aε 0

As C increases, so F increases. 11. A → (q, s) B → (r, s) C → (p, r, s) D → (p, r, s)

For A: If V = E, i.e. potential difference across the capacitor is same as that of e.m.f. of battery, no charge will be supplied by battery and hence work done by battery = 0.

U i = U f so ΔH = 0 In all cases, zero charge appears on outer surfaces of the plates of capacitor.

9/20/2019 11:46:31 AM

Hints and Explanations H.181



For B: If V > E, so capacitor charges the battery and hence thermal energy will be developed in circuit.



For C: If V < E so battery performs some work on the capacitor.



For D: Combination of B and C .

14. A → (p, s) B → (r, s) C → (p, q) D → (q) For A: Charge is constant, C increases, so energy stored decreases. Since system is isolated, so

13. A → (p) B → (p) C → (r) D → (q)

For B: Since potential is constant, so C increases, so energy stored increases



⇒ ΔU > 0 and We = −ΔU < 0

Since charge of capacitor increases, which means that work done by battery is greater then zero. For C: Potential is constant, C decreases, Q decreases and U also decreases. So, Wbattery < 0, ΔU < 0, Wext > 0

a

(A)



C

C/2

C/2

Ceq = 2C

15. A → (p, r) B → (q) C → (s, q) D → (q)

b a

(B)

For D: Charge is constant, C decreases, U increases So, ΔU > 0, Wext > 0

CHAPTER 2

ΔU < 0 and Wext = ΔU < 0

12. A → (q) B → (p) C → (r) D → (s)

C1

C2

2C

2C

C 2C

Ceq = 2C

2C

V

a

(C)

a, 1, 2

1



3

2C C

C

C/2

4 μF

V1

V2

4

C/2Ceq = 3C

b, 3, 4

b

a

C

C

1 C b

Ceq = C

02_Ch 2_Hints and Explanation_P2.indd 181

4 4 2 × V = V = V = 0.6 V and 6 6 3 V 2 × V = = 0.3 V 6 3

1 μF

4 μF

V1

V2

2 = 1 μF 2

C 1, 2

2 C

So, V1 =

When d is double d , then C becomes C ′ =

a

C

V

V2 =

Ceq = 3C (D)

2 μF

C

2C

2

ε0 A d

C =

b

C

Ceq = C

V

C

b

V1′ =

4V = 0.8 V 5

V2′ =

1V = 0.2 V 5

9/20/2019 11:46:43 AM

H.182  JEE Advanced Physics: Electrostatics and Current Electricity So, potential difference across C1 increases and that across C2 decreases. Initial energy stored in C1 is ( U1 )i =

16 1 1 16 C1V12 = × 2 × V 2 = V 2 2 2 36 36

C2 is ( U 2 )i =

2

V 1 1 8 C2V22 = × 4 × = V2 2 2 9 9

Final energy stored in C1 is ( U1 ) f =

1 16 8 × 1 × V2 = V2 2 25 25

C2 is ( U 2 ) f =

V2 1 2 ×4× = V2 2 25 25

18 So, potential energy of C1 decreases by a factor of 25 and potential energy of C2 decreases.

Integer/Numerical Answer Type Questions 1.

Let us consider an infinitesimal strip of width dx and length b to approximate a differential capacitor ⎛y ⎞ of area bdx and separation d = y0 + ⎜ 0 ⎟ x . All such ⎝ a ⎠ differential capacitors are in parallel arrangement.

2.

The original kinetic energy of the particle is

K =

(

1 1 mv 2 = 2 × 10 −16 kg 2 2

) ( 2 × 106 ms−1 )

2

⇒ K = 4 × 10 −4 J

The potential difference across the capacitor is ΔV =

Q 1000 μC = = 100 V C 10 μF

For the particle to reach the negative plate, the particle-capacitor system would need energy U = qΔV = ( −3 × 106 C ) ( −100 V )

⇒ U = 3 × 10 −4 J

Since its original kinetic energy is greater than this, the particle will reach the negative plate As the particle moves, the system keeps constant total energy ( K + U )at + plate = ( K + U )at-plate 1( 2 × 10 −16 ) v 2f + 0 2



4 × 10 −4 J + ( −3 × 10 −6 C ) ( +100 V ) =



⇒ vf =

3.

Initially (capacitors charged in parallel),

2 ( 1 × 10 −4 J ) = 106 ms −1 = 1000 kms −1 2 × 10 −16 kg

q1 = C1 ( ΔV ) = ( 6 μF ) ( 250 V ) = 1500 μC b

2y0

q2 = C2 ( ΔV ) = ( 2 μF ) ( 250 V ) = 500 μC After reconnection (positive plate to negative plate),

y0 dx

x a

ε 0 ( bdx ) dC = ⎛y ⎞ y0 + ⎜ 0 ⎟ x ⎝ a ⎠ a

C = ε 0b

dx y ⎞ + 0x ⎝ 0 a ⎟⎠

∫ ⎛⎜ y 0

y ⎡ ⎛ ⎞⎤ y + 0 ×a ⎥ ε 0b ⎢ ⎜ 0 a ⎟ C = ⎢ log e ⎜ ⎟⎠ ⎥ y ⎛ y0 ⎞ ⎢⎣ ⎝ ⎥⎦ 0 ⎜⎝ ⎟⎠ a

ε ab C = 0 log e 2 y0

⇒ α=2

02_Ch 2_Hints and Explanation_P2.indd 182

qtotal ′ = q1 − q2 = 1000 μC and ΔV ’ =

q ’total 1000 μC = = 125 V Ctotal 8 μF

Therefore, q1′ = C1 ( ΔV ′ ) = ( 6 μF ) ( 125 V ) = 750 μC q ’2 = C2 ( ΔV ′ ) = ( 2 μF ) ( 125 V ) = 250 μC 4.

Assume a potential difference across a and b , and notice that the potential difference across the 8 μF capacitor must be zero by symmetry. Then the equivalent capacitance can be determined from the following circuit. 4 μF

4 μF

a

b 2 μF

⇒

a

6 μF6 μF

b

⇒

a

3 μF

b

2 μF

Cab = 3 μF

9/20/2019 11:46:57 AM

Hints and Explanations H.183

2π rE =



So E =

If a slab of thickness t is introduced between the plates with new separation d′ , then its new capacitance,

qin ε0

λ 2π rε 0

C ′ =

l

As Q = CV , the charge on the capacitor is same in both cases, therefore to maintain the same potential difference the capacitance C and C ′ must be same i.e., from (1) and (2).

r2

r2

∫

ΔV = − E ⋅ dr = r1



λ

∫ 2π rε r1

dr = 0

λ ⎛r ⎞ log e ⎜ 1 ⎟ 2πε 0 ⎝ r2 ⎠

λ max = Emax rinner 2πε 0

⎛ 25 ⎞ ΔV = ( 1.2 × 106 Vm −1 ) ( 0.1 × 10 −3 m ) log e ⎜ ⎝ 0.2 ⎟⎠ ΔVmax = 579 V 6.

Capacitance of metal surface of radius

r = 0.5 × 10 −3 m 0.5 × 10 −3 1 = × 10 −12 farad 18 9 × 109 Rate of escape of charge from surface is R is

C = 4πε 0 r =

R =

80 × ne 100 80 × 6.25 × 1010 × 1.6 × 10 −19 Cs −1 100



⇒ R=



⇒ R = 8 × 10 −9 Cs −1



∴ If t is the required time, then charge escaped

q = Rt = ( 8 × 10 −9 ) t From relation q = CV , we have 1 × 10 −12 × 1    {∵ V = 1 V } 8 × 10 t = 18 −9



10 −12 10 −3 = ⇒ t= −9 8 × 10 × 18 144 1000 × 10 144

−6

s = 6.95 μs = 6950 ns



⇒ t=

7.

If d is the separation between the plates (each of area A0 ) of a parallel plate condenser in air, then its capaciε A tance C = 0 …(1) d

02_Ch 2_Hints and Explanation_P2.indd 183

ε0 A …(2) ⎛ ⎛ 1⎞⎞ d′ − t ⎜ 1 − ⎜ ⎟ ⎟ ⎝ K⎠⎠ ⎝



ε0 A = d

CHAPTER 2

5.

ε0 A ⎛ ⎛ 1⎞⎞ d ’− t ⎜ 1 − ⎜ ⎟ ⎟ ⎝ K⎠⎠ ⎝

⎛ ⎛ 1⎞⎞ ⇒ d = d′ − t ⎜ 1 − ⎜ ⎟ ⎟ ⎝ K⎠⎠ ⎝

Here, d′ = d + 2.4 mm = d + 2.4 × 10 −3 m t = 3 mm = 3 × 10 −3 m

⎧ ⎛ 1⎞⎫ ⇒ d = d + 2.4 × 10 −3 − 3 × 10 −3 ⎨ 1 − ⎜ ⎟ ⎬ ⎩ ⎝ K⎠⎭



⎧ ⎛ 1⎞⎫ ⇒ 3 ⎨ 1 − ⎜ ⎟ ⎬ = 2.4 ⎩ ⎝ K⎠⎭



⇒ K=5

8.

(a) The capacitors of 3, 5 and 4 μ F are in parallel, therefore their equivalent capacitance

   C1 = 3 + 5 + 4 = 12 μF Again the capacitor of 4 and 2 μF are equivalent to a single capacitor of capacitance    C3 = 4 + 2 = 6 μF For equivalent capacitance between x and y, C1 and C2 are in series. If C ′ is the equivalent capacitance of C1 and C2 , then    C =

C1C2 12 × 4 48 = = = 3 μF C1 + C2 12 + 4 16

Similarly, C3 and C4 are in series and their equivalent capacitance C" is given by    C ′′ =

C3C 4 6 × 3 18 = = 2 μF = C3 + C 4 6 + 3 9

For equivalent capacitance between x and y, C’ and C ′′ are in parallel, therefore the equivalent capacitance between x and y    C = C ′ + C ′′ = 3 + 2 = 5 μF

(b) The charge of 5 μF capacitor

   = 120 μC = 120 × 10 −6 C

9/20/2019 11:47:18 AM

H.184  JEE Advanced Physics: Electrostatics and Current Electricity The p.d. across 5 μF capacitor =

In series the charge on each capacitor remains same, but maximum charge on first capacitor will be 6000 μC .

−6

120 × 10 = 24 V 5 × 10 −6

As the capacitor 3, 5 and 4 μF are in parallel, the p.d. across each of these capacitors is 24 V.

∴ The charge on 3 μF capacitor

Therefore charge on second capacitor must also be 6000 μC The potential across second capacitor q 6000 μC = 3000 V = 3 kV = C2 2 μF

= 3 × 10 −6 × 24 C

V2 =

= 72 × 10 −6 C

So, maximum voltage across system will be

and the charge on 4 μF capacitor = 4 × 10

−6

× 24 = 96 × 10

−6

C

= V1 + V2 = 6 kV + 3 kV = 9 kV 10. Capacitance of parallel plate capacitor.

Therefore, the total charge flowing through C1 and C2 is

C =

= ( 72 + 120 + 96 ) × 10 −6 C

Charge on capacitor q = CV =

= 288 × 10 −6 C The potential difference across C2 =

288 × 10 −6 = 72 V 4 × 10 −6

∴ Net potential difference (p.d.) between x and y

ε0 A d

If the separation is doubled, the capacitance becomes C half, i.e., here capacitance C ′ = . If battery is discon2 nected the charge remains same i.e., q = constant.

∴ Net potential V ′ is given by CV = C ′V ′

= 24 + 72 = 96 V

V ′ =

As equivalent capacitance of C3 and C4 is 2 μF , the charge flowing through C3 and C4

As

C′ =

= 96 × 2 × 10 −6 C V ′ =

= 192 × 10 −6 C The potential difference between x and z is Vxz = 9.

192 × 10 −6 = 64 V 3 × 10 −6

=

C2

=

q V2

V1

q1 = C1V1 = 1 μF × ( 6 × 10 3 V ) ⇒ q1 = 6000 μC

Charge on second capacitor q2 = C2V2

CV = 2V = 2 × 10 3 V ⎛ C⎞ ⎜⎝ ⎟⎠ 2

( CV )2 2C

=

1 1 ε0 A 2 CV 2 = V 2 2 d

Here A = 0.2 m 2 , d = 10 −2 m

⇒ W=

1⎛ 1 ⎞ 0.2 ( 3 )2 × × 10 ⎜⎝ 7 ⎟ 2 36π × 10 ⎠ 10 −2



⇒ W=

1 × 10 −2 J = 8.8 × 10 −5 J 36π

Final voltage ( V ′ ) across capacitor is given by

⇒ q2 = ( 2 μF ) × ( 4 × 10 V ) = 8000 μC

02_Ch 2_Hints and Explanation_P2.indd 184

C 2

q2 q2 q2 ⎡ 1 1 ⎤ q2 − = − = 2C ′ 2C 2 ⎢ C C ⎥ 2C ⎢ ⎥ ⎣2 ⎦

Charge on first capacitor



CV C′

Work required, W = U 2 − U1

Given V1 = 6 kV, V2 = 4 kV C1

ε0 A V d

3

q = CV = C ′V ′

9/20/2019 11:47:37 AM

Hints and Explanations H.185

1 1 1 C1 + C2 = + = C ′ C1 C2 C1C2 C1C2 C1 + C2

⇒ C′ =



C × ηC1 ηC1 = ⇒ C′ = 1 …(1) C1 + ηC1 1 + η

Now C ′ and capacitor (4) are in parallel. Hence



⇒

ηC1 + ηC1 1+η

( η2 + 2η ) C1 …(2) C ′′ = 1+η

Further C ′′ and capacitor (3) are in series. So the equivalent capacitance C is given by

1 1 ( 1 + η ) = ( η2 + 2η ) + ( 1 + η ) = + 2 C C1 ( η + 2η ) C1 ( η2 + 2η ) C1

( η2 + 2η ) C1 …(3)



⇒ C=



⇒ q = CE =

η2 + 3η + 1

( η2 + 2η ) C1E …(4) η2 + 3η + 1

The voltage V across equivalent capacitor is given by V =

⇒ V=

q′ q′ 1 ⎡ ηC1E ⎤ = = C2 ηC1 ηC1 ⎢⎣ η2 + 3η + 1 ⎥⎦ E ⇒ V= 2 …(7) η + 3η + 1

V =

Substituting the values, we get V =

110 110 = = 10 V . 4 + 6 + 1 11

12. Charge on 10 μF condenser



C ′′ = C ′ + C2 =

Voltage across C2 is given by

( η2 + 2η ) C1E × ( 1 + η ) q = C ′′ η2 + 3η + 1 ( η2 + 2η ) C1 ( 1 + η ) E …(5) η2 + 3η + 1

(∵ The charge q will be across q1 and equivalent capacitor C ′′ )

q1 = 10 × 10 −6 × 150 C = 2.5 × 10 −3 C Charge on 20 μF condenser q2 = 20 × 10 −6 × 300 C = 6 × 10 −3 C When the two condensers are connected in parallel, the total charge q = ( 1.5 × 10 −3 + 6 × 10 −3 ) = 7.5 × 10 −3 C = 7500 μC The equivalent capacitance = 30 μF If V be the potential, then 30 × 10 −6 V = 7.5 × 10 −3 7.5 × 10 −3 = 250 V 30 × 10 −6 Energy of first condenser before connection

⇒ V=

U1 =

1 ( × 1.5 × 10 −3 ) 150 2

Energy of second condenser before connection U 2 =

1 ( × 6 × 10 −3 ) 300 2

Total energy of condensers before connection U initial =

1 1 ( × 1.5 × 10 −3 ) 150 + ( 6 × 10 −3 ) 300 2 2

⇒ U i = 1.0125 J

Energy after connection U f =

1 ( × 7.5 × 10 −3 ) 250 = 0.9375 J 2

The voltage V will be across C2 and equivalent capacitor C ′

Energy lost

⎛ ηC1 ⎞ ⎡ ( 1 + η ) E ⎤ q′ = C ′V = ⎜ ⎝ 1 + η ⎟⎠ ⎢⎣ η2 + 3η + 1 ⎥⎦

13. The capacitance C0 before the slab is introduced



⇒ q′ =

ηC1E …(6) 2 η + 3η + 1

CHAPTER 2

CV C = V = 2V = 2 × 10 3 V C C′ 2 1 1. First of all we shall calculate the equivalent capacitance of this arrangement. The condensers (1) and (2) are in series. Their equivalent capacitance is give by i.e., V ′ =

− ΔU = 1.0125 − 0.9375 = 0.075 J = 75 mJ

V0 =  

ε 0 A ( 8.9 × 10 −12 )( 10 −2 ) = = 8.9 × 10 −12 farad d 10 −2

The charge will be distributed across C1 and C2 .

02_Ch 2_Hints and Explanation_P2.indd 185

9/20/2019 11:47:56 AM

H.186  JEE Advanced Physics: Electrostatics and Current Electricity

q = C0V0 = ( 8.9 × 10 −12 ) × 100 = 890 × 10 −12 C = 890 pC

Now, Electric field intensity



V0 100 = = 1 × 10 4 Vm −1 = 10 kVm −1 d 10 −2

E0 =

E =

E0 1 × 10 4 = = 1.43 × 10 3 Vm −1 = 1430 Vm −1 K 7

Potential difference between the plates with dielectric present is given by V = E0 ( d − b ) + Eb



⇒ V = ( 1 × 10

4

) ( 10

−2

− 0.6 × 10

−2

)+

( 1.43 × 103 ) ( 0.5 × 10 −2 )

⇒ V = 57 V

The free charge on the plate is the same as before. The capacitance with dielectric present is C =

∴

Cf =

ε0 A farad (5 − x) 100

Again CV = C0V0 = C f V f

Electric field intensity in dielectric



Here it should be remembered that the metal sheet splits the single capacitor into two in series.

∴ Free charge

q 8.9 × 10 −10 C = = 16 × 10 −12 farad = 16 pF V 57 V

14. Potential difference without introducing the slab is given by



⇒ V f = V0

According to the problem V f = 1260 V ⇒ 1500 ×



⇒ ( 5 − x ) = 4.2



⇒ x = 0.8 cm = 8 mm

15. The 3 μF and 5 μF capacitors are in series, the potential difference across them will be in inverse ratio of capacitance. Thus P.D. across 5 μF capacitor = 10 × 60 V

The capacitance C0 is given by





⇒ C=

ε0 A d−t+

t K

where t = thickness of slab

0.05C0 25C0 = …(2) 0.01 21 0.05 + 5

Now, C0V0 = CV C0 × 1500 =

q

⇒ ε 0 A = 0.05C0 …(1)

C =

⇒ V=

1500 × 21 = 1260 V 25

Let x be the thickness of the metal plate. Now the effective distance in air between the plates is ( 5 − x ) cm.

02_Ch 2_Hints and Explanation_P2.indd 186

3

4.8 μ F 63 μF 8 a

P.D. across p and b = ( 10 + 6 )V = 16 V Charge on 6 μF capacitor = ( 16 × 6 ) μC = 96 μC Total charge in enclosed portion 1 of the circuit = ( 96 + 30 ) μC = 126 μC    15 μ F 10 μ F

25C0 ×V 21

p 9 μF

3 =6V 5

C

r

ε0 A ε0 A = farad 0.05 d

When the slab is inserted, the capacitance C is given by

(5 − x) ε0 A = 1260 × 0.05 ε 0 A × 100



V0 = E × d = 300 × 5 = 1500 V

C0 =

(5 − x) C0 ε A = 1500 × 0 × Cf 0.05 ε 0 A × 100

3 μF

6 μF

3 μF

The two enclosed portions (1) and (2) are in series combination as shown in figure. Charge on 9 μF capacitor = 126 μC . If Q1 and Q2 are charges on 3 μF and 6 μF branches respectively.

9/20/2019 11:48:14 AM

Hints and Explanations H.187

Q1 Q2 = …(1) 3 6

and Q1 + Q2 = 126 μC …(2) Now solving equations (1) and (2) simultaneously, we get

For loop ABCEA , traversing clockwise, we have −

q1 ( q1 − q3 ) − + 10 = 0 6 3

3 q1 − 2q3 = 60 …(3) Solving equations (1), (2) and (3), we get

Q1 = 42 μC ,

q1 = 24 μC ;

Q2 = 84 μC

q2 = 18 μC ;

The P.D. across 3 μF capacitor (that is between points q and r )

q3 = 6 μC

=

Q1 42 = = 14 V C 3

CHAPTER 2



Charge on 6 μF capacitor is 24 μC Charge on 3 μF capacitor is 18 μC Q ΔV

Thus the P.D. between p and r = ( 16 + 14 ) V=30 V

17. (a) C =

The total charge on 4.8 μF capacitor

⇒   6 × 10 −6 =

= ( 4.8 × 30 ) μC=144 μC Total charge in the enclosed region 3



Q and Q = 120 μC 20

(b) Since Q = 120 μC

= 144 μC + 126 μC = 270 μC

⇒   Q1 + Q2 = 120 μC

The capacitor C is in series with enclosed portion 3, so charge on it is 270 μC and potential ( 60 − 30 ) V

⇒   Q1 = 120 − Q2 …(1)

⎛ 270 ⎞ Hence C = ⎜ μF = 9 μF ⎝ 60 − 30 ⎟⎠

Also,

16. Let the charge on each capacitor be as shown in figure. Applying KVL in the loop (1), traversing clock-wise,



⇒



⇒ Q2 = 40 μC and Q1 = 80 μC

6 μF

B

q1 – q3 3 μF 2 6 μF

q1 q3 1 3 μF A q2 3

3 μF

D



120 − Q2 Q2 = 6 3

18. With switch closed, distance d′ = C

q2 + q3

E = 10 V

−

Q1 Q2 = C1 C2

q1 q3 q2 − + =0 6 3 3

⇒ − q1 − 2q3 + 2q2 = 0 …(1)

C ’ =

d and capacitance 2

ε 0 A 2ε 0 A = = 2C d’ d

(a) Q = C ′ ( ΔV ) = 2C ( ΔV ) = 2 ( 2 × 10 −6 F ) ( 100 V ) ⇒ Q = 400 μC

(b) The force stretching out one spring is

F =

For loop (2), traversing clockwise, we have

2

4C 2 ( ΔV ) 2C 2 ( ΔV ) Q2 = = 2ε 0 A 2ε 0 A ⎛ ε0 A ⎞ ⎜⎝ ⎟d d ⎠

⎛ q − q ⎞ q + q3 q3 − ⎜ 1 3 ⎟ + 2 + =0 ⎝ 3 ⎠ 6 3

F =

−2q1 + 5q3 + q2 = 0 …(2)

One spring stretches by distance x =

02_Ch 2_Hints and Explanation_P2.indd 187

2C ( ΔV ) d

2

2

d 4

9/20/2019 11:48:34 AM

H.188  JEE Advanced Physics: Electrostatics and Current Electricity

2







F 2C ( ΔV ) ⎛ 4 ⎞ 8C ( ΔV ) ⇒ k= = ⎜⎝ ⎟⎠ = x d d d2 ⇒ k=

8 ( 2 × 10 −6 F ) ( 100 V )

21. Let C1 = C and C2 = 2C and the charges on the different capacitors be as shown in figure.

2

C

M

2

2C q2

q1

( 8 × 10 −3 m )2

q3

⇒ k = 2500 Nm −1

N

E

19. The energy transferred is

2C

E

Net charge on isolated system should be zero.

1 1 H ET = QΔV = ( 50 C ) ( 1 × 108 V ) 2 2

Hence, q1 − q2 − q3 = 0 …(1)

9

H ET = 2.5 × 10 J

Applying Loop’s Law in two loops we have,

and 1% of this is absorbed by the tree So, energy absorbed is given by

E −

q3 q1 − = 0 …(2) 2C C

E −

q2 q + 3 = 0 …(3) 2C 2C

ΔEabsorbed = 2.5 × 107 J If m is the amount of water boiled away, then

(

M

)

ΔEint = m 4186 Jkg −1 °C ( 100 °C − 30 °C ) +

(



)

q3 = 10 EC

m 2.26 × 106 Jkg −1 = 2.5 × 107 J



N

⇒ m = 9.79 kg

Solving these three equations, we get

So, m ≅ 10 kg 20. The given combination of capacitors is equivalent to the circuit diagram shown in the figure A

B

C

q3 = −10EC ⇒ VMN =



⇒ VMN = 550 V

22.

Wext =



⇒ Wext =



⇒ Wext = 0.4 μ J = 400 nJ

23.

W = U f − Ui =

D

40 μ F 10 μ F 40 μ F

Assume the charge on point A to be Q . Then, Q = ( 40 μF ) ΔVAB = ( 10 μF ) ΔVBC = ( 40 μF ) ΔVCD

⇒ ΔVBC = 4 ΔVAB = 4 ΔVCD

 So, the centre capacitor will break down first, at ΔVBC = 15 V and when this occurs, we have ΔVAB = ΔVCD =

1 ( ΔVBC ) = 4 V 4

⇒ VAD = VAB + VBC + VCD = 4 V + 16 V + 4 V = 24 V

02_Ch 2_Hints and Explanation_P2.indd 188

q3 = 5E 2C



q2 q2 q2 ( 2d − d ) = x f − xi ) = d ( 2ε 0 A 2ε 0 A 2ε 0 A q2 q2 ( 20 × 10 −12 × 200 ) = = ⎛ ε A ⎞ 2C 2 × ( 20 × 10 −12 ) 2⎜ 0 ⎟ ⎝ d ⎠

2

1 ⎞ V2 ε 0 AV 2 ⎛ 1 − = ( C f − Ci ) 2 ⎜⎝ x f xi ⎟⎠ 2 ⇒ W = 160 mJ

9/20/2019 11:48:49 AM

Hints and Explanations H.189

ARCHIVE: JEE MAIN C=

kε 0 A d

V Since, E = d k × 8.86 × 10 −12 × 10 −4 × 106 500



⇒ 15 × 10 −12 =



k = 8.5 Hence, the correct answer is (C).

2.

Q = Q′







⇒ CV ( 1 + n ) = ( K + n ) CV ′



⇒ V′ =

V ( n + 1) K+n



⇒ V=

CV ( 1 + n ) V ( 1 + n ) = ( K + n )C (K + n)



Hence, the correct answer is (A).

5.

Equivalent capacitance for series combination is

–1 μ C 3 μC

3 μC

C′ =

CHAPTER 2

1.

C1C2 C1 + C2

For parallel combination, C ′′ = C1 + C2

Since, C ′′ > C ′ –1 μ C

Since V =

q 1 μC = C 1 μF

⇒ C1 + C2 =

and

⇒ V =1V



Hence, the correct answer is (B).

3.

Ui =

Q2 Q2 and U f = 2C1 2C2

Since W = U f − U i Q2 ⎛ 1 1 ⎞ − ⎟ ⎜ 2 ⎝ C2 C1 ⎠



⇒ W=



⇒ W=



⇒ W = 3.75 μ J Hence, the correct answer is (C).

4.

Initially, we have Ceq = C + nC = ( 1 + n ) C

( 5 )2 ⎡ 1

1⎤ − 2 ⎢⎣ 2 5 ⎥⎦

⇒ Q = CV = CV ( 1 + n ) …(1) Finally, Ceq ′ = KC + nC = ( K + n ) C



⇒ Q′ = C ′V ′ = ( K + n ) CV ′ …(2) CV

500 = 50 μF …(1) 10

C1C2 80 = = 8 μF C1 + C2 10

⇒ C1C2 = 400 μF …(2)



Solving equations (1) and (2), we get



C1 = 40 μF C2 = 10 μF Hence, the correct answer is (A).

6.

d ⎛ 1 1 1 1 ⎞ = + + C1 3 Aε 0 ⎜⎝ K1 K 2 K 3 ⎟⎠



⇒ C1 =



⇒ C2 =



⇒

E1 C1 3 K 1K 2 K 3 3 = = × E2 C2 ( K1K 2 + K 2 K 3 + K 3 K1 ) ( K1 + K 2 + K 3 )



⇒

E1 9 K 1K 2 K 3 = E2 ( K1 + K 2 + K 3 ) ( K1K 2 + K 2 K 3 + K 3 K1 )



Hence, the correct answer is (C).

7.

3 Aε 0 ( K1K 2 K 3 )

d ( K 1K 2 + K 2 K 3 + K 3 K 1 ) Aε 0 ( K1 + K 2 + K 3 ) 3d

4 μF

6 μF

nCV 3 μF

Since Q = constant because system is disconnected from battery, so

02_Ch 2_Hints and Explanation_P2.indd 189

10 V

9/20/2019 11:49:09 AM

H.190  JEE Advanced Physics: Electrostatics and Current Electricity ⎛ 4×6⎞ Q = Q( 4 μF ) = ⎜ × 10 μC ⎝ 4 + 6 ⎟⎠

1 × 12 × 10 −12 × 100 2

11.

U in =

⇒ U in = 600 pJ







⇒ Q = 24 μC





Hence, the correct answer is (D).

Since Q = CV

8.

Since Ceq =

K1 K2 where K 2 = K and K1 = 1 ( K1 − K 2 ) d

ε 0 K1K 2 a 2 ln

d

⇒ Q = 120 pC



U final =



⇒ U final = 92.3 pJ



K1 K2

⇒ Wcap = U initial − U final ≈ 508 pJ



Hence, the correct answer is (C).

12. Let the charges be Q1 and Q2 , then

a

Q1 Q2 = 6 4



⎛ 1⎞ ε 0 Ka 2 ln ⎜ ⎟ ⎝ K⎠ ⇒ Ceq = ( d 1− K)



⇒ Ceq =



Hence, the correct answer is (A).

9.

Ceq =

Also, Q1 + Q2 = 30

ε 0 Ka 2 ln K d( K − 1)

⇒ Q1 = 18 μC , Q2 = 12 μC



Hence, the correct answer is (A).

13. Since for combination in OPTION (A), we have

C1C2 CC + 3 4 C1 + C2 C3 + C4

where C1 = K1C0 and C0 =

1 120 × 120 × 10 −24 × 2 × 2 12 × 10 −12 × 13

A 2 d 2

ε0

Similarly



⎛ 2⎞ 6×⎜ ⎟ ⎝ 4⎠ Ceq = 2 6+ 4



6 = μF 1 13 6+ 2 Hence, the correct answer is (A). ⇒ Ceq =



3



C 2 = K 2 C0







C 3 = K 3 C0

14. Heat produced is given by





C 4 = K 4 C0







⇒ ΔH =



A K3K4 ⎞ ε0 2 ⎛ ε 0 A ⎞ ⎛ K 1K 2 ⇒ K eq ⎜ = + ⎝ d ⎟⎠ ⎜⎝ K1 + K 2 K 3 + K 4 ⎟⎠ d 2

1 C × 3C ( 2 ) × V 2 C + 3C



⇒ ΔH =

1 3 3 × CV 2 = CV 2 2 4 8



3 Q2 8 C Hence, the correct answer is (A).



K 1K 2 K3K4 + K1 + K 2 K 3 + K 4



⇒ K eq =



*No given option is correct.

10.

C = C1 + C2 + C3

ε 0 KA ε 0 K1A ε 0 K 2 A ε 0 K 3 A = + + d 3d 3d 3d







K + K 2 + K 3 10 + 12 + 14 ⇒ K= 1 = = 12 3 3



Hence, the correct answer is (C).

02_Ch 2_Hints and Explanation_P2.indd 190

ΔH = ΔU

⇒ ΔH =

15.

C

2

2

A C

2

A

7/3 B

4 B

9/20/2019 11:49:28 AM

Hints and Explanations H.191 1 V2 ( ε0 A ) 2 2 d

Since, F =

2F ε0 A



⇒ V=d



⇒ V = 1.5 × 10 −2



⇒ 14C = 3C + 7



⇒ C=



Hence, the correct answer is (A).

16.

Ι=



⇒ V ≈ 250 V



⇒ I=0



Hence, the correct answer is (C).

Hence, the correct answer is (B).

20. Initially, potential on C1 is V0 = 60 V

7 μF 11

⇒

dq = Slope of q − t graph. dt

1⎞ ⎛ 17. Induced charge on dielectric, Qind = Q ⎜ 1 − ⎟ ⎝ K⎠ Final charge on capacitor, Q = K ( C0V )

2 3⎞ ⎛ ⇒ Qind = 3 ⎜ 1 − ⎟ = 3 × = 1.2 nC ⎝ 5⎠ 5



Hence, the correct answer is (A). 6 μF A

D

D

E

E

5 μF

2 μF

C1

6 μF

2 μF

D

E



4 μF B 2 μF

5 μF



q1 q2 = C C1 2C 3 C2 + C3

Since, C1 = 1 μF , C2 = 3 μF , C3 = 6 μF q2 2



⇒ q1 =



⇒ q2 = 2q1 …(1)

From (1) and (2), we get q1 = 20 μC and q2 = 40 μC

12 = 2.4 μF 5



⇒ Ceq =



Hence, the correct answer is (A).



2

19. Given A = 200 cm = 2 × 10

02_Ch 2_Hints and Explanation_P2.indd 191

⇒

Also, q1 + q2 = 60 …(2)

1 1 1 1 5 = + + = Ceq 6 12 6 12

d = 1.5 cm = 1.5 × 10

q1

Charge starts flowing from C1 till the potential difference across C1 is equal to potential difference across series combination of C2 and C3 .

4 μF

E

5 μF

q2

C3

B

A

Finally, circuit can be redrawn as shown.

Y

E

2 μF

⇒ q0 = C1V0 = 1( μ F ) ( 60 V ) = 60 μC

C2

E

5 μF

25 8.85

X

5 ⇒ Q = × 90 × 10 −12 × 20 = 3 × 10 −9 C = 3 nC 3

18.



V = 1.5 × 10 2

2 × 25 × 10 −6 8.85 × 10 −12 × 2 × 10 −2

CHAPTER 2



7 C 1 ⇒ 3 = 7 2 C+ 3

−2

−2

m

2

m , F = 25 × 10 −6 N , V = ?

Hence, the correct answer is (C).

21. F  ollowing arrangement will meet the required condition. Eight capacitors of 1 μF in parallel with four such branches in series.

9/20/2019 11:49:45 AM

H.192  JEE Advanced Physics: Electrostatics and Current Electricity 1 2

1 2

1 2

1 2

3

3

3

3

8

8

8

8

Cafter

⎛ kε 0 A ⎞ ⎛ ε 0 A ⎞ ε 0 A ⎜⎝ 3 ⎟⎠ ⎜⎝ 2.4 ⎟⎠ From (1) and (2), = ε A ε A 3 k 0 + 0 3 2.4

1000 V 8 μF 8 μF 8 μF 8 μF 250 V 250 V 250 V 250 V 1000 V



Hence, the correct answer is (D).

22. T  he energy stored in the electric field produced by a metal sphere is 4.5 J Q2 = 4.5 2C



⇒



⇒ C=

⎛ kε 0 A ⎞ ⎛ ε 0 A ⎞ ⎟ ⎜⎝ ⎟⎜ 3 ⎠ ⎝ 2.4 ⎠ …(2) = kε 0 A ε 0 A + 3 2.4

Q2 …(1) 2 × 4.5

Capacitance of spherical conductor is C = 4πε 0 R Q2  2 × 4.5



⇒ C = 4πε 0 R =



( 4 × 10 −6 ) 1 16 = 9 × 109 × × 10 −12 ⇒ R= × 4πε 0 2 × 4.5 9



⇒ R = 16 × 10 −3 m = 16 mm



Hence, the correct answer is (C).

{From (1)}



⇒ 3 k = 2.4 k + 3



⇒ 0.6 k = 3



⇒ k=



⇒ k=5



Hence, the correct answer is (B).

3 0.6

24. 3 μ F and 9 μF are in parallel combination, so the equivalent capacitance is Ceq = ( 3 + 9 ) = 12 μF Now, 4 μF and 12 μF are in series, so their equivalent capacitance is Ceq ′ =

4 × 12 = 3 μF 16

Charge on 3 μF is q1 = ( 3 μF ) × ( 8 V ) = 24 μC

2

3 μF

4 μF

9 μF

2 μF

8V

23. A

C3

C D

C1

E

C2

4 μF

B

12 μ F

3 μF

2 μF

2 μF

8V

8V

C1 C2

A

B

C3

As the capacitors are in parallel combination so they have equal potential differences. C before =

ε0 A …(1) 3

02_Ch 2_Hints and Explanation_P2.indd 192

So charge on 4 μF and 12 μF are same (i.e. 24 μC ) as they are in series. ⎛ 9 ⎞ Charge on 9 μ F is q2 = ⎜ × 24 μC = 18 μC ⎝ 9 + 3 ⎟⎠ So, required charge Q = Charge on 4 μ F + Charge on 9 μF

9/20/2019 11:49:59 AM

Hints and Explanations H.193 ⇒  Q = ( 24 + 18 ) μC = 42 μC and hence the required electric field is

E =



Q 1 × 4πε 0 r 2

⇒ E = ( 9 × 10

−6 ) ( 9 ) 42 × 10

( 30 )2

4 1 1 1 1 3 = + + = ⇒ C a = μF 3 Ca 4 4 4 4

For (B), Cb = 4 + 4 + 4 = 12 μF For (C), Cc =

(4 + 4)× 4 8 = μF (4 + 4)+ 4 3

A

C

1μF

8μF

6μF

2μF

⇒ C+



⇒



⇒ C=



Hence, the correct answer is (A).

27.

Q2 =

23 32 C= 9 9 32 μF 23

2Q 2Q = 2+1 3

Now, Q = Cnet E

4×4 + 4 = 6 μF For (D), Cd = 4+4 Hence, the correct answer is (D). 26.

32 32C = 9 9

= 420 NC −1

Hence, the correct answer is (C).

25. For (A) ,

Here C AB

32 9 = 1 μF = 32 C+ 9 C×

CHAPTER 2





⎛ 3C ⎞ ⇒ Q=⎜ E ⎝ 3 + C ⎟⎠



⇒ Q2 =

2 ⎛ 3CE ⎞ 2CE ⎜ ⎟= 3⎝ 3+C⎠ 3+C

4μF

Q2

12 μ F 2μF B

C

1μF

A 8μF

1 μF

4μF

4μF



Hence, the correct answer is (B).

4μF

28. For upper and lower links, Ceq =

B C

1μF



A 8 μF 3

8μF B 8 μF 9

C A

8μ F 3 A

C

02_Ch 2_Hints and Explanation_P2.indd 193

32 μF 9

B

6 μF 5

⇒ Qupper = Qlower = 12 μC 3 μF 2 μF +12 –12 a 12 –12 12

B

C

3 μF

12

–12

3 μF

b

–12

2 μF

2 μF 10 –10 a

3 μF 15 –15

15 –15 b 10 –10 3 μF 2 μF

10 V

10 V

On closing the switch, charge on 2 μF is 10 μC and that on 3 μF is 15 μC At a , we have qi = −12 + 12 = 0 and

9/20/2019 11:50:11 AM

H.194  JEE Advanced Physics: Electrostatics and Current Electricity q f = 15 − 10 = 5 μC



⇒ σ = K ε 0 ( 3 × 10 4 ) = 6 × 10 −7 Cm −2

So, charge flowing from b to a is 5 μC .



Hence, the correct answer is (A).



Hence, the correct answer is (A).

30. For this, charge must be same Q = C1V1 = C2V2

29. Inside the dielectric, E =

E0 σ = = 3 × 10 4 K Kε 0



⇒ 120C1 = 200C2



⇒ 3C1 = 5C2



Hence, the correct answer is (B).

2.

 otal charge of 80 μC will distribute between 2 μF T and 2 μF capacitors (in parallel) in direct ratio of capacitances. So

ARCHIVE: JEE advanced Single Correct Choice Type problems 1.

d/2

C



C′ E1

E2

+

A/2 80 μ C A/2

E1

q3 3 = q2 2

– q3, 3 μ F

2 μ F, q2

C″ d C

C′

+

– C″

ε A C1 = 0 , C = d

⎛ A⎞ 2ε 0 ⎜ ⎟ ⎝ 2 ⎠ 2ε 0 A = d d 2



⎛ 3 ⎞( ) ⇒ q3 = ⎜ 80 = 48 μC ⎝ 3 + 2 ⎟⎠



Hence, the correct answer is (C).

3.

qi = C1V = 2V = q

This charge remains constant after the switch is shifted from position 1 to position 2. So, U i =

⎛ A⎞ 4ε 0 ⎜ ⎟ ⎝ 2 ⎠ 4ε 0 A C ′ = = d d 2

q2 q2 1 q2 = = and 2 Ci 2 × 2 4

U f =

q2 q2 1 q2 = = 2 C f 2 × 10 20

⎛ A⎞ 2ε 0 ⎜ ⎟ ⎝ 2 ⎠ ε0 A and C ′′ = = d d

− ΔU = U i − U f =

Since, C2 =

CC ′ 4 ε0 A ε0 A + C ′′ = + C + C′ 3 d d

7 ε 0 A C2 7 = 3 d C1 3



⇒ C2 =



Hence, the correct answer is (D).

02_Ch 2_Hints and Explanation_P2.indd 194

So, energy dissipated is q2 5

⎛ q2 ⎞ This energy dissipated ⎜ = ⎟ is 80% of the initial ⎝ 5⎠ ⎛ q2 ⎞ stored energy ⎜ = ⎟ ⎝ 4⎠

Hence, the correct answer is (D).

9/20/2019 11:50:24 AM

Hints and Explanations H.195

Loss =



⇒



Hence, the correct answer is (C).

5.

 hen the switch is closed, the inner plates of the two W capacitors get connected whereas the outer plates still are not connected and hence the circuit is not complete.

1⎛ C⎞ 1 2 2 ⎜ ⎟ ( V1 − V2 ) = C ( V1 − V2 ) 2⎝ 2 ⎠ 4

Loss =



Hence, the correct answer is (A).

6.

k1 in series with half of k3 and hence equivalent dielectric constant is

k1k3 k1 + k3

k2 in series with half of k3 and hence equivalent  k2 k3 and then both in parallel dielectric constant is k2 + k3 to give k =

k1k3 k k + 2 3 k1 + k3 k2 + k3



q1 = ( 30 ) ( 2 ) = 60 pC



⇒ q2 = ( 20 )( 3 ) = 60 pC



⇒ q1 = q2 = q (say)

The situation is similar to the two capacitors in series which are first charged with a battery of emf 50 V and then disconnected. So, when S3 is closed, V1 = 30 V and V2 = 20 V Other way of looking at the thing is, when S1 and S3 both are closed, due to attraction with opposite charge, no flow of charge takes place through S3 . Therefore, potential difference across capacitor plates remains unchanged or V1 = 30 V and V2 = 20 V.

Hence, the correct answer is (D).

9.

∫ E ⋅ dA = ε





1

q

0



⇒ E ( 2π r ) =

1 ( λ ) ε0

λ 2πε 0 r



Hence, the correct answer is (D).



⇒ E=

7.

 lectric E field at point   E = EQ1 + EQ2



Hence, the correct answer is (C).

P within the plates is

10. Total Initial Charge

Q1 Q2 E = E1 − E2 = − 2 Aε o 2 Aε o

⇒ E=

Qi = ( 2C )( 2V ) − CV = 3CV Total Final Charge

Q1 − Q2 2 Aε o

Q f = 2CV ′ + CV ′ where V ′ is common potential. By Law of Conservation of Charge

E P +Q1

+Q2

1

E2

2

1

E1

2

So, potential difference between the plates is Q − Q2 Q1 − Q2 ⎛ Q − Q2 ⎞ = VA − VB = Ed = ⎜ 1 d= 1 ⎟ 2C ⎛ Aε o ⎞ ⎝ 2 Aε o ⎠ 2⎜ ⎟ ⎝ d ⎠

Hence, the correct answer is (D).

8.

Charges on the capacitors are q

Qi = Q f

⇒ 3CV = 3CV ′

⇒ V′ = V So, final energy of combination is U f =

1 3 ( C + 2C )V 2 = CV 2 2 2

Hence, the correct answer is (B).

11. 10 μ F 1 μ F

q

2 pF

3 pF

CHAPTER 2

1 ⎛ C1C2 ⎞ ( V1 − V2 )2 2 ⎜⎝ C1 + C2 ⎟⎠

4.

q = 60 pC

q = 60 pC

V1 = 30 V

V2 = 20 V

10 μ F 11



Hence, the correct answer is (A).

50 V

02_Ch 2_Hints and Explanation_P2.indd 195

9/20/2019 11:50:39 AM

H.196  JEE Advanced Physics: Electrostatics and Current Electricity

Multiple Correct Choice Type Problems 1.

C

C = C1 + C2

C1 =

kε 0 d

A 3

d



⇒ C=



⇒

2d

3d

B x

A V0

ε0 2A C2 = 3 d

O

d

2d

x

3d

OA  BC and ( Slope )OA > ( Slope )AB

( K + 2 ) ε0 A

So, E0→ d = E2 d→ 3 d and E0→ d > Ed→ 2 d

3d



C K+2 = C1 K

Hence, (B) and (C) are correct.

4.

V , where V is potential difference d between the plates.

Also, E1 = E2 =



Hence, (A) and (D) are correct.

2.

 fter pressing S1 charge on upper plate of C1 is A +2CV0 . After pressing S2 this charge equally distributes in two capacitors. Therefore charge an upper plates of both capacitors will be +CV0 .

Let C0 be the capacitance initially and C be the capacε A itance finally. Then C0 = 0 d Since, Q = C0V

When S2 is released and S3 is pressed, charge on upper

⇒

ε 0 AV d

Q=

Further E0 =

⇒

E=

V E and E = 0 d K

V Kd 1 C0V 2 2

plate of C1 remains unchanged ( = + CV0 ) but charge

Also, if U i is the initial energy, then U i =

on upper plate of C2 is according to new battery

After the introduction of slab if U f be the final energy, then

( = −CV0 ) .



V

E

Hence, (B) and (D) are correct.

3.

 he magnitude and direction of electric field at difT ferent points are shown in figure. The direction of the electric field remains the same. Hence, answer (B) is correct.  Similarly, electric lines always flow from higher to lower potential, therefore, electric potential increases continuously as we move from x = 0 to x = 3 d. qi

q – – – –

E0

– x=0

qi

+

–

+

–

+

–

+ x=d

E0/k

– x = 2d

1 C0V 2 2 K



⇒

Uf =



⇒

ΔU = U 2 − U1



⇒

ΔU =

1 ⎛ 1 ⎞ C0V 2 ⎜ − 1 ⎟ ⎝ ⎠ K 2

+



⇒

W = −ΔU

+



⇒

W=



Hence, (A), (C) and (D) are correct.

5.

Charging battery is removed. Therefore, q = constant.

+ + + x = 3d

Therefore, answer (C) is also correct. The variation of electric field ( E ) and potential ( V ) with x is shown.

02_Ch 2_Hints and Explanation_P2.indd 196

2

Since work done = Decrease in Potential Energy

q

E0

1 1 ⎛V⎞ 2 CVslab = ( KC0 ) ⎜ ⎟ ⎝ K⎠ 2 2

U f =

1 ε 0 AV 2 ⎛ 1⎞ ⎜ 1 − ⎟⎠ K 2 d ⎝

Distance between the plates is increased. Therefore, C decreases. Now, V =

q , q is constant and C is decreasing. C

9/20/2019 11:50:55 AM

Hints and Explanations H.197

U =



1q again q is constant and C is decreasing. 2C

Therefore U should increase.

Hence, (B) and (D) are correct.

6.

Since battery is still in connection. So,





⇒

⇒ Q = kQ0







⇒

Since k > 1

⇒ Q > Q0

1 Also U 0 = Q0V0 and 2



1 U = QV = kU 0  2

{∵Q = kQ0 and V = V0 }

Hence, (A) and (D) are correct.

Integer/Numerical Answer Type Questions 1.

dC =

∫ dC

1

1 = C

∫ Kε A ⎛⎜ 1 + m ⎞⎟

dx

⎝

0

N⎠

⎛ d⎞ x = mδ = m ⎜ ⎟ ⎝ N⎠ 1 = C

dx

∫ Kε A ⎛⎜ 1 + x ⎞⎟ ⎝

0

d 1 ⇒ = C Kε 0 A

x

d⎠

dx

∫d+x 0

1 d ln ( 2 ) = C Kε 0 A



⇒



⇒ C=



⇒ α =1

Hence U > U 0

1 = C

Distance of mth layer from plate 1 is

⇒ Q0 = C0V0 and

Q = kC0V0





V = V0

Since all capacitors connected in series, so

2

CHAPTER 2

Therefore, V should increase.

Kε 0 A d ln ( 2 )

Kε 0 A dx d

x

dx

1

2 x = mδ = m

02_Ch 2_Hints and Explanation_P2.indd 197

d N

9/20/2019 11:51:07 AM

Chapter 3: Electric Current and Circuits

Test Your Concepts-I (Based on Current Definition) 1.

1 e 2 mv 2 = , we get 4πε 0 r 2 r

(a) Using

q ⎞ ⎛q I1 ( t ) = − v ⎜ 1 + 2 ⎟ ⎝ a b ⎠

1 e2 = 2.19 × 106 ms −1 4πε 0 mr

v=  

q ⎞ ⎛q Hence, q ( t ) = − v ⎜ 1 + 2 ⎟ t , ⎝ a b ⎠

(b) The time for the electron to revolve around the proton once is:

  t=

2π r 2π ( 5.29 × 10 −11 m ) = = 1.52 × 10 −16 s . v 2.19 × 106 ms −1

The total charge flow in this time is 1.60 × 10 −19 C , so the current is I=   2.

1.60 × 10 −19 C = 1.05 × 10 −3 A = 1.05 mA 1.52 × 10 −16 s

q = 4t 3 + 5t + 6 2

1m ⎞ = 2 × 10 −4 m 2 ⎝ 100 cm ⎟⎠

⎛ A = ( 2 cm 2 ) ⎜ dq I (1 s ) = (a) dt

t =1 s

= ( 12t 2 + 5 ) t = 1 s = 17 A

I 17 A (b) J= = = 85 kAm −2 A 2 × 10 −4 m 2 3.

Finally, as soon as the sphere absorbs the two point charges q1 and q2 , the current will stop flowing through the earthing conductor, and we can write I3 ( t ) = 0 . Thus, ⎧ ⎛ q1 q2 ⎞ ⎪ − v ⎜⎝ a + b ⎟⎠ , ⎪ ⎪ q −v 2 , I ( t ) = ⎨ b ⎪ ⎪ 0, ⎪ ⎩ (a) I =

∫

1 s 240

∫

⇒ 120π t ⎞ dt ⎝ s ⎟⎠

( 100 A ) sin ⎛⎜

0

−100 C ⎡ ⎤ +100 C ⎛π⎞ q = cos ⎜ ⎟ − cos 0 ⎥ = = 0.265 C ⎝ 2⎠ 120π ⎢⎣ 120π ⎦ Let us write the condition of the zero potential of the sphere, and hence, of any point inside it (in particular, its centre) , at any instant of time t . We shall find out three times intervals:

(1) t
E by Ir , i.e., ( VB − VA ) − E =

r + r’ when a current is flowing in any battery opposite to its emf, then terminal voltage is given by V = E + Ir 3. Two batteries are connected in series. The effective emf in the circuit is therefore 2E because both push the charge in the same direction. Hence emf’s are added. Net resistance in the circuit is ( r1 + r2 + R )

Therefore, current in the circuit

I =

2E r + ( 1 r2 + R ) (E, r1)

(a) Here E = I ( R + r ) ,

so I =

( E ’− E ) r

A

E 12.6 V = = 2.48 A R + r ( 5 W + 0.08 W )

CHAPTER 3

12 V − ( 4 ) I 3 − ( 6 ) I 2 − 4 V = 0 8 = ( 4 ) I 3 + ( 6 ) I 2

(E, r2) B

C

I

Then, DV = IR = ( 2.48 A )( 5 W ) = 12.4 V R

The potential difference between the terminals of first battery is ( VA − VB ) , terminal potential difference is given by

(b) Let I1 and I 2 be the currents flowing through the battery and the headlights, respectively. Then, I1 = I 2 + 35 A , and E − I1r − I 2 r = 0

⇒ E = ( I 2 + 35 A ) ( 0.08 W ) + I 2 ( 5 W ) = 12.6 V giving I 2 = 1.93 A

Thus, DV2 = ( 1.93 A )( 5 W ) = 9.65 V 2. Two batteries AB and CD emf’s E and E ’( E ’ > E ) and internal resistances r and r ’ respectively are connected in series as shown in the figure. I

( VA − VB ) = E − Ir1 where E is the emf of the battery and r1 is its internal resistance. Substituting the value of I , we get VA − VB = E −

For ( VA − VB ) to be zero, we must have

R = ( r1 − r2 )

This gives meaningful result only if r1 > r2 . Otherwise, if r2 > r1 , then R = r2 − r1 will produce terminal voltage across second cell to be zero ( VBC = 0 ). 4.

E

A

03_Ch 3_Hints and Explanation_P1.indd 215

E′ r ′

r B

C

2Er1 ( R + r2 − r1 ) =E r1 + r2 + R ( R + r2 + r1 )

The voltage supplied by the charging plant is here constant which is equal to,

V = EΙ + I Ι r = ( 90 ) + ( 10 ) ( 2 ) D

V = 110 V Let If be the current at the end of charging

9/20/2019 11:49:48 AM

H.216  JEE Advanced Physics: Electrostatics and Current Electricity Then, V = E f + I f r

⇒ If =

V − Ef

′ = E + IR VAB ⇒ 11 = E + 2R …(2) Solving (1) and (2), we get, E = 10 V and r = 0.5 W

110 − 100 2

=



r ⇒ If = 5 A

5.

In the first case current will be maximum when, Total external resistance = total internal resistance



⇒ R=



nr mR …(1) ⇒ r= m n

nE ⇒ I= …(2) 2R In the second case,

I ’ =

mE mE = mr m2 R+ R+ 2 R n n

( mE ) n2

=

nE ⎛ 2mn ⎞ ⎛ 2mn ⎞ ⎜ ⎟ =⎜ ⎟I 2R ⎝ n 2 + m 2 ⎠ ⎝ m 2 + n 2 ⎠

⇒ I’=

6.

To get 4 V, two cells must be connected in series, and to get 4 A two cells must be connected in ­parallel. Following two arrangements of cells are possible, which can supply 4 A at 4 V. 2V

2A 2V 2V 4V

2V

U i −1 should be true for any cell. For k this reason the resistance of the last cell, of the last two, of the last three, etc., cells should also be R3 . Therefore, 1 1 1 = + R3 R2 R1 + R3



R2 =

R3 ( R1 + R3 )

2A

2A 2A

2A

2A 4A 2A 2A

k k −1

2

⇒ R1 : R2 : R3 = ( k − 1 ) : k : ( k − 1 )

1.

DV = I g rg = ( I − I g ) Rp



⇒ Rp =

Igr g

=

I g ( 60 W )

( I − Ig ) ( I − Ig )

0.5 mA 2A

2V

60 Ω R

2V

 Therefore, to I g = 0.5 mA:

Let A, B represent the terminals of the cell. A +

+ –

r

A +

+ –

r

I

B –

I

B –

⇒ 8.5 = E − 3 R …(1)

Rp = 2.

have

100 mA

I = 0.1 A = 100 mA

( 0.5 mA )( 60 W ) 99.5 mA

when

= 0.302 W

Consider the circuit diagram, realizing that I g = 1 mA. For the 25 mA scale, we have ( I − I g ) S = I gG G Ig I

I – Ig R 1

Common

03_Ch 3_Hints and Explanation_P1.indd 216

= R3

R1

G

VAB = E − IR

R1 = k −1 R3

⇒

2V

4V (B)

7.



Un −1 U R3 = n −1 R1 + R3 k

Test Your Concepts-VIII (Based on Galvanometer, Voltmeter and Ammeter)

(A)

4A

U n =

2V

2A

4A

The last cell is a voltage divider that reduces the potential of the nth point k times as compared with the ( n − 1 ) th point. Hence,

The relation U i =



R ( n2 + m2 )

8.

25 Ω R2

R3

100 mA 50 mA Terminal Terminal

25 mA Terminal

9/20/2019 11:50:05 AM

Hints and Explanations H.217 ⇒

( 24 mA ) ( R1 + R2 + R3 ) = ( 1 mA ) ( 25 W )

⎛ 25 ⎞ ⇒ R1 + R2 + R3 = ⎜ W …(1) ⎝ 24 ⎟⎠ For the 50 mA scale: ( 49 mA ) ( R1 + R2 ) = ( 1 mA ) ( 25 W + R3 )

⇒ R=





(b) From the expression for R , we get





⇒ 49 ( R1 + R2 ) = 25 W + R3 …(2) For the 100 mA scale:

( 99 mA ) R1 = ( 1 mA ) ( 25 W + R2 + R3 )

⇒ 99R1 = 25 W + R2 + R3 …(3) Solving (1), (2) and (3) simultaneously yields

R1 = 0.26 W , R2 = 0.261 W, R3 = 0.521 W 3.

DV = IR and for a voltmeter we have V = I g ( G + R )

(a) 20 V = ( 1 × 10 −3 A ) ( R1 + 60 W ) R1 = 1.994 × 10 4 W = 19.94 kW G

60 Ω

20 V 30 kΩ

Common 19.94 kΩ

50 kΩ

(c) 100 V = ( 1 × 10 −3 A ) ( R3 + R1 + 60 W ) R3 = 50 kW



For ammeter, we have I g r = ( 0.5 A − I g ) ( 0.22 W ) …(1) ⇒ I g ( r + 0.22 W ) = 0.11 V G r

Ig

0.22 Ω 0.5 A – Ig G

r

2500 Ω

ΔV = 2 V



For voltmeter, we have 2 V = I g ( r + 2500 W ) …(2)



Solve (1) and (2) simultaneously, we get

I g = 0.756 mA and r = 145 W (a)  Potential difference across R = ( I meas − I1 ) R ; potential difference across voltmeter = I1Rv = Vmeas and this is measured potential difference. The two branches are in parallel.

5.

  

( Imeas − I1 ) R = Vmeas

V ⎛ ⎞ ⇒   ⎜ I meas − meas ⎟ R = Vmeas Rv ⎠ ⎝

03_Ch 3_Hints and Explanation_P1.indd 217

Vmeas I meas



I I  meas Rv Vmeas

⇒

1 I meas  R Vmeas

⇒ R =

Vmeas I meas

(a) T he voltmeter and ammeter branches are in ­parallel arrangement. Hence,

   Vmeas = ( I − I meas ) RV = I meas ( RA + R ) Hence, RA + R =

Vmeas I meas

⎛V ⎞    R = ⎜ meas − RA ⎟ ⎝ I meas ⎠ Vmeas V  RA , R  meas I meas I meas



(b) For

7.

The given parameters are

IS = 1 mA = 1 × 10 −3 A

Ig 0.5 A

1 I meas 1 = − R Vmeas Rv

If Rv 

6.

(b) 50 V = ( 1 × 10 −3 A ) ( R2 + R1 + 60 W ) R2 = 30 kW

4.



100 V

50 V

Vmeas Vmeas ⎞ ⎛ ⎜⎝ I meas − R ⎟⎠ v



CHAPTER 3



G = 20 W I = I max = 50 × 10 −3 A I GG



Since, S =



⇒ S=



⇒ S = 0.408 W The equivalent resistance of the instrument is



( I − IG ) ( 1 × 10 −3 ) ( 20 )

50 × 10 −3 − 1 × 10 −3

1 1 1 1 1 = + = + Req G S 20 0.408

Req = 0.4 W Note that shunt resistance is so small in comparison to the galvanometer resistance that the equivalent resistance is very nearly equal to the shunt resistance. 8.

The maximum current through the galvanometer is V IG = G

9/20/2019 11:50:26 AM

H.218  JEE Advanced Physics: Electrostatics and Current Electricity With a multiplier resistor in series with galvanometer the potential difference across the entire branch is nV = IGG + IG R

⎛V⎞ ⎛V⎞ ⇒ nV = ⎜ ⎟ G + ⎜ ⎟ R ⎝ G⎠ ⎝ G⎠



⎛V⎞ ⇒ nV = V + ⎜ ⎟ R ⎝ G⎠



⇒ R = ( n − 1 )G

Usually n is referred to as multiplication factor, e.g., a galvanometer with a range 0 to 1 mV is to be converted to a voltmeter of range 0 to 1 V. 1V = 10 3 We have n = 1 mV The multiplier resistor R = ( 10 3 − 1 ) G = 999G . So you can guess how large the multiplier resistor is in comparison to the galvanometer resistance G. 9.

The currents in the ammeters are proportional to deflections produced, let α 1 and α 2 be the proportionality constants. I1 = α 1n1 , I 2 = α 2n2

The ammeters are in series, hence I1 = I 2 ⇒ α 1n1 = α 2n2 …(1) In the second arrangement the potential drops across resistors are equal as they are in parallel arrangement. Resistances of ammeters have been ignored assuming them to be ideal.

Thus we have I1′ R1 = I 2′ Rx

Also I1′ = α 1N1 and I 2′ = α 2 N 2

V0 − V , where V0 is the V0 ­voltage across the resistance R before the voltmeter is switched on and V the voltage after it is switched on. ⎛ RR0 ⎞ , According to Ohm’s Law, V0 = IR and V = I ⎜ ⎝ R + R0 ⎟⎠ where R0 is the resistance of the voltmeter. Hence, the error is R R R0 = ε = R R R0 + 1+ R0 11. The sought error is ε =

This is determined only by the ratio between the resistances of the section of the circuit and the voltmeter. When R0  R , the error may be neglected. 12. A reduction in the sensitivity n times means that the galvanometer carries a current I1 which is n times smaller than the current in the remaining circuit before the branching off, Hence the current I 2 through the ⎛ n − 1⎞ I in the remaining circuit. Hence, shunt is ⎜ ⎝ n ⎟⎠

r I1 1 = = R I2 n − 1

⇒ r=

R ≅ 204 W n−1

13. Let R1 = resistance of ammeter,

R2 = combined resistance of ammeter and voltmeter



V = Voltage across voltmeter and



VA = Voltage across ammeter

⇒ R1α 1N1 = Rxα 2 N 2 …(2)

A

From equations (1) and (2), eliminating α 1 and α 2 , we get

I

R1N1 Rx N 2 = n1 n2

⇒ Rx =

R1n2 N1 n1N 2

V 10. Before the ammeter is connected, I 0 = , and after it R V , where R0 is the resistance of is connected, I = R + R0 the ammeter. The error is given by ε =

I0 − I R0 1 = = R R I0 + R0 1+ R0

When R0  R , then the error may be neglected.

03_Ch 3_Hints and Explanation_P1.indd 218

V

6V



In the first case current in the circuit,

I =

6 …(1) R2

and voltage across voltmeter V = 6 − voltage across ammeter

⇒ V = 6 − VA



⇒ V = 6 − IR1



⇒ V =6−

6 R1 …(2) R2

9/20/2019 11:50:48 AM

Hints and Explanations H.219

6

12 = …(3) I ’ = R2 2 R2 and V ′ = 6 − ( I ’) R1

12R1 ⇒ V′ = 6 − …(4) R2 V Further, it is given that V ′ = 2 12R1 3R R 1 =3− 1 ⇒ 1 = R2 R2 R2 3



⇒ 6−



Substituting this value in equation (4), we have



⇒ RV = 400 W

Further,

100 RV = 80 W 100 + RV

Hence, reading of voltmeter = ( 0.04 )( 80 ) = 3.2 V

Had it been an ideal voltmeter ( RV → α ), its reading would had been, ⎛ 3.4 ⎞ ( 100 ) = 3.238 V ⎜ ⎝ 100 + 5 ⎟⎠ 16. (a) W  hen the switch is open, no current flows through the branch that contains R1 and R2 . Hence E = 250 V

⎛ 1⎞ V ′ = 6 − ( 12 ) ⎜ ⎟ ⎝ 3⎠

I

⇒ V′ = 2 V

14. The two conductors ace and bdf have different potentials, with the result that when the switch S is closed a current will flow from c to d, while the currents will flow from a to c, from e to c, from d to f and from d to b. a, e

b, f



a, e b, f A

A

Earlier

Later

So, the closing of switch leads to an increase in the current flowing through the ammeter. Please note that the sections ab and ef of the resistors are shorted. 15. Net resistance =

250 1 A = 10000 40

According to Ohm’s Law

  V1 =

1 ( 6000 ) = 150 V and 40

  V2 = ( 250 − 150 ) V = 100 V

(b) When the switch is closed the equivalent circuit is shown. E = 250 V

E 3.4 = W = 85 W I 0.04 E

r

A



⇒ RA + r +



⇒ 2+3+

V1

V2

R1

R2

R

Since we observe that V1 and R1 are in parallel and V2 and R2 are in parallel and both in series and also simultaneously we have

V Rv

R1 ( RV1 ) R2 ( RV2 ) = = 2.4 kW   R1 + RV1 R2 + RV2

RRV = 85 R + RV

100 RV = 85 100 + RV

03_Ch 3_Hints and Explanation_P1.indd 219

Rv2

Rv1

I=  

CHAPTER 3

In the second case reading of ammeter becomes two times, i.e., the total resistance becomes half while the resistance of ammeter remains unchanged. Hence,

Hence the potential of 250 V must be equally divided. So, reading of both the voltmeters is   V1 = V2 =

E = 125 V 2

9/20/2019 11:51:01 AM

H.220  JEE Advanced Physics: Electrostatics and Current Electricity

17. (a)

1000 Ω A

RRV 50 × 200 = = 40 W R + RV 50 + 200 4.3 = 0.1 A 1 + 40 + 2 and voltage across voltmeter = 40 I = 4 V

400 Ω

Current in the circuit, I =

( R + RA ) RV = 52 × 200 = 41.27 W (b) R + RA + RV 52 + 200

b

I

4.3 = 0.102 A 1 + 41.27 Reading of voltmeter = 41.27 I = 4.2 V

I=   18.

RV ⎛ Reading of ammeter = ⎜ ⎝ R + RA + RV

⎞( ) ⎟⎠ I = 0.08 A

Vab = ( 0.005 ) ( 20 − 10 −3 ) = 10 −3 ( 20 + R )

6V (b)



(

22.

Ammeter 10–3 Ω

(20 – 10–3) A



Reading of voltmeter

= Vab = I ′R = 5.53 × 10 −3

0.005 Ω

G2

G1

G

20 A a

800 Ω

a

10–3 A R

)( 285.71 ) = 1.58 V

106 Ω

Voltmeter

100 Ω

b

⇒ R = 79.995 W GS G2 = G+S G+S

19. DR = G − 20.

⎛ 2500 × r ⎞ 100 = ⎜ ×5 ⎝ 2500 + r ⎟⎠



⇒ r = 20.16 W

Variable D.C. voltage

23. The usage of voltmeters and ammeters is shown in the figure. A

400 Ω

V

V

21. Refer figure (a): Current through ammeter, 800 Ω

A

V V 10 Ω A

I

V

V

V A

6V (a)

I =

Test Your Concepts-IX (Based on Heating Effects and Power Consumption)

net emf 6 = = net resistance 400 + 800 + 10 4.96 × 10

−3

A = 4.96 mA

 Refer figure (b): Combined resistance of 1000 W ­voltmeter and 400 W resistance is, R =

1000 × 400 = 285.71 W 1000 + 400

6 = 5.53 × 10 −3 A ⇒ I= . + 800 ) 285 71 (

03_Ch 3_Hints and Explanation_P1.indd 220

1.

1 ⎛ 1 1⎞ =⎜ + ⎟ Rp ⎝ 3 1 ⎠



⇒ Rp = 0.75 W

Rs = ( 2 + 0.75 + 4 ) W = 6.75 W I battery =

DV 18 V = = 2.67 A RS 6.75 W

9/20/2019 11:51:11 AM

Hints and Explanations H.221 3.

2Ω

2Ω

1Ω

3Ω

0.75 Ω

18 V

Then, Rs = x + y =

Pp xy Ps = 2 and Rp = 2 x + y I I y

x 4Ω

4Ω

x

6.75 Ω

18 V

CHAPTER 3

18 V

Let the two resistances be x and y.

y 2 Since, P = I 2 R ⇒ P2 = ( 2.67 A ) ( 2 W )



⇒ P2 = 14.2 W in 2 W

2 P4 = ( 2.67 A ) ( 4 A ) = 28.4 W in 4 W

DV2 = ( 2.67 A ) ( 2 W ) = 5.33 V

DV4 = ( 2.67 A ) ( 4 W ) = 10.67 V DVp = 18 V − DV2 − DV4 = 2 V ( = DV3 = DV1 )

So, P3

( DV3 )2 = R3

P1 = 2.

=

( 2 V )2 3W

( DV1 ) = ( 2 V )

2

1W

R1

= 1.33 W in 3 W

From the first equation, y =



⎛ P ⎞ x⎜ 2 s ⎟ Pp ⎝ I −x⎠ = the second becomes ⎛ Ps ⎞ I 2 x+⎜ 2 ⎝ I − x ⎟⎠



Ps Pp ⎛P ⎞ ⇒ x 2 − ⎜ 2s ⎟ x + 4 = 0 ⎝I ⎠ I



Using the quadratic formula, x =

= 4 W in 1 W

A certain quantity of energy DEint = P ( time ) is required to raise the temperature of the water to 100°C. For the power delivered to the heaters we have P = I DV =

( DV )2

Then, y =  The

where DV is a constant. R Thus comparing coils 1 and 2, we have for the energy H=

2

( DV ) Dt R1

=

2

( DV ) 2Dt



(a)  When connected in parallel, the coils present equivalent resistance

( DV )2 Dt R1

=

( DV )2 Dtp 2R1 3

2 Dt 3

(b) For the series connection, Rs = R1 + R2 = R1 + 2R1 = 3R1 and

( DV )2 Dt R1

⇒ Dts = 3 Dt

03_Ch 3_Hints and Explanation_P1.indd 221

Ps2 2I

2I 2

Ps ∓ Ps2 − 4 Ps Pp Ps − x gives y = I2 2I 2 resistances

are

− 4 Ps Pp

Ps + Ps2 − 4 Ps Pp 2I 2

and

2

(a) DV1 = DV2

=

( DV )2 Dts 3 R1

R1

I1

R2

I2

I

1 1 2R = = 1 1 1 1 1 3 + + R1 R2 R1 2R1

⇒ Dtp =

two

Ps ± Ps2 − 4 Ps Pp

I1R1 = I 2 R2

⇒ R2 = 2R1 .

H= and

Ps − 4.

R2



Rp =

Ps − x and I2



⇒ I = I1 + I 2 = I1 + ⇒ I1 =

I1R1 R + R1 = I1 2 R2 R2

IR2 IR IR1 and I 2 = 1 1 = = I2 R1 + R2 R2 R1 + R2

(b)  The power delivered to the pair is 2 P = I12 R1 + I 22 R2 = I12 R1 + ( I − I1 ) R2. For minimum dP =0 power let us find I1 such that dI1 dP ⇒ = 2I1R1 + 2 ( I − I1 ) ( −1 ) R2 = 0 dI1

9/20/2019 11:51:22 AM

H.222  JEE Advanced Physics: Electrostatics and Current Electricity

⇒

I1R1 − IR2 + I1R2 = 0



⇒ I1 =



This is the same condition as that derived in part (a).

5.

For the resistors in series, from KVL we have

IR2 R1 + R2

V0 − I ( R1 + R2 ) = 0 I =

P ’ ( R1 + R2 ) R R = 2+ 1 + 2 >1 = P R1R2 R2 R1 2



 The ratio is greater than unity. Therefore, power absorbed in a parallel connection of two resistors is greater than in a series connection of the same resistors for a given voltage. 6.

V0 ( R1 + R2 )

⇒

(a) Power consumed in resistor 1 is

   P1 =

Req = R1 + R2

Equivalent series circuit



The power absorbed by equivalent resistor,



⇒ I=

Similarly power consumed in resistor 2 is

   P2 =

V0 V0 = Rs R1 + R2

V02

⎛ V0 ⎞ P = IV = ⎜ V0 = R1 + R2 ⎝ R1 + R2 ⎟⎠

V 2 π V 2D22 = R2 4 ρ

Since D1 = 2D2 ⇒

Rs = R1 + R2

V2 V2 V 2 A1 π ⎛ V 2D12 ⎞ = = ⎜ = 4 ⎝ ρ ⎟⎠ ρ R1 ⎛ ρ ⎞ ⎜⎝ A ⎟⎠ 1

P1 D12 4D22 = = 2 =4 P2 D22 D2

(b) Power consumed in resistor 1 is

   P1 = I 2 R1 = I 2

Power consumed in resistor 2 is

P2 = I 2 R2 =    R1R2 R1 + R2

V0

I′ Equivalent parallel circuit



R1R2 R1 + R2

I ’ =

7.

V0 V0 ( R1 + R2 ) = RP R1R2

⎛ R + R2 ⎞ ⎛ R + R2 ⎞ P ′ = I ′V = ⎜ V0 1 V0 = V02 ⎜ 1 R1R2 ⎟⎠ ⎝ ⎝ R1R2 ⎟⎠ The ratio of the power absorbed in parallel and series circuit is ⎛ ( R + R2 ) ⎞ V02 ⎜ 1 ⎝ R1R2 ⎟⎠ V02 ( R1 + R2 )

03_Ch 3_Hints and Explanation_P1.indd 222

=



⇒

R20 1 + 20 α 0 1 + 20 × 0.0055 = = R2000 1 + 2000 α 0 1 + 2000 × 0.0055 R20 = 9.25 × 10 −2 R2000

At 2000 °C : R2000 =

The power absorbed by equivalent resistor is,

P’ = P

The ratio of resistances at the two temperatures is given by



If I ’ be the current in the circuit , then

( R1 + R 2 )2 R1R2

4 I 2 ρ π D22

1 1 P1 D12 D2 If D1 = 2D2 , then = = 22 = 1 P2 4D2 4 D22

For a parallel connection the equivalent resistance is

Rp =

ρ 4 I 2 ρ = A1 π D12

2

V 2 ( 110 ) = 121 W = P 100

At 20 °C : R20 = 121 × 9.25 × 10 −2 = 11.2 W P 110 = = 9.82 A V 11.2



Current at 20 °C, I 20 =



Current at 2000 °C, I 2000 =

P 100 = = 0.91 A V 110

Thus we can see that at the instant the bulb is turned on and when temperature of filament is 20 °C , the current is very large ( 9.82 A ) as compared to current at 2000 °C ( I 2000 = 0.91 A )

9/20/2019 11:51:31 AM

Hints and Explanations H.223 (a) W  hen a current is passed through a wire, its temperature rises with time due to heat produced. Since the fuse wire is exposed to the surroundings it also loses heat mainly due to radiation, heat loss depends upon temperature and it increases as temperature rises. The temperature attains a steady value when the rate at which heat is produced in the wire is equal to the rate at which heat is lost due to radiation. Let the steady temperature be equal to the melting point of the wire and I be the current at the melting point. The rate at which heat is produced in the wire is given by

2

⎛ I ⎞3 ⇒ r2 = r1 ⎜ 2 ⎟ ⎝ I1 ⎠

Substituting numerical values r1 = 0.1 mm, I1 = 10 A and I 2 = 20 A, we have

9.

   P = I 2 R where R is the resistance of the wire given by ρ    R = 2 πr  where ρ = resistivity,  = length and r = radius of the wire. Thus I 2 ρ …(1) πr2  Now let h be the rate of loss of heat per unit surface area of the wire. The rate of loss of heat from the wire is    P’ = 2π r × h …(2) In the steady state, P = P ’ . Therefore equating equations (1) and (2) we have

The circuit has asymmetry about line XY , i.e., the current in left and right are not mirror images. Just imagine the central junction of wires in the form of two junctions connected by wire ef as shown. Then it follows from symmetry consideration that there is no current in the wire ef , thus we can remove it from the circuit. The resistors of the wires will be proportional to their lengths.

I 2 ρ = 2π rh πr2

⇒ I =

2π 2 r 3 h …(3) ρ

Thus the current at which the melting point is reached is independent of the length,  of the wire, it depends upon the values of r, ρ and h. If the two wires have the same value of r, they are made of the same material ( ρ is the same for both wires), the rate of loss of heat per unit surface area ( h ) depends upon the material of the wire and the melting point which is same for both the wires, both fuses will melt at the same value of current, inspite of different lengths. (b) From equation (3), we have I =

2 3

2π r h ρ

⇒

I 2 2π 2 h = constant = ρ r3

⇒

I12 I 22 = r13 r23

03_Ch 3_Hints and Explanation_P1.indd 223

X

c

   P =

  

2

2 20 ⎞ 3 = ( 0.1 ) ( 2 ) 3 ⎟ ⎝ 10 ⎠ ⇒ r2 = 0.159 mm ≈ 0.16 mm

⎛ r2 = ( 0.1 ) ⎜

f a

CHAPTER 3

8.

d e b

Y V

Rab = Rac = Rcd = Rbd = r r 2

Rae = Rbe = Rcf = Rdf =

The power dissipated in resistor ab is

Pab =

V2 r

By Applying Ohm’s Law to the upper part, we can determine current through cd , I cd =

(

V

2 + 3 )r

{∵ Vab = Vaeb = Vshaded = V }



The power dissipated in the conductor cd , c r r r2 2 a

r f

e

r

d r 2 r r 2 b

c I r

I1 r 2

r

I2

2 P2 = I cd r=

  

r b

a V

d r 2

V

V2

(

2 + 3) r 2

9/20/2019 11:51:41 AM

H.224  JEE Advanced Physics: Electrostatics and Current Electricity

Thus the required ratio is

2 P 1 = ( 2 + 3 ) = 11 + 6 2 P2

10. (a) T  he current in the circuit is related to the voltage across the capacitor as

P = 2Pc   

C

The battery supplies twice as much energy to the capacitor as is stored in the capacitor. The difference is the work done by the capacitor against the external agent which is changing the capacitance.

Vc

11. Net emf of the circuit = ( 10 − 4 ) V = 6 V Total resistance of the circuit = 3 W

dV I = C c dt

I

P = PB − Pr = IV − I 2 r = I ( V − Ir ) ,    which is twice as large as that used in charging the capacitor. So,

Battery r

10 V

S

I

V

As the capacitor charges up, VC rises toward its limiting value and its rate of change goes to zero. Therefore, in order to keep I constant, capacitance, C, should compensate for this drop in dVc dVC (observe that, as time t → ∞ , → 0 and dt dt we should have C∞ → ∞) . Ignoring the practical difficulties of maintaining a constant I, asymptotically we have a continuous charge flow in the circuit dq I = dt VC = V and The battery supplies a constant power since its emf is constant, PB = VI

(b)  The capacitor is being continuously charged. Once the voltage across the capacitor stabilizes at t → ∞ , we have V = Ir + VC , ∞

⇒ VC, ∞ = V − Ir

3Ω

⇒ Current in the circuit I =



(a) Power supplied by 10 V battery

  = EI = ( 10 ) ( 2 ) = 20 W (b) Power consumed by 4 V battery   = EI = ( 4 )( 2 ) = 8 W (c) Power consumed by 3 W resistance 2

  = I 2 R = ( 2 ) ( 3 ) = 12 W Here we can see that total power supplied by 10 V battery (i.e., 20 W) = power consumed by 4 V b ­ attery and 3 W resistance. Which proves that conservation of energy holds good in electric ­circuits also. 12. The 6 W and 3 W resistances are in parallel. So their combined resistance is, 1 1 1 1 = + = or R = 2 W R 6 3 2 The equivalent simple circuit can be drawn as shown. Current in the circuit,



The energy stored in the capacitor becomes

3Ω

dq dE 1 1 = Vc , ∞ ∞ = I ( V − Ir ) dt 2 dt 2 The rate of dissipation of energy in the resistor r is Pr = I 2 r,  Therefore the power supplied directly to the capacitor is

03_Ch 3_Hints and Explanation_P1.indd 224

2Ω V 5Ω

I

The rate of change of W with time is

   Pc =

net emf 6 = =2 A total resistance 3



1 W = q∞VC , ∞ 2 1 ⇒ W = q∞ ( V − Ir ) 2

4V

20 V

I =

net emf 20 =2A = total resistance 3 + 2 + 5

V = IR = ( 2 )( 2 ) = 4 V i.e., potential difference across 6 W and 3 W resistances are 4 V. Now,

9/20/2019 11:51:50 AM

Hints and Explanations H.225

H 6 W = H 3 W

14. Applying Kirchhoff’s Laws to the two loops we have, R

( 4 )2 V2 ( 2 ) = 16 J t= R 6 3

1

2

( 4 ) ( 2 ) 32 V = t= = J R 3 3 2

I1 + I2

and H 5 W = I Rt = ( 2 ) ( 5 ) ( 2 ) = 40 J 13. Suppose, we insert the battery with 2 W resistance. Then we can take 2 W as the internal resistance ( r ) of the battery and combined resistance of the other two as the external resistance ( R ) . The circuit in that case shown in figure.

R r

6Ω

10 − I1R − 6 + 3 I 2 = 0 …(1) 6 − 6 ( I1 + I 2 ) − 3 I 2 = 0 …(2)

Solving these two equations we get, 6 R+2 Power developed in R,



P = I12 R =

E2 Now power, P = R+r This power will be minimum where R + r is maximum and we can see that ( R + r ) will be maximum when the battery is inserted with 6 W resistance as shown in figure. I1

I I2

2Ω

4Ω

6Ω



36

( R + 2 )2

R …(3)

For power to be maximum,



dP =0 dR 2

⇒ ( R + 2 ) ( 36 ) − ( 36 R ) ( 2 ) ( R + 2 ) = 0 ⇒ R + 2 − 2R = 0 ⇒ R=2W For maximum power from equation (3), we have Pmax = 4.5 W 15. (a) At I = 0.4 A, voltage across the rod,

11 V

Rod

Net resistance in this case is

6 +

2 × 4 22 = W 2+ 4 3

⇒ I=

11 = 1.5 A 22 3



I2

2

I1 =

E



I1

3Ω

6V

2

2

10 V

CHAPTER 3

2 ( )2 ( ) ( ) H 3 W = I Rt = 2 3 2 = 24 J

This current will be distributed in 2 W and 4 W in the inverse ratio of their resistances. I1 4 = =2 I2 2



⇒



⎛ 2 ⎞( ) ⇒ I1 = ⎜ 1.5 = 1 A ⎝ 2 + 1 ⎟⎠

R 2

⎛ 0.4 ⎞ V = ⎜ =4V ⎝ 0.2 ⎟⎠ ⇒ VR = 6 − 4 = 2 V Now, IR = VR = 2 V 2 =5W 0.4 = EI = 6 I

⇒ R = Psupplied (b)

2

PR = I 2 R

2

⇒ Prod = 6 I − I 2 R

So, P6 = ( 1.5 ) 6 = 13.5 W P4 = ( 0.5 ) 4 = 1 W 2 P2 = ( 1 ) ( 2 ) = 2 W

03_Ch 3_Hints and Explanation_P1.indd 225

6V

Given that, 6 I − I 2 R = 2I 2 R ⇒ 6 − IR = 2IR

9/20/2019 11:52:01 AM

H.226  JEE Advanced Physics: Electrostatics and Current Electricity ⇒ IR = 2 …(1) Further, 6 = VR + Vrod = IR +

{

1

∵ I = 0.2 V 2 ⇒ I 2 = 0.04Vrod



I2 …(2) 0.04

}

Solving equations (1) and (2), we get

R = 5 W 16.

V = E − Ir



⇒ 2 = 2.6 − ( 1 )( r ) ⇒ r = 0.6 W

Power generated by battery = VI = 2 W  and power developed in it by electric forces = I 2 r = 0.6 W 17. (a) By the 4 V battery DU = ( DV ) I Dt = ( 4 V ) ( −0.462 A ) 120 s = −222 J By the 12 V battery ( 12 V )( 1.31 A ) 120 s = 1.88 kJ (b) By the 8 W resistor 2 I 2 RDt = ( 0.846 A ) ( 8 W ) 120 s = 687 J By the 5 W resistor 2 I 2 RDt = ( 0.462 A ) ( 5 W ) 120 s = 128 J By the 1 W resistor 2 I 2 RDt = ( 0.462 A ) ( 1 W ) 120 s = 25.6 J By the 3 W resistor 2 I 2 RDt = ( 1.31 A ) ( 3 W ) 120 s = 616 J By the 1 W resistor 2

I 2 RDt = ( 1.31 A ) ( 1 W ) 120 s = 205 J −222 J + 1.88 kJ = 1.66 kJ from chemical to (c)  electrical 687 J + 128 J + 25.6 J + 616 J + 205 J = 1.66 kJ from  electrical to internal

Test Your Concepts-X (Based on Wheatstone Bridge and Potentiometer) 1.

(a) T  he value of E is given by ( E1 = 1.02 V is the emf of the standard cell C1 ) 1.02 × 82.3 = 1.25 V 67.3 (b) The high resistance R keeps the current drawn from the standard cell within permissible limit and prevents a large current to flow through the galvanometer when far away from the balance point. (c) At null point no current flows through R , hence no effect on null point.

E=  



03_Ch 3_Hints and Explanation_P1.indd 226



(d) The null point depends on the terminal voltage of C and the emf of C1 only, hence no effect. (e)  The method would not work if the potential difference across AB due to the driver cell C became less than the emf of cell C1 , because there would be no null point on the wire.   (i)  In this case the potential difference across AB due to cell C of emf 2 V will be less than 1 V . Since the emf of the standard cell is greater than this value, there will be no null point on the wire. So, it will not work. (ii) Similarly, the method would not work if the emf of C were 1 V instead of 2 V . (f)  Suppose the emf E to be measured is, say, 1 mV = 1 × 10 −3 V, the potential difference across AB due to C is 2 V , the length AB of the wire is 100 cm and its resistance per centimetre is λ (say). Then the balance length will be

A

B

R′

G R



 1 × 10 −3 V = P.D. across AJ due to C = current in the main circuit × resistance of AJ 2 2 AJ λ AJ = V 100 λ 100 ⇒ AJ = 0.05 cm ⇒ 1 × 10 −3 =

Thus the null point will be very close to A and there is an extremely large percentage error in its measurement. To have a large balance length, the circuit shown in figure is modified by putting a suitable resistor R’ . The balance length will then be measurable and the percentage error will be much smaller. 2.

Let E be the emf of the cell C’ and r its internal resistance. Let  = AJ be the balance length when switch S is open. When a resistance R is introduced by closing the switch, a current begins to flow through the cell C’ and resistance R . The potential difference between the terminals of the cell falls and the balance length decreases to ′ = AJ ′ . The internal resistance of the cell is given by

E−V I where V is the terminal voltage of C’ and I is the current in the circuit involving C’ and R . Also V I = R

r =

9/20/2019 11:52:15 AM

Hints and Explanations H.227 ⎛E ⎞ ⇒ r = ⎜ − 1⎟ R ⎝V ⎠

But

E  = . Hence V ′

⎛  − ′ ⎞ ⎛ 76.3 − 60 ⎞ ⇒ r = R⎜ = 4⎜ ⎟⎠ = 1.1 W ⎝ ′ ⎟⎠ ⎝ 60

3.

The resistances of the two segments of the wire AD and DC are in the ratio of their lengths. If R0 is the resistance of Y in melting ice ( 0 °C ) , the balance condition of Wheatstone bridge gives X λ  = = R0 λ ( 100 −  ) 100 −  where λ is the resistance per centimetre of wire AC. Now,  = 40 cm and X = 4 W . Substituting these values, we get R0 = 6 W . Let Rt be the resistance of Y when heated to a temperature t = 100 °C . When it is connected in parallel with 100 W resistor as shown in figure, the net resistance becomes R ’ =

100 Rt Rt + 100

⇒ r = 0.3 R = ( 0.3 )( 5 ) = 1.5 W

5.

Emf of cell = (potential gradient) × (balancing length)



⇒ E1 =



G

VAB ×x L V ⇒ 2.4 = AB × 8 10 ⇒ VAB = 3 V

Consider the loop containing E and applying potential divider concept, we get VAB = E

⇒

3=4



⇒

r=

RAB RAB + r

( 1.6 )( 10 )  ∵ RAB = ( 1.6 Wm −1 ) ( 10 m ) ( 1.6 )( 10 ) + r

{

60 cm

C G

Since the null point remains unchanged, we have

X 40 = R’ 60 ⇒ R’ = 6 W

⇒ 6=

}

16 W 3

X



E R+r

(a) T  here are no positive and negative terminals on the galvanometer because only zero deflection is needed. (b)

Y

D 40 cm



∵I=

6. B

A

}

⎛ Er ⎞ ⇒ 1.3 ⎜ = 0.3E  ⎝ R + r ⎟⎠

Note that as there is no current through the cell and galvanometer, the battery E , the internal resistance r and the potentiometer wire AB are in series.

100 Ω

X=4Ω

{



100 Rt Rt + 100

J

A

B

12 C

D

(c) AJ = 60 cm ⇒ BJ = 40 cm



⇒ Rt = 6.38 W

 If no deflection is taking place. Then, the Wheatstone bridge is said to be balanced.



Temperature coefficient of resistance of the coil Y is

Hence,

α =

Rt − R0 6.38 − 6 = 6.3 × 10 −4 K −1 = 6 × 100 R0t

4.

E 0.52 13 = = = 1.3 V 0.4 10



⇒



⇒ 1.3 Ir = 0.3E

E = 1.3 E − Ir

03_Ch 3_Hints and Explanation_P1.indd 227

CHAPTER 3



⇒

X RBJ = 12 RAJ

X 40 2 = = 12 60 3

⇒ x = 8 W 7.

Slide wire bridge is most sensitive when the resistance of all the four arms of bridge is same. Hence, for the resistor B the value of X is the most accurate answer.

9/20/2019 11:52:27 AM

H.228  JEE Advanced Physics: Electrostatics and Current Electricity

8. Resistance of x metre length of wire, R0 x L Resistance R and R1 are in parallel, so their effective resistance, RR1 R′ = R + R1 R1 =

This is in series with length ( L − x ) of wire. Resistance of length ( L − x ) of wire is R0 (L − x) L Hence, total resistance is given by

R2 =

Reff

RR1 = R′ + R2 = + R2 R + R2

So, current in the main circuit is given by

V V0 I = 0 = Reff ⎛ RR1 ⎞ ⎜⎝ R + R + R2 ⎟⎠ 1

Potential difference across the length x is

RR1 V = IR′ = I R + R1

⇒ V=

V0 ⎛ RR1 ⎞ RR ⎛ ⎞ ⎜⎝ R + R1 ⎟⎠ 1 ⎜⎝ R + R + R2 ⎟⎠ 1



⎛ 75 ⎞ ⇒ r=⎜ − 1⎟ 6 ⎝ 60 ⎠



⎛5 ⎞ ⇒ r = ⎜ − 1⎟ 6 ⎝4 ⎠



⇒ r = 1.5 W

Test Your Concepts-XI (Based on RC Circuit)

⇒ V=



V0 RR1 V0 RR1 = ⇒ V= R ( R1 + R2 ) + R1R2 R0 R + R1R2



⎛R ⎞ V0 R ⎜ 0 x ⎟ ⎝ L ⎠ ⇒ V= ⎛ R0 ⎞ ⎛ R0 ( ⎞ x L − x )⎟ R0 R + ⎜ ⎝ L ⎟⎠ ⎜⎝ L ⎠ V0 Rx ⇒ V= x⎞ ⎛ RL + R0 x ⎜ 1 − ⎟ ⎝ L⎠

For R  R0 , we get V ≅

V0 x L

9.

⎛  key open ⎞ Since r = ⎜ − 1⎟ R ⎝  key closed ⎠



⎛ 60 ⎞ ⇒ r=⎜ − 1⎟ 4 ⎝ 40 ⎠



⇒ r=2W

03_Ch 3_Hints and Explanation_P1.indd 228

(

t

1.

∫

Q ( t ) = Idt = I 0τ 1 − e

−

t τ

)

0



(a) Q ( τ ) = I 0τ ( 1 − e −1 ) = ( 0.632 ) I 0τ

Q ( 10τ ) = I 0τ ( 1 − e −10 ) = ( 0.99995 ) I 0τ (b) Q ( ∞ ) = I 0τ ( 1 − e −∞ ) = I 0τ (c) 2.

(a)  The

resistance d ρ R= = A σA



of

the

dielectric

The capacitance of the capacitor is C =

block

is

κε 0 A d

d κε 0 A κε 0 = is a characteristic of σA d σ the material only.

 Then RC =

V0 RR1 RR1 + RR2 + R1R2





⎛  key open ⎞ 10. Since r = ⎜ − 1⎟ R ⎝  key closed ⎠

(b) R=

κε 0 ρκε 0 75 × 1016 ( 3.78 ) 8.85 × 10 −12 = = C σC 14 × 10 −9

R = 1.79 × 1015 W 3.

(a) I ( t ) =

(

t

)

t

− − dq d = Qe RC = − I 0 e RC dt dt

where, I 0 =

Q 5.1 × 10 −6 = = 1.96 A RC ( 1300 ) ( 2 × 10 −9 )

⎡ ⎤ −9 × 10 −6 ⇒ I ( t ) = − ( 1.96 A ) exp ⎢ ⎥ − 9 ⎣ ( 1 300 ) ( 2 × 10 ) ⎦ I ( t ) = −61.6 mA t RC

⎡ ⎤ −8 × 10 −6 = ( 5.1 μC ) exp ⎢ ⎥ − 9 ⎣ ( 1 300 ) ( 2 × 10 ) ⎦ = 0.235 μC (b) q ( t ) = Qe



−

(c)  The magnitude of the maximum current is I 0 = 1.96 A

9/20/2019 11:52:38 AM

Hints and Explanations H.229

5.

U=

(a) L  et us denote the potential at the left junction VL and at the right VR . After a long time, the capacitor is fully charged.

6.

The total resistance between points b and c is

R =

( 2 )( 3 )

= 1.2 kW 2+3 The total capacitance between points d and e is C = 2 μF + 3 μF = 5 μF The potential difference between point d and e in this series RC circuit at any time is 1000 t ⎤ t ⎤ ⎡ ⎡ − − DV = E ⎣⎢ 1 − e RC ⎦⎥ = ( 120 V ) ⎣⎢ 1 − e 6 ⎦⎥

C1 = 2 μ F

2 kΩ 1Ω

8Ω

4Ω

2Ω

b

10 V

CHAPTER 3

Q 1 2 C ( DV ) and DV = 2 C Q2 Therefore, U = and when the charge decreases to 2C half its original value, the stored energy is one-quarter 1 its original value and so, U f = U 0 . 4

4.

c

d

e

3 kΩ C = 3 μ F 2 a

S

f 120 V

   VL = 8 V because of voltage divider    I L =

10 V =2A 5W

   VL = 10 V − ( 2 A )( 1 W ) = 8 V 2W ⎞ ⎛ ( 10 V ) = 2 V Similarly, VR = ⎜ ⎝ 2 W + 8 W ⎟⎠ 10 V ⇒ I R = =1A 10 W

Therefore, the charge on each capacitor between points d and e is 1000 t ⎤ ⎡ − q1 = C1DV = ( 2 μF ) ( 120 V ) ⎣⎢ 1 − e 6 ⎦⎥



1000 t ⎤ ⎡ − ⇒ q1 = ( 240 μC ) ⎢⎣ 1 − e 6 ⎥⎦

1000 t ⎤ ⎡ − q2 = C2 ( DV ) = ( 3 μF ) ( 120 V ) ⎣⎢ 1 − e 6 ⎦⎥

VR = ( 10 V ) − ( 8 W ) ( 1 A ) = 2 V



1000 t ⎤ ⎡ − ⇒ q2 = ( 360 μC ) ⎣⎢ 1 − e 6 ⎦⎥

Therefore, DV = VL − VR = 8 − 2 = 6 V

7.

The current in a charging series RC circuit is given by



(b) Redraw the circuit

I = I 0 e

9Ω



−

t RC

when the current is halved, t



1 μF

− I0 = I 0 e RC 2 t



6Ω

R =

1 ⎛ 1 ⎞ ⎛1 ⎞ W⎟ + ⎜ W⎟ ⎜⎝ ⎠ ⎝6 ⎠ 9

= 3.6 W

RC = 3.6 × 10 −6 s t and e RC −

1 = 10

⇒ t = RClog e 10 = 8.29 μs

03_Ch 3_Hints and Explanation_P1.indd 229

− 1 = e RC 2 Taking the natural logarithms on both sides,

⇒

log e 1 − log e 2 = −

t RC

On solving for t , we get

t = RClog e 2 = τ log e 2 = 0.693τ 8.

The charge at time t in a discharging circuit is given by

q = Q0 e

−

t RC

= Q0 e

−

t τ

9/20/2019 11:52:48 AM

H.230  JEE Advanced Physics: Electrostatics and Current Electricity

When the charge is halved,



Q0 = Q0 2

t − e τ

t

− 1 =e τ 2



⇒



Taking the natural logarithms on both sides, we get t τ

−0.693 = −

⇒ t = 0.693τ

9.

The electric potential energy stored in the capacitor is 2t

U =

1 Q2 1 2 −τ = Q0 e 2 C 2C

Since U 0 =

Q02 is initial potential energy, we have 2C

U = U 0 e

−

2t τ

When the potential energy is half its initial value, we have

 Observation: The result is independent of R. Therefore, when a capacitor is charged by a battery with a constant emf, half the energy provided by the battery is stored in the capacitor and half goes into thermal energy, independent of resistance; the thermal energy includes the energy that goes into the internal resistance of the battery. While charging, the charge flowing through battery = CV0 Work done by battery = ( CV0 )V0 = CV02

Power dissipated in the resistor =



Energy stored in the capacitor =



− 1 =e τ 2

R =

d  σA

C= and

Kε o A d



∵ R=

∴ Time constant, τ C = CR =

10. (a) Rate at which energy is dissipated in the resistor is P ( t ) = I 2 R 2

⇒ P ( t ) =

V R



Substituting the values, we have 5 × 8.86 × 10 −12 = 5.98 s 7.4 × 10 −12 Charge at any time decrease exponentially as

Here qo = 8.85 × 10 −6 C (Charge at time t = 0 )

∞

∫

V02

∫Re

−

2t RC

dt

Therefore, discharging (leakage) current at time t will be given by ⎛ dq ⎞ q I = ⎜ − ⎟ = o e − t/τ C ⎝ dt ⎠ τ C

or current at t = 12 s is

0

2t

V 2C − RC ⇒ W = 0 e 2

)

∞

( 8.85 × 10 ) e −6



I=

5.98

0

V02C ( −0 + 1 ) 2 1 ⇒ W = CV02 2

03_Ch 3_Hints and Explanation_P1.indd 230

R

q = q0 e − t/τ C

2t − e RC

W = P ( t ) dt =

⇒ W =

C



(b) We find the total joule heat by integrating from t = 0 to t → ∞

(

}

τ 1 =

⎛V − t ⎞ ⇒ P ( t ) = ⎜ 0 e RC ⎟ R ⎝ R ⎠ 2

 σA

Kε o σ

2t τ

⇒ t = 0.34τ The potential energy decreases to half its value in half the time it takes the charge to decrease to half its value.



{

Taking natural logarithms of both sides, we get

−0.693 = −

1 CV02 2

11. The problem is basically of discharging of CR circuit, because between the plates of the capacitor, there is capacitor as well as resistance.

2t



1 CV02 2

−12/5.98

= 0.198 × 10 −6 A = 0.198 μA

12. In SITUATION 1, the capacitor and resistor form a series RC circuit. Charge at time t is

(

q = q0 1 − e

−

t CR

)

9/20/2019 11:52:58 AM

Hints and Explanations H.231

(

⇒ CV = CV0 1 − e

(

−

t CR

)

t − e CR

)





⇒

V = 1− V0



⇒

0.75 = 1− e 1.5



⇒ e



Taking natural logarithm on both sides, we have



(

−

t CR

−

t CR

)



⇒ t = ( 10 × 10



⇒ t = log e 2



⇒ t = 0.693 s

) ( 0.1 × 10 ) log e 2 6

8.846 × 10 −12 × 5 7.4 × 10 −12 ⇒ RC = 6 12



8.85 × 10 −6 − 6 ⇒ I= e 6 ⇒ I = 0.24 μA {where e = 2.718 and e 2 = 7.39 }

13. The charge on a capacitor at time t is

(

q = q0 1 − e

−

t RC

(

) −

t RC



⇒ CV = CV0 1 − e



So, the potential difference at time t is given by

(

V = V0 1 − e

t RC

V f

RC

= V0 − Ve …(4)

e

= V0 − V f …(5)

03_Ch 3_Hints and Explanation_P1.indd 231

T RC

…(6)

⎛ V0 − Ve ⎞ ⎜⎝ V0 − V f ⎟⎠

40 V

500 Ω

– +

+ –

I=

dq dt



Since time constant, τ = RC



⇒ τ = ( 500 ) ( 0.5 × 10 −6 ) = 250 μs



⇒

0.5 μ F

Position 2

1 106 = = 4000 s −1 τ 250

At t = 250 μs , we have q = 10 ( 1 − e −1 ) = 6.3 μC 40 14.7 O

t t = 250 μ S

I(mA)

–105

When the switch is shifted to position 2 at t = 250 μs , applying KVL, we get − R

dq q − − 40 = 0 dt C

1 Putting R = 500 W , C = 0.5 μF and = 4000 s −1 , we τ get q



( t+T ) RC

−

Position 1

From equation (3), −

=e

0.5 μ F

( t+T )

From equation (2),

t − e RC



t RC

−

0

t RC

20 V –

) …(1)

( ) …(2) ) …(3) = V (1 − e −

+

500 Ω

+

Let the potential difference Ve be attained at time t and V f at time t + T , we have Ve = V0 1 − e

RC

⇒ q = 10 ( 1 − e −4000t ) μC …(1)

)



−

( t+T )

q = ( 0.5 × 10 −6 ) ( 20 ) ( 1 − e −4000t )

RC =



−

14. When the switch is shifted to position 1, the capacitor gets charged and the instantaneous charge is

In SITUATION 2, the capacitor is connected in parallel with the battery. Initially it acts like a short circuit and will be charged instantaneously. Now,



V0 − Ve

=e

T = RC log e

t = CRlog e 2 −6

V0 − V f

Now taking logarithm of both sides of equation (6), we get

1 2

=

Dividing equation (5) by (4), we get

CHAPTER 3



∫

6.3 μ C

dq = −4000 q + 20

t

∫

dt

250 μ s

9/20/2019 11:53:09 AM

H.232  JEE Advanced Physics: Electrostatics and Current Electricity q + 20 = − ( 4000t − 1 ) 26.3



⇒ log e



⇒ q = 26.3 e −

( 4000 t − 1 )

−4000 t

(

)

− 20 = 26.3 e −4000t ( 2.718 ) − 20



⇒ q = 71.5e



Differentiating equations (1) and (2), we get

I ( in mA ) =

−4000 t dq ⎡ 40 e =⎢ dt ⎣⎢ −286 e −4000t

0 < t < 250 μs

The potential across capacitor will reach to 20 V (or half its steady state value) after a time, t = τ log e 2 = ( 0.693 ) ×

40 × 10 −4 s = 9.24 × 10 −4 s 3

Current through R2 at time t would be,

2 −t τ I0e 3 E 40 Here I 0 = = =3A R ⎛ 40 ⎞ ⎜⎝ ⎟⎠ 3 ⇒ I=

9.24 × 10



∫

−4

∫

⇒ H = I 2 R2 dt = 0

⇒ DU =

1 3 C [ 4E2 − E2 ] = CE2 2 2 3 1 ⇒ ( DH )2 = 2CE2 − CE2 = CE2 2 2



⇒ DH = CE2



17. Applying Loop Law to abcda, we get

C

E

( 4e −1.5×10 t ) ( 20 ) dt



⇒ H = 40 × 10 −3 J = 40 mJ

( DH ) = I 2 R dt =

∫(I e 0

0

2E and τ c = CR R

(b) DH = ( DH )1 + ( DH )2 Here ( DH )1 =

1 2 CE 2

−t τ c

)

R dt = 2 CE2

C

–



c

At t = t

Q0 2q − − IR = 0 C C

{

2q Q ⎛ dq ⎞  ⇒ ⎜ ⎟R= E+ 0 − ⎝ dt ⎠ C C q

2

+

I



∞

b

+ – q (q – Q0)

d

At t = 0

E +

( DH ) = 1 ( 2E )2 C = 2CE2   2  Alternate Method:

C0

R

E

3

16. (a) H  alf of the energy supplied by the battery is stored in the capacitor and rest half is lost as a heat. So,

∞

a

+ – Q0

−1 80 ( 1 − e −1.5 × 9.24 ×10 ) J 3 1.5 × 10

0

q ⎛ q − Q0 ⎞ −⎜ ⎟ =0 C0 ⎝ C ⎠

C0

R

⇒ H=

03_Ch 3_Hints and Explanation_P1.indd 232



Substituting C0 = C , we get



Here I 0 =

So, energy supplied by the battery = ( Dq ) ( 2E ) = 2CE2

0

∫

E



3 × 10 3 t − 2e 4

=

t

E

− IR + E −

I =



qi = EC

t > 250 μs

40 ⎛ 20 × 40 ⎞ τ = CR = ( 10 −4 ) ⎜ s= × 10 −4 s ⎝ 20 + 40 ⎟⎠ 3

3 × 10 4 t − 2e 40

qf = 2EC

− 20 …(2)

15. Time constant of the circuit,



Calculation of (DH)2: Charge transferred from the battery = Dq = q f − qi = EC 

∵I =

dq dt

}

t

dq 1 dt ⇒ = Q0 2q R − Q0 E + 0 C C

∫

∫

Solving this, we get

q =

2t ⎤ − 1⎡ ⎢ ( EC + Q0 ) − ( EC − Q0 ) e CR ⎦⎥ ⎣ 2

and q − Q0 =

2t ⎤ − 1⎡ ⎢⎣ ( EC − Q0 ) − ( EC − Q0 ) e CR ⎥⎦ 2 2t

So, I =

− dq 1 = ( EC − Q0 ) e CR dt CR 2t



⎛E Q ⎞ − ⇒ I = ⎜ − 0 ⎟ e CR ⎝ R CR ⎠

9/20/2019 11:53:19 AM

Hints and Explanations H.233 Substituting the given values, we have



2t

⎛ 4 2 × 10 −6 ⎞ − 2 ×10 −6 ×100 I = ⎜ − e ⎝ 100 2 × 10 −6 × 100 ⎟⎠

(

or I = 0.03 e −10

4

t

VCD

)A

In this case, though the capacitors are in series yet the charges on the capacitors are not equal (even in steady state). 18. From O to A: V = at , where a is a positive constant q ⇒ V = at = + IR C

A

t

⇒

t

dI

∫a− I =∫ 0

C

{

∵I =

dq dt

}

) ⇒ I = aC ( 1 − e

VCD

q

V = at

t CR



⎛ dq ⎞ ⎛ η⎞ ⇒ ⎜ ⎟R=E−⎜ ⎟q ⎝ dt ⎠ ⎝ C⎠

VCD also increases exponentially. i.e., From A onwards: When V = constant (say V0 ), then V0 = at



(

⇒ VCD = aCR 1 −

03_Ch 3_Hints and Explanation_P1.indd 233

)

d

q =0 Cη

t

dq 1 dt = η R q q0 E − 0 C

∫

∫



⇒



⇒ q=



⇒ I=−

ηt ⎤ − EC ⎡ ⎢⎣ 1 + ( η − 1 ) e CR ⎥⎦ η ηt

I

R

−V0 e aCR

I

Applying Loop Law to abcda, we get, E + IR −

− dq E = ( η − 1 ) e CR dt R

20. Charge on capacitor q = CV …(1)

V ⇒ t= 0 a

c

At t = t



i.e., current in the circuit increases exponentially and hence

) = IR = aCR ( 1 − e

⇒ E a

q

t CR

−

q

R

C/η

At t = 0

E

dt R −

b

q0

1 ⎛ dq ⎞ ⎛ dI ⎞ ⎜ ⎟ +⎜ ⎟R C ⎝ dt ⎠ ⎝ dt ⎠

0



C/η

R

I ⎛ dI ⎞ ⇒ ⎜ ⎟R= a−  ⎝ dt ⎠ C I



dq dt

Differentiating w.r.t. time, we get

a =



19. At t = 0 , charge stored in the capacitor is, q0 = EC

I = −

B

O



t

Since, the capacity of the capacitor has been decreased η fold, the charge stored in the capacitor will decrease. Hence, the current in the circuit will flow anticlockwise, with

V



After this VCD will decrease exponentially. Hence, a rough graph is as shown in figure.

CHAPTER 3





But from Gauss Theorem   ⇒ q = KE0 E ⋅ dA …(2)

∫ S

∫     ⇒ I = σ E ⋅ dA = σ E ⋅ dA …(3) ∫ ∫

Also, I =

  J ⋅ dA



  {∵ J = σ E where σ = conductivity}

9/20/2019 11:53:27 AM

H.234  JEE Advanced Physics: Electrostatics and Current Electricity

Now using (2) and (3), we get

I =

σq 1 ⎛ CV ⎞ =  KE0 ρ ⎜⎝ KE0 ⎟⎠

1⎫ ⎧ ⎨∵ σ = ⎬ ρ ⎩ ⎭

( 4 × 10 −9 ) ( 2 × 103 ) ( 1011 ) ( 6 ) ( 8.85 × 10 −12 )



⇒ I=



⇒ I = 1.5 × 10 −6 A = 1.5 μA

J =

I 2π r

Further E = ρ J =

Single Correct Choice Type Questions

ρI  2π r

{∵ J = σ E }

with both J and E directed radially outwards. Since dV = − E dr dV = −

ρI dr 2π  r

I = nqv A where v is the drift velocity of charge carriers each with charge q. Hence, the correct answer is (D).



⇒



⇒

h (Just for the sake of knowledge) 2π Charge Circulating I = Time for one revolution



⇒



⇒



Hence, the correct answer is (B).

6.

Let V be the voltage of the source and R be the resistance of each bulb. Then, we have

1.

2.

j=

=

⎛ e2 ⎞ e⎜ ⎟ ⎝ ⎠ e me 5 I= = = ⎛ 2π r ⎞ 2 2π  3 ⎜⎝ ⎟⎠ 2π 2 v me



⇒



Hence, the correct answer is (A).

3.

In parallel, current distributes in inverse ratio of resistance. 9 Ω 0.9 Ω 0.01 A (I – 0.01)A

I A



⇒

0.1 Ω

B

0.01 0.1 = I − 0.01 9 + 0.9

b

∫

b

dV = −

a

ρI dr 2π  r

∫ a

ρI ⎛ b⎞ log e ⎜ ⎟ ⎝ a⎠ 2π  2π V I= ⎛ b⎞ ρ log e ⎜ ⎟ ⎝ a⎠ −V = −

V2 R= when N bulbs are joined in series across V ,  P V current in each bulb is given by I = . Power drawn NR ⎛ V2 ⎞ V2 P by each bulb is I 2 R = ⎜ 2 2 ⎟ R = 2 = 2 ⎝N R ⎠ N R N

⎛ P ⎞ P Hence, total power drawn is Np = N ⎜ 2 ⎟ = ⎝N ⎠ N



Hence, the correct answer is (C).

7.

Whenever a resistance is joined in parallel with the voltmeter, the total resistance of the circuit decreases so that the current increases and hence ammeter reading also increases. The equivalent resistance across the voltmeter decreases and hence its reading will decrease.



Solving we get, I = 1 A



Hence, the correct answer is (C).

4.

Q = Q0 e − t τ =



⇒ e −t τ =



⇒ et τ = η



t ⇒ = log e η τ



Hence, the correct answer is (B).



Hence, the correct answer is (A).

5.

Let us assume that charge flow between the rod and cylinder be radial. Then at point P at distance r from centre of rod the current density J is

8.

Since the resistivity of a current carrying conductor carrying charge carriers each of charge q, mass m is given by

Q0 η

+ –

1 η

03_Ch 3_Hints and Explanation_P1.indd 234

V

A

9/20/2019 11:53:35 AM

Hints and Explanations H.235

m nq2τ

 where n is number density of charge carriers and τ is the average relaxation time. Hence, the correct answer is (B). 9.

Convectional current is the current which is developed due to the transportation of charge.



⇒



qtransported

= 2λ v t Hence, the correct answer is (B). I=

From (1) and (2), we get 7 1 W, r = W 3 3

R =

Hence, the correct answer is (C).

16. In steady state the branch containing the capacitor can be omitted and hence current in the circuit is 2V − V I= R + 2R

⇒

I=

V 3R V

10. This is basically a RC circuit, being charged from a battery. The resistance ( R ) of the voltmeter is the resistance in the circuit. The voltage across R is

I

V

6

V = ( Circuit Current )( Resistance ) = Reading of the voltmeter ( V ) Hence the nature of the V -t curve is the same as that of the I -t curve. Hence, the correct answer is (C).

R

1 C –

5 2V



I

+

2R

R1 + R2 = Constant

−VC − V + 2V − I ( 2R ) = 0



⇒ D ( R1 + R2 ) = 0



⇒

VC = −V + 2V −



⇒ DR1 = DR2



⇒ R1αDT = R2βDT



⇒

VC = V −



⇒



Hence, the correct answer is (C).

17.

G = 25 W



Hence, the correct answer is (D).

14.

R 4

A

R 4 R 4

RAB =

3R = 16

3 ( 16 )

B

⇒



Hence, the correct answer is (C).

15.

I=



⇒ R + 2r = 3 …(1)



=3W

2 ( 1.5 ) R + 2r

0.6 =

V ( 2R ) 3R

2V V = 3 3

⇒ I g = 0.01 A

Since



16

4

I g = ( 2 × 10 −4 ) ( 50 )

R 4

3

For loop 36543

13.

R1 β = R2 α

2 I

CHAPTER 3

ρ =

R =

V −G Ig 25 − 25 0.01



⇒ R=



⇒ R = 2500 − 25



⇒ R = 2475 W



Hence, the correct answer is (C).

18. On short circuiting

1.5 R + (r 2)

⇒ 2R + r = 5 …(2)

03_Ch 3_Hints and Explanation_P1.indd 235

E R 1.5 ⇒ 3 = R ⇒ R = 0.5 W Hence, the correct answer is (A).

I =

9/20/2019 11:53:45 AM

H.236  JEE Advanced Physics: Electrostatics and Current Electricity 19. When K is open Rnet =

⇒ i1 =

22.

3R 2

3

A

E 2E = ⎛ 3R ⎞ 3R ⎜⎝ ⎟ 2 ⎠

8

4

2

6 6 3

When K is closed 1.6

⎡ R × 2R ⎤ 4 Rnet = 2 ⎢ = R ⎣ R + 2R ⎥⎦ 3

⇒ i2 =



Hence, the correct answer is (C).

7×3 4×6 + 7+3 4+6



⇒

Req =



⇒

Req = 2.1 + 2.4



⇒

Req = 4.5 W



Hence, the correct answer is (A).

21. Let E be the e.m.f. and r be the internal resistance of battery. Then



⇒

E 20 + r

E = 10 + 0.5 r …(1)

E 0.8 = and 10 + r

⇒



From (1) and (2)

E = 8 + 0.8 r …(2)

2 = 0.3 r

⇒

r=

E =

B 6 3 4 A

20. 7 Ω and 3 Ω are in parallel; 6 Ω and 4 Ω are in parallel and both in series. So

0.5 =

2.4

A

3E E = ⎛ 4R ⎞ 4R ⎜⎝ ⎟ 3 ⎠

i 8 ⇒ 1 = i2 9

20 W and 3 40 V 3

Hence, the correct answer is (D).

03_Ch 3_Hints and Explanation_P1.indd 236

B

B 6



3, 4 and 6 W all in parallel. Hence



1 1 1 1 = + + RP 3 4 6



⇒

1 4+3+2 = RP 12



⇒

RP =



Hence, the correct answer is (A).

24.

0.2 =



⇒

1 + 0.2 r = 1.5



⇒

r=



Hence, the correct answer is (C).

12 4 W= W 9 3

1.5 5+r 5 = 2.5 W 2

25. In parallel, the potential across all the resistances is the same. Hence, the correct answer is (D). 26. Since I =

⇒ I=

E E = Req R + 10

2 10 + R

100 cm has a resistance of 10 W

10 ⇒ 40 cm has a resistance of × 40 100 ⇒ R′ = 4 W

9/20/2019 11:53:53 AM

Hints and Explanations H.237 So, potential difference across 4 W is

V ′ = IR′ = 10 mV

⎛ 2 ⎞ ⇒ V′ = ⎜ 4 = 10 −2 ⎝ 10 + R ⎟⎠



⇒ 800 = 10 + R



⇒ R = 790 W



Hence, the correct answer is (B).

27. If A is fused, then complete circuit is broken. Hence, the correct answer is (C). R1R2 =4 R1 + R2

28.

R1 + R2 = 18 and



Hence, the correct answer is (C).

I total =

V = 1.5 A Rnet



⇒



Voltage drop across 2.8 W is V1 = ( 2.8 )( 1.5 )



⇒

V1 = 4.2 V

So, potential across parallel combination of 2 W and 3 W is V2 = 6 − 4.2 = 1.8 V If I 2 is the current in 2 W resistor, then I 2 =

1.8 = 0.9 A 2

Hence, the correct answer is (B).

31. Let E = emf of the cell E 4E ×4W= 4 W+6 W 10

VA − VB =

29. If a wire is stretched to n times its original length, then new resistance is n2 times original resistance. Stretching to make the wire 0.1% longer implies its new length is

A 6

4 + –

D

0.1 1001 0 + 0 = 0 100 1000

B

4

2

6 C

⎛ 1001 ⎞ Rnew = ⎜ R So, ⎝ 1000 ⎟⎠ 0 ⇒ %age increment is

VA − VD =

E 6E ×6 W= 6 W+4 W 10

⎡ ⎛ 1001 ⎞ 2 ⎤ ⎢ ⎜ ⎟⎠ − 1 ⎥ × 100% ⎝ ⎣ 1000 ⎦

VB − VD =

E 5



2 ⎤ ⎡⎛ 1 ⎞ = ⎢ ⎜ 1 + − 1 ⎥ × 100% ⎟ 1000 ⎠ ⎣⎝ ⎦

2 ⎡ ⎤ − 1 ⎥ × 100 = 0.2% = ⎢ 1 + 1000 ⎣ ⎦



Hence, the correct answer is (C). NE E = Nr r The equivalent circuit of one cell is shown in the figure.

32. Current in the circuit is I =

(Since according to Binomial Theorem

A

( 1 + x )  1 + nx For x T2

1 μF



Hence, the correct answer is (A).

10 V

95. The circuit is a balanced Wheatstone Bridge with net resistance

VAB ( C )BC 3 1 = = = VBC ( C )AB 6 2

Rtotal =



⎛ 1 ⎞( ) ⇒ VAB = ⎜ 10 V ⎝ 1 + 2 ⎟⎠



⇒ VAB =



Hence, the correct answer is (D).

89.

B 7A

2A

C

3A

5Ω 2A

R

100 V

Applying loop equation in closed loop we have,

+100 − 30 − 35 − 2R = 0

⇒ 2R = 35 V = VR

V5 W = 7 × 5 = 35 V

V5 W =1 VR Hence, the correct answer is (D). ⇒

90. At t = 0 when capacitors are initially uncharged, their equivalent resistance is zero. Hence, whole current passes through these capacitors. Hence, the correct answer is (D). 91. Since the balancing length is at the mid point, each wire has a resistance equal to the known resistance value R . When they are in series, if  is the balancing length measured from the left, we have resistance R in the left gap and 2R in the right gap. R  = Thus 2R 100 − 

( 3 R )( 6 R ) 3R + 6R

⇒

 = 33.3 cm Hence, the correct answer is (B).

For maximum power



⇒ 4 = 2R ⇒ R=2W Hence, the correct answer is (B).

96.

E1 + E2 204 51 = = E1 − E2 36 9



⇒ 9 E1 + 9 E2 = 51 E1 − 51 E2



⇒ 42 E1 = 60 E2



⇒ E1 =



60 ( 1.4 ) 42 ⇒ E1 = 2 V



Hence, the correct answer is (D).

97. According to Kirchhoff’s Junction Law ICO + I BO + I AO = IOD

⇒

VC − VO VB − VO VA − VO VO − VD + + = R R R R



⇒

⎛ V − VO ⎞ VO − VA + VA − VD 3⎜ A ⎟⎠ = ⎝ R R



⇒

⎛ V − VO ⎞ ⎛ VA − VO ⎞ ⎛ VA − VD ⎞ 3⎜ A ⎟⎠ = − ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ ⎝ R R R



⇒

⎛ V − VO ⎞ VA − VD 4⎜ A ⎟⎠ = ⎝ R R



⇒

4 ( VA − VO ) = 40



⇒

VA − VO = 10 V



Hence, the correct answer is (A).

98. e.m.f. is the value when no current is withdrawn from the circuit. So E = 2 V

93.

1 1 1 1 = + + RAB 2R 2R R

Also r = slope =



Hence, the correct answer is (B).



03_Ch 3_Hints and Explanation_P1.indd 244

= 2R

Rinternal = Rtotal external

10 V 3

5A



V = Resistance I So, R1 > R2 94. Slope =

B

2 = 0.4 W 5

Hence, the correct answer is (B).

9/20/2019 11:55:05 AM

Hints and Explanations H.245

R

1I 2

R (I1 – I2) 4 I2 R R (I – I1 – I2) 5

3 I1

R

(I – I1) 7



R

6

R

I

8

− I1R − I 2 R + 2 ( I − I1 ) R = 0

⇒   3 I1 + I 2 = 2I …(1)



Loop 34563

−2 ( I1 − I 2 ) R + ( I − I1 + I 2 ) R + I 2 R = 0

⇒ Rmin =



⇒



Hence, the correct answer is (C).

⇒   −3 I1 + 4 I 2 = − I …(2) Add (1) and (2), we get



⇒   I 2 =

I 5





−2R ( I − I1 ) − R ( I − I1 + I 2 ) + E = 0

3I I ⎞ 3I ⎞ ⎛ ⎛ ⇒   E = 2R ⎜ I − ⎟ + R ⎜ I − + ⎟ ⎝ ⎝ 5⎠ 5 5⎠



⇒   E =

4 IR 3 IR + 5 5



⇒   E =

7 IR 5



Finally, we have

E = I total Rnet

⇒   IRnet =



⇒   Rnet =

7 IR 5

7 R=7 W 5



Hence, the correct answer is (B).

101.

Rmax =



⇒ Rmax =



⇒  0 = Minimum Length

ρ max Amin

Rmin

15 × 10 60 5 ⇒ R = = 2.5 W 2 Hence, the correct answer is (B). ⇒ R=

20 2 times or times. Therefore, 30 3 3 times. net resistance should become 2 3 ⇒ R + 50 = ( 2950 + 50 ) 2 Solving we get, R = 4450 W Hence, the correct answer is (A).

104.

2R

2R

r

r

2R

2R

P



 20

ρ = min Amax

Q

Hence, the correct answer is (A).

105. Out of 5 one is wrongly connected i.e. 4 are rightly connected Net EMF = 4E − E = 3E

Net resistance = 5r Hence, the correct answer is (B).

106.

I = I1 + I 2 …(1)



r=

r1r2 …(2) r1 + r2

Further I =

ρ ( 3 0 )

03_Ch 3_Hints and Explanation_P1.indd 245

⎛ 75 ⎞ ⇒ r=⎜ − 1 ⎟ × 10 ⎝ 60 ⎠

103. Current decreases



3I and  I1 = 5 For 127658EI

Rmax =9 Rmin

⎛  ⎞ r = ⎜ − 1 ⎟ × R ⎝ ′ ⎠

5I 2 = I

3 20

102. Since



Loop 23672

ρ ( 0 )



CHAPTER 3

99. L  et a fictitious battery E be connected across the circuit.

E1 E2 + r1 r2



⇒

E E1 E2 = + r r1 r2



⇒

E=



Hence, the correct answer is (C).

E1r2 + E2 r1 r1 + r2

9/20/2019 11:55:16 AM

H.246  JEE Advanced Physics: Electrostatics and Current Electricity 107. The circuit reduces to the following

(a ­resistance) is connected in parallel across a galvanometer. Hence voltage drop across shunt = (30)(12)

C



G E

A

B

F H

Hence, the correct answer is (D).

108. Resistance between diagonal corners of cube 6I 1

2I 2I

2

3

I

2I

2I

I

5

6 8

2I

2I

6I 7



⇒

V17 = ( V1 − V5 ) + ( V5 − V8 ) + ( V8 − V7 )



⇒

V17 = 2IR0 + IR0 + 2IR0



⇒

V17 = 5IR0

Also, V17 = x ( 6 I )

⇒

x ( 6 I ) = 5IR0



⇒

x=



Hence, the correct answer is (D).

5R0 6

5R 6 R + =R 11 11 V Since I = Rtotal 109.

Rtotal =

12 R R = 6W



⇒

2=



⇒



Hence, the correct answer is (C).

110. In a galvanometer deflection is directly proportional to the current. So 20 divisions correspond to current through galvanometer and 30 divisions for current through shunt. So, voltage drop a cross galvanometer = 20 G where G is resistance of galvanometer. Since a fall of 30 divisions is registered when a 12 W shunt

03_Ch 3_Hints and Explanation_P2.indd 246

Total Momentum = Nmv I neA



⇒ p = ( nA ) m



⇒ p=



⇒



Hence, the correct answer is (B).

I

I

I

I

4

1 11. Since I = neAv If N is total number of electrons in the wire then N = nA Momentum possessed by a single electron is mv.

D



⇒ 20 G = ( 30 ) ( 12 ) ⇒ G = 18 W Hence, the correct answer is (A).

mI e

p I I = =  em s

112. Since E ∝ . So, for E1 > E2 we have  1 >  2 and hence null point will be obtained at shorter length i.e. to left of C . Hence, the correct answer is (A). 113. Voltage drop across 50 W and 30 W (n series) is V ′ = ( 80 )( 0.1 ) = 8 V Voltage across 20 W branch is also 8 V . So, current in 20 W branch is 8 I 2 = = 0.4 A 20 Hence total current in the circuit is 0.5 A. Also voltage drop across R is 12 – 8 = 4 V.



⇒

4 = I total R



⇒ ⇒

4 = ( 0.5 ) R R=8W



Hence, the correct answer is (B).

114. Net resistance of the balanced Wheatstone Bridge is R which is connected in parallel with another resistance of same value. Hence RAB = R 2

Hence, the correct answer is (A).

115.

Req = 220 W



⇒

I total = 1 A

 Reading of ammeter is the current which flows through 3 branches.

9/20/2019 11:49:05 AM

Hints and Explanations H.247 119.

3 A 5



Hence reading =



Hence, the correct answer is (C).

116.

Shunt = 0.1 W = S

Rtotal = R1 + r



(Since capacitor is a dc blocking element) VC = VR1

I g = 10 mA

Since

I = ?

Q = CVC

Since S =

I gG I − Ig

( 10 × 10 −3 ) ( 9 ⋅ 9 )



⇒ 0.1 =



⇒ I − 10 −2 = 99 × 10 −2



⇒ I = 100 × 10 −2



⇒ I =1A



Hence, the correct answer is (C).

117.

⇒ RAB =



⇒ RAB  22 ⋅ 26 W

128.63

2Ω

100 Ω B A

30 Ω

6Ω

B

20 Ω

3Ω

9Ω 100 Ω

⇒ Q=



Hence, the correct answer is (B).

N = 24 = mn …(1) For current to be MAXIMUM Rinternal = Rexternal

( 28.63 )( 100 )

A

CER1 R1 + r



120.

I − 10 × 10 −3

9Ω



⇒

mr =3 n



⇒

m ( 0.5 ) = 3 n



⇒



m =6 n ⇒ m = 6n …(2)



Substituting (2) in (1), we get

24 = 6n2 ⇒ n=2 ⇒ m = 12

Hence, the correct answer is (A).

121. Initial current, i1 = 750 Ω 55

A

E1 R + r1

i2 > i1 if 28.63 Ω

B



Hence, the correct answer is (C).

118. Taking the total current as 6I, current through 4R is 2I and current through 2R is 4I. Applying Kirchhoff’s Second Law, we get current through 2R is 2E 7 R .

Hence, the correct answer is (C).

03_Ch 3_Hints and Explanation_P2.indd 247

E1 + E2 R + r1 + r2

Final current, when second battery is short circuited is i2 =

6Ω

100 Ω

E ⎫ ⎧ ⎬ ⎨∵ I = R 1+r⎭ ⎩

ER1  R1 + r

CHAPTER 3

VC =

Gnet = 9.9 W



E1 E + E2 > 1 R + r1 R + r1 + r2

⇒ E1R + E1r1 + E1r2 > E1R + E1r1 + E2 R + E2 r1



⇒ E1r2 > E2 ( R + r1 )



Hence, the correct answer is (B).

122.

V0 = i0 R = ( 10 )( 10 ) = 100 V

1 th . Therefore, after 1 s, 4 current will remain half also called half-life.

After 2 s , current becomes

t 1 = ( ln 2 )τ C = ( ln 2 ) CR 2

9/20/2019 11:49:18 AM

H.248  JEE Advanced Physics: Electrostatics and Current Electricity

(t ) 1 2



1 ⇒ C= = F ( ln 2 ) R 10 ln 2



Total heat ΔH =



1 CV02 2 1 1 ( 100 )2 ⇒ ΔH = × 2 10 ln 2

At the given time, given current is 1 A of half of the above value. Hence, at this is instant capacitor is also charged to half of the final value of 5 V. Now, it is shifted to position-2 wherein steady state it is again charged to 5 V but with opposite polarity. U i = U f =

1 CV 2  2

{∵ V = 5 V }

500 J ln 2 Hence, the correct answer is (D).

So, total energy supplied by the lower battery is converted into heat. But double charge transfer (from the normal) takes place from this battery.

123.

E1 + E2 400 = E1 − E2 200



⇒ Heat produced = Energy supplied by the battery ⇒ ΔH = ( Δq )V = ( 2CV )( V ) = 2CV 2



⇒ E1 + E2 = 2 ( E1 − E2 )





⇒ E1 = 3E2



⇒ ΔH = 2 × 2 × 10 −6 × ( 5 )



E 3 ⇒ 1 = E2 1



⇒ ΔH = 100 × 10 −6 J



⇒ ΔH = 100 m J



Hence, the correct answer is (B).



Hence, the correct answer is (C).

126.

R1 + 10 50 = R2 50



⇒ R1 + 10 = R2 …(1)



Now, when 10 W is removed, then



124.

⇒ ΔH =

dQ dt ⇒ I = α − 2βt I=

α Since at t0 = t = , I becomes zero. Hence we shall 2β calculate the heat produced from α t = 0 to t0 = 2β t0



⇒

∫

H = I 2 R dt 0

α 2β



⇒

H=R

∫ ( α − 2βt ) dt 2

0



⇒

( α − 2β t ) 2 + 1 H=R −6β

α 2β 0



⇒

⎛ α3 ⎞ H = R⎜ 0 − −6β ⎟⎠ ⎝



⇒

R=



Hence, the correct answer is (D).

α 3R 6β

125. In position-1, initial maximum current is i0 =

V 10 = =2A R 5

03_Ch 3_Hints and Explanation_P2.indd 248

2

R 40 1 = R2 60 R1 2 = R2 3



⇒



⇒ R2 =



Put (2) in (1), we get

3 R1 …(2) 2

R1 + 10 =

3 R1 2



⇒ 2R1 + 20 = 3 R1



⇒ R1 = 20 W



Hence, the correct answer is (A).

127. Resistance between A and B can be removed due to balanced Wheatstone bridge concept. Now, RDE and RGH are in series and they are connected in parallel with 10 V battery. 10 10 = RDE + RHG 2 + 2



⇒ I DE =



⇒ I DE = 2.5 A



Hence, the correct answer is (B).

9/20/2019 11:49:30 AM

Hints and Explanations H.249



V = iR ⇒ V ∝R



⇒

129.

⇒



( as i = constant )

VA ⎛ ρ A ⎞ ⎛ π rB2 ⎞ = VB ⎜⎝ π rA2 ⎟⎠ ⎜⎝ ρ B ⎟⎠

3 1 1 × = 2 6 2



Hence, the correct answer is (B).

130.

τ C = CR

{

⎛ Kε 0 A ⎞ ⎛ d ⎞ ⇒ τC = ⎜  ⎝ d ⎟⎠ ⎜⎝ Aσ ⎟⎠ Kε 0 ⇒ τC = σ



∴ 27 mC charge flows from Y to X Hence, the correct answer is (C).

134. Initially, the rate of charging is fast. Hence, the correct answer is (B). 1 dR R0 dT

135.

α (T ) =



⇒

( 3T 2 + 2T ) =



⇒

dR = R0 ( 3T 2 + 2T ) dT



⇒

T ⎡ T ⎤ dR = R0 ⎢ 3 T 2 dT + 2 T dT ⎥ ⎢ ⎥ R0 0 ⎣ 0 ⎦



⇒

R = R0 [ 1 + T 2 + T 3 ]



Hence, the correct answer is (C).

R

rB VA  B = × rA VB  A

=



∵ R=

 σA

}

∫

1 dR R0 dT

∫

∫

136. Let R = at + b At t = 10 s, R = 20 W ⇒ 20 = 10 a + b …(1) At t = 30 a + b …(2) Solving these two equations, we get a = 1.0 Ws −1 and b = 10 W



5 × 8.86 × 10 −12 7.4 × 10 −12 ⇒ τC = 6 s



Hence, the correct answer is (D).

and Δq =

∫

30

131.

⎛ ⎞ ⎛ 70 ⎞ r = R ⎜ 1 − 1 ⎟ = 132.4 ⎜ − 1 ⎟ ≈ 22.1 W ⎝ 60 ⎠ ⎝ 2 ⎠



Hence, the correct answer is (A).



⇒ Δq =

∫

30



⇒ Δq = 10 log e ( 2 )



Hence, the correct answer is (D).

⇒ τC =



132. Since





⇒ X = 0.25 W



Hence, the correct answer is (A).

133. From Y to X charge flows to plates a and b q ( a + qb )i = 0, ( qa + qb ) f = 27 mC 6 μF b + – 18 μ C

9 μC + –

X

9V Initial figure (When switch was open)

03_Ch 3_Hints and Explanation_P2.indd 249

36 μ C + –

3Ω

1A Y

6Ω

9V Final figure (When switch is closed)

E 10 = R t + 10 10

10

idt ⎛ 10 ⎞ ⎜⎝ ⎟ dt t + 10 ⎠

137. During charging capacitor and resistance of its wire are independently connected with the battery. Hence, τ C = CR During discharging capacitor is discharged through both resistors (in series). Hence, τ C = C ( 2R ) = 2CR

1A 3Ω 1A 6Ω

⇒ R = ( t + 10 )

Since, i =

X 1 = 20 80

3 μF a + – 18 μ C

CHAPTER 3

128. RAB = 2 [Net resistance of infinite series] +1 In parallel net resistance is always less than the smallest one. Hence, net resistance of infinite series is less than 1 W . ⇒ 1 W < RAB < 3 W Hence, the correct answer is (C).

Hence, the correct answer is (C).

138. Mean free path is the average distance travelled by the charge carriers between two successive collisions jτ ⇒  = vdτ = nq

Hence, the correct answer is (B).

9/20/2019 11:49:41 AM

H.250  JEE Advanced Physics: Electrostatics and Current Electricity 139. The given time is the half-life time of the exponentially decreasing equation. So, t = t 1 = ( ln 2 )τ C = ( ln 2 ) CRnet 2

t = ( ln 2 ) C



⇒ Rnet



⇒ Rnet =



So, resistance of ammeter is 2 W Hence, the correct answer is (C).

9 − 3 × 3 = VB 1



⇒ VA − VB = 12 V



Hence, the correct answer is (A).

142.

E

E

b

C

i

[ E − E0 ] i = R + R0

Now, Va − E + E0 + iR0 = Vb

⇒ Va − Vb = ( E − E0 ) − iR0



R0 ⎤ ⎡ ⇒ Va − Vb = ( E − E0 ) ⎢ 1 − R + R0 ⎥⎦ ⎣



⇒ Va − Vb =



Hence, the correct answer is (D).

143.

I=



P = I 2 R = Power consumed ⇒  across external resistance



⇒

R + R0

E2 R

( R + r )2

03_Ch 3_Hints and Explanation_P2.indd 250

144.

⎛ V ⎞ − 6CR i1 = ⎜ e ⎝ 2R ⎟⎠



⎛V⎞ − ⇒ i2 = ⎜ ⎟ e CR ⎝ R⎠



i e 6CR ⇒ 1 = 2 i2



We can see that this ratio is increasing with time. Hence, the correct answer is (B).

5t

nr m



⇒

R=



⇒

n = 10 …(2) m



⇒

n = 10 m



⇒

10 m2 = 24



⇒

m = 2.4 Option-1



⇒

I1 =



⇒

I1 =

nE nr R+ m 24 E 34

Option-2

m = 2 ; n = 12 Since

E R+r

P=

Hence, the correct answer is (A).

Since

CR ( E − E0 )

⇒ q = C ( Va − Vb ) =



⇒ m  1.55  m = 1 ; n = 24

R( E − E ) R + R0



⇒

External Total Internal ⎛ Resistance ⎞ = ⎛ Resistance ⎞ ⎝ ⎠ ⎝ ⎠

R0

E0



145. mn = 24 …(1)  For power consumption across the load to be maximum

R

a

⇒

t

140. 9, 9 and 9 are in parallel to give 3 Ω. This 3 Ω is further in series with the other 3 Ω to give 6Ω Hence, the correct answer is (B). VA − 6 − 3 × 2 +

dP =0 dR R=r



t

2 ( ln 2 ) ms =4W ( ln 2 ) ( 0.5 m F )

141.

For P to be Maximum

⇒

I2 =



⇒

I2 =



From above we observe that I 2 > I1

R I

E r

nE 12 = E nr 16 R+ m



24 E 32

9/20/2019 11:49:51 AM

Hints and Explanations H.251 Hence n = 12, m = 2 for the power consumption across the load to be maximum. Hence, the correct answer is (B).



In this case, 6 W and 3 W are parallel



( equivalent = 2 W )

146. In steady state condition, current flows from outermost loop.



⇒ Rnet ( 1 + 2 ) W = 3 W



⇒ Current from battery =

12 =4A 3 = Current through 1 W resistor

Hence, the correct answer is (B).

152.

τ C = CR = 6 s

12 = 1.5 A 6+2 Now, VC = V6 W = iR

i =



⇒ VC = 1.5 × 6 = 9 V



⇒ q = CVC = 18 mC



Hence, the correct answer is (D).

Now, q = q0 e

2R R

A

B 2R

2

t τC

149.

E − ir = 0



⎛ 3 + 15 ⎞ ( ) 1 =0 ⇒ 3−⎜ ⎝ 1 + 2 + R ⎟⎠



Solving this equation, we get



⇒ q = ( 0.37 ) ( 10 mC )



Hence, the correct answer is (B).

( 10 mC ) 2

S

G

IG

In parallel, current distributers in inverse ratio of resistance. r 0.03 − IG G = = =4 IG S ⎛ r⎞ ⎜⎝ ⎟⎠ 4

R = 3 W

Solving this equation, we get

IG = 0.006 A

Hence, the correct answer is (C).



150.

Hence, the correct answer is (C).

156.

1A

B

2Ω

2Ω A

3A

A R

RR R 2

R RR

C

R R 2

B R



Let each resistance be having a value R



⇒

2V1Ω



Hence, the correct answer is (B).



⇒ VA − VB = 17 V



Hence, the correct answer is (D).

157.

B

151. At t = 0, when capacitor is under charged equivalent resistance of capacitor = 0

R R/2

R/2

C

RAB = RAC + RCB

03_Ch 3_Hints and Explanation_P2.indd 251

6A

4A

VA − 3 × 2 − 3 − 1 × 4 + 2 − 1 × 6 = VB

R 2

R

2A

3V 1Ω

Line of symmetry R 2

12 6

0.03 – IG

Now 2 W, 2 W and R are in parallel. Hence, the correct answer is (A).

R

−

⎛ 1⎞ ⇒ q=⎜ ⎟ ⎝ e⎠

154.

A

= ( 10 mC ) e



⎛ 4 ⎞( ) 4 R = ⎜ 2r = ⎝ 2π r ⎟⎠ π



−

CHAPTER 3

147.

q0 = CV = 10 mC

D R/2

R/2 A



R

B

R/2 A

R

R/2 B

Hence, the correct answer is (D).

9/20/2019 11:49:59 AM

H.252  JEE Advanced Physics: Electrostatics and Current Electricity 159.

V1 W = 5 + 2 = 7 V



⇒ i1 W =

V2 mF

V =7 A R =6V



⇒ q2 mF = CV = 12 mC



Hence, the correct answer is (B).

161.

H1 = H 2 = U i − U f

165.

qg + qd = q0 = 10 mC



⇒



Hence, the correct answer is (B).

qd = 7 mC

166. Total heat produced ΔH = H = 1 ( 2 m F ) ( 5 )2 2



⇒ ΔH =

The only change is by increasing the resistance τ C increase. Hence, process of redistribution of charge slows down. Hence, the correct answer is (A).



⇒ ΔH = 25 m J

162. When K1 and K 2 both are closed R1 is short-circuited,



⇒



⎛ R ⎞ ⇒ H5 W = ⎜  ⎝ R + 5 ⎠⎟



⎛ R ⎞( ) ⇒ 10 = ⎜ 25 ⎝ R + 5 ⎟⎠



Solving this equation, we get

Rnet

= ( 50 + r ) W

When K1 is open and K 2 is closed, current remains half. Therefore, net resistance of the circuit becomes two times. ⇒ ( 50 + r ) + R1 = 2 ( 50 + r ) Of the given options, the above equation is satisfied if r = 0 and R1 = 50 W

Hence, the correct answer is (D).

163. Practically, the steady state is attained when five time constants have elapsed. As soon as the key is opened the peak value of charge also decays in five time constants. Hence pulse period is

Now, this should distribute in inverse ratio of resistors, as they are in parallel.



A R R

10 RC

t

Hence, the correct answer is (C).

164. Potential drop across potentiometer wire = ( 0.2 × 10 −3 ) ( 100 ) = 0.02 V Now given resistance and potentiometer wire are in series with given battery. So, potential will drop in direct ratio of resistance.

R R

R



0.02 R = 2 − 0.02 490 ⇒ R = 4.9 W



Hence, the correct answer is (A).

R B

Connection can be removed from centre. 3R and 3R from two sides of AB are in parallel. Hence, the correct answer is (B). 168.

1 = 1 − e −t τ 10



⇒

e −t τ =



⇒

et τ =



⇒

t ⎛ 10 ⎞ = n ⎜ ⎟ ⎝ 9 ⎠ τ



⇒

⎛ 10 ⎞ t = τ n ⎜ ⎟ ⎝ 9 ⎠



Hence, the correct answer is (B).

⇒

03_Ch 3_Hints and Explanation_P2.indd 252

R

R

R

q0



R

R R

5 RC

(Total heat)

Hence, the correct answer is (C).

167.

q



H5 W R = 5 HR

⎛ 10 ⎞ R = ⎜ ⎟ W ⎝ 3 ⎠

5RC + 5RC = 10 RC

O

1 CV 2 2

9 10

10 9

9/20/2019 11:50:09 AM

Hints and Explanations H.253 169. Applying Kirchhoff’s loop law in outermost loop, we have



Current through either of the resistance is

3 μF + –



⇒

5Ω

1Ω 3A

−

15 V

4Ω

2r

q – + 2 μF



2.5 A

q q + 15 − 2 × 2.5 − − 3 × 1 + 18 = 0 2 3

Solving this equation, we get

⎛ 1⎞ ⇒ V = iR = ⎜ ⎟ ( 9 ) = 3 V ⎝ 3⎠ Hence, the correct answer is (A).

175. Since capacitors are identical, hence they share equal Q amount of charge 0 . 2 Ei =

q = 30 mC

Hence, the correct answer is (A).

171. Suppose n ( < 1 ) fraction of length is stretched to m times. Then, ( 1 − n )  + ( n ) m = 1.5

⇒ nm − n = 0.5 …(1)

R =

ρ ρ =  A ⎛V⎞ ⎜⎝ ⎟⎠ 

( V = volume )

ρ V

⇒ R=



⇒ R ∝ 2 



Now, the second condition is

( if V = constant )



⇒ nm2 − n = 3 …(2)



Solving these two equations, we get



1 8

Hence, the correct answer is (B).

172. Total potential of 10 V equally distributes between 50 W and other parallel combination of 100 W and voltmeter. Hence, their net resistance should be same.

100 × R = 50 ⇒ 100 + R



⇒ R = 100 W = resistance of voltmeter



Hence, the correct answer is (B).

174. All these resistors are in parallel.

R +r=4W 3 Hence. the main current

⇒



03_Ch 3_Hints and Explanation_P2.indd 253

Q02 4C



Heat produced = Ei − E f =



Hence, the correct answer is (C).

Let us assume that charge flow between the rod and cylinder be radial. Then at point P at distance r from centre of rod the current density J is I J = 2π r Further E = ρ J =

ρI  2π r

{∵ J = σ E }

with both J and E directed radially outwards. Since dV = − E dr

⇒

dV = − b



⇒

∫ a

ρI dr 2π  r b

ρI dr dV = − 2π  r

∫ a

ρI ⎛ log e ⎜ ⎝ 2π 

b⎞ ⎟ a⎠



⇒

−V = −



⇒

I=



Hence, (A) and (B) are correct.

2.

R0 =

⇒ Rnet =

E =1A i = Rnet

{Due to sharing the new capacitance is 2C}}

1.

( − n ) R + ( nR ) m3 = 4 R 1

n =

⎡ ⎛ Q ⎞2 ⎤ ⎢ ⎜⎝ 0 ⎟⎠ ⎥ Q 2 ⎥= 0 Ef = 2⎢ 2 ⎢⎣ 2C ⎥⎦ 4C

Multiple Correct Choice Type Questions

2





Q02 2C

CHAPTER 3

q

18 V

1 A 3

i 3

2π V ⎛ b⎞ ρ log e ⎜ ⎟ ⎝ a⎠

V R2 R2 R3 and V0 = R2 + R3 R2 + R3

9/20/2019 11:50:20 AM

H.254  JEE Advanced Physics: Electrostatics and Current Electricity R3



R2 S

R1

R0

R1

C

V0

C

V



⇒

RR ⎞ ⎛ τ = C ( R1 + R0 ) = C ⎜ R1 + 2 3 ⎟ R2 + R3 ⎠ ⎝



⇒

q = q0 1 − e



⇒

q = CV0 1 − e



⇒

q=



Hence, (B) and (C) are correct.

3.

Let E2 > E1 , then

(

I =

−

(

t τ

)

−

t τ

(

) t

− CVR2 1− e τ R2 + R3

)

⇒ VA − VB =

Also, we could have thought of equating equation (1) and (2), to get E − E1 I= 2 , which also satisfies the same result  r1 + r2 obtained. Also, from equation (1), we get that potential difference across 1 is greater than its emf E1 . Now, since the current I flows from the positive plate to the negative plate inside the battery 2, hence energy is absorbed by it or in other words the battery will continuously get the energy supplied by the other battery. Hence all the statements A, B, C and D are correct. Hence, (A), (B), (C) and (D) are correct. 4.

In steady state, no current passes through the branches containing the capacitors. So,

I =

E2 − E1 r1 + r2

4 4 1 = = A 8 + 2 + 10 20 5

For potential difference across battery 1, apply KVL to A1B , we get A

E1, r1

8Ω 1A 5 C + – C1 = 8 μ F

1

I 2

A 1A 5 2Ω B

E2, r2 B

VA − E1 − Ir1 − VB = 0

⇒ VA − VB = E1 + Ir1 …(1)



⎛ E − E1 ⎞ ⇒ VA − VB = E1 + ⎜ 2 r1 ⎝ r1 + r2 ⎟⎠



⇒ VA − VB =



⇒ VA − VB =



Loop ABCA

⎛ − ⎜ ⎝

1⎞ ⎛ ⎟ 8 − ⎜⎝ 5⎠

⇒ ΔV1 =

1⎞ ⎟ 2 + ΔV1 = 0 5⎠

10 =2V 5 A

E1r1 + E1r2 + E2 r1 − E1r1 r1 + r2 E1r2 + E2 r1 …(2) r1 + r2

VA + Ir2 − E2 − VB = 0

⇒ VA − VB = E2 − Ir2 …(3)



⎛ E − E1 ⎞ ⇒ VA − VB = E2 − ⎜ 2 r2 ⎝ r1 + r2 ⎟⎠



⇒ VA − VB =

+ C2 = 10 μ F –

1A 5 2Ω

For potential difference across battery 2, apply KVL to A 2B , we get

03_Ch 3_Hints and Explanation_P2.indd 254

E1r2 + E2 r1 …(4) r1 + r2

B



D 1A 5 10 Ω

Loop ABDA

⎛ 1⎞ ⎛ 1⎞ − ⎜ ⎟ 2 − ⎜ ⎟ ( 10 ) + ΔV2 = 0 ⎝ 5⎠ ⎝ 5⎠

⇒ ΔV2 = 2 + 0.4

E2 r1 + E2 r2 − E2 r2 + E1r2



⇒ ΔV2 = 2.4 V

r1 + r2



Hence, (B) and (D) are correct.

9/20/2019 11:50:29 AM

Hints and Explanations H.255 Let VD be the potential at the point D, then

70 − VD = I1 × 10 W VD − 0 = I 2 × 20 W VD − 10 = ( I1 − I 2 ) × 30 W B (VB = 0)

VD I1

10 Ω

A (VA = 70 V)

I1 – D I2

I2

20 Ω 30 Ω



⇒

⎛R ⎞ E = I ⎜ total ⎟  ⎝  ⎠



⇒

E=



Hence, (A) and (C) are correct.

8.

I=



⇒ RA + RV =

2πα I A2

E RA + RV

C (VC = 10 V)



{By Ohm’s Law}

E …(1) I V

A

I

Solve for I1 , I 2 and VD , we get

VD = 40 V

I1 = 3 A , I 2 = 2 A and ( I1 − I 2 ) = 1 A

Ptotal

2 2 2 = ( 3 ) ( 10 ) + ( 2 ) ( 20 ) + ( 1 ) ( 30 )

E

Since E = V + IRA

⇒ V = E − IRA < E …(2)



⇒ Ptotal = 90 + 80 + 30 = 200 W



⇒ I1 : I 2 : ( I1 − I 2 ) :: 3 : 2 : 1 and Ptotal = 200 W



Hence, (A), (B) and (D) are correct.



⇒ RV =

7.

Consider a cylindrical element of radius r , thickness dr. If dR is the resistance of this element then



Potential difference across the ammeter is ΔVA = IRA



⇒ ΔVA = IRA = E − V …(4)

dR =

Total resistance of the cylinder is given by





ρ ( r ) 2π r dr

⇒

1 Rtotal 1 Rtotal

=

=

∫

a

1 2π 3 r dr = dR α 

∫

dr

I

Resistance for the upper half between A and D is ⎛ R⎞⎛ n⎞ R and that for the lower half between A ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = n 2 2 R and D is also . 2 F

2α  π a4

⇒

Rtotal =



⇒

Rtotal 2πα =  ( π a 2 )2 R=

So, from (1), (2), (3) and (4), we observe all the options to be correct. Hence, (A), (B), (C) and (D) are correct.

RMAX = RAD = RBE = RCF

r

⇒

R n

V (in magnitude) 

03_Ch 3_Hints and Explanation_P2.indd 255

R n

E R n

A

D R n

2πα A2

Since E =

V …(3) I

For the points that face each other like ( A , D ) , ( B, E ) and ( C , F ) , we have

0

a



Also, V = IRV

10. Consider the ‘ n ’ sided polygon, shown below

2π ⎛ a 4 ⎞ ⎜ ⎟ α ⎝ 4 ⎠



CHAPTER 3

6.

R n B



R n

C

These both are in parallel across A and D , hence

9/20/2019 11:50:40 AM

H.256  JEE Advanced Physics: Electrostatics and Current Electricity

RMAX = RAD = RBE = RCF

12. In steady state,

⎛ R⎞⎛ R⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ R 2 = = 2 R R 4 + 2 2

qC = EC and q2C = 2EC τ C = 2CR of both circuits

We shall get the minimum value of resistance between the adjacent points of the polygon i.e., AB, BC , CD, DE, EF or FA . Between two adjacent points we have two resistors connected in parallel one having resisR R tance and other having resistance ( n − 1 ) n n So, RMIN = RADJACENT POINTS



⎛ R⎞ ⎛ R⎞( ⎜⎝ ⎟⎠ n − 1 ) ⎜⎝ ⎟⎠ n n = ⎛ n − 1⎞ R = ⎜⎝ 2 ⎟⎠ R R n + ( n − 1) n n

Hence, (A) and (C) are correct.



q2C



RV +

RA 2

New ammeter reading is

V

Initial reading of ammeter was

I initial =

E RV + RA

So, we observe that

1⎛ E E ⎞ > 2RV + RA 2 ⎜⎝ RV + RA ⎟⎠

⇒ I final >



qC 1 = q2C 2

Hence, (B), (C) and (D) are correct. 2

14. A

D A

3

Hence, (B) and (D) are correct.

03_Ch 3_Hints and Explanation_P2.indd 256

B

4

C 2

4

A

A

B

5

A

B 6

3 D E Initially

Rearrangement of the circuit as shown gives a balanced Wheatstone bridge, and no current flows through the 5 W resistor. It can thus be removed from the circuit to find the net resistance. When the current in the 5 W branch is zero, then ΔV = 0 . Hence, (B) and (D) are correct. 15. In steady state, no current passes through the branch that contains a fully charged capacitor, because a fully charged capacitor is a dc blocking element. Hence the circuit becomes R

1

7

+ –

2

3

I1

I1

E ⎛ ⎞ =⎜ ⎟ RV R + R ⎝ V A⎠

E ⎛ ⎞ = R > Vinitial ⎜ RA ⎟ V ⎜⎝ RV + ⎟⎠ 2

B

C

5

6

R

R

I

Final reading of the voltmeter is

Vfinal

1 ( Iinitial ) 2

t τC



Initial reading of the voltmeter is

Vinitial

−

⇒

I

I E final = 2 2RV + RA

) = 2EC ( 1 − e

E …(1) RV + RA

E

)

t



Finally, when another resistance of value RA is connected in parallel with the ammeter, then we have I final =

(

qC = EC 1 − e τ C

11. Let the respective resistance of the voltmeter and ammeter be RV and RA . Then, initially I initial =

At time t,

8

(I – I1)

6

R

I

4

5 E



For Loop 1284561

−2I1R − IR + E = 0

9/20/2019 11:50:49 AM

Hints and Explanations H.257 ⇒



For Loop 784567

6 I1 + 3 I = 15 …(1)

ρ=I

md 2eVxA 2



⇒

− ( I − I1 ) R − IR + E = 0



Hence, (A) and (B) are correct.

−2IR + I1R + E = 0

18. Case-1: Net resistance of the circuit is



⇒ 3 I1 − 6 I = −15



15 3 I1 ⇒ − 3 I = − …(2) 2 2



Add (1) and (2), we get

R1 = 600 Ω V1

I1

600 Ω

15 ⎛3 ⎞ ⎜ + 6 ⎟ I1 = 15 − ⎝2 ⎠ 2

E = 120 V



30 − 15 ⎛ 3 + 12 ⎞ ⇒ ⎜ I = ⎝ 2 ⎟⎠ 1 2 ⇒ 15I1 = 15



⇒ I1 = 1 A



⇒ 6 + 3 I = 15 ⇒ 3I = 9 ⇒ I=3A

V1 =



For Loop 23482





R =

( 600 )( 600 ) 600 + 600

⇒ VC = 9 + 3 = 12 V



Hence, (C) and (D) are correct.

Please note that though no current flows through the branch containing the fully charged capacitor, yet potential difference across the capacitor will be finite q non zero value given by ΔV = ± . C R1 ⎞ ⎛R ⎞ V = ⎜ 1 ⎟V ⎟ R1 + R2 + R3 + ..... ⎠ ⎝ Rs ⎠

16.

⎛ V1 = ⎜ ⎝



⇒



Hence, (A), (B) and (C) are correct.

R2 R ⎛ ⎞ V2 = ⎜ V = 2V Rs ⎝ R1 + R2 + R3 + .... ⎟⎠

17. Acceleration of each electron is a =

eE eV = m md



⇒

v 2 − 0 2 = 2 ax



⇒

v=

2eVx md

Since I = neAv and ρ = en = e

I veA

03_Ch 3_Hints and Explanation_P2.indd 257

+ 300

R = 600 W

⇒ I1 =

120 1 = A 600 5 1 ( 300 ) = 60 V 5

Case-2: Net resistance of circuit is

−VC + I ( 3 ) + I1 ( 3 ) = 0

R2 = 600 Ω

CHAPTER 3



R1 = 600 Ω

R2 = 600 Ω V2

I2

600 Ω E = 120 V

R′ = 600 +

( 300 )( 600 ) 300 + 600



⇒ R′ = 600 + 200 = 800 W



⇒ I2 =

E 120 3 = = A R′ 800 20

⎛ 3 ⎞ ⎛ 300 × 600 ⎞ Now V2 = ⎜ = 30 V ⎝ 20 ⎟⎠ ⎜⎝ 300 + 600 ⎟⎠

Hence, (C) and (D) are correct.

20. For conductors, the resistance decreases with the increase in temperature as random motion of the charge carriers decreases, so α 2 > 0 For semiconductors, the resistance increases with the decrease in temperature and hence α 1 < 0 . Hence, (B) and (C) are correct. 21. ⇒ I =

9 =1A 9

At A a current of 1 A divides into 0.5 A and 0.5 A. At B the current of 0.5 A divides into 0.25 A and 0.25 A Hence, (A) and (D) are correct.

9/20/2019 11:50:58 AM

H.258  JEE Advanced Physics: Electrostatics and Current Electricity 22. See Theory Hence, (A) and (C) are correct. 23. The electrical paths AC1D2B and AD1C2B are in ­parallel. The resistance of each of them is 10 W , and hence the current is 1 A. When C1C2 and D1D2 are joined, the ammeters are in parallel combination. As they are of a same resistance, their readings will be equal. Hence, (B), (C) and (D) are correct. 24. Since R = C= and

I=

E R+r

R I E, r

Kε 0 A 

P = EI E2 R+r  So, heat produced per second across the external ­circuit is

Since q = q0 e − t τ q = q0 e

−

⇒



dq Leakage current = I = dt



q0σ e ε0K −

I = I0e

tσ ε0 K

⇒ P=

E ( IR ) R+r and that across the internal circuit is

H = I 2 R = I ( IR ) =

σt Kε0



I=−

28.

⇒ I 0 − 2I1 = I 0 −

ρ  = A σA

ε K ⇒ τ = RC = 0 σ

⇒



2I 0 3 I 0 = which is the current 5 5 flowing from D to C.  ence, (A), (B) and (C) are correct. H



−



h = I 2 r =

tσ ε0 K



where I 0 = −

Hence, (A), (B), (C) and (D) are correct.

q0σ ε0K

⇒



Hence, (A), (C) and (D) are correct.

R

x =

R ( 2R + x ) 3R + x



⇒

3 Rx + x 2 = 2R2 + Rx



⇒

x 2 + 2Rx − 2R2 = 0



⇒

x=−



⇒

x=



⇒

x = ( 3 − 1)R

C A

B

D 20 Ω

5Ω

20 Ω I1 C 5 Ω I0 – I1 I0 I0 – I1 5Ω

D

I0 – 2I1 I1

B

20 Ω

⇒ 20 I1 = 5 ( I 0 − I1 )



I ⇒ I1 = 0 5

2R ± 4 R 2 + 8 R 2 2

−2R + 2 3 R 2

1

2R R

VA − VD = VA − VC

x

2

5Ω

20 Ω

2R

1

25. As C and D are joined, they must be at the same potential, and may be treated as the same point. This gives the equivalent resistance as 8 W. When we distribute current in the network using symmetry,

03_Ch 3_Hints and Explanation_P2.indd 258

( R + r )2

29. CIRCUIT A



A

E2 r

y

2

CIRCUIT B yR + 2R y = y+R

9/20/2019 11:51:06 AM

Hints and Explanations H.259 V17 = 5IR0

⇒

y 2 + Ry = yR + 2Ry + 2R2





⇒

y 2 − 2 yR − 2R2 = 0

Also,



⇒

y = ( 3 + 1)R

V17 = x ( 6 I )



Hence, (A), (B), (C) and (D) are correct.

31. Since voltmeter is a device connected in parallel across the circuit, hence RV R0 RV + R0

Requivalent = For R0  RV

⇒



⇒

x ( 6 I ) = 5IR0



⇒

x=



⇒

V18 = ( V1 − V5 ) + ( V5 − V8 )



⇒

V18 = 4 IR0 + 5IR0



⇒

V18 = 9IR0

5R0 6

Also V18 = y ( 12I )

Requivalent  R0

{i.e., resistance of the circuit remains unaltered when a voltmeter of extremely high resistance is applied across the circuit} Hence, (A) and (B) are correct. 33.

1

12I

I

I – Ig

I



I′

B  

A

⇒ I=

I′ – Ig

0.01 A × 10 =1A 0.1

Similarly ( I ′ − I g ) ( 0.9 ) = I g ( 9.1 )

I

5I

6 5I

4I

2I

8

⇒

( 12I ) y = 9IR0



⇒

y=



⇒

⎛ x ⎜⎝ = y ⎛ ⎜⎝



⇒

x−y=

5 3 R0 − R0 6 4



⇒

x−y=

R0 12

Hence, (A) and (B) are correct.

B

A

2I



0.9 Ω

0.1 Ω

3

4

5

9 Ω 0.1 Ω

Ig

2I

2I I

⇒ I × 0.1 = I g × 10 9 Ω 0.9 Ω

2

4I

I g = 10 mA = 0.01 A

Ig

4I

4I

VA − VB = ( I − I g ) ( 0.1 A ) = I g ( 9.9 )

CHAPTER 3

⇒



7 12I

3 R0 4 5⎞ ⎟ 6 ⎠ = 20 = 10 3 ⎞ 18 9 ⎟ 4⎠



⇒ I ′ ( 0.9 ) = I g ( 10 )





1 ⇒ I ′ = A = 111 mA 9

35. Let V = 220 V and R1 and R2 be the resistances of the 25 W and 100 W bulbs.



Hence, (A) and (C) are correct.

R1

34. Resistance between diagonal corners of cube

⇒

V17 = ( V1 − V5 ) + ( V5 − V8 ) + ( V8 − V7 )



⇒

V17 = 2IR0 + IR0 + 2IR0 6I 1

2I 2I

2

4

03_Ch 3_Hints and Explanation_P2.indd 259

I

2I 6

8



I

I I

5

3

I

2I

2I

I V

Since, PA = 25 W = I

2I

6I 7

R2

⇒ R1 =

V2 V2 and PB = 100 W = R1 R2

V2 V2 and R2 = 25 W 100 W

When the bulbs are joined in series then the current is V I= R1 + R2

9/20/2019 11:51:17 AM

H.260  JEE Advanced Physics: Electrostatics and Current Electricity Power in the 25 W bulb is R1I2 and that in the 100 W bulb is R2I2. Hence, (A) and (C) are correct. 36. When X is joined to Y for a long time (charging), the energy stored in the capacitor is equal to the heat produced in R i.e., H1. When X is joined to Z (discharging), the energy stored in C ( = H1 ) reappears as heat ( H 2 ) in R. So, H1 = H 2 . Also, we observe that energy supplied by the battery is H = H1 + H 2 . C

R

⎛ V2 ⎞ ⎛ V2 ⎞ tA = ⎜ tB …(1) H = ⎜ ⎟ ⎝ R1 ⎠ ⎝ R2 ⎟⎠

When used in series,

⎛ V2 ⎞ H = ⎜ T …(2) ⎝ R1 + R2 ⎟⎠

When used in parallel

⎛ V2 V2 ⎞ H = ⎜ + t …(3) ⎝ R1 R2 ⎟⎠

Using (1), (2) and (3), we get the desired results. Hence, (B) and (D) are correct.

39. When steady state is reached no current will flow in the branch having the capacitor. So Y Z X



Hence, (A), (C) and (D) are correct.

37.

I=

2E R + r1 + r2

E

E

r2

r1 R

⇒

E − Ir1 = 0



⇒

2Er1 E− =0 R + r1 + r2



⇒

2r1 = R + r1 + r2



⇒

R = r1 − r2

V2 = 0 , for terminal potential difference to be zero  across source 2.

⇒

E − Ir2 = 0



⇒

E−



⇒

2r2 = R + r1 + r2



⇒

R = r2 − r1



Hence, (A) and (C) are correct.

2 Er2 =0 R + r1 + r2

38. Let R1 and R2 be the resistances of the two heaters. Let, initially, the heat produced be H , then

03_Ch 3_Hints and Explanation_P2.indd 260



⇒

Rtotal = 4 W



⇒

I=

( 2 )( 3 ) 2+3

6 = 1.5 A 4



⎛ 12 ⎞ Current in 2 W resistor = I 2 = ⎜ 1.5 ⎝ 1 2 + 1 3 ⎟⎠ ⎛ 3⎞ ⇒ I 2 = ⎜ ⎟ ( 1.5 ) ⎝ 5⎠ ⇒ I 2 = 0.9 A



Potential drop across 2.8 W is



 V1 = 0 , for terminal potential difference to be zero across source 1.



Rtotal = 2.8 +



V = ( 2.8 )( 1.5 ) ⇒ V = 4.2 V

Hence, (A), (B) and (D) are correct.

40.

VA − VB = 9 V

 15 V battery and 24 V battery form the dominating combination and hence I must flow from D to A through R . 24 + 15 − 6 R +1+ 2 +1 33 ⇒ I = R+4 Since inside the 6 V battery the current is going from positive terminal to the negative terminal so

⇒

I=

VA − VB = 6 + Ir

⇒



⇒ ⇒

33 R+4 R + 4 = 11 R=7W



⇒

VBC = 15 − 3 ( 2 ) = 9 V 

9=6+

{

∵I=

33 =3A 11

}

9/20/2019 11:51:28 AM

Hints and Explanations H.261 ⇒

VBD = 39 − 3 ( 3 )



⇒

VBD = 30 V



Hence, (B) and (C) are correct.

41.

V = V0 ( 1 − e − t τ )

RC 2 E V0 = (as both R and R are in series) 2 τ =

V=

(

)

2t

− E 1 − e RC 2



⇒



Hence, (C) and (D) are correct.

42.

H=



V 2t1 ⇒ R1 = H

V2 t1 R1

2

V t2 H

⎛ V2 ⎞ t In series, H = ⎜ ⎝ R1 + R2 ⎟⎠

t =

H ( R1 + R2 )

V2 Substituting the values of R1 and R2 , we get t = t1 + t2

In parallel, H =



C discharges through R + R in series. Hence, (A) and (B) are correct.

Q = Q0 e − t1 π and potential difference across C is proportional to Q . For the potential difference to fall by 10%, Q must fall by 10%. So, 46.

Q = 0.9 Q0 = Q0 e − t1 τ

⇒ e t1



⇒



⎛ 10 ⎞ ⇒ t1 = τ log e ⎜ ⎟ ⎝ 9 ⎠

=

V2 1 ⎞ ⎛ 1 t = V 2t ⎜ + Rnet ⎝ R1 R2 ⎟⎠

H ⎞ ⎛ H = V 2t ⎜ 2 + 2 ⎟ ⎝ V t1 V t2 ⎠ t1t2 t1 + t2



Solving we get, t =



Hence, (A) and (C) are correct.

43.

I=



I = 0 for t =

dQ = a − 2bt , which decreases linearly with t. dt a dI and rate of charging is . So, 2b dt

dI = −2b dt Hence the correct options are (A), (C) and (D) Hence, (A), (C) and (D) are correct. 45. The resistance in the middle plays no part in the charging process of C, as it does not alter either the potential difference across the RC combination or the current through it.

03_Ch 3_Hints and Explanation_P2.indd 261

τ

10 9



t1 ⎛ 10 ⎞ = log e ⎜ ⎟ ⎝ 9 ⎠ τ

Q = 0.1 Q0 = Q0 e − t2

Similarly, R2 =





⇒ e t2



⇒

τ

τ

= 10



t2 = log e 10 = 2.303 τ ⇒ t2 = 2.303 τ



Hence, (B) and (C) are correct.

47.

Q = Q0 ( 1 − e − t1



⇒ e − t1



⇒ e t1



⎛ 10 ⎞ ⇒ t1 = τ log e ⎜ ⎟ ⎝ 9 ⎠

τ

τ

τ

) = 0.1Q0

= 0.9

=

10 9

Q = Q0 ( 1 − e − t2 ⇒ e

− t2 τ



⇒ e

t2 τ



⇒ t2 = τ log e ( 10 )



⇒



CHAPTER 3



τ

) = 0.9 Q0

= 0.1

= 10

( t1 + t2 ) = τ log e ⎛⎜⎝

100 ⎞ ⎛ 10 ⎞ ⎟ = 2τ log e ⎜⎝ ⎟⎠ 9 ⎠ 3

and ( t2 − t1 ) = 2τ log e ( 3 )

Hence, (B) and (D) are correct.

49. By closing S1, net external resistance will decrease. So, main current will increase. By closing S2 , net emf will remain unchanged but net internal resistance will decrease. Hence, main current will increase. Hence, (A) and (C) are correct. 50.

11.4 = I1 ( 9.5 )



⇒

I1 = 1.2 A

9/20/2019 11:51:42 AM

H.262  JEE Advanced Physics: Electrostatics and Current Electricity Since, E − 11.4 = I1r



⇒ E − 11.4 = 1.2r …(1) Similarly I 2 = 1 A

V2 = E2 − i2 r2  

⇒ E − 11.5 = 1r …(2) 0.1 = 0.2r

which is less than E2 . Hence, null point length will decrease. Hence, (A), (B), (C) and (D) are correct.



⇒

r = 0.5 W and E = 12 V



Hence, (A) and (B) are correct.

51.

Cs =

10 mF 7

Q = CsV =

Q 20 = =4V C1 5

Similarly VC2 =

Q 20 = = 10 V C2 2

Hence, (B) and (C) are correct.

52. At t = 0, emf of the circuit = PD across the capacitor = 6 V. 6 ⇒ i= =2A 1+ 2 Half-life of the circuit = ( ln 2 )τ C ( ln 2 ) CR = ( 6 ln 2 ) s.

In half-life time, all values get halved. For example

6 VC = = 3 V 2 2 i = = 1 A 1 ⇒ V1 W = iR = 1 V and V2 W = iR = 2 V

55. Current through A is the main current passing through the battery. So, this current is more than the current passing through B. Hence, during charging more heat is produced in A. In steady state, iC = 0 and iA = iB

⇒ Q = 20 mC

Further VC1 =



10 × 14 7

Hence, (A), (B), (C) and (D) are correct.

54. If switch S is open,



where, i1 = current in upper circuit and λ is resistance per unit length of potentiometer wire. E ⇒ Null point length,  = 2 i1λ (a) If jockey is shifted towards right, resistance in upper circuit will increase. So, current i1 will decrease. Hence,  will increase. (b) If E1 is increased, i1 will also increase, So,  will decrease.  ∝ E2 (c)

Hence, heat is produced at the same rate in A and B. Further, in steady state

VC = VB =

ε 2

1 1 CVC2 = Cε 2 2 8



⇒ U=



Hence, (A), (B) and (D) are correct.

Reasoning Based Questions 2.

3.

4. 5.

The electrons are in motion which constitute electric current in a conductor but no. of positive and negative charges are same. Hence, the correct answer is (D). When current flows through a conductor, it always remains uncharged. Hence no electric field is produced outside it. Hence, the correct answer is (A). Direction of flow of current is from higher potential to lower potential. Hence, the correct answer is (D).



Since current arises due to continuous flow of charged particles. There is no free charge in insulator. Hence no flow of charges is possible. Therefore current does not flow through insulators. Hence, the correct answer is (A).

6.

V = IR Hence, the correct answer is (D).

i1λ  = E2

03_Ch 3_Hints and Explanation_P2.indd 262

(d) If switch is closed, then null point will be obtained corresponding to

8.

Since, in the case of stretching the length n times, resistance becomes n2 times the original value. So, Statement 1 is false, however Statement 2 is true.  Because R = ρ A Hence, the correct answer is (D).

9/20/2019 11:51:53 AM

9.



On increasing temperature of wire, the kinetic energy of the electrons increases and so they collide more rapidly with each other as a result of which their drift velocity decreases and resistivity increases which is inversely proportional to the conductivity of material. Hence, the correct answer is (B).

1 0. Voltmeter gives terminal potential ( V ) thought it can

give emf if internal resistance of the cell is zero. Hence, the correct answer is (D).

12.

I1 =

E R

Req =

19.

P = I 2R



⎛ dP ⎞ ( ⎛ dI ⎞ ⇒ ⎜ 100% ) = 2 ⎜ ⎟ 100% ⎝ P ⎟⎠ ⎝ I ⎠



Hence, the correct answer is (B).

20. If either the e.m.f. of the driver cell or potential difference across whole potentiometer wire is lesser than the e.m.f. of the experimental cell, then balance point will not obtained. Hence, the correct answer is (A). 21. V = E + Ir when charging of cell take place, i.e., V > E Hence, the correct answer is (D).

RRV R + RV

22. The resistance of galvanometer is fixed. In metre bridge experiments, to protect the galvanometer from a high current, high resistance is connected to the galvanometer in order to protect it from damage (because then less current can pass through the galvanometer). Hence, the correct answer is (C).

E I 2 = ⎛ RRV ⎞ ⎜⎝ R + R ⎟⎠ V V

23.

Rv

R 2 =  1x G

R

R



l1

E

Hence, the correct answer is (A).

13. In a simple battery circuit the point at lowest potential is the negative terminal of battery. The current flows in the circuit from positive terminal to negative terminal. Hence, the correct answer is (D). 15. Internal resistance is always in series with the battery. Hence, the correct answer is (D). V R

25. Charge on capacitor q = CE ( 1 − e

16.

P=



Hence, the correct answer is (D).

At t = 0

17.

VA − E − Ir = VB

I =



⇒ VA − VB = E + Ir



E

B

A

E

I

Similarly VA − VB = E − Ir

Hence, the correct answer is (A).

18. Power used = I 2 R Hence, power is consumed and not the current Hence, the correct answer is (D).

03_Ch 3_Hints and Explanation_P2.indd 263

B

l2

To get null point at the same position, means  1 and  2 are still the same. As temperature increases, value of unknown resistance increases. To get the same null point, R must be increased. So Statement-1 is wrong. Statement-2 is True. Hence, the correct answer is (D).

2

A

X

R I2

I1

CHAPTER 3

Hints and Explanations H.263



⇒ I=

− t CReq

)

dq E − t CR e = dt Req E E = Req R + r

⇒ Resistance offered by capacitor is zero. Hence, the correct answer is (A).

Linked Comprehension Type Questions 1.

To find the charge per pulse let us integrate the expression dQ = Idt. Since the current to be constant for the entire duration of the pulse, so

∫

Qpulse = I dt = I Δt = ( 250 × 10 −3 A ) ( 200 × 10 −9 s )

9/20/2019 11:51:58 AM

H.264  JEE Advanced Physics: Electrostatics and Current Electricity

⇒

Qpulse = 5 × 10 −8 C

Dividing this quantity of charge per pulse by the electronic charge gives the number of electrons per pulse. So, we have Electrons per pulse =

5 × 10 −8 C Pulse = 1.6 × 10 −19 C electron



3.13 × 1011 electrons pulse



Hence, the correct answer is (C).

2.

Average current is given by Equation, I av =

ΔQ . Δt Because the time interval between pulses is 4 ms, and also we have already calculated the charge per pulse, so we obtain Qpulse 5 × 10 −8 C = = 12.5 μΑ I av = Δt 4 × 10 −3 s  This represents only 0.005% of the peak current, which is 250 mA. Hence, the correct answer is (A). 3.

Ppeak

pulse energy …(1) pulse duration

( 3.13 × 1011 electrons pulse ) ( 40 MeV electron ) = 2 × 10 −7 s pulse

⎛ 1.6 × 10 −13 J ⎞ ⎜⎝ ⎟ 1 MeV ⎠



⇒ Ppeak = 1 × 107 W = 10 MW

Also we could have calculated the power directly by assuming that each electron has zero energy before being accelerated. Thus, by definition, each electron must go through a potential difference of 40 MV to acquire a final energy of 40 MeV. Hence, we have Ppeak = I peak ΔV …(2)

⇒ Ppeak = ( 250 × 10 −3 A ) ( 40 × 106 V )



⇒ Ppeak = 10 MW



Hence, the correct answer is (B).

4.

To calculate the average power, we shall make use of the time interval between pulses rather than the duration of a pulse. So, the average power is given by



Pav =

( 3.13 × 1011 electrons pulse ) ( 40 MeV electron ) × 4 × 10 −3 s pulse

⎛ 1.6 × 10 −13 J ⎞ = 500 W ⎝⎜ 1 MeV ⎠⎟



 Otherwise we could have calculated the average power using the similar formula where we have the average power given by the relation Pav = I av ΔV = ( 12.5 × 10 −6 A ) ( 40 × 106 V )

⇒ Pav = 500 W

 Notice that these two calculations agree with each other and that the average power is much lower than the peak power. Hence, the correct answer is (D). 5.

For ammeter 99I g = ( I − I g ) 1

⇒ I = 100 I g …(1)

 I g is the full scale deflection current of the galvanometer I and the range of ammeter

Since, power is defined as the energy delivered per unit time interval. Thus, the peak power is equal to the energy delivered by a pulse divided by the pulse duration.

Ppeak =





Pav =

pulse energy time interval between pulses

03_Ch 3_Hints and Explanation_P2.indd 264

I

Ig 99 Ω G

1Ω



For the circuit in figure, given in the question



12V =3A 99 × 1 2+r+ 99 + 1



⇒ r = 1.01 W



Hence, the correct answer is (C).

6.

The correct answer is (C).

7.

Hence, the correct answer is (B).



Combined Solution to 6 & 7 For voltmeter, range

V = I g ( 99 + 101 ) V = 200 I g …(2) Also resistance of the voltmeter is RV = 99 + 101 = 200 W. In figure resistance across the terminals of the battery is given by R1 = r +

200 × 2 = 2.99 W 200

⇒ Current drawn from the battery,

I1 =

12 = 4.01 A 2.99

9/24/2019 11:38:05 AM

Hints and Explanations H.265 12.

⇒ Voltmeter reading Ig

99 Ω 101 Ω G

The equivalent circuit is as shown in figure. I1

A

2

8

16

16

1

24 V

Net resistance of circuit R = 6 W . Hence,

I =

When switch S is closed, a current,

I =

18 = 2 A, flows in the circuit 6+3

Vb − 0 = 3 × 2 = 6 V

⇒ Vb = 6 V



Hence, the correct answer is (C).

13.

Δq = ( Δq )3 μF = ( 18 − 54 ) μC = −36 μC



and Δq′ = ( Δq )6 μF = ( 72 − 108 ) μC = −36 μC



Hence, the correct answer is (A).

15. The correct answer is (B). 16. Hence, the correct answer is (D).

B





14. The correct answer is (C).

3

5

+ – 6 Ω a 18 μ C

I

4 V = 12 − I1r 5 4 ⇒ V = 12 − 4.01 × 1.01 5 5 ⇒ V = 7.96 × = 9.95 V 4 9.95 Using (2), I g = = 0.05 A 200 Using (1) range of the ammeter I = 100 I g = 5 A

I

0V

18 V

V

8.

72 μ C 3Ω + – b

CHAPTER 3





Combined Solution to 14, 15 & 16 Applying Junction Law, we get

I 3 W = 5 + 3 = 8 A

V 24 = =4A R 6



Hence, the correct answer is (B).

9.

I1 =



Hence, the correct answer is (C).

VAB 4 1 = = A 16 16 4

1A

g

= E − Ir = 24 − ( 4 ) ( 5 ) = 4 V

10.

VAB



Hence, the correct answer is (A).

11. When switch S is open circuit is as shown in figure. 108 μ C + –

R

a 2A

E1

h 7A

4Ω

3Ω

3A

8A f



b E2

e

c 5A 6Ω d

Applying Kirchhoff’s Loop Law, we have for

Loop ghefg, E1 − 12 − 24 = 0

⇒ E1 = 36 V

Loop chedc, − E2 + 24 + 30 = 0 +18 V

0V + – 54 μ C

Va = +18 V and Vb = 0

⇒ Vab = 18 V



Hence, the correct answer is (B).

03_Ch 3_Hints and Explanation_P2.indd 265



⇒ E2 = 54 V

Loop cbagc, −2R − E1 + E2 = 0

⇒ R=

E2 − E1 =9W 2

17. Time constant of circuit is ⎛ K ε 0 A ⎞ ⎛ ρd ⎞ τ = CR = ⎜ = Kε 0 ρ ⎝ d ⎟⎠ ⎜⎝ A ⎟⎠

9/24/2019 11:38:13 AM

H.266  JEE Advanced Physics: Electrostatics and Current Electricity



1 ⎛ ⎞( ⇒ τ = 18 ⎜ 4π × 10 3 ) ⎝ 4π × 9 × 109 ⎟⎠



⇒ τ = 2 × 10 −6 s = 2 ms



Hence, the correct answer is (B).

18. Since, q = q0

The capacitor is connected across points a and b. From KVL we have Va + 10 I1 − 40 I 2 = Vb Va − Vb = 40 I 2 − 10 I1 = 20 V

t − e CR −

2 2

21. When battery is disconnected, the equivalent circuit is shown here.

2 × 10 −6 = e



⇒ q = 2 × 10 −6 × e



⇒ q = 0.37 × 2 × 10 −6 = 0.74 mC



Hence, the correct answer is (D).

10 Ω

40 Ω 10 μ F

80 Ω

t − e CR

19.

q = q0



⇒ I=−



⇒ I=



1 1 2 − ⇒ I= e 2 = 2 = (As, e = 2.718 and e 2 = 7.39) 2 7.39 e



⇒ I = 0.13 mA



Hence, the correct answer is (D).



t

− dq ⎛ 1 ⎞ = − q0 e CR ⎜ − ⎝ CR ⎟⎠ dt

q0 − CR e CR 4

20. In steady state there is no current in the capacitor, the equivalent circuit is shown. Req = 36 W The current I is given by

⎛ 36 ⎞ I = ⎜ A=1A ⎝ 36 ⎟⎠



⇒ Q0 = 200 mC



The time constant of circuit is

⎛ 100 ⎞ ⎛ 1000 ⎞ × 10 ⎟ ms = ⎜ ms τ = RC = ⎜ ⎝ 3 ⎠ ⎝ 3 ⎟⎠ The charge on a capacitor, Q as a function of time in a discharging circuit is given by the relation Q = Q0 e

t RC

−

−

t ⎛ 1000 ⎞ ⎟ ⎜⎝ 3 ⎠



⇒ Q = 200 e



⇒ Q = 200 e



Hence, the correct answer is (C).

−

3t 1000

40 Ω

a

I2

b

CV = CV0 e 36 Ω

36 V

20 Ω

80 Ω I

The current divides into two parallel branches, as I1 and I 2 . In parallel branches current divides in inverse ratio of resistance, thus ⎛ 60 ⎞ I = 0.4 A  I1 = ⎜ ⎝ 90 + 60 ⎟⎠

{∵

⎛ 90 ⎞ I = 0.6 A  I 2 = ⎜ ⎝ 90 + 60 ⎟⎠

{∵ I = 1 A }

03_Ch 3_Hints and Explanation_P2.indd 266

3t ⎞ ⎛ = 200 exp ⎜ − mC ⎝ 1000 ⎟⎠

22. Voltage, V across capacitor as a function of time is

10 Ω I1

100 3

The initial charge on capacitor is

I

36 V

20 Ω

10 μ F

Q0 = ( 10 × 20 ) mC

t



Thus, the voltage across capacitor is 20 V Hence, the correct answer is (C).

I = 1 A}

−

⇒ V = V0 e



⇒ V = 200 e



t RC

t RC



I = 200 e

−

−

−

t RC

t RC

Taking log both sides, we get

t = RC log e 200 =

1000 log e 200 3

t = 1766.1 ms



⇒



⇒ t = 1.766 ms



Hence, the correct answer is (B).

9/20/2019 11:52:26 AM

Hints and Explanations H.267 τ 1 = RC = ( 1.5 × 10 5 W ) ( 10 × 10 −6 F ) = 1.5 s



Hence, the correct answer is (C).

24.

τ 2 = ( 1 × 10 5 W ) ( 10 × 10 −6 F ) = 1 s



Hence, the correct answer is (B).





10 V = 200 mA 50 × 10 3 W



10 V ⎛ ⎞ =⎜ ⎝ 100 × 10 3 W ⎟⎠

t − e 1s

= 100 e − t mA

So the switch carries downward current, I ′ given by

I ′ = 200 mA+ ( 100 mA ) e

−

t 1s



⇒ I ′ = 100 ( 2 + e − t ) mA



Hence, the correct answer is (C).

26. Taking loop equation, we get

E−

dI I =− dt Cf

q + RI = 0  [C f is the capacitance at t > 0]…(1) Cf



∫



⇒ I = I0e





t ⎞ ⎛ − RC f ⎟ ⎜ ⇒ q = C f ⎝ E + RI 0 e ⎠



⇒ q = ( 5 ) 10 + 10 e



⇒ q = ( 5 × 10 −5 ) 1 + e

(

−

(

t 2.5 × 10 −4

−

)

t 2.5 × 10 −4

q = 5 × 10 −5 ( 1 + e −4000t )

)C



⇒ q = 50 ( 1 + e −4000t ) mC



Hence, the correct answer is (C).

29. Power in resistor = I 2 R Hence, heat produced in the wire as function of time is H , given by t

H=

t

∫ ( I R ) dt = I R∫ e 2

2 0

0

{ Ci = initial capacitance }

−

2t RC f

0

2t − I 02 R2C f ⎛⎜ − RC f ⎞⎟ ⎝ ⎠ dt = e 2



⇒ H=+

5 × 10 1 = A 5 × 50 5 Hence, the correct answer is (A).



⇒ H=



⇒ H = ( 2.5 × 10 −4 )[ 1 − e −8000t ] J



⇒ H = 250 ( 1 − e −8000t ) m J



Hence, the correct answer is (C).

27. Differentiating equation (1), we get R

dI I dq = dt C f dt

03_Ch 3_Hints and Explanation_P2.indd 267

t 0

2t ⎞ − I 02 R2C f ⎛⎜ ⎝ 1 − e RC f ⎟⎠ 2

putting the values, we get

I 0 =

t

t



RC f

…(2)

− q RC f + RI 0 e =0 E − Cf

where q0 is charge on capacitor at t = 0 , q0 − EC f

t RC f

Hence, the correct answer is (B).

⇒

q E − 0 + RI 0 = 0 Cf

−

28. Substituting I in equation ( 1 ) we get

E

As charge on the capacitor cannot change abruptly, the current I 0 , just after the capacitance has been changed, will be given by

0

−

I

As q0 = Ci E 

∫

−4 Hence, I = ( 0.2 A ) e ( 2.5 ×10 ) = 0.2 exp ( −4000t ) A where t is in seconds.

R

Hence, I 0 =

t

dI dt =− ⇒ R I Cf Ι0

The 100 kW carries current of magnitude I = I o

⇒ R

Ι

25. The battery carries current

t − e RC

dq dt

I= as

CHAPTER 3

23.

( 0.2 × 50 )2 × ( 5 × 10 −6 ) 2

[ 1 − e −8000t ] J

9/20/2019 11:52:37 AM

10 Ω

10 Ω

10 Ω

10 Ω

No current

H.268  JEE Advanced Physics: Electrostatics and Current Electricity 10 Ω

30. Finally, the capacitors are in parallel and total charge ( = q0 ) distributes between them in direct ratio of capacity.

⎛ C2 ⎞ q0 → in steady state. ⇒ qC2 = ⎜ ⎝ C1 + C2 ⎟⎠



But this charge increases exponentially. Hence, charge on C2 at any time t is

qC2

(

t

− ⎛ C2 q0 ⎞ =⎜ 1 − e τC ⎟ ⎝ C1 + C2 ⎠



10 Ω 20 Ω 3

10 Ω

Hence, the correct answer is (B).

31. Common potential in steady state when they finally come in parallel is V =

20 Ω 3

)

Initially, C2 is uncharged so, whatever is the charge on C2 , it is charge flown through switches.





q2 1 ⎛ q0 ⎞ ⇒ ΔH = 0 − ( C1 + C2 ) ⎜ 2C1 2 ⎝ C1 + C2 ⎟⎠



⎛ q2 ⎞ ⎛ C C ⎞ ⇒ ΔH = ⎜ 0 ⎟ ⎜ 1 2 ⎟ ⎝ 2C1 ⎠ ⎝ C1 + C2 ⎠



Hence, the correct answer is (A).

I =

5 + 10 2

70 25 A and I1 = A 33 33

10 Ω

2

10 Ω

10 Ω

20 Ω I1 I

(2I1 – I)

20 Ω 3 (I – I1) 10 Ω

I

I1 20 Ω

10 Ω

No current

10 Ω

10 Ω

10 Ω S

20 Ω 3

10 Ω

10 Ω

(I – I1)

10 Ω

10 Ω

10 Ω

10 Ω

=4A

10 Ω

10 Ω

10 Ω

32. Immediately after closing the switch a capacitor behaves as a short circuit. Figure shows the effective circuit from the point of view of current. Figure shows the equivalent circuit, current Ι from battery is I =

S

Hence, the correct answer is (B).

10 Ω

Total heat dissipated is ΔH = U i − u f

50

50 V

33. A long time after closing the switch a capacitor behaves as an open circuit. The equivalent circuits are shown. Using Kirchhoff’s Laws, we get

Total charge q0 = Total capacity C1 + C2



S

50 V

10 Ω 10 Ω

10 Ω

Hence, the correct answer is (C).

34. The 10 mF capacitor is connected across the 20 W resistor of figure (see question)

No current

50 V



50 V

S

P.D. across 20 W resistor = 20 × I1 =

500 V 33

So, charge on 10 mF capacitor is given by

20 Ω 3 10 Ω 20 Ω

03_Ch 3_Hints and Explanation_P2.indd 268 3

9/20/2019 11:52:42 AM

Hints and Explanations H.269 C

500 ⎞ 5000 ⎛ q10 = ⎜ 10 × mC ⎟⎠ mC = ⎝ 33 33

Hence, the correct answer is (C).

35. Immediately after closing switch S1 the capacitor C1 short-circuits the part of circuit to the right of it. So, the current will flow through the loop ABGHA only, with a value V 12 = 0.12 A I 0 = I ( t = 0 ) = = R 100

Hence, the correct answer is (C).

36. Long time after closing the switches the capacitors do not allow the current to pass through them. So, the current will now flow in the loop ABCFGHA (with no capacitor in the branch BG ). So, at t → ∞, the net resistance offered to the flow of current is given by Req = 100 + 50 + 150 = 300 W V 12 = = 0.04 A Req 300



⇒ I = I (t → ∞ ) =



Hence, the correct answer is (B).

37.

C1 is connected across points B and G . We traverse the circuit along BAHGB and apply KVL, we have VB + 100 I − 12 − VG = 0 VB − VG = 12 − 100 I = 12 − ( 100 )( 0.04 ) = 8 V



Final voltage across C1 is 8 V Hence, the correct answer is (D).

38. Capacitor C2 is connected across the 150 W resistor, hence voltage across C2 is V2 = ( 150 )( 0.04 ) = 6 V

Hence, the correct answer is (C).

39. When switch S2 is opened, the loop CDEF is a discharging CR circuit. The charge as function of time is Q = Q0 e

⎛ t ⎞ −⎜ ⎝ RC2 ⎠⎟

⇒ Q = ( C2 × 6 ) e

⎛ t ⎞ −⎜ ⎝ RC2 ⎠⎟

dQ Current I = dt ⎛



⇒ I=−

t ⎞

6C2 − ⎜⎝ RC2 ⎟⎠ e RC2

03_Ch 3_Hints and Explanation_P2.indd 269



{∵ Q0 = C2V2 = 6C2 }

F

E

t ⎞

⎛



⇒ I=−

6 − ⎝⎜ RC2 ⎠⎟ e R



⇒ I=−

6 − ( 50 ×10 −6 )( 150 ) e 150



⇒ I = −0.04 e



⎛ 400t ⎞ ⇒ I = −0.04 exp ⎜ − ⎟ ⎝ 3 ⎠



Hence, the correct answer is (C).

t

−

400 t 3

Matrix Match/Column Match Type Questions

CHAPTER 3



C2 = 50 μ F

50 Ω

The 5 mF capacitor is connected in parallel to the 20 W resistor hence charge on the 5 mF capacitor, is 2500 ⎛ 5 × 500 ⎞ mC = mC q5 = ⎜ ⎝ 33 ⎟⎠ 33

D

1. A → (q) B → (s) C → (p) D → (r) 2. A → (s) B → (q) C → (q) D → (p)  ρ or R ∝ A A ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⇒ R1 : R2 : R3 = ⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ ⎝ A ⎠1 ⎝ A ⎠2 ⎝ A ⎠3

Since R =

1 1 1 : : = 1:1: 6 2 2 3



⇒ R1 : R2 : R3 =



Hence (A) → (s) As V is constant so

I ∝

1 1 and P ∝ , so P ∝ I R R

⎛ 2⎞ ⎛ 2⎞ ⎛ 1⎞ Hence, I1 : I 2 : I 3 = P1 : P2 : P3 = ⎜ ⎟ : ⎜ ⎟ : ⎜ ⎟ ⎝ 1⎠ ⎝ 1⎠ ⎝ 3⎠ ⇒ I1 : I 2 : I 3 = 6 : 6 : 1 So (B) → (q), (C) → (q) Since wires are of same material, so ρ1 : ρ2 : ρ3 = 1 : 1 : 1

Hence (D) → (p)

3. A → (q, r, s) B → (q, s) C → (p, s) D → (p, s)

9/20/2019 11:52:53 AM

H.270  JEE Advanced Physics: Electrostatics and Current Electricity Since,



q = q0 ( 1 − e

− t RC

)

I ae + I be + I ce + I de = 0

So, the charging current is

I =

dq V − t RC = e dt R



V − t RC = I 0 e − t RC e R At t = 0, we get

Let potential of point e is V volts. Then,

⇒ I=

V I = I 0 = R Hence current is maximum and capacitor acts as a short circuit. So, (A) → (q, r, s) The decay charge is given by q = q0 e − t RC

⎛ 2 −V ⎞ ⎛ 4 −V ⎞ ⎛ 6 −V ⎞ ⎛ 4 −V ⎞ ⇒ ⎜ + + + =0 ⎝ 1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 1 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⇒ V=4V

Now current through any wire can be obtained by the equation, I =

ΔV R

5. A → (q) B → (q) C → (s) D → (r) For series combination Rext = R , Rint = 2r and Enet = 2E



Hence, the discharging current is dq I = − = I 0 e − t RC dt

E

r

⇒ I = I 0 e − t RC

V (maximum) R Hence, (B) → (q, s)  Now, when the capacitor is fully charged, energy stored is I = I 0 =

1 CV 2 2 2 1 ⇒ U = C ⎡⎣ V0 ( 1 − e − t RC ) ⎤⎦ 2 2 1 ⇒ U = CV02 ( 1 − e − t RC ) 2

U =





Acc to Maximum Power Transfer Theorem

R = 2r ⎡ 4E2 ⎤ E2 E2 4E2 ⇒ P=⎢ R= R= = 2 ⎥ 2 R 2r 4R ⎣ ( R + 2r ) ⎦ For parallel combination r Enet = E , Rext = R, Rint = 2 Acc to Maximum Power Transfer Theorem,

Rext = Rint

Hence, U is an exponential function of t So, (C) → (p, s) Energy dissipated as heat is dH = I 2 R dt ∞



⇒

∫ dH = ∫ ( I e 0



1 ⇒ H = CV 2 2 Hence, (D) → (p, s)

4. A → (q) B → (s) C → (q) D → (s)

03_Ch 3_Hints and Explanation_P2.indd 270

r

E

r

E

R

)

− t RC 2

R dt



0



r

R

at t = 0 , we have



E



r 2 ⇒ r = 2R ⇒ R=

Now P =



⇒ P=

⎛ E2 ⎞ R=⎜ R ⎝ 4 R2 ⎟⎠ r⎞ ⎛ ⎜⎝ R + ⎟⎠ 2 E2

2

E2 E2 E2 = = 4R ⎛ r ⎞ 2r 4⎜ ⎟ ⎝ 2⎠

9/20/2019 11:53:03 AM

Hints and Explanations H.271





i1 = i2 or i is same at both sections.

RA =

VA2 P

and RB =

VB2 P

A1 < A2 i 1 (A) Current density = ∝ A A Resistance ρ 1 (C) = ∝ A A length

(D) and (B) E or potential difference per unit length = ( i ) (Resistance per unit length)



⇒

P.D. ⎛ ρ⎞ 1 = ( i )⎜ ⎟ ∝ ⎝ A⎠ A 

7. A → (q) B → (r) C → (t) D → (p)



⇒

RA VA2 = RB VB2

 When the bulbs are connected in series, current through them will be same. Hence, potential difference across them will be proportional to their resistances. Therefore, ratio of potential difference across them will be equal to ratio of their resistances, RA : RB which is VA2 : VB2

So, (A) → (p, q) Power consumed will be

PB = I 2 RB and

R(V ) 3 × 6 18 = = = 1.8 V ( R + S ) 3 + 7 10

Hence (A) → (q)

Similarly, VA − VB =

X (V ) 2×6 = =2V (X + Y) 2 + 4

( VA − VB ) − ( VA − VD ) = VD − VB = 2 − 1.8 = 0.2 V Hence (B) → (r) Energy stored in the capacitor will be zero if VB = VD . So for condition of a balanced wheatstone bridge, we have 2 3 = Y 7 14 ⇒ Y= W 3





PA = I 2 RA

As 3 W and 7 W are in series across 6 V and in series potential divides in proportion to resistance So, VA − VD =

Resistance of bulbs A and B are

CHAPTER 3

6. A → (p) B → (p) C → (p) D → (p)

So (C) → (t)  Also, in steady state no current flows through the branch containing the capacitor. Hence (D) → (p) 8. A → (r) B → (s) C → (q) D → (p) 9. A → (p, q) B → (p, q) C → (r, s) D → (r, s)

03_Ch 3_Hints and Explanation_P2.indd 271



⇒

PA RA VA2 = = PB RB VB2

Hence, (B) → (p, q) If bulbs are connected in parallel, then I1RA = I 2 RB I1 RB VB2 = = I 2 RA VA2



⇒



So, (C) → (r, s) and power consumed is given by

V2 P R V2 R A = A2 = B = B2 PB V RA VA RB

Hence, (D) → (r, s)

10. A → (s) B → (r) C → (p) D → (q) 11. A → (s) B → (r) C → (s) D → (r) 4−1 = 1 A (anti-clockwise) 1+1+1 VA = E − ir = 4 − 1 × 1 = 3 V (A)



i=

VB = E + ir = 1 + 1 × 1 = 2 V (B)

9/20/2019 11:53:10 AM

H.272  JEE Advanced Physics: Electrostatics and Current Electricity 2 (C) PA = Ei − i 2 r = ( 4 × 1 ) − ( 1 ) ( 1 ) = 3 W



2

(D) PB = Ei + i r = ( 1 )( 1 ) + ( 1 ) ( 1 ) = 2 W 2



12. A → (s) B → (q) C → (r) D → (p)

7R 12 So, (A) → (s) ⇒ Rab =

Resistance between b and c can be calculated as shown. Due to symmetry condition, network can be folded along the line bc. The circuit looks like e, f

e

R/2

g R/2 a

b

c

d

R/2

h

a f

Symmetry of circuit shows that e and f are at same potential and g and h are another same potential. So, this circuit can be redrawn as shown.

R/2

g, h R/2

c

b

R

R

d

 Finally it becomes balanced wheatsone bridge, so R resistance of between g and e can be neglected. 2 Equivalent resistance between b and c is the parallel combination of R and 3R g, h

a

e f

d

c

g

b

R/2

R/2

h b

Equivalent resistance in the middle line between e and h is

The total equivalent resistance between e and h is R′ such that 1 1 1 1 = + + R′ R R 2 R 2R ⇒ R′ = 5



The equivalent resistance between a and b along path aeb

R 2R R 7 R + + = 2 5 2 5

The total equivalent resistance between a and b is the 7R resistance of R and in parallel. 5

7 R 7 R2 5 = 5 = 7R ⇒ Rab = 7 R 12R 12 R+ 5 5 R×

03_Ch 3_Hints and Explanation_P2.indd 272

3R/2

3R/2

R R + R + = 2R 2 2

c

R/2

e, f

So, Rbc =

R × 3R 3R = 4R 4



So (B) → (q)



Similarly, we can get Rac =



So, (C) → (r)



when c and h an shorted, Rch = 0



So, (D) → (p)

5R 6

13. A → (q) B → (p) C → (r) D → (s)

Apply Kirchhoff’s Law

Va =

q q + 12 + − 24 = Va 2 4

9/20/2019 11:53:15 AM

Hints and Explanations H.273 −1



α ⎞ ⎛ ⇒ Rc = R ⎜ 1 − c ⎟ α ⎝ n⎠



⎡ ( 0.4 × 10 −3 °C −1 ) ⎤ ⇒ Rn = 10 kW ⎢ 1 − ⎥ ( −0.5 × 10 −3 °C −1 ) ⎦ ⎣



⇒ Rn ≅ 5560 W and Rc = 4440 W



3.

The drift velocity is I = nqvd A = nqvd ( π r 2 )





⇒ vd =



⇒ vd = 2.34 × 10 −4 ms −1

+q 2 μF –q

c –q 4 μF +q

a

d 24 V

3 q = 12 4 ⇒ q = 16 mC

16 Va + = Vb 2

⇒ Va − Vb = −8V

Similarly Vd − Vc = −4 V

Work done by cell 12 V is −12 × 16 = 192 m J



Work done by cell 24 V is 24 × 16 = 384 m J

Integer/Numerical Answer Type Questions 1.

Since, I = nqAvd , where n is the number of charge carriers per unit volume, and is identical to the number of atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationship above. For aluminium, which has a molar mass of 27, we know that Avogadro’s number of atoms, N A , has a mass of 27 g . Thus, the mass per atom is



27 g 27 g = = 4.49 × 10 −23 gatom −1 NA 6.02 × 10 23

Thus, n =

⇒ vd = 130 × 10 −6 ms −1 = 130 mms −1

2.

R = Rc + Rn = Rc ⎡⎣ 1 + α c ( T − T0 ) ⎤⎦ + Rn ⎡⎣ 1 + α n ( T − T0 ) ⎤⎦



0 = Rcα c ( T − T0 ) + Rnα n ( T − T0 )



α ⇒ Rc = − Rn n αc

So, R = Rc + Rn = − Rn

α ⎞ ⎛ ⇒ Rn = R ⎜ 1 − n ⎟ αc ⎠ ⎝

03_Ch 3_Hints and Explanation_P2.indd 273

−1

αn + Rn αc

) (

)

2

x t

x 200 × 10 3 m = = 8.54 × 108 s = 27 yr v 2.34 × 10 −4 ms −1

⇒ t=

4.

R = R0 ⎡⎣ 1 + α ( T − T0 ) ⎤⎦



⇒ T = T0 +



In this case, I =



⇒ T = T0 +

5.

Label the currents in the branches as shown in the first figure. Reduce the circuit by combining the two parallel resistors as shown in the second figure, we get the resistance in the middle branch as (1.71) R. b

Therefore, I 5A vd = = nqA ( 6.02 × 10 28 m −3 ) ( 1.6 × 10 −19 C ) ( 4 × 10 −6 m 2 )



(



⇒ n = 6.02 × 10 22 atomscm −3 = 6.02 × 10 28 atomsm −3

⇒ vd = 1.3 × 10 −4 ms −1

1000 A I = nqπ r 2 8.49 × 10 28 m −3 1.6 × 10 −19 C π 10 −2 m

Since v =

density of aluminium 2.7 gcm −3 = mass per atom 4.49 × 10 −23 gatom −1



−1

R 250 V I4

I0 10

1 9 ( 9 ) = 20° + = 2020 °C α 0.00450 °C −1

500 V

4R I1 a

1⎡ R 1⎡I ⎤ ⎤ − 1 ⎥ = T0 + ⎢ 0 − 1 ⎥ α ⎢⎣ R0 α I ⎣ ⎦ ⎦

2R

c

3R I

CHAPTER 3

12 V

b

d

b

I2 e

I1 a

R

c

2R 500 V

250 V

d

1.71R I3

Fig. (a)

I1 + I2 a′

I2 e

Fig. (b)

Apply Kirchhoff’s Loop Rule to both loops in figure (b) to obtain ( 2.71R ) I1 + ( 1.71R ) I 2 = 250

( 1.71R ) I1 + ( 3.71R ) I 2 = 500 and With R = 1000 W, simultaneous solution of these ­equations yields I1 = 10 mA and I 2 = 130 mA

From figure (b), Vc − Va = ( I1 + I 2 ) ( 1.71R ) = 240 V

9/20/2019 11:53:27 AM

H.274  JEE Advanced Physics: Electrostatics and Current Electricity Vc − Va 240 V = = 60 mA 4R 4000 W Finally, Applying Kirchhoff’s Point Rule at point a in figure (a) gives :



I = I 4 − I1 = 60 mA − 10 mA = +50 mA,



Thus, from figure (a), I 4 =

Then, the three equations become

ΔVab = 2 x − y ,

⇒ y = 2 x − ΔVab

ΔVab = −4 x + 6 y + 5

⇒ I = 50 mA from point a to point e.





⇒ ΔVab = 8 − 8 x + 5 y

6.



Substituting the first into the last two gives

Name the currents as shown in the figure. Then I1 + I 2 + I 4 = I 3. Loop equations are

7 ΔVab = 8 x + 5

Loop abgha

−200 I1 − 40 + 80 I 2 = 0

Loop bcfgb

−80 I 2 + 40 + 360 − 20 I 3 = 0

Loop cfedc

+360 − 20 I 3 − 70 I 4 + 80 = 0 a I1

b

40 V 200 Ω

c 360 V

80 Ω

I3

20 Ω g

h

d

I2

I4 80 V

27 Solving these simultaneously yields ΔVab = V 17 27 ΔVab 17 V = Then, Rab = I 1A 27 x ⇒ Rab = W = 1+ 17 17 ⇒ x = 10 8.

(a) RC = ( 1 × 106 W ) ( 5 × 10 −6 F ) = 5 s

(b) Q = CE = ( 5 × 10 −6 C ) ( 30 V ) = 150 mC

e

I (t )= (c)

Eliminate I 3 by substitution I 2 = 2.5I1 + 0.5 ⎧ ⎪ 400 100 I 2 − 20 I1 − 200 I 4 = 0 − ⎨ ⎪ 440 − 20 I − 20 I − 90 I = 0 1 2 4 ⎩



⇒ 6 ΔVab = 2 x + 8



70 Ω f



t

E − RC ⎛ 30 ⎞ −10 ⎡ ⎤ e =⎜ exp ⎢ 6⎟ 6 −6 ⎥ R ⎝ 1 × 10 ⎠ × 1 10 5 10 × ⎣ ⎦

(

t ⎤ ⎡ − q ( t ) = Q ⎢⎣ 1 − e RC ⎥⎦



⇒

430 − 70 I1 − 1575 + 1215I1 = 0



⇒ 0.6 = 1 − e

70 = 1 A upward in 200 W 70 Now I 4 = 4 A upward in 70 W



⇒ e



0.9 = log e ( 0.4 ) ⇒ − RC



⇒ RC =

t

Eliminate I 4 = 17.5 − 13.5I1 to obtain

I1 =

I 2 = 3 A upward in 80 W I 3 = 8 A downward in 20 W

and for the 200 W , ΔV = IR = ( 1 A ) ( 200 W ) = 200 V

7.

ΔVab = ( 1 ) I1 + ( 1 ) ( I1 − I 2 )

1 Ω I1 I1 – I2 1 Ω

I

1Ω

3Ω I – I1

I2

I

5Ω

I – I1 + I2

ΔVab = ( 3 ) ( I − I1 ) + ( 5 ) ( I − I1 + I 2 )

Let I = 1 A, I1 = x , and I 2 = y

03_Ch 3_Hints and Explanation_P2.indd 274

− q(t ) = 1 − e RC Q

−

0.9 RC

−

0.9 RC

= 1 − 0.6 = 0.4

−0.9 = 0.982 s  1 s log e ( 0.4 )

10. The potential difference across the capacitor

ΔVab = ( 1 ) I1 + ( 1 ) I 2 + ( 5 ) ( I1 − I1 + I 2 ) a

)

= 4.06 mA  4 mA 9.

⎧ 350 − 270 I1 − 20 I 4 = 0 Eliminate I 2 ⋅ ⎨ ⎩ 430 − 70 I1 − 90 I 4 = 0

)(

b

(

ΔV ( t ) = ΔVmax 1 − e

−

t RC

)

−( 3 s ) ⎡ ⎤ ⎡⎣ R( 10 × 10 −6 sW −1 ) ⎤⎦ ⎥ ⎢ ⇒ 4 V = ( 10 V ) ⎣ 1 − e ⎦

Therefore, 0.4 = 1 − e −

−

( 3 ×105 W ) R

( 3 ×105 W )



⇒ e



Taking the natural logarithm of both sides,

R

= 0.6

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Hints and Explanations H.275



⇒ R=−

3 × 10 5 W = + 5.87 × 10 5 W = 587 kW log e ( 0.6 )

11. Let the two resistances be x and y. Then, Rs = x + y =

Ps 225 W = =9W I 2 ( 5 A )2 y

x

y = 9 W − x



⇒

Pp xy 50 W = 2 = =2W x+y I ( 5 A )2

x (9 W − x ) =2W x + (9 W − x )

Factoring the second equation, ( x − 6 ) ( x − 3 ) = 0



⇒ x=6W ⇒ x=3W

⇒ I =

= 40 W

P = Req

40 W 1 = A 360 W 3

The potential difference across R1 is

P 2.4 VA = = 18.5 mA V R2 7000 A I



⇒ y=6W



The two resistances are found to be 6 W and 3 W

12. (a) First determine the resistance of each light bulb:

( ΔV )2 R

( ΔV )

360 W

13. (a) W  ith the switch closed, current exists in a simple series circuit as shown. The capacitors carry no current. For R2 we have

Then, y = 9 W − x gives y = 3 W

R =

( 120 V )2

(b)  The current through the network is given by P = I 2 Req

⇒ I =



P=

Req

=

P = I 2 R2

x 2 − 9 x + 18 = 0





( ΔV )2

1 ⎛ 1 ⎞⎛ ⎞ ΔV23 = IR23 = ⎜ A ⎟ = 40 V   ⎝ 3 ⎠⎜ 1 1 ⎟ + ⎜⎝ ⎟⎠ 240 W 240 W

y

⇒ RP =

  P=

⎛1 ⎞ ΔV1 = IR1 = ⎜ A ⎟ ( 240 W ) = 80 V   ⎝3 ⎠ The potential difference ΔV23 across the parallel combination of R2 and R3 is

x



The total power dissipated in the 360 W is

CHAPTER 3

5 − 3 × 10 W = log e ( 0.6 ) R

2

P

=

( 120 V ) 60 W

2

= 240 W

I

C1 = 3 μ F R1 = 4 kΩ R2 = 7 kΩ



C2 = 6 μ F

The potential difference across R1 and C1 is

V⎞ ⎛   ΔV = IR1 = ( 1.85 × 10 −2 A ) ⎜ 4000 ⎟ = 74.1 V ⎝ A⎠ The charge on C1 is Q1 = C1 ΔV = ( 3 × 10 −6 CV −1 ) ( 74.1 V ) = 222 mC  

R1 120 V

R2

R3

  ΔV = IR2 = ( 1.85 × 10 −2 A ) ( 7000 W ) = 130 V

We obtain the equivalent resistance Req of the network of light bulbs by identifying series and parallel equivalent resistances Req = R1 +

03_Ch 3_Hints and Explanation_P2.indd 275

1 ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜⎝ R ⎟⎠ + ⎜⎝ R ⎟⎠ 2 3

= 240 W + 120 W = 360 W

The potential difference across R2 and C2 is

The charge on C2 is

C⎞ ⎛ Q2 = C2 ΔV = ⎜ 6 × 10 −6   ⎟ ( 130 V ) = 778 mC ⎝ V⎠ The battery emf is given by IReq = I ( R1 + R2 ) −2 ⇒ IReq = 1.85 × 10 ( 4000 + 7000 ) = 204 V

9/20/2019 11:53:54 AM

H.276  JEE Advanced Physics: Electrostatics and Current Electricity

(b) In equilibrium, after the switch has been opened for a long time, no current can flow in the c­ ircuit. The potential difference across each resistor is zero. The full 204 V appears across both ­capacitors. The new charge C2 is

 ′ Q2 = C2 ΔV = ( 6 × 10

−6

CV

−1

I0 = ⇒  I 0 =

) ( 204 V ) = 1222 mC

( ΔV )C IR2 = + R R R ( 2 3 ) ( 2 + R3 )

( 333 mA ) ( 15 kW ) = 278 mA ( 15 kW + 3 kW )

 Thus, when the switch is opened, the current through R2 changes instantaneously from 333 mA (downward) to 278 mA (downward) as shown in the graph. I2 ( μ A)

C2 = 6 μ F

Switch opened

333



Change in the value of charge on Q2 is given by

278

  ΔQ2 = Q2′ − Q2 = 1222 mC − 778 mC = 444 mC

t=0

14. (a) A  fter steady state conditions have been reached, there is no DC current through the capacitor Thus, for R3 , we have I R3 = 0 (steady state)



E

9V

R1 15 kΩ

C R2 R3

10 μ F

Afterwards, it decays according to

( ) −t R + R C I = I 0 e ( 2 3 ) = ( 278 mA ) e − t 0.18 s ( for t > 0 )  

12 kΩ S

t



(d) The charge q on the capacitor decays from Q0 to Q0 according to 5

3 kΩ

−t R + R C q = Q0 e ( 2 3 )

For the other two resistors, the steady state current is simply determined by the 9 V emf across the 12 kW and 15 kW resistors in series For R1 and R2 , we have I( R1 + R2 ) =  

E 9V = = 333 mA R1 + R2 ( 12 kW + 15 kW )

 (steady state) (b)  After the transient currents have ceased, the potential difference across C is the same as the potential difference across R2 ( = IR2 ) because there is no voltage drop across R3 . Therefore, the charge Q on C is Q = C ( ΔV )R2 = C ( IR2 ) = ( 10 mF )( 333 mA ) ( 15 kW ) ⇒ Q = 50 mC

(c) When the switch is opened, the branch containing R1 is no longer a part of the circuit. The capacitor discharges through ( R2 + R3 ) with a time constant of ( R2 + R3 ) C = ( 15 kW + 3 kW ) ( 10 mF ) = 0.18 s  The initial current I 0 in this discharge circuit is determined by the initial potential difference across the capacitor applied to ( R2 + R3 ) in series. So,

03_Ch 3_Hints and Explanation_P2.indd 276

⇒

Q0 − t 0.18 s ) = Q0 e ( 5

⇒ 5 = e t 0.18 s ⇒ log e 5 =

t 180 ms

⇒ t = ( 0.18 s ) ( log e 5 ) = 290 ms 15. Since α can vary at different temperatures, so it is advisable to make a rough estimate R = R20 ( 1 + αΔT ) R − R20 α R20



⇒ ΔT =



⇒ T − 20 °C=



⇒ T = 2612 °C



⇒ T = 2612 + 273 = 2885 K

( 190 − 15 )

( 4.5 × 10 −3 ) ( 15 )

= 2592 °C

16. Let Rx and Ry be the respective resistances at a given temperature. The total resistance after a charge ΔT becomes R = Rx ( 1 + 0.0025 ΔT ) + Ry ( 1 + 0.00075 ΔT ) …(1)

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Hints and Explanations H.277

R = ( Rx + Ry ) ( 1 + 0.001 ΔT ) …(2)



From equations (1) and (2), we get

18. Heat produced in a resistance R in time t is H = Pt =

For coil 1,

Rx ( 1 + 0.0025 ΔT ) + Ry ( 1 + 0.00075 ΔT )

H1 =

= ( Rx + Ry ) ( 1 + 0.001 ΔT )





⇒ Rx ( 0.0015 ΔT ) = Ry ( 0.00025 ΔT )



Thus we get 5 Ry …(3) 3

Rx =

Also, Rx + Ry = 1000 W …(4)

On solving equations (3) and (4), we get

Rx = 625 W and Ry = 375 W

The respective lengths are

Lx =

1 × 625 = 6.25 km = 6250 m 100

and Ly =

1 × 375 = 7.5 km = 7500 m 50

17. The power consumed ( P ) in time t is given by dU V2 P = = dt R ( t )

⇒ U=

V

⇒ U=

∫ R ( t ) dt ( 110 )

∫e

0.5

⇒ R2 = 2R1 …(3)



(a) When both the coils are used in series, then

   Hs =

V2 V2 ts = ×t  ( R1 + R2 ) 3R1 s

{∵

R2 = 2R1 }

But as here, H s = H1 = H 2 = H

⇒

V2 V2 ( 15 × 60 ) = ts 3 R1 R1

⇒ ts = ( 45 × 60 ) s = 45 min (b) When both the coils are used in parallel, then

3V 2 V2 × tp = × ( 15 × 30 ) 2R1 R1

⇒ tp = ( 10 × 60 ) s = 10 min −2t

dt

0

( 110 )2



⇒

0



15 30 = R1 R2

⎛ V2 V2 ⎞ ⎛ 3V 2 ⎞ H P = H = ⎜ + tp = ⎜ tp {∵ R2 = 2R1 } ⎟ ⎝ R1 R2 ⎠ ⎝ 2R1 ⎟⎠

2

2 t

V2 ( 30 × 60 ) …(2) R2

But according to given problem H1 = H 2 = H i.e.,





V2 dt ⇒ dU = R(t ) t

V2 ( 15 × 60 ) …(1) R1

For coil 2,

H 2 =

V2 t R

19. Resistance of the wire at 300 K is R0 =

( e −2t )t0

ρL ρL π 2 × 10 −8 × 1 = π × 10 −2 W = = −3 )2 A πr2 ( π × 10



⇒ U=



⇒ U = ( 110 ) ( 1 − e −2t ) J

Rt = R0 ( 1 + α t )



According to problem,



⇒ Rt = π × 10 −2 [ 1 + 7.8 × 10 −3 × ( 900 − 300 ) ]



⇒ Rt = 5.68 π × 10 −2 W



The rate at which the wire radiates energy is

2 × 0.5 2

U = 7644 J 7644



⇒ 1 − e −2t =



⇒ e −2t = 0.367 ⇒ −2t log e e = log e 0.367



⇒ −2t = −1 ⇒ t = 0.5 s ⇒ t = 500 ms

( 110 )2

03_Ch 3_Hints and Explanation_P2.indd 277

CHAPTER 3

−1

As 0.001( °C ) is the temperature coefficient of the combination, we also have

= 0.632



Resistance at 900 K is

W = P = I 2 Rt = 5.68π × 10 −2 I 2 watt …(1) The rate at which a black body radiates energy is given by Stefan’s Law

(

W = σ As T 4 − T04

)

where As = surface area of the wire = ( 2π r ) L , T = 900 K, T0 = 300 K and σ = 5.68 × 10 −8 Wm −2K −4

9/20/2019 11:54:21 AM

H.278  JEE Advanced Physics: Electrostatics and Current Electricity

(

)

4 4 Thus W = 5.68 × 10 −8 × 2π × 10 −3 × 1 ⎡⎣ ( 900 ) − ( 300 ) ⎤⎦ ⇒ W = 5.68 × 12.96 π watt …(2) Equating ( 1 ) and ( 2 ) , we get

5.68 π × 10 −2 × I 2 = 5.68 × 12.96 π

⇒ I 2 = 1296 ⇒ I = 36 A

V

3 μF

   Vc − 12I − 12 = Vd

6Ω

  

(b) Without current the resistors act as conducting wires. Hence potential difference across capacitor is 12 V . Charge on capacitor, 3 mF , is   Q = ( 3 × 12 ) mC = 36 mC (c)  When all the switches are closed, the branch containing capacitor will not allow current after steady state is reached. The equivalent circuit from point of view of current is redrawn. Let the current in the single loop circuit be I. We begin at left lower corner and traverse the circuit anticlockwise while applying KVL. We get

−20 I + 20 − 12I − 12 = 0

1 A 4 20 V

c

b

2

⇒ P =

d

2

⎛ 1⎞ P = I 2 Req = ⎜ ⎟ ( 32 ) = 2 W    ⎝ 4⎠

R⎞ ⎛ N2 ⎜ r + ⎟ ⎝ N⎠

( 200 )2 ⎛⎜ 0.5 + 300 ⎞⎟

2

=

200 ⎠

2

( 200 )2 ( 300 ) ( 200 )2 ( 2 )2

300 = 75 W 4

(b) In the next stage, one of the bulb burns out, i.e., now only ( N − 1 ) resistors are left out. Power generated by any single resistor can be obtained by just replacing N by ( N − 1 ) in the previous equation. So, we get P′ =

E2

R

( N − 1 )2 ⎛ r + ⎜⎝

20 Ω

In the steady state power is consumed only in 12 W and 20 W resistors. All the resistors in a single loop circuit are in series, thus power consumed is

E2 R

( 200 )2 ( 300 ) ⎝



12 V

E r + Re

I 2R ⎛ I ⎞ P = ⎜ ⎟ R = 2 = ⎝ N⎠ N



I

R N

This current will be equally divided between all the N resistors (so that potential drop across each resistor is same). Hence power generated in any single resistor will be

⇒ P =

8 ⇒ I = 32

12 Ω

1 1 1 = + + … N terms Re R R

   I =



03_Ch 3_Hints and Explanation_P2.indd 278

21. (a) I f there are N resistors of resistance R each, connected in parallel, the net effective Re of such a combination is given by

Now, this combination is connected to a source of emf E and internal resistance r. Hence current in the main circuit would be

20 Ω

a

Charge on 2 mF capacitor = ( 15 ) × 2 = 30 mC

⇒ Re =

V

⇒ I =

(d) The capacitor is connected between points c and d. In order to determine the potential difference across it we begin at point c and traverse along any path to d , to get

Vc − Vd = 12I + 12 = 3 + 12 = 15 V   

20. (a) S  ince, a capacitor is an open circuit to DC in steady state. The effective circuit through which current will pass initially after closing switch S1 is shown. When steady state is reached the capacitor will not allow current. Thus there is no current from battery and hence the reading of voltmeter is zero.







R ⎞ N − 1 ⎟⎠

= 2

E2 R

( N − 1 )2 ⎛⎜ r + N − 1 ⎞⎟ R

⎝

2

⎠

The relative change f, by definition, is given by

   f = ⇒  

P ’− P P

⎤ ⎡ ( rN + R )2 f =⎢ − 1⎥ 2 ⎢⎣ ( r ( N − 1 ) + R ) ⎦⎥

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Hints and Explanations H.279



⇒ f =

1 r ⎞ ⎛ ⎜⎝ 1 − ⎟ rN + R ⎠

2

−1



Now, ( Nr + R )  r , since there are many bulbs ( N = 200 ) . Hence, we can approximate, r ⎞ ⎛ ⎜ 1 − ⎟ ⎝ rN + R ⎠

−2

2r ⎞ ⎛ ≈ ⎜ 1+ ⎟ ⎝ rN + R ⎠

⎛ 2 ⎞( ) ⇒ I1 = ⎜ 1.5 = 1 A ⎝ 2 + 1 ⎟⎠

23. Applying Kirchhoff’s Junction Law at E current in wire DE is 8 A from D to E . Now further applying Junction Law at D. The current in 3 W resistance will be 3 A towards D.

f =

⇒

C

10000 f = 25

r

E2 R+r This power will be minimum where ( R + r ) is maximum and we observe that ( R + r ) will be maximum only when the battery is inserted with 6 W resistance as shown in figure. Now power, P =

I1

4Ω

6Ω 11 V

Net resistance in this case is 2 × 4 22 = W 2+ 4 3

11 = 1.5 A ⇒ I= 22 3

⇒

3Ω

6Ω

I1 4 = =2 I2 2

03_Ch 3_Hints and Explanation_P2.indd 279

B



For path CDEB , we have

VC − ( 5 ) ( 1 ) + 12 − ( 8 ) ( 2 ) − 3 − ( 4 )( 2 ) = VB

⇒ VC − VB = 5 − 12 + 16 + 3 + 8



⇒ VC − VB = 20 V



and voltage across voltmeter is given by A

V

I

6V

This current will be distributed in 2 W and 4 W in the inverse ratio of their resistances.

6A

2 = I 2 R = ( 3 ) ( 3 ) = 27 W

V = 6 − ( voltage across ammeter )

⇒ V = 6 − IR1

6 R1 …(2) R2 In the second case reading of ammeter becomes two times, i.e., the total resistance becomes half whereas the resistance of ammeter remains unchanged. Hence,



3A 8A

I I2

Req = 6 +

4Ω 2A

24. Let R1 = resistance of ammeter and R2 = combined resistance of ammeter and voltmeter In the first case, current in the circuit is given by 6 …(1) I = R2

R

2Ω

E

Power dissipated in 3 W resistance

E



5A

3V

D 2Ω

2 ( 0.5 ) 1 = = 0.0025 ( 0.5 )( 200 ) + 300 400

22. Suppose, we insert the battery with 2 W resistance. Then we can take 2 W as the internal resistance ( r ) of the battery and combined resistance of the other two as the external resistance ( R ) . The circuit in that case shown in figure.



12 V

1Ω

Substituting r = 0.5 W, R = 300 W and N = 200, we get

CHAPTER 3





⇒ V =6−



I’=

6 R2 2

=

12 …(3) R2

and V ′ = 6 − ( I ’) R1

12R1 …(4) R2 V Further, it is given that V ′ = 2 ⇒ V′ = 6 −

9/20/2019 11:54:42 AM

H.280  JEE Advanced Physics: Electrostatics and Current Electricity V = 110 V

12R1 3R =3− 1 R2 R2



⇒ 6−



⇒



Substituting this value in equation (4), we have

Let I f be the current at the end of charging process. Then,

R1 1 = R2 3

V = E f + I f r

⎛ 1⎞ V ′ = 6 − ( 12 ) ⎜ ⎟ ⎝ 3⎠



⇒ V′ = 2 V

25. Let E be the emf of the battery In the first case let I be the current in the circuit, then E = I ( R1 + R2 ) …(1) In the second case main current increases three times I while current through voltmeter will reduce to . 3 I 8I passes through R Hence, the remaining 3 I − = 3 3 as shown in figure. E

E

A I

⇒ If =

V − Ef

r ⇒ If = 5 A

V R1

B

110 − 100 2

27. Let us represent the central junction of wires in the form of two junctions connected by the wire 5-6 as shown in figure. Then it follows from symmetry that there is no current through this wire. Therefore, the central junction can be removed from the initial circuit. Further, R12 = R13 = R34 = R24 = r r 2

R15 = R25 = R36 = R46 = and 3

G

4

3

4

6

3I A R2

=

A

R2

I C 3 8I 3

V

D R1

6

5 F

5 2

1

1

2

R

⎛I⎞ ⎛ 8I ⎞ VC − VD = ⎜ ⎟ R1 = ⎜ ⎟ R ⎝ 3⎠ ⎝ 3⎠

⇒ R1 = 8 R = 8 ( 3 W ) = 24 W



Applying Kirchhoff’s Second Law in loop ABFGA,

R ⎞ ⎛ ⎛I⎞ E = 3 I ( R2 ) + ⎜ ⎟ ( R1 ) = I ⎜ 3 R2 + 1 ⎟ …(2) ⎝ ⎝ 3⎠ 3 ⎠

From equations (1) and (2),

R R1 + R2 = 3 R2 + 1 3 2R1 3



⇒ 2R2 =



⇒ R2 =

R1 3



⇒ R2 =

8R 8 = (3 W) = 8 W 3 3

26. The voltage supplied by the charging plant is here constant which is equal to, V = Ei + I i r = ( 90 ) + ( 10 ) ( 2 )

03_Ch 3_Hints and Explanation_P2.indd 280

Let V be the voltage between 1 and 2. Then the amount of heat liberated in conductor 1-2. V2 …(1) r V Current through 3-4, I 3 − 4 = r( 2 + 3)

H12 =

V2



⇒ H 3 − 4 = I 32− 4 r =



⇒



Comparing with 11x + y 2x , we get x = 1, y = 6

r( 2 + 3)

2

2 H1 − 2 = ( 2 + 3 ) = 11 + 6 2 H3 − 4

28. (a) W  hen the switch is open, no current flows through the branch that contains R1 and R2 . Hence E = 250 V I

Rv1

Rv2

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Hints and Explanations H.281



30. Before connecting B with D and C with E,

250 1 A = 10000 40

RAF = 5R = 5 ( 2 W ) = 10 W

According to Ohm’s Law

   V1 =

1 ( 6000 ) = 150 V and 40

After connecting B with D and C with E , a balanced Wheatstone bridge is formed between B and E. So, RBE = R

V2 = ( 250 − 150 ) V = 100 V   

and RAF ′ = 3R = 3 ( 2 W ) = 6 W





(b) When the switch is closed the equivalent circuit is shown. E = 250 V

⇒ ΔR = 10 − 6 = 4 W i.e., the new resistance decreases by 4 W

31. Points ( A and C ) and ( D and B ) are symmetrically located with respect to points O and P. Hence, the circuit can be drawn as shown in figure.

V1

V2

R1

R2

P r1/2

Since we observe that V1 and R1 are in parallel and V2 and R2 are in parallel and both in series and also simultaneously we have

B

r1/2

r2/2 O r2/2

C

r3

R1 ( RV1 ) R2 ( RV2 ) = = 2.4 kW    R1 + RV1 R2 + RV2

This is a balanced Wheatstone bridge between P and O Hence, r3 can be removed. And,

Hence the potential of 250 V must be equally divided. So, reading of both the voltmeters is

RPO =

E = 125 V 2 2 9. Let N be the total number of cells required to be grouped in m rows, each row carrying n cells, then N = mn    V1 = V2 =







nE mnE NE We have, I = = = nr mR + nr mR + nr R+ m ⇒ 8=

2N 0.5 N 5m + m

dN = 0, For least value of N , dm Differentiating equation (1), we have

40 m + 2

dN dN =m +N dm dm

{

}

r1 + r2 4

Here r1 = RPB = RPD =

(πa )λ 2

and r2 = ROB = ( a ) λ

( 2 + π ) aλ

⎛ 2 + π ⎞ ⎛ 64 ⎞ 1 =⎜ a =8W ⎝ 8 ⎟⎠ ⎜⎝ 2 + π ⎟⎠ a



⇒ RPO =

32.

vd =



⇒ vd = 0.735 × 10 −6 ms −1



⇒ vd = 0.735 mms −1

2

⇒ 20 m + 2 N = Nm …(1)

I 1 = neA 8.5 × 10 28 × 1.6 × 10 −19 × 10 −4

Since, t =

8

 10 4 year = −6 vd 0.735 × 10 × 365 × 24 × 3600

⇒ t = 431 year

33. The equivalent circuit is as shown in figure. 4Ω



⇒ N = 40 m    ∵

dN = 0 …(2) dm But N = mn …(3)

6Ω

8Ω



Solving equations (1), (2) and (3), we get

4Ω

2Ω

n = 40 and m = 4

⇒ N = mn = 160

12 V 4Ω

2.4 Ω 03_Ch 3_Hints and Explanation_P2.indd 281

CHAPTER 3

   I=

1.6 Ω 9/20/2019 11:55:04 AM

4Ω 6Ω

8Ω

Ω 2Ω H.282  JEE Advanced4 Physics: Electrostatics and Current Electricity 12 V



4Ω

Solving, we get n = 1 Hence one cell of the battery is wrongly connected.

36. Leakage current I is given by 2.4 Ω



⇒ Rseries =

Rtotal =

⇒ I=

1.6 Ω

6×4 8×2 + =4W 6+2 8+2 4 ( 2.4 + 1.6 ) ( 4 )( 4 ) = =2W 4 + 2.4 + 1.6 4+4

V 12 = =6A R 2 net emf 6 + 12 = =3A net resistance 1 + 2 + 3

Now, VG = 0 V VA = 12 V, VB = 12 − 1 × 3 = 9 V ,   VC = 12 − 3 × 3 = 3 V and VD = 12 − 3 × 6 = −6 V (b) I=



⇒ I=

12 − 6 =1A 1+ 2 + 3

⇒ I = 6 × 10 −13 A So, ∗ = 6

ρ ( x ) dx 1 dx = A σ (x) A Since σ varies linearly with distance from the first plate, so we have dR =





1 ⎛ σ − σ1 ⎞ ⎡ ⎤ dx σ = σ1 + ⎜ 2 ⎟ x ⇒ dR = ⎢ ⎝ d ⎠ ⎛ σ 2 − σ1 ⎞ ⎥ A ⎢ σ1 + ⎜ ⎟x⎥ ⎝ d ⎠ ⎦ ⎣ Hence, total resistance R is given by

R =

  VG = 0  ⇒ VA = 12 V, VB = 12 − 1 × 1 = 11 V,

  VC = 12 − 3 × 1 = 9 V and VD = 12 − 6 × 1 = 6 V

35. Let n cells be wrongly connected in the battery. Let e.m.f. of each cell be E and R be the total resistance of the circuit ( 12 − n ) E − nE = ( 12 − 2n ) E



( 12 − 2n ) E + 2E R

…(1)



I 2 = 2 =

( 12 − 2n ) E − 2E

R Dividing (1) by (2), we get

The current I is given by

I =

V AV ( σ 2 − σ 1 ) AV ⎡ σ 2 − σ 1 = = R d ⎢ ⎛σ ⎞ ⎛σ ⎞ d log e ⎜ 2 ⎟ ⎢ log e ⎜ 2 ⎟ ⎝ σ1 ⎠ ⎝ σ1 ⎠ ⎢⎣



3 ( 12 − 2n ) + 2 = 2 ( 12 − 2n ) − 2

03_Ch 3_Hints and Explanation_P2.indd 282

⎤ ⎥ ⎥ ⎥⎦

( 230 × 10 −4 ) ( 300 ) ( 2 − 1 ) × 10 −12 ( 2 × 10 −3 ) ( 2.3 log10 2 )



⇒ I=



⇒ I = 5 × 10 −9 A = 5 nA

38. Resistance of bulb is given by

…(2)

0

dx ⎛ σ 2 − σ1 ⎞ ⎟x 1+⎜ ⎝ d ⎠



R0 =

So, current in circuit is

∫σ

d ⎛σ ⎞ log e ⎜ 2 ⎟ A (σ 2 − σ1 ) ⎝ σ1 ⎠

When the cells oppose the battery, the net e.m.f. is

E2 = ( 12 − 2n ) E − 2E

d

⇒ R=

So, current in circuit is

I1 = 3 =

1 A



When the cells aid the battery, the net e.m.f. is

E1 = ( 12 − 2n ) E + 2E

( 10 × 10 −4 ) ( 6 ) AV = ρd ( 1013 ) ( 1 × 10 −3 )

37. The resistance of a layer of medium of thickness dx at a distance x from the first plate of capacitor is

34. (a) Current in the circuit, I=  

⎛ Kε 0 A ⎞ ⎜⎝ ⎟V CV d ⎠ = I = K ρε 0 Kε 0 ρ

V02 100 × 100 = = 20 W P 500

Current in bulb is I =

V0 100 = =5A R0 20

For same power dissipation in bulb, the current in ­circuit must be 5 A When bulb is in a circuit having supply voltage 220 V, the safe resistance of the circuit

9/20/2019 11:55:16 AM

Hints and Explanations H.283 V ′ 200 = = 40 W I 5 Hence, required resistance in series is

R′ =

R = R′ − R0 = 40 − 20 = 20 W 39. The resistance R of fuse wire of length , cross sectional area A and resistivity ρ is

ρ R = A

From (1) and (2), we get

⎛ ρ ⎞ t ( Ad ) c × 307 = I 2 ⎜ ⎝ AJ ⎟⎠

⇒ t=

Here

( A2cd ) ( 307 ) J I 2ρ

density

…(3)

d = 11.34 gcm −3 = 11.34 × 10 3 kgm −3 ,

current I = 20 A , ρ = 22 × 10 −6 Wcm = 22 × 10 −8 Wm and specific heat c = 0.032 calg −1 ( °C )

Here ρ = 22 × 10 −6 W cm = 22 × 10 −8 W m

−1

⇒ c = ( 0.032 )( 4200 ) Jkg −1 ( °C ) ,

A = 0.2 mm 2 = 0.2 × 10 −6 m 2 = 2 × 10 −7 m 2



Since, Q = I 2 Rt (in joule)

A = 0.2 mm 2 = 2 × 10 −7 m 2

⎛ ρ ⎞ Q = I 2 ⎜ ⎟ t …(1) ⎝ A⎠



Again if m is mass of wire, d is its density, c its specific heat and ΔT , the rise of temperature, then

t= 

Q = mcΔT



⇒ t = 0.095 s = 95 × 10 −3 s = 95 ms



⇒ R=r Hence, the correct answer is (B).

4.

Resistance of potentiometer wire is



⇒ Q = ( Volume × density ) × c × ( 327 − 20 )



⇒ Q = ( Adc )( 307 ) …(2)

−1

Substituting given values in (3), we get

( 2 × 10 −7 )2 × 4.2 × 103 × 11.34 × 103 × 0.032 × 307

CHAPTER 3





( 20 )2 × 22 × 10 −8

ARCHIVE: JEE MAIN 1.

From Colour Code of Carbon Resistors table, we get 200 W = Red + Black + Brown

Since, Green ≡ 5

2.

⇒ Green + Black + Brown ≡ 500 W Hence, the correct answer is (B).

Eeq

E1 E E + 2 + 3 2R1 R2 2R1 = 1 1 1 + + 2R1 R2 2R1 2 4 + + = 2 2 1 1 + + 2 2

4 2 1 2

Rp = 400 × 0.01 ⇒ Rp = 4 W 3 ⇒ I = = 0.5 A 6 Reading of voltmeter is VAJ = IRAJ ⇒ VAJ = 0.5 × 50 × 0.01 ⇒ VAJ = 0.25 V Hence, the correct answer is (D).



⇒ Eeq



⇒ Eeq =



Hence, the correct answer is (D).



3.

For maximum power consumption in external resistance





5 10 = = 3.3 V 3 3 2

Internal resistance = External resistance

03_Ch 3_Hints and Explanation_P2.indd 283

5.



Equivalent resistance of the given circuit is Req = 45 +

50 ⎞ 160 10 × 50 ⎛ = ⎜ 45 + ⎟ W = W 10 + 50 ⎝ 6 ⎠ 3

15 9 = A 160 32 3 Hence, the correct answer is (B). ⇒ I=

9/20/2019 11:55:27 AM

H.284  JEE Advanced Physics: Electrostatics and Current Electricity 6.

At steady state, current through capacitor is zero.



⇒ VC = V1

where, V1 =

5 3 × V0 × 6 9 6Ω 4Ω

72 V



⇒ V1 =

2Ω 10 Ω

V1

5 × 72 × 3 = 20 V 6×9

Since Q1 = CV1 ⇒ Q1 = 200 mC



Hence, the correct answer is (A).

7.

Between D and C, the resistance is



⎛ ⇒ R1 = ⎜ ⎝



⇒ R1 = 2 W

⎛ 12 ⎞ ⎛ 5π ⎞ and R2 = ⎜ a ⎝ 2π a ⎟⎠ ⎜⎝ 3 ⎟⎠

⇒ R2 = 10 W



⇒



⇒ Req =



Hence, the correct answer is (A).

1 1 1 6 = + = Req 2 10 10



⇒

1 8 8 = + Req R 7 R

8.

V = I g ( Rs + Rg )



⇒ V = 4 × 10 −3 [ 5050 ]



Hence, the correct answer is (B).

9.

R0 = ρ

⇒ Req =

⇒ V ≈ 20 V  =3W A

When wire is stretched to twice its original length, then

2

R′ = n R0

2 ⇒ R′ = ( 2 ) ( 3 ) = 12 W

We have to find resistance between A and B. B R1 A

a 60° a

O



⇒ ( 0.002 )( 50 ) = ( 0.5 − I g ) S



0.002 × 50 = 0.2 W 0.5 Hence, the correct answer is (A).



⇒ S≈

11. Since ρ =

m ne 2τ 9.1 × 10 −31



⇒ ρ=



⇒ ρ=



⇒ ρ = 1.67 × 10 −8 W m Hence, the correct answer is (D).

12.

nR ( T ) = a −



a, b are constant

8.5 × 10 28 × ( 1.6 × 10 −19 ) × 25 × 10 −15 2

Wm

9.1 ( ) × 10 −59 + 53 8.5 × 2.56 × 25

a 1 − b T2

⇒ R ( T ) = R0 e

− T02 T2



Hence, the correct answer is (B).

13.

I g = 10 −4 A , R = 2 × 106 W

and V = 5 volt R2

03_Ch 3_Hints and Explanation_P2.indd 284

(0.5 – Ig)S

Since, I g Rg = ( 0.5 − I g ) S

7R 64 Hence, the correct answer is (D).



Rg = 50 Ω I = 0.002 A g G Ig

I = 0.5 A

and between E and C, the resistance is 7R 8

5 W 3

10.

R R1 = 8

R2 =

π 3 12 ⎞ ⎛ π ⎞ ⎟ a⎜ ⎟ 2π a ⎠ ⎝ 3 ⎠

∠AOB = 60° =

Since V = I g ( R + G )

9/20/2019 11:55:39 AM

Hints and Explanations H.285



⇒ 5 = 10 −4 ( 2 × 106 + G ) ⇒ G = NEGATIVE Which is NOT POSSIBLE. Data Contradictory. *No given option is correct.



⇒ m = 1.0 m 2 V −1s −1



Hence, the correct answer is (D).

17. a

b

14. For the given circuit, it is given that 15 Ω A

B

x

2Ω

dx

10 Ω r 1.5 V





r 1.5 V



VAB = 2 V

2 2 1 + = A ⇒ I= 15 10 3

Also I ( 2r + 2 ) = 1.5 + 1.5 − VAB

⇒ 2r + 2 = ( 3 − 2 ) 3



1 W 2 Hence, the correct answer is (A).

15.

R  = X 100 − 



⇒ X=



For 1, X =

100 × 40 2000 = W = 667 W 60 3



For 2, X =

100 × 87 8700 = W = 669 W 13 13



For 3, X =

10 × 98.5 1970 = W = 656 W 1.5 3



1 × 99 = 99 W For 4, X = 1

∫



⎛ ρ ⎞ ⇒ R=⎜ ⎝ 4π ⎟⎠

vd E



E EA and resistivity ρ = = j i



⇒ m=



1.1 × 10 −3 × π × ( 5 × 10 −3 ) ⇒ m= 5 × 1.7 × 10 −8

b

dx

∫x

2

a



⎛ ρ ⎞ ⎛ 1 1⎞ ⇒ R=⎜ − ⎝ 4π ⎟⎠ ⎜⎝ a b ⎟⎠ Hence, the correct answer is (A).

18.

Req = 2R + R + 4 R + R = 8 R E2 Req

16 × 16 = 4 watt 8R



⇒



⇒ R=8W Hence, the correct answer is (A).

19. For R1 , Since I g = 10 −3 Amp

Clearly, we can see that the value of X calculated in CASE-4 is inconsistent with the other values. Hence, the correct answer is (D).

ρdx 4π x 2

⇒ R = dR

Since, P =

R ( 100 −  ) 

⇒ 10 −3 ( R1 + 100 ) = 2 V

⇒ R1 = 1900 W For R2 , 10 −3 ( R1 + R2 + 100 ) = 10 V ⇒ R1 + R2 + 100 = 10000 ⇒ R2 = 8000 W For R3 ,

vd A iρ

03_Ch 3_Hints and Explanation_P2.indd 285

dR =





⇒ r=

16. Mobility is m =

Resistance of infinitesimal element is given by

CHAPTER 3



2

10 −3 ( R1 + R2 + R3 + 100 ) = 20 V ⇒ R1 + R2 + R3 + 100 = 20 × 1000

⇒ R3 = 10000 W Hence, the correct answer is (A).

9/20/2019 11:55:50 AM

H.286  JEE Advanced Physics: Electrostatics and Current Electricity ER R+r For R → ∞ , V = E = 1.5 V 20.

V=

For R = 0 , I =

E =1 r



⇒ r = 1.5 W Hence, the correct answer is (A).

21.

V = I g ( RV + G ) = GI 0 …(1)

Also, ( I 0 − I g ) RA = I gG …(2)

G ( I0 − I g )

Ig From (2), we get RA =

I gG

I0 − I g ⇒ RA RV = G 2 RA ⎛ I g ⎞ = RV ⎜⎝ I 0 − I g ⎟⎠

2



⇒



Hence, the correct answer is (A).

22. Since A = Constant R=ρ and

 A



⇒

ΔR Δ ΔA = + R  A



⇒

ΔR 2 Δ = = 2 × 0.5%  R



ΔR ⇒ % = 1% R Hence, the correct answer is (D).

23.



From (1) and (2), we get

20 = 2i1 + 2i1 + 2i2 ⇒ 20 = 4i1 + 5 − i1

⇒ 3i1 = 15



⇒ i1 = 5 and i2 = 0



⇒ i = i1 + i2 = 5 A Hence, the correct answer is (B).

25.

vd =



⇒ vd =



⇒ vd = 2.08 × 10 −5 ms −1



⇒ vd = 0.02 mms −1 Hence, the correct answer is (A).

26.

G O Y Golden

From (1), we get RV =





I neA 1.5

( 9 × 1028 ) ( 1.6 × 10 −19 ) ( 5 × 10 −6 )

Since G = 5 , O = 3 , Y = 4 and Golden = 5%

⇒ R = 53 × 10 4 ± 5% ⇒ R = ( 530 kW ± 5% ) Hence, the correct answer is (D).

27.

I1 100 Ω A

400 Ω

R3

500 Ω R4

I2

R2

18 V



Since voltmeter is ideal, so I1 =

V 5 = = 0.01 A R4 500

R V O S ↓ ↓ ↓ ↓ 2 7 3 ±10%

⇒ VB − VC = 600 I1 = 6 V

⇒ R = 27 × 10 3 ± 10% Hence, the correct answer is (C).



⇒ ( VA − VB ) + ( VB − VC ) = 18 V



⇒ VA − VB = 12 V

24. Applying Kirchhoff’s Loop Law, we get

Since VA − VC = 18 V

So, I1 + I 2 =

12 = 0.03 A 400

20 − 2i1 − 2 ( i1 + i2 ) = 0 ⇒ 20 − 2i1 = 10 − 4i2 …(1)



⇒ I 2 = 0.03 − 0.01 = 0.02 A

Also, 10 − 4i1 − 2 ( i1 + i2 ) = 0



⇒ R2 =



Hence, the correct answer is (D).



⇒

10 − 2i1 = i2 …(2) 4

03_Ch 3_Hints and Explanation_P2.indd 286

C

R1 B

6 = 300 W 0.02

9/20/2019 11:56:04 AM

Hints and Explanations H.287

–10 V

–10 V

–10 V

32. From colour code table, we get

R1 20 Ω I1 –10 V 10 V 0 V

R1 ( O, R, B ) = 320 W

R2 20 Ω

G

10 V –10 V R3

The potentials of various points is shown in figure after assigning 0 V potential to the junction. So,

I1 =

0 − ( −10 ) 1 = A = 0.5 A and 20 2

−10 − ( −10 ) I2 = =0A 20

Hence, the correct answer is (C). 29.

G B R Br

Resistance = 50 × 10 ± 1%

P = I 2R



⇒ 2 = ( 5000 ) Ι 2



⇒



⇒ I=



⇒ I = 20 mA Hence, the correct answer is (A).

30.

RAB =



Hence, the correct answer is (A).

4 = I2 10000 2 100

31. For no deflection, I =

ε ε = r + 12r 13 r

12ε ⎛ ε ⎞ 12r = ⇒ VAB = IRAB = ⎜ ⎝ 13 r ⎟⎠ 13 r 12ε potential is developed across L 13 13 L L 1 unit potential is developed across = ⎛ 12ε ⎞ 12ε ⎜⎝ ⎟ 13 ⎠ ε is developed So, potential of 2 across a length given by

So,



⎛ 13 L ⎞ ε  AJ = ⎜ ⎝ 12ε ⎟⎠ 2

13 L ⇒  AJ = 24 Hence, the correct answer is (D).

03_Ch 3_Hints and Explanation_P2.indd 287

R4 = 40 W For a Balanced Wheatstone Bridge, we have R1 R3 = R2 R4 320 × 40 80



⇒ R3 =



⇒ R3 = 160 W = 16 × 101 W So, colour for R3 is Brown, Blue, Brown



Hence, the correct answer is (D).

33.

R′ =

RRV R + RV

Since, R′ =

12 × 6 =4W 12 + 6

R4= 40 Ω

Since R2 = 80 W

2

R2= 80 Ω

R1

I2

CHAPTER 3

28.

95 × 30 = 28.5 W 100 30 RV 30 + RV



⇒ 28.5 =



⇒ 855 + 28.5RV = 30 RV



⇒ RV =



⇒ RV = 570 W Hence, the correct answer is (A).

34.

I 2 R = 4.4 watt



⇒ R=

855 1.5

4.4 W 4 × 10 −6

Since P( dissipated ) at voltage V is 11 volt

⇒ P=

V 2 11 × 11 × 4 × 10 −6 = R 4.4



⇒ P=

11 × 11 × 4 × 10 −6 × 10 44



⇒ P = 1.1 × 10 −4 watt



⇒ P = 11 × 10 −5 watt Hence, the correct answer is (A).

9/20/2019 11:56:16 AM

H.288  JEE Advanced Physics: Electrostatics and Current Electricity

35. When in series, PS =

P1P2 = 60 W P1 + P2

Since P1 = P2 = P

P ⇒ PS = 2



⇒ P = 120 watt When in parallel,



Hence, the correct answer is (D).



P Q = R X



P Q = …(1) 400 X On interchanging P and Q, we have Q P = 405 X

⇒ X = 402.5 W Hence, the correct answer is (C).

37. At null point, E = VAJ = IRAJ ⎞ ⎟⎠ RAJ

Since AB is uniform wire, so if λ is resistance per unit length of AB, then RAJ = 4 λ

⎛ 6 ⎞( ) ⇒ E1 = 0.5 = ⎜ 4λ ⎝ 4 + 2 ⎟⎠

⎛ 6 ⎞( ) Also, E2 = ⎜ 4λ ⎝ 4 + 6 ⎟⎠ E2 6 = 0.5 10



⇒



⇒ E2 = 0.3 V Hence, the correct answer is (C).

38. Full scale deflection current is I g = 30 × 0.005 = 0.15 A Since γ = I g ( G + R )

⇒ 15 = 0.15 ( 20 + R )

03_Ch 3_Hints and Explanation_P2.indd 288

Eeq req

1 W 3 =

E1 E2 E3 + + r1 r2 r3

Eeq × 3



1 2 3 = + + =6 1 1 1 1 ⇒ Eeq = 2 V



Hence, the correct answer is (C).

⇒

40. For Condition of Balance, we have R1 40 2 = = …(1) R2 60 3

Now,

R1 + 10 = 1 …(2) R2

From (1) and (2), we get R1 2 = R1 + 10 3

⇒ X = 400 × 405

6 ⎛ where I = ⎜ ⎝ RAB + RH

1 =3 req

⇒ req =



QX ⇒ P= 405 Substitute in (1), we get QX Q = 400 × 405 X



Hence, the correct answer is (C).

Also,



⇒



⇒ R = 80 W

39. Since

PP = 2P = 2 × 120 = 240 W

36.



⇒ R1 = 20 W and R2 = 30 W

Let resistance to be connected in parallel to

R1 + 10 i.e., 30 W be R, then for balance point to shift to the initial situation, we have



30 R 30 + R = 40 30 60 30 × R = 20 30 + R



⇒



⇒ 30 R = 600 + 20 R



⇒ R = 60 W Hence, the correct answer is (A).

41. Since deflection θ is proportional to i , so

θ ∝i Let RG = R

⇒ i1 = i2 =



V = kθ 0 …(1) 220 + R V 5 θ × = k × 0 …(2) 5×R R+5 5 220 + 5+R

9/20/2019 11:56:29 AM

Hints and Explanations H.289 From (1) and (2), we get 25 1 = 220 × ( 5 + R ) + 5R ( 220 + R )



25 W

⇒ R = 22 W

⇒

⇒ P = 20 W Since Power ∝ R

dR K = d  R



0

0

∫ dR = K ∫

⇒ R = 2K 



⇒

d 

2K   R = = R′ 2 K ( 1 −  ) 1 − 

Since R = R′ ⇒

 1− 

PR1 = 16 W R1 + R2



⇒ P1 =



⇒ P2 = 4 W Hence, the correct answer is (D).

45.

I g = 25 × 4 × 10 −4



⇒ I g = 10 −2 A

Since V = I g ( R + 50 )

=1



⇒ R = 200 W Hence, the correct answer is (C).



⇒ 2  =1



⇒ =



Hence, the correct answer is (C).

I4 = I3 + I5 ⇒ I 4 = I 3 + 0.4

43.

VAJ = 5 × 10 −3



⇒ 0.8 − 0.4 = I 3



⇒ I 3 = 0.4 A

⇒ i=

46. At Junction S, we have

1 = 0.25 m 4

5 10 = × = 0.5 W 1 100

RAJ



100 W

220 V

Hence, the correct answer is (A).





1 1 1 = + P 25 100

1 1 = ⇒ 45R + 220 5 × ( 220 + R )

42. Since



44.

VAJ RAJ

= 10

4V

A

Since, i =

CHAPTER 3



−2

I1 + I 2 = I 4 ⇒ −0.3 + I 2 = 0.8

A

R

5Ω

4V

B

A



⇒ R = 395 W Hence, the correct answer is (D).

03_Ch 3_Hints and Explanation_P2.indd 289



R

J 10 cm

4 = 10 −2 R+5

⇒ R + 5 = 400 W

At Junction R, we have

B

⇒ I 2 = 1.1 A At Junction Q, we have

I 3 + I 6 = I1 + I 2 ⇒ 0.4 + I 6 = −0.3 + 1.1

⇒ I 6 = 0.4 A Hence, the correct answer is (A).

47. Equivalent e.m.f. of parallel batteries E1 E2 12 13 + + r r2 2 = 37 V E = 1 = 1 1 1 1 1 3 + + 1 2 r1 r2

9/20/2019 11:56:41 AM

H.290  JEE Advanced Physics: Electrostatics and Current Electricity

Equivalent resistance of parallel batteries, 13 V

1Ω

12 V

2Ω

37/3 V

10 Ω

2×1 2 = W 2+1 3

Now, its equivalent circuit is as drawn.



37 37 3 Current in the circuit, i = = ⎛ 2 ⎞ 32 10 + ⎜ ⎟ ⎝ 3⎠

Hence, the correct answer is (B).

Then,

R1  = …(1) R2 ( 100 −  )

Also, R1 + R2 = 1000 W …(2)  On interchanging the resistances, balancing length becomes (  − 10 ) , so R  − 10  − 10 = 2 = R1 100 − (  − 10 ) 110 −  100 −   − 10 =   110 − 



⇒



⇒ 11000 +  2 − 210 =  2 − 10 ⇒ 200 = 11000 ⇒  = 55 cm



From equation (1), we get



⇒ R1 =

55 R2 45



⇒ R1 =

55 ( 1000 − R1 )  45

55 55 R1 = 1000 × ⇒ R1 + 45 45 ⇒ 100 R1 = 1000 × 55

03_Ch 3_Hints and Explanation_P2.indd 290

12.5 × 39.5 = 8.16 W 60.5

When X and Y are interchanged so  1 and ( 100 −  1 ) will also interchange and hence  2 = 60.5 cm. Hence, the correct answer is (B). 50. R

A

R

B

R

R

A

R

R

R

R

R

C

R C

R

R C

C

R

C

≡ B

R

R

RAB = R + R = 2R

Hence, the correct answer is (A).

51. Rate of heat developed, P =

V2 R

For a given V , we have

P ∝

1 A πr2 = = R ρ ρ

P1 ⎛ r12 ⎞ ⎛  2 ⎞ = P2 ⎜⎝ r22 ⎟⎠ ⎜⎝  1 ⎟⎠



⇒



Since it is given that,  2 =



⇒



⇒ P2 = 8 P1



Hence, the correct answer is (A).

P1 1 1 1 = × = P2 4 2 8

1 and r2 = 2r1 2

52. Current flowing through copper rod is given by

R1 55 = R2 45





(From equation (1))

⇒ X=



37 370 = i × 10 = × 10 = = 11.56 V 32 32

48. Let R1 (left slot) and R2 (right slot) be two resistances in two slots of a meter bridge. Initially  be the balancing length



Hence, the correct answer is (C).

Voltage across the load,

V10 W



Y ( 39.5 ) = X ( 100 − 39.5 )





⇒ R1 = 550 W

49. For a balanced meter bridge i

10 Ω

req =

2/3 Ω



I = neAvd = ρ Avd 

(From equation (2))

{∵ ρ = ne }

I ρA



⇒ vd =



Time taken by charges to travel distanced d is

t =



d d ρ Ad = = vd ⎛ I ⎞ I ⎜⎝ ρ A ⎟⎠

Hence, the correct answer is (A).

9/20/2019 11:56:52 AM

Hints and Explanations H.291 53. Equivalent circuit is shown in figure. Charging of capacitor is given by

6V

4V 2V

Ceq

2V

1Ω

6V

t ⎞ ⎛ − ⎜ RCeq ⎟ ⎠ , where qmax = CeqE q = qmax ⎝ 1 − e



t ⎞ ⎛ − ⎜⎝ RCeq ⎟ ⎠ ⇒ q = CeqE 1 − e

Both capacitors will have same charge as they are connected in series. Hence, the correct answer is (A).



2V 4V



⇒ α=



Hence, the correct answer is (B).

10 = 2 × 10 −4 °C −1 100 × 500

55. In steady state, flow of current through capacitor will be zero. E I

r

C

r1

r2

57. In a balanced Wheatstone bridge, the null point remains unchanged even if cell and galvanometer are interchanged. Hence, the correct answer is (B). 58. Let equivalent resistance of the infinite network be x . Then, equivalent resistance between points A and B is also x . A

1Ω 4Ω

4x +2 4+x



⇒ x=



⇒ x 2 − 2x − 8 = 0

x =

1Ω

2 ± 4 − 4 ( 1 ) ( −8 ) 2 ± 36 = 2 2



2±6 =4W 2 (Since negative value of resistance is not accepted)



⇒ I1 =



So, reading of A1 is 2 A .



Hence, the correct answer is (B).



⇒ x=

9 =2A 4 + 0.5

59. When key is plugged between 2 and 1, then where k is the potential gradient of wire PQ When key is plugged between 3 and 1 , then

Er2 r + r2



r ⇒ qC = CVC = CE 2 r + r2



Hence, the correct answer is (C).

56. Using Nodal Analysis method, we observe that the potential difference across each resistor is zero.

03_Ch 3_Hints and Explanation_P2.indd 291

B

x

V1 = iR1 = k 1 …(1)

E I = r + r2 VC = Ir2 =

0V

So, no current will flow through each resistor. Hence, the correct answer is (D).

Voltage applied 220 RT = = = 110 W Current 2

⇒ 110 = 100 ( 1 + α × 500 )

2V 2V

54. Resistance of element at 500 °C is

Since RT = R0 ( 1 + αΔT )

1Ω

CHAPTER 3

R

E

2V

1Ω

2V q

0V

2V

V2 = i ( R1 + R2 ) = k 2 …(2)

On dividing equation (2) by equation (1), we get



 R1 = 1 R1 + R2  2 1 R1 = R2  2 −  1



⇒



Hence, the correct answer is (B).

9/20/2019 11:57:00 AM

H.292  JEE Advanced Physics: Electrostatics and Current Electricity 62. Resistance of a wire of length  and radius r is given by

60. For P = 4 W ,  1 = 60 cm ⇒



2 8 ⇒ Q= P= W 3 3

Now, P ′ = P + R and ′1 = 80 cm 80 P′ ′1 = = =4 Q 100 − ′1 20



⇒



P+R ⇒ =4 Q



⇒

ρ ρ A ρV ρV = × = 2 = 2 4  A A A A π r

R =

1 P 60 3 = = = Q 100 −  1 40 2



1 r4



⇒ R∝



⇒



According to the problem, we have

R1 ⎛ r2 ⎞ = R2 ⎜⎝ r1 ⎟⎠



⎛r ⎞ ⇒ R2 = R1 ⎜ 1 ⎟ = 16 R1 = 1600 W ⎝ r2 ⎠



Hence, the correct answer is (D).

63.

Cu is conductor, so with increase in temperature, resistance will increase. Si is semiconductor, so with increase in temperature  resistance will decrease. Hence, the correct answer is (D).

32 3

⇒ 4+R=



⇒ R=



Hence, the correct answer is (A).

20 32 −4= W 3 3

61. The given three circuits are equivalent to the following three simpler circuits.

64. Let the source voltage be V. Then, equivalent resistance of the circuit when r = fR is Req = R +

1Ω

1Ω 1Ω

1Ω 3V

1Ω

1Ω

3V

1Ω

3V

1Ω

R2

R2 = 1/2 Ω

R1

R1 = 1 Ω

1Ω

1Ω 3V

3V

V ( f + 1) V = Req R ( 2 f + 1 )

Current through the resistance r ( = fR ) is

I 2 = 3V

fR r×R ( 2 f + 1)R = R+ = r+R f +1 ( f + 1)

⇒ Current in the circuit is

I =

R3

R3 = 2 Ω

1Ω

1Ω

1Ω

I V = f + 1 R( 2 f + 1)

So, heat generated per unit time in the resistor r is

H = I 22 r =

V2 f R( 2 f + 1)

2

32 So, P1 = =9W 1



For maximum H , we have

32 = 18 W and P2 = 1 2



⇒



⇒ 2f +1− 4 f = 0

2



3 = 4.5 W 2 ⇒ P2 > P1 > P3



Hence, the correct answer is (B).

P3 =

r , R2 = ? 2

4



1Ω

4

R1 = 100 W , r1 = r, r2 =

4+R =4 8 3

1Ω

{∵ V = A }

03_Ch 3_Hints and Explanation_P2.indd 292



dH =0 df

4f V2 ⎛ 1 ⎞ − =0 R ⎜⎝ ( 2 f + 1 )2 ( 2 f + 1 )3 ⎟⎠

1 2 Hence, the correct answer is (A). ⇒ f =

9/20/2019 11:57:11 AM

Hints and Explanations H.293 67. Let us redraw the circuit and assign potential at P to be zero and that at Q to be x. Then at the junction Q, ΣIQ = 0

T0 = 300 K, R0 = 100 W and At at T = 500 K, R = 120 W 120 = 100 ( 1 + α ( 200 ) )



⇒



⇒ 200α =



⇒ α = 10 −3 °C −1

6 1 −1= 5 5

x−6 x−0 x+9 + + =0 3 1 5



⇒



9 ⎛ 5 + 15 + 3 ⎞ ⇒ x⎜ ⎟⎠ = 2 − ⎝ 15 5



⇒ x=

Temperature of the toaster is raised at constant rate from 300 K to 500 K is 30 s. So, increment in the temperature in time t is ΔT =

( 500 − 300 ) 30

W =

∫ 0

V 2 dt = R(t )

30



⇒ W=

∫ 0

∫ 100 ⎛⎜ 1 + 10 0

⎝

−3

Q

×

20 ⎞ t⎟ 3 ⎠

3Ω

x

–9

5Ω

x−0 3 = A 1 23



⇒ I=



⇒ I = 0.13 A from Q to P



Hence, the correct answer is (C).

68. Given, vd ∝ E 30

( 200 )2 dt

–9

P

6

V 2 dt R0 ( 1 + αΔT )

9V

1Ω

t

20 t 3 Total work done during this time in raising the temperature is t

6V 0

6

ΔT =

t

3 V 23

CHAPTER 3

65. Here, R ( T ) = R0 ⎡⎣ 1 + α ( T − T0 ) ⎤⎦

= 400

∫ ⎛⎜ 1 + 0

Since, I = neAvd

dt

⎝

t ⎞ ⎟ 150 ⎠



⇒ I ∝ E …(1) V

30



t ⎞⎤ ⎡ ⎛ ⇒ W = 400 × 150 ⎢  n ⎜ 1 + ⎟ 150 ⎠ ⎥⎦ 0 ⎣ ⎝



30 ⎞ ⎤ ⎡ ⎛ ⎛ ⇒ W = 60000 ⎢ n ⎜ 1 + ⎟ − n1 ⎥ = 60000n ⎜⎝ 150 ⎠ ⎣ ⎝ ⎦

6⎞ ⎟ J 5⎠

I

V 

Since E =



⎛ 6⎞ ⇒ W = 60n ⎜ ⎟ kJ ⎝ 5⎠ Hence, the correct answer is (C).

66.

vd = 2.5 × 10 −4 ms −1 , n = 8 × 10 28 m −3

I ∝ V



Since, I = neAvd

⇒

V = neAvd R

V = neAvd ⇒ ⎛ ρ ⎞ ⎜⎝ ⎟⎠ A V nevd



⇒ ρ=



Substituting values, we get

ρ = 1.6 × 10 −5 Wm

Hence, the correct answer is (D).

03_Ch 3_Hints and Explanation_P2.indd 293



⇒ E ∝V So from (1), we get ⇒ I2 ∝ V Hence, the correct answer is (C).

69. Current in the circuit, I =

5 1.6

50 25 = A 16 8



⇒ I=



Reading of the voltmeter is 10 V

1Ω

I 0.6 Ω

15 V V

9/20/2019 11:57:23 AM

H.294  JEE Advanced Physics: Electrostatics and Current Electricity

V = 15 −

25 × 0.6 8



Initial current is i =

120 120 = 240 + 6 246

15 = 13.1 V 8



New current is i′ =

120 60 120 1 24 × = × = 48 + 6 60 + 240 54 5 54



⎛ 120 24 ⎞ − Decrease in voltage = 240 × ⎜ = 10.4 V ⎝ 246 54 ⎟⎠



So the nearest approximate answer is (D) Hence, the correct answer is (D).



⇒ V = 15 −



Hence, the correct answer is (C).

70.

2Ω

F

D

E

2V

4Ω

2Ω

R

73. During discharging of a capacitor, we have V = V0 e − t/τ

4V



Current in 4 W is zero.

where τ ( = RC ) is the time constant of RC circuit. At V t = τ , we have V = 0 = 0.37 V0 e



Start from C and go to D (via B and E ), we get



From the graph, t = 0, V0 = 25 V



⇒ V = 0.37 × 25



⇒ V = 9.25 V

A

9V

B

3V

C

VC − 3 + 4 − VD = 0

⇒ VC − VD = −1 V …(1)



Starting from A and going to C (via B ), we get

VA − 9 + 3 − VC = 0

⇒ VA − VC = 6 V …(2)



Add (1) and (2), we get

This voltage occurs at time that lies between 100 sec and 500 sec. Hence, time constant τ of this circuit lies between 100 sec and 150 sec. Hence, the correct answer is (C). 74. 25 W – 220 V 100 W – 220 V

VA − VD = 5 V

Hence, the correct answer is (C).

71. Total power P is given by

440 V

P = ( 15 )( 40 ) + ( 5 )( 100 ) + ( 5 )( 80 ) + ( 1 ) ( 1000 )

⇒ P = 600 + 500 + 400 + 1000 = 2500 W

Since, P = VI

⇒ I=



2500 A 220

125 ⇒ I= A = 11.3 A 11 So, minimum capacity of main fuse of building must be 12 A. Hence, the correct answer is (C). 72. Assuming that both bulb and the heater have a rating of 120 V , then Rbulb

Since R =

120 × 120 = = 240 W 60

and Rheater = 60 W

Rated power

So, the resistance of 25 W- 220 V bulb is

R1 =

( Rated voltage )2

( 220 )2 25

and resistance of 100 W- 200 V bulb is

R2 =

( 220 )2 100

2 1 ⎞ ( 220 ) 2⎛ 1 + W RS = R1 + R2 = ( 220 ) ⎜ = ⎟ ⎝ 25 100 ⎠ 20



60 W

So, the current, I =

440

( 220 )

2

=

2 A 11

20 Heater (240 W)



Potential difference across 25 W bulb is

V1 = IR1 =

03_Ch 3_Hints and Explanation_P2.indd 294

W

When these two bulbs are connected in series, the total resistance is

6Ω 120 V

W

2

2 ( 220 ) × = 352 V 11 25

9/20/2019 11:57:34 AM

Hints and Explanations H.295 Potential difference across 100 W bulb is

V2 = IR2 =

2

2 ( 220 ) = 88 V × 11 100

Thus the bulb 25 W will be fused, because it can tolerate only 220 V while the voltage across it is 352 V. Hence, the correct answer is (B).

ρ …(1) 7 5. Resistance of wire R = A On stretching, volume ( V ) remains constant. So V = A

⇒ A=

V 



⇒ R=

ρ 2  V



⇒ R ∝ 2

{From (1)}

ΔR 2 Δ {∵ V and ρ are constants} = R  Hence, when wire is stretched by 0.1% its resistance will increase by 0.2%. Hence, the correct answer is (B).

⇒

76. Since, by definition ΔR = RαΔT For series combination, we have



⇒ ΔRs = ΔR1 + ΔR2



⇒ 2R0α s ΔT = R0α 1ΔT + R0α 2 ΔT



α1 + α 2 2 For parallel combination, we have ⇒ αs =

RP =

R1R2 RR = 1 2 R1 + R2 Rs

ΔRP ΔR1 ΔR2 ΔRs = + − Rp R1 R2 Rs



⇒ α P ΔT = α 1ΔT + α 2 ΔT − α s ΔT



⇒ α P = α1 + α 2 − α s



⎛ α + α2 ⎞ ⇒ α P = α1 + α 2 − ⎜ 1 ⎟ ⎝ 2 ⎠



CHAPTER 3



α1 + α 2 2 Hence, the correct answer is (A). ⇒ αP =

77. From the statement given, α = 2.5 × 10 −3°C −1 . The resistance of a wire change from 100 W to 150 W when the temperature is increased from 27 °C to 227 °C . Hence, the correct answer is (A).

Rs = R1 + R2

ARCHIVE: JEE ADVANCED Single Correct Choice Type Problems 1.

For infinite line charge, we have

E =

λ 2πε r

⇒



⇒

⎛ λ ⎞ ⇒ j=σ⎜ ⎝ 2πε r ⎟⎠

But i = jA = j ( 2π r ) dq σλ ( 2π r ) = j ( 2π r ) = dt 2πε r



⇒



⇒ i=−



(Negative sign because i is flowing radially outwards).



⇒

σλ  ε

dq σλ  =− dt ε

03_Ch 3_Hints and Explanation_P2.indd 295

d ( λ ) σλ  =− dt ε λ

∫

λ0

Also, we know that

j = σ E



dλ σ =− λ ε

log e λ

λ λ0

⇒ λ = λ0 e

Since, j =

t

∫ dt 0

⎛σ⎞ = −⎜ ⎟ t ⎝ε⎠

⎛σ ⎞ −⎜ t ⎟ ⎝ε ⎠

σλ 2πε r ⎛σ ⎞

2.

⎛ λ σ ⎞ −⎜ t ⎟ ⇒ j=⎜ 0 ⎟e ⎝ε ⎠ ⎝ 2πε r ⎠ Hence, the correct answer is (C). A ⎞1 1 1 1 ⎛A = + = ⎜ Al + Fe ⎟ R RAl RFe ⎝ ρ Al ρFe ⎠ 

⎡ ( 7 2 − 22 ) 22 ⎤ 10 −6 1 = ⎢ + ⎥ −8 × 27 10 ⎦ 10 ⎣ 50 × 10 −3

9/20/2019 11:57:47 AM

H.296  JEE Advanced Physics: Electrostatics and Current Electricity

Solving we get,

1875 × 10 −6 W R = 64

1875 mW ⇒ R= 64



Hence, the correct answer is (B).

3.

For balanced meter bridge

X  = {where, R = 90 W } R ( 100 −  )

R=



⇒ R∝



⇒



Hence, the correct answer is (D).

8.

The given three circuits R1, R2 and R3 are equivalent to the following three circuits.

X 40 = 90 100 − 40



⇒



⇒ X = 60 W



⇒ X=R



ΔX Δ Δ = + ⇒ X  100 − 



⇒



⇒ ΔX = 0.25

1 P

1 1 1 > > R100 R60 R40

1Ω



( 100 −  )

V2 P

7.

1Ω

1Ω



X

( 52 + 1 )

=

10

( 48 + 2 )

1Ω

R2

R1

ΔX 0.1 0.1 = + X 40 60

1Ω

1Ω

1Ω

1Ω

3V

1Ω

Using the concept of balanced Wheat stone bridge, we have

⇒

3V

1Ω

1Ω

R3

P R = Q S

1Ω

1Ω

3V

1Ω

So, X = ( 60 ± 0.25 ) W Hence, the correct answer is (C). 4.

1Ω

1Ω

10 × 53 = 10.6 W 50 Hence, the correct answer is (B).

1/2 Ω

3V

3V

⇒ X=

R1

R2

5.

We will require a voltmeter, an ammeter, a test resistor and a variable battery to verify Ohm’s law. Voltmeter which is made by connecting a high resistance with a galvanometer is connected in parallel with the test resistor.  Further, an ammeter which is formed by connecting a low resistance in parallel with galvanometer is required to measure the current through test resistor. Hence, the correct answer is (C).

ρ A

6.

R=



⇒ R=



Independent of L Hence, the correct answer is (C).

ρL ρ = tL t

03_Ch 3_Hints and Explanation_P2.indd 296

2Ω

3V

R3

P1 = P2 =

P3 =

32 =9W 1 32 = 18 W 1 2 32 = 4.5 W 2

9/20/2019 11:57:54 AM

Hints and Explanations H.297 3 ( 2Rε 0 )

6 Rε 0 5d + 3Vt



Hence, the correct answer is (C).



⇒ τ=

9.

R>2W



Hence, the correct answer is (A).



⇒ 100 − x > x

11. From Y to X charge flows to plates a and b

Applying

P R = Q S

( qa + qb )i = 0 , ( qa + qb ) f

2Ω

d − 3Vt + 6 d − 2d + 6Vt

3 μF a

6 μF b

18 μ C

18 μ C

R

=

= 27 mC 9 μC

G 3Ω 1 A 6Ω

G x + 20

We have

80 – x

2 x = …(1) R 100 − x

R = 3 W Hence, the correct answer is (A).

10. Time constant = τ = RC , where



9V Initial figure (When switch is closed)

27 mC charge flows from Y to X



∴



Hence, the correct answer is (C).

H = I 2 Rt

⇒ H ∝R



⇒



⇒ H BC = 4 H AB



Hence, the correct answer is (A).

13.

τ = CR

H AB RAB ( 1 / 2r ) 1 = = =  2 H BC RBC (1/ r ) 4 2

{

as R ∝

1 1 ∝ A r2

}

τ 1 = ( C1 + C2 ) ( R1 + R2 ) = 18 ms

⎛ Aε 0 ⎞ ⎛ KAε 0 ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟ C1C2 x ⎠ C= = d−x Aε 0 Aε 0 C1 + C2 + d−x x

⎛ C C ⎞⎛ R R ⎞ 8 2 8 τ 2 = ⎜ 1 2 ⎟ ⎜ 1 2 ⎟ = × = ms ⎝ C1 + C2 ⎠ ⎝ R1 + R2 ⎠ 6 3 9 ⎛ RR ⎞ ⎛ 2⎞ τ 3 = ( C1 + C2 ) ⎜ 1 2 ⎟ = ( 6 ) ⎜ ⎟ = 4 ms ⎝ 3⎠ ⎝ R1 + R2 ⎠

C d

6Ω

12. Current flowing through both the bars is equal. Now, the heat produced is given by

R x + 20 = …(2) 2 80 − x Solving equations (1) and (2), we get

Y

3Ω

9V Initial figure (When switch was open)

2Ω

R



36 μ C

1A 1A 100 – x

x

X

CHAPTER 3

P2 > P1 > P3

d/3

R



Hence, the correct answer is (B).

14.

S (i – ig)



⇒ C=



τ= ⇒ 

KAε 0 d where x = − Vt x + K(d − x) 3

RKAε 0 where A = 1 m 2 and d d ⎛ ⎞ − Vt + K ⎜ d − + Vt ⎟ ⎝ ⎠ 3 3 K=2

03_Ch 3_Hints and Explanation_P2.indd 297

i a ig

G

b

(

)

Vab = ig .G = i − ig S

∴

G⎞ ⎛ i = ⎜ 1 + ⎟ ig ⎝ S⎠

9/20/2019 11:58:04 AM

H.298  JEE Advanced Physics: Electrostatics and Current Electricity

Substituting the values, we get

i = 100.1 mA

22. The ratio

AC will remain unchanged. CB



Hence, the correct answer is (A).



Hence, the correct answer is (A).

15.

W = 0. Therefore, from First Law of Thermodynamics,

23.

P = I 2R



Current is same, so P ∝ R .

ΔU = ΔQ = I 2 Rt = ( 1 ) ( 100 ) ( 5 × 60 ) J = 30 kJ 2



Hence, the correct answer is (D).

16. Current in the respective loop will remain confined in the loop itself. Therefore, current through 2 W resistance = 0

Hence, the correct answer is (C).

2 r , in third 3 r 3r case it is and in fourth case the net resistance is . 2 3

In the first case it is 3r, in second case it is

RIII < RII < RIV < RI

17. Given: VC = 3VR = 3 ( V − VC )

⇒ PIII < PII < PIV < PI



Hence, the correct answer is (A).

Here, V is the applied potential.

24.



3 ⇒ VC = V 4

(

⇒ V 1− e ⇒ e

− t/τ C

− t/τ C

)

P

3 = V 4

1 = …(1) 4

Here, τ C = cR = 10 sec Substituting this value of τ C in equation (1) and solving for t we get t = 13.86 sec

Hence, the correct answer is (A).

18.

RPQ =



⇒ RPQ is maximum.



Hence, the correct answer is (A).

5 4 3 r , RQR = r and RPR = r 11 11 11

t

E − 2 0. Charging current, I = e RC R t ⎛ E⎞ Taking log both sides, log I = log ⎜ ⎟ − ⎝ R ⎠ RC

When R is doubled, slope of curve increases. Also at t = 0, the current will be less. Graph Q represents the best. Hence, the correct answer is (B). 21. Ammeter is always connected in series and voltmeter in parallel. Hence, the correct answer is (A).

03_Ch 3_Hints and Explanation_P2.indd 298

2R

r

r

2R

2R

Q



Hence, the correct answer is (A).

25.

P=



⇒ R1 =

V2 V2 so, R = R P V2 V2 and R2 = R3 = 100 60

Now, W1 = and W3 =

1 9. BC , CD and BA are known resistance. The unknown resistance is connected between A and D. Hence, the correct answer is (C).



2R

( 250 )2

R , W2 = 2 1

( R1 + R2 )

( 250 )2

( R1 + R2 )2

R2

( 250 )2 R3

W1 : W2 : W3 = 15 : 25 : 64

⇒ W1 < W2 < W3



Hence, the correct answer is (D).

26. Current I can be independent of R6 only when R1 , R2 , R3 , R4 and R6 from a balanced Wheatstone bridge. R R Therefore, 1 = 3 or R1R4 = R2 R3 . R2 R4 27.

Hence, the correct answer is (C). V2 t R In the first case (wire of length  ) H=

mc ΔT =

( 3 E )2 R

t …(1)

9/20/2019 11:58:16 AM

Hints and Explanations H.299 In the second case (wire of length 2 )

( 2m ) c ΔT =

( nE )2 2R

t …(2)

{Because doubling the length will simply double the mass and the resistance of the wire} so, from (1) and (2) we get, n2 = 36

⇒



Hence, the correct answer is (B).

n=6

31.

i = neAvd

1 1 ∝ neA A Therefore, for non-uniform cross-section (different values of A ) drift speed will be different at different sections. Only current (or rate of flow of charge) will be same. Hence, the correct answer is (D).

Drift speed, vd =

32. Since, the capacitor plates are directly connected to the battery, it will take no time in charging. C

28. In steady state the branch containing the capacitor can be omitted and hence current in the circuit is I =

⇒

I=

2V − V R + 2R

R V

V 3R



V

R

1 I

V

6 5 2V



I

2

C

I

– +

3

2R

Hence, the correct answer is (D).

33. The given circuit is that of a Wheatstone bridge. R

⇒



⇒

VC = V −



Hence, the correct answer is (C).

R

6R

4Ω R

4R C

−VC − V + 2V − I ( 2R ) = 0

R

R

E

4

For loop 36543

V ( 2R ) VC = −V + 2V − 3R

R

2R R

A

B

2R

2V V = 3 3

4R D E

4Ω

29. Reading of Galvanometer remains same whether S is opened or closed i.e. no current must be flowing through this branch. Hence P and Q are in series and R and G are in series too.



So, I R = IG

Thus, no current will flow across 6R of the side CD. The given circuit will now be equivalent to



Hence, the correct answer is (A).

The circuit is a balanced one since,



Resistance across AC resistance across CB = resistance across AD resistance across BD

9 3 0. ⇒ I = = 1 A 9 At A a current of 1 A divides into 0.5 A and 0.5 A. At B the current of 0.5 A divides into 0.25 A and 0.25 A

CHAPTER 3



R

2R

2R

4R E

Hence, the correct answer is (D).

4Ω 3R 2R 6R E 03_Ch 3_Hints and Explanation_P2.indd 299

4Ω

E 4Ω

9/20/2019 11:58:22 AM

R

2R

2R

4R

E Electrostatics and Current Electricity H.300  JEE Advanced Physics: 4Ω 3R

2

2R 6R

P5 = ( i ) ( 5 )

E 4Ω

4Ω

For maximum power, net external resistance = Total internal resistance. ⇒ 2R = 4



⇒ R=2W



Hence, the correct answer is (B).

P4 1 = P5 5



⇒



⇒ P4 =



Hence, the correct answer is (B).

35. Cooling increases the resistance of a semiconductor and decreases the resistance of a conductor. Hence, the correct answer is (D).

1.

S1

2V

60 Ω

20 Ω

5 V 30 Ω 70 Ω I1 100 Ω

5 = 25 mA 200



⇒ I1 =



At steady state, for S1 closed, we have



Ceq = 8 mF and Q = 40 mC

i

Therefore, current i =

2 1 = A 20 10

P C1

4Ω

i/2 6 Ω

5Ω

i

80 μ F 5 V C4

10 μ F

C3

Hence, the correct answer is (C).

37. Since, resistance in upper branch of the circuit is twice the resistance in lower branch. Hence, current there will be half.

100 Ω

When switch S1 is closed, so the equivalent circuit at t = 0 is shown in figure.

30 Ω

30 Ω

03_Ch 3_Hints and Explanation_P2.indd 300

C3 10 V

30 Ω

i



S2

Q

30 Ω



C4 5 V 30 Ω

P

10 μ F C2 30 Ω

C1 70 Ω

36. The simplified circuit is shown in the figure.

2V

P5 10 = = 2 cals −1 5 5

Multiple Correct Choice Type Problems

34. Resistivity of conductors increases with increase in temperature because rate of collisions between free electrons and ions increase with increase of temperature. However, the resistivity of semiconductors decreases with increase in temperature, because more and more covalent bonds are broken at higher temperatures. Hence, the correct answer is (C).

2V

2

2

E



(P = i R)

⎛ i⎞ Now, P4 = ⎜ ⎟ ( 4 )  ⎝ 2⎠

80 μ F

Q

So, V ( C1 ) = 4 volt and V ( C3 ) = V ( C4 ) = 0.5 volt At steady state potential difference between P and Q is 4 volt. Now when switch S2 is closed, then equivalent circuit is shown in figure.

9/20/2019 11:58:28 AM

Hints and Explanations H.301

4V

4V

30 Ω

70 Ω

130 Ω

10 V

30 Ω

4V 91 2

10 V

Q

Q

⇒ I PQ =



Hence, (A) and (C) are correct.

2.

Maximum coil current is

( 10 + R0 ) × 2 × 10 −6 = 100 × 10 −3 ⇒ R0 = 49990 W G Rs

Rs × 9.98 × 10 ⇒ Rs = 0.02 W

= 2 × 10

−5

A

1000 Ω

W

980.39 Ω

0.099 Ω

If Req is the equivalent resistance of the circuit, then Req = 980.48 W V0 980.48

If V ′ be the reading of the voltmeter, then

V′ =

V0 × 50000 980.48 × 51

03_Ch 3_Hints and Explanation_P2.indd 301



i2 –q2 +q2 20 μ F

Just after pressing key,

5 − 25000i1 = 0



{As charge on both capacitors is zero} ⇒ i1 = 0.2 mA and i2 = 0.1 mA

Also, VB + 25000i1 = VA ⇒ VB − VA = −5V

After a long time, steady state is achieved, so i1 and i2 = 0 q1 =0 40



⇒ 5−



⇒ q1 = 200 mC



V0

⇒ I=

V

50 kΩ

and 5−

Now, Ohm’s Law setup is shown in the figure. V

25 kΩ

5V



When converted into an ammeter of range 1 mA, then we have

−4

i1

–q1



When converted into voltmeter of range 100 mV, then we have



40 μ F +q1 i i1

5 − 50000i2 = 0 

R0

G



3.

I c = 2 × 10 −6 A and coil resistance is

Vmax = 2 × 10 −6 × 10 = 2 × 10 −5 volt

V ′ 50000 = = 980.4 W I 51

Hence, (A) and (B) are correct.

A

So, maximum potential difference across coil is



R( measured ) =

i2

Rc = 10 W



So, the measured value of resistance is

6×2 = 0.08 A 151







P

CHAPTER 3

P

q2 =0 20

⇒ q2 = 100 mC

So, VB −

q2 = VA 20



⇒ VB − VA = +5 V



Hence, (A) is correct



For capacitor 1, q1 = 200 ( 1 − e − t 1 ) mC



⇒ i1 =



For capacitor 2, q2 = 100 [ 1 − e − t 1 ] mC



⇒ i2 =

dq1 1 − t 1 = e mA dt 5

dq2 1 −t 1 = e mA dt 10

VB −

q2 + i1 × 25 − VA = 0 20

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H.302  JEE Advanced Physics: Electrostatics and Current Electricity



At t = ln 2 , VB − VA = 5 ( 1 − 1 ) = 0 

So, (B) is correct

At t = 1, i = i1 + i2 =

{

∵ e − ln 2 = −

For (C)



⇒ VB − VA = 5 ( 1 − e − t ) − 5e − t = 5 ( 1 − 2e − t ) 1 2

ig

}

ig



I1 = 2ig + 2i = 2ig + 2ig



For (D)

Hence, (D) is correct. Hence, (A),(B), (C) and (D) are correct.

I

4.

When the filament breaks, the temperature of the filament will be high. Since according to Wein’s Law, 1 λ m ∝ , so T 1 ⎫ ⎧ ν m ∝ T  ⎨∵ ν m ∝ ⎬ λm ⎭ ⎩ Hence the filament emits more light at higher band of frequencies.  Towards the end of life of the bulb, R (resistance) increases as temperature increases. Since V = con2



V . So, less power is consumed towards R the end of life of bulb. Hence, (C) and (D) are correct.

5.

For (A),

stant, so P =

R

ig

Rc

R



V1 = 2ig R + 2ig RC



For (B), R

R

R R

i

So, i1 = i2 = 0

Rc

i

1 −1 1 −1 ⎛ 3 ⎞ 1 e + e =⎜ ⎟ ⎝ 10 ⎠ e 5 10

1 1 3 At t = 0, i = i1 + i2 = + = 5 10 10 So, (C) is correct After a long time, q1 = q2 = constant

Rc

RC R

ig

ig

Rc

Rc

i

R

i

R

I 2 = ig + 2i = ig + 2 ×

⇒ I 2 = I1 +



⇒ I 2 < I1 as RC
V1 as > RC 2

Hence, 2R > RC

So, (B) is correct

03_Ch 3_Hints and Explanation_P2.indd 302

R

R R ⇒ I 2 = 2ig + 2ig C + 2ig C − ig R R

2ig

V2 = V1 + ig ( 2R − RC )

2ig RC

T 4Ω

4Ω

12 V

I 2 =

12 =2A 2+2+2



⇒ I3 =

12 =1A 4+4+4



⇒ I1 = I 2 + I 3 = 3 A

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Hints and Explanations H.303 ⇒ I PQ = 0



⇒ VP = VQ

3 ⎡ ( 2 ) ( 1.5 ) ⎤ R′ = R1  R2 = ⎢ = ⎣ 2 + 1.5 ⎥⎦ 3.5

Potential drop (from left to right) across each resistance is

VL′ =

12 =4V 3

R′ × 24 V = 3 V R2 + R′

Now power dissipated in RL is

PL′ =

VL′ 2 3 2 = = 6 mW RL 1.5



⇒ VMS = 2 × 4 = 8 V



⇒ VNQ = 1 × 4 = 4 V





⇒ VS < VQ



Hence, (A), (B), (C) and (D) are correct.

7.

16 ⎛ 6 × 1.5 ⎞ 6 Req = ⎜ + 2⎟ = + 2 = kW ⎝ 7.5 ⎠ 5 5



I=

8. At 0 K, a semiconductor becomes a perfect insulator. Therefore, at 0 K, if some potential difference is applied across an insulator or semiconductor, current is zero. But a conductor will become a super conductors at 0 K. Therefore, current will be infinite. In reverse biasing at 300 K through a p -n junction diode, a small finite current flows due to minority charge carriers. Hence, (A), (B) and (D) are correct.

V 24 3 = × 5 mA = × 5 = 7.5 mA Req 16 2

9. I

24 V

2 kΩ

R1

6 kΩ

R2

RL

for potential difference across R1



⇒ V1 = 7.5 × 2 = 15 V



for potential difference across R2



⇒ V2 = 24 − 15 = 9 V



for power

PR1 : PR2 = PRL =

:

V22

R1 R2

1.5 kΩ



( 15 )

2

2

:

⎛ ⎞ 50 × 10 −6 100 ) ≈ 1 W For option (C): S = ⎜ −3 −6 ⎟ ( ⎝ 5 × 10 − 50 × 10 ⎠

To change it in voltmeter, a high resistance R is put in series, where R is given by R =

=

To increase the range of ammeter a parallel resistance (called shunt) is required which is given by

⎛ ig ⎞ S = ⎜ ⎟G ⎝ i − ig ⎠



V12

Hence, (A) and (D) are correct.

CHAPTER 3



For option (B):

R =

2

9 225 6 25 = × = 6 2 81 3

V22 92 = = 54 mW RL 1.5

If R1 and R2 are interchanged

V −G ig 10 − 100 ≈ 200 kW 50 × 10 −6



Hence, (B) and (C) are correct.

10.

q = q 0e



⇒ q = ( CV0 ) e

−

t RC −

t RC t

I

6 kΩ

I =

R2

dq CV − = − 0 e RC dt RC

V0 R which we observe is independent of the capacitance and only depends upon resistance. Hence, the correct answer is (B). I ( t = 0 ) = −

24 V

03_Ch 3_Hints and Explanation_P2.indd 303

2 kΩ

R1

RL

1.5 kΩ

9/20/2019 11:58:58 AM

H.304  JEE Advanced Physics: Electrostatics and Current Electricity

Integer/Numerical Answer Type Questions 1.

R=2Ω

6.5 V

2Ω

G

V

ig ( G + 4990 ) = V 4Ω

2Ω

10 Ω 4Ω

12 Ω

R=2Ω

ig 4990 Ω

1Ω 8Ω

6Ω

3.

6 ( G + 4990 ) = 30 1000



⇒



⇒ G + 4990 =



⇒ G = 10 W

30000 = 5000 6

Vab = Vcd

1Ω

6Ω 6.5 V

4Ω

2Ω

12 Ω

1.5 A a 4Ω

R=2Ω

6Ω

12 Ω

2Ω 10 Ω

4Ω

R=2Ω

6Ω

2Ω

6.5 V 12 Ω

4Ω

d

(1.5 – ig)

10 Ω

6.5 V

S

c

2Ω

G

ig



⇒ igG = ( 1.5 − ig ) S



⇒



⇒ S=

60 2n = 1494 249



⇒ n=

249 × 30 1494



⇒ n=

2490 =5 498

4.

In series, i =



⎛ 2E ⎞ ⇒ J1 = i 2 R = ⎜ ⋅R ⎝ 2 + R ⎟⎠



In parallel, i =



⎛ E ⎞ ⇒ J 2 = i2R = ⎜ ⋅R ⎝ 0.5 + R ⎟⎠

b

6 6 ⎞ ⎛ × 10 = ⎜ 1.5 − ⎟S ⎝ 1000 1000 ⎠

2E 2+R 2

E 0.5 + R 2

R=2Ω

Since,

1A 4.5 Ω 6.5 V

VAB = Equivalent emf of two batteries in parallel

E1 E2 6 3 + + r1 r2 = = 1 2 =5V 1 1 1 1 + + r1 r2 1 2

03_Ch 3_Hints and Explanation_P2.indd 304

2 J1 4 ( 0.5 + R ) = J2 ( 2 + R )2



⇒



⇒ 1.5 =



Solving we get, R = 4 W

5.

Voltage across the capacitors will increase from 0 to 10 V exponentially. The voltage at time t will be given by

2.

J1 = 2.25 J2

2 ( 0.5 + R ) (2 + R)

9/20/2019 11:59:05 AM

Hints and Explanations H.305

(

−t

V = 10 1 − e τ C

)

Assertion and Reasoning Type Problems 1.

Here τ c = Cnet Rnet = ( 1 × 106 ) ( 4 × 10 −6 ) = 4 s

∴

(

V = 10 1 −

−t e4

R 2 =  1x G

R

)

l1

X l2

(

−t

4 = 10 1 − e 4

) or e

−t 4

= 0.6 =

Taking log both sides we have,

−

t = In 3 − In 5 4

or t = 4 ( In 5 − In 3 ) = 2 s

03_Ch 3_Hints and Explanation_P2.indd 305

3 5

To get null point at the same position, means  1 and  2 are still the same. As temperature increases, value of unknown resistance increases. To get the same null point, R must be increased. So Statement-1 is wrong. Statement-2 is True. Hence, the correct answer is (D).

CHAPTER 3

Substituting V = 4 volt , we have

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03_Ch 3_Hints and Explanation_P2.indd 306

9/20/2019 11:59:09 AM