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Introduction to Electrodynamics David J Grif ths Reed College Prentice Hall A Prentice Hall Upper Saddle River New Jersey 07458 Library of Congress CataloginginPublication Data Grif ths David J David Jeffrey Introduction to electrodynamics David J Grif ths 7 3rd ed p cm Includes bibliographical references and index ISBN 0 13A805326 X l Electrodynamics I Title OC680G74 1999 53746dc21 9850525 ClP Executive Editor Alison Reeves Production Editor Kim Dellas Manufacturing Manager Trudy Pisciotti Art Director Jayne Conte Cover Designer Bruce Kenselaar Editorial Assistant Gillian Keiff Composition PreTEX Inc From ice IIall 1999 1989 1981 by Prentice Hall Inc 39A39 Upper Saddle River New Jersey 07458 All rights reserved No part of this book may be reproduced in any form or by any means without permission in writing from the publisher Reprinted with corrections September 1999 Printed in the United States of America 10 9 8 7 6 5 ISBN DlEEDSEEEX Prentice Hall International UK Limited London Prentice Hall of Australia Pty Limited Sydney Prentice Hall Canada Inc Toronto Prentice Hall Hispanoamericana SA Mexico City Prentice Hall of India Private Limited New Delhi Prentice Hall of Japan Inc Tokyo Prentice Hall Asia Pte Ltd Singapore Editora Prentice Hall do Brasil Ltda Rio de Janeiro Contents Preface ix Advertisement Xi 1 Vector Analysis 1 11 Vector Algebra 1 111 Vector Operations 1 112 Vector Algebra Component Form 4 113 Triple Products 7 114 Position Displacement and Separation Vectors 8 115 How Vectors Transform 10 12 Differential Calculus 13 121 Ordinary Derivatives 13 122 Gradient 13 123 The Operator V 16 124 The Divergence 17 125 The Curl 19 126 Product Rules 20 127 Second Derivatives 22 13 Integral Calculus 24 131 Line Surface and Volume Integrals 24 132 The Fundamental Theorem of Calculus 28 133 The Fundamental Theorem for Gradients 29 134 The Fundamental Theorem for Divergences 31 135 The Fundamental Theorem for Curls 34 136 Integration by Parts 37 14 Curvilinear Coordinates 38 141 Spherical Polar Coordinates 38 142 Cylindrical Coordinates 43 15 The Dirac Delta Function 45 151 The Divergence of fr2 45 152 The OneDimensional Dirac Delta Function 46 iii CONTENTS 153 The ThreeDimensional Delta Function 50 16 The Theory of Vector Fields 52 161 The Helmholtz Theorem 52 162 Potentials 53 Electrostatics 58 21 The Electric Field 58 211 Introduction 58 212 Coulomb s Law 59 213 The Electric Field 60 214 Continuous Charge Distributions 61 22 Divergence and Curl of Electrostatic Fields 65 221 Field Lines Flux and Gauss s Law 65 222 The Divergence of E 69 223 Applications of Gauss s Law 70 224 The Cur1 of E 76 23 Electric Potential 77 231 Introduction to Potential 77 232 Comments on Potential 79 233 Poisson s Equation and Laplace s Equation 83 234 The Potential of a Localized Charge Distribution 83 235 Summary Electrostatic Boundary Conditions 87 24 Work and Energy in Electrostatics 90 241 The Work Done to Move a Charge 90 242 The Energy of a Point Charge Distribution 91 243 The Energy of a Continuous Charge Distribution 93 244 Comments on Electrostatic Energy 95 25 Conductors 96 251 Basic Properties 96 252 Induced Charges 98 253 Surface Charge and the Force on a Conductor 102 254 Capacitors 103 Special Techniques 110 31 Laplace s Equation 110 311 Introduction 1 10 312 Laplace s Equation in One Dimension 1 11 313 Laplace s Equation in Two Dimensions 112 314 Laplace s Equation in Three Dimensions 114 315 Boundary Conditions and Uniqueness Theorems 116 316 Conductors and the Second Uniqueness Theorem 118 32 The Method of Images 121 321 The Classic Image Problem 121 322 Induced Surface Charge 123 CONTENTS v 323 Force and Energy 123 324 Other Image Problems 124 33 Separation of Variables 127 331 Cartesian Coordinates 127 332 Spherical Coordinates 137 34 Multipole Expansion 146 341 Approximate Potentials at Large Distances 146 342 The MonOpole and Dipole Terms 149 343 Origin of Coordinates in Multipole Expansions 151 344 The Electric Field of a Dipole 153 4 Electric Fields in Matter 160 41 Polarization 160 411 Dielectrics 160 412 Induced Dipoles 160 413 Alignment of Polar Molecules 163 414 Polarization 166 42 The Field of a Polarized Object 166 421 Bound Charges 166 422 Physical Interpretation of Bound Charges 170 423 The Field Inside a Dielectric 173 43 The Electric Displacement 175 431 Gauss s Law in the Presence of Dielectrics 175 432 A Deceptive Parallel 178 433 Boundary Conditions 178 44 Linear Dielectrics 179 441 Susceptibility Permittivity Dielectric Constant 179 442 Boundary Value Problems with Linear Dielectrics 186 443 Energy in Dielectric Systems 191 444 Forces on Dielectrics 193 5 Magnetostatics 202 51 The Lorentz Force Law 202 511 Magnetic Fields 202 512 Magnetic Forces 204 513 Currents 208 52 The BiotSavart Law 215 521 Steady Currents 215 522 The Magnetic Field of a Steady Current 215 53 The Divergence and Curl of B 221 531 StraightLine Currents 221 532 The Divergence and Curl of B 222 533 Applications of Ampere s Law 225 534 Comparison of Magnetostatics and Electrostatics 232 vi CONTENTS 54 Magnetic Vector Potential 234 541 The Vector Potential 234 542 Summary Magnetostatic Boundary Conditions 240 543 Multipole Expansion of the Vector Potential 242 Magnetic Fields in Matter 255 61 Magnetization 255 611 Diamagnets Paramagnets Ferromagnets 255 612 Torques and Forces on Magnetic Dipoles 255 613 Effect of a Magnetic Field on Atomic Orbits 260 614 Magnetization 262 62 The Field of a Magnetized Object 263 621 Bound Currents 263 622 Physical Interpretation of Bound Currents 266 623 The Magnetic Field Inside Matter 268 63 The Auxiliary Field H 269 631 Amp re s law in Magnetized Materials 269 632 A Deceptive Parallel 273 633 Boundary Conditions 273 64 Linear and Nonlinear Media 274 641 Magnetic Susceptibility and Permeability 274 642 Ferromagnetism 278 Electrodynamics 285 71 Electromotive Force 285 711 Ohm s Law 285 712 Electromotive Force 292 713 Motional emf 294 72 Electromagnetic Induction 301 721 Faraday s Law 301 722 The Induced Electric Field 305 723 Inductance 310 724 Energy in Magnetic Fields 317 73 Maxwell s Equations 321 731 ElectrodynamiCs Before Maxwell 321 732 How Maxwell Fixed Amp re s Law 323 733 Maxwell s Equations 326 734 Magnetic Charge 327 735 Maxwell s Equations in Matter 328 736 Boundary Conditions 331 CONTENTS vii 8 Conservation Laws 345 81 Charge and Energy 345 811 The Continuity Equation 345 812 Poynting s Theorem 346 82 Momentum 349 821 Newton s Third Law in Electrodynamics 349 822 Maxwell s Stress Tensor 351 823 Conservation of Momentum 355 824 Angular Momentum 358 9 Electromagnetic Waves 364 91 Waves in One Dimension 364 911 The Wave Equation 364 912 Sinusoidal Waves 367 913 Boundary Conditions Re ection and Transmission 370 914 Polarization 373 92 Electromagnetic Waves in Vacuum 375 921 The Wave Equation for E and B 375 922 Monochromatic Plane Waves 376 923 Energy and Momentum in Electromagnetic Waves 380 93 Electromagnetic Waves in Matter 382 931 Propagation in Linear Media 382 932 Re ection and Transmission at Normal Incidence 384 933 Re ection and Transmission at Oblique Incidence 386 94 Absorption and DiSpersion 392 941 Electromagnetic Waves in Conductors 392 942 Re ection at a Conducting Surface 396 943 The Frequency Dependence of Permittivity 398 95 Guided Waves 405 951 Wave Guides 405 952 TE Waves in a Rectangular Wave Guide 408 953 The Coaxial Transmission Line 411 10 Potentials and Fields 416 101 The Potential Formulation 416 1011 Scalar and Vector Potentials 416 1012 Gauge Transformations 419 1013 Coulomb Gauge and Lorentz Gauge 421 102 Continuous Distributions 422 1021 Retarded Potentials 422 1022 Je menko s Equations 427 103 Point Charges 429 1031 Li nardWiechert Potentials 429 1032 The Fields of a Moving Point Charge 435 CONTENTS viii 11 Radiation 443 111 Dipole Radiation 443 1111 What is Radiation 443 1112 Electric Dipole Radiation 444 1113 Magnetic Dipole Radiation 451 1114 Radiation from an Arbitrary Source 454 112 Point Charges 460 1121 Power Radiated by a Point Charge 460 1 122 Radiation Reaction 465 1123 The Physical Basis of the Radiation Reaction 469 12 Electrodynamics and Relativity 477 121 The Special Theory of Relativity 477 1211 Einstein s Postulates 477 1212 The Geometry of Relativity 483 1213 The Lorentz Transformations 493 1214 The Structure of Spacetime 500 122 Relativistic Mechanics 507 1221 Proper Time and Proper Velocity 507 1222 Relativistic Energy and Momentum 509 1223 Relativistic Kinematics 51 1 1224 Relativistic Dynamics 516 123 Relativistic Electrodynamics 522 1231 Magnetism as a Relativistic Phenomenon 522 1232 How the Fields Transform 525 1233 The Field Tensor 535 1234 Electrodynamics in Tensor Notation 537 1235 Relativistic Potentials 541 A Vector Calculus in Curvilinear Coordinates 547 A1 Introduction 547 A2 Notation 547 A3 Gradient 548 A4 Divergence 549 A5 Curl 552 A6 Laplacian 554 B The Helmholtz Theorem 555 C Units 558 Index 562 Preface This is a textbook on electricity and magnetism designed for an undergraduate course at the junior or senior level It can be covered comfortably in two semesters maybe even with room to spare for special t0pics AC circuits numerical methods plasma physics transmission lines antenna theory etc A one semester course could reasonably st0p after Chapter 7 Unlike quantum mechanics or thermal physics for example there is a fairly general consensus with respect to the teaching of electrodynamics the subjects to be included and even their order of presentation are not particularly controversial and textbooks differ mainly in style and tone My approach is perhaps less formal than most I think this makes dif cult ideas more interesting and accessible For the third edition I have made a large number of small changes in the interests of clarity and grace I have also modi ed some notation to avoid inconsistencies or ambiguities Thus the Cartesian unit vectors i j and I have been replaced with 2 y and i so that all vectors are bold and all unit vectors inherit the letter of the corresponding coordinate This also frees up k to be the pr0pagation vector for electromagnetic waves It has always bothered me to use the same letter r for the spherical coordinate distance from the origin and the cylindrical coordinate distance from the z axis A common alternative for the latter is p but that has more important business in electrodynamics and after an exhaustive search I settled on the underemployed letter 5 I hope this unorthodox usage will not be confusing Some readers have urged me to abandon the script letter the vector from a source point r to the eld point r in favor of the more explicit r r But this makes many equations distractingly cumbersome especially when the unit vector 22 is involved I know from my own teaching experience that unwary students are tempted to read 4 as r it certainly makes the integrals easier I have inserted a section in Chapter 1 explaining this notation and I hope that will help If you are a student please take note 4 2 r r which is not the same as r If you re a teacher please warn your students to pay close attention to the meaning of L I think it s good notation but it does have to be handled with care The main structural change is that I have removed the conservation laws and potentials from Chapter 7 creating two new short chapters 8 and 10 This should more smoothly accommodate onesemester courses and it gives a tighter focus to Chapter 7 I have added some problems and examples and removed a few that were not effective And I have included more references to the accessible literature particularly the American Journal of Physics I realize of course that most readers will not have the time or incli ix x PREFACE nation to consult these resources but I think it is worthwhile anyway if only to emphasize that electrodynamics notwithstanding its venerable age is very much alive and intriguing new discoveries are being made all the time I h0pe that occasionally a problem will pique your curiosity and you will be inSpired to look up the reference some of them are real gems As in the previous editions 1 distinguish two kinds of problems Some have a speci c pedagogical purpose and should be worked immediately after reading the section to which they pertain these I have placed at the pertinent point within the chapter In a few cases the solution to a problem is used later in the text these are indicated by a bullet o in the left margin Longer problems or those of a more general nature will be found at the end of each chapter When I teach the subject I assign some of these and work a few of them in class Unusually challenging problems are agged by an exclamation point in the margin Many readers have asked that the answers to problems be provided at the back of the book unfortunately just as many are strenuously opposed I have compromised supplying answers when this seems particularly appr0priate A complete solution manual is available to instructors from the publisher I have bene tted from the comments of many colleagues I cannot list them all here But I would like to thank the following peOple for suggestions that contributed Speci cally to the third edition Burton Brody Bard Steven Grimes Ohio Mark Heald Swarth more Jim McTavish Liverpool Matthew Moelter Puget Sound Paul Nachman New Mexico State Gigi Quartapelle Milan Carl A Rotter West Virginia Daniel Schroeder Weber State Juri Silmberg Ryerson Polytechnic Walther N Spjeldvik Weber State Larry Tankersley Naval Academy and Dudley Towne Amherst Practically everything I know about electrodynamics certainly about teaching electrodynamics I owe to Edward Purcell David J Grif ths Advertisement What is electrodynamics and how does it t into the general scheme of physics Four Realms of Mechanics In the diagram below I have sketched out the four great realms of mechanics Classical Mechanics Quantum Mechanics Newton Bohr Heisenberg Schrodinger et al Special Relativity Quantum Field Theory Einstein Dirac Pauli Feynman Schwinger et al Newtonian mechanics was found to be inadequate in the early years of this century it s all right in everyday life but for objects moving at high speeds near the speed of light it is incorrect and must be replaced by special relativity introduced by Einstein in 1905 for objects that are extremely small near the size of atoms it fails for different reasons and is superseded by quantum mechanics developed by Bohr Schrodinger Heisenberg and many others in the twenties mostly For objects that are both very fast and very small as is common in modern particle physics a mechanics that combines relativity and quantum principles is in order this relativistic quantum mechanics is known as quantum eld theory it was worked out in the thirties and forties but even today it cannot claim to be a completely satisfactory system In this book save for the last chapter we shall work exclusively in the domain of classical mechanics although electrodynamics extends with unique simplicity to the other three realms In fact the theory is in most respects automatically consistent with special relativity for which it was historically the main stimulus xi xii ADVERTISEMENT Four Kinds of Forces Mechanics tells us how a system will behave when subjected to a given force There are just four basic forces known presently to physics 1 list them in the order of decreasing strength 1 Strong 2 Electromagnetic 3 Weak 4 Gravitational The brevity of this list may surprise you Where is friction Where is the normal force that keeps you from falling through the oor Where are the chemical forces that bind molecules together Where is the force of impact between two colliding billiard balls The answer is that all these forces are electromagnetic Indeed it is scarcely an exaggeration to say that we live in an electromagnetic world for virtually every force we experience in everyday life with the exception of gravity is electromagnetic in origin The strong forces which hold protons and neutrons together in the atomic nucleus have extremely short range so we do not feel them in spite of the fact that they are a hundred times more powerful than electrical forces The weak forces which account for certain kinds of radioactive decay are not only of short range they are far weaker than electromagnetic ones to begin with As for gravity it is so pitifully feeble compared to all of the others that it is only by virtue of huge mass concentrations like the earth and the sun that we ever notice it at all The electrical repulsion between two electrons is 1042 times as large as their gravitational attraction and if atoms were held together by gravitational instead of electrical forces a single hydrogen atom would be much larger than the known universe Not only are electromagnetic forces overwhelmingly the dominant ones in everyday life they are also at present the only ones that are completely understood There is of course a classical theory of gravity Newton s law of universal gravitation and a relativistic one Einstein s general relativity but no entirely satisfactory quantum mechanical theory of gravity has been constructed though many people are working on it At the present time there is a very successful if cumbersome theory for the weak interactions and a strikingly attractive candidate called chromodynamics for the strong interactions All these theories draw their inspiration from electrodynamics none can claim conclusive experimental veri cation at this stage So electrodynamics a beautifully complete and successful theory has become a kind of paradigm for physicists an ideal model that other theories strive to emulate The laws of classical electrodynamics were discovered in bits and pieces by Franklin Coulomb Ampere Faraday and others but the person who completed the job and packaged it all in the compact and consistent form it has today was James Clerk Maxwell The theory is now a little over a hundred years old xiii The Uni cation of Physical Theories In the beginning electricity and magnetism were entirely separate subjects The one dealt with glass rods and cat s fur pith balls batteries currents electrolysis and lightning the other with bar magnets iron lings compass needles and the North Pole But in 1820 Oersted noticed that an electric current could de ect a magnetic compass needle Soon afterward Ampere correctly postulated that all magnetic phenomena are due to electric charges in motion Then in 1831 Faraday discovered that a moving magnet generates an electric current By the time Maxwell and Lorentz put the nishing touches on the theory electricity and magnetism were inextricably intertwined They could no longer be regarded as separate subjects but rather as two aspects of a single subject electromagnetism Faraday had speculated that light too is electrical in nature Maxwell s theory provided spectacular justi cation for this hypothesis and soon optics the study of lenses mirrors prisms interference and diffraction was incorporated into electromagnetism Hertz who presented the decisive experimental con rmation for Maxwell s theory in 1888 put it this way The connection between light and electricity is now established In every ame in every luminous particle we see an electrical process Thus the domain of electricity extends over the whole of nature It even affects ourselves intimately we perceive that we possess an electrical organ the eye By 1900 then three great branches of physics electricity magnetism and optics had merged into a single uni ed theory And it was soon apparent that visible light represents only a tiny window in the vast spectrum of electromagnetic radiation from radio though microwaves infrared and ultraviolet to x rays and gamma rays Einstein dreamed of a further uni cation which would combine gravity and electrody namiCs in much the same way as electricity and magnetism had been combined a century earlier His uni ed eld theory was not particularly successful but in recent years the same impulse has spawned a hierarchy of increasingly ambitious and speculative uni cation schemes beginning in the 1960s with the electroweak theory of Glashow Weinberg and Salam which joins the weak and electromagnetic forces and culminating in the 1980s with the superstring theory which according to its proponents incorporates all four forces in a single theory of everything At each step in this hierarchy the mathematical dif culties mount and the gap between inspired conjecture and experimental test widens nevertheless it is clear that the uni cation of forces initiated by electrodynamics has become a major theme in the progress of physics The Field Formulation of Electrodynamics The fundamental problem a theory of electromagnetism hopes to solve is this I hold up a bunch of electric charges here and maybe shake them around what happens to some other charge over there The classical solution takes the form of a eld theory We say that the space around an electric charge is permeated by electric and magnetic elds the electromagnetic odor as it were of the charge A second charge in the presence of these elds experiences a force the elds then transmit the in uence from one charge to the other they mediate the interaction xiv ADVERTISEMENT When a charge undergoes acceleration a portion of the eld detaches itself in a sense and travels off at the speed of light carrying with it energy momentum and angular momentum We call this electromagnetic radiation Its existence invites if not compels us to regard the elds as independent dynamical entities in their own right every bit as real as atoms or baseballs Our interest accordingly shifts from the study of forces between charges to the theory of the elds themselves But it takes a charge to produce an electromagnetic eld and it takes another charge to detect one so we had best begin by reviewing the essential properties of electric charge Electric Charge 1 Charge comes in two varieties which we call plus and minus because their effects tend to cancel if you have q and q at the same point electrically it is the same as having no charge there at all This may seem too obvious to warrant comment but I encourage you to contemplate other possibilities what if there were 8 or 10 different species of charge In chromodynamics there are in fact three quantities analogous to electric charge each of which may be positive or negative Or what if the two kinds did not tend to cancel The extraordinary fact is that plus and minus charges occur in exactly equal amounts to fantastic precision in bulk matter so that their effects are almost completely neutralized Were it not for this we would be subjected to enormous forces a potato would explode violently if the cancellation were imperfect by as little as one part in 1010 2 Charge is conserved it cannot be created or destroyed what there is now has always been A plus charge can annihilate an equal minus charge but a plus charge cannot simply disappear by itself somethin g must account for that electric charge So the total charge of the universe is xed for all time This is called global conservation of charge Actually Ican say something much stronger Global conservation would allow for a charge to disappear in New York and instantly reappear in San Francisco that wouldn t affect the total and yet we know this doesn t happen If the charge was in New York and it went to San Francisco then it must have passed along some continuous path from one to the other This is called local conservation of charge Later on we ll see how to formulate a precise mathematical law expressing local conservation of charge it s called the continuity equation 3 Charge is quantized Although nothing in classical electrodynamics requires that it be so the fact is that electric charge comes only in discrete lumps integer multiples of the basic unit of charge If we call the charge on the proton e then the electron carries charge e the neutron charge zero the pi mesons e 0 and e the carbon nucleus 6e and so on never 7392e or even 1 2e1 This fundamental unit of charge is extremely small so for practical purposes it is usually appropriate to ignore quantization altogether Water too really consists of discrete lumps molecules yet if we are dealing with reasonably large large quantities of it we can treat it as a continuous uid This is in fact much closer to Maxwell s own view he knew nothing of electrons and protons he must have pictured lActually protons and neutrons are composed of three quarks which carry fractional charges 1 e and l e However free quarks do not appear to exist in nature and in any event this does not alter the fact that charge is quantized it merely reduces the size of the basic unit XV charge as a kind of jelly that could be divided up into portions of any size and smeared out at will These then are the basic properties of charge Before we disci1ss the forces between charges some mathematical tools are necessary their introduction will occupy us in Chap ter 1 Units The subject of electrodynamics is plagued by competing systems of units which sometimes render it dif cult for physicists to communicate with one another The problem is far worse than in mechanics where Neanderthals still speak of pounds and feet for in mechaniCs at least all equations look the same regardless of the units used to measure quantities Newton s second law remains F 2 ma whether it is feetpoundsseconds kilograms meters seconds or whatever But this is not so in electromagnetism where Coulomb s law may appear variously as 11422 1 QICIZA 47160 a2 ILZ 1 HL 1 2A g Gauss1an or a 51 Of the systems in common use the two most popular are Gaussian cgs and SI mks Ele mentary particle theorists favor yet a third system HeavisideLorentz Although Gaussian units offer distinct theoretical advantages most undergraduate instructors seem to prefer SI I suppose because they incorporate the familiar household units volts amperes and watts In this book therefore I have used SI units Appendix C provides a dictionary for converting the main results into Gaussian units Chapter 1 Vector Analysis 11 Vector Algebra 111 Vector Operations If you walk 4 miles due north and then 3 miles due east Fig 11 you will have gone a total of 7 miles but you re not 7 miles from Where you set out you re only 5 We need an arithmetic to describe quantities like this which evidently do not add in the ordinary way The reason they don t of course is that displacements straight line segments going from one point to another have direction as well as magnitude length and it is essential to take both into account when you combine them Such objects are called vectors velocity acceleration force and momentum are other examples By contrast quantities that have magnitude but no direction are called scalars examples include mass charge density and temperature I shall use boldface A B and so on for vectors and ordinary type for scalars The magnitude of a vector A is written AI or more simply A In diagrams vectors are denoted by arrows the length of the arrow is proportional to the magnitude of the vector and the arrowhead indicates its direction Minus A A is a vector with the 3 mi A 4 5 mi A m1 Figure 11 Figure 12 2 CHAPTER I VECTOR ANALYSIS same magnitude as A but of opposite direction Fig 12 Note that vectors have magnitude and direction but not location a displacement of 4 miles due north from Washington is represented by the same vector as a displacement 4 miles north from Baltimore neglecting of course the curvature of the earth On a diagram therefore you can slide the arrow around at will as long as you don t change its length or direction We de ne four vector operations addition and three kinds of multiplication i Addition of two vectors Place the tail of B at the head of A the sum A B is the vector from the tail of A to the head of B Fig 13 This rule generalizes the obvious procedure for combining two displacements Addition is commutative ABBx 3 miles east followed by 4 miles north gets you to the same place as 4 miles north followed by 3 miles east Addition is also associative ABCABC To subtract a vector Fig 14 add its opposite A BA B BA A A B B Figure 13 Figure 14 ii Multiplication by a scalar Multiplication of a vector by a positive scalar a mul tiplies the magnitude but leaves the direction unchanged Fig 15 If a is negative the direction is reversed Scala r multiplication is distributive aA B aA aB iii Dot product of two vectors The dot product of Mo vectors is de ned by ABEABcosd 11 Where 9 is the angle they form when placed tail to tail Fig 16 Note that A B is itself a scalar hence the alternative name scalar product The dot product is commutative ABBA 11 VECTOR ALGEBRA 5 Figure 19 The numbers Ax Ay and AZ are called components of A geometrically they are the projections of A along the three coordinate axes We can now reformulate each of the four vector operations as a rule for manipulating components A B Ax Ayy A22 3 39 32 Ax Bx Ay 8 AZ Bz 17 i Rule To add vectors add like components aA my aAy aAz 18 ii Rule To multiply by a scalar multiply each component Because 2 9 and i are mutually perpendicular unit vectors aayy221 ayza2y2o 19 Accordingly A B Ax Ayy A22 Bx Byy 322 AXBXAyByAZBZ 110 iii Rule To calculate the dot product multiply like components and add In particular AAA A A A A A A 111 This is if you like the threedimensional generalization of the Pythagorean theorem Note that the dot product of A with any unit vector is the component of A along that direction thusAx AxAy Ay andAi AZ SO 6 CHAPTER I VECTOR ANALYSIS Similarlyl xxx 9x9 2 ixizO x x y y x 2 i y x i i x y x ixxz xxi 112 Therefore A x B Axx l Ayy l Azi x B By 32 113 AyBZ A119 AZBX Aszy 1 ACBy AyBx This cumbersome expression can be written more neatly as a determinant x y i AxB A A Az 114 B By BZ iv Rule To calculate the cross product form the determinant whose rst row is a y 2 whose second row is A in component form and whose third row is B Example 12 Find the angle between the face diagonals of a cube Solution We might as well use a cube of side 1 and place it as shown in Fig 110 with one corner at the origin The face diagonals A and B are A1 0571Z B0 1571i z 0071 B 9 A 010 V xlt100 Figure 110 1These signs pertain to a righthanded coordinate system xaxis out of the page y axis to the right z axis up or any rotated version thereof In a lefthanded system zaxis down the signs are reversed 2 x y i and so on We shall use right handed systems exclusively 11 VECTOR ALGEBRA 7 So in component form AB100ll11 On the other hand in abstract form A B ABcos6 xExEcose 2cos6 Therefore cos6 12 or 6 60 Of course you can get the answer more easily by drawing in a diagonal across the top of the cube completing the equilateral triangle But in cases where the geometry is not so simple this device of comparing the abstract and component forms of the dot product can be a very ef cient means of nding angles Problem 13 Find the angle between the body diagonals of a cube Problem 14 Use the cross product to nd the components of the unit vector 11 perpendicular to the plane shown in Fig 11 1 113 Triple Products Since the cross product of two vectors is itself a vector it can be dotted or crossed with a third vector to form a triple product i Scalar triple product A B x C Geometrically A B x C is the volume of the parallelepiped generated by A B and C since B X Cl is the area of the base and IA cos 9 is the altitude Fig 112 Evidently ABxCBCxACAXB 115 for they all correspond to the same gure Note that alphabetical order is preserved in view of Eq 16 the nonalphabetical triple products ACxBBAxCCBXA Figure 111 Figure 112 8 CHAPTER I VECTOR ANALYSIS have the opposite sign In component form A A AZ ABxC B B B 116 C C CZ Note that the dot and cross can be interchanged ABXCAXBC this follows immediately from Eq 115 however the placement of the parentheses is critical A B x C is a meaningless expression you can t make a cross product from a scalar and a vector ii Vector triple product A x B x C The vector triple product can be simpli ed by the socalled BACCAB rule AXBXCBAC CAB 117 Notice that AxBxC CxAxB ABCBAC is an entirely different vector Incidentally all higher vector products can be similarly reduced often by repeated application of Eq 117 so it is never necessary for an expression to contain more than one cross product in any term For instance AxBCxD ACBDADBC AxBxCxD BA CxD ABCXD 118 Problem 15 Prove the BACCAB rule by writing out both sides in component form Problem 16 Prove that AXBXCBXCgtltACgtltAXB0 Under what conditions does A x B X C A x B x C 114 Position Displacement and Separation Vectors The location of a point in three dimensions can be described by listing its Cartesian coor dinates x y z The vector to that point from the origin Fig 113 is called the position vector rEx yyzL 119 11 VECTOR ALGEBRA 9 source point eld point Figure 113 Figure 114 I will reserve the letter r for this purpose throughout the book Its magnitude r x2 y2z2 120 is the distance from the origin and E WZZ 121 r xJCZer2z2 is a unit vector pointing radially outward The in nitesimal displacement vector from x y z t0 x dx y dy z dz is dldx dy dzi 122 We could call this dr since that s what it is but it is useful to reserve a special letter for in nitesimal displacements In electrodynamics one frequently encounters problems involving two points typically a source point r where an electric charge is located and a eld point r at which you are calculating the electric or magnetic eld Fig 114 It pays to adopt right from the start some shorthand notation for the separation vector from the source point to the eld point I shall use for this purpose the script letter IL ear r 123 Its magnitude is Ii lr r l 124 and a unit vector in the direction from r to r is A It r r IL a lr r l 125 10 CHAPTER 1 VECTOR ANALYSIS In Cartesian coordinates e x m y my z m 126 altx x2lty y 2ltz z 2 127 2 x x y y z z2 128 xx x2 y y2 z 22 from which you can begin to appreciate the advantage of the scriptIi notation Problem 17 Find the separation vectora from the source point 287 to the eld point 468 Determine its magnitude IL and construct the unit vector 2 115 How Vectors Transform The de nition of a vector as a quantity with a magnitude and direction is not altogether satisfactory What precisely does direction mean2 This may seem a pedantic question but we shall shortly encounter a species of derivative that looks rather like a vector and we ll want to know for sure whether it is one You might be inclined to say that a vector is anything that has three components that combine properly under addition Well how about this We have a barrel of fruit that contains Nx pears Ny apples and NZ bananas Is N Nxfr Nyy N22 a vector It has three components and when you add another barrel with M x pears My apples and Mz bananas the result is N x Mx pears Ny M y apples NZ M2 bananas So it does add like a vector Yet it s obviously not a vector in the physicist s sense of the word because it doesn t really have a direction What exactly is wrong with it The answer is that N does not transform prOperly when you change coordinates The coordinate frame we use to describe positions in space is of course entirely arbitrary but there is a speci c geometrical transformation law for converting vector components from one frame to another Suppose for instance the f 7 2 system is rotated by angle 45 relative to x y z about the common x Y axes From Fig 115 AyAcos9 AZAsin9 while Ky A cosg A cos9 Acos 9 cos 43 sin6 sin cos Ay sin Az AZ Asin Asin9 Asin9cos cos95in sin Ay cos 45142 2This section can be skipped without loss of continuity 11 VECTOR ALGEBRA 11 Figure 115 We might express this conclusion in matrix notation Ky cos 1 sin I Ay ZZ sin cos Az 39 1 29 More generally for rotation about an arbitrary axis in three dimensions the transfor mation law takes the form E R Rty Rig Ax y Ryx RVy Ryz Ay 130 AZ RZX Rzy RZZ AZ or more compactly 3 Z1 ERAA 131 391 where the index 1 stands for x 2 for y and 3 for z The elements of the matrix R can be ascertained for a given rotation by the same sort of geometrical arguments as we used for a rotation about the x axis Now Do the components of N transform in this way Of course not it doesn t matter what coordinates you use to represent positions in space there is still the same number of apples in the barrel You can t convert a pear into a banana by choosing a different set of axes but you can turn Ax into Ky Formally then a vector is any set of three components that transforms in the same manner as a displacement when you change coordinates As always displacement is the model for the behavior of all vectors By the way a second rank tensor is a quantity with nine components T T sz Tyx TZZ which transforms with two factors of R 7xx 2 RxxRxxTxx ny Txy sz Trz nyRxx Tyx ny Tyy szTyz sz Rxx sz 1 nysz sz Tzza 12 CHAPTER 1 VECTOR ANALYSIS or more compactly 3 TU ZZRikleTk1 132 In general an nthrank tensor has n indices and 3quot components and transforms with n factors of R In this hierarchy a vector is a tensor of rank 1 and a scalar is a tensor of rank zero Problem 18 a Prove maLthe two dimensional rotation matrix 129 preserves dot products That is show that AyBy AZBZ AyBy AZBZ b What constraints must the elements Rtj of the threedimensional rotation matrix 130 satisfy in order to preserve the length of A for all vectors A Problem 19 Find the transformation matrix R that describes a rotation by 120 about an axis from the origin through the point 1 1 1 The rotation is clockwise as you look down the axis toward the origin Problem 110 a How do the components of a vector transform under a translation of coordinates f x y y aE 2 Fig 1163 b How do the components of a vector transform under an inversion of coordinates J x y y Z z Fig 116b c How does the cross product 113 of two vectors transform under inversion The cross product of two vectors is properly called a pseudovector because of this anomalous be havior Is the cross product of two pseudovectors a vector or a pseudovector Name two pseudovector quantities in classical mechanics d How does the scalar triple product of three vectors transform under inversions Such an object is called a pseudoscalar Figure 116 I 2 DIFFERENTIAL CALC UL US l 3 12 Differential Calculus 121 Ordinary Derivatives Question Suppose we have a function of one variable f x What does the derivative d f dx do for us Answer It tells us how rapidly the function f x varies when we change the argument x by a tiny amount dx df g dx 133 X In words If we change x by an amount dx then f changes by an amount d f the derivative is the proportionality factor For example in Fig 117a the function varies slowly with x and the derivative is correspondingly small In Fig 117b f increases rapidly with x and the derivative is large as you move away from x O Geometrical Interpretation The derivative d f dx is the slope of the graph of f versus x f f a b Figure 117 122 Gradient Suppose now that we have a function of three variables say the temperature Tx y z in a room Start out in one corner and set up a system of axes then for each point x y z in the room T gives the temperature at that spot We want to generalize the notion of derivative to functions like T which depend not on one but on three variables Now a derivative is supposed to tell us how fast the function varies if we move a little distance But this time the situation is more complicated because it depends on what direction we move If we go straight up then the temperature will probably increase fairly rapidly but if we move horizontally it may not change much at all In fact the question How fast does T vary has an in nite number of answers one for each direction we might choose to explore Fortunately the problem is not as bad as it looks A theorem on partial derivatives states that 8T 8T 8T dT dx dy dz 134 8x 8y BZ 14 CHAPTER I VECTOR ANALYSIS This tells us how T changes when we alter all three variables by the in nitesimal amounts dx dy dz Notice that we do not require an in nite number of derivatives three will suf ce the partial derivatives along each of the three coordinate directions Equation 134 is reminiscent of a dot product 8T 8T ET A A A dT x y zdxxdyydzz 8x 8y Bz VT dl 135 where 8T 8T 8T VT E X y z 136 8x 8y Bz is the gradient of T VT is a vector quantity with three components it is the generalized derivative we have been looking for Equation 135 is the threedimensional version of Eq 133 Geometrical Interpretation of the Gradient Like any vector the gradient has magnitude and direction To determine its geometrical meaning let s rewrite the dot product 135 in abstract form dTVTdl VTdlc0s9 137 where 9 is the angle between VT and d1 Now if we x the magnitude dl and search around in various directions that is vary 9 the maximum change in T evidentally occurs when 9 O for then cos 9 1 That is for a xed distance dll d T is greatest when I move in the same direction as VT Thus The gradient VT points in the direction of maximum increase of the function T Moreover The magnitude VT gives the slope rate of increase along this maximal direction Imagine you are standing on a hillside Look all around you and nd the direction of steepest ascent That is the direction of the gradient Now measure the s10pe in that direction rise over run That is the magnitude of the gradient Here the function we re talking about is the height of the hill and the coordinates it depends on are positions latitude and longitude say This function depends on only two variables not three but the geometrical meaning of the gradient is easier to grasp in two dimensions Notice from Eq 137 that the direction of maximum descent is opposite to the direction of maximum ascent while at right angles 6 90 the slope is zero the gradient is perpendicular to the contour lines You can conceive of surfaces that do not have these properties but they always have kinks in them and correspond to nondifferentiable functions What would it mean for the gradient to vanish If VT O at x y z then d T O for small displacements about the point x y z This is then a stationary point of the function Tx y 2 It could be a maximum a summit a minimum a valley a saddle 12 DIFFERENTIAL CALCULUS 15 point a pass or a shoulder This is analogous to the situation for functions of one vanable where a vanishing derivative signals a maximum a minimum or an in ection In particular if you want to locate the extrema of a function of three variables set its gradient equal to zero Example 13 Find the gradient of r yx2 y2 22 the magnitude of the position vector Solution Vr 3ri3rA 3r M ay y az z I 2x 1 2y I 2z A Zxxz l yZdZZ ZxJ62y2z2 xxyyz2ri xXzy2z2 r Does this make sense Well it says that the distance from the origin increases most rapidly in the radial direction and that its rate of increase in that direction is l just what you d expect 9 Z Problem 111 Find the gradients of the following functions a fx y z x2 y3 Z4 b x V z x2y3z4 C fx y z 6quot siny 1nz Problem 112 The height of a certain hill in feet is given by hxy102xy 3x2 4y2 18x 28y 12 where y is the distance in miles north x the distance east of South Hadley a Where is the top of the hill located b How high is the hill 0 How steep is the slope in feet per mile at a point 1 mile north and one mile east of South Hadley In what direction is the slope steepest at that point Problem 113 Letabe the separation vector from a xed point x y z to the point x y z and let a be its length Show that a vnz 2a b V 1m 42212 c What is the general formula for Vaquot 16 CHAPTER 1 VECTOR ANALYSIS Problem 114 Suppose that f is a function of tw0 variables y and z only Show that the gradient V f Elf3y 3 f am transforms as a vector under rotations Eq 129 Hint Elf8y 3f3y3y3y 3f8z3z3 and the analogous formula for Elf32 We know that y y cos 45 z sin 45 and E y sin 45 z cos 45 solve these equations for y and z as functions of y and Z and compute the needed derivatives ByBy azay etc 123 The Operator V The gradient has the formal appearance of a vector V multiplying a scalar T a a a VT 23 y 2 T 138 Z For once 1 write the unit vectors to the le just so no one will think this means 312 Bx and so on which would be zero since 2 is constant The term in parentheses is called del A3 139 Zaz 8 3 Vx A Bx y3y Of course del is not a vector in the usual sense Indeed it is without speci c meaning until we provide it with a function to act upon Furthermore it does not multiply T rather it is an instruction to dz39 erentiate what follows To be precise then we should say that V is a Vector operator that acts upon T not a vector that multiplies T With this quali cation though V mimics the behavior of an ordinary vector in virtually every way almost anything that can be done with other vectors can also be done with V if We merely translate multiply by act upon So by all means take the vector appearance of V seriously it is a marvelous piece of notational simpli cation as you will appreciate if you ever consult Maxwell s original work on electromagnetism written without the bene t of V Now an ordinary vector A can multiply in three ways 1 Multiply a scalar a Au 2 Multiply another vector B via the dot product A B 3 Multiply another vector via the cross product A x B Correspondingly there are three ways the operator V can act 1 On a scalar function T VT the gradient 2 On a vector function v via the dot product V V the divergence 3 On a vector function v via the cross product V x V the curl We have already discussed the gradient 1n the following sections we examine the other two vector derivatives divergence and curl 12 DIFFERENTIAL CAL C UL US 17 124 The Divergence From the de nition of V we construct the divergence 13 3 A Vv x y z Z vxxvyyvzi 140 Observe that the divergence of a vector function v is itself a scalar V v You can t have the divergence of a scalar that s meaningless Geometrical Interpretation The name divergence is well chosen for V 39v is ameasure of how much the vector v spreads out diverges from the point in question For example the vector function in Fig 118a has a large positive divergence if the arrows pointed in it would be a large negative divergence the function in Fig 118b has zero divergence and the function in Fig 118c again has a positive divergence Please understand that V here is a function there s a different vector associated with every point in space In the diagrams WWW 731 WWW WW 1111 Figure 118 18 CHAPTER I VECTOR ANALYSIS of course I can only draw the arrows at a few representative locations Imagine standing at the edge of a pond Sprinkle some sawdust or pine needles on the surface If the material spreads out then you dropped it at a point of positive divergence if it collects together you dropped it at a point of negative divergence The vector function v in this model is the velocity of the water this is a twodimensional example but it helps give one a feel for what the divergence means A point of positive divergence is a source or faucet a point of negative divergence is a sink or drain Example 14 Suppose the functions in Fig 118 are va r xx yy zi V 2 and VC 2 zi Calculate their divergences Solution 8 3 3 VVa x y zl113 dx 3y 3z As anticipated this function has a positive divergence a a a V39Wz 70 0 10000 dx 3y dz as expected 3 3 Vv 0 x 3y 3 0 0 0 121 c a 3Zz l l Problem 115 Calculate the divergence of the following vector functions 21 Va 2 x2 x 3sz9 sz i b v xyx2yzy 3zxi c vc yzx 2xy z2y 2yzi Problem 116 Sketch the vector function and compute its divergence The answer may surprise you can you explain it Problem 117 In two dimensions show that the divergence transforms as a scalar under rota tions Hint Use Eq 129 to determine By and E and the method of Prob 114 to calculate the derivatives Your aim is to show that any3y 35232 2 ivy3y avg3L 12 DIFFERENTIAL CALC UL US 19 125 The Curl From the de nition of V we construct the curl it y 2 V X V BBx BBy BBz vx vy vz A Big Bi A Bvx sz A Buy Bvx Xay azyaz 5 3 13941 Notice that the curl of a vector function V is like any cross product a vector You cannot have the curl of a scalar that s meaningless Geometrical Interpretation The name curl is also well chosen for V X V is a measure of how much the vector V curls around the point in question Thus the three functions in Fig 118 all have zero curl as you can easily check for yourself whereas the functions in Fig 119 have a substantial curl pointing in the z direction as the natural righthand rule would suggest Imagine again you are standing at the edge of a pond Float a small paddlewheel a cork with toothpicks pointing out radially would do if it starts to rotate then you placed it at a point of nonzero curl A whirlpool would be a region of large curl Z Figure 119 Example 15 Suppose the function sketched in Fig 119a is Va 2 yi xy and that in Fig 119b is v 2 x37 Calculate their curls Solution it 9 i V X Va 2 33x 33y 33z 22 y x 0 and 9 vabl 38x 88y BBz x 20 CHAPTER 1 VECTOR ANALYSIS As expected these curls point in the z direction Incidentally they both have zero divergence as you might guess from the pictures nothing is spreading out it just curls around Problem 118 Calculate the curls of the vector functions in Prob 115 Problem 119 Construct a vector function that has zero divergence and zero curl everywhere A constant will do the job of course but make it something a little more interesting than that 126 Product Rules The calculation of ordinary derivatives is facilitated by a number of general rules such as the sum rule d df dg 5f g dx dx 9 the rule for multiplying by a constant d df k 2 k dx f dx the product rule dg df d f fdx gdx and the quotient rule df dg g r gd x f a dx g 22 39 Similar relations hold for the vector derivatives Thus Vfg VfVg VAB V39AV39B Vgtlt ABVgtltAVgtltB and Vkfka VkAkVA V x kAkV xA as you can check for yourself The product rules are not quite so simple There are two ways to construct a scalar as the product of two functions f g product of two scalar functions A B dot product of two vector functions and two ways to make a vector f A scalar times vector A x B cross product of two vectors 1 2 DIFFERENTIAL CALC UL US 2 1 Accordingly there are six product rules two for gradients i Vfg ng gi ii VABAgtltVxBBgtltVXAAVBBVA two for divergences iii VfAfV39AAVf iv VAxBBVgtltA AVgtltB and two for curls v V x M m7 x A A x Vf vi VxAxBBVA AVBAVB BVA You will be using these product rules so frequently that I have put them on the inside front cover for easy reference The proofs come straight from the product rule for ordinary derivatives For instance 3 3 3 V fA fofAya ZfAz af 8A if 8A 8f aAz MAX163yAyf3yEAZfEgt Vf AfV A It is also possible to formulate three quotient rules V1 gi ng g g2 V A gV39AA39Vg f g g VXltAgt gltVgtltAgt2AgtltltVgi g g However since these can be obtained quickly from the corresponding product rules I haven t bothered to put them on the inside front cover 22 CHAPTER 1 VECTOR ANALYSIS Problem 120 Pr0ve product rules i iv and v Problem 121 21 HA and B are two vector functions what does the expression A VB mean That is what are its x y and z components in terms of the Cartesian components of A B and V b Compute f39 Vr where r is the unit vector de ned in Eq 121 c For the functions in Prob 115 evaluate Va Vvb Problem 122 For masochists only Prove product rules ii and vi Refer to Prob 121 for the de nition of A VB Problem 123 Derive the three quotient rules Problem 124 a Check product rule iv by calculating each term separately for the functions Ax 2yy3zi B23yfr 2xy b Do the same for product rule ii c The same for rule vi 127 Second Derivatives The gradient the divergence and the curl are the only first derivatives we can make with V by applying V twice we can construct ve species of second derivatives The gradient VT is a vector so we can take the divergence and curl of it 1 Divergence of gradient V VT 2 Curl of gradient V x VT The divergence V V is a scalar all we can do is take its gradient 3 Gradient of divergence VV V The curl V x V is a vector so we can take its divergence and curl 4 Divergence of curl V V X v 5 Curl of curl V x V X V This exhausts the possibilities and in fact not all of them give anything new Let s consider them one at a time ABA323 BT BTABTi X A ax yBy az ax Byy az BZT azr azr 3x2 Byz Bz2 1 v VT 142 12 DIFFERENTIAL CALCULUS 23 This object which we write VZT for short is called the Laplacian of T we shall be studying it in great detail later on Notice that the Laplacian of a scalar T is a scalar Occasionally we shall speak of the Laplacian of a vector V2V By this we mean a vector quantity whose xcomponent is the Laplacian of vx and so on3 VZV E Vzvx r Vzvyy Vzvz 143 This is nothing more than a convenient extension of the meaning of V2 2 The curl of a gradient is always zero V x VT 0 144 This is an important fact which we shall use repeatedly you can easily prove it from the de nition of V Eq 139 Beware You might think Eq 144 is obviously true isn t it just V x VT and isn t the cross product of any vector in this case V with itself always zero This reasoning is suggestive but not quite conclusive since V is an operator and does not multiply in the usual way The proof of Eq 144 in fact hinges on the equality of cross derivatives 3 3T 3 3T 145 ix 3y 3y 3x 39 If you think I m being fussy test your intuition on this one VT x VS Is that always zero It would be of course if you replaced the V s by an ordinary vector 3 V V 39v for some reason seldom occurs in physical applications and it has not been given any special name of its own it s just the gradient of the divergence Notice that VV V is not the same as the Laplacian ofa vector Vzv V 39 VV 5e VV V 4 The divergence of a curl like the curl of a gradient is always zero VV Xv 0 146 You can prove this for yourself Again there is a fraudulent shortcut proof using the vector identity A B x C A X B 39 C 5 As you can check from the de nition of V V x V x v VV v Vzv 147 So curlofcurl gives nothing new the rst term is just number 3 and the second is the Laplacian of a vector In fact Eq 147 is often used to de ne the Laplacian of a vector in preference to Eq 143 which makes speci c reference to Cartesian coordinates Really then there are just two kinds of second derivatives the Laplacian which is of fundamental importance and the gradientofdivergence which we seldom encounter 3In curvilinear coordinates where the unit vectors themselves depend on position they too must be differentiated see Sect 141 24 CHAPTER 1 VECTOR ANALYSIS We could go through a similar ritual to work out third derivatives but fortunately second derivatives suf ce for practically all physical applications A nal word on vector differential calculus It all ows from the operator V and from taking seriously its vector character Even if you remembered only the de nition of V you should be able in principle to reconstruct all the rest Problem 125 Calculate the Laplacian of the following functions a Ta x2 2xy 3z 4 b T sinx siny sin z c Tc 2 e5x sin4y cos 3z d v x2 3xz2 7 2xz2 Problem 126 Prove that the divergence of a curl is always zero Check it for function Va in Prob 115 Problem 127 Prove that the curl of a gradient is always zero Check it for function h in Prob 111 13 Integral Calculus 131 Line Surface and Volume Integrals In electrodynamics we encounter several different kinds of integrals among which the most important are line or path integrals surface integrals or ux and volume integrals a Line Integrals A line integral is an expression of the form h f V d 148 377 where V is a vector function d l is the in nitesimal displacement vector Eq 122 and the integral is to be carried out along a prescribed path 73 from point a to point b Fig 120 If the path in question forms a closed loop that is if b a I shall put a circle on the integral sign yiv d1 149 At each point on the path we take the dot product of V evaluated at that point with the displacement d to the next point on the path To a physicist the most familiar example of a line integral is the work done by a force F W f F dl Ordinarily the value of a line integral depends critically on the particular path taken from a to b but there is an important special class of vector functions for which the line integral is independent of the path and is determined entirely by the end points It will be our business in due course to characterize this special class of vectors A force that has this property is called conservative 13 INTEGRAL CALCULUS 27 iiiy2 dazdxdzy vdax2dxdz so 2 2 fvdax2dx dz12 0 0 ivy0 da dxdzy vda x2dxdz so 2 2 fVda x2dx dzz l2 0 0 vz2 dadxdy2 vdayz2 3dxdyydxdy so 2 2 fvdazf dx ydy4 0 0 Vdal6012 12420 surface Evidently the total ux is c Volume Integrals A volume integral is an expression of the form Tdr 151 V where T is a scalar function and dr is an in nitesimal volume element In Cartesian coordinates drzdxdydz 152 For example if T is the density of a substance which might vary from point to point then the volume integral would give the total mass Occasionally we shall encounter volume integrals of vector functions Vd1 vxxvyyvzidr vxdryvydrivzdr 153 because the unit vectors are constants they come outside the integral Example 18 Calculate the volume integral of T xyz2 over the prism in Fig 124 Solution You can do the three integrals in any order Let s do x rst it runs from 0 to 1 y then y it goes from 0 to 1 and nally z 0 to 3 deIfo3z2AlyA1 yxdxldydz 1 3 2 f1 2 1 1 a V Zdz l d 9 20 0 y y y 2 12 8 28 CHAPTER 1 VECTOR ANALYSIS Figure 124 Problem 128 Calculate the line integral of the function v x2 x 2yz y y2 2 from the origin to the point 111 by three different routes a 000 gt 100 gt 1 10 gt 1 1 1 b000 gt 001 gt 011 gt 111 c The direct straight line 1 What is the line integral around the closed loop that goes out along path a and back along path b Problem 129 Calculate the surface integral of the function in Ex 17 over the bottom of the box For consistency let upward be the positive direction Does the surface integral depend only on the boundary line for this function What is the total ux over the closed surface of the box including the bottom Nata For the closed surface the positive direction is outward and hence down for the bottom face Problem 130 Calculate the volume integral of the function T Z2 over the tetrahedron with comers at 000 100 010 and 001 132 The Fundamental Theorem of Calculus Suppose f x is a function of one variable The fundamental theorem of calculus states b d d fdx fb fa 154 a x In case this doesn t look familiar let s write it another way 12 FXdx f0 fa a where d f dx F x The fundamental theorem tells you how to integrate F x you think up a function f x whose d rivative is equal to F 13 INTEGRAL CALCULUS 29 Geometrical Interpretation According to Eq 133 d f d f dxdx is the in nitesi mal change in f when you go from x to x dx The fundamental theorem 154 says that if you Chop the interval from a to b Fig 125 into many tiny pieces dx and add up the increments d f from each little piece the result is not surprisingly equal to the total change in f f b f a In other words there are two ways to determine the total change in the function either subtract the values at the ends or go stepbystep adding up all the tiny increments as you go You ll get the same answer either way Notice the basic format of the fundamental theorem the integral of a derivative over an interval is given by the value of the function at the end points boundaries In vector calculus there are three species of derivative gradient divergence and curl and each has its own fundamental theorem with essentially the same format I don t plan to prove these theorems here rather I shall explain what they mean and try to make them plausible Proofs are given in Appendix A Figure 125 Figure 126 133 The Fundamental Theorem for Gradients Suppose we have a scalar function of three variables Tx y z Starting at point a we move a small distance dll Fig 126 According to Eq 137 the function T will change by an amount dT VT d11 Now we move a little further by an additional small displacement dlz the incremental change in T will be VT d12 In this manner proceeding by in nitesimal steps we make the journey to point b At each step we compute the gradient of T at that point and dot it into the displacement dl this gives us the change in T Evidently the total change in T in going from a to b along the path selected is b VT d1 Tb Ta 7 155 30 CHAPTER 1 VECTOR ANALYSIS This is called the fundamental theorem for gradients like the ordinary fundamental theorem it says that the integral here a line integral of a derivative here the gradient is given by the value of the function at the boundaries a and b Geometrical Interpretation Suppose you wanted to determine the height of the Eiffel Tower You could climb the stairs using a ruler to measure the rise at each step and adding them all up that s the left side of Eq 155 or you could place altimeters at the top and the bottom and subtract the two readings that s the right side you should get the same answer either way that s the fundamental theorem Incidentally as we found in Ex 16 line integrals ordinarily depend on the path taken from a to b But the right side of Eq 155 makes no reference to the path only to the end points Evidently gradients have the special property that their line integrals are path independent Corollary 1 f V T dl is independent of path taken from a to b Corollary 2 f V T dl 0 since the beginning and end points are identical and hence Tb Ta O Example 19 Let T xyz and take point a to be the origin 0 00 and b the point 2 10 Check the fundamental theorem for gradients Solution Although the integral is independent of path we must pick a speci c path in order to evaluate it Let s go out along the x axis step i and then up step ii Fig 127 As always d dx dy 7dzi VT 2 yzfr 2xyy i y 0 d1 dx VTdly2dx 0 so fVTdlzo 1 iix 2 dldyy VTdl2xydy4ydy so 1 fVlef 4ydy2y2 i39 0 1 1 2 0 39 Figure 1 27 13 INTEGRAL CALCULUS 31 Evidently the total line integral is 2 Is this consistent with the fundamental theorem Yes Tb Ta2 02 Now just to convince you that the answer is independent of path let me calculate the same integral along path iii the straight line from a to b iii y x dy dx VT d1 yz dx 2xydy x2 dx so 2 2 fVTdl x2dxx3 iii 0 0 2 Problem 131 Check the fundamental theorem for gradients using T x2 4x y 2 yz3 the points a 000 b 1 1 1 and the three paths in Fig 128 a 000 gt 100 gt 1 10 gt 1 1 1 b000 gt 001 gt 01l gt 111 c the parabolic path z x2 y x Figure 128 134 The Fundamental Theorem for Divergences The fundamental theorem for divergences states that Vvdr vda V s 156 In honor I suppose of its great importance this theorem has at least three special names Gauss s theorem Green s theorem or simply the divergence theorem Like the other fundamental theorems it says that the integral of a derivative in this case the divergence over a region in this case a volume is equal to the value of the function at the boundary 32 CHAPTER 1 VECTOR ANALYSIS in this case the surface that bounds the volume Notice that the boundary term is itself an integral speci cally a surface integral This is reasonable the boundary of a line is just two end points but the boundary of a volume is a closed surface Geometrical Interpretation If v represents the ow of an incompressible uid then the ux of v the right side of Eq 156 is the total amount of uid passing out through the surface per unit time Now the divergence measures the spreading out of the vectors from a point a place of high divergence is like a faucet pouring out liquid If we have lots of faucets in a region lled with incompressible uid an equal amount of liquid will be forced out through the boundaries of the region In fact there are two ways we could determine how much is being produced a we could count up all the faucets recording how much each puts out or b we could go around the boundary measuring the ow at each point and add it all up You get the same answer either way faucets within the volume f ow out through the surface This in essence is what the divergence theorem says Example 110 Check the divergence theorem using the function v yz an z239 2yzi and the unit cube situated at the origin Fig 129 Solution In this case V v 2x y 1 1 1 2xydr2 fxydxdydz V 0 0 0 l l 1 xydxy ydyl fldzzl 0 0 0 and z 1 T ii lt gt iv 0 iii 1 1 y x 1 vi Figure 129 13 INTEGRAL CALCULUS 33 Evidently iV vdrzl V So much for the left side of the divergence theorem To evaluate the surface integral we must consider separately the six sides of the cube 1 l fvda yzdydz 0 0 i l l 2 1 vdaz ff dydz n f 0 0 y 3 1 1 2 4 iii fvdazf 2xzdxdz 0 0 l l 1 iv fvdaz f zzdxdz 0 0 i l l v fvdazf Zydxdyzl 0 0 l 1 vi fvdaz f 0dxdy0 0 0 So the total ux is as expected Problem 132 Test the divergence theorem for the function v 2 xy 2 2yz 3zx 2 Take as your volume the cube shown in Fig 130 with sides of length 2 34 CHAPTER I VECTOR ANALYSIS V Figure 130 135 The Fundamental Theorem for Curls The fundamental theorem for curls which goes by the special name of Stokes theorem states that fvada vdl S 72 157 As always the integral of a derivative here the curl over a region here a patch of surface is equal to the value of the function at the boundary here the perimeter of the patch As in the case of the divergence theorem the boundary term is itself an integral speci cally a closed line integral 39 Geometrical Interpretation Recall that the curl measures the twist of the vectors v a region of high curl is a whirlpool if you put a tiny paddle wheel there it will rotate NOw the integral of the curl over some surface or more precisely the ux of the curl through that surface represents the total amount of swirl and we can determine that swirl just as well by going around the edge and nding how much the ow is following the boundary Fig 131 You may nd this a rather forced interpretation of Stokes theorem but it s a helpful mnemonic if nothing else You might have noticed an apparent ambiguity in Stokes theorem concerning the boundary line integral which way are we supposed to go around clockwise or counter clockwise If we go the wrong way we ll pick up an overall sign error The answer is that it doesn t matter which Way you go as long as you are consistent for there is a com pensating sign ambiguity in the surface integral Which way does d a point For a closed surface as in the divergence theorem d a points in the direction of the outward normal but for an open surface which way is out Consistency in Stokes theorem as in all such matters is given by the right hand rule If your ngers point in the direction of the line integral then your thumb xes the direction of d a Fig 132 Now there are plenty of surfaces in nitely many that share any given boundary line Twist a paper clip into a loop and dip it in soapy water The soap lm constitutes a surface with the wire loop as its boundary If you blow on it the soap lm will expand making a larger surface with the same boundary Ordinarily a ux integral depends critically on what surface you integrate over but evidently this is not the case with curls For Stokes 39 13 INTEGRAL CALCULUS Figure 131 theorem says that f V x v da is equal to the line integral of v around the boundary and da 9 D dl Figure 132 the latter makes no reference to the speci c surface you choose Corollary 1 Corollary 2 These corollaries are analogous to those for the gradient theorem We shall develop the f V X v da depends only on the boundary line not on the particular surface used 55 V x v da 2 0 for any closed surface since the boundary line like the mouth of a balloon shrinks down to a point and hence the right side of Eq 157 vanishes parallel further in due course 35 Example 111 Suppose v 2xz 3 y2y 4yz22 Check Stokes theorem for the square surface shown in Fig 133 Solution Here VXv4z2 2xfr2zi and dadydzfr Figure 133 36 CHAPTER I VECTOR ANALYSIS In saying that da points in the x direction we are committing ourselves to a counterclockwise line integral We could as well write da 2 dy dz fr but then we would be obliged to go clockwise Since x 0 for this surface 1 l 2 4 vada A 4z dydz Now what about the line integral We must break this up into four segments i x0 z0 vdl3y2dy fvd1f03y2dy1 ii x0 y1 vdl4z2dz fvd1f0 4z2dz iii x0 z1 vdl3y2dy fvdlzflo3y2dyz l fvdlf100dz0 So 4 4 jgvdl13 10 It checks A point of strategy notice how I handled step iii There is a temptation to write d d y 3 here since the path goes to the left You can get away with this if you insist by running the integral from 0 gt 1 Personally I prefer to say dl dx 1 d y y dz 2 always never any minus signs and let the limits of the integral take care of the direction Problem 133 Test Stokes theorem for the function v xy 1 2yz y 3zx i using the triangular shaded area of Fig 134 Problem 134 Check Corollary 1 by using the same function and boundary line as in Ex 11 1 but integrating over the ve sides of the cube in Fig 135 The back of the cube is open fV z 1 z 2 lt gt iv iii 1 1 y 2 x 1 x y ii Figure 134 Figure 135 I 3 INTEGRAL CALCULUS 37 136 Integration by Parts The technique known awkwardly as integration by parts exploits the product rule for derivatives d d df g W fltdx gdx Integrating both sides and invoking the fundamental theorem b b dg b df d d Md Hf Mr 7 b d d 7 ff g dx g f dxfg a dx a dx a That s integration by parts It pertains to the situation in which you are called upon to integrate the product of one function f and the derivative of another g it says you can transfer the derivative from g to f at the cost of a minus sign and a boundary term b d d fgdx fg a x Of 158 Example 112 Evaluate the integral 14 00 xeix dx 0 Solution The exponential can be expressed as a derivative d e XE e x in this case then fx x gx ex and dfdx 1 so 00 00 gt ace X dx 2 ex dx xex 0 0 We can exploit the product rules of vector calculus together with the appropriate fun damental theorems in exactly the same way For example integrating OO 2 9 0 V39fAfVAAVf over a volume and invoking the divergence theorem yields VfAdtfVAdrAVfdty fAda 0139 ffVAdr AVfdtffAda 159 V V S 38 CHAPTER 1 VECTOR ANALYSIS Here again the integrand is the product of one function f and the derivative in this case the divergence of another A and integration by parts licenses us to transfer the derivative from A to f where it becomes a gradient at the cost of a minus sign and a boundary term in this case a surface integral You might wonder how often one is likely to encounter an integral involving the product of one function and the derivative of another the answer is surprisingly often and integration by parts turns out to be one of the most powerful tools in vector calculus Problem 135 a Show that ffVxAdaAXVfday fAdl 160 s s 72 b Show that fBVxAdtAVXBdry AgtltBda 161 V V S Curvilinear Coordinates 141 Spherical Polar Coordinates The spherical polar coordinates r 6 13 of a point P are de ned in Fig 136 r is the distance from the origin the magnitude of the position vector 9 the angle down from the z axis is called the polar angle and a5 the angle around from the x axis is the azimuthal angle Their relation to Cartesian coordinates x y z can be read from the gure Z rcos9 162 xrsin6cos yrsin65in Figure 136 14 C URVILINEAR COORDINATES 39 Figure 136 also shows three unit vectors f39 5 cf pointing in the direction of increase of the corresponding coordinates They constitute an orthogonal mutually perpendicular basis set just like x y 2 and any vector A can be expressed in terms of them in the usual way AAr2Ag A 163 A A9 and A are the radial polar and azimuthal components of A In terms of the Cartesian unit vectors 2 sin9cos xsin6sin ycos92 9 cos9cos xcos6sin y sin62 164 d sin xcos y as you can easily check for yourself Prob 137 I have put these formulas inside the back cover for easy reference But there is a poisonous snake lurking here that I d better warn you about f39 5 and 43 are associated with a particular point P and they change direction as P moves around For example 2 always points radially outward but radially outwar can be the x direction the y direction or any other direction depending on where you are In Fig 137 A y and B y and yet both of them would be written as f39 in spherical coordinates One could take account of this by explicitly indicating the point of reference f399 as 5 9 63 439 ab but this would be cumbersome and as long as you are alert to the problem I don t think it will cause dif culties4 In particular do not nai39vely combine the spherical components of vectors associated with different points in Fig 137 A B 0 not 22 and A B 1 not 1 Beware of differentiating a vector that is expressed in spherical coordinates since the unit vectors themselves are functions of position 8289 5 for example And do not take f39 9 and f outside an integral as we did with x y and 2 in Eq 153 In general if you re uncertain about the validity of an operation reexpress the problem in Cartesian coordinates where this dif culty does not arise Figure 137 41 claimed on the very rst page ihat vectors haVe no location and I ll stand by that The vectors themselves live out there completely independent of our choice of coordinates But the notation we use to represent them does depend on the point in question in curvilinear coordinates 40 CHAPTER 1 VECTOR ANALYSIS An in nitesimal displacement in the 1quot direction is simply dr Fig 138a just as an in nitesimal element of length in the x direction is dx d1 dr 165 On the other hand an in nitesimal element of length in the 5 direction Fig 138b is not just d9 that s an angle it doesn t even have the right units for a length but rather r d 6 dlg rd6 166 Similarly an in nitesimal element of length in the 4 direction Fig 138c is r sin 6 d 2 dl r sin 6 d 167 Thus the general in nitesimal displacement d is d1drfrde rsin9d 168 This plays the role in line integrals for example that d1 dx x d y y dz 2 played in Cartesian coordinates d rsine dq r d6 6 r r d6 rsine a b C Figure 138 The in nitesimal volume element d L39 in spherical coordinates is the product of the three in nitesimal displacements dt d1 dig dl r2 sin 9 dr d9 d 169 I cannot give you a general expression for surface elements d 3 since these depend on the orientation of the surface You simply have to analyze the geometry for any given case this goes for Cartesian and curvilinear coordinates alike If you are integrating over the surface of a sphere for instance then r is constant whereas 6 and as change Fig 139 so da dlg at r r2 sin 9 d9 d t On the other hand if the surface lies in the xy plane say so that 6 is constant to wit nZ while r and as vary then A A dag 2 d1 dl 9 rdr d 9 14 C URVILINEAR COORDINATES 41 Figure 139 Notice nally that r ranges from 0 to 00 d from 0 to Zn and 9 from 0 to It not Zn that would count every point twice5 Example 113 Find the volume of a sphere of radius R Solution R 7T 271 v fdr f r2sinedrded r0 60 1150 R 2 7 271 f r dr sin6d6gt dd 0 0 0 R3 7 22n g nR3 Not a big surprise I So far we have talked only about the geometry of spherical coordinates Now I would 11ke to translate the vector derivatives gradient divergence curl and Laplacian into r 6 notation In principle this is entirely straightforward in the case of the gradient 8T 3T aT VT x ax 8yy azz for instance we would rst use the chain rule to reexpress the partials Br Br ar ar 39 ar a Bx 8r 8x 39 8x Bab 8x 5Alternatively you could run 1 from 0 to 71 the eastern hemisphere and cover the western hemisphere by extendrng 6 from 71 up to 271 But this is very bad notation since among other things sin 6 will then run negative and you ll have to put absolute value signs around that term in volume and surface elements area and volume berng 1ntr1nsically positive quantities 42 CHAPTER 1 VECTOR ANALYSIS The terms in parentheses could he worked out from Eq 162 or rather the inverse of those equations Prob 136 Then we d do the same for BTBy and BTBz Finally we d substitute in the formulas for x and 2 in terms of r 5 and Prob 137 It would take an hour to gure out the gradient in spherical coordinates by this bruteforce method I suppose this is how it was rst done but there is a much more ef cient indirect approach explained in Appendix A which has the extra advantage of treating all coordinate systems at once I described the straightforward method only to show you that there is nothing subtle or mysterious about transforming to spherical coordinates you re expressing the same quantity gradient divergence or whatever in different notation that s all Here then are the vector derivatives in spherical coordinates Gradient 3T 1 8T 1 3T T A 9 170 V arrr39 rsin68 Divergence a 3 V 2 r 6 171 V 2 r vrrsin986sm v9rsin9 3 Curl 1 a ave A 1 1 Mr 3 A V 9 quot 9 X V r sin9 39 gm W 33 l r r Sing 3 3 0 1 a 8Dr A 172 r Brmm 66 ii Laplacian 1 3 3T 1 3 3T 1 azr V2T 2 n6 173 r2 8r r 3r r2 sin6 89 SI 89 r2 sin29 3 2 For reference these formulas are listed inside the front cover Problem 136 Find formulas for r 6 in terms of x y z the inverse in other words of Eq 162 0 Problem 137 Express the unit vectors f39 5 3 in terms of x y i that is derive Eq 164 A A A A 7 A A 7 A Check your answers several ways r r l 0 139 X 0 d Also Work out the inverse formulas giving x y 2 in terms of r 0 13 and 6 ab 0 Problem 138 a Check the divergence theorem for the function V rzf39 using as your volume the sphere of radius R centered at the origin b Do the same for v2 1 r2r If the answer surprises you look back at Prob 116 V 14 C URVILINEAR COORDINATES 43 Figure 140 Figure 141 Problem 139 Compute the divergence of the function v r cos6 r r sin6 r sin6 cos q3 Check the divergence theorem for this function using as your volume the inverted hemispher ical bowl of radius R resting on the x y plane and centered at the origin Fig 140 Problem 140 Compute the gradient and Laplacian of the function T rcos 6 sin 6 cos db Check the Laplacian by converting T to Cartesian coordinates and using Eq 142 Test the gradient theorem for this function using the path shown in Fig 141 from 0 0 0 to 0 0 2 142 Cylindrical Coordinates The cylindrical coordinates s z of a point P are de ned in Fig 142 Notice that lt15 has the same meaning as in spherical coordinates and z is the same as Cartesian s is the distance to P from the z axis whereas the spherical coordinate r is the distance from the origin The relation to Cartesian coordinates is xscos yssin zzz 174 The unit vectors Prob 141 are 6 cos 2sin g sin 2cos y 175 2 2 The in nitesimal displacements are dl y 2 ds dl s d dlZ dz 176 44 CHAPTER I VECTOR ANALYSIS Figure 142 so d1ds sd dzi 177 and the volume element is drsdsd dz 178 The range ofs is 0 gt oo lt15 goes from 0 gt 2n and z from 00 to 00 The vector derivatives in cylindrical coordinates are Gradient 8T IBT A 8T VT 2 A A 179 as 8 s 8 BZ z Divergence 1 8 181 av 180 V V s3ssv s3 3z Curl 13v 81 A Bus av A l 8 Buy A 181 VXV s 8 azsltaz as s 5 g z Laplacian 1 3 3T 1 azr BZT V2T 182 s as S as s2 a 2 az2 These formulas are also listed inside the front cover Problem 141 Express the cylindrical unit vectors s 43 2 inAterms of 2 y 2 that is derive Eq 175 Invert your formulas to get i y 2 in terms of g 2 and ab 15 THE DIRAC DELTA FUNCTION 45 Figure 143 Figure 144 Problem 142 a Find the divergence of the function vs2sin2 ssin cosq 3z 2 b Test the divergence theorem for this function using the quartercylinder radius 2 height 5 shown in Fig 143 c Find the curl of v The Dirac Delta Function 151 The Divergence of f r2 Consider the vector function 1 V h r2 r 183 At every location v is directed radially outward Fig 144 if ever there was a function that ought to have a large positive divergence this is it And yet when you actually calculate the divergence using Eq 171 you get precisely zero 1 3 2 1 1 3 vvr23rr 72r 2510 184 You will have encountered this paradox already if you worked Prob 116 The plot thickens if you apply the divergence theorem to this function Suppose we integrate over a sphere of radius R centered at the origin Prob 138b the surface integral is yivaa R2sin6d6 d f 7r Zn 2 sin6d6 lei 471 185 o o 46 CHAPTER 1 VECTOR ANALYSIS But the volume integral f V vdr is zero if we are really to believe Eq 184 Does this mean that the divergence theorem is false What s going on here The source of the problem is the point r 0 where v blows up and where in Eq 184 we have unwittingly divided by zero It is quite true that V v 0 everywhere except the origin but right at the origin the situation is more complicated Notice that the surface integral 185 is independent of R if the divergence theorem is right and it is we should get f V v dr 471 for any sphere centered at the origin no matter how small Evidently the entire contribution must be coming from the point r 0 Thus V v has the bizarre property that it vanishes everywhere except at one point and yet its integral over any volume containing that point is 471 N0 ordinary function behaves like that On the other hand a physical example does come to mind the density mass per unit volume of a point particle It s zero except at the exact location of the particle and yet its integral is nite namely the mass of the particle What we have stumbled on is a mathematical object known to physicists as the Dirac delta function It arises in many branches of theoretical physics Moreover the speci c problem at hand the divergence of the function f r2 is not just some arcane curiosity it is in fact central to the whole theory of electrodynamics So it is worthwhile to pause here and study the Dirac delta function with some care 152 The OneDimensional Dirac Delta Function The one dimensional Dirac delta function 6x can be pictured as an in nitely high in nitesimally narrow spike with area 1 Fig 145 That is to say 0 ifx 75 0 8x 00 ifx 0 186 and 00 6xdx 1 187 Technically 6x is not a function at all since its value is not nite at x 0 In the mathematical literature it is known as a generalized function or distribution It is if you Figure 145 15 THE DIRACDELTA FUNCTION 47 2 R2x 2 T2x FRI 1 sz 12 14 14 12 x l l2 12 1 x a b Figure 146 like the limit of a sequence of functions such as rectangles Rn x of height n and width 1n or isosceles triangles Tn x of height n and base 2n Fig 146 If f x is some ordinary function that is not another delta function in fact just to be on the safe side let s say that f x is continuous then the product f x6x is zero everywhere except at x 0 It follows that fx6x f06x 188 This is the most important fact about the delta function so make sure you understand why it is true since the product is zero anyway except at x 0 we may as well replace f x by the value it assumes at the origin In particular fx6xdx f0 8xdx f0 189 Under an integral then the delta function picks out the value of f x at x 0 Here and below the integral need not run from 00 to 00 it is suf cient that the domain extend across the delta function and e to 6 would do as well Of course we can shift the spike from x 0 to some other point x a Fig 147 6x a 0 in with 00 6x adx 1 190 00 ifxa 00 Equation 188 becomes fx5x a fa5x a 191 and Eq 189 generalizes to 00 fx6x adx fa 192 Example 114 Evaluate the integral 3 x3606 2 dx 0 48 CHAPTER I VECTOR ANALYSIS a X Figure 147 Solution The delta function picks out the value of x3 at the point x 2 so the integral is 23 8 Notice however that if the upper limit had been 1 instead of 3 the answer would be 0 because the spike would then be outside the domain of integration Although 6 itself is not a legitimate function integrals over 6 are perfectly acceptable In fact it s best to think of the delta function as something that is always intended for use under an integral sign In particular two expressions involving delta functions say D1 x and D2x are considered equal if 6 OO 00 fxD1x dx fxD2xdx 193 00 00 for all ordinary functions f x Example 115 Show that 1 8kx 2 mm 194 where k is any nonzero constant In particular 6 x 8x Solution For an arbitrary test function f x consider the integral 00 fx8kxdx 00 Changing variables we let y E kx so that x yk and dx lkdy If k is positive the integration still runs from 00 to 00 but if k is negative then x 00 implies y 2 00 and 6This is not as arbitrary as it may sound The crucial point is that the integrals must be equal for any f x Suppose D1x and D2x actually di ered say in the neighborhood of the point x 17 Then we could pick a function f x that was sharply peaked about x 17 and the integrals would not be equal 15 THE DIRAC DELTA FUNCTION 49 vice versa so the order of the limits is reversed Restoring the proper order costs a minus sign Thus 00 0 dy 1 l fx5kxdx i fyk5y if0 f0 OO 00 k k kl The lower signs apply when k is negative and we account for this neatly by putting absolute value bars around the nal k as indicated Under the integral sign then 8kx serves the same purpose as lk6 x 00 fx8 xdx 00 fx mam dx According to criteriOn 193 therefore 8kx and 1 lk6x are equal Problem 143 Evaluate the following integrals a f263x2 2x 15x 3dx b cosx 8x ndx c f03 x3806 1dx d ffooo lnx 3 6x 2 dx Problem 144 Evaluate the following integrals a 32m 383xdx b f02x3 3x 261 xdx cf119x283x ldx d ffoo 8x b dx Problem 145 a Show that d 6 x dx 1 5 x Himt Use integration by parts b Let 9x be the step function 1 ifx gt 0 9x E 195 0 ifx 5 0 Show that dQdx 6x 50 CHAPTER 1 VECTOR ANALYSIS 153 The ThreeDimensional Delta Function It is an easy matter to generalize the delta function to three dimensions 639 69 6o 6z 196 As always r E x x y y z i is the position vector extending from the origin to the point x y z This three dimensional delta function is zero everywhere except at 000 where it blows up Its volume integral is 1 00 00 00 63r dr f 5x 6y 8zdx dy dz 1 197 all space oo oo 00 And generalizing Eq 192 fr63r 3 dr fa 198 all space As in the one dimensional case integration with 6 picks out the value of the function f at the location of the spike We are now in a position to resolve the paradox introduced in Sect 151 As you will recall we found that the divergence of fr2 is zero everywhere except at the origin and yet its integral over any volume containing the origin is a constant to wit 471 These are precisely the de ning conditions for the Dirac delta function evidently V 471530 199 r More generally 2 3 V 2 4718 a 1100 0 where as always 4 is the separation vector 4 E r r Note that differentiation here is with respect to r while r is held constant Incidentally since 1 22 Prob 113 it follows that 21 3 v 47154 1102 0 16 THE THEORY OF VECTOR FIELDS 53 Maxwell s formulation raises an important mathematical question To what extent is a vector function determined by its divergence and curl In other words if I tell you that the divergence of F which stands for E or B as the case may be is a speci ed scalar function D V F D and the curl of F is a speci ed vector function C V X F C for consistency C must be divergenceless V C 0 because the divergence of a curl is always zero can you then determine the function F Well not quite For example as you may have discovered in Prob 119 there are many functions whose divergence and curl are both zero everywhere the trivial case F 0 ofcourse but alsoF yz xzx yl xy i F sin x cosh y x cos x sinh y 9 etc To solve a differential equation you must also be supplied with appropriate boundary conditions In electrodynamics we typically require that the elds go to zero at in nity far away from all charges8 With that extra information the Helmholtz theorem guarantees that the eld is uniquely determined by its divergence and curl A proof of the Helmholtz theorem is given in Appendix B 162 Potentials If the curl of a vector eld F vanishes everywhere then F can be written as the gradient of a scalar potential V VxF0ltgtF VV 1103 The minus sign is purely conventional That s the essential burden of the following theorem Theorem 1 Curlless or irrotational elds The following con ditions are equivalent that is F satis es one if and only if it satis es all the others a V x F 0 everywhere b fab F d is independent of path for any given end pornts c 55 F d1 0 for any closed loop d F is the gradient of some scalar F 2 VV 8In some textbook problems the charge itself extends to in nity We speak for instance of the electric eld of an in nite plane or the magnetic eld of an in nite wire In such cases the normal boundary conditions do not apply and one must invoke symmetry arguments to determine the elds uniquely 54 CHAPTER I VECTOR ANALYSIS The scalar potential is not unique any constant can be added to V with impunity since this will not affect its gradient If the divergence of a vector eld F vanishes everywhere then F can be expressed as the curl of a vector potential A VF0ltgtFVgtltA 1104 That s the main conclusion of the following theorem Theorem 2 Divergenceless or solenoidal elds The following conditions are equivalent a V F 0 everywhere b f Fdais independent of surface for any given bound ary line c 55F da 2 0 for any closed surface d F is the curl of some vector F V X A The vector potential is not unique the gradient of any scalar function can be added to A without affecting the curl since the curl of a gradient is zero You should by now be able to prove all the connections in these theorems save for the ones that say a b or c implies d Those are more subtle and will come later Incidentally in all cases whatever its curl and divergence may be a vector eld F can be written as the gradient of a scalar plus the curl of a vector F VVV gtltA always 1105 Problem 149 a Let F1 2 x2 2 and F2 2 x 2 y 7 z 2 Calculate the divergence and curl of F1 and F2 Which one can be written as the gradient of a scalar Find a scalar potential that does the job Which one can be written as the curl of a vector Find a suitable vector potentral b Show that F3 2 vz 2 zx 7 xy 2 can be written both as the gradient of a scalar and as the curl of a vector Find scalar and vector potentials for this function Problem 150 For Theorem 1 show that d a a gt c c gt b b gt c and 0 gt a Problem 151 For Theorem 2 show that d gt a a gt c c gt b b gt c and c gt a 16 THE THEORY OF VECTOR FIELDS 57 Problem 161 The integral a E da 1106 S is sometimes called the vector area of the surface S If S happens to be at then a is the ordinary scalar area obviously a Find the vector area of a hemispherical bowl of radius R b Show that a 0 for any closed surface Hint39 Use Prob l60a c Show that a is the same for all surfaces sharing the same boundary d Show that a I rxdl 1107 where the integral is around the boundary line Hints One way to do it is to draw the cone subtended by the loop at the origin Divide the conical surface up into in nitesimal triangu lar wedges each with vertex at the origin and opposite side dl and exploit the geometrical interpretation of the cross product Fig 18 e Show that 39 crdlaxc 1108 for any constant vector c Hint39 let T c r in Prob l60e Problem 162 a Find the divergence of the function 1quot V r First compute it directly as in Eq 184 Test your result using the divergence theorem as in Eq 185 Is there a delta function at the origin as there was for r r2 What is the general formula for the divergence of rnr Answer V rnr n 2r 1 unless n 2 in which case it is 47163r b Find the curl of rquotr Test your conclusion using Prob l60b Answers V X M 0 Chapter 2 Electrostatics 21 The Electric Field 211 Introduction The fundamental problem electromagnetic theory hopes to solve is this Fig 21 We have some electric charges q q2 q3 call them source charges what force do they exert on another charge Q call it the test charge The positions of the source charges are given as functions of time the trajectory of the test particle is to be calculated In general both the source charges and the test charge are in motion The solution to this problem is facilitated by the principle of superposition which states that the interaction betWeen any two charges is completely unaffected by the presence of others This means that to determine the force on Q we can rst compute the force F1 due to q 1 alone ignoring all the others then we compute the force F2 due to qz alone and so on Finally we take the vector sum of all these individual forces F F1 F2 F3 Thus if we can nd the force on Q due to a single source charge q we are in principle done the rest is just a qUestion of repeating the same operation over and over and adding it all up1 Well at rst sight this sounds very easy Why don t I just write down the formula for the force on Q due to q and be done with it I could and in Chapter 10 I shall but you would be shocked to see it at this stage for not only does the force on Q depend on the separation distance a between the charges Fig 22 it also depends on both their velocities and on the acceleration of q Moreover it is not the position velocity and acceleration of q right now that matter Electromagnetic news travels at the speed of light so what concerns Q is the position velocity and acceleration q had at some earlier time when the message left l The principle of superposition may seem obvious to you but it did not have to be so simple if the electromag netic force were proportional to the square of the total source charge for instance the principle of superposition would not hold since q1 q2 a ql2 q there would be cross terms to consider Superposition is not a logical necesstty but an experimental fact 58 21 THE ELECTRIC FIELD 59 0 Q 51239 4139 Q 0 gt 4 quotSourcequot charges quotTestquot charge q Figure 21 Figure 22 Therefore in spite of the fact that the basic question What is the force on Q due to q is easy to state it does not pay to confront it head on rather we shall go at it by stages In the meantime the theory we develop will permit the solution of more subtle electromagnetic problems that do not present themselves in quite this simple format To begin with we shall consider the special case of electrostatics in which all the source charges are stationary though the test charge may be moving 212 Coulomb s Law What is the force on a test charge Q due to a single point charge q which is at rest a distance 2 away The answer based on experiments is given by Coulomb s law 1 A E F 47T 0 22 21 The constant 60 is called the permitivity of free space In SI units where force is in Newtons N distance in meters m and charge in coulombs C 2 60 885 x 10 12 C N m2 In words the force is proportional to the product of the charges and inversely proportional to the square of the separation distance As always Sect 114 4 is the separation vector from r the location of q to r the location of Q 4 r r 22 4 is its magnitude and I is its direction The force points along the line from q to Q it is repulsive if q and Q have the same sign and attractive if their signs are opposite Coulomb s law and the principle of superposition constitute the physical input for electrostatics the rest except for some special properties of matter is mathematical elab oration of these fundamental rules 60 CHAPTER 2 ELECTROSTATICS Problem 21 a Twelve equal charges q are situated at the comers of a regular 12sided 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 the one at 6 o clock What is the force on Q Explain your reasoning carefully c Now 13 equal charges q are placed at the corners of a regular l3sided polygon What is the force on a test charge Q at the center 1 If one of the 13 q s is removed what is the force on Q Explain your reasoning 213 The Electric Field If we have several point charges q1 q2 qn at distances 21 22 2 from Q the total force on Q is evidently 1 A F F1F2 41gazm 47T 0 21 212 21 22 3 23 i q1 2q2 2q 2 47T60 21 22 23 or F QE 23 where 1 n 1139 E r E 2 24 4m 2 a i lt i1 E is called the electric eld of the source charges Notice that it is a function of position r because the separation vectors 4i depend on the location of the eld point P Fig 23 But it makes no reference to the test charge Q The electric eld is a vector quantity that varies Source point Figure 23 21 THE ELECTRIC FIELD 61 from point to point and is determined by the con guration of source charges physically E r is the force per unit charge that would be exerted on a test charge if you were to place one at P What exactly is an electric eld I have deliberately begun with what you might call the minimal interpretation of E as an intermediate step in the calculation of electric forces But I encourage you to think of the eld as a real physical entity lling the space in the neighborhood of any electric charge Maxwell himself came to believe that electric and magnetic elds represented actual stresses and strains in an invisible primordial jellylike ether Special relativity has forced us to abandon the notion of ether and with it Maxwell s mechanical interpretation of electromagnetic elds It is even possible though cumbersome to formulate classical electrodynamics as an actionatadistance theory and dispense with the eld concept altogether I can t tell you then what a eld is only how to calculate it and what it can do for you once you ve got it Problem 22 a Find the electric eld magnitude and direction a distance z above the midpoint between two equal charges q a distance d apart Fig 24 Check that your result is consistent with what you d expect when z gtgt d b Repeat part a only this time make the right hand charge q instead of 61 P a Continuous distribution z da 4 39P 1 12 12 1 c Surface charge 6 d Volume charge p Figure 24 Figure 25 214 Continuous Charge Distributions Our de nition of the electric eld Eq 24 assumes that the source of the eld is a collection of discrete point charges qi If instead the charge is distributed continuously over some region the sum becomes an integral Fig 25a 1 E 4d 25 1 47m 2 q 62 CHAPTER 2 ELECTROS TATI CS If the charge is spread out along a line Fig 25b with chargeperunitlength A then dq A dl where dl is an element of length along the line if the charge is smeared out over a surface Fig 25c with chargeperunitarea a then dq a da where da is an element of area on the surface and if the charge lls a volume Fig 25d with chargeperunit volume 0 then dq o d r where d T is an element of volume dq gt Adl 0da pdr Thus the electric eld of a line charge is l Ar A 4dl 26 130 4M0 2 73 for a surface charge I a r A Er f 2 4da 27 47T 0 4 S and for a volume charge 1 r A Er f 2 4dr 28 47T 0 4 V Equation 28 itself is often referred to as Coulomb s law because it is such a short step from the original 21 and because a volume charge is in a sense the most general and realistic case Please note carefully the meaning of 4 in these formulas Originally in Eq 24 In stood for the vector from the source charge qi to the eld point r Correspondingly in Eqs 25 28 4 is the vector from dq therefore from dl da or d r to the eld point r2 Example 21 Find the electric eld a distance z above the midpoint of a straight line segment of length 2L which carries a uniform line charge A Fig 26 Solution It is advantageous to chop the line up into symmetrically placed pairs at ix for then the horizontal components of the two elds cancel and the net eld of the pair is l A dx A dE 2 cos 6 2 47160 42 2 Warning The unit vector2 is not constant its direction depends on the source point r and hence it cannot be taken outside the integrals 25 28 In practice you must work with Cartesian components a 539 2 are constant and do come out even if you use curvilinear coordinates to perform the integration 2 1 THE ELECTRIC FIELD 63 Figure 26 Here cos6 z z2 x2 and x runs from 0 to L 1 L 2A E de 47T60 0 22x232 L 2A2 x 47T 0 Z2 Z2 x2 1 2AL 471 0Z Z2L2 O and it aims in the zdirection For points far from the line 2 gtgt L this result simpli es N 1 2AL E 47160 2 which makes sense From far away the line looks like a point charge q 2AL so the eld reduces to that of point charge q 47160Z2 In the limit L gt 00 on the other hand we obtain the eld of an in nite straight wire 1 2139 E 47160 z or more generally 1 2A 47160 T 29 where s is the distance from the wire Problem 23 Find the electric eld a distance z above one end of a straight line segment of length L Fig 27 which carries a uniform line charge A Check that your formula is consistent with what you would expect for the case z gtgt L 64 CHAPTER 2 ELECTROSTATICS P P T P T l i I Z i z I Z I I 39 E C L l I a 39 Figure 27 Figure 28 Figure 29 Problem 24 Find the electric eld a distance z above the center of a square loop side a carrying uniform line charge A Fig 28 Hinli39 Use the result of Ex 21 Problem 25 Find the electric eld a distance 2 above the center of a circular loop of radius r Fig 29 which carries a uniform line charge A Problem 26 Find the electric eld a distance 2 above the center of a at circular disk of radius R Fig 210 which carries a uniform surface charge a What does your formula give in the limit R gt 00 Also check the case z gtgt R Problem 27 Find the electric eld a distance z from the center of a spherical surface of radius R Fig 211 which carries a uniform charge density 0 Treat the case z lt R inside as well as z gt R outside Express your answers in terms of the total charge 1 on the sphere Him Use the law of cosines to write 4 in terms of R and 6 Be sure to take the positive square root xR2z2 2Rz R zifR gtzbutit sz RifR ltz Problem 28 Use your result in Prob 27 to nd the eld inside and outside a sphere of radius R which carries a uniform volume charge density p Express your answers in terms of the total charge of the sphere q Draw a graph of E as a function of the distance from the center Figure210 Figure 211 22 DI VERGENCE AND CURL OF ELECTROSTATIC FIELDS 65 22 Divergence and Curl 0f Electrostatic Fields 221 Field Lines Flux and Gauss s Law In principle we are done with the subject of electrostatics Equation 28 tells us how to compute the eld of a charge distribution and Eq 23 tells us what the force on a charge Q placed in this eld will be Unfortunately as you may have discovered in working Prob 27 the integrals involved in computing E can be formidable even for reasonably simple charge distributions Much of the rest of electrostatics is devoted to assembling a bag of tools and tricks for avoiding these integrals It all begins with the divergence and curl of E I shall calculate the divergence of E directly from Eq 28 in Sect 222 but rst I want to show you a more qualitative and perhaps more illuminating intuitive approach Let s begin with the simplest possible case a single point charge q situated at the origm 1 210 E r 47 60 r2 To get a feel for this eld I might sketch a few representative vectors as in Fig 212a Because the eld falls off like 1r2 the vectors get shorter as you go farther away from the origin they always point radially outward But there is a nicer way to represent this eld and that s to connect up the arrows to form eld lines Fig 212b You might think that I have thereby thrown away information about the strength of the eld which was contained in the length of the arrows But actually I have not The magnitude of the eld is indicated by the density of the eld lines it s strong near the center where the eld lines are close together and weak farther out where they are relatively far apart In truth the eldline diagram is deceptive when I draw it on a twodimensional surface for the density of lines passing through a circle of radius r is the total number divided by the circumference n 2nr which goes like 1 r not 1 r2 But if you imagine the model in three dimensions a pincushion with needles sticking out in all directions then the density of lines is the total number divided by the area of the sphere n 47ir2 which does go like lrz v a b Figure 212 66 CHAPTER 2 ELECTROSTATICS Equal but opposite charges Figure 213 Such diagrams are also convenient for representing more complicated elds Of course the number of lines you draw depends on how energetic you are and how sharp your pencil is though you ought to include enough to get an accurate sense of the eld and you must be consistent If charge q gets 8 lines then 2q deserves 16 And you must space them fairly they emanate from a point charge symmetrically in all directions Field lines begin on positive charges and end on negative ones they cannot simply terminate in midair though they may extend out to in nity Moreover eld lines can never cross at the intersection the eld would have two different directions at once With all this in mind it is easy to sketch the eld of any simple con guration of point charges Begin by drawing the lines in the neighborhood of each charge and then connect them up or extend them to in nity Figs 213 and 214 Equal charges Figure 214 22 DI VERGENCE AND CURL OF ELECTROSTATIC FIELDS 67 Figure 215 In this model the ux of E through a surface S ltIgtEEfEda 211 S is a measure of the number of eld lines passing through 5 Iput this in quotes because of course we can only draw a representative sample of the eld lines the total number would be in nite But for a given sampling rate the ux is proportional to the number of lines drawn because the eld strength remember is proportional to the density of eld lines the number per unit area and hence E da is proportional to the number of lines passing through the in nitesimal area da The dot product picks out the component of da along the direction of E as indicated in Fig 215 It is only the area in the plane perpendicular to E that we have in mind when we say that the density of eld lines is the number per unit area This suggests that the ux through any closed surface is a measure of the total charge inside For the eld lines that originate on a positive charge must either pass out through the surface or else terminate on a negative charge inside Fig 216a On the other hand a charge outside the surface will contribute nothing to the total ux since its eld lines pass in one side and out the other Fig 26b This is the essence of Gauss s law Now let s make it quantitative b Figure 216 68 CHAPTER 2 ELECTROSTATICS In the case of a point charge q at the origin the ux of E through a sphere of radius r is 1 1 gEda 12fr2sineded f q 212 4739l60 r 60 Notice that the radius of the sphere cancels out for while the surface area goes up as r2 the eld goes down as 1 r2 and so the product is constant In terms of the eldline picture this makes good sense since the same number of eld lines passes through any sphere centered at the origin regardless of its size In fact it didn t have to be a sphere any closed surface whatever its shape would trap the same number of eld lines Evidently the ux through any surface enclosing the charge is q 60 Now suppose that instead of a single charge at the origin we have a bunch of charges scattered about According to the principle of superposition the total eld is the vector sum of all the individual elds n E E E 1 The ux through a surface that encloses them all then is For any closed surface then l E39da QBHC7 213 60 S where QEmc is the total charge enclosed within the surface This is the quantitative state ment of Gauss s law Although it contains no information that was not already present in Coulomb s law and the principle of superposition it is of almost magical power as you will see in Sect 223 Notice that it all hinges on the l r2 character of Coulomb s law without that the crucial cancellation of the r s in Eq 212 would not take place and the total ux of E would depend on the surface chosen not merely on the total charge enclosed Other 1 r2 forces 1 am thinking particularly of Newton s law of universal gravitation will obey Gauss s laws of their own and the applications we develop here carry over directly As it stands Gauss s law is an integral equation but we can readily turn it into a di erential one by applying the divergence theorem E39daVEdr V S Rewriting QgnC in terms of the charge density 0 we have QenCZPdr39 V 22 DI VERGENCE AND CURL OF ELECTROSTATIC FIELDS 71 Gauss1an surface Figure 218 and the magnitude of E is constant over the Gaussian surface so it comes outside the integral flElda E da E47tr2 s 8 Thus 2 l lEl4 r q 60 or 1 q A r 4760 r2 Notice a remarkable feature of this result The eld outside the sphere is exactly the same as it would have been if all the charge had been concentrated at the center Gauss s law is always true but it is not always useful If 0 had not been uniform or at any rate not spherically symmetrical or if I had chosen some other shdpe for my Gaussian surface it would still have been true that the ux of E is 1 60q but I would not have been certain that E was in the same direction as da and constant in magnitude over the surface and without that I could not pull E out of the integral Symmetry is crucial to this application of Gauss s law As far as I know there are only three kinds of symmetry that work A Spherical symmetry Make your Gaussian surface a concentric sphere 2 Cylindrical symmetry Make your Gaussian surface a coaxial cylinder Fig 219 3 Plane symmetry Use a Gaussian pillbox which straddles the surface Fig 220 Although 2 and 3 technically require in nitely long cylinders and planes extending to in nity in all directions we shall often use them to get approximate answers for long cylinders or large plane surfaces at points far from the edges 72 CHAPTER 2 ELECTROSTATICS Gaussian pillbox Gaussian surface Figure 219 Figure 220 Example 23 A long cylinder Fig 221 carries a Charge density that is proportional to the distance from the axis 0 ks for some constant k Find the electric eld inside this cylinder Solution Draw a Gaussian cylinder of length l and radius s For this surface Gauss s law states 1 E39da Qenc 6O S The enclosed charge is S Qenc fpdr kssdsd dz 2nkl s2 ds nkls3 O I used the volume element appropriate to cylindrical coordinates Eq 178 and integrated d from 0 to 27139 d z from 0 to l I put a prime on the integration variable s to distinguish it from the radius s of the Gaussian surface Gaussian surface Figure 221 22 DI VERGENCE AND CURL OF ELECTROSTATIC FIELDS 73 Now symmetry dictates that E must point radially outward so for the curved portion of the Gaussian cylinder we have fEdaEdaEldaE2n sl while the two ends contribute nothing here E is perpendicular to da Thus 1 2 E2nsl Jrkls3 60 3 or nally Example 24 An in nite plane carries a uniform surface charge 0 Find its electric eld Solution Draw a Gaussian pillbox extending equal distances above and below the plane Fig 222 Apply Gauss s law to this surface 1 thda Qenc 60 In this case Qenc 0A where A is the area of the lid of the pillbox By symmetry E points away from the plane upward for points above downward for points below Thus the top and bottom surfaces yield fE da 2 2AE Figure 222 74 CHAPTER 2 ELECTROSTATICS whereas the sides contribute nothing Thus 1 2A 2 UA 60 Or a E 217 260 where 3 is a unit vector pointing away from the surface In Prob 26 you obtained this same result by a much more laborioirs method It seems surprising at rst that the eld of an in nite plane is independent of haw far away you are What about the 1 r2 in Coulomb s law Well the point is that as you move farther and farther away from the plane more and more charge comes into your eld of view a cone shape extending out from your eye and this compensates for the diminishing in uence of any particular piece The electric eld of a sphere falls off like 1 r2 the electric eld of an in nite line falls off like 1 r and the electric eld of an in nite plane does not fall off at all Although the direct use of Gauss s law to compute electric elds is limited to cases of spherical cylindrical and planar symmetry we can put together combinations of objects possessing such symmetry even though the arrangement as a whole is not symmetrical For example invoking the principle of superposition we could nd the eld in the vicinity of two uniformly charged parallel cylinders or a sphere near an in nite charged plane Example 25 Two in nite parallel planes carry equal but opposite uniform charge densities i0 Fig 223 Find the eld in each of the three regions i to the left of both ii between them iii to the right of both Solution The left plate produces a eld 1260O39 which points away from it Fig 224 to the left in region i and to the right in regions ii and iii The right plate being negatively charged produces a eld 1 260Q which points toward it to the right in regions i and ii and to the left in region iii The two elds cancel in regions i and iii they conspire in region ii Conclusion The eld is 1600 and points to the right between the planes elsewhere it is zero i ii iii 6 5 Figure 223 Figure 224 22 DI VERGENCE AND CURL OF ELECTROSTATIC FIELDS 75 Problem 211 Use Gauss s law to nd the electric eld inside and outside a spherical shell of radius R which carries a uniform surface charge density 7 Compare your answer to Prob 27 Problem 212 Use Gauss s law to nd the electric eld inside a uniformly charged sphere charge density p Compare your answer to Prob 28 Problem 213 Find the electric eld a distance s from an in nitely long straight wire which carries a uniform line charge A Compare Eq 29 Problem 214 Find the electric eld inside a sphere which carries a charge density proportional to the distance from the origin p kr for some constant k Hint This charge density is not uniform and you must integrate to get the enclosed charge Problem 215 A hollow spherical shell carries charge density 10 r 2 in the region a 5 r E b Fig 225 Find the electric eld in the three regions i r lt a ii a lt r lt 17 iii r gt 7 Plot El as afunction of r Problem 216 A long coaxial cable Fig 226 carries a uniform volume charge density p on the inner cylinder radius a and a uniform surface charge density on the outer cylindrical shell radius 19 This surface charge is negative and of just the right magnitude so that the cable as a whole is electrically neutral Find the electric eld in each of the three regions i inside the inner cylinder s lt a ii between the cylinders a lt s lt 19 iii outside the cable 3 gt 9 Plot lE as afunction of s Problem 217 An in nite plane slab of thickness 2d carries a uniform volume charge density p Fig 227 Find the electric eld as a function of y where y 0 at the center Plot E versus y calling E positive when it points in the y direction and negative when it points in the y direction Problem 218 Two spheres each of radius R and carrying uniform charge densities p and p respectively are placed so that they partially Overlap Fig 228 Call the vector from the positive center to the negative center 1 Show that the eld in the region of Overlap is constant and nd its value Hints Use the answer to Prob 212 Figure 225 Figure 226 76 CHAPTER 2 ELECTROSTATICS X Figure 227 Figure 228 224 The Curl of E I ll calculate the curl of E as I did the divergence in Sect 221 by studying rst the simplest possible con guration a point charge at the origin In this case 1 q A I39 47160 r2 Now a glance at Fig 212 should convince you that the curl of this eld has to be zero but I suppose we ought to come up with something a little more rigorous than that What if we calculate the line integral of this eld from some point a to some other point b Fig 229 b fEdl a In spherical coordinates d1 2 dr f39 r d6 5 r sine d gt if so 1 q E l d d 471607392 r b 1 b 1 b 1 Edl 12m 1 1 1 218 a 47160 a r 47T 0 r M 47160 ra r1 where ra is the distance from the origin to the point a and r1 is the distance to b The integral around a closed path is evidently zero for then rd 2 rb Therefore Edl0 219 23 ELECTRIC POTENTIAL 77 Figure 229 and hence applying Stokes theorem 220 Now 1 proved Eqs 219 and 220 only for the eld of a single point charge at the origin but these results make no reference to what is after all a perfectly arbitrary choice of coordinates they also hold no matter where the charge is located Moreover if we have many charges the principle of superposition states that the total eld is a vector sum of their individual elds EE1E2 so VXEVXEE2VXE1VXE2O Thus Eqs 219 and 220 hold for any static charge distribution whatever Problem 219 Calculate V x E directly from Eq 28 by the method of Sect 222 Refer to Prob 162 if you get stuck 23 Electric Potential 231 Introduction to Potential The electric eld E is not just any old vector function it is a very special kind of vector function one whose curl is always zero E yx for example could not possibly be an electrostatic eld no set of charges regardless of their sizes and positions could ever produce such a eld In this section we re going to exploit this special property of electric elds to reduce a vector problem nding E down to a much simpler Scalar problem The rst theorem in Sect 162 asserts that any vector whose curl is zero is equal to the gradient of some scalar What I m going to do now amounts to a proof of that claim in the context of electrostatics 78 CHAPTER 2 ELECTROSTATICS ii Figure 230 Because V x E 0 the line integral of E around any closed loop is zero that follows from Stokes theorem Because f E d1 0 the line integral of E from point a to point b is the same for all paths otherwise you could go out along path i and return along path ii Fig 230 and obtain f E dl 7s 0 Because the line integral is independent of path we can de ne a function4 Vr E rEdl 221 O Here 9 is some standard reference point on which we have agreed beforehand V then depends only on the point r It is called the electric potential Evidently the potential di erence between two points a and b is b a Vb Va EdH Edl O O b O b Edl f Edl Edl 222 O a 2 Now the fundamental theorem for gradients states that b Vb Va VVdl b b fVVdl Edl Since nally this is true for any points a and b the integrands must be equal lt22 Equation 223 is the differential version of Eq 221 it says that the electric eld is the gradient of a scalar potential which is what we set out to prove SO 4To avoid any possible ambiguity I should perhaps put a prime on the integration variable r Vr Er dl 9 But this makes for cumbersome notation and I prefer whenever possible to reserve the primes for source points However when as in Ex 26 we calculate such integrals explicitly I shall put in the primes 23 ELECTRIC POTENTIAL 79 Notice the subtle but crucial role played by path independence or equivalently the fact that V X E O in this argument If the line integral of E depended on the path taken then the de nition of V Eq 221 would be nonsense It simply would not de ne a function since changing the path would alter the value of Vr By the way don t let the minus sign in Eq 223 distract you it carries over from 221 and is largely a matter of convention Problem 220 One of these is an impossible electrostatic eld Which one a E kxyf 2yzy 3xz i b E My2 1 2xy ZZ 9 2yz 2 Here k is a constant with the appropriate units For the possible one nd the potential using the origin as your reference point Check your answer by computing V V Hint You must select a speci c path to integrate along It doesn t matter what path you choose since the answer is path independent but you simply cannot integrate unless you have a particular path in mind 232 Comments on Potential i The name The word potential is a hideous misnomer because it inevitably reminds you of potential energy This is particularly confusing because there is a connection between potential and potential energy as you will see in Sect 24 I m sorry that it is impossible to escape this word The best I can do is to insist once and for all that potential and potential energy are completely different terms and should by all rights have different names Incidentally a surface over which the potential is constant is called an equipotential ii Advantage of the potential formulation If you know V you can easily get E just take the gradient E 2 VV This is quite extraordinary when you stop to think about it for E is a vector quantity three components but V is a scalar one component How can one function possibly contain all the information that three independent functions carry The answer is that the three components of E are not really as independent as they look in fact they are explicitly interrelated by the very condition we started with V x E 0 In terms of components 3E 3E 3E 35 3E 3EZ ay ax E az az W39 This brings us back to my observation at the beginning of Sect 231 E is a very special kind of vector What the potential formulation does is to exploit this feature to maximum advantage reducing a vector problem down to a scalar one in which there is no need to fuss with components 80 CHAPTER 2 ELECTROSTATICS iii The reference point 9 There is an essential ambiguity in the de nition of potential since the choice of reference point 9 was arbitrary Changing reference points amounts to adding a constant K to the potential r O r Vr E dlz f Edl EdlKVr I 0 where K is the line integral of E from the old reference point 9 to the new one 9 Of course adding a constant to V will not affect the potential di erence between two points Vb Va Vb Va since the K s cancel out Actually it was already clear from Eq 222 that the potential difference is independent of 9 because it can be written as the line integral of E from a to b with no reference to O Nor does the ambiguity affect the gradient of V VV VV since the derivative of a constant is zero That s why all such V s differing only in their choice of reference point correspond to the same eld E Evidently potential as such carries no real physical signi cance for at any given point we can adjust its value at will by a suitable relocation of O In this sense it is rather like altitude If I ask you how high Denver is you will probably tell me its height above sea level because that is a convenient and traditional reference point But we could as well agree to measure altitude above Washington DC or Greenwich or wherever That would add or rather subtract a xed amount from all our sealevel readings but it wouldn t change anything about the real world The only quantity of intrinsic interest is the di erence in altitude between two points and that is the same whatever your reference level Having said this however there is a natural spot to use for O in electrostatics analogous to sea level for altitude and that is a point in nitely far from the charge Or dinarily then we set the zero of potential at in nity Since VO 0 choosing a reference point is equivalent to selecting a place where V is to be zero But I must warn you that there is one special circumstance in which this convention fails when the charge distribution itself extends to in nity The symptom of trouble in such cases is that the potential blows up For instance the eld of a uniformly charged plane is a2eo as we found in Ex 24 if we na39r vely put 9 00 then the potential at height z above the plane becomes Z l l Vz adz crz oo 00 260 260 The remedy is simply to choose some other reference point in this problem you might use the origin Notice that the dif culty occurs only in textbook problems in real life there is no such thing as a charge distribution that goes on forever and we can always use in nity as our reference point 2 3 ELECTRIC POTENTIAL 81 iv Potential obeys the superposition principle The original superposition princi ple of electrodynamics pertains to the force on a test charge Q It says that the total force on Q is the vector sum of the forces attributable to the source charges individually F F1 F2 Dividing through by Q we nd that the electric eld too obeys the superposition principle E E1 E2 Integrating from the common reference point to r it follows that the potential also satis es such a principle VV1V2 That is the potential at any given point is the sum of the potentials due to all the source charges separately Only this time it is an ordinary sum not a vector sum which makes it a lot easier to work with v Units of Potential In our units force is measured in newtons and charge in coulombs so electric elds are in newtons per coulomb Accordingly potential is measured in newton meters per coulomb or joules per coulomb A joule per coulomb is called a volt Example 26 Find the potential inside and outside a spherical shell of radius R Fig 231 which carries a uniform surface charge Set the reference point at in nity Figure 231 82 CHAPTER 2 ELECTROSTATICS Solution From Gauss s law the eld outside is 1 E 11 47160 r2 where q is the total charge on the sphere The eld inside is zero For points outside the sphere r gt R r l q 00 47160 r r 1 r l Vr Edl q dr 1 9 47160 00 r 2 47160 r To nd the potential inside the sphere r lt R we must break the integral into two sections using in each region the eld that prevails there R 1 q Vr 47160 1 R r 1 idr 0dr 1 47160 00 r2 R 47160 r 00 Notice that the potential is not zero inside the shell even though the eld is V is a constant in this region to be sure so that V V O that s what matters In problems of this type you must always work your way in from the reference point that s where the potential is nailed down It is tempting to suppose that you could gure out the potential inside the sphere on the basis of the eld there alone but this is false The potential inside the sphere is sensitive to what s going on outside the sphere as well If I placed a second uniformly charged shell out at radius R gt R the potential inside R would change even though the eld would still be zero Gauss s law guarantees that Charge exterior to a given point that is at larger r produces no net eld at that point provided it is spherically or39cylindrically symmetric but there is no such rule for potential when in nity is used as the reference point Problem 221 Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q Use in nity as your reference point Compute the gradient of V in each region and check that it yields the correct eld Sketch Vr Problem 222 Find the potential a distance s from an in nitely long straight wire that carries a uniform line charge A Compute the gradient of your potential and check that it yields the correct eld Problem 223 For the charge con guration of Prob 215 nd the potential at the center using in nity as your reference point Problem 224 For the con guration of Prob 216 nd the potential difference between a point on the axis and a point on the outer cylinder Note that it is not necessary to commit yourself to a particular reference point if you use Eq 222 23 ELECTRIC POTENTIAL 83 233 Poisson s Equation and Laplace s Equation We found in Sect 231 that the electric eld can be written as the gradient of a scalar potential E VV The question arises What do the fundamental equations for E V1223 and VxE0 60 look like in terms of V Well V E V V V V2V so apart from that persisting minus sign the divergence of E is the Laplacian of V Gauss s law then says that WV 224 50 This is known as Poisson s equation In regions where there is no charge so that 0 0 Poisson s equation reduces to Laplace s equation vzv 0 225 We ll explore these equations more fully in Chapter 3 So much for Gauss s law What about the curl law This says that VXEVX VV must equal zero But that s no condition on V curl of gradient is always zero Of course we used the curl law to show that E could be expressed as the gradient of a scalar so it s not really surprising that this works out V x E 0 permits E V V in return E VV guarantees V x E 0 It takes only one differential equation Poisson s to determine V because V is a scalar for E we needed two the divergence and the curl 234 The Potential of a Localized Charge Distribution I de ned V in terms of E Eq 221 Ordinarily though it s E that we re looking for if we already knew E there wouldn t be much point in calculating V The idea is that it might be easier to get V rst and then calculate E by taking the gradient Typically then we know where the charge is that is we know 0 and we want to nd V Now Poisson s equation relates V and 0 but unfortunately it s the wrong way around it would give us 0 if we knew V whereas we want V knowing 0 What we must do then is invert Poisson s equation That s the program for this section although I shall do it by roundabout means beginning as always with a point charge at the origin 84 CHAPTER 2 ELECTROSTATICS 4 q O Ii 12 g Figure 232 Setting the reference point at in nity the potential of a point charge q at the origin is r r W f2 12 g 47reo 00 r 2 47reo r 00 47160 r You see here the special virtue of using in nity for the reference point it kills the lower limit on the integral Notice the sign of V presumably the conventional minus sign in the de nition of V Eq 221 was chosen precisely in order to make the potential of a positive charge come out positive It is useful to remember that regions of positive charge are potential hills regions of negative charge are potential valleys and the electric eld points downhill from plus toward minus In general the potential of a point charge q is 1 V i 226 47TEQ L where a as always is the distance from the charge to r Fig 232 Invoking the superpo sition principle then the potential of a collection of charges is quot 611 V r 227 43160 I M or for a continuous distribution V 1 f 1a39 2 28 r 47160 L q In particular for a volume charge it s 1 r Vr M dr 229 47160 t This is the equation we were looking for telling us how to compute V when we know 0 it is if you like the solution to Poisson s equation for a localized charge distribution5 1 5Equation 229 is an example of the Helmholtz theorem Appendix B in the context of electrostatics where the curl of E is zero and its divergence is peo 23 ELECTRIC POTENTIAL 85 invite you to compare Eq 229 with the corresponding formula for the electric eld in terms of p Eq 28 l r A Er f 4dt 42160 a The main point to notice is that the pesky unit vector 2 is now missing so there is no need to worry about components Incidentally the potentials of line and surface charges are 1 Mr 1 0r dl 47160 and 47160 a da 230 I should warn you that everything in this section is predicated on the assumption that the reference point is at in nity This is hardly apparent in Eq 229 but remember that we got that equation from the potential of a point charge at the origin 1 4Ir60 q r which is valid only when 0 00 If you try to apply these formulas to one of those arti cial problems in which the charge itself extends to in nity the integral will diverge Example 27 Find the potential of a uniformly charged spherical shell of radius R Fig 233 Solution This is the same problem we solved in Ex 26 but this time we shall do it using Eq 230 VI39 l J da 4ne0 a Let s set the point r on the z axis and use the law of cosines to express a in terms of the polar angle 0 2 R2 z2 2Rz cos 0 Figure 233 86 CHAPTER 2 ELECTROSTATICS An element of surface area on this Sphere is R2 sin 0 de dqb so R2 sin 9 de d 471 0Vz a R2 z2 2Rz 0030 Sine 7T d9 0 R2 z2 2Rz cos6 1 7T 271R20 v R2 z2 2chos0 Rz 0 271R xR2z22Rz R2z2 2Rz Z 2 R 7 Z a Rz2 R z2 At this stage we must be very careful to take the positive root For points outside the sphere z is greater than R and hence v R z2 z R for points inside the sphere V R z2 R z Thus R0 R20 VZ R Z z R outside 2 0z EOZ R R Vz 0R z R z inside 2 0Z 60 In terms of the total charge on the shell q 4nR20 Vz 147160qz or in general Vr 147t60qr for points outside the sphere and 147160qR for points inside Of course in this particular case it was easier to get V by using 221 than 230 because Gauss s law gave us E with so little effort But if you compare Ex 27 with Prob 27 you will appreciate the power of the potential formulation Problem 225 Using Eqs 227 and 230 nd the potential at a distance z above the center of the charge distributions in Fig 234 In each case compute E 2 VV and compare your answers with Prob 22a Ex 21 and Prob 26 respectively Suppose that we changed the righthand charge in Fig 234a to q what then is the potential at P What eld does that suggest Compare your answer to Prob 22b and explain carefully any discrepancy TP T P I z I z I I I I I I I I I I I q d q 2L I a Two point charges b Uniform line charge c Uniform surface charge Figure 234 23 ELECTRIC POTENTIAL 87 Problem 226 A conical surface an empty ice cream cone carries a uniform surface charge 039 The height of the cone is h as is the radius of the top Find the potential difference between points a the vertex and b the center of the top Problem 227 Find the potential on the axis of a uniformly charged solid cylinder a distance z from the center The length of the cylinder is L its radius is R and the charge density is p Use your result to calculate the electric eld at this point Assume that z gt L 2 Problem 228 Use Eq 229 to calculate the potential inside a uniformly charged solid sphere of radius R and total charge q Compare your answer to Prob 221 Problem 229 Check that Eq 229 satis es Poisson s equation by applying the Laplacian and using Eq 1102 235 Summary Electrostatic Boundary Conditions In the typical electrostatic problem you are given a source charge distribution p and you want to nd the electric eld E it produces Unless the symmetry of the problem admits a solution by Gauss s law it is generally to your advantage to calculate the potential rst as an intermediate step These then are the three fundamental quantities of electrOStatics p E and V We have in the course of our discussion derived all six formulas interrelating them These equations are neatly summarized in Fig 235 We began with just two exper imental observations 1 the principle of superposition a broad general rule applying to all electromagnetic forces and 2 Coulomb s law the fundamental law of electrostatics From these all else followed Figure 235 88 CHAPTER 2 ELEC TROSTATI CS EJ Figure 236 You may have noticed in studying Exs 24 and 25 or working problems such as 27 211 and 216 that the electric eld always undergoes a discontinuity when you cross a surface charge a In fact it is a simple matter to nd the amount by which E changes at such a boundary Suppose we draw a wafer thin Gaussian pillbox extending just barely over the edge in each direction Fig 236 Gauss s law states that 1 1 E da Qenc 039A 60 S 60 where A is the area of the pillbox lid If a varies from point to point or the surface is curved we must pick A to be extremely small Now the sides of the pillbox contribute nothing to the ux in the limit as the thickness 6 goes to zero so we are left with 1 1 EL Emw 6 00 231 above where E 5 6 denotes the component of E that is perpendicular to the surface immediately above and E6210 is the same only just below the surface For consistency we let upward be the positive direction for both Conclusion The normal component of E is discontinuous by an amount 060 at any boundary In particular where there is no surface charge E L is continuous as for instance at the surface of a uniformly charged solid sphere The tangential component of E by contrast is always continuous For if we apply Eq 219 fE d1 0 to the thin rectangular loop of Fig 237 the ends give nothing as e gt 0 and the sides 1 E I so below give E above Ebelow Ell above 232 24 WORK AND ENERGYIN ELECTROSTATICS 93 represents energy stored in the configuration potential energy if you like though for obvious reasons Iprefer to avoid that word in this context Problem 231 a Three charges are situated at the corners of a square side a as shown in Fig 241 How much work does it take to bring in another charge q from far away and place it in the fourth corner b How much work does it take to assemble the whole con guration of four charges q 1 4 Figure 241 243 The Energy of a Continuous Charge Distribution For a volume charge density p Eq 242 becomes 1 W E pV dr 243 The corresponding integrals for line and surface charges would be f AV dl and f 0V da respectively There is a lovely way to rewrite this result in which p and V are eliminated in favor of E First use Gauss s law to express p in terms of E p60VE so WVEVdr Now use integration by parts Eq 159 to transfer the derivative from E to V W6 EVVdry VEda W2 EzererEda 244 V 8 But VV 2 E so 94 CHAPTER 2 ELECTROSTATICS But what volume is this we re integrating over Let s go back to the formula we started with Eq 243 From its derivation it is clear that we should integrate over the region where the charge is located But actually any larger volume would do just as well The extra territory we throw in will contribute nothing to the integral anyway since p 0 out there With this in mind let s return to Eq 244 What happens here as we enlarge the volume beyond the minimum necessary to trap all the charge Well the integral of E 2 can only increase the integrand being positive evidently the surface integral must decrease correspondingly to leave the sum intact In fact at large distances from the charge E goes like 1 r2 and V like 1 r while the surface area grows like r2 Roughly speaking then the surface integral goes down like 1 r Please understand that Eq 244 gives you the correct energy W whatever volume you use as long as it encloses all the charge but the contribution from the volume integral goes up and that of the surface integral goes down as you take larger and larger volumes In particular why not integrate over all space Then the surface integral goes to zero and we are left with W E2dr 245 all space Example 28 Find the energy of a uniformly charged spherical shell of total charge 11 and radius R Solution 1 Use Eq 243 in the version appropriate to surface charges 1 W70Vda 2 Now the potential at the surface of this sphere is 147160qR a constant so 14 1q 2 0 da 8713960 R 8713960 R Solution 2 Use Eq 245 Inside the sphere E 0 outside 2 E 4760 SO E2 Therefore 60 12 2 Wlul 24n 02 r smedr 16 14gt outside 2 47t d 3271260q R r2 r 8713960 R 24 WORK AND ElWERGYIN ELECTROSTATICS 95 Problem 232 Find the energy stored in a uniformly charged solid sphere of radius R and charge q Do it three different ways a Use Eq 243 You found the potential in Prob 221 b Use Eq 245 Don t forget to integrate over all space c Use Eq 244 Take a spherical volume of radius a Notice what happens as a gt 00 Problem 233 Here is a fourth way of computing the energy of a uniformly charged sphere Assemble the sphere layer by layer each time bringing in an in nitesimal charge dq from far away and smearing it uniformly over the surface thereby increasing the radius How much work at W does it take to build up the radius by an amount dr Integrate this to nd the work necessary to create the entire sphere of radius R and total charge 1 244 Comments on Electrostatic Energy i A perplexing inconsistency Equation 245 clearly implies that the energy of a stationary charge distribution is always positive On the other hand Eq 242 from which 245 was in fact derived can be positive or negative For instance according to 242 the energy of two equal but opposite charges a distance d apart would be 147T 0q2L What s gone wrong Which equation is correct The answer is that both equations are correct but they pertain to slightly different situations Equation 242 does not take into account the work necessary to make the point charges in the rst place we started with point charges and simply found the work required to bring them together This is wise policy since Eq 245 indicates that the energy of a point charge is in fact in nite 2 2 oo 60 q 2 q l W n6d d6d d 2471602ltr4gtr S r p 87139600 r2 r 00 Equation 245 is more complete in the sense that it tells you the total energy stored in a charge con guration but Eq 242 is more appropriate when you re dealing with point charges because we prefer for good reason to leave out that portion of the total energy that is attributable to the fabricatiOn of the point charges themselves In practice after all the point charges electrons say are given to us readymade all we do is mow them around Since we did not put them together and we cannot take them apart it is immaterial how much work the process would involve Still the in nite energy of a point charge is a recurring source of embarrassment for electromagnetic theory af icting the quantum version as well as the classical We shall return to the problem in Chapter 11 Now you may wonder where the inconsistency crept into an apparently watertight derivation The aw lies between Eqs 242 and 243 In the former Vri represents the potential due to all the other charges but not qi whereas in the latter Vr is the ll potential For a continuous distribution there is no distinction since the amount of charge right at the point r is vanishingly small and its contribution to the potential is zero 96 CHAPTER 2 ELECTROSTATICS ii Where is the energy stored Equations 243 and 245 offer two different ways of calculating the same thing The rst is an integral over the charge distn39bution the second is an integral over the eld These can involve completely different regions For instance in the case of the spherical shell Ex 28 the charge is con ned to the surface whereas the electric eld is present everywhere outside this surface Where is the energy then Is it stored in the eld as Eq 245 seems to suggest or is it stored in the charge as Eq 243 implies At the present level this is simply an unanswerable question I can tell you what the total energy is and I can provide you with several different ways to compute it but it is unnecessary to worry about where the energy is located In the context of radiation theory Chapter 11 it is useful and in General Relativity it is essential to regard the energy as being stored in the eld with a density 60 2 3 E 2 energy per unit volume 246 But in electrostatics one could just as well say it is stored in the charge with a density oV The difference is purely a matter of bookkeeping iii The superposition principle Because electrostatic energy is quadratic in the elds it does not obey a superposition principle The energy of a compound system is not the sum of the energies of its parts considered separately there are also cross terms Wtot 6 0 Ezdr 6 0E1E22dr 2 2 60 2 2 3EE22E1E2dr 2 W1 W2 60E1E2dr 247 For example if you double the charge everywhere you quadruple the total energy Problem 234 Consider two concentric spherical shells of radii a and b Suppose the inner one carries a charge 1 and the outer one a charge q both of them uniformly distributed over the surface Calculate the energy of this con guration a using Eq 245 and b using Eq 247 and the results of Ex 28 25 Conductors 251 Basic Properties In an insulator such as glass or rubber each electron is attached to a particular atom In a metallic conductor by contrast one or more electrons per atom are free to roam about at will through the material In liquid conductors such as salt water it is ions that do the moving A perfect conductor would be a material containing an unlimited supply of completely free 25 CONDUCTORS 97 charges In real life there are no perfect conductors but many substances come amazingly close From this de nition the basic electrostatic properties of ideal conductors immediately follow i E 0 inside a conductor Why Because if there were any eld those free charges would move and it wouldn t be electr0statics any more Well that s hardly a satisfactory explanation maybe all it proves is that you can t have electrostatics when conductors are present We had better examine what happens when you put a conductor into an external electric eld E0 Fig 242 Initially this will drive any free positive charges to the right and negative ones to the left In practice it s only the negative charges electrons that do the moving but when they depart the right side is left with a net positive charge the stationary nuclei so it doesn t really matter which charges move the effect is the same When they come to the edge of the material the charges pile up plus on the right side minus on the left Now these induced charges produce a eld of their own E which as you can see from the gure is in the opposite direction to E0 That s the crucial point for it means that the eld of the induced charges tends to cancel o the original eld Charge will continue to ow until this cancellation is complete and the resultant eld inside the conductor is precisely zero7 The whole process is practically instantaneous 11 E0 Figure 242 ii p 0 inside a conductor This follows from Gauss s law V E 060 If E 0 so also is 0 There is still charge around but exactly as much plus charge as minus so the net charge density in the interior is zero iii Any net charge resides on the surface That s the only other place it can be iv A conductor is an equipotential For if a and b are any two points within or at the surface of a given conductor Vb Va j E d1 0 and hence Va Vb 7Om side the conductor the eld is not zero for here E0 and E1 do not cancel 98 CHAPTER 2 ELECTROSTATICS E Conductor E 0 Figure 243 v E is perpendicular to the surface just outside a conductor Otherwise as in i charge will immediately ow around the surface until it kills off the tangential component Fig 243 Perpendicular to the surface charge cannot ow of course since it is con ned to the conducting object I think it is strange that the charge on a conductor ows to the surface Because of their mutual repulsion the charges naturally spread out as much as possible but for all of them to go to the surface seems like a waste of the interior space Surely we could do better from the point of view of making each charge as far as possible from its neighbors to sprinkle some of them throughoUt the volume Well it simply is not so You do best to put all the charge on the surface and this is true regardless of the size or shape of the conductors The problem can also be phrased in terms of energy Like any other free dynamical system the charge on a conductor will seek the con guration that minimizes its potential energy What property iii asserts is that the electrostatic energy of a solid object with speci ed shape and total charge is a minimum when that charge is spread over the surface For instance the energy of a sphere is 1 87160 q2 R if the charge is uniformly distributed over the surface as we found in EX 28 but it is greater 3 2071 60q2 R if the charge is uniformly distributed throughout the volume Prob 232 252 Induced Charges If you hold a charge q near an uncharged conductor Fig 244 the two will attract one another The reason for this is that q will pull minus charges over to the near side and repel plus charges to the far side Another way to think of it is that the charge moves around in such a way as to cancel off the eld of q for points inside the conductor where the total eld must be zero Since the negative induced charge is closer to q there is a net force of attraction In Chapter 3 we shall calculate this force explicitly for the case of a spherical conductor 8By the way the one and two dimensional analogs are quite different The charge on a conducting disk does not all go to the perimeter R Friedberg Am J of Phys 61 1084 1993 nor does the charge on a conducting needle go to the ends D J Grif ths and Y Li Am J ofPhys 64 706 1996 See Prob 252 25 CONDUCTORS 99 Gaussian surface Conductor Figure 2 Figure 2 By the way when I speak of the eld charge or potential inside a conductor I mean in the meat of the conductor if there is some cavity in the conductor and within that cavity there is some charge then the eld in the cavity will not be zero But in a remarkable way the cavity and its contents are electrically isolated from the outside world by the surrounding conductor Fig 245 No external elds penetrate the conductor they are canceled at the outer surface by the induced charge there Similarly the eld due to charges within the cavity is killed off for all exterior points by the induced charge on the inner surface However the compensating charge left over on the outer surface of the conductor effectively communicates the presence of q to the outside world as we shall see in Ex 29 Incidentally the total charge induced on the cavity wall is equal and opposite to the charge inside for if we surround the cavity with a Gaussian surface all points of which are in the conductor Fig 245 315 E da 2 0 and hence by Gauss s law the net enclosed charge must be zero But QenC q qinduced so qinduced q Example 29 An uncharged spherical conductor centered at the origin has a cavity of some weird shape carved out of it Fig 246 Somewhere within the cavity is a charge q Question What is the eld outside the sphere Conductor Figure 246 100 CHAPTER 2 ELECTROSTATICS Solution At rst glance it would appear that the answer depends on the shape of the cavity and on the placement of the charge But that s wrong The answer is l if 47r60 r2 regardless The conductor conceals from us all information concerning the nature of the cavity revealing only the total charge it contains How can this be Well the charge q induces an oppOsite charge q on the wall of the cavity which distributes itself in such a way that its eld cancels that of q for all points exterior to the cavity Since the conductor carries no net charge this leaves q to distribute itself uniformly over the surface of the sphere lt s uniform because the asymmetrical in uence of the point charge q is negated by that of the induced charge q on the inner surface For points outside the sphere then the only thing that survives is the eld of the leftover q uniformly distributed over the outer surface It may occur to you that in one respect this argument is open to challenge There are actually three elds at work here Eq Einduced and Eleflover All we know for certain is that the sum of the three is zero inside the conductor yet I claimed that the rst two alone cancel while the third is separately zero there Moreover even if the rst two cancel within the conductor who is to say they still cancel for points outside They do not after all cancel for points inside the cavity Icannot give you a completely satisfactory answer at the moment but this much at least is true There exists a way of distributing q over the inner surface so as to cancel the eld of 1 at all exterior points For that same cavity could have been carved out of a huge spherical conductor with a radius of 27 miles or light years or whatever In that case the leftover q on the outer surface is simply too far away to produce a signi cant eld and the other two elds would have to accomplish the cancellation by themselves So we know they can do it but are we sure they choose to Perhaps for small spheres nature prefers some complicated threeway cancellation Nope As we ll see in the uniqueness theorems of Chapter 3 electrostatics is very stingy with its options there is always precisely one way no more of distributing the charge on a conductor so as to make the eld inside zero Having found a possible way we are guaranteed that no alternative exists even in principle If a cavity surrounded by conducting material is itself empty of charge then the eld within the cavity is zero For any eld line would have to begin and end on the cavity wall going from a plus charge to a minus charge Fig 247 Letting that eld line be part of a closed loop the rest of which is entirely inside the conductor where E 0 the integral Figure 247 25 CONDUCTORS 101 315 E dl is distinctly positive in violation of Eq 219 It follows that E 0 within an empty cavity and there is in fact no charge on the surface of the cavity This is why you are relatively safe inside a metal car during a thunderstorm you may get cooked if lightning strikes but you will not be electrocuted The same principle applies to the placement of sensitive apparatus inside a grounded Faraday cage to shield out stray electric elds In practice the enclosure doesn t even have to be solid conductor chicken wire will often suf ce Problem 235 A metal sphere of radius R carrying charge 1 is surrounded by athick concentric metal shell inner radius a outer radius 2 as in Fig 248 The shell carries no net charge a Find the surface charge density a at R at a and at b b Find the potential at the center using in nity as the reference point C Now the outer surface is touched to a grounding wire which lowers its potential to zero same as at in nity How do your answers to a and b change Problem 236 Two spherical cavities of radii a and b are hollowed out from the interior of a neutral conducting sphere of radius R Fig 249 At the center of each cavity a point charge is placed call these charges qa and 11 a Find the surface charges 0a 0 and OR b What is the eld outside the conductor c What is the eld within each cavity d What is the force on qa and qb e Which of these answers would change if a third charge 16 were brought near the conductor Figure 248 Figure 249 102 CHAPTER 2 ELECTROSTATICS 253 Surface Charge and the Force on a Conductor Because the eld inside a conductor is zero boundary condition 233 requires that the eld immediately outside is 0 A E n 248 60 consistent with our earlier conclusion that the eld is normal to the surface In terms of potential Eq 236 yields 0 60 K 249 an These equations enable you to calculate the surface charge on a conductor if you can determine E or V we shall use them frequently in the next chapter In the presence of an electric eld a surface charge will naturally experience a force the force per unit area f is oE But there s a problem here for the electric eld is discontinuous at a surface charge so which value are we supposed to use Eabove Ebelow or something in between The answer is that we should use the average of the two 1 f 0E average 2 300 above Ebelow Why the average The reason is very simple though the telling makes it sound complicated Let s focus our attention on a small patch of surface surrounding the point in question Fig 250 Make it tiny enough so it is essentially at and the surface charge on it is essentially constant The total eld consists of two parts that attributable to the patch itself and that due to everything else other regions of the surface as well as any external sources that may be present E Epalch l Eother Now the patch cannot exert a force on itself any more than you can lift yourself by standing in a basket and pulling up on the handles The force on the patch then is due exclusively to Ember and this suffers no diSContinuity if we removed the patch the eld in the hole would be perfectly smooth The discontinuity is due entirely to the charge on the patch 680 1 Figure 250

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