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Classical Electrodynamics

by JOHN DAVID JACKSON
~)}~CLAS'SiCAL
ELECTRODYNAMICS
@ John Wiley & Sons, Inc., New York. London. Sydne,

9 10
Copyright © 1962 by John Wiley & Sons, Inc. All rights reserved.
This book or any part thereof must not be reproduced in any form without the written permission o/ the publisher.
Printed in the United States of America Library of Congress Catalog Card Number: 62-8774

To the memory of my father, Walter David Jackson

Preface
Classical electromagnetic theory, together with classica] and quantum mechanics, forms the core of present-day theoretical training for undergraduate and graduate physicists. A thorough grounding in these subjects is a requirement for more advanced or specialized training.
Typically the undergraduate program in electricity and magnetism involves two or perhaps three semesters beyond elementary physics, with the emphasis on the fundamental laws, laboratory verification and elaboration of their consequences, circuit analysis, simple wave phenomena, and radiation. The mathematical tools utilized include vector calculus, ordinary differential equations with constant coefficients, Fourier series, and perhaps Fourier or Laplace transforms, partial differential equations, Legendre polynomials, and Bessel functions.
As a general rule a two-semester course in electromagnetic theory is given to beginning graduate students. It is for such a course that my book is designed. My aim in teaching a graduate course in electromagnetism is at least threefold. The first aim is to present the basic subject matter as a coherent whole, with emphasis on the unity of electric and magnetic phenomena, both in their physical basis and in the mode of mathematical description. The second, concurrent aim is to develop and utilize a number of topics in mathematical physics which are useful in both electromagnetic theory and wave mechanics. These include Green's theorems and Green's functions, orthonormal expansions, spherical harmonics, cylindrical and spherical Bessel functions. A third and perhaps most important purpose is the presentation of new material; especially on the interaction of
vii

'Viii

Preface

relativistic charged particles with electromagnetic fields. In this last area personal preferences and prejudices enter strongly. My choice of topics is governed by what I feel is important and useful for students interested in theoretical physics, experimental nuclear and high-energy physics, and that as yet ill-defined field of plasma physics.
The book begins in the traditional manner with electrostatics. The first six chapters are devoted to the development of Maxwell's theory of electromagnetism. Much of the necessary mathematical apparatus is constructed along the way, especially in Chapters 2 and 3, where boundaryvalue problems are discussed thoroughly. The treatment is initially in terms of the electric field E and the magnetic induction Bt with the derived macroscopic quantities, D and H, introduced by suitable averaging over ensembles of atoms or molecules. In the discussion of dielectrics, simple classical models for atomic polarizability are described, but for magnetic materials no such attempt is made. Partly this omission was a question of space, but truly classical models of magnetic susceptibility are not possible. Furthermore, elucidation ofthe interesting phenomenon offerromagnetism needs almost a book in itself.
The next three chapters (7-9) illustrate various electromagnetic phenomena, mostly of a macroscopic sort. Plane waves in different media, including plasmas, as well as dispersion and the propagation of pulses, are treated in Chapter 7. The discussion ofwave guides and cavities in Chapter 8 is developed for systems of arbitrary cross section, and the problems of attenuation in guides and the Q of a cavity are handled in a very general way which emphasizes the physical processes involved. The elementary theory of multipole radiation from a localized source and diffraction occupy Chapter 9. Since the simple scalar theory of diffraction is covered in many optics textbooks, as well as undergraduate books on electricity and magnetism, I have presented an improved, although still approximate, theory of diffraction based on vector rather than scalar Green's theorems.
The subject of magnetohydrodynamics and plasmas receives increasingly more attention from physicists and astrophysicists. Chapter 10 represents a survey of this complex field with an introduction to the main physical ideas involved.
The first nine or ten chapters constitute the basic material of classical electricity and magnetism. A graduate student in physics may be expected to have been exposed to much of this material, perhaps at a somewhat lower level, as an undergraduate. But he obtains a more mature view ofit, understands it more deeply, and gains a considerable technical ability in analytic methods of solution when he studies the subject at the level of this book. He is then prepared to go on to more advanced topics. The advanced topics presented here are predominantly those involving the

Preface

ix

interaction of charged particles with each other and with electromagnetic fields, especially when moving relativistically.
The special theory ofrelativity had its origins in classical electrodynamics. And even after almost 60 years, classical electrodynamics still impresses and delights as a beautiful example of the covariance ofphysical laws under Lorentz transformations. The special theory of relativity is discussed in Chapter 11, where all the necessary formal apparatus is developed, various kinematic consequences are explored, and the covariance ofelectrodynamics is established. The next chapter is devoted to relativistic particle kinematics and dynamics. Although the dynamics of charged particles in electromagnetic fields can properly be considered electrodynamics, the reader may wonder whether such things as kinematic transformations of collision problems can. My reply is that these examples occur naturally once one has established the four-vector character ofa particle's momentum and energy, that they serve as useful practice in manipulating Lorentz transformations, and that the end results are valuable and often hard to find elsewhere.
Chapter 13 on collisions between charged particles emphasizes energy loss and scattering and develops concepts of use in later chapters. Here for the first time in the book I use semiclassical arguments based on the uncertainty principle to obtain approximate quantum-mechanical expressions for energy loss, etc., from the classical results. This approach, so fruitful in the hands of Niels Bohr and E. J. Williams, allows one to see clearly how and when quantum-mechanical effects enter to modify classical considerations.
The important subject of emission of radiation by accelerated point charges is discussed in detail in Chapters 14 and 15. Relativistic effects are stressed, and expressions for the frequency and angular dependence of the emitted radiation are developed in sufficient generality for all applications. The examples treated range from synchrotron radiation to bremsstrahlung and radiative beta processes. Cherenkov radiation and the Weizsa.cker-Williams method of virtual quanta are also discussed. In the atomic and nuclear collision processes semiclassical arguments are again employed to obtain approximate quantum-mechanical results. I lay considerable stress on this point because I feel that it is important for the student to see that radiative effects such as bremsstrahlung are almost entirely classical in nature, even though involving small-scale collisions. A student who meets bremsstrahlung for the first time as an example of a calculation in quantum field theory will not understand its physical basis.
Multipole fields form the subject matter of Chapter 16. The expansion of scalar and vector fields in spherical waves is developed from first principles with no restrictions as to the relative dimensions of source and

X

Preface

wavelength. Then the properties of electric and magnetic multipole radiation fields are considered. Once the connection to the multipole moments of the source has been made, examples of atomic and nuclear multipole radiation are discussed, as well as a macroscopic source whose dimensions are comparable to a wavelength. The scattering of a plane electromagnetic wave by a spherical object is treated in some detail in order to illustrate a boundary-value problem with vector spherical .waves.
In the last chapter the difficult problem of radiative reaction is discussed. The treatment is physical, rather than mathematical, with the emphasis on delimiting the areas where approximate radiative corrections are adequate and on finding where and why existing theories fail. The original AbrahamLorentz theory of the self-force is presented, as well as more recent classical considerations.
The book ends with an appendix on units and dimensions and a bibliography. In the appendix I have attempted to show the logical steps involved in setting up a system of units, without haranguing the reader as to the obvious virtues of my choice of units. I have provided two tables which I hope will be useful, one for converting equations and symbols and the other for converting a given quantity of something from so many Gaussian units to so many mks units, and vice versa. The bibliography lists books which I think the reader may find pertinent and useful for reference or additional study. These books are referred to by author's name in the reading lists at the end of each chapter.
This book is the outgrowth of a graduate course in classical electrodynamics which I have taught off and on over the past eleven years, at both the University of Illinois and McGill University. I wish to thank my colleagues and students at both institutions for countless helpful remarks and discussions. Special mention must be made of Professor P. R. Wallace of McGill, who gave me the opportunity and encouragement to teach what was then a rather unorthodox course in electromagnetism, and Professors H. W. Wyld and G. Ascoli of Illinois, who have been particularly free with many helpful suggestions on the treatment of various topics. My thanks are also extended to Dr. A. N. Kaufman for reading and commenting on a preliminary version of the manuscript, and to Mr. G. L. Kane for his zealous help in preparing the index.
J. D. JACKSON
Urbana, Illinois January, 1962

Contents

chapter 1. Introduction to Electrostatics

1

1.1 Coulomb's law, 1. 1.2 Electric field, 2. 1.3 Gauss's law, 4. 1.4 Differential form of Gauss's law, 6. _ 1.5 Scalar potential, 7. 1.6 Surface distributions of charges and dipoles, 9. 1. 7 Poisson's and Laplace's equations, 12. 1.8 Green's theorem, 14. 1.9 Uniqueness theorem, 15. 1.10 Formal solution of boundary-value problem, Green's functions, 18. 1.11 Electrostatic potential energy, 20.
References and suggested reading, 23. Problems, 23.

chapter 2. Boundary-Value Problems in Electrostatics, I

26

2.1 Method of images, 26. 2.2 Point charge and a grounded conducting sphere, 27. 2.3 Point charge and a charged, insulated, conducting sphere, 31. 2.4 Point charge and a conducting sphere at fixed potential, 33. 2.5 Conducting sphere in a uniform field, 33. 2.6 Method of inversion, 35. 2.7 Green's function for a sphere, 40.
xi

xii

Contents

2.8 Conducting sphere with hemispheres at different potentials, 42. 2.9 Orthogonal functions and expansions, 44. 2.10 Separation of variables in rectangular coordinates, 47.
References and suggested reading, 50. Problems, 5l.

chapter 3. Boundary-Value Problems in Electrostatics, II

54

3.1 Laplace's equation in spherical coordinates, 54. 3.2 Legendre polynomials, 56. 3.3 Boundary-value problems with azimuthal symmetry, 60. 3.4 Spherical harmonics, 64. 3.5 Addition theorem for spherical harmonics, 67. 3.6 Cylindrical coordinates, Bessel functions, 69. 3.7 Boundary-value problems in cylindrical coordinates, 75. 3.8 Expansion of Green's functions in spherical coordinates, 77. 3.9 Use of spherical Green's function expansion, 81. 3.10 Expansion of Green's functions in cylindrical coordinates, 84. 3.11 Eigenfunction expansions for Green's functions, 87. 3.12 Mixed boundary conditions, charged conducting disc, 89.
References and suggested reading, 93. Problems, 94.

chapter 4. Multipoles, Electrostatics of Macroscopic Media,

Dielectrics

98

4.1 Multipole expansion, 98. 4.2 Multipole expansion of the energy of a charge distribution in an
external field, 101. 4.3 Macroscopic electrostatics, 103. 4.4 Simple dielectrics and boundary conditions, 108. 4.5 Boundary-value problems with dielectrics, 110. 4.6 Molecular polarizability and electric susceptibility, 116. 4.7 Models for molecular polarizability, 119. 4.8 Electrostatic energy in dielectric media, 123.
References and suggested reading, 127. Problems, 128.

chapter 5. Magnetostatics

132

5.1 Introduction and definitions, 132. 5.2 Biot and Savart law, 133. 5.3 Differential equations of magnetostatics, Ampere's law, 137. 5.4 Vector potential, 139. 5.5 Magnetic induction of a circular loop of current, 141. 5.6 Localized current distribution, magnetic moment, 145.

Contents

xiii

5.7 Force and torque on localized currents in an external field, 148. 5.8 Macroscopic equations, 150. 5.9 Boundary conditions, 154. 5.10 Uniformly magnetized sphere, 156. 5.11 Magnetized sphere in an external field, permanent magnets, 160. 5.12 Magnetic shielding, 162.
References and suggested reading, 164. Problems, 165.

chapter 6. Time-Varying Fields, Maxwell's Equations, Con-

servation Laws

169

6.1 Faraday's law of incmction, 170. 6. 2 Energy in the magnetic field, 173. 6.3 Maxwell's displacement current, Maxwell's equations, 177. 6.4 Vector and scalar potentials, wave equations, 179. 6.5 Gauge transformations, 181. 6.6 Green's function for the time-dependent wave equation, 183. 6.7 Initial-value problem, Kirchhoff's integral representation, 186. 6.8 Poynting's theorem, 189. 6.9 Conservation laws, 190. 6.10 Macroscopic equations, 194.
References and suggested reading, 198. Problems, 198.

chapter 7. Plane Electromagnetic Waves

202

7.1 Plane waves in a nonconducting medium, 202. 7.2 Linear and circular polarization, 205. 7.3 Superposition of waves, group velocity, 208. 7.4 Propagation of a pulse in a dispersive medium, 212. 7.5 Reflection and refraction, 216. 7.6 Polarization by reflection, total internal reflection, 220. 7.7 Waves in a conducting medium, 222. 7.8 Simple model for conductivity, 225. 7.9 Transverse waves in a tenuous plasma, 226.
References and suggested reading, 231. Problems, 231.

chapter 8. Wave Guides and Resonant Cavities

235

8.1 Fields at the surface of and within a conductor, 236. 8.2 Cylindrical cavities and wave guides, 240. 8.3 Wave guides, 244. 8.4 Modes in a rectangular wave guide, 246. 8.5 Energy flow and attenuation in wave guides, 248.

8.6 Resonant cavities, 252. 8.7 Power losses in a cavity, 255. 8.8 Dielectric wave guides, 259.
References and suggested reading, 264. Problems, 264.

chapter 9. Simple Radiating Systems and Diffraction

268

9.1 Fields and radiation of a localized source, 268. 9.2 Oscillating electric dipole, 271. 9.3 Magnetic dipole and quadrupole fields, 273. 9.4 Center-fed linear antenna, 277. 9.5 Kirchhoff's integral for diffraction, 280. 9.6 Vector equivalents of Kirchhoff's integral, 283. 9. 7 Babinet's principle, 288. 9.8 Diffraction by a circular aperture, 292. 9.9 Diffraction by small apertures, 297. 9. IO Scattering by a conducting sphere at short wavelengths, 299.
References and suggested reading, 304. Problems, 305.

chapter 10. Magnetohydrodynamics and Plasma Physics

309

10.1 Introduction and definitions, 309. 10.2 Magnetohydrodynamic equations, 311. 10.3 Magnetic diffusion, viscosity, and pressure, 313. 10.4 Magnetohydrodynamic flow, 316. 10.5 Pinch effect, 320. 10.6 Dynamic model of the pinch effect, 322. 10.7 Instabilities, 326. 10.8 Magnetohydrodynamic waves, 329. 10.9 High-frequency plasma oscillations, 335. 10.10 Short-wavelength limit, Debye screening distance, 339.
References and suggested reading, 343. Problems, 343.

chapter 11. Special Theory of Relativity

347

11.1 Historical background and key experiments, 347. 11.2 Postulates of special relativity, Lorentz transformation, 352. 11.3 FitzGerald-Lorentz contraction and time dilatation, 357. 11.4 Addition of velocities, Doppler shift, 360. 11.5 Thomas precession, 364. 11.6 Proper time and light cone, 369. 11.7 Lorentz transformations as orthogonal transformations, 371. 11.8 4-vectors and tensors, 374.

Contents

xv

11.9 Covariance of electrodynamics, 377. 11.10 Transformation of electromagnetic fields, 380. 11.11 Covariance of the force equation and the conservation laws, 383.
References a11d suggested reading, 386. Problems, 387.

chapter 12. Relativistic-Particle Kinematics and Dynamics 391
12.1 Momentum and energy of a particle, 391. 12.2 Kinematics of decay of an unstable particle, 394. 12.3 Center of momentum transformation, 397. 12.4 Transformation of momenta from the center of momentum frame
to the laboratory, 400. 12.5 Covariant Lorentz force equation, Lagrangian and Hamiltonian,
404. 12.6 Relativistic corrections to the Lagrangian for interacting charged
particles, 409. 12.7 Motion in a uniform, static, magnetic field, 411. 12.8 Motion in combined uniform, static, electric and magnetic fields,
412. 12.9 Particle drifts in nonuniform magnetic fields, 415. 12.10 Adiabatic invariance of flux through an orbit, 419.
References a11d suggested reading, 424. Problems, 425.

chapter 13. Collisions between Charged Particles, Energy Loss,

~&~~

ffi

13.1 Energy transfer in a Coulomb collision, 430. 13.2 Energy transfer to a harmonically bound charge, 434. 13.3 Classical and quantum-mechanical energy loss, 438. 13.4 Density effect in collision energy loss, 443. 13.5 Energy loss in an electronic plasma, 450. 13.6 Elastic scattering of fast particles by atoms, 451. 13. 7 Mean square angle of scattering, multiple scattering, 456. 13. 8 Electrical conductivity of a plasma, 459.

References and suggested_ reading, 462. Problems, 462.

chapter 14. Radiation by Moving Charges

464

14.l Lienard-Wiechert potentials and fields, 464. 14.2 Larmor's radiated power formula and its relativistic
generalization, 468. 14.3 Angular distribution-of radiation, 472. 14.4 Radiation by an extremely relativistic charged particle, 475.

xvi

Contents

14.5 General angular and frequency distributions of radiation from accelerated charges, 477.
14.6 Frequency spectrum from relativistic charged particle in an instantaneously circular orbit, synchrotron radiation, 481.
14.7 Thomson scattering, 488. 14.8 Scattering by quasi-free charges, 491. 14.9 Cherenkov radiation, 494.
References and suggested reading, 499. Problems, 500.

chapter 15. Bremsstrahlung, Method of Virtual Quanta, Radia-

tive Beta Processes

505

15.1 Radiation emitted during collisions, 506. 15.2 Bremsstrahlung in nonrelativistic Coulomb collisions, 509. 15.3 Relativistic bremsstrahlung, 513. 15.4 Screening, relativistic radiative energy loss, 516. 15.5 Weizsacker-Williams method of virtual quanta, 520. 15.6 Bremsstrahlung as the scattering of virtual quanta, 525. 15.7 Radiation emitted during beta decay, 526. 15.8 Radiation emitted in orbital-electron capture, 528.
References and suggested reading, 533. Problems, 534.

chapter 16. Multipole Fields

538

16.1 Scalar spherical waves, 538-. 16.2 Multipole expansion of electromagnetic fields, 543. 16.3 Properties of multipole fields, energy and angular momentum of
radiation, 546. 16.4 Angular distributions, 550. 16.5 Sources of multipole radiation, multipole moments, 553. 16.6 Multipole radiation in atoms and nuclei, 557. 16.7 Radiation from a linear, center-fed antenna, 562. 16.8 Spherical expansion of a vector plane wave, 566. 16.9 Scattering by a conducting sphere, 569. 16.10 Boundary-value problems with multipole fields, 574.

References and suggested reading, 514. Problems, 574.

chapter 17. Radiation Damping, Self-Fields of a Particle,

Scattering and Absorption of Radiation by a Bound

System

578

17.1 Introductory considerations, 578. 17.2 Radiative reaction force, 581.

Contents

xvii

17.3 Abraham-Lorentz evalmttion of the self-force, 584. 17.4 Difficulties with the Abraham-Lorentz model, 589. 17.S Lorentz transformation properties of the Abraham-Lorentz model,
Poincare stresses, 590. 17.6 Covariant definitions of self-energy and momentum, 594. 17.7 Integrodifferential equation of motion, including damping, 597. 17.8 Line breadth and level shift of an oscillator, 600. 17.9 Scattering and absorption of radiation by an oscillator, 602.
References and suggested reading, 601. Problems, 608.

appendix. Units and Dimensions

611

Bibliography

622

Index

625

1
Introduction to Electrostatics
Although amber and lodestone were :known by the ancient Greeks, electrodynamics developed as a quantitative subject in about 80 years. Coulomb's observations on the forces between charged bodies were made around 1785. About 50 years later, Faraday was studying the effects of currents and magnetic fields. By 1864, Maxwell had published his famous paper on a dynamical theory of the electromagnetic field.
We will begin our discussion with the subject ofelectrostatics-problems involving time-independent electric fields. Much of the material will be covered rather rapidly because it is in the nature of a review. We will use electrostatics as a testing ground to devefop and use mathematical techniques of general applicability.
1.1 Coulomb's Law
All of electrostatics stems from the quantitative statement of Coulomb's law concerning the force acting between charged bodies at rest with respect to each other. Coulomb (and, even earlier, Cavendish) showed experimentally that the force between two small charged bodies separated a distance large compared to their dimensions
( l) varied directly as the magnitude of each charge, (2) varied inversely as the square of the distance between them, (3) was directed along the line joining the charges, (4) was attractive if the bodies were oppositely charged and repulsive
if the bodies had the same type of charge. Furthermore it was shown experimentally that the total force produced
1

2

Classical Electrodynamics

on one sma11 charged body by a number of the other smalJ charged bodies placed around it was the vector sum of the individual two-body forces of Coulomb.

1.2 Electric Field

Although the thing that eventually gets measured is a force, it is useful to introduce a concept one step removed from the forces, the concept of an electric field due to some array of charged bodies. At the moment, the electric field can be defined as the force per unit charge acting at a given point. It is a vector function of position, denoted by E. One must be careful in its definition, however. It is not necessarily the force that one would observe by placing one unit of charge on a pith ball and placing it in position. The reason is that one unit of charge (e.g., 100 strokes of cat's fur on an amber rod) may be so large that its presence alters appreciably the field configuration of the array. Consequently one must use a limiting process whereby the ratio of the force on the small test body to the charge on it is measured for smaller and smaller amounts of charge. Experimentally, this ratio and the direction of the force will become constant as the amount of test charge is made smaller and smaller. These Jimiting values of magnitude and direction define the magnitude and direction of the electric field E at the point in question. In symbols we may write

F=qE

(1.1)

where F is the force, E the electric field, and q the charge. In this equation it is assumed that the charge q is located at a point, and the force and the

electric field are evaluated at that point. Coulomb's law can be written down similarly. If F is the force on a
point charge q1, located at x1, due to another point charge q2, located at x2, then Coulomb's law is

F = kq1q2 (x1 - ~)a

(1.2)

IX1 - X2I

Note that q1 and q2 are algebraic quantities which can be positive or negative. The constant of proportionality k depends on the system of units

used. The electric field at the point x due to a point charge q1 at the point x1
can be obtained directly:

E(x) = kq1 (x - X1)

(1.3)

Ix - X1ia

as indicated in Fig. 1.1. The constant k is determined by the unit of charge

[Sect. 1.2]

Introduction to Electrostatics

3

E

Fig. 1.1

chosen. In electrostatic units (esu), unit charge is chosen as that charge
which exerts a force of one dyne on an equal charge located one centimeter
= away. Thus, with cgs units, k I and the unit of charge is called the = "stat-coulomb." In the mks system, k (41rc0)-1, where £0 ( = 8.854 x
10-12 farad/meter) is the permittivity of free space. We will use esu. *
The experimentally observed linear superposition of forces due to many
charges means that we may write the electric field at x due to a system of
point charges qt, located at~, i = 1, 2, ... , n, as the vector sum:

fl,
E(x) = Lqi (x - xt\
i""l Ix - xii

(1.4)

If the charges are so sma11 and so numerous that they can be described by

a charge density p(x') [if dq is the charge in a small volume dx dy dz at
= the point x', then dq p(x') dx dy dz], the sum is replaced by an

integral:

f E(x) = p(x') (x - x') d3x'

(1.5)

jx - x'l3

= where d3x' dx' dy' dz' is a three-dimensional vo1ume element at x'.

At this point it is worth while to introduce the Dirac delta function. In one dimension, the delta function, written c5(x - a), is a mathematically improper function having the properties:

J (1) ci(x - a) = 0 for x =fa a, and
(2) b(x - a) dx = I if the region of integration includes x = a, and is zero

otherwise.
The delta function can be given rigorous meaning as the limit of a peaked curve such as a Gaussian which becomes narrower and narrower, but higher and higher, in such a way that the area under the curve is always constant. L. Schwartz's theory of distributions is a comprehensive rigorous mathematical approach to delta functions and their manipulations. t

* The question of units is discussed in detail in the Appendix.
t A useful. rigorous account of the Dirac delta function is given by Lighthill. (Full
references for items cited in the text or footnotes by author only will be found in the
Bibliography.)

4

Classical Electrodynamics

JFrom the definitions above it is evident that, for an arbitrary function/(x),
(3) f(x) <Xx - a) dx = f(a)~ and
f(4) f(x) ~'(x - a) dx = -['(a),

where a prime denotes differentiation with respect to the argument.

If the delta function has as argument a function f(x) of the independent

L I~ (" - variable x, it can be transformed according to the rule,

~ (SJ J(/(x))

x,),

i dx(xi)

where /(x) is assumed to have only simple zeros, located at x = xt.

In more than one dimension. we merely take products of delta functions in

each dimension. In three dimensions, for example,

= (6) !5(x - X) 6(x1 - X1) 6(x2 - X2) 6(x3 - X3)

( i is a function which vanishes everywhere except at x = X, and is such that

7)

d(x -
Av

= X) u-x ..I'll. {1
O

if ll.Vcontains x = X,
if 6. V does not contain x = X.

Note that a delta function has the dimensions of an inverse volume in whatever number of dimensions the space has.
A discrete set of point charges can be described with a charge density by means of delta functions. For example,

! n
p(x) = q, o{x - xi)

(1.6)

i=l

represents a distribution of n point charges qi, located at the points xi. Substitu-

tion of this charge density (1.6) into (1.5) and integration, using the properties of

the delta function, yields the discrete sum (1.4).

1.3 Gauss's Law

The integral (I.5) is not the most suitable form for the evaluation of electric fields. There is another integral result, called Gauss's law, which is often more useful and which furthermore leads to a differential equation for E(x). To obtain Gauss's Jaw we first consider a point charge q and a closed surface S, as shown in Fig. 1.2. Let r be the distance from the charge to a point on the surface, n be the outwardly directed unit nonnal to the surface at that point, da be an element of surface area. If the electric
field E at the point on the surface due to the charge q makes an angle (}
with the unit normal, then the normal component of E times the area

element is:

= cos 0
E • n da q - - da

(1.7)

r2

Since E is directed along the line· from the surface element to the charge q,

[Sect. 1.3]

Intl'oduction to Electrostatics

5

.E

q insides

E n

q outside S

E

E

Fig. 1.2 Gauss's law. The normal component of electric field is integrated over the closed surface S. If the charge is inside (outside) S, the total solid angle subtended at the charge by the inner side of the surface is 4rr (zero).

cos 0 da = r2 dO., where dQ. is the element of solid angle subtended by da
at the position of the charge. Therefore

E • n da = q dD.

(1.8)

If we now integrate the normal component of E over the whole surface, it is easy to see that

,( E . d _ {41rq if q lies inside S Ys O a - 0 if q lies outside S

(1.9)

6

Classical Electrodynamics

This result is Gauss's law for a single point charge. For a discrete set of charges, it is immediately apparent that

Yi.s

E

• n

da

=

4n- 2 qi i

(1.10)

where the sum is over only those charges inside the surface S. For a

continuous charge density p(x), Gauss's law becomes:

fs E -nda = 4n- fv p(x) d3x

(1.11)

where Vis the volume enclosed by S. Equation (1.11) is one of the basic equations of electrostatics. Note that
it depends upon (1) the inverse square law for the force between charges, (2) the central nature of the force, (3) the linear superposition of the effects of different charges.
Clearly, then, Gauss's law holds for Newtonian gravitational force fields, with matter density replacing charge density.
It is interesting to observe that before Coulomb's observations Cavendish, by what amounted to a direct application of Gauss's law, did an experiment with two concentric conducting spheres and deduced that
= the power law of the force was inverse nth power, where n 2.00 ± 0.02.
By a refinement of the technique, Maxwell showed that n = 2.0 ± 0.00005.
(See Jeans, p. 37, or Maxwell, Vol. 1, p. 80.)

1.4 Differential Form of Gauss's Law

Gauss's law can be thought of as being an integral formulation of the

law of electrostatics. We can obtain a differential form (i.e., a differential

equation) by using the divergence theorem. The divergence theorem states

that for any vector field A(x) defined within a volume V surrounded by

the closect surface S the relation

Lv . fs A. n da =

A cPx

holds between the volume integral of the divergence of A and the surface integral of the outwardly directed normal component of A. The equation in fact can be used as the definition of the divergence (see Stratton, p. 4).
To apply the divergence theorem we consider the integral relation expressed in Gauss's theorem:
f f 8 E • n da = 4n- vp(x) d3x

[Sect. 1.5]

Introduction to Electrostatics

7

Now the divergence theorem allows us to write this as:

I,,, (V • E - 4"p) d'x = 0

(1.12)

for an arbitrary volume V. We can, in the usual way, put the integrand

equal to zero to obtain

V • E = 4,rp

(1.13)

which is the differential form of Gauss's law of electrostatics. This equation can itself be used to solve problems in electrostatics. However, it is often.simpler to deal with scalar rather than vector functions of position, and then to derive the vector quantities at the end if necessary (see below).

1.5 Another Equation of Electrostatics and the Scalar Potential

The single equation (1.13) is not enough to specify completely the three components of the electric field E(x). Perhaps some readers know that a vector field can be specified completely if its divergence and curl are given everywhere in space. Thus we look for an equation specifying curl E as a function of position. Such an equation, namely,

V><E=O

(1.14)

follows direct]y from our generalized Coulomb's ]aw (1.5):

f E(x) = p(x') (x - x') d3x'

.

Ix - x'l 3

The vector factor in the integrand, viewed as a function ofx, is the negative
gradient of the scalar lflx - x'l :

(x - x') _ -V ( 1 )

Ix - x'f3 -

Ix - x'I

Since the gradient operation involves x, but not the integration variable x', it can be taken outside the integral sign. Then the field can be written

E(x) = -VJ p(x') d3x'
Ix- x'I

(1.15)

Since the curl of the gradient of any scalar function of position vanishes (V x V"P = 0, for all VJ), (1.14) follows immediately from (1.15).
Note that V x E = 0 depends on the central_ nature of the force
between charges, and on the fact that the force is a function of relative distances only, but does not depend on the inverse square nature.

8

Classical Electrodynamics

Fig. 1.3

In (1.15) the electric field (a vector) is derived from a scalar by the

gradient operation. Since one function of position is easier to deal with than three, it is worth while concentrating on the scalar function and giving it a name. Consequently we define the "scalar potential" ©(x) by the

equation:

E = -V(J)

(1.16)

Then (1.15) shows that the sca1ar potential is given in terms of the charge

density by

=J (J)(x)

p(x') d3x'

Ix - x'I

( 1.17)

where the integration is over all charges in the universe, and (f) is arbitrary

to the extent that a constant can be added to the right side of (I .17).

The scalar potential has a physical interpretation when we consider the
work done on a test charge q in transporting it from one point (A) to

another point (B) in the presence of an electric field E(x), as shown in Fig.

1.3. The force acting on the charge at any point is

F = qE

I: so that the work done in moving the charge from A to Bis W = - F • di = - q rE • di

(1.18)

The minus sign appears because we are calculating the work done on the

charge against the action of the field. With definition (1.16) the work can

be written

J, J W = q V<D • di = lJ d(J) = q[<I>u - <I>A]

A

,I

(1.19)

which shows that q© can be interpreted as the potential energy of the test

charge in the electrostatic field.

From (I. 18) and (1.19) it can be seen that the line integral of the electric

field between two points is independent of the path and is the negative of

the potential difference between the points:

J~B
E. dl = -(<I>JJ - (J)A)
A

(1.20)

[Sect. 1.6]

Introduction to Electrostatics

This follows directly, of course, from definition (1.16). If the path is closed, the line integral is zero,

(1.21)

a resu1t that can also be obtained directly from Coulomb's law. Then application of Stokes's theorem [if A(x) is a vector field, S is an open surface, and C is the closed curve bounding S,
f J c A • di = s (V x A) • n da

where di is a line element of C, n is the normal to S, and the path C is
traversed in a right-hand screw sense relative to n] leads immediately back
to \7 X E = 0.

1.6 Surface Distributions of Charges and Dipoles and Discontinuities in the Electric Field and Potential
One of the common problems in electrostatics is the determination of electric field or potential due to a given surface distribution of charges. Gauss's law (1.11) allows us to write down a partial result directly. If a surface S, with a unit normal n, has a surface-charge density of a(x) (measured in statcoulombs per square centimeter) and electric fields E1 and E2 on either side of the surface, as shown in Fig. 1.4, then Gauss's law tells us immediately that
(1.22)
This does not determine E1 and E2 unless there are no other sources of field and the geometry and form a are especially simple. All that (1.22) says is that there is a discontinuity of 41Ta in the normal component of electric fie]d in crossing a surface with a surface-charge density a, the crossing being made from the "inner" to the "outer" side of the surface.
n

Fig. 1.4 Discontinuity in the normal component of electric field across a surface layer
of charge.

10

Classical Electrodynamics

The tangential component ofelectric field can be shown to be continuous across a boundary surface by using (1.21) for the line integral of E around a closed path. It is only necessary to take a rectangular path with negligible ends and one side on either side of the boundary.
A general result for the potential (and hence the field, by differentiation) at any point in space (not just at the surface) can be obtained from (1.17) by replacing p d3x by Cf da:

<t>(x)

= J(s

a(x')
Ix~ x'I

da'

(1.23)

Another problem of interest is the potential due to a dipole-layer distribution on a surface S. A dipole layer can be imagined as being formed by letting the surface S have a surface-charge density a(x) on it, and another surface S', lying close to S, have an equal and opposite surfacecharge density on it at neighboring points, as shown in Fig. 1.5. The dipole-layer distribution of strength D(x) is formed by letting S' approach infinitesimally close to S while the surface-charge density a(x) becomes infinite in such a manner that the product of a(x) and the local separation d(x) of Sand S' approaches the limit D(x):

lim a(x) d(x) = D(x)
d(x)-0

(1.24)

The direction of the dipole moment of the layer is normal to the surface S and in the direction going from negative to positive charge.
To find the potential due to a dipole layer we can consider a single dipole and then superpose a surface density of them, or we can obtain the same result by performing mathematicalJy the limiting process described in words above on the surface-density expression (1.23). The first way is perhaps simpler, but the second gives useful practice in vector calculus. Consequently we proceed with the limiting process. With n, the unit normal to

s
s·

d(x)

s Fig. 1.5 Limiting process involved in

S'

creating a dipole layer.

[Sect. 1.6]

Introduction to Electrostatics

11

s
0 Fig. 1.6 Dipole-layer geometry.

the surface S, directed away from S', as shown in Fig. 1.6, the potential

due to the two dose surfaces is

-I. d>(x) = f a(x') da'

o-(x') da"

Js Ix - x'I

S' Ix - x' + ndf

For small d we can expand Ix - x' + ndJ-1. Consider the general expression Ix + a1-1, where la! ~ lxl. Then we write

J - -1 - -

1

Ix + al x2 + a2 + 2a • x

=;(1 _a~x+··)

=;+a -v(;) + •••

This is, of course, just a Taylor"s series expansion in three dimensions. ln this way we find that the potential becomes [upon taking the limit (1.24)]:

J: <l>(x) = D(x')n • V' ( 1 , ) da'

s

Ix-xi

(1.25)

Equation (1.25) has a simple geometrical interpretation. We note that

n. V'( 1 ) da' = _ cos 0 da' = -dO.

Ix - x'I

Ix - x'l 2

where dQ. is the element of solid angle subtended at the observation point by the area clement da', as indicated in Fig. 1.7. Note that dQ has a positive sign if Ois an acute angle, i.e., when the observation point views the •'inner" side of the dipole layer. The potential can be written:

J <I>(x) = - sD(x') dQ

(1.26)

12

Classical Electrodynamics

Fig. 1.7 The potential at P due to the
dipole layer D on the area ele~ent da' is
just the negative product of D and the
solid angle element dfl subtendFd by da'
atP.

For a constant surface-dipole-moment density D, the potential -is just the

product of the moment and the solid angle subtended at the observation

point by the surface, regardless of its shape.

1

There is a discontinuity in potential in crossing a double layer. This

can be seen by letting the observation point come infinitesimally close to

the double layer. The double layer is now imagined to consist of two

parts, one being a small disc directly under the observation point. The

disc is sufficiently small that it is sensibly flat and has constant surface-

dipole-moment density D. Evidently the total potential can be obtained

by linear superposition of the potential of the disc and that of the remain-

der. From (1.26) it is clear that the potential of the....d. isc alone has a

discontinuity of 41TD in crossing from the inner to the outer side, being

-21TD on the inner side and +21TD on the outer. The potential of the

remainder alone, with its hole where the disc fits in, is continuous across

the plane of the hole. Consequently the total potential jump in crossing

the surface is:

= <1>2 _ <I>i 477D

(1.27)

This result is analogous to (1.22) for the discontinuity of electric field in crossing a surface-charge density. Equation (1.27) can be interpreted "physically" as a potential drop occurring "inside" the dipole layer, and can be calculated as the product of the field between the two layers of surface charge times the separation before the limit is taken.

1.7 Poisson's and Laplace's Equations

In Sections 1.4 and 1.5 it was shown that the behavior of an electrostatic field can be described by the two differential equations:

V •E = 4-?rp

(1.13)

and
V><E=O

(1.14)

the latter equation being equivalent to the statement that Eis the gradient of a scalar function, the scalar potential <I>:

E = -V<D

(1.16)

[Sect. 1.7]

Introduction to Electrostatics

13

Equations (1.13) and (1.16) can be combined into one partial differential equation for the single function <l>(x):

V2<1> = -41rp

(1.28)

Thjs equation is called Poisson's equation. In regions of space where there is no charge density, the scalar potential satisfies Laplace's equation:

(1.29)

We already have a solution for the scalar potential in expression (1.17):

f <l>(x) = p(x') d3x' Ix - x'I

(1.17)

To verify that this does indeed satisfy Poisson's equation (1.28) we operate

with the Laplacian on both sides:

v J = V2<1>

2

p(x') d3x' =Jp(x')V2 (

1 ) d3x'

[x - x'j

•

]x - x'I

(1.30)

We must now calculate the value of V2(1/jx - x'[). It is convenient (and

allowable) to translate the origin to x' and so consider V2(1/r), where r is
the magnitude of x. By direct calculation we find that V2(1/r) = 0 for

r -::fa O:

v2(!) = ! ~(r. !) = !~(1) = o r r dr2 r r dr2

At r = 0, however, the expression is undefined. Hence we must use a
limiting process. Since we anticipate something like a Dirac delta function,. we integrate V2(1/r) over a small volume V containing the origin. Then we use the divergence theorem to obtain a surface integral:
L fv v2(;) d3x =~CV. v(;) d8x = n. v(;) da

=i = (!) i

r2 dD. -41T

sor .r

It has now been established that V2(1/r) = 0 for r -::fa O, and that its volume
integral is -47T. Consequently we can write the improper (but mathe-
matically justifiable) equation, V2(1/r) = -47T~(x), or, more generally,

v2 ( 1 ) = -4n-~(x - x') Ix - x'!

(1.31)

Having established the singular nature of the Laplacian of 1/r, we can now complete our check on (1.17) as a solution of Poisson's equation.

14

Classical Electrodynamics

Equation (1.30) becomes
J V2<1> = p(x')[ -'4-n-b(x - x')] d3x' = -41rp(x)

verifying the correctness of our solution (LI 7).

1.8 Green's Theorem

If electrostatic problems always involved localized discrete or continuous distributions of charge with no boundary surfaces, the general solution (1.17) would be the most convenient and straightforward solution to any problem. There would be no need of Poisson's or Laplace's equation. In actual fact, of course, many, if not most, of the problems of electrostatics involve finite regions of space, with or without charge inside, and with prescribed boundary conditions on the bounding surfaces. These boundary conditions may be simulated by an appropriate distribution of charges outside the region of interest (perhaps at infinity), but (1.17) becomes inconvenient as a means of calculating the potential, except in simple cases (e.g., method of images).
To handle the boundary conditions it is necessary to develop some new mathematical tools, namely, the identities or theorems due to George Green (1824). These follow as simple applications of the divergence theorem. The divergence theorem:
J f v V • A d3x = 8 A • n da

applies to any vector field A defined in the volume V bounded by the closed
surface S. Let A = cpV'f/J, where cf, and VJ are arbitrary scalar fields. Now

(1.32) and

cf,V'P

• n

=

cf,

0'P on

(1.33)

where o/on is the normal derivative at the surface S (directed outwards
from inside the volume V). When (1.32) and (1.33) are substituted into

the divergence theorem, there results Green's first identity:

Ys J(v (cf,V2VJ + Vcf,. VVJ) d3x = ,! cf, OoVnJ da

(1.34)

lfwe write down (1.34) again with cf, and V' interchanged, and then subtract it from (1.34), the V,f, • VVJ terms cancel, and we obtain Green's second

[Sect. 1.9]

Introduction to Electrostatics

15

identity or Green's theorem:

r (Jv 4,v21P -

?pV2ef,)

d3x

=

Y!s lr<p

Otp
on

-

tp O<p] da
on

(1.35)

Poisson's differential equation for the potential can be converted into an
integral equation if we choose a particular tp, namely 1/R = 1/jx - x'I,

where xis the observation point and x' is the integration variable. Further,
we put 4, = <!>, the scalar potential, and make use of V2<1> = -4n-p. From

(1.31) we know that V2(1/ R) = -41rb(x - x'), so that (1.35) becomes

i (l) - J(v[-4n-<I>(x') b(x -

x') + 47T p(x')] d3x' = [<t> _E__

R

"8 on'

R

_! o<l>] da'
Ron'

If the point x lies within the volume V, we obtain:

r J.~ (l)] ! Jv <l>(x) = p(x') d3x' + _1

R

41T

[!_ o<l> - <l> _E_

Ron'

on' R

da'

(1.36)

If x lies outside the surface S, the left-hand side of (1.36) is zero. [Note that this is consistent with the interpretation of the surface integral as being the potential due to a surface-charge density a= (1/47T)(o<l>/on') and a
dipole layer D = -(l/411)<1>. The discontinuities in electric field and
potential (1.22) and (1.27) across the surface then lead to zero field and zero potential outside the volume V.]
Two remarks are in order about result (1.36). First, if the surface S goes to infinity and the electric field on S falls off faster than R-1, then the surface integral vanishes and (1.36) reduces to the familiar result (1.17). Second, for a charge-free volume the potential anywhere inside the volume (a solution of Laplace's equation) is expressed in (1.36) in terms of the potential and its normal derivative only on the surface of the volume. This rather surprising result is not a solution to a boundary-value problem, but only an integral equation, since the specification of both <I> and o<l>/on (Cauchy boundary conditions) is an overspecification of the problem. This will be discussed in detail in the next sections, where techniques yielding solutions for appropriate boundary conditions will be developed using Green's theorem (1.35).

1.9 Uniqueness of the Solution with Dirichlet or Neumann Boundary Conditions
The question arises as to what are the boundary conditions appropriate for Poisson's (or Laplace's) equation in order that a unique and wellbehaved (i.e., physically reasonable) solution exist inside the bounded

16

Classical Electrodynamics

region. Physical experience leads us to believe that specification of the potential on a closed surface (e.g., a system of conductors held at different potentials) defines a unique potential problem. This is called a Dirichlet problem, or Dirichlet boundary conditions. Similarly it is plausible that specification of the electric field (normal derivative of the potential) everywhere on the surface (corresponding to a given surface-charge density) also defines a unique problem. Specification of the normal derivative is known as the Neumann boundary condition. We now proceed to prove these expectations by means of Green's first identity (I .34).
We want to show the uniqueness of the solution of Poisson's equation,
V2<1> = -41rp, inside a volume V subject to either Dirichlet or Neumann
boundary conditions on the closed bounding surface S. We suppose, to the contrary, that there exist two solutions (1')1 and <1>2 satisfying the same boundary conditions. Let
(1.37)

Then V2 U = 0 inside V, and U = 0 or au/on= 0 on S for Dirichlet and
Neumann boundary conditions, respectively. From Green's first identity
(1.34), with cp = "P = U, we find

Jv[ (UV2U

+VU·

VU)

d3x

=

J.
rs

U

aanu

da

(1.38)

With the specified properties of U, this reduces (for both types of boundary conditions) to :

fv IVl/12 d3x = 0

which implies VU= 0. Consequently, inside V, U is constant. For
Dirichlet boundary conditions, U = 0 on S so that, inside V, <I\ = <1>2 and
the solution is unique. Similarly, for Neumann boundary conditions, the solution is unique, apart from an unimportant arbitrary additive constant.
From the right-hand side of (1.38) it is clear that there is also a unique
solution to a problem with mixed boundary conditions (i.e.. Dirichlet over part of the surface S, and Neumann over the remaining part).
It should be clear that a solution to Poisson's equation with both <I> and
o<l>/on specified on a cJosed boundary (Cauchy boundary conditions) does
not exist, since there are unique solutions for Dirichlet and Neumann conditions separately and these will in general not be consistent. The question of whether Cauchy boundary conditions on an open surface define a unique electrostatic problem requires more discussion than is warranted here. The reader may refer to Morse and Fesnbach, Section 6.2, pp. 692-
706, or to Sommerfeld, Partial Differential Equations in Physics, Chapter

[Sect. 1.9]

Introduction to Electrostatics

17

II, for a detailed discussion of these questions. Morse and Feshbach base
their treatment on the replacement of the partial differential equation by appropriate difference equations which they then solve by an iterative procedure. On the other hand, Sommerfeld bases his discussion on the method of characteristics where possible. The result of these investigations on which boundary conditions are appropriate is summarized in the table below (based on one given in Morse and Feshbach), where different types

Type of Equation

Type of Boundary Condition

Elliptic (Poisson's eq.)

Hyperbolic
(wave eq.)

Parabolic (heat-conduction eq.)

Dirichlet
Open surface

Not enough

Not enough

Closed surface

Unique, stable solution

Too much

Unique, stable solution in one direction
Too much

Neumann
Open surface

Not enough

Not enough

Closed surface

Unique, stable solution in general

Too much

Unique, stable solution in one direction
Too much

Cauchy
Open surface
Closed surface

Unphysical results
Too much

Unique, stablel Too much solution

Too much

Too much

A stable solution is one for which small changes in the boundary conditions cause appreciable changes in the solution only in the neighborhood of the boundary.

of partial differential equations and different kinds of boundary conditions are listed.
Study of the table shows that electrostatic problems are specified only by Dirichlet or Neumann boundary conditions on a closed surface (part or all of which may be at infinity, of course).

18

Classical Electrodynamics

1.10 Formal Solution -of Electrostatic Boundary-Value Problem with Green's Function

The solution of Poisson's or Laplace's equation in a finite volume Vwith

either Dirichlet or Neumann boundary conditions on the bounding surface

S can be obtained by means of Green's theorem (1.35) and so-called

"Green's functions."

In obtaining result (1.36)-not a solution-we chose the function 'P to

be 1/lx - x'I, it being the potential of a unit point charge, satisfying the

equation:

V'2( 1 ) = -41r~(x - x')
Ix - x'I

(1.31)

The function 1/Jx - x'I is only one of a class of functions depending on the

variables x and x', and called Green's functions, which satisfy (1.31). In

general,

V'2G(x, x') = -4m5(x - x')

(1.39)

where

G(x, x') = 1 + F(x, x') jx - x')

(1.40)

with the function F satisfying Laplace's equation inside the volume V:

V'2F(x, x') = 0

(1.41)

In facing the problem of satisfying the prescribed boundary conditions
on <I> or o<b/on, we can find the key by considering result (1.36). As has

been pointed out already, thls is not a solution satisfying the correct type
of boundary conditions because both <I> and o(f)/on appear in the surface

integral. It is at best an integral equation for <I>. With the generalized

concept of a Green's function and its additional freedom [via t.he function

F(x, x')], there arises the possibility that we can use Green's theorem with

tp = G(x, x') and choose F(x, x') to eliminate one or the other of the two

surface integrals, obtaining a result which involves only Dirichlet or

Neumann boundary conditions. Of course, if the necessary G(x, x')

depended in detail on the exact form of the boundary conditions, the

method would have little generality. As will be seen immediately, this is

not required, and G(x, x') satisfies rather simple boundary conditions on S.
With Green's theorem (1.35), rt, = <I>, 1P = G(x, x'), and the specified

properties of G (1.39), it is simple to obtain the generalization of (1.36):

f. f, [ + - <I>(x) = p(x')G(x, x') d3x' 1 G(x, x') -o<I> - <I>(x') oG(x' x')] da'

V

47T S

an'

on'

(1.42)

[Sect. 1.1 OJ

Introduction to Electrostatics

19

The freedom available in the definition of G (l .40) means that we can make

the surface integral depend only on the chosen type of boundary con-

ditions. Thus, for Dirichlet boundary conditions we demand:

Gn(x, x 1) = 0 for x' on S

(1.43)

Then the first term in the surface integral in (1.42) vanishes and the

solution is
=i Jis cI>(x)

p(x')Gp(X, x') d3x' - _l (f>(x') aGv da'

V

41T

an'

(1.44)

For Neumann boundary conditions we must be more careful. The obvious choice of boundary condition on G(x, x') seems to be

-aG-N(.x, x') = 0 for x' on S
on'

since that makes the second term in the surface integral in (1.42) vanish, as desired. But an application of Gauss's theorem to (1.39) shows that

J: aa. da' = -41r
Yson'
Consequently the simplest allowable boundary condition on GN is

oG,v
= - - __ i

(x,

x')

s on 1

41T for x' on S

(1.45)

where S is the total area of the boundary surface. Then the solution is

J <l>(x) = (<1>)8 + ( p(x')Giv(x, x') d3x' + _!_ a<I>.' GN da' (1.46)
Jv Ys 41r an
where (<l>)s is the average value of the potential over the whole surface. The customary Neumann problem is the so-cal1ed "exterior problem" in which the volume Vis bounded by two surfaces~ one closed and finite, the other at infinity. Then the surface area S is infinite; the boundary condition (1.45) becomes homogeneous; the average value (<P)s vanishes.
We note that the Green's functions satisfy simple boundary conditions (1.43) or (1.45) which do not depend on the detailed form of the Dirichlet (or Neumann) boundary values. Even so, it is often rather involved (if not impossible) to determine G(x, x') because of its dependence on the shape of the surface S. We will encounter such problems in Chapter 2 and 3.
The mathematical symmetry property G(x, x') = G(x', x) can be proved
for the Green's functions satisfying the Dirichlet boundary condition
= (1.43) by means of Green~s theorem with ,f, = G(x, y) and '1/J G(x', y),

20

Classical Electrodynamics

where y is the integration variable. Since the Green's function, as a function of one of its variables, is a potential due to a unit point charge, this symmetry merely represents the physical interchangeability of the source and the observation points. For Neumann boundary conditions the symmetry is not automatic, but can be imposed as a separate requirement.
As a final, important remark we note the physical meaning of F(x, x'). It is a solution of Laplace's equation inside V and so represents the potential of a system of charges external to the volume V. It can be thought of as the potential due to an external distribution of charges so chosen as to satisfy the homogeneous boundary conditions of zero potential (or zero normal derivative) on the surface S when combined with the potential of a point charge at the source point x'. Since the potential at a point x on the surface due to the point charge depends on the position of the source point, the external distribution of charge F(x, x') must also depend on the "parameter" x'. From this point of view, we see that the method of images (to be discussed in Chapter 2) is a physical equivalent of the determination of the appropriate F(x, x') to satisfy the boundary conditions (1.43) or (1.45). For the Dirichlet problem with conductors, F(x, x') can also be interpreted as the potential due to the surface-charge distribution induced on the conductors by the presence of a point charge at the source point x'.

1.11 Electrostatic Potential Energy and Energy Density

In Section 1.5 it was shown that the product of the scalar potential and the charge of a point object could be interpreted as potential energy. More precisely, if a point charge qi is brought from infinity to a point xi in a region of localized electric fields described by the scalar potential <l> (which vanishes at infinity), the work done on the charge (and hence its potential energy) is given by
(1.47)

The potential ct> can be viewed as produced by an array of (n - 1) charges
qij = l, 2, ... , n - 1) at positions xj. Then

2n-1

ct>(~)=

qi

1=1 lxi - X1I

so that the potential energy of the charge qi is

(1.48)

(1.49)

[Sect. 1.11]

Introduction to Electrostatics

21

It is clear that the total potential energy of all the charges due to all the forces acting between them is:

(1.50)

as can be seen most easily by adding each charge in succession. A more symmetric form can be written by summing over i and j unrestricted, and then dividing by 2:

(1.51)

lt is understood that i = j terms (infinite "self-energy" terms) are omitted

in the double sum.

For a continuous charge distribution [or, in general, using the Dirac

delta functions (1.6)] the potential energy takes the form:

II W = l

p(x)p(x') d3x d3x'

(1.52)

2 lx-x'I

Another expression, equivalent to (1.52), can be obtained by noting that

one of the integrals in (1.52) is just the scalar potential (1.17). Therefore

I W = } p(x)<P(x) d3x

(1.53)

Equations (1.51), (1.52), and (1.53) express the electrostatic potential energy in terms of the positions of the charges and so emphasize the interactions between charges via Coulomb forces. An alternative, and very fruitful, approach is to emphasize the electric field and to interpret the energy as being stored in the electric field surrounding the charges. To obtain this latter form, we make use of Poisson's equation to eliminate the charge density from (1.53):

= W -1 f(J)V2(J) d3x
81r

Integration by parts leads to the result:

f I W = _s!~_

IV<Pl 2 d3x = __!__
81r

IEl2 d3x

(1.54)

where the integration is over all space. In (1.54) all explicit reference to

charges has gone, and the energy is expressed as an integral of the square of the electric field over all space. This leads naturally to the identification of the integrand as an energy density w:

w = _.!_ /El 2
817'

(1.55)

22

Classical Electrodynamics

Q2

0
Fig. 1.8
This expression for energy density is intuitively reasonable, since regions of high fields "must" contain considerable energy.
There is perhaps one puzzling thing about (1.55). The energy density is positive definite. Consequently its volume integral is necessarily nonnegative. This seems to contradict our impression from (1.51) that the potential energy of two charges of. opposite sign is negative. The reason for this apparent contradiction is that (1.54) and (1.55) contain "selfenergy" contributions to the energy density, whereas the double sum in (1.51) does not. To illustrate this, consider two point charges q1 and q2
located at x1 and x2, as in Fig. 1.8. The electric field at the point P with
coordinate x is
E = qi(x - X1) + qlx - x2)
Ix - X113 Ix - xil
so that the energy density (1.55) is

w = ql

+ q22

+ q1qlx - X1) • (x - X2) (1.S6)

87TIX - x1 14 81Tlx - x2l 4 4-n-lx - x113 Ix - x213

Clearly the first two terms are self-energy contributions. To show that the
third term gives the proper result for the interaction potential energy we

= I integrate over all space: Wi t q1q2 (x - X1) • (x - X2) dax m 411 Ix - x1l3 Ix - l X 2 3

(1.57)

f A change of integration variable to p = (x - x1)/lx1 - x21 yields

= Wint

q 1q 2

X _.!_ P • (p + n) d3 p

lx1 - X2I 4-n- llP + nl3

(1.58)

where n is a unit vector in the direction (x1 - x2).. By straightforward integration the dimensionless volume integral can be shown to have the

value 47T, so that the interaction energy reduces to the expected value.

Forces acting between charged bodies can be obtained by calculating

the change in the total electrostatic energy of the system under small virtual displacements. Examples of this are discussed in the problems.

Care must be taken to exhibit the energy in a form showing clearly those

[Probs. l]

Introduction to Electrostatics

23

factors which vary with a change in configuration and those which are kept constant.
As a simple illustration we calculate the force per unit area on the surface of a conductor with a surface-charge density o{x). In the immediate neighborhood of the surface the energy density is

w = _!_ 1El2 = 27Ta2
81r

(1.59)

If we now imagine a small outward displacement .l\x of an elemental area

.l\a of the conducting surface, the electrostatic energy decreases by an

amount which is the product of energy density wand the excluded volume

.l\x .l\a:

.l\ W = - 21ra2.l\a .l\x

(l.60)

This means that there is an outward force per unit area equal to 2m;2 = w
at the surface of the conductor. This result is normally derived by taking the product of the surface-charge density and the electric field, with care taken to eliminate the electric field due to the element of surface-charge density itself.

REFERENCES AND SUGGESTED READING
On the mathematical side, the subject of delta functions is treated simply but rigorously by
Lighthill. For a discussion of different types of partial differential equations and the appropriate boundary conditions for each type, see
Morse and Feshbach, Chapter 6, Sommerfeld, Partial Differential Equations in Physics, Chapter II, Courant and Hilbert, Vol. II, Chapters III-VI. The general theory of Green's functions is treated in detail by Friedman, Chapter 3, Morse and Feshbach, Chapter 7.
The general theory of electrostatics is discussed extensively in many of the older books.
Notable, in spite of some old-fashioned notation, are Maxwell, Vol. 1, Chapters II and IV, Jeans, Chapters II, VI, VII.
Of more recent books, mention may be made of the treatment of the general theory by Stratton, Chapter III, and parts of Chapter II.

PROBLEMS
1.1 Use Gauss's theorem to prove the following statements: (a) Any excess charge placed on a conductor must lie entirely on its
surface. (A conductor by definition contains charges capable of moving freely under the action of app1ied electric fields.)

24

Classical Electrodynamics

(b) A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from the fields due to charges placed inside it.
(c) The electric field at the surface of a conductor is normal to the surface and has a magnitude 4·mr, where a is the charge density per unit area on the surface.
1.2 Two infinite, conducting, plane sheets of uniform thicknesses t1 and t2, respectively, are placed parallel to one another with their adjacent faces separated by a distance L. The first sheet has a total charge per unit area (sum of the surface.charge densities on either side) equal to q1, while the second has q2. Use symmetry arguments and Gauss's law to prove that (a) the surface.charge densities on the adjacent faces are equal and opposite; (b) the surface-charge densities on the outer faces of the two sheets are the same; (c) the magnitudes of the charge densities and the fields produced are independent of the thicknesses t1 and t2 and the separation L. Find the surface-charge densities and fields explicitly in terms of q1 and
q2, and apply your results to the special case q1 = -q2 = Q.
1.3 Each of three charged spheres of radius a, one conducting, one having a uniform charge density within its volume, and one having a spherically symmetric charge density which varies radially as rn (n > -3), has a total charge Q. Use Gauss's theorem to obtain the electric fields both inside and outside each sphere. Sketch the behavior of the fields as a function of radius for the first two spheres, and for the third with n = -2, +2.
1.4 The time-average potential of a neutral hydrogen atom is given by

where q is the magnitude of the electronic charge, and cc-1 = a0/2. Find the distribution of charge (both continuous and discrete) which will give this potential and interpret your result physically.
1.S A simple capacitor is a device formed by two insulated conductors adjacent to each other. If equal and opposite charges are placed on the conductors, there will be a certain difference of potential between them. The ratio of the magnitude of the charge on one conductor to the magnitude of the potential difference is called the capacitance (in electrostatic units it is measured in centimeters). Using Gauss's law, calculate the capacitance of (a) two large, flat, conducting sheets of area A, separated by a small distanced; (b) two concentric conducting spheres with radii a, b (b > a); (c) two concentric conducting cylinders of length L, large compared to their radii a, b (b > a). (d) What is the inner diameter of the outer conductor in an air-filled coaxial cable whose center conductor is B&S #20 gauge wire and whose capacitance is 0.5 micromicrofarad/cm? 0.05 micromicrofarad/cm?
1.6 Two long, cylindrical conductors of radii a1 and a2 are parallel and separated by a distance d which is large compared with either radius.

[Probs. 1]

Introduction to Electrostatics

25

Show that the capacitance per unit length is given approximately hy

C ~ ( 4 Jn~d)-1

where a is the geometrical mean of the two radii. Approximately what B&S gauge wire (state diameter in millimeters
as well as gauge) would be necessary to make a two-wire transmission line with a capacitance of 0.1 ,,,,f/cm if the separation of the wires was 0.5 cm 1 1.5 cm? 5.0 cm?
1.7 (a) For the three capacitor geometries in Problem l .5 calculate the total electrostatic energy and express it alternatively in terms of the equal and opposite charges Q and - Q placed on the conductors and the potential difference between them. (b) Sketch the energy density of the electrostatic field in each case as a function of the appropriate linear coordinate.
1.8 Calculate the attractive force between conductors in the parallel plate capacitor (Problem l.5a) and the parallel cylinder capacitor (Problem 1.6) for
(a) fixed charges on each conductor; (b) fixed potential difference between conductors.
1.9 Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.
1.10 Use Gauss's theorem to prove that at the surface of a curved charged conductor the normal derivative of the electric field is given by

!_ uE = -(_!_ + _!_)

£ on

R1 R2

where R1 and R2 are the principal radii of curvature of the surface.
1.11 Prove Green's reciprocation theorem: If ct> is the potential due to a volumecharge density p within a volume V and a surface-charge density a on the surface S bounding the volume V, while ct>' is the potential due to another charge distribution p' and a', then
J Iv p<l>' d3x +L a©' da = v p'<l> d3x +LG'<l> da

1.12 Prove Thomson's theorem: If a number of conducting surfaces are fixed in position and a given total charge is placed on each surface, then the electrostatic energy in the region bounded by the surfaces is a minimum when the charges are placed so that every surface is an equipotential.
1.13 Prove the following theorem: If a number of conducting surfaces are fixed in position with a given total charge on each, the introduction of an uncharged, insulated conductor into the region bounded by the surfaces lowers the electrostatic energy.

2
Boundary-Value Problems in Electrostatics: I
Many problems in electrostatics involve boundary surfaces on which either the potential or the surface-charge density is specified. The formal solution of such problems was presented in Section 1.10, using the method of Green's functions. In practical situations (or even rather idealized approximations to practical situations) the discovery of the correct Green's function is sometimes easy and sometimes not. Consequently a number of approaches to electrostatic boundary-value problems have been developed, some of which are only remotely connected to the Green's function method. In this chapter we will examine two of these special techniques: (1) the method of images, which is closely related to the use of Green's functions; (2) expansion in orthogonal functions, an approach directly through the differential equation and rather remote from the direct construction of a Green's function. Other methods of attack, such as the use of conformal mapping in two-dimensional problems~ will be omitted. For a discussion of conformal mapping the interested reader may refer to the references cited at the end of the chapter.
2.1 Method of Images
The method of images concerns itself with the problem of one or more point charges in the presence of boundary surfaces, e.g., conductors either grounded or held at fixed potentials. Under favorable conditions it is possible to infer from the geometry of the situation that a small number of suitably placed charges of appropriate magnitudes, external to the region of interest, can simulate the required boundary conditions. These charges
26

[Sect. 2.2]

Boundary- Value Problems in Electrostatics: I

27

Fig. 2.1 Solution by method of images. The original potential problem is on the left, the equivalent-image problem on
the right.

t=O
- - - - - -q

I
I
k-t=O I I I
-q---,I-----q
I
I I
I
l
I
I

are called image charges, and the replacement of the actual problem with boundaries by an enlarged region with image charges but no boundaries is called the method of images. The image charges must be external to the volume of interest, since their potentials must be solutions of Laplace's equation inside the volume; the "particular integral" (i.e., solution of Poisson's equation) is provided by the sum of the potentials of the charges inside the volume.
A simple example is a point charge located in front of an infinite plane conductor at zero potential, as shown in Fig. 2.1. It is clear that this is equivalent to the problem of the original charge and an equal and opposite charge located at the mirror-image point behind the plane defined by the position of the conductor.

2.2 Point Charge in the Presence of a Grounded Conducting Sphere
As an illustration of the method of images we consider the problem illustrated in Fig. 2.2 of a point charge q located at y relative to the origin around which is centered a grounded conducting sphere of radius a.* We
seek the potential <l>(x) such that Cl>(lxl = a) = 0. By symmetry it is
evident that the image charge q' (assuming that only one image is needed) will lie on the ray from the origin to the charge q. If we consider the charge q outside the sphere, the image position y' will lie inside the sphere. The
* The term grounded is used to imply that the surface or object is held at the same potential as the point at infinity by means of some fine conducting connector. The connection is assumed not to disturb the potential distribution. But arbitrary amounts of charge of either sign can flow onto the object from infinity in order to maintain its potential at "ground" (usually taken to be zero potential). A conductor held at a fixed potential is essentially the same situation, except that a voltage source is interposed between the object and "ground."

28

Classical Electrodynamics

q

Fig. 2.2 Conducting sphere of radius a, with chargeq and image
charge q'.

potential due to the charges q and q' is:

I

«l>(x) = q + q

(2.1)

Ix - YI Ix - y'I

We now must try to choose q' and ly'I such that this potential vanishes at
lxl = a. If n is a unit vector in the direction x, and n' a unit vector in the

direction y, then

<l>(x) = q

+ q'

(2.2)

lxn - yn'I lxn - y'n'I

If x is factored out of the first term and y' out of the second, the potential
at x = a becomes:

<I>(x = a) = ~ u I" -q ~ u I+~ y l" q-' y' 111

(2.3)

From the form of (2.3) it will be seen that the choices:

q

q'

-=--;,

a

y

-y =a-
a y'

make 4>(x =a)= 0, for all possible values of n • n'. Hence the magnitude and position of the image charge are

qI = - -Qq,

yI

a2
=-

(2.4)

y

y

[Sect. 2.2]

Boundary-Value Problems in Electrostatics: I

29

We note that, as the charge q is brought closer to the sphere, the image
charge grows in magnitude and moves out from the center of the sphere. When q is just outside the surface of the sphere, the image charge is equal and opposite in magnitude and lies just beneath the surface.
Now that the image charge has been found, we can return to the original problem of a charge q outside a grounded conducting sphere and consider various effects. The actual charge density induced on the surface of the sphere can be calculated from the normal derivative of <I> at the surface:

(a) ( 1 o<I>

q

y ( = (J - 41r ax x=a = - 41ra2

1 - ~)

1 + -a2 -

a )¾
2-cos y

(2.5)

y2

y

where y is the angle between x and y. This charge density in units of
-q/41ra2 is shown plotted in Fig. 2.3 as a function of y for two values of y/a. The concentration of charge in the direction of the point charge q is
evident, especially for y/a = 2. It is easy to show by direct integration
that the total induced charge on the sphere is equal to the magnitude of the
image charge, as it must according to Gauss's law.

t 2

Fig. 2.3 Surface-charge density a

1

induced on the grounded sphere

of radius a due to the presence

of a point charge q located a dis-

tance y away from the center of

the sphere. <1 is plotted in units of

-q/4rra1 as function of the angular

position -y away from the radius
to the charge for y = 2a, 4a.

, . _11" ~

30

Classical Electrodynamics

= dF 21ru2 da

---------eq

Fig. 2.4

The force acting on the charge q can be calculated in different ways.
One (the easiest) way is to write down immediately the force between the
charge q and the image charge q'. The distance between them is y - y' =
y(l - a2/ y2). Hence the attractive force, according to Coulomb's law, is:

IF[=qa-22 (a-y)a ( 1- -ay22)-2

(2.6)

For large separations the force is an inverse cube law, but close to the sphere it is proportional to the inverse square of the distance away from the surface of the sphere.
The alternative method for obtaining the force is to calculate the total force acting on the surface of the sphere. The force on each element of area da is 27Ta2 da, where O' is given by (2.5), as indicated in Fig. 2.4. But from symmetry it is clear that only the component parallel to the radius vector from the center of the sphere to q contributes to the total force. Hence the total force acting on the sphere (equal and opposite to the force acting on q) is given by the integral:

)2( IF[= _q2_ ( ~
(1 81ra2 y

1 - ~2)2f
'112

cosy

dD.

+ a2 - 2a cos y)a

(2.7)

y2

y

Integration immediately yields (2.6). The whole discussion has been based on the understanding that the
point charge q is outside the sphere. Actually, the results apply equally for the charge q inside the sphere. The only change neces_sary is in the surfacecharge density (2.5), where the normal derivative out of the conductor is now radially inwards, implying a change in sign. The reader may transcribe
all the formulas, remembering that now y < a. The angu]ar distributions
of surface charge are similar to those of Fig. 2.3, but the total induced surface charge is evidently equal to -q, independent of y.

[Sect. 2.3]

Boundary-Value Problems in Electrostatics: I

31

2.3 Point Charge in the Presence of a Charged, Insulated, Conducting Sphere

In the previous section we considered the problem of a point charge q near a grounded sphere and saw that a surface-charge density was induced
on the sphere. This charge was of total amount q' = -aq/y, and was
distributed over the surface in such a way as to be in equilibrium under all forces acting.
If we wish to consider the problem of an insulated conducting sphere with total charge Q in the presence of a point charge q, we can build up the solution for the potential by linear superposition. In an operational sense, we can imagine that we start with the grounded conducting sphere (with its charge q' distributed over its surface). We then disconnect the ground wire and add to the sphere an amount of charge (Q - q'). This brings the total charge on the sphere up to Q. To find the potential we merely note that the added charge (Q - q') will distribute itself uniformly over the surface, since the electrostatic forces due to the point charge q are already balanced by the charge q'. Hence the potential due to the added charge (Q - q') will be the same as if a point charge of that magnitude were at the origin, at least for points outside the sphere.
The potential is the superposition of (2.1) and the potential of a point charge (Q - q') at the origin:

Q +-aq

<l>(x) = q -

aq +

y

(2.8)

Ix - YI ~yx- -2y

lxl

y2

The force acting on the charge q can be written down directly from Coulomb's law. It is directed along the radius vector to q and has the magnitude:

(2.9)

In the limit of y ► a, the force reduces to the usual Coulomb's law for two small charged bodies. But close to the sphere the force is modified because of the induced charge distribution on the surface of the sphere. Figure 2.5 shows the force as a function of distance for various ratios of Q/q. The force is expressed in units of q2/y2 ; positive (negative) values correspond to a repulsion (attraction). If the sphere is charged oppositely to q, or is

32
5 4 3 --
2 Fyz 1
q2 0
-1
-2

Classical Electrodynamics

I

I

I
I

I
I

= Q/q 3

I

I
I I I

I

1

I

I

11

3

4 y/a~s

0

-1

-3

-4

-5
Fig. 2.S The force on a point charge q due to an insulated, conducting sphere of radius a carrying a total charge Q. Positive values mean a repulsion, negative an attraction. The asymptotic dependence of the force has been divided out. Fy2/q2 is plotted versus
y/a for Q/q = -1,0, 1, 3. Regardless of the value of Q, the force is always attractive
at close distances because of the indueed surface charge.
uncharged, the force is attractive at all distances. Even if the charge Q is the same sign as q, however, the force becomes attractive at very close distances. In the limit of Q ► q, the point of zero force (unstable equili-
brium point) is very close to the sphere, namely, at y ~ a(l + ½Vq/ Q).
Note that the asymptotic value of the force is attained as soon as the charge
q is more than a few radii away from the sphere.
This example exhibits a general property which explains why an excess of charge on the surface does not immediately leave the surface because of mutual repulsion of the individual charges. As soon as an element of charge is removed from the surface, the image force tends to attract it back. If sufficient work is done, of course, charge can be removed from the surface to infinity. The work function of a metal is in large part just the work done against the attractive image force in order to remove an electron from the surface.

[Sect. 2.5]

Boundary-Value Problems in Electrostatics: I

33

2.4 Point Charge near a Conducting Sphere at Fixed Potential

Another problem which can be discussed easily is that of a point charge

near a conducting sphere held at a fixed potential V. The potential is the

same as for the charged sphere, except that the charge (Q - q') at the

center is replaced by a charge (Va). This can be seen from (2.8), since at
[x[ = a the first two terms cancel and the last term will be equal to Vas

required. Thus the potential is <I>(x) = q [x-y[

_l__aq a_2 _

+

Va -[xi

yx--y
y2

(2.10)

The force on the charge q due to the sphere at fixed potential is

F = !L[va - qaif ]Y

y2

(y2 _ a2)2 y

(2.11)

For corresponding values of Va/q and Q/q this force is very similar to that
of the charged sphere, shown in Fig. 2.5, although the approach to the asymptotic value (Vaq/y2) is more gradual. For Va ► q, the unstable
equilibrium point has the equivalent location y ,...._, a(l + ½Vq/ Va).

2.5 Conducting Sphere in a Uniform Electric Field by Method of Images

As a final example of the method of images we consider a conducting
sphere of radius a in a uniform electric field £ 0. A uniform field can be thought of as being produced by appropriate positive and negative charges
at infinity. For example, if there are two charges ± Q, located at positions
z = =FR, as shown in Fig. 2.6a, then in a region near the origin whose
dimensions are very small compared to R there is an approximately
constant electric field £0 ,...._, 2Q/R2 parallel to the z axis. In the limit as R, Q ~ co, with Q/R2 constant, this approximation becomes exact.
If now a conducting sphere of radius a is placed· at the origin, the
potential will be that due to the charges ± Q at =FR and their images
=f-Qa/Ratz = -=t=a2/R:

$=

Q

Q

+ + + (r2 R 2 2rR cos 0)½ (r2 R2 - 2rR cos 0)½

aQ

aQ

+ - - - - + - + ( R ( r2

a4 + -2a-2r cos ()•)"1lL!

R r2

a4

2a 2r

cos

J½ fJ

R2 R

R2 R

(2.12)

34

Classical Electrodynamics

p
+Q ---------------
z=-R

:: •Eo
-- -- --------- -Q z=R

(a)
p

+Q

-Q

z= -R

(b)
Fig. 2.6 Conducting sphere in a uniform electric field by the method of images.

where <I> has been expressed in terms of the spherical coordinates of the observation point. In the first two terms R is much larger than r by assumption. Hence we can expand the radicals after factoring out R2. Similarly, in the third and fourth terms, we can factor out r2 and then expand. The result is:

o] + <I> = [- 2Q r cos (j R2

2Q
R2

ar23

cos

+ ...

(2.13)

where the omitted terms vanish in the limit R---+ oo. In that limit 2Q/R2 becomes the applied uniform field, so that the potential is

(r - ~) <I> = -E0

cos 0

(2.14)

The first term (-Er,Z) is, of course, just the potential of a uniform field £ 0 which could have been written down directly instead of the first two terms

in (2.12). The second is the potential due to the induced surface charge

density or, equivalently, the image charges. Note that the image charges

= form a dipole of strength D = Qa/R x 2a2 R E a3 The induced

/

•

0

surface-charge density is

a = - -1 a-<1> = -3 E0 cos ()
47T dr r=a 4w

(2.15)

[Sect. 2.6]

Boundary- Value Problems in Electrostatics: I

35

We note that the surface integral of this charge density vanishes, so that there is no difference between a grounded and an insulated sphere.

2.6 Method of Inversion

The method of images for a sphere and related topics discussed in the previous sections suggest that there is some sort of equivalence of solutions of potential problems under the reciprocal radius transformation,

a2
r---+ r' = -
r

(2.16)

This equivalence forms the basis of the method of inversion, and transformation (2.16) is called inversion in a sphere. The radius of the sphere is called the radius of inversion, and the center of the sphere, the center of inversion. The mathematical equivalence is contained in the foJlowing theorem:

Let <I>(r, 0, </,) be the potential due to a set of point charges qi at the points (ri, ()i, <pi). Then the potential

(2.17)
is the potential due to charges, (2.18)
located at the points (a2/ri, ()i, <pi). The proof of the theorem is as follows. The potential <l>(r, 0, cf>) can be written as

where 'Yi is the angle between the radius vectors x and xi. Under transformation (2.16) the angles remain unchanged. Consequently the new potential <I>' is

_2 ¢'(r' 0, </>) = ~ --,====q=i====

r i

- + a4

2 2a2

ri - - ri cos y,

r2

r

36

Classical Electrodynamics

p

Fig. 2.7
By factoring (r?/r2) out of the square root, this can be written
This proves the theorem. Figure 2. 7 shows a simple configuration of charges before and after
inversion. The potential <I>' at the point P due to the inverted distribution of charge is related by (2.17) to the original potential <I> at the point P' in the figure.
The inversion theorem has been stated and proved with discrete charges. It is left as an exercise for the reader to show that, if the potential <I> satisfies Poisson's equation,
y'2<1> = -41rp
the new potential <I>' (2.17) also satisfies Poisson's equation, (2.19)
where the new charge density is given by
(2.20)
The connection between this transformation law for charge densities and the law (2.18) for point charges can be established by considering the charge density as a sum of delta functions:
I p(x) = qiocx - x,)
i

[Sect. 2.6]

Boundary-Value Problems in Electrostatics: I

37

In terms of spherical coordinates centered at the center of inversion the

charge density can be written

2 p(r, 0, cf,) = qi<5(0 - Qi)~ o(r - r;,)

i

ri

where J(Q - Qi) is the angular delta function whose integral over solid angle gives unity, and ()(r - ri) is the radial delta function.* Under inversion the angular factor is unchanged. Consequently we have

2 p(a2 , 0, cf,) = qib(Q - .Qi) \ 6(a2 - ri)

r

i

ri r

The radial delta function can be transformed according to rule 5 at the end of Section 1.2 as

Then

p ( -ar2 , 0, cf,)

= ~ "". ' qit}(Q i

() ( r - a- 2)

ni) sa6 ~

('

a2)

r-
2 i

-

ri
and the inverted charge density (2.20) becomes

where x/ = (a2/ri, 0, cf,) and q/ = (a/ri)q1• as required by (2.18).
With the transformation laws for charges and volume-charge densities given by (2.18) and (2.20), it will not come as a great surprise that the transformation of suiface-charge densities is according to
(2.21)
Before treating any examples of inversion there are one or two physical and geometrical points which need discussion. First, in regard to the physical points, if the original potential problem is one where there are conducting surfaces at fixed potentials, the inverted problem will not in general involve the inversions of those surfaces held at fixed potentials. This is evident from (2.17), where the factor a/r shows that even if¢ is constant on the original surface the potential <I>' on the inverted surface is
* The factor ri- 2 multiplying the radial delta function is present to cancel out the r2
which appears in the volume element d3x = r2 dr dO..

38

Classical Electrodynamics

Fig. 2.8 Geometry of inversion. Center of inversion is at 0. Radius of inversion is a. The inversion of the surface S is the surface S', and
vice versa.
not. The only exception occurs when <I) vanishes on some surface. Then <I>' also vanishes on the inverted surface.
One might think that, since <D is arbitrary to the extent of an additive constant, we could make any surface in the original problem have zero potential and so also be at zero potential in the inverted problem. This brings us to the second physical point. The inverted potentials corresponding to two potential problems differing only by an added constant potential <1>0 represent physically different charge configurations, namely, charge distributions which differ by a point charge a¢0 located at the center of inversion. This can be seen from (2.17), where a constant term Cl>0 in <D is transformed into (a<D0/r). Consequently care must be taken in applying the method of inversion to remember that the mapping of the point at infinity into the origin may introduce point charges there. If these are not wanted, they must be separately removed by suitable linear superposition.
The geometrical considerations involve only some elementary points which can be proved very simply. The notation is shown in Fig. 2.8. Let 0 be the center of inversion, and a the radius of inversion. The intersection of the sphere of inversion and the plane of the paper is shown as the dotted circle. A surface S intersects the page with the curve AB. The inverted surface S', obtained by transformation (2.16), intersects the page in the curve A'B'. The following facts are stated without proof:
(a) Angles of intersection are not altered by inversion. (b) An element of area da on the surface S is related to an element of
area da' on the inverted surface S' by da/da' = r2/r'2•
(c) The inverse of a sphere is always another sphere [perhaps of infinite radius; see (d)].
(d) The inverse of any plane is a sphere which passes through the center of inversion, and conversely.
Figure 2.9 illustrates the possibilities involved in (c) and (d) when the center of inversion lies outside, on the surface of, or inside the sphere.

[Sect. 2.6]

Boundary-Value Problems in Electrostatics: I

39

As a very simple example of the solution of a potential problem by inversion we consider an isolated conducting sphere of radius R with a total charge Q on it. The potential has the constant value Q/ R inside the sphere and falls off inversely with distance away from the center for points outside the sphere. By a suitable choice of center of inversion and associated parameters we can obtain the potential due to a point charge q a distanced away from an infinite, grounded, conducting plane. Evidently, if the center of inversion O is chosen to lie on the surface of the sphere of radius R, the sphere will invert into a plane. This geometric situation is shown in Fig. 2.10. Furthermore, if we choose the arbitrary additive constant potential <1>0 to have the value - Q/R, the sphere and its inversion, the plane, will be at zero potential, while a point charge -aQ/R will appear
at the center of inversion. In order that we end up with a point charge q a
distance d away from the plane it is necessary to choose the radius of
inversion to be a = (2Rd)½ and the initial charge, Q = -(R/2df2q. The
surface-charge density induced on the plane can be found easily from (2.21). Since the charge density on the sphere is uniform over its surface, the induced charge density on the plane varies inversely as the cube of the distance away from the origin (as can be verified from the image solution; see Problem 2.1).
If the center of inversion is chosen to lie outside the isolated uniformly charged sphere, it is clear from Fig. 2.9 that the inverted problem can be

/ / C'
/
,,, ,_~-~..... V / / S'

I/ s(/)"'\\

I

O A BI

B'

\

'-

I

\

D I

\

\. ,

......

_____

'\,..,,I ,.,.'

' '- D'

' '

~ig. 2.9 Various possibilities for the inversion of a sphere. If the center of inversion 0 hes on the surface S of the sphere, the inverted surface S' is a plane; otherwise it is
another sphere. The sphere of inversion is shown dotted.

40

Classical Electrodynamics

Fig. 2.10 Potential due to isolated, charged, conducting sphere of radius R is inverted to give the potential of a point charge a distance d away from an infinite, flat, conducting surface.
made that of a point charge near a grounded conducting sphere, handled by images in Section 2.2. The explicit verification of this is left to Problem 2.9.
A very interesting use of inversion was made by Lord Kelvin in 1847. He calculated the charge densities on the inner and outer surfaces of a thin, charged, conducting bowl made from a sphere with a cap cut out ofit. The potential distribution which he inverted was that of a thin, flat, charged, circular disc (the charged disc is discussed in Section 3.12). As the shape of the bowl is varied from a shallow watch glass-like shape to an almost closed sphere, the charge densities go from those of the disc to those of a closed sphere, in the one limit being almost the same inside and out, but concentrated at the edges of the bowl, and in the other limit being almost zero on the inner surface and uniform over the outer surface. Numerical values are given in Kelvin's collected papers, p. 186, and in Jeans, pp. 250-251.
2.7 Green's Function for the Sphere; General Solution for the Potential
ln preceding sections the problem of a conducting sphere in the presence of a point charge has been discussed by the method of images. As was mentioned in Section 1.10, the potential due to a unit charge and its image (or images), chosen to satisfy homogeneous boundary conditions, is just

[Sect. 2.7]

Boundary-Value Problems in Electrostatics: I

41

the Green's function (1.43 or 1.45) appropriate for Dirichlet or Neumann

boundary conditions. In G(x, x') the variable x' refers to the location P'

of the unit charge, while the variable xis the point Pat which the potential

is being evaluated. These coordinates and the sphere are shown in Fig.

2.11. For Dirichlet boundary conditions on the sphere of radius a the
potential due to a unit charge and its image is given by (2.1) with q = I

and relations (2.4). Transforming variables appropriately, we obtain the

Green's function:

G(x, x') = - -1 -
fx - x'I

a
a2
x' x - -x'
x'2

(2.22)

In terms of spherical coordinates this can be written:

+ G(x, x') =
(x2

1

x'2 -

2xx'

cos

y f1 , 2

1
( ~:'2 + a' - 2xx' cos yr

(2.23)
where y is the angle between x and x'. The symmetry in the variables x
and x' is obvious in the form (2.23), as is the condition that G = 0 if either x
or x' is on the surface of the sphere.

z

p

y

X
Fig. 2.11

42

Classical Electrodynamics

For solution (1.44) of Poisson's equation we need not only G, but also 'aG/on'. Remembering that n' is the unit normal outwards from the volume of interest, i.e., inwards along x' toward the origin, we have

oG

=

on' a:'==a

-

(xz - a2)

---
a(x2

-+ -

--a2 -

------,,2ax cosy)¾

(2.24)

[Note that this is essentially the induced surface-charge density (2.5).] Hence the solution of Laplace's equation outside a sphere with the potential specified on its surface is, according to (1.44),

(J)(x) = l._ f<l>(a, ()', cf,')

a(x2 - a2) a dO.'

4rr

+ (x2 a2 - 2ax cos yfA.

(2.25)

where dO.' is the element of solid angle at the point (a, ()', ef,') and cos y = cos () cos ()' + sin 0 sin 01 cos (cf, - cf,'). For the interior problem, the
normal derivative is radially outwards, so that the sign of oG/on' is opposite to (2.24). This is equivalent to replacing the factor (x2 - a2) by (a2 - x2) in (2.25). For a prob]em with a charge distribution, we must add to (2.25) the appropriate integral in (1.44), with the Green's function (2.23).

2.8 Conducting Sphere with Hemispheres at Different Potentials
As an example of general solution for the potential outside a sphere with prescribed values of potential on its surface, we consider the con-
ducting sphere of radius a made up of two hemispheres separated by a small
insulating ring. The hemispheres are kept at different potentials. Jt will
suffice to consider the potentials as ± V, since arbitrary potentials can be
handled by superposition of the solution for a sphere at fixed potential
over its whole surface. The insulating ring lies in the z = 0 plane, as shown in Fig. 2.12, with the upper (lower) hemisphere at potential + V
(-V).
z

-v
Fig. 2.12

[Sect. 2.8]

Boundary-Value Problems in Electrostatics: I

43

From (2.25) the solution for <l>(x, 0, cf,) is given by the integral:

i {J: = - <D(x, 0, </>)

V

2
ir def,'

4rr o

1d(cos 0') -
o

Jo d(cos
-1

0') )

(a 2

+

a(x2 x2 -

- a2)

%

2ax cosy)

(2.26)

By a suitable change of variables in the second integral (0' -. 1r - (}',
cf,' - </>' + 1r), this can be cast in the form:

<I>(x, 0, ¢,)=-Vaa.(.x.2..---a-2'-) i2ird</>'11d(cos 0')[(a2 + x2 - 2axcos y)-½s

41r

O

0

- (a2 + x2 + 2axcos y)-½] (2.27)

Because of the complicated dependence of cosy on the angles (O', cf,') and

(0, ¢,), equation (2.27) cannot in general be integrated in closed form.

As a special case we consider the potential on the positive z axis. Then

cos y = cos 0' since () = 0. The integration is elementary, and the

potential can be shown to be

[1 - <l>(z) = V

(z2 - a2) ]

z✓z2 + a2

(2.28)

At z = a, this reduces to <I> = Vas required, while at large distances it
goes asymptotically as <I> ~ 3 Va2/2z,2•
In the absence of a closed expression for the integrals in (2.27), we can
expand the denominator in power series and integrate term by term.
Factoring out (a2 + x2) from each denominator, we obtain

<V(x,

0,

cf,)=

Va(x2 -
41r(x2 +

a2)
a )2 3 , 72

L2irdef,'11d(cos0')[(1-

o

o

2a.cosy)-%a

- (1 + 2e<cosy)-%] (2.29)

where oc = ax/(a2 + x2). We observe that in the expansion of the radicals
only odd powers of oc co~ y will appear:

[(1 - 2oc cosy)-½ - (l + 2cx cosy)-½] = 6oc cosy + 35oc3 cos3 y + · · ·
(2.30)

It is now necessary to integrate odd powers of cosy over def,' d(cos O'):

i2ir

(1

0 d</,'Jo d(cos 0') cosy = 1T' cos 0

(2.31)

J:2irdef,' ild(cos 0') cos3 y = -71' cos 0(3 - cos2 0)

0

0

4

44

Classical Electrodynamics

If (2.30) and (2.31) are inserted into (2.29), the potential becomes

<l>(x,

0,

</>)

=

-3 V a2
2x-2

(x(3x(x2 2+-a2a)2%))

cos

(J

x

[ 1

+

-35
24

-(a-2a+-2x-2x2() 23

-

cos2 0) + · · ·]

(2.32)

We note that only odd powers ofcos 0 appear, as required by the symmetry
of the problem. If the expansion parameter is (a2/x2), rather than !X2, the
series takes on the form:
t;: <l>(x,0, cp) = 3~~2[cosO- 2 (icos3 0- ~cos0) + ••·] (2.33)

For large values of x/a this expansion converges rapidly and so is a useful
representation for the potential. Even for x/a = 5, the second term in the
series is only of the order of 2 per cent. It is easily verified that, for
cos 0 = 1, expression (2.33) agrees with the expansion of (2.28) for the
potential on the axis. [The particular choice of angular factors in (2.33) is
dictated by the definitions of the Legendre polynomials. The two factors
are, in fact, Pi(cos 0) and Pa(cos 0), and the expansion of the potential is
one in Legendre polynomials of odd order. We shall establish this in a systematic fashion in Section 3.3.]

2.9 Orthogonal Functions and Expansions

The representation of solutions of potential problems (or any mathematical physics problem) by expansions in orthogonal functions forms a powerful technique that can be used in a large class of problems. The particular orthogonal set chosen depends on the symmetries or near symmetries involved. To recall the general properties of orthogonal functions and expansions in terms of them, we consider an interval (a, b)
= in a variable ~ with a set of real or complex functions U,,l~), n 1, 2, ... ,
orthogonal on the interval (a~ b). The orthogonality condition on the functions Un(~) is expressed by

d; f.b Un*(;)Um(;) = 0, m =I= n

(2.34)

If n = m, the integral is finite. We assume that the functions are normal-
ized so that the integral is unity. Then the functions are said to be orthonormal, and they satisfy

d~ f.b Un'''(~)Um(~) = dnm

(2.35)

[Sect. 2.9]

Boundary- Value Problems in Electrostatics: I

45

An arbitrary functionf(~), square integrable on the interval (a, b), can be expanded in a series of the orthonormal functions Un(;). If the numher of terms in the series is finite (say N),

N
Ja) 2 f----', anUn(~) n=l

(2.36)

then we can ask for the "best" choice of coefficients an so that we get the "best" representation of the function f(~). If "best" is defined as minimizing the mean square error M~,T :

MN = f.b f(~) - nitnu nC~)l 2 d~

(2.37)

it is easy to show that the coefficients are given by

an = f.b Un*(;) f(fl d~

(2.38)

where the orthonormality condition (2.35) has been used. This is the standard result for the coefficients in an orthonormal function expansion.
If the number of terms Nin series (2.36) is taken larger and larger, we intuitively expect that our series representation off(~) is Hbetter" and "better." Our intuition will be correct provided the set of orthonormal functions is complete, completeness being defined by the requirement that
there exist a finite number N0 such that for N > N0 the mean square error
MN can be made smaller than any arbitrarily small positive quantity. Then the series representation
(2.39)

with a11 given by (2.38) is said to converge in the mean to/(;). Physicists generally leave the difficult job of proving completeness of a given set of functions to the mathematicians. All orthonormal sets of functions normally occurring in mathematical physics have been proved to be complete.
Series (2. 39) can be rewritten with the explicit form (2.38) for the coefficients an:
(2.40)

Since this represents any function J(t) on the interval (a, b), it is clear that
the sum of bilinear terms Un*(t) UnC~) must exist only in the neighborhood
of ,;' = ; . In fact, it must be true that

I00
Un*(r)Un(;) = o(t - t)
n=l

(2.41)

46

Classical Electrodynamics

This is the so-called completeness or closure relation. It is analogous to the

orthonormality condition (2.35), except that the roles of the continuous

variable~ and the discrete index n have been interchanged.

The most famous orthogonal functions are the sines and cosines, an

expansion in terms of them being a Fourier series. If the interval in xis

( - a/2, a/2), the orthonormal functions are

A 2

sin

(

":x),

A 2":~

cos

(

where m is an integer. The series equivalent to (2.39) is customarily written in the form:

where

f(x) = ½A0 + ~L [ Am cos (2-1rm-x) + Bm sm• (2-1r- mx)J (2.42)

m=l

a

a

(2 Am= -2 Ja/2 .f(x) cos -17m-x) dx

a -a/2

a

~ Bm = Ja/2 f(x) sin ( 21TmX) dx

a -a/2

a

(2.43)

If the interval spanned by the orthonormal set has more than one dimension, formulas (2.34)-(2.39) have obvious generalizations. Suppose that the space is two dimensional, and that the variable ~ ranges over the interval (a, b) while the variable 'Y/ has the interval (c, d). The orthonormal functions in each dimension are Un(~) and Vm('YJ). Then the expansion of

an arbitrary function/(~, 7/) is

where

L L /(~, 1/) =

llnmUnC~)Vm(1J)

n m

(2.44)

(2.45)

If the interval (a, b) becomes infinite, the set of orthogonal functions U,,.(~) may become a continuum of functions, rather than a denumerable set. Then the Kronecker delta symbol in (2.35) becomes a Dirac delta function. An important example is the Fourier integral. Start with the orthonormal set of complex exponentials,

,.fo Um(x) = ••(Z,=/•l

(2.46)

m = 0, ± 1, ±2, ... , on the interval (-a/2, a/2), with the expansion:

(2.47)

[Sect. 2.1 O]

Boundary- Value Problems in Electrostatics: I

47

where

.Ja Am = -1= fa/2 e-i(2,nnx'Ju) f(x') dx' -a/2

(2.48)

Then let the interval become infinite (a----+ co), at the same time transforming

2-f"° f dm = .!:_ 00 dk

m

-oo

21r -co

fo- Am----+

A(k)

a

(2.49)

The resulting expansion, equivalent to (2.47), is

= f ( x)

l

f00 A(k)e1b dk

..j2Tr -a,

where

(2.50)

A(k) =--;1= Jco e-ik. x f(x) dx
✓ 2Tr -oo

(2.51)

The orthogonality condition is

f_!_

dx 00 ei(k-k');e

=

o(k -

k')

2Tr - 00

(2.52)

while the completeness relation is
f = _!_ 00 eik(:r-x') dk d(x - x')
21r - co

(2.53)

These last integrals serve as convenient representations of a delta function. We note in (2.50)-(2.53) the complete equivalence of the two continuous variables x and k.

2.10 Separation of Variables; Laplace's Equation in Rectangular Coordinates
The partial differential equations of mathematical physics are often solved conveniently by a method called separation of variables. In the process, one often generates orthogonal sets of functions which are useful in their own right. Equations involving the three-dimensional Laplacian operator are known to be separable in eleven different coordinate systems

48

Classical Electrodynamics

(see Morse and Feshbach, pp. 509, 655). We will discuss only three of these

in any detail-rectangular, spherical, and cylindrical-and will begin with

the simplest, rectangular coordinates.

•

Laplace's equation in rectangular coordinates is

a2<1> + a2<I> + a2<I> = o
ax2 oy2 oz2

(2.54)

A solution of this partial differential equation can be found in terms of

three ordinary differential equations, all ofthe same form, by the assumption

that the potential can be represented by a product of three functions, one

for each coordinate:
<I>(x, y, z) = X(x) Y(y)Z(z)

(2.55)

Substitution into (2.54) and division of the result by (2.55) yields

-1-d+ 2X - -1 + d2Y - -1 = d2Z 0
X(x) dx2 Y(y) dy2 Z(z) dz2

(2.56)

where total derivatives have replaced partial derivatives, since each term

involves a function of one variable only. If (2.56) is to hold for arbitrary

values of the independent coordinates, each of the three terms must be

separately constant:

a x - 1

2
-

=

-oc2

X dx2

l_ d2 Y = -{32
Yd'!/'
-1 -d2= Z y2 Z dz2

(2.57)

where
If we arbitrarily choose oc2 and p2 to be positive, then the solutions of the
three ordinary differential equations (2.57) are exp (±iocx); exp (±i{Jy),
exp (±Voc2 + /J2z). The potential (2.55) can thus be built up from the
product solutions:
(2.58)
At this stage oc and {J are completely arbitrary. Consequently (2.58), by linear superposition, represents a very large class of solutions to Laplace's equation.
To determine oc and {J it is necessary to impose specific boundary conditions on the potential. As an example, consider a rectangular box, located as shown in Fig. 2.13, with dimensions (a, b, c) in the (x, y, z)

[Sect. 2, 10]

Boundary- Value Problems in Electrostatics: I

49

z

Fig. 2.13 Hollow, rectangular box with five sides at zero
potential, while the sixth (z = c)
bas the specified potential <I) =
V(x, y).

7

directions. All surfaces of the box are kept at zero potential, except the
surface z = c, which is at a potential V(x, y). It is required to find the

potential everywhere inside the box. Starting with the requirement that
cf>= 0 for x = 0, y = 0, z = 0, it is easy to see that the required forms of

X, Y, Z are

X = sin ocx

Y = sin {Jy

}

+ Z = sinh (Voc2 f32z)

(2.59)

= In order that <I> = 0 at x a and y = b, it is necessary that oca = mr and
{Jb = m1r. With the definitions,

f3

-
m-

m1r b

(2.60)

We can write the partial potential <l>nm; satisfying all the boundary conditions except one,

(2.61)

The potential can be expanded in terms ofthese (f>nm with initially arbitrary coefficients (to be chosen to satisfy the final boundary condition):

.200
<l>(x, y, z) = Anm sin (ocnx) sin (µmy) sinh (Ynmz)
n,m=l

(2.62)

50

Classical Electrodynamics

There remains only the boundary condition <I> = V(x, y) at z = c:

(X)
V(x, y) = ! Anm sin (~nx) sin (PmY) sinh (ynmc) n,m=l

(2.63)

This is just a double Fourier series for the function V(x, y). Consequently the coefficients Anm are given by:

A 11 m

=

ab

. 4
smh (y11mc)

fadx (1>dyV(x, Jo Jo

y)

sin

(oc11 x)

sin

(/JmY)

(2.64)

If the rectangular box has potentials different from zero on all six sides,

the required solution for the potential inside the box can be obtained by a

linear superposition of six solutions, one for each side, equivalent to (2.62)

and (2.64). The problem of the solution of Poisson's equation, i.e., the

potential inside the box with a charge distribution inside, as well as

prescribed boundary conditions on the surface, requires the construction of

the appropriate Green's function, according to (1.43) and (1.44). Discus-

sion of this topic will be deferred until we have treated Laplace's equation

in spherical and cylindrical coordinates. For the moment, we merely note

that solution (2.62) and (2.64) is equivalent to the surface integral in the

Green's function solution {1.44).

REFERENCES AND SUGGESTED READING
Images and inversion are treated in many books; among the better or more extensive discussions are those by
Jeans, Chapter VIII, Maxwell, Vol. 1, Chapter XI, Smythe, Chapters IV and V. A truly encyclopedic source of examples with numerous diagrams is the book by Durand, especially Chapters III and IV. Durand discusses inversion on pp. 107-114. Conformal mapping techniques for the solution oftwo-dimensional potential problems are discussed by Durand, Chapter X, Jeans, Chapter VIII, Sections 306-337, Maxwell, Vol. 1, Chapter XII, Smythe, Chapter IV, Sections 4.09-4.29. There are, in addition, many engineering books devoted to the subject, e.g., Rothe, Ollendorff, and Polhausen. Elementary, but clear, discussions of the mathematical theory of Fourier series and integrals, and orthogonal expansions, can be found in Churchill, Hildebrand, Chapter 5. A somewhat old-fashioned treatment of Fourier series and integrals, but with many examples and problems, is given by Byerly.

[Probs. 2]

Boundary-Value Problems in Electrostatics: I

51

PROBLEMS

2.1 A point charge q is brought to a position a distance d away from an infinite plane conductor held at zero potential. Using the method of images, find: (a) the surface-charge density induced on the plane, and plot it; (b) the force between the plane and the charge by using Coulomb's law for the force between the charge and its image; (c) the total force acting on the plane by integrating 211a2 over the whole plane; (d) the work necessary to remove the charge q from its position to infinity; (e) the potential energy between the charge q and its image [compare the answer to (d) and discuss]. (f) Find answer (d) in electron volts for an electron originally one angstrom from the surface.
2.2 Using the method of images, discuss the problem of a point charge q
inside a hollow, grounded, conducting sphere of inner radius a. Find (a) the potential inside the sphere; (b) the induced surface-charge density; (c) the magnitude and direction of the force acting on q.
Is there any change in the solution if the sphere is kept at a fixed potential V? If the sphere has a total charge Q on it?
2.3 Two infinite, grounded, conducting planes are located at x = a/2 and x = -a/2. A point charge q is placed between the planes at the point
(x', y', z'), where -(a/2) < x' < (a/2). (a) Find the location and magnitude of all the image charges needed to
satisfy the boundary conditions on the potential, and write down the Green's function G(x, x').
(b) If the charge q is at (x', 0, 0), find the surface-charge densities induced on each conducting plane and show that the sum of induced charge on the two planes is -q.
2.4 Consider a potential problem in the half-space defined by z > 0, with
Dirichlet boundary conditions on the plane z = 0 (and at infinity).
(a) Write down the appropriate Green's function G(x, x').
(b) If the potential on the plane z = 0 is specified to be <I> = V inside a circle of radius a centered at the origin, and ~ = 0 outside that circle, find
an integral expression for the potential at the point P specified in terms of cylindrical coordinates (p, rf,, z).
(c) Show that, along the axis of the circle (p = 0), the potential is given by

<I> = V ( 1 - -V--a:=2==z+==•=z;2:)

>- (d) Show that at large distances (p2 + z2 ► a2) the potential can be
expanded in a power series in (p2 + z2 1, and that the leading terms are

r Va2 z

3a2

5(3p2a2 + a4)

]

L cl> = 2 (p2 + z2)¾ 1 - 4(p2 + z2) + 8(p2 + z2)2 + • • •

52

Classical Electrodynamics

Verify that the results of (c) and (d) are consistent with each other in their common range of validity.
2.5 An insulated, spherical, conducting shell of radius a is in a uniform electric field £ 0• If the sphere is cut into two hemispheres by a plane perpendicular to the field, find the force required to prevent the hemispheres from separa• ting (a) if the shell is uncharged; (b) if the total charge on the shell is Q.
2.6 A large parallel plate capacitor is made up of two plane conducting sheets, one of which has a small hemispherical boss of radius a on its inner surface. The conductor with the boss is kept at zero potential, and the other conductor is at a potential such that far from the boss the electric field between the plates is £ 0 . (a) Calculate the surface.charge densities at an arbitrary point on the pJane and on the boss, and sketch their behavior as a function of distance (or angle). (b) Show that the total charge on the boss has the magnitude 3E,p2/4. (c) If, instead of the other conducting sheet at a different potential, a point charge q is placed directly above the hemispherical boss at a distance d from its center, show that the charge induced on the boss is

q' = -q[1 -

d~:;:

2 a2]

2.7 A line charge with linear charge density -r is placed paraUel to, and a distance R away from, the axis of a conducting cylinder of radius b held at fixed voltage such that the potential vanishes at infinity. Find (a) the magnitude and position of the image charge(s); (b) the potential at any point (expressed in polar coordinates with the line from the cylinder axis to the line charge as the x axis), including the asymptotic form far from the cylinder; (c) the induced surface.charge density, and plot it as a function of angle
for R/b = 2, 4 in units of -r/27Tb;
(d) the force on the charge.
2.8 (a) Find the Green's function for the two•dimensionaJ potential problem with the potential specified on the surface of a cylinder of radius b, and show that the solution inside the cylinder is given by Poisson's integral:

1 f2,,.

b2 - 2

+ ll'.l(r, 0) = 21r Jo ll'.l(h, 0') b2 r2 - 2hr ;os (0' - 0) dO'

(b) Two halves of a long conducting cylinder of radius b are separated

by a small gap, and are kept at different potentials V1 and V2. Show that

the potential inside is given by

e) <I>(r, (J)

-_

V1 +
2

V2

+-V--1 --"-V--2tan_ 1
7T

(

b22hr -

r

2

cos

where (J is measured from a plane perpendicular to the plane through the gap.
(c) Calculate the surface-charge density on each half of the cylinder. (d) What modification is necessary in (a) if the potential is desired in the region of space bounded by the cylinder and infinity?

[Probs. 2]

Boundary-Value Problems in Electrostatics: 1

53

2.9 (a) An isolated conducting sphere is raised to a potential V. Write down the (trivia]) solution for the electrostatic potential everywhere in space. (b) Apply the inversion theorem, choosing the center of inversion outside the conducting sphere. Show explicitly that the solution obtained for the potential is that of a grounded sphere in the presence of a point charge of magnitude - VR, where R is the inversion radius. (c) What is the physical situation described by the inverted solution jf the center of inversion li~s inside the conducting sphere?
2.10 Knowing that the capacitance of a thin, fl.at, circular, conducting disc of radius a is (2/7T)a and that the surface-charge density on an isolated disc raised to a given potential is proportional to (a2 - r 2)-112, where r is the distance from the center of the disc, (a) show that by inversion the potential can be found for the problem of an infinite, grounded, conducting plane with a circular hole in it and a point charge lying anywhere in the opening; (b) show that, for a unit point charge at the center of the opening, the induced charge density on the plane is

u(r0,1.)=-

a

,,.,.,

1212rAv1r2 - a2

(c) show that (a) and (b) are a special case of the general problem, obtained by inversion of the disc, of a grounded, conducting, spherical bowl under the influence of a point charge located on the cap which is the complement of the bowl.
2.11 A hollow cube has conducting walls defined by six planes x = y = z = 0,
and x = y = z = a. The walls z = 0 and z = a are held at a constant
potential V. The other four sides are at zero potential. (a) Find the potential ct>(x, y, z) at any point inside the cube. (b) Evaluate the potential at the center of the cube numerically, accurate
to three significant figures. How many terms in the series is it necessary to keep in order to attain this accuracy? Compare your numerical result with the average value of the potential on the walls.
(c) Find the surface-charge density on the surface z = a.

3

Boundary-Value Problems in Electrostatics: II

In this chapter the discussion of boundary-value problems is continued. Spherical and cylindrical geometries are first considered, and solutions of Laplace's equation are represented by expansions in series of the appropriate orthonormal functions. Only an outline is given of the solution of the various ordinary differential equations obtained from Laplace's equation by separation of variables, but an adequate summary of the properties of the different functions is presented.
The problem of construction of Green's functions in terms of orthonormal functions arises naturally in the attempt to solve Poisson's equation in the various geometries. Explicit examples of Green's functions are obtained and applied to specific problems, and the equivalence of the various approaches to potential problems is discussed.

3.1 Laplace's Equation in Spherical Coordinates

In spherical coordinates (r, 0, cp), shown in Fig. 3.1, Laplace's equation can be written in the form:

!
r

~
or2

(r<l>)

+

r2

1 sin

Oio_0(sin

(}

oa<oP)

+

r2

1 sin2

(}

0o2cp<12>

=

0

(3.1)

If a product form for the potential is assumed, then it can be written:

<I) = U(r) P(O)Q(cp)

(3.2)

r

54

[Sect. 3.1]

Boundary-Value Problems in Electrostatics: II

55

z

X

I

y

' ,' Il
'..J Fig. 3.1

When this is substituted into (3 .1), there results the equation:

PQ d2U + UQ i.(sin 0 dP) + UP d2Q = 0
dr2 r2 sin Od0 , d0 r2 sin2 0 d,f}
If we multiply by r2 sin2 0/UPQ, we obtain:

r2 sin2 0[ -1 d- 2U + - -1 - -d ( sin () d- P)] + -1 -d2Q = 0 (3.3)

U dr2 r2 sin OP d0

d(}

Q dcf,2

+ The dependence of the equation has now been isolated in the last term.

Consequently that term must be a constant which we call (-m2):

(3.4)

This has solutions

Q = e±im<,i,

(3.5)

In order that Q be single valued, m must be an integer. By similar considerations we find separate equations for P(O) and U(r):

_l_ !!._(sin 0 dP) + [1(l + 1) - ~ J P = 0

(3.6)

~0~

~

~~0

d2U - l(l + l) U = 0

(3.7)

dr2

r2

where /(/ + 1) is another real constant.
From the form of the radial equation it is apparent that a single power of r (rather than a power series) will satisfy it. The solution is found to be:

-U = ArZ+l + B,-z

(3.8)

but / is as yet undetermined.

56

Classical Electrodynamics

3.2 Legendre EcJ.uation and Legendre Polynomials

The () equation for P(O) is customarily expressed in terms of x = cos 0,
instead of () itself. Then it takes the form:

)p ·.!!._ (c1 - x2) dP) + (10 + 1) - m2 = 0

(3.9)

dx

dx

1 - x2

This equation is called the generalized Legendre equation, and its solutions are the associated Legendre functions. Before considering (3.9) we will
outline the solution by power series of the ordinary Legendre differential
equation with m2 = O:

.E._((l - x2) dP) + l(l + l)P = 0

dx

dx

(3.10)

The desired solution should be single valued, finite, and continuous on the
interval -1 < x < 1 in order that it represents a physical potential. The
solution will be assumed to be represented by a power series of the form:

00
P(x) = xi%! a1xi
:i=O

(3.11)

where oc is a parameter to be determined. When this is substituted into (3.10), there results the series:

!0C
{(ri. + j)(ri. + j - l)a;xll+:i-2
:i=O
- [(oc + j)(oc + j + 1) - l(l + l)]a:i~+ 1} = 0 (3.12)

In this expansion the coefficient of each power of x must vanish separately.
For j = 0, I we find that

if a0 -=I= 0, then oc(rx - 1) = 0 } if a1 -::/= 0, then oc(rx + 1) = 0
while for a general j value

(3.13)

a;+2

=

[ (oc

+ j)(oc + j + 1) (oc + j + l)(oc + j

l(l + + 2)

1)]
a1

(3.14)

A moment's thought shows that the two relations (3.13) are equivalent and
that it is sufficient to choose either a0 or a1 different from zero, but not both.
Making the former choice, we have :x = 0 or :x = 1. From (3.14) we see that the power series has only even powers of x(~ = 0) or only odd powers of x(oc = I).

[Sect. 3.2]

Boundary-Value Problems in Electrostatics: II

57

For either of the series oc = 0 or oc = 1 it is possible to prove the
following properties:
(a) the series converges for x2 < I, regardless of the value of l; (b) the series diverges at x = ± 1, unless it terminates.
Since we want a solution that is finite at x = ± 1, as well as for x2 < 1, we
demand that the series terminate. Since oc and j are positive integers or zero, the recurrence relation (3.14) will terminate only if l is zero or a
positive integer. Even then only one of the two series converges at x = ± 1. = If/ is even (odd), then only the <X. = 0 (ix I) series terminates.* The
polynomials in each case have xi as their highest power of x, the next
highest beingx1- 2, and so on, down to :tJ(x)for /even(odd). By convention
these polynomials are normalized to have the value unity at x = + 1 and
are called the Legendre polynomials of order /, Pi(x). The first few
Legendre polynomials are:

P0(x) = 1
Pi(x) = x
= Plx) ½(3x2 - 1) = P3(x) l(5x3 ~ 3x)
= Plx) l(35x4 - 30x2 + 3)

(3.15)

By manipulation of the power series solutions (3.11) and (3.14) it is possible to obtain a compact representation of the Legendre polynomials, known as Rodrigues' formula:

Pi(x) = J.. ..E:.. (x2 - J)1
211! dx1

(3.16)

[This can be obtained by other, more elegant means, or by direct /-fold integration of the differential equation (3. I0).]
The Legendre polynomials form a complete orthogonal set of functions on the interval -1 < x s:;; 1. To prove the orthogonality we can appeal directly to the differential equation (3. JO).. We write down the differential equation for P1(x), multiply by Pr(x), and then integrate over the interval:

f1 P -1

1,(x)[.dEx._((t

-

x2) dPi) + l(l + 1)Pi(x)] dx = 0
dx

(3.17)

• For example, if l == 0 the ex == 1 series has a geneml coefficient a; = a0/j + 1 for
/ = 0, 2, 4, .... Thus the series is a0(x + -j-x3 + ¼xJ + •• •.) This is just the power

series expansion of a function

Q0(x)

==

½ln

( 1 + x), which clearly diverges atx
1-x

==

±1.

For each / value there is a similar function Qi(x) with logdl'ithms in it as the partner to

the well-behaved polynomial solution. See Magnus and Oberhettinger, p. 59.

58

Classical Electrodynamics

Integrating the first term by parts, we obtain

J1 [ (x2 - 1)d-Pz _ dPi, + l(l + l)Pi,(x)Pz(x)] dx = 0

-1

dx dx

(3.18)

ff we now write down (3.18) with / and l' interchanged and subtract it from

(3.18), the result is the orthogonality condition:
J~ [l([ + 1) - l'(l1 + 1)] 1Pz,(x)Pi(x) dx = 0

(3.19)

For /-;/= I', the integral must vanish. For I = l', the integral is finite. To
determine its value it is necessary to use an explicit representation of the Legendre polynomials, e.g., Rodrigues' formula. Then the integral is

explicitly:

J J 1 [Pz(x)]2 dx = 1 1 .!!:._ (x2 - 1)1-~-\ (x2 - ll dx

-1

22 z(l!)2 -1 dxi

dx

Integration by parts/ times yields the result:

J J tY = 1 [Pi(x)]2 dx

(-l)z

(x 1 2 -

-1

221(/ !)2 -1

d2 z (x2 -
dx2i

1Ydx

The differentiation of (x2 - l)z 2/ times yields the constant (2/)!, so that

J f 1 [Pi(x)J2 dx = (~l)! 1 (1 - x2)i dx

(3.20)

-1

22 (/!)2 -I

The remaining integral is easily shown to be 22 H-1 (1!)2 /(2/ + 1) ! Con-

sequently the orthogonality condition can be written:

f1 P1,(x)Pi(x) -1

dx

=

21

2
+

1

bn

(3.21)

and the orthonormal functions in the sense of Section 2.9 are

(3.22)

Since the Legendre polynomials form a complete set of orthogonal functions, any function f (x) on the interval -1 < x < l can be expanded in

I r

+11------f

I I

I I

-1 I

0

I 1

-------1-l Fig. 3.2

[Sect. 3.2]

Boundary-Value Problems in Electrostatics: II

59

terms of them. The Legendre series representation is:

where

00
f(x) = IAiPi(x)
l=O
Ai= - 2[ +-1J1 j(x)P.(x) dx
2 -1

As an example, consider the function shown in Fig. 3.2:

f (x) = + I for x > 0

= -1 for X < 0

Then

[i f~ Ai= 21 ; I 1Pi(x) dx - 1Pl(x) dx]

(3.23) (3.24)

Since Pi(x) is odd (even) about x = 0 if l is odd (even), only the odd l
coefficients are different from zero. Thus, for / odd,

(3.25)

By means of Rodrigues' formula the integral can be evaluated, yielding

Ai= (- ~r-1)12 (21 + l)(l ~ 2)!!

(3.26)

2

2([: 1)1

where (2n + I)!! - (2n + 1)(2n - 1)(2n - 3) • • • X S X 3 X I. Thus the
series for f (x) is:
(3.27)
Certain recurrence relations among Legendre polynomials of different order are useful in evaluating integrals, generating higher-order polynomials from lower-order ones, etc. From Rodrigues' formula it is a straightforward matter to show that

dPl+l - dPi-1 - (21 + l)Pi = 0

dx

dx

(3.28)

This result, combined with differential equation (3.10), can be made to yield various recurrence formulas, some of which are:

(l + 1)P1+1 - (21 + 1)xP1 + lPz-i = 0

dPl+l - X dPi - (l + l)Pz = 0

dx

dx

(3.29)

60

Classical Electrodynamics

As an illustration of the use of these recurrence formulas consider the

f~ evaluation of the integral:

= 11

1 xPi(x)Pi,(x) dx

(3.30)

From the first of the recurrence formulas (3.29) we obtain an expression

for xP(x). Therefore (3.30) becomes

+ J + + = / 1 l

1 P1,(x)[(l l)P1+i(x) IP1_ 1(x)] dx

21 1 -1

The orthogonality integral (3.21) can now be employed to show that the
integral vanishes unless l' = l ± I, and that, for those values,

J lr l

(2[ +2(l1)+(2[1)+ 3)'

xPi(x)Pz-(x) dx =

-1

2l

(21 - 1)(21 + 1) '

L' = l + 1
= l' 1 - 1

(3.31)

These are really the same result with the roles of land/' interchanged. In a similar manner it is easy to show that

f = 1 x2Pi(x)Pr(x) dx
., -1

2(1 + 1)(/ + 2)
(2/ + 1)(2/ + 3)(2[ + 5) '
+ 2(212 2! - 1) (21 - 1)(21 + 1)(2l + 3)'

l'=l+2
l' = l

(3.32)

where it is assumed that /' > /.

3.3 Boundaey-Value Problems with Azimuthal Symmetry

From the form of the solution of Laplace's equation in spherical coordinates (3.2) it will be seen that, for a problem possessing azimuthal
symmetry, m = 0 in (3.5). This means that the general solution for such
a problem is:

= I + 00

<I>(r, 0)

[A 1ri B1r-(z+l>]Pi(cos 0)

z=o

(3.33)

The coefficients Ai and BI can be determined from the boundary conditions. Suppose that the potential is specified to be V(O) on the surface of a
sphere of radius a, and it is required to find the potential inside the sphere.
If there are no charges at the origin, the potential must be finite there.
Consequently Bi = 0 for all/. The coefficients Ai are found by evaluating

[Sect. 3.3]

Boundary-Value Problems in Electrostatics: II

61

(3.33) on the surface of the sphere:

I00
V(0) = Aia1Pi(cos 0)
z=o

(3.34)

This is just a Legendre series of the form (3.23), so that the coefficients Ai

are:

+ A1 =

- 21 -1J1TV(O)Pi(cos 0) sm• 0 d0 2al 0

(3.35)

If, for example, V(O) is that of Section 2.8, with two hemispheres at equal and opposite potentials,

!+V, 0~0<!

V(0) =

2

-V, 1-T<0<71'

2

(3.36)

then the coefficients are proportional to those in (3.27). Thus the potential inside the sphere is :

r = cl>(r, 0} [ 3 Pi(cos

V

-

-

0)

-

2 a

r) r) + - - 7( P (cos 3

-

-

3

0)

11 ( 5Ps(cos

0)

-

8 a

16 a

]

•

•

•

(3.37)

To find the potential outside the sphere we merely replace (r/a)' by (a/r)'+I. The resulting potential can be seen to be the same as (2.33), obtained by another means.
Series (3.33), with its coefficients determined by the boundary conditions, is a unique expansion of the potential. This uniqueness provides a means of obtaining the solution of potential problems from a knowledge of the potential in a limited domain, namely on the symmetry axis. On the
symmetry axis (3.33) becomes (with z = r):

= I [ 00
<l>(z r) = Air' + B,r-U+I>]
l=O

(3.38)

valid for positive z. For negative z each term must be multiplied by (-l)i. Suppose that, by some means, we can evaluate the potential ct>(z) at an arbitrary point z on the symmetry axis. If this potential function can be expanded in a power series in z = r of the form (3.38), with known coefficients, then the solution for the potential at any point in space is obtained by multiplying each power of rl and ,-<lH) by Pi{cos 0).

62

Classical Electrodynamics

z
x'

X

I
I

I

I y

--..,_ ---....,

I

"- -----1--..

I

'- I ----...J

X

''-1

Fig. 3.3

At the risk of boring the reader we return to the problem of the hemi-

spheres at equal and opposite potentials. We have already obtained the

series solution in two different ways, (2.33) and (3.37). The method just

stated gives a third way. For a point on the axis we have found the closed

form (2.28):

+ <D(z = r) = V [ 1 - r2 - a2 ] r.J r2 a2

This can be expanded in powers of a2/r2:
! <l>(z = r) = ~ (-1)1-1 (2] ✓ 'TT' j=I

½)_~1(j -
] •

½) (~)21 r

(3.39)

Comparison with expansion {3.38) shows that only odd l values
= (1 2j - 1) enter. The solution, valid for all points outside the sphere,

! r is consequently: <l>(r, 0) = ~-

~ (-1y-iC2j - ½)_~u - ½) ( P:H-1(cos 0) (3.40)

✓ 1T j=l

}•

r,

This is the same solution as already obtained, (2.33) and (3.37). An important expansion is that of the potential at x due to a unit point
charge at x':

(3.41)

where r < (r>) is the smaller (larger) of !xi and lx'I, and y is the angle
between x and x', as shown in Fig. 3.3. This can be proved by rotating axes so that x' lies along the z axis. Then the potential satisfies Laplace's
equation, possesses azimuthal symmetry, and can be expanded according
to (3.33), except at the point x = x':

_L + 00

1 ,=

(A 1rz B1r-F+I))Pi(cos y)

Ix - x I z=o

(3.42)

[Sect. 3.3]

Boundary-Value Problems in Electrostatics: II

63

If the point xis on the z axis, the right-hand side reduces to (3.38), while
the left-hand side becomes:

1

1

1

= jx - x'l + (r2 r'2 - 2rr' cosy)½ - Ir - r'I

Expanding (3.43), we find

(3.43) (3.44)

For po1nts off the axis it is only necessary, according to (3.33) and (3.38),
to multiply each term in (3.44) by Pz(cos y). This proves the general result
(3.41).
Another example is the potential due to a total charge q uniformly distributed around a circular ring of radius a, located as shown in Fig. 3.4,
with its axis the z axis and its center at z = b. The potential at a point P on the axis of symmetry with z = r is just q divided by the distance AP:

<l>(z

=

r)

= {r2 + c2 -

q

•

2cr cos ix)½

(3.45)

where c2 = a2 + b2 and ix = tan-1 (a/b). The inverse distance AP can be
expanded using (3.41). Thus, for r > c,

oo
<l>(z = r) = q2

~~l 1 Pi(cos ix)

l=O f

(3.46)

z
z=r P
I / / I I / ------1-----,,...__
A

Fig. 3.4 Ring of charge of radius a and total
charge q located on the z axis with center at
z = b.

64

Classical Electrodynamics

For r < c, the corresponding form is:

<X)

i

<I>(z = r) = qL ;+I P,(cos ex)

t=O C

(3.47)

The potential at any point in space is now obtained by multiplying each member of these series by Pz(cos 0):

2 :: 00 z

<l>(r, 0) = q

1 Pz(cos oc)Pi(cos 0)

z=o r>

(3.48)

where r < (r>) is the smaller (larger) of rand c.

3.4 Associated Legendre Polynomials and the Spherical Harmonics Yim(0,rf,)
So far we have dealt with potential problems possessing azimuthal symmetry with solutions of the form (3.33). These involve only ordinary Legendre polynomials. The general potential problem can, however, have azimuthal variations so that m =/:- 0 in (3.5) and (3.9). Then we need the generalization of Pi(cos 0), namely, the solution of (3.9) with land m both arbitrary. In essentially the same manner as for the ordinary Legendre functions it can be shown that in order to have finite solutions on the
interval - I < x < 1 the parameter / must be zero or a positive integer and
that the integer m can take on only the values -/, -(/ - 1), ... , 0, ... , (/ - 1), /. The solution having these properties is ca11ed an associated Legendre function Pim(x). For positive m it is defined by the formula*:
(3.49)
If Rodrigues' formula is used to represent Plx), a definition valid for both positive and negative m is obtained:
(3.50)
* The choice of phase for P1m(x) is that of Magnus and Oberhettinger, and of E. U. Condon and G. H. Shortley in Theory of Atomic Spectra, Cambridge University Press (1953). For explicit expressions and recursion formulas, see Magnus and Oberhettinger,
p. 54.

[Sect. 3.4]

Boundary- Value Problems in Electrostatics: I/

65

Pi-m(x) and Pr(x) are proportional, since differential equation (3.9) depends only on m2 and m is an integer. It can be shown that

P -m(x)
L

=

(-l)m

(l (l

-+

m)! m)!

p m(x)
l

(3.51)

For fixed m the functions P,m(x) form an orthogonal set in the index I

on the interval -1 < x < 1. By the same means as for the Legendre

functions the orthogonality relation can be obtained:

f1 Pi-m(x)P,m(x) dx -1

=

- 21 2+-•1

(-l-+-
(l -

-m-)-! m)l

bn

(3.52)

The solution of Laplace's equation was decomposed into a product of factors for the three variables,, 0, and r/,. It is convenient to combine the angular factors and construct orthonormal functions over the unit sphere. We will call these functions spherical harmonics, although this terminology is often reserved for solutions of the generalized Legendre equation (3.9).
Our spherical harmonics are sometimes called "tesseral harmonics'' in
older books. The functions Qm(</>) = eimr/> form a complete set of ortho-
gonal functions in the index m on the interval 0 < r/, < 21r. The functions
P,""(cos 0) form a similar set in the index I for each m value on the interval
-1 <cos(} < I. Therefore their product PrQ,n will form a complete
orthogonal set on the surface of the unit sphere in the two indices l, m.
From the normalization condition (3.52) it is clear that the suitably normalized functions, denoted by Y1m(O, r/,), are:

y; (8
zm ,

..J..)
~

=

J2l +
4rr

1 (l
( l

-+

m)! m) !

p m(cos z

0)ei·m,t,

(3.53)

From (3.51) it can be seen that

(3.54)

The normalization and orthogonality conditions are
J2irdr/,f.1rsin 0 dOYtm,(O, </>)Yzm(0, r/,) = !5nc5m'm 0

(3.55)

The completeness relation, equivalent to (2.41), is

2 2 ro l
Yi!(O', ¢,') Yzm(0, r/,) = 6(r/, - ¢/)/J(cos 0 - cos 0') (3.56)
i=om==-l
For a few small I values and m > 0 the table shows the explicit form of the Y1m(0, cf>). For negative m values (3.54) can be used.

66

Classical Electrodynamics

Spherical harmonics Yim(O, cp)

l=O

l
YooV=4-rr

l= 1

Y11 = - ✓{s3; sin Oei<f,
- f l Y10 - _ cos 0 4rr

/=2

l = 3

! = Y33 - 4 ✓{~ii sin3 0e3i<f,
Ya2 = -1 ✓-105 sin2 0 cos ee2i<.1,
4 2rr

! Y31 = - 4 ✓{7i;i; sin 0(5 cos2 0 - l)ei</,

R ( Y30 = _ -5 cos3 0 - -3 cos 0)

7T 2

2 ,

Note that, for m = 0,
YioC0, cp) = ;J1~ 21P+z(co1s 0.) 41r

(3.57)

An arbitrary function g(0, cp) can be expanded in spherical harmonics:

L 2 ro l

g(0, <p) =

Aim Yzm(0, <p)

Z=o m= -l

where the coefficients are

(3.58)

[Sect. 3.5]

Boundary-Value Problems in Electrostatics: II

67

A point of interest to us in the next section is the form of the expansion
for() = 0. With definition (3.57), we find:

where

[g(O,

cp)]o=o

=~ ~J

2

1

+
477

1 Aio

(3.59) (3.60)

All terms in the series with m =I= 0 vanish at 0 = 0.
The general solution for a boundary-value problem in spherical coordinates can be written in terms of spherical harmonics and powers of , in a generalization of (3.33):

+ rJJ

l

<P(r, 0, <p) = ,2 _2 [Azmri Bzm,-(i+l)J Yzm(O, <p)

l"'O m= -L

(3.61)

If the potential is specified on a spherical surface, the coefficients can be determined by evaluating (3.61) on the surface and using (3.58).

3.5 Addition Theorem for Spherical Harmonics
A mathematical result of considerable interest and use is called the addition theorem for spherical harmonics. Two coordinate vectors x and x', with spherical coordinates (r, 0, cp) and (,1, ()', cp'), respectively, have an angle y between them, as shown in Fig. 3.5. The addition theorem expresses a Legendre polynomial of order l in the angle y in terms of
2
x'

X
Fig. 3.5

68

Classical Electrodynamics

products of the spherical harmonics of the angles fJ, cp and 0', cf/:

I Pi(cos y) = 417

l
Yi!(0', cp')Yim(0, cp)

+ 2/ 1 rn= -i

(3.62)

where cos y = cos (J cos 0' + sin fJ sin ()1 cos (cp - cf/). To prove this
theorem we consider the vector x' as fixed in space. Then Pi(cos y) is a function of the angles 0, cp, with the angles 0', cp' as parameters. It may be

expanded in a series (3.58):

I I 00

(

Pz(cos y) =

Ai'm(0', <//)Yi,m(0, cp)

l'=O m=-l'

(3.63)

Comparison with (3.62) shows that only terms with /' = l appear. To see
why this is so, note that, if coordinate axes are chosen so that x' is on the z
axis, then y becomes the usual polar angle and Pi(cos y) satisfies the

equation:

V'-,,Pz(cos y) + -l(-l -+P1i)(cos y) = 0 r2

(3.64)

where V'2 is the Laplacian referred to these new axes. If the axes are now
rotated to the position shown in Fig. 3.5, V'2 = V2 and r is unchanged.*
Consequently Pi(cos y) still satisfies an equation of the form (3.64); i.e., it is a spherical harmonic of order l. This means that it is a linear combination of Ylm's of that order only:

I l
Pi(cos y) = Am(0', cp')Yzm(0, cp) m= -l

(3.65)

The coefficients Am(0', cf/) are given by:

f Am(0', </>') = Y1!(0, ef,)Pz(cos y) dQ "

(3.66)

To evaluate this coefficient we note that it may be viewed, according to
(3.60), as the m' = 0 coefficient in an expansion of the function
v' 41r/(2l + 1) Yi:i(e, cp) in a series of Ylm'(y, /3) referred to the primed
axis of (3.64). From (3.59) it is then found that, since only one l value is

present, coefficient (3.66) is

Am(0', </>') = 2/ : l [Yz!(0(y, /3), rp(y, P))J,.=o

(3.67)

In the limit y ~ 0, the angles (0, </>), as functions of (y, {3), go over into
* The proof that V'2 = V2 under rotations follows most easily from noting that V2tp = V • V1 is an operator scalar product, and that all scalar products are invariant
under rotations.

[Sect. 3.6]

Boundary-Value Problems in Electrostatics: II

69

(O', ef,'). Thus addition theorem (3.62) is proved. Sometimes the theorem is written in terms of Pim(cos 0) rather than Yzm• Then it has the form:
= Pz(cos y) Pi(cos 0)Pi(cos B')

l
+2 _2 (l - m)! Pim(cos 0)Pt(cos 0') cos [m(¢ - ¢')] (3.68)
m=I (l + m)!

It the angle y goes to zero, there results a "sum rule" for the squares of

Yim's:

z

_L IYzm(0, <fo)l2 = 21 + l

m~-z

4tt

(3.69)

The addition theorem can be used to put expansion (3.41) of the potential at x due to a unit charge at x' into its most general form. Substituting (3.62) for Pz(cos y) into (3.41) we obtain

+ 1 _
Ix -, x/1 -

41r~ ~ro

~ l
m~l

21

1

l
r < y* (B, 1 rZf l lm ,

,1..,) y: (O .J.)
't' lm , 't'

(3.70)

Equation (3.70) gives the potential in a completely factorized form in the coordinates x and x'. This is useful in any integrations over charge densities, etc., where one variable is the variable of integration and the other is the coordinate of the observation point. The price paid is that there is a double sum involved, rather than a single term.

3.6 Laplace's Equation in Cylindrical Coordinates; Bessel Functions

In cylindrical coordinates (p, ¢,, z), as shown in Fig. 3.6, Laplace's

equation takes the form:

a-2<t+ >
op2

-1 oC-D
pop

+p12-aa42->ct2>

+ao2-zc2t>

=

0

(3.71)

The separation of variables is accomplished by the substitution:

<P(p, <fo, z) = R(p)Q(<fo)Z(z)

(3.72)

In the usual way this leads to the three ordinary differential equations:

a2z - k2Z = o
dz2

(3.73)

(3.74)

(3.75)

70

Classical Electrodynamics

z

y I ...... I '...J
X
Fig. 3.6

The solutions of the first two equations are elementary:
Z(z) = e±kz
= Q( cp) e±iv<f,

(3.76)

In order that the potential be single valued, 11 must be an integer. But
barring some boundary-condition requirement in the z direction, the
parameter k is arbitrary. For the present we will assume that k is real.
The radial equation can be put in a standard form by the change of
variable x = k p. Then it becomes

(3.77)

This is Bessel's equation, and the solutions are called Bessel functions of order v. If a power series solution of the form:

= 100
R(x) :if aix'
i=O
is assumed, then it is found that
and 1
a21 = - 4j(j + oc) G21-2

(3.78) (3.79) (3.80)

for j = 0, 1, 2, 3, . . . . All odd powers of xi have vanishing coefficients.
The recursion formula can be iterated to obtain

(-1)1r(oc + 1) = + + a2i 221j! r(j fX 1) Go

(3.81}

[Sect. 3.6]

Boundary-Value Problems in Electrostatics: II

71

= + It is conventional to choose the constant a0 [2~r(a J)J-1. Then the
two solutions are

J.,(x)

=

(

x)v
2

~ ~

(-J)i (x\2i + j!r(j V + 1) 7_/

(3.82)

2 + 2 J - v(X)

=

( x)-v ~~

(-1); j !r{j - V

(x)2;
1)

(3.83)

These solutions are called Bessel functions of the first kind of order ±v.
The series converge for all finite values of x. If v is not an integer, these two solutions J ±,,(x) form a pair of linearly independent solutions to the second-order Bessel's equation. However, if vis an integer, it is wel1 known
that the solutions are linearly dependent. In fact, for v = m, an integer,
it can be seen from the series representation that

(3.84)

Consequently it is necessary to find another linearly independent solution when mis an integer. It is customary, even if vis not an integer, to replace

the pair J ±,,(x) by Jv(x) and N,,(x), the Neumann function (or Bessel's

function of the second kind):

N..,(x) = Jv(x) cos ~11 - J _v(x)

(3.SS)

Slil V1T

For v not an integer, N,,(x) is clearly linearly independent of J1,(x). In the limit v _,_ integer, it can be shown that Nix) is still linearly independent

of J,,(x). As expected, it involves log x. Its series representation is given

in the reference books.

The Bessel functions of the third kind, called Hankel functions, are

defined as linear combinations of J11(x) and N,,(x):
1 H~1)(x) = Jv(x) + iNv(x)
H~2)(x) = Jix) - iNv(x) J

(3.86)

The Hankel functions form a fundamental set of solutions to Bessel's equation, just as do Jv(x) and Nv(x).
The functions J,,, N,,, H~1>, H~2>all satisfy the recursion formulas:

(3.87)

(3.88)

72

Classical Electrodynamics

where .Oix) is any one of the cylinder functions of order v. These may be verified directly from the series representation (3.82).
For reference purposes, the limiting forms of the various kinds of Bessel functions will be given for small and large values of their argument. Only the leading terms will be given for simplicity:

~ J,(x)- r(v l)w

(3.89)

y l r In (~) + 0.5772. •
l- r~\iJ, Nv(x)-+

(3.90)

In these formulas v is assumed to be real and nonnegative.

Ex l ►X I, V J,(X) -

cos ( X - ' ; - ~)

-l) j (x - •; N,(x)-J!sin

(3.91)

The transition from the small x behavior to the large x asymptotic form

occurs in the region of x ,__, v.

From the asymptotic forms (3.91) it is clear that each Bessel function

has an infinite number of roots. We will be chiefly concerned with the

roots of Jv(x):

= Jv(xvn-) 0, n = 1, 2, 3, ...

(3.92)

xvn is the nth root of Jv(x). For the first few integer values of v, the first
three roots are:
V = 0, x0n = 2.405, 5.520, 8.654, .. . v = I, X1n = 3.832, 7.016, 10.173, .. . 'V = 2, X2n = 5.136, 8.417, 11.620, .. .

For higher roots, the asymptotic formula
+ Xvn ,.._, mr (v - ½) :!:
2

gives adequate accuracy (to at least three figures). Tables of roots are given in Jahnke and Emde, pp. 166-168.
Having found the solution of the radial part of Laplace's equation in terms of Bessel functions, we can now ask in what sense the Bessel functions form an orthogonal, complete set of functions. We will consider

[Sect. 3.6]

Boundary- Value Problems in Electrostatics: II

73

Vp only Bessel functions of the first kind, and will show that Jix.mpfa), for
:fixed 11 > 0, n = l, 2, ... , form an orthogonal set on the interval O <
p ~ a. The demonstration starts with the differential equation satisfied by
JJx,,npfa):

e)) ( l !!_ ( p dJ,,(x,,n a

p dp

dp

+ = 2 Xvn a2

~2) p2

J,,

(

X 11 n

)
f!.
a

0

(3.93)

If we multiply the equation by pJ,,(xvn'p/a) and integrate from Oto a, we obtain

Integration by parts, combined with the vanishing of (pJ,,.I/) at p = 0
(for v > 0) and p = a, leads to the result:

la dJv( Xvn' ~) dJv( Xvn ~)
- p--------dp

o

dp

dp

l p) ( P) + oa

(x;n
a2

-

pv22) pl,, ( x,,n•~ J,, xvn~

dp=O

If we now write down the same expression, with n and n' interchanged, and subtract, we obtain the orthogonality condition:

= (x!n - X~n') I.a pJ11 ( Xvn' ~)J11 ( Xvn ~) dp 0

(3.94)

By means of the recursion formulas (3.87) and (3.88) and the differential equation, the normalization integral can be found to be:

La ( p) ( p) _ l 0 pJv Xvn' ~ Jv Xvn ~

dp -

a2

2

[Jv+1(X11n)] bn'n

(3.95)

Assuming that the set of Bessel functions is complete, we can expand an arbitrary function of p on the interval O < p ~ a in a Bessel-Fourier series:

(3.96)

74
where

Classical Electrodynamics

(3.97)

Our derivation of (3.96) involved the restriction v > 0. Actually it can be proved to hold for all v > -1.
Expansion (3.96) and (3.97) is the conventional Fourier-Bessel series
and is particularly appropriate to functions which vanish at p = a (e.g.,
homogeneous Dirichlet boundary conditions on a cylinder; see the following section). But it will be noted that an alternative expansion is
possible in a series of functions VpJ,,(y,,npfa) where Yvn is the nth root of
the equation [dl,,(x)]/dx = 0. The reason is that, in proving the ortho-
gonality of the functions, all that is demanded is that the quantity
[pJ,,(lp)(d/dp)J,,().'p)] vanish at the end points p = 0 and p = a. The
requirement is met by either).= xvn/a or). = Yvn/a, where J,,(x,,J = 0 and
= J,,'(y,,J 0. The expansion in terms of the set VpJ,,(Y,,nPIa) is especially useful for functions with vanishing slope at p = a. (See Problem 3.8.)
A Fourier-Bessel series is only one type of expansion involving Bessel
functions. Neumann series [nio anJ,,+nCz)], Kapteyn series [nioan X
Jv+nC(v + n)z)], and Schlomilch series [ni/nJv(nx)] are some of the other
possibilities. The reader may refer to Watson, Chapters XVI-XIX, for a detailed discussion of the properties of these series. Kapteyn series occur in the discussion of the Kepler motion of planets and of radiation by rapidly moving charges (see Problems 14.7 and 14.8).
Before leaving the properties of Bessel functions it should be noted that if, in the separation of Laplace's equation, the separation constant k2 in (3.73) had been taken as -k2, then Z(z) would have been sin kz or cos kz and the equation for R(p) would have been:

With k p = x, this becomes

(3.98)

(3.99)

The solutions of this equation are called modified Bessel pmctions. It is evident that they are just Bessel functions of pure imaginary argument.

[Sect. 3.7]

Boundary-Value Problems in Electrostatics: II

75

The usual choices of linearly independent solutions are denoted by fix) and K.(x). They are defined by

(3.100)

(3.101)

and are real functions for real x. Their limiting forms for small and large
z are, assuming real v > 0:

2 Iv(x) -

r(v

1 +

(X)v 1)

(3.102)

-(In(~) + 0.5772 • • -), ,, = 0
(3.103)

X ► 1, P

(3.104)

3.7 Boundary-Value Problems in Cylindrical <;oordinates

The solution of Laplace's equation in cylindrical coordinates is <I> =
R(p)Q(cp)Z(z), where the separate factors are given in the previous section.
Consider now the specific boundary-value problem shown in Fig. 3.7. The cylinder has a radius a and a height L, the top and bottom surfaces
being at z =Land z = 0. The potential on the side and the bottom of
the cylinder is zero, while the top has a potential <I> = V(p, cp). We want
to find the potential at any point inside the cylinder. In order that <I> be
single valued and vanish at z = 0,

} Q(cp) = A sin m¢, + B cos mcp
Z(z) = sinh kz

(3.105)

where 11 = m is an integer and k is a constant to be determined. The radial
factor is
(3.106)

76

Classical Electrodynamics

z

a

L

Fig. 3.7

If the potential is finite at p = 0, D = 0. The requirement that the

potential vanish at p = a means that k can take on only those special

values:

= k mn

Xmn
a ~

n = 1, 2, 3, ...

(3.107)

where Xmn are the roots of Jm(xmJ = 0.

Combining all these conditions, we find that the general form of the

solution is

00

(X)

<I> (p, cp, z) = L L Jm(kmnP) sinh (kmnz)[Amn sin mcp + Bmn cos mcp]

m=O n=l

(3.108)

= At z L, we are given the potential as V(p, <f,). Therefore we have

= L V(p, cp)

sinh m,n

(km11L)Jm(kmnP)[Amn

sin

mcp

+

Bmn

COS

mef>]

This is a Fourier series in cp and a Bessel-Fourier series in p. The coefficients are, from (2. 43) and (3.97),

= A mn

2

cosech 2J2

((kkmnL))

i2ITd.'Jt.'.J,adPP Vi(P,

,I.\ 'f'J

Jm(

k

mnP )

sin

m.J't.'.

'TT'Q m+l mna O

0

f J and

Bmn

=

2 cosech (k L) 1ra2J 2m+i(kmmnna)

2,,.dcp (adp pV(p, cp)Jm(kmnP) COS mcf,

o

o11

(3.109)

with the proviso that, for m = 0, ~e use ½Bon in the series.

The particular form of expansion (3.108) is indicated by the requirement
that the potential vanish at z = 0 for arbitrary p and at p = a for arbitrary

z. For different boundary conditions the expansion would take a different

[Sect. 3.8]

Boundary-Value Problems in Electrostatics: II

77

form. An example where the potential is zero on the end faces and equal to V(cp, z) on the side surface is left as Problem 3.6 for the reader.
The Fourier-Bessel series (3.108) is appropriate for a finite interval in
p, O < p :S: a. If a ---+ co, the series goes over into an integral in a manner entirely analogous to the transition from a trigonometric Fourier series
to a Fourier integral. Thus, for example, if the potential in charge-free
space is finite for z > 0 and vanishes for z---+ oo, the general form of the solution for z > 0 must be

! <I>(p, </,, z) = f.dk e-kzJm(kp)[Am(k) sin m¢, + Bm(k) cos mef,J (3.110)

m==O O
If the potential is specified over the whole plane z = 0 to be V(p, rp) the
coefficients are determined by

! f. V(p, </,) =

+ 00dkJ m(kp)[Am(k) sin mcf, Bm(k) cos m<f,]

m=O O
The variation in q, is just a Fourier series. Consequently the coefficients
Am(k) and Bm(k) are separately specified by the integral relations:

¾f.2irV(p, </,)(sin mq,) def, = iooJm(k'p)(Am(k')) dk,

0

cos mcf,

0

Bm(k')

(3.111)

These radial integral equations of the first kind can be easily solved, since they are Hankel transforms. For our purposes, the integral relation,

J:00 xlm(kx)J.,,,,(k'x)

dx

=

-1 d(k'

-

k)

o

k

(3.112)

can be exploited to invert equations (3.111). Multiplying both sides by

plm(kp) and integrating over p, we find with the help of (3.112) that the

coefficients are determined by integrals over the whole area of the plane
z = 0:

J. = - mq, A,n(k)) kf. 00 dp p 211'd<J, V(p, cf,)Jm(kp)(sin mcp

Bm(k) 1T o

o

cos

(3,113)

As usual, form = 0, we must use ¼Bo(k) in series (3.110).

3.8 Expansion of Green's Functions in Spherical Coordinates
In order to handle problems involving distributions of charge as well as ~oundary values for the potential (i.e., solutions of Poisson's equation) it IS necessary to determine the Green's function G(x, x') which satisfies the

78

Classical Electrodynamics

appropriate boundary conditions. Often these boundary conditions are specified on surfaces of some separable coordinate system, e.g., spherical or cylindrical boundaries. Then it is convenient to express the Green's function as a series of products of the functions appropriate to the coordinates in question. We first illustrate the type of expansion involved by considering spherical coordinates.
For the case of no boundary surfaces, except at infinity, we already have the expansion of the Green's function, namely (3. 70):

~ ~ m4'_, ~ oo

l

Ix x'I = 4,,,

21 1 ;~, Y~(O', ,f/) Y,m(B, ,f,)

Suppose that we wish to obtain a similar expansion for the Green's
function appropriate for the "exterior" problem with a spherical boundary
at r = a. The result is readily foun:d from the image form of the Green's
function (2.22). Using expansion (3. 70) for both terms in (2.22), we obtain:

To see clearly the structure of (3.114) and to verify that it satisfies the
boundary conditions, we exhibit the radial factors separately for r < r'
and for r > r' :

r < r' r > r'

(3.115)

First of all, we note that for either r or r' equal to a the radial factor vanishes, as required. Similarly, as r or r'--+ oo, the radial factor vanishes. It is symmetric in r and r'. Viewed as a function of r, for fixed r', the radial factor is just a linear combination of the solutions ,i and r-(i+l> ofthe radial part (3. 7) of Laplace's equation. It is admittedly a different linear
combination for r < r' and for r > r'. The reason for this will become
apparent below, and is connected with the fact that the Green's function is a solution of Poisson's equation with a delta function inhomogeneity.
Now that we have seen the general structure of the expansion of a Green's function in separable coordinates we turn to the systematic construction of such expansions from first principles. A Green's function for a potential problem satisfies the equation

V,/G(x, x') = -477 J(x - x')

(3.116)

[Sect. 3.8]

Boundary~ Value Problems in Electrostatics: II

79

subject to the boundary conditions G(x, x1) = 0 for either x or x' on the
boundary surface S. For spherical boundary surfaces we desire an expansion of the general form (3.114). Accordingly we exploit the fact that the delta function can be written*

b(x -

x') =

1
-

o(r

-

r') J( ef> -

ef>') b(cos 0 -

cos 0')

r2

and that the completeness relation (3.56) can be used to represent the

angular delta functions:

! I J(x - x') = : 2 O(r - r')

Y;!,.(0', cf,')Y,m(0, cf,)

(3.117)

lzOm=~l

Then the Green's function, considered as a function of x, can be expanded

as

l 2 O'.l

l

G(x, x') =

Aim(0 ', rp ')gi(r, r') Yzm(0, </>)

l=O m= -l

(3.118)

Substitution of (3.117) and (3.118) into (3.116) leads to the results

(3.119)

and

+ -l
r

-ddr22

(rg1(r,

r '))

-

l(l r2 1) gi(r, r ') = -

-4T;;T u""('r -
ru

r ')

(3.120)

The radial Green's function is seen to satisfy the homogeneous radial equation (3. 7) for r =I= r 1• Thus it can be written as:

for r < r1 for r > r'

The coefficients A, B, A', B' are functions of r' to be determined by the boundary conditions, the requirement implied by c5(r - r') in (3.120), and the symmetry of gi(r, r') in rand r'. Suppose that the boundary sufaces are concentric spheres at r = a and r = b. The vanishing of G(x, x') for x on

= * To express o(x - x') o(x1 - xi') ci(:r: 2 - x/) c5(x3 - xa') in terms of the coordi-
nates (l= 1, ~ 2, ~ 3), related to (xi. x 2, 1:3) via the Jacobian J(x,-, ;;), we note that the mean-
ingful quantity is o(x - x') d3x. Hence

80

Classical Electrodynamics

the surface implies the vanishing of gi(r, r') for r = a and r = b. Con-
sequently g1(r, r') becomes

r < r' r > r'

(3.121)

The symmetry in r and r' requires that the coefficients A(r') and B'(r') be

such that gz(r, r') can be written

C~, - g,(r, r') = c(,:s - ::::)

;,~+,)

(3.122)

where r < (r >) is the smaller (larger) of rand r'. To determine the constant
Cwe must consider the effect ofthe delta function in (3.120). Ifwe multiply
both sides by r and integrate over the interval from r = r' - e to r = r' +
e, where Eis very small, we obtain

7 [ !!.._ (rgz(r, r'))] - [!!__ (rgi(r, r'))] = - 4

dr

r·+e dr

r'-E

r

(3.123)

Thus there is a discontinuity in slope at r = r', as indicated in Fig. 3.8. For r = r' + E, r> = r, r< = r'. Hence

(1 ,- [ -dd (rg1(r, r'))]

= C (1 r'1 -

r

r' + ~

\

~ a2z+1 r

)'

[

-ddr

-ri -

1+1 )] b21+1 r =r'

c( (a b = - r' 1 -

)21+1)' (
r'

l

+

( l

+

(r')2l+1·) 1)

Similar]y

[:r ~ (rgi(r, r'))l'-E = ( l + 1 + 1(;,)21+1) ( 1 - ( ~')21+1)

Substituting these derivatives into (3.123), we find:

C = (21 + 1)[ 1 - (~)21+1]

(3.124)

Combination of (3.124), (3.122), (3.119), and (3.118) yields the expansion
of the Green's function for a spherical shell bounded by r = a and r = b:

* G(x, x') =

4 ~

Y/!',J0', </>') Yim((), cp) (ri

L 7T z=o

L
m=

-z

(21

+

1) [ 1

-

(~)21+ 1]

<

a2l+I)' ( 1 r':t1 r;:1

r~ )
b2i+1 (3.125)

[Sect. 3.9]

Boundary-Value Problems in Electrostatics: II

81

I

I

\ I

\ I

,,.,,,,,

\ I .,,.,,,,,,

1

\ l ,,,,,,,.
/

Fig. 3.8 Discontinuity in slope of the radial Green's function.

I I \ I
r'

r____.

For the special cases a---+ 0, b---+ oo, and b---+ oo, we recover the previous expansions (3.70) and (3.114), respectively. For the "interior" problem with a sphere of radius b we merely let a ~ 0. Whereas the expansion for a single sphere is most easily obtained from the image solution, the general result (3.125) for a spherical shell is rather difficult to obtain by the method of images, since it involves an infinite set of images.

3.9 Solution of Potential Problems with the Spherical Green's Function Expansion

The general solution to Poisson's equation with specified values of the potential on the boundary surface is (see Section 1.10):

J 1! (IJ(x) = p(x')G(x, x') d3x' --- -1 c.b(x') -aG da'

V

4rr S

on'

(3.126)

For purposes of illustration let us consider the potential inside a sphere of

radius b. First we will establish the equivalence of the surface integral in

(3.126) to the previous method of Section 3.4, equations (3.61) and (3.58).
With a = 0 in (3.125), the normal derivative, evaluated at r' = b, is:

I (!.)z ~ = ::i 1
Un

aG =
u'.3 r, r' ==b

--- 4b12r

b

Y* (0' tm ,

'..fl.')Y,I m<O ,

A-)
'f

l,m

(3.127)

Consequently the solution of Laplace's equation inside r = b with
«t> = V(0', </,') on the surface is, according to (3.126):

~[J <b(x) =

V(0', <f,')Y~(0', <p') dn'] (!Y Yzm(0, </>) (3.128)

lm

b

'

For the case considered, this is the same form of solution as (3.61) with

(3.58). There is a third form of solution for the sphere, the so-called

82

Classical Electrodynamics

z

Fig. 3.9 Ring of charge of radius a and total charge Q inside a grounded, conduct-
ing sphere of radius b.

Poisson integral (2.25). The equivalence of this solution to the Green's function expansion solution is implied by the fact that both were derived from the general expression (3.126) and the image Green~s function. The explicit demonstration of the equivalence of (2.25) and the series solution (3.61) will be left to the problems.
We now tum to the solution of problems with charge distributed in the volume, so that the volume integral in (3. 126) is involved. It is sufficient to consider problems in which the potential vanishes on the boundary surfaces. By linear superposition of a solution of Laplace's equation the general situation can be obtained. The first illustration is that of a hollow grounded sphere of radius b with a concentric ring of charge of radius a and total charge Q. The ring of charge is located in the x-yplane, as shown in Fig. 3.9. The charge density of the ring can be written with the help of delta functions in angle and radius as

= p(x') __g_ o(r' - a) o(cos 0') 21ra 2

(3.129)

In the volume integral over the Green's function only terms in (3.125) with
m = 0 will survive because of azimuthal symmetry. Then, using (3.57)
and remembering that a--+ 0 in (3.125), we find
J (!)(x) = p(x')G(x, x') d3x'

i: :f:. )P = Q!Pi(O)r!::( 1 -

1 1(cos 0)

i=o

r> b

(3.130)

where now r < (r >) is the smaller (larger) of r and a. Using the fact that

P2n+l(O)

=

0

and

P2n{O)

=

-( -1-)n-(22nn-n!--1-)!!,

(3.130)

can

be

. wntten

as:

)p <l>(x)

=

QL~

(-lt(2n
2nn

!

1)!! r2n(_l_ -
< r~+l

r~
b<ln+l

(cos 0)
2n

(3.131)

n=O

[Sect. 3.9]

Boundary-Value Problems in Electrostatics: II

83

z

Ffg. 3.10 Uniform line charge of length 2b and total charge Q inside a grounded, conducting sphere of
radius b.

In the limit b-+ oo, it will be seen that (3.130) or (3.131) reduces to

expression (3.48) for a ring of charge in free space. The present result can

be obtained alternatively by using (3.48) and the images for a sphere.

A second example of charge densities, illustrated in Fig. 3.10, is that of

a hollow grounded sphere with a uniform line charge of total charge Q

located on the z axis between the north and south poles of the sphere.

Again with the help of delta functions the volume-charge density can be

written:

p(x') = Q - 1-[b(cos 0' - 1) + C,(cos fJ' + l)] 2b 21rr'2

(3.132)

The two delta functions in cos 6 correspond to the two halves of the line

charge, above and below the x-y plane. The factor 21Tr12 in the denominator

assures that the charge density has a constant linear density Q/2b. With

this density in (3.126) we obtain

! J:\~( :f+ g_ ~x) =

[P,(1) + Pz(-l)]Pz(cos 0)

2b i=o

o

z~i -

r>

b

1) dr'

(3.133)

The integral must be broken up into the intervals O :S: r' < r and
r < r' s b. Then we find

(!.)z) =

(2l l(l

+
+

1)(1
1)

_

b

(3.134)

For I = 0 this result is indeterminate. Applying L'Hospital's rule, we have, for I = 0 only,

!!(1 - (!r)

lb= lim dl •

I = b. = lim (- i 10 <rlb>) In (~) (3.135)

0 Z-tO

!i_ (l)

z-o dl

r

dl

84

Classical Electrodynamics

This can be verified by direct integration in (3.133) for l = 0. Using the
= fact that Pi(-1) (-l)i, the potential (3.133) can be put in the form:

! \ ("!.) + <l>(x) = Q (1n (~) +

4j _+ l) [1 - 21]P.u(cos 0)) (3.136)

b r ; =1 2J(2J 1)

b

The presence of the logarithm for I = 0 reminds us that the potential

diverges along the z axis. This is borne out by the series in (3.136), which
diverges for cos 0 = ± 1, except at r = b exactly.

The surface-charge density on the grounded sphere is readily obtained

from (3.136) by differentiation :

I ! ( or + + a(O) = _!_ o<I> = - ~[1

41r

1'=b

411b2

j=l

4j (2j

+

l) 1)

P2 .(cos
J

0)]

(3.137)

The leading term shows that the total charge induced on the sphere is -Q, the other terms integrating to zero over the surface of the sphere.

3.10 Expansion of Green's Functions in Cylindrical Coordinates

The expansion of the potential of a unit point charge in cylindrical coordinates affords another useful example of Green's function expansions. We will present the initial steps in general enough fashion that the procedure can be readily adapted to finding Green's functions for potential problems with cylindrical boundary surfaces. The starting point is the equation for the Green's function:

= - V:z:2G(x, x')

41r IJ(p -

p') J( cf> -

cf>') ~(z -

z')

p

(3.138)

where the delta function has been expressed in cylindrical coordinates. The cf, and z delta functions can be written in terms of orthonormal functions:

/J(z -

z') = -1 J<X>

dk

e

.
ik(z-z

, >

=

-1 J:aodk cos [k(z -

z')]

21r -oo
'1)

1T 0

L d( cp - cf/) = _!__ ' 21r

eim(t/J - tt,')

m=-oo

(3.139)

We expand the Green's function in similar fashion:

= ! G(x, x') 2~

J: 00dk eim(,t,-t/J') cos [k(z - z')]gm(P, p')

m=-oo O

(3.140)

[Sect. 3.1O] Boundary- Value Problems in Electrostatics: II

85

Then substitution into (3.138) leads to an equation for the radial Green's function gm(P, p'):

-1 -d ( P -dgm) -

(k<:> w

+

-m2) gm

=

-

-477 us(P -

P,.)

p dp dp

p2

p

(3.141)

for p -:f p' this is just equation (3.98) for the modified Bessel functions,
J,,n(kp) and Km(kp). Suppose that 'lf}i(kp) is some linear combination of
Jm and Km which satisfies the correct boundary conditions for p < p', and
that 1.plkp) is a linearly independent combination which satisfies the
proper boundary conditions for p > p'. Then the symmetry of the Green's
function in p and p' requires that

(3.142)

The normalization of the product 'l{J1tp2 is determined by the discontinuity in slope implied by the delta function in (3.141):

dgm - dgm = -4-7T

dp + dp -

pl

(3.143)

where '± means evaluated at p = p' ± £. From (3.142) it is evident that

-1; _] [d:; +

= k(,p1,p2 - 'l','1'1') = kW[,p,, '!',] (3.144)

where primes mean differentiation with respect to the argument, and W[tp1, ~ 2] is the Wronskian of 1Pi and 7P2, Equation (3.141) is of the Sturm-Liouville type

!_ (p(x) dy) + g(x)y = 0

dx

dx

(3.145)

and it is well known that the Wronskian of two linearly independent solutions of such an equation is proportional to [1/p(x)]. Hence the possibility of satisfying (3.143) for all values of p' is assured. Clearly we must demand that the normalization of the product V'i''P2 is such that the Wronskian has the value :

W[1P1('C), "P2(x)] = - -41T
X

(3.146)

If there are no boundary surfaces, the requirement is that gm(P, p') be
= finite at p 0 and vanish at p -c>- oo. Consequently 1.pi(kp) = Alm(kp) and
"Plkp) = Km(kp). The constant A is to be determined from the Wronskian
condition (3.146). Since the Wronskian is proportional to (1/x) for all Values of x, it does not matter where we evaluate it. Usin_g the limitin~