Answers To a Selection of Problems from Classical Electrodynamics John David Jackson by Kasper van Wijk Center for Wave Phenomena Department of Geophysics ColoradoSchool of Mines Golden, CO 80401 Samizdat Press Published by the Samizdat Press Center for Wave Phenomena Department of Geophysics ColoradoSchool of Mines Golden, Colorado80401 and New England Research 76 Olcott Drive White River Junction, Vermont 05001 c Samizdat Press, 1996 (cid:13) Release 2.0, January1999 Samizdat Press publications are available via FTP or WWW from samizdat.mines.edu Permission is given to freely copy these documents. Contents 1 Introduction to Electrostatics 7 1.1 Electric Fields for a Hollow Conductor . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Charged Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.5 Charge Density for a Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.7 Charged Cylindrical Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.13 Green’s Reciprocity Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2 Boundary-Value Problems in Electrostatics: 1 15 2.2 The Method of Image Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.7 An Exercise in Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.9 Two Halves of a Conducting Spherical Shell . . . . . . . . . . . . . . . . . . . . . . 19 2.10 A Conducting Plate with a Boss . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.11 Line Charges and the Method of Images . . . . . . . . . . . . . . . . . . . . . . . . 24 2.13 Two Cylinder Halves at Constant Potentials . . . . . . . . . . . . . . . . . . . . . . 26 2.23 A Hollow Cubical Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 8 Waveguides, Resonant Cavities and Optical Fibers 31 8.1 Time Averaged Forces Per Unit Area on a Conductor . . . . . . . . . . . . . . . . 31 8.2 TEM Waves in a Medium of Two Concentric Cylinders . . . . . . . . . . . . . . . 33 8.3 TEM Waves Between Metal Strips . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3 4 CONTENTS 8.4 TE and TM Waves along a Brass Cylinder. . . . . . . . . . . . . . . . . . . . . . . 38 8.4.1 a. Cuto(cid:11) Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 10 Scattering and Di(cid:11)raction 41 10.3 Scattering Due to a Solid Uniform Conducting Sphere . . . . . . . . . . . . . . . . 41 10.14Di(cid:11)raction from a Rectangular Opening . . . . . . . . . . . . . . . . . . . . . . . . 42 11 Special Theory of Relativity 47 11.3 The Parallel-velocityAddition Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 11.5 The Lorentz TransformationLaw for Acceleration . . . . . . . . . . . . . . . . . . 48 11.6 The Rocket Ship . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 12 Practice Problems 53 12.1 Angle between Two Coplanar Dipoles . . . . . . . . . . . . . . . . . . . . . . . . . 53 12.2 The Potential in Multipole Moments . . . . . . . . . . . . . . . . . . . . . . . . . . 54 12.3 Potential by Taylor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 A Mathematical Tools 57 A.1 Partial integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 A.2 Vector analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 A.3 Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 A.4 Euler Formula. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 A.5 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 CONTENTS 5 Introduction This is a collection of my answers to problems from a graduate course in electrodynamics. These problemsaremainlyfromthe bookbyJackson[4], but appended aresomepractice problems. My answers are by no means guaranteed to be perfect, but I hope they will provide the reader with a guideline to understand the problems. Throughoutthese notes Iwill refertoequationsandpagesof Jacksonand Du(cid:14)n [2]. Thelatteris a textbook in electricity and magnetism that I used as an undergraduate student. References to equations starting with a \D" are from the book by Du(cid:14)n. Accordingly, equations starting with the letter \J" refer to Jackson. Ingeneral,primedvariablesdenotevectorsorcomponentsofvectorsrelatedtothedistancebetween source and origin. Unprimed coordinates refer to the location of the point of interest. The text will be a work in progress. As time progresses,I will add more chapters. 6 CONTENTS Chapter 1 Introduction to Electrostatics 1.1 Electric Fields for a Hollow Conductor a. The Location of Free Charges in the Conductor Gauss’ law states that (cid:26) = E; (1.1) (cid:15) r(cid:1) 0 where (cid:26) is the volume charge density and (cid:15) is the permittivity of free space. We know that 0 conductors allow charges free to move within. So, when placed in an external static electric (cid:12)eld charges move to the surface of the conductor, canceling the external (cid:12)eld inside the conductor. Therefore, a conductor carrying only static charge can have no electric (cid:12)eld within its material, whichmeansthevolumechargedensityiszeroandexcesschargeslieonthesurfaceofaconductor. b. The Electric Field inside a Hollow Conductor When the free charge lies outside the cavity circumferenced by conducting material (see (cid:12)gure 1.1b), Gauss’ law simpli(cid:12)es to Laplace’s equation in the cavity. The conducting material forms a volume of equipotential, because the electric (cid:12)eld in the conductor is zero and E= (cid:8) (1.2) (cid:0)r Since the potential is a continuous function acrossa charged boundary, the potential on the inner surface of the conductor has to be constant. This is now a problem satisfying Laplace’s equation with Dirichletboundaryconditions. Insection1.9ofJackson,itisshownthatthe solutionforthis problem is unique. The constant value of the potential on the outer surface of the cavity satis(cid:12)es Laplace’s equation and is therefore the solution. In other words, the hollow conductor acts like a electric (cid:12)eld shield for the cavity. 7 8 CHAPTER 1. INTRODUCTION TO ELECTROSTATICS a b q q q Figure 1.1: a: point charge in the cavity of a hollow conductor. b: point charge outside the cavity of a hollow conductor. With a point charge q inside the cavity (see (cid:12)gure 1.1a), we use the following representation of Gauss’ law: q E dS= (1.3) (cid:1) (cid:15) I 0 Therefore,theelectric(cid:12)eldinsidethehollowconductorisnon-zero. Note: the electric(cid:12)eldoutside the conductordue to a point sourceinside is in(cid:13)uenced by the shape of the conductor,as you can see in part c. c. The Direction of the Electric Field outside a Conductor An electrostatic (cid:12)eld is conservative. Therefore, the circulation of E around any closed path is zero E dl=0 (1.4) (cid:1) I This is called the circuital law for E (E4.14 or J1.21). I have drawn a closed path in four legs 2 1 4 3 Figure 1.2: Electric (cid:12)eld near the surface of a charged spherical conductor. A closed path crossing the surface of the conductor is divided in four sections. through the surface of a rectangular conductor ((cid:12)gure 1.2). Sections 1 and 4 can be chosen negligible small. Also, we have seen earlier that the (cid:12)eld in the conductor (section 3) is zero. For 1.4. CHARGED SPHERES 9 the total integral around the closed path to be zero, the tangential component (section 2) has to be zero. Therefore, the electric (cid:12)eld is described by (cid:27) E= r^ (1.5) (cid:15) 0 where(cid:27)isthesurfacechargedensity,since{asshownearlier{freechargeinaconductorislocated on the surface. 1.4 Charged Spheres Here we have a conducting, a homogeneously charged and an in-homogeneously charged sphere. Their total charge is Q. Finding the electric (cid:12)eld for each case in- and outside the sphere is an exercise in using Gauss’ law q E dS= (1.6) (cid:1) (cid:15) I 0 For all cases: Problem 1.1c showed that the electric (cid:12)eld is directed radially outward from the center of (cid:15) the spheres. For r > a, E behaves as if caused by a point charge of magnitude of the total charge Q of (cid:15) the sphere, at the origin. Q E= ^r 4(cid:25)(cid:15) r2 0 As we have seen earlier, for a solid spherical conductor the electric (cid:12)eld inside is zero (see (cid:12)gure 1.3). For a sphere with a homogeneous charge distribution the electric (cid:12)eld at points inside the sphere increases with r. As the surface S increases, the amount of charge surrounded increases (see equation 1.6): Qr E= ^r (1.7) 4(cid:25)(cid:15) a3 0 For points inside a sphere with an inhomogeneous charge distribution, we use Gauss’ law (once again) 2(cid:25) (cid:25) r E4(cid:25)r2 =1=(cid:15) (cid:26)(r)r2sin(cid:18) dr d(cid:30)d(cid:18) (1.8) 0 0 0 0 0 0 0 Z0 Z0 Z0 Implementing the volume charge distribution (cid:26)(r )=(cid:26) rn; 0 0 0 the integration over r for n> 3 is straightforward: 0 (cid:0) (cid:26) rn+1 0 E= ^r; (1.9) (cid:15) (n+3) 0 10 CHAPTER 1. INTRODUCTION TO ELECTROSTATICS homogeneously charged inhomogeneously charged: n=2 inhomogeneously charged: n=−2 conductor h gt n e str d el c fi ctri e el 0 a distance from the origin Figure 1.3: Electric (cid:12)eld for di(cid:11)erently charged spheres of radius a. The electric (cid:12)eld outside the spheres is the same for all, since the total charge is Q in all cases. where a 4(cid:25)(cid:26) an+3 Q = 4(cid:25)(cid:26) rn+2dr = 0 0 n+3 , Z0 Q(n+3) (cid:26) = (1.10) 0 4(cid:25)an+3 It can easily be veri(cid:12)ed that for n = 0, we have the case of the homogeneously charged sphere (equation1.7). Theelectric(cid:12)eldasafunctionofdistanceareplottedin(cid:12)gure1.3fortheconductor, the homogeneously charged sphere and in-homogeneously charged spheres with n= 2;2. (cid:0) 1.5 Charge Density for a Hydrogen Atom The potential of a neutral hydrogen atom is q e (cid:11)r (cid:11)r (cid:0) (cid:8)(r)= 1+ (1.11) 4(cid:25)(cid:15) r 2 0 (cid:16) (cid:17) where (cid:11) equals 2 divided by the Bohr radius. If we calculate the Laplacian, we obtain the volume charge density (cid:26), through Poisson’sequation (cid:26) = 2(cid:8) (cid:15) r 0