Answers To a Selection of Problems from Classical Electrodynamics John David Jackson by Kasper van Wijk Center for Wave Phenomena Department of Geophysics Colorado School of Mines Golden, CO 80401 Samizdat Press Published by the Samizdat Press Center for Wave Phenomena Department of Geophysics Colorado School of Mines Golden, Colorado 80401 and New England Research 76 Olcott Drive White River Junction, Vermont 05001 c Samizdat Press, 1996 Release 2.0, January 1999 Samizdat Press publications are available via FTP or WWW from samizdat.mines.edu Permission is given to freely copy these documents Contents Introduction to Electrostatics 1.1 Electric Fields for a Hollow Conductor 1.4 Charged Spheres 1.5 Charge Density for a Hydrogen Atom 10 1.7 Charged Cylindrical Conductors 11 1.13 Green’s Reciprocity Theorem 12 Boundary-Value Problems in Electrostatics: 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 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 CONTENTS 8.4 TE and TM Waves along a Brass Cylinder 8.4.1 a Cutoff Frequencies 10 Scattering and Diffraction 38 38 41 10.3 Scattering Due to a Solid Uniform Conducting Sphere 41 10.14Diffraction from a Rectangular Opening 42 11 Special Theory of Relativity 47 11.3 The Parallel-velocity Addition Law 47 11.5 The Lorentz Transformation Law 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 Introduction This is a collection of my answers to problems from a graduate course in electrodynamics These problems are mainly from the book by Jackson [4], but appended are some practice 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 Throughout these notes I will refer to equations and pages of Jackson and Duffin [2] The latter is 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 Duffin Accordingly, equations starting with the letter “J” refer to Jackson In general, primed variables denote vectors or components of vectors related to the distance between 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 CONTENTS Chapter Introduction to Electrostatics 1.1 Electric Fields for a Hollow Conductor a The Location of Free Charges in the Conductor Gauss’ law states that ρ · E, = (1.1) where ρ is the volume charge density and is the permittivity of free space We know that conductors allow charges free to move within So, when placed in an external static electric field charges move to the surface of the conductor, canceling the external field inside the conductor Therefore, a conductor carrying only static charge can have no electric field within its material, which means the volume charge density is zero and excess charges lie on the surface of a conductor b The Electric Field inside a Hollow Conductor When the free charge lies outside the cavity circumferenced by conducting material (see figure 1.1b), Gauss’ law simplifies to Laplace’s equation in the cavity The conducting material forms a volume of equipotential, because the electric field in the conductor is zero and E=− Φ (1.2) Since the potential is a continuous function across a 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 Dirichlet boundary conditions In section 1.9 of Jackson, it is shown that the solution for this problem is unique The constant value of the potential on the outer surface of the cavity satisfies Laplace’s equation and is therefore the solution In other words, the hollow conductor acts like a electric field shield for the cavity CHAPTER 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 figure 1.1a), we use the following representation of Gauss’ law: q (1.3) E · dS = Therefore, the electric field inside the hollow conductor is non-zero Note: the electric field outside the conductor due to a point source inside is influenced 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 field is conservative Therefore, the circulation of E around any closed path is zero E · dl = (1.4) This is called the circuital law for E (E4.14 or J1.21) I have drawn a closed path in four legs Figure 1.2: Electric field 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 (figure 1.2) Sections and can be chosen negligible small Also, we have seen earlier that the field in the conductor (section 3) is zero For 1.4 CHARGED SPHERES the total integral around the closed path to be zero, the tangential component (section 2) has to be zero Therefore, the electric field is described by σ E= r ˆ (1.5) where σ is the surface charge density, since – as shown earlier – free charge in a conductor is located 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 field for each case in- and outside the sphere is an exercise in using Gauss’ law q (1.6) E · dS = For all cases: • Problem 1.1c showed that the electric field is directed radially outward from the center of the spheres • For r > a, E behaves as if caused by a point charge of magnitude of the total charge Q of the sphere, at the origin E= Q ˆ r 4π r2 As we have seen earlier, for a solid spherical conductor the electric field inside is zero (see figure 1.3) For a sphere with a homogeneous charge distribution the electric field 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= ˆ r (1.7) 4π a3 For points inside a sphere with an inhomogeneous charge distribution, we use Gauss’ law (once again) E4πr2 = 1/ 2π π r 0 ρ(r )r sinθ dr dφ dθ (1.8) Implementing the volume charge distribution ρ(r ) = ρ0 r n , the integration over r for n > −3 is straightforward: E= ρ0 rn+1 ˆ, r (n + 3) (1.9) 10 CHAPTER INTRODUCTION TO ELECTROSTATICS electric field strength homogeneously charged inhomogeneously charged: n=2 inhomogeneously charged: n=−2 conductor a distance from the origin Figure 1.3: Electric field for differently charged spheres of radius a The electric field outside the spheres is the same for all, since the total charge is Q in all cases where a Q = 4πρ0 rn+2 dr = ρ0 = 4πρ0 an+3 n+3 ⇔ Q(n + 3) 4πan+3 (1.10) It can easily be verified that for n = 0, we have the case of the homogeneously charged sphere (equation 1.7) The electric field as a function of distance are plotted in figure 1.3 for the conductor, the homogeneously charged sphere and in-homogeneously charged spheres with n = −2, 1.5 Charge Density for a Hydrogen Atom The potential of a neutral hydrogen atom is Φ(r) = q 4π e−αr αr 1+ r (1.11) where α equals divided by the Bohr radius If we calculate the Laplacian, we obtain the volume charge density ρ, through Poisson’s equation ρ = Φ Chapter 11 Special Theory of Relativity 11.3 The Parallel-velocity Addition Law Two reference frames K ,K move in the same direction from K with different velocities v1 and v2 , respectively The frames can be oriented such that the direction of propagation with respect to K is x The Lorentz transform can then be written in matrix form x = AK→K x ˆ v1 K v2 K’’ K’ Figure 11.1: Three reference frames K and K move in the same direction from K with velocity v1 and v2 , respectively    x0 γ  x1   −γβ     x2  =  x3 −γβ γ 0 0  x0   x1    x2 x3   ,  (11.1) where β= v c and γ= 47 1 − β2 (11.2) 48 CHAPTER 11 SPECIAL THEORY OF RELATIVITY with c equal to the speed of light The transform from K to K multiplication of the transform matrix AK→K and AK →K Here same as a direct Lorentz transform AK→K   γ1 −γ1 β1 0 γ2 −γ2 β2  −γ1 β1 γ1 0   −γ2 β2 γ2    0  0 0 0  γ1 γ2 + γ1 γ2 β1 β2 −γ1 γ2 β2 − γ1 γ2 β1  −γ2 γ1 β1 − γ2 γ1 β2 γ1 γ2 β1 β2 + γ1 γ2   0 0 We can define γ and β in the total transformation AK→K velocity: γ = γ γ + γ γ β1 β2 and can then be written as the we will show that that is the  0 0  =   0    (11.3) and find the total transformation βγ = γ1 γ2 β2 − γ1 γ2 β1 (11.4) Therefore β v 11.5 = = v1 + vc2 γ γ β2 − γ γ β1 β2 + β v γ γ β2 − γ γ β1 = = = c v1 v2 = ⇔ γ γ γ + γ γ β1 β2 + β β2 + c2 c v1 + v v + v1 2 c (11.5) The Lorentz Transformation Law for Acceleration Reference frame K moves from K with velocity v We arrange our reference frame that v = vˆ x To find the acceleration transform we first find how velocity u and time t transform, so that we can define a = du From equation (J11.18) it is clear that dt dt = dx0 γ v = (dx0 + βdx1 ) = γ dt + dx c c c (11.6) The derivative of the transformed displacement with respect to time (11.6) gives us (J11.31) which states u +v u⊥ u = and u⊥ = (11.7) + v·u γ + v·u c2 c2 The acceleration parallel to the relative velocity between the reference frames is a = v + v·u du − u + v c2 du du c2 = dt + v·u γ dt + cv dx 2 c (11.8) 11.6 THE ROCKET SHIP Use dx1 = u1 c dx0 49 (Jackson page 531) and a = du /dt to conclude that 1+ a a = γ 1+ v2 c2 vu 1− a = 1+ c2 v2 c2 3/2 vu (11.9) c2 The acceleration perpendicular to the direction of propagation is u d γ 1+⊥ ( v·u ) du⊥ c2 a⊥ = = v dt γ dt + c2 dx (11.10) After computing the differential of the numerator with the chain rule, we obtain du⊥ γ + v·u c2 du⊥ = − v·du c2 v·u + c2 u⊥ γ (11.11) (11.12) Use v · du = vdu and we have 1+ a⊥ = Realizing that du⊥ dt = a⊥ , du dt a⊥ = v·u c2 v c2 u ⊥ v·u c2 du⊥ − γ dt + du = a and 1/γ = − v /c2 simplifies equation 11.12 to 1− 1+ v2 c2 v·u c2 a⊥ + v·u v·a a⊥ − u ⊥ c c (11.13) Writing out the outer products v × (a × u ) shows that a⊥ = 11.6 1− 1+ v2 c2 v·u c2 a⊥ + v × (a × u ) c2 (11.14) The Rocket Ship a How long are they gone? There a four legs to the trip We’ll the calculations on the first and then use symmetry to find the total answer The first leg is five years to the people in the rocket ship: t1 = years They experience an acceleration of a1 = g in the direction which I will call x What we want to know is ˆ 50 CHAPTER 11 SPECIAL THEORY OF RELATIVITY how long are those five years in the rocket ship to the observers (the other part of the twins) on earth In other words, what is t? They are related by t=γ t + where β x c , (11.15) (11.16) − β2 The twin in the rocket ship feels a force but does not move in its reference frame: x = 0, so v c β= and γ= dt = γdt (11.17) I wrote this in differentials because γ is function of velocity and the velocity of the space ship as seen from earth increases with time We’ll find this velocity via the acceleration a We know a = g and that a=a x= ˆ 1+ gt = 1− v 1− v2 c2 vu x=g 1− ˆ v2 c2 3/2 x⇔ ˆ c2 x⇔ ˆ −3/2 v2 c2 3/2 3/2 dv v2 =g 1− dt c gdt = v2 c2 1− a dv ⇔ (11.18) Rewriting this result as a function of velocity leads to v(t) = gt 1+ (11.19) gt c Plugging this result into equation 11.16 gives dt = dt/γ = dt 1+ gt c (11.20) This we integrate to c gt sinh−1 (11.21) g c The time on the clock in the space ship records τ = years for the first leg Equation 11.21 written as a function of time on earth t is gτ c t = sinh = 83.7612 years (11.22) g c τ= Since all four stages are symmetric the total time spent away from earth is · 83.7612 = 335.05 years, while 20 years passed on the rocket The space ship left in 2100, so the year of return is 2435.05 AD 11.6 THE ROCKET SHIP 51 b How far did the rocket ship get? From Earth dx = vdt, where the velocity for leg one is calculated in equation 11.19 Therefore the distance traveled is merely an integration over time from zero to the t1 seconds from part a (83.7612 years):   t1 t1 gt gt1 d1 = dt = f racc2 g  + v(t)dt = − 1 = 7.83×1017 m = 82.77 light years c gt 0 1+ c (11.23) The total is distance is just twice the first leg distance: d = 2d1 = 165.55 light years 52 CHAPTER 11 SPECIAL THEORY OF RELATIVITY Chapter 12 Practice Problems 12.1 Angle between Two Coplanar Dipoles p1 p2 θ Eθ r Er θ’ Figure 12.1: Coplanar dipoles separated by a distance r Two dipoles are separated by a distance r Dipole p1 is fixed at an angle θ as defined in figure 12.1, while p2 is free to rotate The orientation of the latter is defined by the angle θ as defined in figure 12.1 Let us calculate the angular dependence between the two dipoles in equilibrium The electric field due to a dipole can be decomposed into a radial and a tangential component (D3.39): psinθ ˆ 2pcosθ ˆ+ r θ, (12.1) E(r, θ) = 4π r3 4π r3 where is the dielectric permittivity Writing p2 in terms of r and θ: ˆ ˆ p2 (r, θ) = (p2 · ˆ)ˆ + (p2 · θ = p2 cosθ ˆ + p2 sinθ θ rr r (12.2) The potential energy of the second dipole in the electric field due to the first dipole is U2 (r, θ, θ ) = −E1 · p2 = − p1 p2 (2cosθcosθ − sinθsinθ ) 4π r3 53 (12.3) 54 CHAPTER 12 PRACTICE PROBLEMS The second dipole will rotate to minimize its potential energy, defining the angular dependence between the two dipoles: ∂U2 =0 ⇔ ∂θ p1 p2 (2cosθsinθ + sinθcosθ ) = ⇔ 4π r3 2cosθsinθ = −sinθcosθ ⇔ tanθ = −(tanθ)/2 12.2 (12.4) The Potential in Multipole Moments z b a y +q -q x Figure 12.2: Concentric rings of radii a and b Their charge is q and −q, respectively This exercise is how to find the potential due to two charged concentric rings (see figure 12.2) in terms of the monopole dipole and quadrupole moments Discarding higher order moments (J4.10): Φ(r) = p·r q + + 4π r 4π r3 8π r5 xi xj Qij + (12.5) i,j The monopole moment is zero since there is no net charge The dipole moment is (J4.8): p= rρ(r)dV, V (12.6) 12.3 POTENTIAL BY TAYLOR EXPANSION 55 where the volume charge density for our case can be written in cylindrical coordinates: ρ(r) = q q δ(r − a)δ(z) − δ(r − b)δ(z) 2πa 2πb (12.7) After the x-component of the dipole moment is also written in cylindrical coordinates, the integral can be easily evaluated: ∞ xρ(r, z)dV = px = −∞ V ∞ 2π cosφρ(r, z)r drdφdz = 0, (12.8) because the cosφ integrated over one period is zero The y-component is zero, because the integration involves a sinusoid over one period Finally, the z-component is zero, because pz ∝ zδ(z)dz = (12.9) if the value is within the integration limits The quadrupole moment is a tensor: Qij = V (3xi xj − r2 )ρ(r)dV (12.10) We use the following properties of the δ-function: δ(z)dz = and rδ(r − a)dr = a, (12.11) where and a are within the respective integral limits Knowing this, the calculations for each element of Q is left to the reader After some algebra, the quadrupole moment turns out to be   0 q  Qij =  a − b2 (12.12) 0 −2 This means the potential of the two charged concentric rings is approximately Φ(r) ≈ 12.3 (a2 − b2 ) (x + y − 2z ) 16π r5 (12.13) Potential by Taylor Expansion Here, I will show that the electrostatic potential Φ(x, y, z) can be approximated by the average of the potentials at the positions perturbed by a small quantity +/ − a by doing a Taylor expansion 56 CHAPTER 12 PRACTICE PROBLEMS This expansion is correct to the third order First, the Taylor expansion around the Φ(x, y, z) perturbed in the positive x-component Φ(x + a, y, z) = Φ(x, y, z) + ∂Φ(x, y, z) ∂ Φ(x, y, z) a+ a + ∂x ∂x2 ∂ Φ(x, y, z) a + O(a4 ) ∂x3 (12.14) Next, the same expansion around Φ(x − a, y, z): Φ(x − a, y, z) = Φ(x, y, z) − ∂ Φ(x, y, z) ∂Φ(x, y, z) a+ a − ∂x ∂x2 ∂ Φ(x, y, z) a + O(a4 ) ∂x3 (12.15) When we add up these two equations, the odd powers of a cancel This is the same for the y- and z-component The a2 -term adds up to the Laplacian Assuming there is no charge within the radius a of (x, y, z), Laplace’s equation holds: Φ=0 (12.16) and thus the a2 term is zero, too Therefore, the potential at (x, y, z) can be given by: Φ(x, y, z) = 1/6(Φ(x + a, y, z) + Φ(x − a, y, z) + Φ(x, y + a, z) +Φ(x, y − a, z) + Φ(x, y, z + a) + Φ(x, y, z − a)) + O(a4 ) (12.17) Appendix A Mathematical Tools A.1 Partial integration b a A.2 b b f (x)g (x)dx = [f (x)g(x)]a − f (x)g(x)dx (A.1) a Vector analysis Stokes’ theorem M · dl = S M · dS = V L curlM · dS (A.2) divM · dV (A.3) Gauss’ theorem S Computation of the curl i hiˆ curlM = h1 h2 h3 j h2 ˆ ˆ h3 k ∂ ∂x1 ∂ ∂x2 ∂ ∂x2 h1 M 57 h2 M h3 M (A.4) 58 APPENDIX A MATHEMATICAL TOOLS hi are the geometrical components that depend on the coordinate system For a Cartesian coordinate system they are one For the cylindrical system: h1 = h2 = r h3 = For the spherical coordinate system: h1 = h2 = r h3 = r sin θ Computation of the divergence divM = h1 h2 h3 ∂ ∂ ∂ (h2 h3 M1 ) + (h1 h3 M2 ) + (h1 h2 M3 ) ∂x1 ∂x2 ∂x3 (A.5) With the same factors hi depending on the coordinate system a × (b × c) = (a · c) b − (a · b) c (A.6) Relations between grad and div × ×M= M= A.3 Φ+ ·M− M ×A (A.7) (A.8) Expansions Taylor series f (x) = f (a) + (x − a)f (a) + (x − a)2 f (a) xn + + (x − a)n f (n) (a) 2! n! (A.9) A.4 EULER FORMULA A.4 59 Euler Formula eiφ = cos φ + i sin φ sin θ cos θ = sinh θ = cosh θ A.5 = = eıθ − e−ıθ 2i ıθ e + e−ıθ eθ − e−θ θ e + e−θ (A.10) (A.11) (A.12) (A.13) (A.14) Trigonometry sin2 θ = cos(φ − θ) − cos(2θ) (A.15) = Re eı(φ−θ) = Re eıφ e−ıθ = Re [cos φ cos θ + sin φ sin θ + i( )] = cos φ cos θ + sin φ sin θ (A.16) 60 APPENDIX A MATHEMATICAL TOOLS Bibliography [1] R Bracewell The Fourier Transform and Its Applications McGraw-Hill Book Company, first edition, 1965 [2] W.J Duffin Electricity and Magnetism McGraw-Hill Book Company, fourth edition, 1990 [3] ?.? Griffith Electricity and Magnetism John Wiley & Sons, Inc., third edition, 1999 [4] J.D Jackson Classical Electrodynamics John Wiley & Sons, Inc., third edition, 1999 61 ... collection of my answers to problems from a graduate course in electrodynamics These problems are mainly from the book by Jackson [4], but appended are some practice problems My answers are by... conductors allow charges free to move within So, when placed in an external static electric field charges move to the surface of the conductor, canceling the external field inside the conductor Therefore,... induced charges on the conductor is equal to the force on q due to the field of the image charge: F = qE (2.8) The electric field at y due to the image charge at y is directed towards the origin and