JD Jackson Solution Chapter 1 and 2

Solution to

John David Jackson

Classical

Electrodynamics

FOR THIRD EDITION

Jackson Consortium

Rev. 41

iii

Introduction

John David Jackson ClassicalElectrodynamics () Creative CommonsBY-NC-SA1)

(y) (z)

Copyright c Jackson Consortium 2010Y.Hotta, M.Hyuga, T.Matsuda, Y.O^no

1) http://creativecommons.org/licenses/by-nc-sa/3.0/deed.ja

vContents

Introduction iii

Chapter 1 Introduction to Electrostatics 1

1.2 Dirac delta function for a general orthogonal coordinate system . . . . 1

1.4 Electric elds produced by a spherically symmetric charge density . . . 2

1.6 The capacitance of simple capacitors . . . . . . . . . . . . . . . . . . 4

1.8 Energy densities of certain capaciters . . . . . . . . . . . . . . . . . . 6

1.10 The mean value theorem (for electrostatics) . . . . . . . . . . . . . . 7

1.11 Normal derivertive of the electric eld at the surface of a curved conductor 9

1.12 Green's reciprocation theorem . . . . . . . . . . . . . . . . . . . . . 10

1.14 The behavior of Green functions for Poisson equation . . . . . . . . . 11

1.16 Energy decreasing with introduction of a conductor . . . . . . . . . . 12

1.18 The variational principal for capacitance . . . . . . . . . . . . . . . . 14

1.20 Theory of the capacitance estimation . . . . . . . . . . . . . . . . . . 15

1.22 The mathematical bases for relaxation method . . . . . . . . . . . . . 17

1.24 Numerical analysis performed by the relaxation method II . . . . . . . 18

Chapter 2 Boundary-Value Problems in Electrostatics: I 23

2.4 A point charge placed outside a conducting sphere . . . . . . . . . . . 23

2.8 Two straight parallel line charges with equal and opposite charge densities 25

2.12 Poisson's integral form of the potential . . . . . . . . . . . . . . . . . 31

2.16 The potential on a unit square area with a uniform charge density . . 33

2.20 Four symmetrically placed line charges . . . . . . . . . . . . . . . . . 35

2.24 The completeness relation for sine functions . . . . . . . . . . . . . . 38

2.28 The potential at the center of a regular polyhedron . . . . . . . . . . 39

Referential Sites 43

vi Contents

About us 47

1Chapter 1

Introduction to Electrostatics

1.2 Dirac delta function for a general orthogonal coordinate system

D(;x; y; z) ! 0

(x; y; z) 6= 0 lim!0D(;x; y; z) = 0 (x; y; z) = 0) lim!+0D(;x; y; z) =1 lim!+0

RD(;x; y; z) = 1

, U; V;W metric coecients

U1 =@~x@u

(1.2.1)V 1 =

@~x@v (1.2.2)

W1 = @~x@w

(1.2.3)x = x(u; v; w); y = y(u; v; w); z = z(u; v; w) 1 1 (u; v; w) (u0; v0; w0)

2 Chapter 1 Introduction to Electrostatics

D(;x x0; y y0; z z0) (1.2.4)

=(2)3=23 exp 122

[(x x0)2 + (y y0)2 + (z z0)2]

(1.2.5)

=(2)3=23 exp 122

@x

@uu+

@x

@vv +

@x

@ww

2+

@y

@uu+

@y

@vv +

@y

@ww

2+

@z

@uu+

@z

@vv +

@z

@ww

2(1.2.6)

(* u = u u0 )

(u; v; w)@~r

@u;@~r

@v;@~r

@w

=(2)3=23 exp 122

(U2u2 + V 2v2 +W2w2)

(1.2.7)

=(2)1=21 expU

2u2

22

(2)1=21 exp

V

2v2

22

(2)1=21 exp

W

2w2

22

(1.2.8)

lim!+0

D(;x x0; y y0; z z0) = (u)(v)(w)UVW (1.2.9)

1.4 Electric elds produced by a spherically symmetric charge density

E = E(r)er (1.4.1)

Gauss ()Z

@V

E dS = 10

ZV

dV (1.4.2)

Gauss 1)

1) r

1.4 Electric elds produced by a spherically symmetric charge density 3

1.1

4r2E(r) =Q

0(1.4.3)

E(r) =Q

4r20(1.4.4)

E(r; ; ) =

Q

4r20er (1.4.5)

rn n = 0

= 0 Gauss

E = 0 (1.4.6)

1.1

rn(n > 3) r Q(r)


1.2 rn

Q(r) = Q

Z r0

4r02r0ndr0Z a0

4r02r0ndr0(1.4.7)

= Q

Z r0

r0n+2dr0Z a0

r0n+2dr0(1.4.8)

= Q

rn+3

n+ 3

r0

rn+3

n+ 3

a0

(1.4.9)

= Q ra

n+3(1.4.10)

Gauss

E(r)4r2 =Q(r)

0(1.4.11)

) E(r) = Qrn+1

40an+3(1.4.12)

E(r; ; ) =Qrn+1

40an+3er (1.4.13)

n = 0;2 1.2

1.6 The capacitance of simple capacitors

(a)

1.6 The capacitance of simple capacitors 5

E =

0=

Q

0A(1.6.1)

V = Ed

C = Q=V = 0A

d(1.6.2)

(b)

r

Q

0= 4r2E (1.6.3)

or

E =Q

40

1

r2(1.6.4)

V =

Z ba

Edr =Q

40(1

b+1

a) (1.6.5)

C =Q

V= 40

ab

b a (1.6.6)

(c)

(b), r

Q

0= 2rLE (1.6.7)

V =

Z ba

Edr =Q

2L0ln

b

a

(1.6.8)

C =Q

V= 20L

ln

b

a

1(1.6.9)


(d)

(c) a = 1mm, C=L = 3 1011F=m, 0 = 8:9 1012F=m , b =6:5mm C=L = 3 1012 b = 1:2 105m

1.8 Energy densities of certain capaciters

(a)

W :=

02

ZE2dV (1.8.1)

1.6

Q V

(a)d

20AQ2

0A

2dV 2

(b)Q2

80

1

a 1b

20V

2

1

a 1b

1(c)

Q2

40Lln

b

a0LV

2

ln

b

a

1

W =QV

2(1.8.2)

A;B +Q;Q

1.10 The mean value theorem (for electrostatics) 7

W =02

ZE2dV (1.8.3)

=02

Zr' r'dV (1.8.4)

=02

['r']

Z'4'dV

(1.8.5)

= 02

Z'4'dV (* 0)

=1

2

Z'dV (1.8.6)

=1

2(+Q'(A) + (Q'(B))) (1.8.7)

=Q

2('(A) '(B)) (1.8.8)

=QV

2(1.8.9)

1)

(b)

()

W /(a) 1

(b) r4

(c) r2

1.3

1.10 The mean value theorem (for electrostatics)

Green V; S R Z

V

(~y)r2yG(~x; ~y)G(~x; ~y)r2y(~y)

dVy

=

ZS

((~y)ryG(~x; ~y)G(~x; ~y)ry(~y)) !dSy (1.10.1)

1) (p. 43 (1.62) )


1.3

G(~x; ~y) = R := 1=(j~x ~yj) r2G = 4;r2(~y) = =0 = 0

() = 4(~x) (1.10.2)

() =ZS

r

1

R

1Rr

!dS (1.10.3)

= 1R2

ZS

dS +1

R

ZV

r ~EdV (*) (1.10.4)

= 1R2

ZS

dS (1.10.5)

y

y(Editor's note) Chapter 2 28 Gauss I

SErdS = 0 (1.10.6)

Er E r

(r) = (0) +

Z R0

Erdr (1.10.7)

IS(r)dS =

IS(0)dS +

ISdS

Z R0

drEr (1.10.8)

= 4R2(0) +

Z R0

dr

ISdSEr| {z }

=0 (*(1:10:6))

(1.10.9)

= 4R2(0) (1.10.10)

1.11 Normal derivertive of the electric eld at the surface of a curved conductor 9

1.4

1.5 1.4 ()

1.11 Normal derivertive of the electric eld at the surface of a curved conductor

R11 R22 n V Gauss ( 1.5)Z

S=@V

E dS = 10

ZV

dV (1.11.1)

= 0 1)E(0) n E(n) 2)

1) n 2) i n


Gauss

E(n) (R1 +n)1 (R2 +n)2 E(0)R11 R22 = 0 (1.11.2)

E(n) E(0)n

R1R2 + E(n)(R1 +R2) + E(n)n = 0 (1.11.3)

n! +0 @E

@n(0)R1R2 + E(0)(R1 +R2) + 0 = 0 (1.11.4)

1

E

@E

@n=

1

R1+

1

R2

(1.11.5)

1.12 Green's reciprocation theorem

(x) =1

40

ZV

(x0)jx x0jd

3x0 +ZS

(x0)jx x0jda

0

(1.12.1)

(LHS) =

ZV

(x)0(x)d3x+ZS

(x)0(x)da (1.12.2)

=1

40

ZV

d3x(x)

ZV

0(x0)jx x0jd

3x0 +ZS

0(x0)jx x0jda

0+Z

S

da(x)

ZV

0(x0)jx x0jd

3x0 +ZS

0(x0)jx x0jda

0

(1.12.3)

=1

40

ZV

d3x00(x0)Z

V

(x)

jx x0jd3x+

ZS

(x)

jx x0jda+Z

S

da00(x0)Z

V

(x)

jx x0jd3x+

ZS

(x)

jx x0jda

(1.12.4)

=

ZV

0(x0)(x0)d3x0 +ZS

0(x0)(x0)da0 (1.12.5)

= (RHS) (1.12.6)

1.14 The behavior of Green functions for Poisson equation 11

1.14 The behavior of Green functions for Poisson equation

(a)

Dirichlet Green

GD(~x; ~y) = 0 (~y S) (1.14.1)

ZV

GD(~x; ~y)r2GD(~x0; ~y)GD(~x0; ~y)r2GD(~x; ~y)

dV

=

ZS

GD(~x; ~y)rGD(~x0; ~y)GD(~x0; ~y)rGD(~x; ~y)

!dSy (1.14.2)

or

4GD(~x; ~x0) + 4GD(~x0; ~x) = 0 (1.14.3)

GD(~x; ~x0) = GD(~x0; ~x) (1.14.4)

(b)

Neumann Green

@GN (~x; ~y)

@ny= 4

S(~y S) (1.14.5)

GreenZV

GN (~x; ~y)r2GN (~x0; ~y)GN (~x0; ~y)r2GN (~x; ~y)

dV

=

ZS

GN (~x; ~y)

@GN@ny

(~x0; ~y)GN (~x0; ~y)@GN@ny

(~x; ~y)

day (1.14.6)

i.e.,

4GN (~x; ~x0) + 4GN (~x0; ~x) = 4S

ZS

GN (~x; ~y)day +4

S

ZS

GN (~x0; ~y)day

(1.14.7)


(c)

Neumann

(~x) = hSi+ 140

ZV

(~x0)GN (~x; ~x0)d3x0 +1

4

ZS

@

@n0GNda

0 (1.14.8)

GN ! GN F

()!hSi+ 140

ZV

(~x0)(GN (~x; ~x0) F (~x))d3x0

+1

4

ZS

@

@n0(GN (~x; ~x0) F (~x))da0 (1.14.9)

= (~x) F (~x)40

ZV

(~x0)d3x0 F (~x)4

ZS

@

@n0da0 (1.14.10)

= (~x) (* ) (1.14.11)

1.16 Energy decreasing with introduction of a conductor

(0)

W =02

ZV

jEj2dV (1.16.1)

W 0 =02

ZV

jE0j2dV (1.16.2)

E := E0 E

W =W 0 +02

ZV

jEj2 dV 2ZV

E0 EdV

(1.16.3)

Z

V

E0 EdV = ZV

r0 EdV (1.16.4)

= ZS

0E da+ZV

0r (E0 E)dV (1.16.5)

1.16 Energy decreasing with introduction of a conductor 13

ZS

0E da =Xi

0i

ISi

(E0i Ei) da

() (1.16.6)

=Xi

0i

qi0 qi0

(1.16.7)

= 0 (1.16.8)

ZV

0r(E0 E)dV = 10

Xi

ZVi

0i(0i i)dV +

ZVins

0(0 )dV!

(1.16.9)

= 10

Xi

0i

ZVi

(0i i)dV +0ZVins

(0 )dV!

(* ) (1.16.10)= 0 (* ) (1.16.11)

ZV

E0 EdV = 0 (1.16.12)

(1.16.3)

W =W 0 +02

ZV

jEj2 dV (1.16.13)

W >W 0 (1.16.14)

y the surfaces lowers the electrostatic energyz W = W 0 = W 0 =W + 02

RVjEj2 dV


RVins

(0 )dV 0

1.18 The variational principal for capacitance

(a)

G(~x; ~x0) ~x0 S 0

(~x) =1

4

ZV

d3x0(~x0)G(~x; ~x0) +1

4

ZS

dS0G(~x; ~x0)

@

@n0 (~x0) @G

@n0

(1.18.1)

=1

4

ZS1

da0G(~x; ~x0)@

@n0 14

ZS1

!dS r0G(~x; ~x0) (1.18.2)

~x0 S r0G(~x; ~x0) = 0 0

W =1

2

ZV

dV (~x)(~x) =1

80

ZS1

da

ZS1

da01(~x0)G(~x; ~x0)1(~x0) (1.18.3)

(b)

dC1[]d

(1.18.4)

=1

40

1RS1da(~x)

4ZS1

da

ZS1

da0G(~x; ~x0)((~x) + (~x0))Z

S1

da(~x)

2

2ZS1

da(~x)

ZS1

da

ZS1

da

ZS1

da0G(~x; ~x0)(~x)(~x0)

(1.18.5)

(a)dC1[1]

d= 0 (1.18.6)

1 1 !

1.20 Theory of the capacitance estimation 15

1.20 Theory of the capacitance estimation

(a)

1 = 01 1, 1' 1 1' (r) ' (r) 1.17

trial function

0(r) 1.17

C 0 6 C 0[] (1.20.1)

= 0

ZV

jrj2d3x (1.20.2)= C (1.20.3)

(b)

() ()1) R E = Er(r)er r > R Gauss

Er(r) =Q

40

1

r2(1.20.4)

Q

V = Z R1

Er(r)dr =Q

40

1

R(1.20.5)

C =Q

V= 40R (1.20.6)

1) ()Jackson


a

p32 a

(1.20.6)

C" := 0:866 40a (1.20.7)

a a2 (1.20.6)

C# := 0:5 40a (1.20.8)

0:683 40a 0:655 40a (4 %)2)

(c)

(a) 0 0 1.7 1.7 y (a)

2)

1.22 The mathematical bases for relaxation method 17

fCijg

Q1Q2

=

C11 C12C12 C22

V1V2

(1.20.9)

1.7 (1 ) +Q (2 ) Q V1 V2

C =Q

V1 V2 (1.20.10)

C =C11C22 C212

C11 + C22 + 2C12(1.20.11)

(1) C12 3)(a) C11 4) C C11

@C

@C11=

(C12 + C22)2

C11 + C22 + 2C12> 0 (1.20.12)

C C11 C11 1.7

1.22 The mathematical bases for relaxation method

f well-behaved

f(x1 + h1; x2 + h2; ; xn + hn) =1Xk=0

(dkf)(x1;x2; ;xn)(h1; h2; ; hn) (1.22.1)

A := fx1; x2; ; xng Sk :=kMi=1

A (1.22.2)

3) y ()4) V1 = V; V2 = 0


(dkf)(x1;x2; ;xn)(h1; h2; ; hn) :=X

(1;2; ;k)2Sk

@kf(x1; ; xn)@1@2 @k hl1hl2 hlk

(1.22.3)

(hli i )

(a) (b)

()

f (a; b) (a; b) R C

f(a; b) =1

2R

ZC

dsf(x; y) (1.22.4)

4 ( 8 )

1.24 Numerical analysis performed by the relaxation method II

p.49 (1.82) :

new(i; j) = hhold(i; j)ii+ h2

5

0+h2

10

0

C

+O(h6) (1.24.1)

hf(i; j)iC := 14

f(i 1; j) + f(i+ 1; j) + f(i; j 1) + f(i; j + 1) (1.24.2)

hf(i; j)iS := 14

f(i1; j1)+f(i+1; j1)+f(i1; j1)+f(i+1; j+1) (1.24.3)

hhf(i; j)ii := 15hfiS + 4

5hfiC (1.24.4)

1) = 1 (0:25; 0:25) 1 (0:25; 0:5) 2 (0:5; 0:5)

1)

1.24 Numerical analysis performed by the relaxation method II 19

1.1 Jacobian () Gauss-Seidel ()PPPPPPPPP

Jacobian Gauss-Seidel 1 2 3 1 2 3

0 1:0000 1:0000 1:0000 1:0000 1:0000 1:0000

1 0:6463 0:9160 1:2356 0:6463 0:7745 0:9845

2 0:6245 0:8132 1:0977 0:5554 0:7125 0:9167

3 0:5765 0:7667 1:0111 0:5272 0:6814 0:8862

4 0:5536 0:7255 0:9642 0:5132 0:6667 0:8716

5 0:5348 0:7028 0:9267 0:5066 0:6596 0:8646

6 0:5238 0:6855 0:9048 0:5034 0:6562 0:8613

7 0:5158 0:6750 0:8888 0:5019 0:6546 0:8597

8 0:5108 0:6676 0:8788 0:5012 0:6539 0:8589

9 0:5073 0:6628 0:8718 0:5008 0:6535 0:8586

10 0:5051 0:6596 0:8673 0:5007 0:6533 0:8584...

......

......

......

100 0:5005 0:6532 0:8583 0:5005 0:6532 0:8583

() 0:5691 0:7205 0:9258 0:5691 0:7205 0:9258

3 40i i : 8>:

new1 = 0:42 + 0:053 +16

new2 = 0:41 + 0:12 + 0:23 +11160

new3 = 0:21 + 0:82 +340

(1.24.5)

i = 1:0 Jackson Java ( 1.1) 1.1 1.6 Jacobian Gauss-Seidel Jacobian (?) 0.07


1.6 Jacobian Gauss-Seidel

n 0 4n () () ( 1.2)2) 1.2

1.1 Relaxation.java

1 import java.io.*;2 import java.awt .*;3 import java.awt.image .*;4 class Relaxation{5 public static final int WIDTH = 500;6 public static final int HEIGHT = WIDTH;7 public static final int TRY = 100;8 public static final double PRECISE [] = {0.5691 , 0.7205 , 0.9258};9 public static void main(String args []){

10 double x[] = new double [3], y[] = new double [3], z[] = new double [3];

2) O(n2) ( x+y 2 O(n) ) 100

1.24 Numerical analysis performed by the relaxation method II 21

1.2 4 8 12 16 20 24 28 32 361 0:50053 0:55254 0:56177 0:56498 0:56646 0:56726 0:56775 0:56806 0:568272 0:65316 0:70401 0:71320 0:71639 0:71787 0:71867 0:71916 0:71947 0:719683 0:85826 0:90929 0:91848 0:92168 0:92316 0:92396 0:92444 0:92476 0:92497

40 44 48 52 56 60 64 681 0:56843 0:56854 0:56863 0:56870 0:56875 0:56879 0:56883 0:568862 0:71984 0:71995 0:72004 0:72010 0:72016 0:72020 0:72024 0:720273 0:92513 0:92524 0:92533 0:92539 0:92545 0:92549 0:92553 0:92556

72 76 80 84 88 92 96 1001 0:56888 0:56890 0:56892 0:56893 0:56895 0:56896 0:56897 0:568982 0:72029 0:72031 0:72033 0:72034 0:72036 0:72037 0:72038 0:720393 0:92558 0:92560 0:92562 0:92563 0:92565 0:92566 0:92567 0:92568

11 double jacobi [][] = new double[TRY +1][3];12 double gauss [][] = new double[TRY +1][3];13 x[0] = x[1] = x[2] = 1.0;14 z[0] = z[1] = z[2] = 1.0;15 for(int i=0; i


49 try{50 System.out.println(51 javax.imageio.ImageIO.write(img , "PNG", new File("relaxation.png")) ?52 "Success" : "Fail"53 );54 }catch(IOException e){55 e.printStackTrace ();56 }57 }58 }

1.2 Relaxation2.java1 class Relaxation2{2 public static void main(String args []){3 int cases = 25;4 for(int i=1; i

23

Chapter 2

Boundary-Value Problems in

Electrostatics: I

2.4 A point charge placed outside a conducting sphere

q Q = rq 1) p.61 (2.9)

F =1

40

q2

d2

r R

3(2d2 R2)d(d2 R2)2

(2.4.1)

(i)

() F = 0

F = 0 (2.4.2)

,rd(d2R2)2 = R3(2d2 R2) (2.4.3),r5 2r3 22 + r + 1 = 0 (2.4.4)

:=

d

R(2.4.5)

1) (a), (b) r = 1:0 (c) r = 0:5; 2:0

24 Chapter 2 Boundary-Value Problems in Electrostatics: I

> 1 r = 0:5; 1:0; 2:0 Mathematica :

2.1 ( = dR)

r

0:5 1.8823

1:0 1.6180

2:0 1.4276

r = 1 2)

(ii)

d := a+R(a R) (2.4.1)

F =1

40

q2

(R+ a)2

r R

3(R2 + 4aR+ 2a2)

(R+ a)(2aR+ a2)2

(2.4.6)

a! 0

F 140

q2

R2

0 R

3 R2R(2aR)2

= q

2

160a2(2.4.7)

Q = rq Q = rq ( r) | | +1 Q

F 140

q (q)(2a)2

(2.4.8)

a ( 2.1)

2) Mathematica


a

( )

a

image charge

2.1

2.2

2.8 Two straight parallel line charges with equal and opposite charge densities

(a)R

2; 0

+

R2; 0

x-y

(x; y; z) =

40 ln

0BBB@x R

2

2+ y2

x+R

2

2+ y2

1CCCA (2.8.1)


(x; y; z) = () (2.8.2)

,

x R

2

2+ y2

x+R

2

2+ y2

= () (2.8.3)

A x R

2

2+ y2

x+R

2

2+ y2

A1 (2.8.4)

(A 1)x2 + y2 +

R2

4

(A+ 1)Rx = 0 (2.8.5)

A = 1 x 0 y-z A 6= 1

x 12

A+ 1

A 1R2

+ y2 =A

(A 1)2R2 (2.8.6)

1

2

A+ 1

A 1R; 0

pA

jA 1jR 1)

(b)

2:3 \"

1)


""

2.3 "" (d > a+ b; )

(a) 8>>>>>>>>>>>>>>>>>>>>>>>>>>>:

1

2

Aa + 1

Aa 1R = xa (2.8.7a)pAa

jAa 1jR = a (2.8.7b)1

2

Ab + 1

Ab 1R = xb (2.8.7c)pAb

jAb 1jR = b (2.8.7d)xa xb = d (2.8.7e)

Aa; Ab (a) A

V :=

40 lnA1a 40 lnA1b (2.8.8)

=

40

ln AbAa (2.8.9)

C :=

V= 40

ln AbAa1 (2.8.10)


(2.8.7a)

Aa = 1 +2R

2xa R =2xa +R

2xa R (2.8.11)

(2.8.7b) ()

a2 =1

4(4x2a R2) (2.8.12)

b2 =

1

4(4x2b R2) (2.8.13)

(2.8.7e)

d2 a2 b2 = (x2a + x2b 2xaxb)1

4(4x2a R2)

1

4(4x2b R2) (2.8.14)

=1

2(R2 4xaxb) (2.8.15)

(2ab)2 =

1

4(4x2a R2)(4x2b R2) (2.8.16)

d > a+ b d2 a2 b2 > 0 2ab > 0

d2 a2 b22ab

=(R2 4xaxb)p

(4x2a R2)(4x2b R2)(2.8.17)

=

vuuuut4xaR

xbR 1

4xaR

2 1

4xbR

2 1

(2.8.18)

2xaR

=Aa + 1

Aa 1 (2.8.19)

4xaR

xbR 1 = 2 Aa +Ab

(Aa 1)(Ab 1) (2.8.20)

4xaR

2 1 = 4 Aa

(Aa 1)2 (2.8.21)


d2 a2 b22ab

=

s2

Aa +Ab(Aa 1)(Ab 1)

2 14

(Aa 1)2Aa

14

(Ab 1)2Ab

(2.8.22)

=1

2

rAbAa

+ 2 +AaAb

(2.8.23)

=1

2

rAbAa

+

rAaAb

!(2.8.24)

= cosh

ln

rAbAa

!(2.8.25)

arccosh

d2 a2 b2

2ab

=

lnr

AbAa

= 12ln AbAa

(2.8.26) (2.8.10)

C = 40

ln AbAa1 = 20

arccosh

d2 a2 b2

2ab

(2.8.27)

(c)

N := d2 a2 b2

2ab(2.8.28)

arccoshN = ln (2.8.29)

cosh

N = cosh(ln) = 12(+1) (2.8.30)

= NpN2 1 (2.8.31)

arccosh > 1 +

= N+pN2 1 (2.8.32)


""

2.4 "" (d < ja bj; )

a d; b d N

2N (2.8.33)

arccoshN ln 2N (2.8.34)

C

20lnd2 (a2 + b2)

2ab

1(2.8.35)

a2 + b2 d2 a2 + b2 Taylor

C 0ln

dpab

1+1

20

ab(a2 + b2)

d4

ln

dpab

2(2.8.36)

1.7

(d)

(b)

2.12 Poisson's integral form of the potential 31

(b)

d2 a2 b22ab

= cosh ln

rAbAa

!(2.8.37)

arccosh

a2 + b2 d2

2ab

=

1

2

ln AbAa (2.8.38)

C

C =20

arccosh

a2 + b2 d2

2ab

(2.8.39)d = 0

arccosha2 + b2

2ab= ln (2.8.40)

, = baor

a

b(2.8.41)

arccosha2 + b2

2ab=

ln ba (2.8.42)

C = 20

ln ba1 (2.8.43)

1.6 (c)

2.12 Poisson's integral form of the potential

77 (2.71)

(; ) = a0 +1Xn=1

(ann cos(n) + bn

n sin(n)) (2.12.1)

= b

(b; ) = a0 +1Xn=1

(anbn cos(n) + bnb

n sin(n)) (2.12.2)

(b; ) Fourier


(b; ) =02+

1Xn=1

(n cosn+ n sinn) (2.12.3)

n =1

Z 20

(b; 0) cosn0d0 (2.12.4)

n =1

Z 20

(b; 0) sinn0d0 (2.12.5)

(2.12.2) (2.12.3) 8>>>:a0 =

02

(2.12.6a)

an = bnn (n > 1) (2.12.6b)

bn = bnn (2.12.6c)

(2.12.1)

(; ) =1

2

Z 20

(b; 0)d0

+1Xn=1

b

2Z 20

(b; 0) cosn0d0cosn

+b

2Z 20

(b; 0) sinn0d0sinn

(2.12.7)

=1

2

Z 20

(b; 0)

1 + 2

1Xn=1

b

n(cosn0 cosn+ sinn0 sinn)

!d0

(2.12.8)

2.16 The potential on a unit square area with a uniform charge density 33

21Xn=1

b

n(cosn0 cosn+ sinn0 sinn) (2.12.9)

= 21Xn=1

b

ncosn(0 ) (2.12.10)

=

1Xn=1

b

n(exp(in(0 )) + exp(in(0 ))) (2.12.11)

=

1Xn=1

bei(

0)n

+bei(

0)n

(2.12.12)

=b e

i(0)

1 b ei(0)+

b ei(0)

1 b ei(0)(2.12.13)

*bei(

0) =

bei(

0) =

b< 1

=

2b cos(0 ) 22b2 + 2 2b cos(0 ) (2.12.14)

(; ) =1

2

Z 20

(b; 0)b2 2

b2 + 2 2b cos(0 )d0 (2.12.15)

2.16 The potential on a unit square area with a uniform charge density

p.39 (1.44) 2.15

(x; y) =1

40

Z 10

dx0Z 10

dy0G(x; y;x0; y0) (2.16.1)

=2

0

1Xn=1

sin(nx)

n sinh(n)

Z 10

sin(nx0)dx0| {z }

(2.16.2)

sinh(n(1 y))

Z y0

sinh(ny0)dy0 + sinh(ny)Z 1y

sinh(n(1 y0))dy0

| {z }F

(2.16.3)


= 1n

cos(nx0)

10

(2.16.4)

= 1n

(cos(n) 1) (2.16.5)

=1 (1)n

n(2.16.6)

F = sinh(n(1 y)) 1n

cosh(ny0)

y0 sinh(ny) 1

n

cosh(n(1 y0))1

y

(2.16.7)

=1

n

sinh(n(1 y)) cosh(ny)| {z } sinh(n(1 y)) sinh(ny) + sinh(ny) cosh(n(1 y))| {z } (2.16.8)

=1

n

sinh(n(1 y) + ny) sinh(ny) + sinh(n(1 y)) (2.16.9)

=1

n

sinh(n) 2 sinh

n2

cosh

n

y 1

2

(2.16.10)

=2

0

1Xn=1

sin(nx)

n sinh(n) 1 (1)

n

n 1n

sinh(n) 2 sinh

n2

cosh

n

y 1

2

(2.16.11)

=2

30

1Xn=1

sin(nx)

n3(1 (1)n)

1 cosh(n(y

12 ))

cosh(n2 )

(2.16.12)

=4

30

1Xm=0

sin[(2m+ 1)x]

(2m+ 1)3

1 cosh[(2m+ 1)(y

12 )]

cosh[ (2m+1)2 ]

!(2.16.13)

2.20 Four symmetrically placed line charges 35

2.5 2.16

2.20 Four symmetrically placed line charges

(a)

(; ) =1

40

ZS

(0; 0)G(; ; 0; 0)da0 (2.20.1)

=1

40

a

3Xn=0

(1)2Z 20

d0Z 10

0d0(0 a)0 n

2

G(; ; 0; 0)

(2.20.2)

=1

40

a

3Xn=0

(1)2aG; ; a;

n

2

(2.20.3)

=

20

3Xn=0

1Xm=1

(1)n 1m

mcos

hm n

2

i(2.20.4)

n m m 4 2 4 cos(m)

(; ) =

20

1Xk=0

1

4k + 2

4k+24 cos[(4k + 2)] (2.20.5)

=

0

1Xk=0

1

2k + 1

4k+2cos[(4k + 2)] (2.20.6)


(b)

(; ) =2

0Re

" 1Xk=0

1

4k + 2

4k+2exp[i(4k + 2)]

#(2.20.7)

=2

0Re

266641Xk=0

1

4k + 2

0BBB@ exp(i)| {z }

1CCCA4k+237775 (2.20.8)

ln(1 + z) = z z2

2+

z3

3 z

4

4+

ln(1 z) = z z2

2 z

3

3 z

4

4

ln(1 + iz) = +iz +z2

2 iz

3

3 z

4

4+

ln(1 iz) = iz + z2

2+ i

z3

3 z

4

4

1

4[ln(1 + iz) + ln(1 iz) ln(1 + z) ln(1 z)] = z

2

2+z6

6+ (2.20.9)

) 14ln

1 + z2

1 z2=

1Xk=0

z4k+2

4k + 2(2.20.10)

(; ) =2

0Re

1

4ln

1 + 2

1 2

(2.20.11)

=

20Re

ln

2 + 1

2 1

(* Re[ln(z)] = Re[ln(z)]) (2.20.12)

=

20Re

ln

2e2i + a2

2e2i a2

(* > a a < )(2.20.13)

=

20Re

ln

(ei ia)(ei + ia)(ei a)(ei + a)

(* < = ; > = a) (2.20.14)

= Rew(ei) (2.20.15)

2.20 Four symmetrically placed line charges 37

(; ) =

20ln

(ei ia)(ei + ia)(ei a)(ei + a) (2.20.16)

=

40lnjei iaj2jei + iaj2jei aj2jei + aj2 (2.20.17)

=

40ln(x2 + (y a)2)(x2 + (y + a)2)((x a)2 + y2)((x+ a)2 + y2) (2.20.18)

2.3

(c)

< a (a)


2.6

k = 1 k = 0 xa

4(2.20.26)

2.24 The completeness relation for sine functions

() :=1Xn=1

An sin

n

(2.24.1)

Z 0

(0)2

1Xm=1

sin

m

sin

m0

d0 = () (2.24.2)

Z 0

sin

n

sin

m

d (2.24.3)

= 12

Z 0

cos

(n+m)

cos

(nm)

d (2.24.4)

= 12(n;m n;m) (2.24.5)

(2.24.2)

2.28 The potential at the center of a regular polyhedron 39

((2.24.2)-LHS) =2

1Xn=1

An

1Xm=1

sin

m

2 (n;mn;m)z }| {Z 0

(0)2

sin

m0

d0

(2.24.6)

= 1Xn=1

An

1Xm=1

sin

m

(n;m n;m) (2.24.7)

=

1Xn=1

An sin

n

(2.24.8)

= () (2.24.9)

(2.24.1)

( 0) = 2

1Xm=1

sin

m

sin

m0

(2.24.10)

2.28 The potential at the center of a regular polyhedron

n 0 k V 0 0(r) k lV l0(r)

(1; 2; ; n) i Vi (r) fgi(r)g

(r) =nXi=1

gi(r)Vi (2.28.1)

i Vi 1 1) V1

1) 0V


(r) = A(r;V2; V3; )V1 + (r;V2; V3; ) (2.28.2)

V1 V2 () V := V1 = V2 A(r;V2; V3; ) V2 2) A V3; V4; A V2

(r) = A(r)V1 +B(r;V3; V4; )V2 + (r;V3; V4; ) (2.28.3)

B V3; V4;

(r) =nXi=1

gi(r)Vi + C(r) (2.28.4)

V1 = V2 = = 0 0 C 0 (2.28.1)

rC

gi(rC) = (const.) (2.28.5)

gC

(rC) = gC

nXi=1

Vi (2.28.6)

gC V (r) V

V = gC

nXi=1

V = gCnV (2.28.7)

2) * V1 V2

2.28 The potential at the center of a regular polyhedron 41

gC = 1=n

(rC) =1

n

nXi=1

Vi (2.28.8)

()

Green

Green Green G(r; r0)

(r) = 14

IS

(r0)@G

@n0dS0 (2.28.9)

Vi(i = 1; 2; ; n) n

(r) = 14

nXi=1

Vi

ISi

@G

@n0dS0| {z }

=:4gi(r)

(2.28.10)

=nXi=1

gi(r)Vi (2.28.11)

(2.28.1)

(the mean value theorem)

V S := @V 0(r) (* Laplace )S 0(r) n n Vi


(r) =nXi=1

gi(r)Vi (2.28.12)

n ! 1 0(r)

(r) =

IS

g(r; r0)(r0)dS0 (2.28.13)

Green

@G

@n0=: 4g(r; r0)

rC

g(rC ; r0) = (const.) (2.28.14)

gC

(r) = gC

IS

(r0)dS0 (2.28.15)

(r0) V

V = gC 4R2 (2.28.16)

( R )

(r) =1

4R2

IS

(r0)dS0 (2.28.17)

43

Referential Sites

Ocial Sites, etc.

F J. D. Jackson Home Page(http://www-theory.lbl.gov/jdj/)

John David Jackson

F Errata(2010).pdf(http://www-theory.lbl.gov/jdj/Errata%282010%29.pdf)

Classical Electrodynamics

F UT2010(https://sites.google.com/site/jacksonut2010/)

Solutions

F Jackson Electrodynamics Solutions(http://www.airynothing.com/jackson/)

"Solutions in the left column are the problems I did myself. Solutions in the

right column were sent to me by Azar Mustafayev, ..." :(

F Jackson Physics Solutions(http://www-personal.umich.edu/~pran/jackson/)

"The only way to survive Jackson E&M is by standing on the

44 Referential Sites

shoulders of those who've gone before."

F Solutions to Jackson's Electrodynamics(http://www-personal.umich.edu/~jbourj/em.htm)

F Jackson's Electrodynamics solutions(http://web.ipac.caltech.edu/staff/turrutia/public_html/jackson/jackson.html)

"These are all I have. Maybe in the vast World Wide Web, the rest are hidden."

F Rudy's Physics Resource Page(http://www.physics.rutgers.edu/~rmagyar/physics/)

PDF http://www.physics.rutgers.edu/~rmagyar/physics/jackson.pdf

F Solutions to problems of Jackson's Classical Electrodynamics byKasper van Wyk

(http://samizdat.mines.edu/jackson/)

Chapter 1, 2, 8+

F Walter Johnson - Electromagnetism(http://www.nd.edu/~johnson/Classes/E&M/probNN.pdf)

"NN" 1-11

http://www.nd.edu/~johnson/Classes/E&M/prob1.pdf http://www.nd.edu/~johnson/Classes/E&M/prob2.pdf http://www.nd.edu/~johnson/Classes/E&M/prob3.pdf http://www.nd.edu/~johnson/Classes/E&M/prob4.pdf http://www.nd.edu/~johnson/Classes/E&M/prob5.pdf http://www.nd.edu/~johnson/Classes/E&M/prob6.pdf http://www.nd.edu/~johnson/Classes/E&M/prob7.pdf http://www.nd.edu/~johnson/Classes/E&M/prob8.pdf

45

http://www.nd.edu/~johnson/Classes/E&M/prob9.pdf http://www.nd.edu/~johnson/Classes/E&M/prob10.pdf http://www.nd.edu/~johnson/Classes/E&M/prob11.pdf

47

About us

Jackson (Y.Hotta, M.Hyuga, T.Matsuda,Y.O^no)

Y.Hotta

4 2 :

M.Hyuga

4 0 : 1.11

T.Matsuda

4 3 :

Y.Ohno

4 1 :

? M E M O ?

Solution to Classical Electrodynamics

Built on 2010/4/20 (Rev. 41)

/ Jackson Consortium

(https://sites.google.com/site/jacksonut2010/)

This document is licensed under

the Creative Commons BY-NC-SA license.

Free to use, but absolute no warranty.

c Jackson Consortium 2010 Edited in Japan

IntroductionChapter 1 Introduction to Electrostatics1.2 Dirac delta function for a general orthogonal coordinate system1.4 Electric fields produced by a spherically symmetric charge density1.6 The capacitance of simple capacitors1.8 Energy densities of certain capaciters1.10 The mean value theorem (for electrostatics)1.11 Normal derivertive of the electric field at the surface of a curved conductor1.12 Green's reciprocation theorem1.14 The behavior of Green functions for Poisson equation1.16 Energy decreasing with introduction of a conductor1.18 The variational principal for capacitance1.20 Theory of the capacitance estimation1.22 The mathematical bases for relaxation method1.24 Numerical analysis performed by the relaxation method II

Chapter 2 Boundary-Value Problems in Electrostatics: I2.4 A point charge placed outside a conducting sphere2.8 Two straight parallel line charges with equal and opposite charge densities2.12 Poisson's integral form of the potential2.16 The potential on a unit square area with a uniform charge density2.20 Four symmetrically placed line charges2.24 The completeness relation for sine functions2.28 The potential at the center of a regular polyhedron

Referential SitesAbout us

Documents

JD Jackson Solution Chapter 1 and 2