jd jackson solution chapter 1 and 2
Post on 20-Nov-2015
1.591 Views
Preview:
DESCRIPTION
TRANSCRIPT
-
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)
-
4 Chapter 1 Introduction to Electrostatics
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)
-
6 Chapter 1 Introduction to Electrostatics
(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) )
-
8 Chapter 1 Introduction to Electrostatics
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
-
10 Chapter 1 Introduction to Electrostatics
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)
-
12 Chapter 1 Introduction to Electrostatics
(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
-
14 Chapter 1 Introduction to Electrostatics
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
-
16 Chapter 1 Introduction to Electrostatics
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
-
18 Chapter 1 Introduction to Electrostatics
(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
-
20 Chapter 1 Introduction to Electrostatics
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
-
22 Chapter 1 Introduction to Electrostatics
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
-
2.8 Two straight parallel line charges with equal and opposite charge densities 25
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)
-
26 Chapter 2 Boundary-Value Problems in Electrostatics: I
(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)
-
2.8 Two straight parallel line charges with equal and opposite charge densities 27
""
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)
-
28 Chapter 2 Boundary-Value Problems in Electrostatics: I
(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)
-
2.8 Two straight parallel line charges with equal and opposite charge densities 29
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)
-
30 Chapter 2 Boundary-Value Problems in Electrostatics: I
""
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
-
32 Chapter 2 Boundary-Value Problems in Electrostatics: I
(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)
-
34 Chapter 2 Boundary-Value Problems in Electrostatics: I
= 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)
-
36 Chapter 2 Boundary-Value Problems in Electrostatics: I
(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)
-
38 Chapter 2 Boundary-Value Problems in Electrostatics: I
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
-
40 Chapter 2 Boundary-Value Problems in Electrostatics: I
(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
-
42 Chapter 2 Boundary-Value Problems in Electrostatics: I
(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 ?
-
? 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
top related