6. electrostatic boundary-value problems · 3 if a solution to laplace’s equation can be found...
TRANSCRIPT
6. Electrostatic Boundary-Value Problems
1
6.2 Poisson’s and Laplace’s Equation
)5.6()equations'Laplace( 0V
0
)4.6()equations'Poisson(V
)3.6()V(
)2.6(VE
)1.6(ED
2
v
v2
v
v
2
)8.6(0V
sinr
1Vsin
sinr
1
r
Vr
rr
1
)7.6(0z
VV1V1
)6.6(0z
V
y
V
x
V
coodinates Sphericalor l,Cylindrica Catesian,in equation sLaplace'
2
2
222
2
2
2
2
2
2
2
2
2
2
2
2
2
3
If a solution to Laplace’s equation can be found
that satisfies the boundary conditions,then the solution is unique.
6.3 Uniqueness Theorem
4
6.4 General Procedure for Solving Poisson’sor Laplace Equation
6.4 Section에서 문제 풀이할 때 사용하는 방정식.
I/VR
SdJI
dSQ
D
ED
VE
S
S
nS
5
EX 6.1 그림 6.1과 같이 EHD Pump에서 ρ0=25 mC/m3 인 Charge로 채워져 있고V0=22 kV 일 때 Pump의 압력을 구하라.
Figure 6.1 An electrohydrodynamic
pump.
BAzz2
V
Azdz
dV
dz
Vd
V
20
0
02
2
02
6
d
V
2
dz
dz
dVE
Vzd
V
2
dz
2V
d
V
2
dA
VAdd2
0
0)dz(V
VB
B00V
V)0z(V
BAzz2
V
000z
00020
00
020
0
0
0
20
Figure 6.1 An electrohydrodynamic
pump.
7
Figure 6.1 An electrohydrodynamic
pump.
00
d
0
002
00
d
0000
0
d
0 z0
v z3
0
z0z
000
z
00020
SVF
zd
V
2
d
2
zS
dzd
V
2
dzS
dzEdS
xEdF
dvEdQEdF
d
V
2
dz
dz
dVE
Vzd
V
2
dz
2V
8
2
33
00
00
d
0
002
00
d
0000
0
d
0 z0
v z3
0
z0z
000z
m/N550
10221025
VS
Fp
SVF
zd
V
2
d
2
zS
dzd
V
2
dzS
dzEdS
xEdF
dvEdQEdF
d
V
2
dzE
Figure 6.1 An electrohydrodynamic
pump.
9
EX 6.2 그림 (a)에서 전극 사이의 전장을 구하라.
복사기 원리.
* 광전도체: 빛을 받으면 전기전도성이 생기는 물질.
(c)
0
a
d
x
(b)
재결합
빛E1
E20
a
d
x
(a)
광전도체
ρsε1
ε20
a
d
x
10
2211S
22
11n22n11n2n1S
211
Sn2n1
21
22
1111
2
1
222
111
v2
22
AA
dx
dV
dx
dVEEDD
aABaA
)ax(DD
)ax(V)ax(V
0BB00
dABBdA0
0)0x(V
0)dx(V
)ax0(BxAV
)dxa(BxAV
BAxV
00dx
VdV
ρsε1
ε20
a
d
x
11
ax0ata
d1/
)1a/d(
A
dx
dVE
dxaata
d1/
A
dx
dVE
B,B,A,A:unknows4&
eqs4ofsetconsistentSelf
AA
aABaA
0B
dAB
2
1
1
2
1
S
2
2x2
2
1
1
2
1
S
1
1x1
2121
2211S
111
2
11
ρs
ε1
ε20
a
d
x
12
EX 6.3 φ=0 평면의 전위가 0 volt, φ=π/6 평면의 전위가 100 volt 일 때 작은간격으로 분리된 두 면 사이의 전위와 전장을 구하라.
600
d
dV1
az
Va
V1a
VV)29.3(VE
600V
/600AB6/A100
0BB00
100)6/(V
0)0(V
BAV
0z
VV1V1)7.6(0
d
Vd1V
z
2
2
2
2
22
2
2
2
φ0 = π/6
0 volt 100 volt
x
y
z
13
aV
sinr
1a
V
r
1a
r
VV)30.3(
az
Va
V1a
VV)39.3(
az
Va
y
Va
x
VV)28.3(
r
z
zyx
14
Bsin
dAV
dsin
AdV
sin
A
d
dV
Ad
dVsin
0d
dVsin
d
d
0d
dVsin
d
d
sinr
1V
sinr
1a
r
1a
ra)30.3(
0V
sinr
1Vsin
sinr
1
r
Vr
rr
1)8.6(
2
2
r
2
2
222
2
2
EX 6.4 θ=π/10 면의 전위가 0 volt, θ=π/6 면의 전위가 50 volt 일 때 작은 간격으로 분리된 두 면 사이의 전위와 전장을 구하라.
V0z
θ1
θ2
Gap
15
tan 1 cot
sin cos
cscsec
)2/tan(ln
)2/tan(
)2/tan(d
)2/tan(
d)2/(sec)2/1(
)2/cos(/)2/sin()2/(cos2
d
)2/sin()2/cos(2
d
sin
d
2
2
2
22
secd
tand
sind
cosd
cosd
sind
tantan1
tantan)tan(
sincos)2cos(
sinsincoscos)cos(
cossin2)2sin(
sincoscossin)sin(
16
B)2/tan(lnA
Bsin
dAV
dsin
AdV
sin
A
d
dV
Ad
dVsin
0d
dVsin
d
d
0d
dVsin
d
d
sinr
1V
2
2
)20/tan()20/tan(
)12/tan(ln/50B
)20/tan(
)12/tan(ln/50A
B)12/tan(lnA50
B)20/tan(lnA0
50)6/(V
0)10/(V
B)2/tan(lnAV
17
)20/tan(
)12/tan(ln/
sinr
50
sinr
A
d
dV
r
1E
)20/tan()20/tan(
)12/tan(ln/50))2/ln(tan(
)20/tan(
)12/tan(ln/50
B))2/ln(tan(AV
)20/tan()20/tan(
)12/tan(ln/50B
)20/tan(
)12/tan(ln/50A
sinr
1a
r
1a
ra)30.3(
0V
sinr
1Vsin
sinr
1
r
Vr
rr
1)8.6(
r
2
2
222
2
2
18
EX 6.5 길이가 무한대인 사각형 관내의 전위 분포를 구하라.
y
a
0b
0VV
0V
x
0
2
2
2
22
V)a(Y)x(X)a,x(V
0)0(Y0)0(Y)x(X)0,x(V
0)b(X0)y(Y)b(X)y,b(V
0)0(X0)y(Y)0(X)y,0(V
0YY
0XX
Y
Y
X
X
0XYYX
)y(Y)x(X)y,x(V
0y
V
x
VV
19
y
a
0b
0VV
0V
x
sMeaningles:0)x(X
0A0Ab0)b(X
0B0)0(X
BAxX
0X
0XX
0:1Case
20
dxAX
dX
AXdx
dX
AXdx
dX
XdX2dx
dXd
0dx
dXX
dx
dX
dx
d
2
1
0Xdx
dX
dx
Xd
dx
dX
0XX
0XX
0:2Case
22
22
222
22
22
2
2
2
2
Bxi
di
cos
dcosi1
1sini
)sini(d1
1XA
XA
d1
BxAX
dX
dxAX
dX
2
2
22
22
y
a
0b
0VV
0V
x
21
it
it
Cit
Cit
etsinitcos
eC
ee
etsinitcos
Cit)tsinitln(cos
xlnx
dx
idttsinitcos
)tsinit(cosd
idt)tsinit(cos)tsinit(cosd
tcositsindt
)tsinit(cosd
참고
22
y
a
0b
0VV
0V
x
x2
x1
xBxB
)Bx(2)Bx(
)Bx()Bx(222
)Bx(2
2
i
eAeA
eeA
eeA
X
1eXeA
XeA
eXA
1XA
eXA
1XA
XA
1XA
ln
siniXA
)siniln(cos
eln
i)Bx(
iBx
23
y
a
0b
0VV
0V
x
sMeaningles:0)x(X
0B
0B
bsinhB000)bx(X
01B00)0x(X
xsinhBxcoshB
2
eexsinh
2
eexcosh
eAeAX
2
1
2
1
21
xx
xx
x2
x1
24
- β=0는해의더하기에서의미가없음.
- n이Minus인경우 Plus인경우의 sine 함수의상수배이기때문에해를고려할필요가없음.
b
ynsinhhY
0YY
b
xnsing)x(X
...,3,2,1n,b
n
nbnsin0bsin
0bsinbsing000)bx(X
0g01g00)0x(X
xsingxcosg
eCeC)x(X
0XX
0XX
0:3Case
nn
2
nn
1
00
10
xi1
xi0
2
1nn
nnn
b
ynsinh
b
xnsinc)y,x(V
b
ynsinh
b
xnsinhg)y,x(V
합해의여러
0YY
0XX
y
a
0b
0VV
0V
x
25
짝수
홀수
n,0
n,
b
ansinhn
V4
c
...,6,4,2m,0
...,5,3,1m,m
V4
)mcos1(m
V2
b
amsinhc
)mcos1(m
bV
2
b
b
amsinhc
dxb
xmsinVdx
b
xmsin
b
amsinhc
dxb
xmsinV
b
ansinh
b
xnsinc
dxb
xmsin)a,x(V
Vb
ansinh
b
xnsinc)a,x(V
b
ynsinh
b
xnsinc)y,x(V
0
n
00
m
0m
b
00b
02
m
b
0 01n
n
b
0
01n
n
1nn
,..5,3,1n
0
b
ansinhn
b
ynsinh
b
xnsin
V4)y,x(V
y
a
0b
0VV
0V
x
26
0
b
x)nm(sin
)nm(
b
b
x)nm(sin
)nm(
b
dxb
x)nm(cos
b
x)nm(cos
dxb
xnsin
b
xmsin2
)nm:1case(
)cos()cos(sinsin2
sinsincoscos)cos(
sinsincoscos)cos(
dxb
xnsin
b
xmsin
b
0
b
0
b
0
b
0
b
0
계산
tan 1 cot
sin cos
cscsec
2
22
secd
tand
sind
cosd
cosd
sind
tantan1
tantan)tan(
sincos)2cos(
sinsincoscos)cos(
cossin2)2sin(
sincoscossin)sin(
27
tan 1 cot
sin cos
cscsec
2
22
secd
tand
sind
cosd
cosd
sind
tantan1
tantan)tan(
sincos)2cos(
sinsincoscos)cos(
cossin2)2sin(
sincoscossin)sin(
2
bdx
b
xmsin
b
b
xm2sin
m2
bx
dxb
xm2cos1
dxb
xmsin2
)nm:2case(
dxb
xmsin
b
xmsin
b
02
b
0
b
0
b
02
b
0
계산
2
22
sin21
sincos)2cos(
28
EX 6.6 Ex 6.5에서 V0가 다음 값을 가질 때 전위 분포를 구하라.
y
a
0b
0VV
0V
xb
a3sinh
b
y3sinh
b
x3sin10)y,x(V
b
a3sinh/10c
b
a3sinhc10
3n,0c
b
ansinh
b
xnsinc
b
x3sin10V)a,x(V
)19.5.6(b
ynsinh
b
xnsinc)y,x(V
bx0,ay,b
x3sin10V)a(
33
n
1nn0
1nn
0
,..5,3,1n
0
b
ansinhn
b
ynsinh
b
xnsin
V4)y,x(V
29b
a5sinh10
b
y5sinh
b
x5sin
b
asinh
b
ysinh
b
xsin2
)y,x(V
b
a5sinh10
1c
b
a5sinhc
10
1
b
a5sinh
2c
b
asinhc2
5,1n,0c
b
ansinh
b
xnsinc
b
x5sin
10
1
b
xsin2)a,x(V
)19.5.6(b
ynsinh
b
xnsinc)y,x(V
bx0,ay,b
x5sin
10
1
b
xsin2V)b(
55
11
n
1nn
1nn
0
y
a
0b
0VV
0V
x
,..5,3,1n
0
b
ansinhn
b
ynsinh
b
xnsin
V4)y,x(V
30
EX 6.7 전하가 없는 영역에서 𝛻2V(ρ, φ, z)의 분리된 미분방정식을 구하라.
zsinhczcoshcZ
)7.7.6(0ZZ
z
Z
Z
1
z
Z
Z
11R
R
1
0z
ZR
RZRZ
)z(Z)()(RV
0z
VV1V1V
21
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
ftBessel:)r(JR
)11.7.6(0R)(RR
sinccosc
)10.7.6(0
1R
R
R
R
1R
R
n
2222
43
2
2
2
222
2
2
22
2
2
31
1. Choose a suitable coordinate system
2. Assume Vo as the potential difference between conductor terminals
3. Solve Laplace’s Equation to obtain V, then E and I in Eq. (6.16)
4. Obtain R as Vo/I
6.5 Resistance and Capacitance
)16.6(SdE
LdE
SdJ
LdE
I
VR
uniform) isty conductivi theifonly (validS
lR
)17.6(LdEVVV1
221
32
1. Assuming Q and determining V in terms of Q (Gauss’s Law): Section A-C
a) Choose a suitable coordinate system
b) Let the two conducting plates carry charges +Q and –Q
c) Determine E (Coulomb’s or Gauss’s Law), then V (a function of Q)
d) Obtain C = Q/V
2. Assuming V and determining Q in terms of V (Laplace’s Eq.) : Ex 6.10/14
a) Choose a suitable coordinate system
b) Assume V0 as the potential difference between conductor terminals
c) Solve Laplace’s Equation to obtain V, then Ed) Obtain Q as a function of V
e) Obtain C = Q/V
)18.6(LdE
SdE
V
QC
33
Figure 6.12 A two-conductor capacitor.
34
A. Parallel-Plate Capacitor
)22.6(d
S
V
QC
)21.6(S
Qd
adxaS
Q
LdEV
)20.6(aS
Q
)a(E
)19.6(S
Q
xd
0 x
1
2
x
xS
S
Dielectric ε
Plate area S x
E
1
2
35
Figure 6.13 (a) Parallel-plate capacitor.
(b) Fringing effect due to a parallel-plate capacitor.
(a)
Dielectric ε
Plate area S x
E
1
2
36
)24.6(C2
QQV
2
1CV
2
1W
QV2
1
C2
Q
S
d
2
Q
S
Q
S
d
d
Q
d
C/Q
d
VE
S2
)Sd(Qdv
S
Q
2
1dvE
2
1W
)23.6(
d
SC
d
S
V
QC
C
C
22
E
22
22
2
v
2
v2
E
00
0r
37
B. Coaxial Capacitor
ρ
L
Dielectric ε
)28.6(
a
bln
L2
V
QC
)b27.6(a
bln
L2
Q
)a27.6()a(daL2
Q
LdEV
)26.6(aL2
QE
)25.6(L2ESdEQ
a
b
1
2
38
C. Spherical Capacitor
)32.6(
b
1
a
1
4
V
QC
)31.6(b
1
a
1
4
Q
)a(drar4
Q
LdEV
)30.6(ar4
QE
)29.6(r4ESdEQ
a
b rr2
1
2
r2
2r
2
1 r
Dielectric ε
a
b
39
)35.6(RC
LdE
SdE
V
QC)17.6(
SdE
LdE
I
VR)16.6(
)34.6(CCC
)33.6(CC
CCC
C
1
C
1
C
1
21
21
21
21
C1 C2
(b)
C1
C2
(a)
40F1007.7
2
104
36
104
)39.6(a4
1R,a4C
)38.6(4
b
1
a
1
R,
b
1
a
1
4C
)37.6(L2
a
bln
R,
a
bln
L2C
)36.6(C
RRCS
dR,
d
SC
4
79
지구
41
)a/ln()a/bln(
VV
)a/bln(/aVlnB
)a/bln(/VA
BblnAVV)b(V
BalnA00)a(V
BlnAV
AddV
Ad
dV
0d
dV
d
d
0d
dV
d
d1
0z
VV1V1V
0
0
0
00
2
2
2
2
2
2
EX 6.8 (a) 90o로 구부러진 도체의 ρ=a와 ρ=b 사이의 저항을 구하라.
ab
t
전도율=σx
y
z
42
t
)a/bln(2
I
VR
)a/bln(
Vt
2
ddz)a/bln(
V
SdJI
EJ
)a/bln(
V
d
dVE
)a/ln()a/bln(
VV
0
0
2/
0
t
00
0
0
aV
sinr
1a
V
r
1a
r
VV)30.3(
az
Va
V1a
VV)29.3(
az
Va
y
Va
x
VV)28.3(
r
z
zyx
ab
t
전도율=σx
y
z
43
EX 6.8 (b) 90o로 구부러진 도체의 z=0와 z=t 사이의 저항을 구하라.
)ab(
t4
I
VR
t4/)ab(Vddt
VdSJI
t
VEJ
t
V
dz
dVE
zt
VV
t/VAV)tz(V
0B0)0z(V
BAzV
0z
V
0z
VV1V1V
220
220
b
a
2/
00
z
0zz
0z
0
00
2
2
2
2
2
2
2
2
ab
t
전도율=σx
y
z
44
)ab(
t4
4
)ab(
t
S
LR
22
22
방법다른
45
EX 6.9 무한 동축 Cable의 단위 길이당 ρ=a, ρ=b 사이의 저항과 단위 길이당Conductance를 구하라. a와 b 사이는 도전율 σ인 물질로 채워져 있다.
)a/ln()a/bln(
VV
)a/bln(/aVlnB
)a/bln(/VA
BblnAVV)b(V
BalnA00)a(V
BlnAV
0d
dV
d
d1
0z
VV1V1V
8.6Ex
0
0
0
00
2
2
2
2
2
2
유사과
46
)a/bln(
2
2
)a/bln(
1
LL2
)a/bln(
1
RL
1G
L2
)a/bln(
I
VR
)a/bln(
VL
2ddz
)a/bln(
VSdJI
EJ
)a/bln(
V
d
dVE
)a/ln()a/bln(
VV
0
02
0
L
00
0
0
47
b
1
a
1b
1
r
1
VV
a
1
b
1/
b
VB
a
1
b
1/VA
0Bb
A)br(V
100Ba
A)ar(V
Br
AV
r
A
dr
dV
0dr
dVr
dr
d
0dr
dVr
dr
d
r
1V
V
sinr
1Vsin
sinr
1
r
Vr
rr
1V)62.3(
00
0
2
2
2
2
2
2
2
222
2
2
2
EX 6.10 도체 이중 구각에서 V(r=a)=100 volt, V(r=b)=0 volt일 때 V, Er 분포를 구하라. 사이 물질은 εr=2.5 일 때 총전하와 Capacitance를 구하라.
Er
a=10 cm
b=30 cm
48
b
1
a
1
4
V
QC
b
1
a
1
V4
ddsinr
b
1
a
1r
V
dSEQ
b
1
a
1/
r
V
r
A
dr
dVE
r0
0
0r0
0
2
02
2
0r0
r
20
2r
반지름 r인 축구공껍데기 적분
aV
sinr
1a
V
r
1a
r
VV)30.3(
az
Va
V1a
VV)29.3(
az
Va
y
Va
x
VV)28.3(
r
z
zyx
Er
a=10 cm
b=30 cm
49
θ는 Zenith Angle이고φ는Azimuthal Angle.
φ
z
x
y
θ
rdφ
r
dr
rsinθdφ
rdθρ=rsinθ
adrdr
addrsinr
addsinrSd r2
50
EX 6.11 V0를 가정하고 Q를 유도하여 평면사이의 C=εS/d 를 유도하라.
S0VV
0V
x
d
d
S
V
QC
d
SV
dSd
VdS
dx
dVdSEdSDdSQ
xd
VV
.volt0
.voltV
BAxV
0dx
VdV
0
0
0nnS
00
2
22
가정라고하판을
가정라고상판을그림의
51
EX 6.12 Capacitance를 구하라. d=5mm, S=30cm2.
41r
62r
2/d
2/d
)a(
1r2r
2/w 2/w)b(
)(d2
SCCC
Capacitor)b(
d
S2
CC
CCC
2/d
SC
2/d
SC
Capacitor)a(
2r1r0
21
2r1r
2r1r0
21
21
2r02
1r01
연결병렬의
연결직렬의
52
EX 6.13 a=1cm, b=2.5cm이고 극판사이에 εr=(10+ρ)/ρ 인 유전체가 있다. ρ는 cm 단위 이다. 단위 길이당 Capacitance를 구하라.
적분을 위한Closed Surface
La
b
L
dL 방향
a10
b10ln
2
LV
Q
L
C
a10
b10ln
L2
Q
10
d
L2
Q
dL2
QV
L2
QE
L2ESdEQ
0
0
a
b0
a
b
53Figure 6.21 Image system: (a) charge configurations above a perfectly conducting plane, (b) image configuration with the conducting
plane replaced by equipotential surface.
The Image Theory states that a given charge configuration above an infinite grounded perfect conducting plane may be replaced by the charge configuration itself, its image, and an equipotential surface in the place of the conducting plane.
6.6 Method of Image
54
310
1
1kv
10
1kv2
)xx(4
)xx(QE
)xx(QE
xx4
QV
)xx(QV
x=-x1
E(x=∞)=0
V(x=∞)=0
Q
δ(x-x0)
x0 x0+Δx
x
1/Δx
Δx
)x(fdx)xx()x(f
)zz()yy()xx()xx(
00
kkkk
55
x1
Metal
E(x=∞)=0
V(x=∞)=0
320
23
10
1
2k1kv
2010
2k1kv2
)xx(4
)xx(Q
)xx(4
)xx(QE
)xx(Q)xx(QE
xx4
Q
xx4
QV
)xx(Q)xx(QV
δ(x-x0)
x0 x0+Δx
x
1/Δx
Δx
x1
x2=-x1
E(x=∞)=0
V(x=∞)=0
56
Metal
x1
Metal
E(x=∞)=0
V(x=∞)=0
x1
x2=-x1
E(x=∞)=0
V(x=∞)=0
57
A. A point Charge Above a Grounded Conducting Plane
)45.6(])hz(yx[
1
])hz(yx[
1
4
QV
r4
Q
r4
Q
VVV
)44.6(
])hz(yx[
a)hz(ayax
])hz(yx[
a)hz(ayax
4
QE
)43.6()hz,y,x()h,0,0()z,y,x(r
)42.6()hz,y,x()h,0,0()z,y,x(r
)41.6(r4
rQ
r4
rQ
)40.6(EEE
2/12222/12220
2010
2/3222
zyx
2/3222
zyx
0
2
1
320
2
31o
1
58
Figure 6.22 (a) Point charge and grounded conducting plane.
(b) Image configuration and field lines.
59
)49.6(Q
|]h[
Qh
)(d2
1]h[2
2
QhQ
)48.6(]h[
dd
2
QhQ
dddxdy
)47.6(]hyx[2
QhdxdydSQ
)46.6(]hyx[2
Qh
|ED
])hz(yx[
a)hz(ayax
])hz(yx[
a)hz(ayax
4
QE
02/122
022/322
i
0 2/322
2
0i
2
0 0
2/3222Si
2/3222
0zn0nS
2/3222
zyx
2/3222
zyx
0
60
2
0 0dddxdy
y
z
dφ
x
ρ
dρdρdφ
dxdy
x
z
y
61
B. A Line Charge above a Grounded Conducting Plane
y
zP(x,y,z)
(0,y,h)
(0,y,-h)x
h
-h
)55.6(ln2
ln2
ln2
VVV
)54.6()hz(x
a)hz(ax
)hz(x
a)hz(ax
2E
)53.6()hz,0,x()h,y,0()z,y,x(
)52.6()hz,0,x()h,y,0()z,y,x(
)51.6(a2
a2
)50.6(EEE
2
1
0
L
20
L1
0
L
22zx
22zx
0
L
2
1
220
L1
10
L
62
)59.6(
dsechdx
tanhx
h
dh
(6.58)length)unit per (charge hx
dxhdx
)57.6()hx(
h|ED
)56.6()hz(x
)hz(xln
2V
ln2
V)55.6(
)hz(x
a)hz(ax
)hz(x
a)hz(ax
2E(6.54)
L
2
2/
2/L
22L
Si
22L
0zz0nS
2/1
22
22
0
L
2
1
0
L
22zx
22zx
0
L
63
EX 6.14 그림 (a)와 같이 두 개의 반 무한 평면 사이에 점전하 Q가 (a,0,b)에 있다.전위를 구하고 Q에 작용하는 힘을 구하라.
x
z
b
a
)a(
Q
x
z
b
aQ
Q2
Q
Q4
a
a2
b2
b
)b(
Q
Q3
Q
Q1
z22/322x22/3220
2
z20
2
2/322zx
0
2
x20
2
4131211
2224
2223
2222
2221
43210
ab
1
)ba(
ba
a
1
)ba(
a
16
Q
a)b2(4
Q
])b2()a2[(
ab2aa2
4
Qa
)a2(4
Q
FFFF
])bz(y)ax[(r
])bz(y)ax[(r
])bz(y)ax[(r
])bz(y)ax[(r
r
1
r
1
r
1
r
1
4
QV
64
Figure 6.25 Point charge between two
semi-infinite conducting walls inclined
at 60° to each other.
1
360N
imageofNumber
o