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(

1r 2r

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=∞)=0V(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=∞)=0V(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=∞)=0V(x=∞)=0

56

Metal

x1

Metal

E(x=∞)=0V(x=∞)=0

x1

x2=-x1

E(x=∞)=0V(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