EECS 117 Homework Assignment 5 Spring 2004

4. At the boundary of the two perfect dielectrics, there is not free charge. So the normal component of D is continuous across the boundary, i.e., nn DD 21 = . So the scalar potential is continuous across the boundary. The tangential component of E is also continuous, i.e.,

. tt EE 21 = 5. Using Gauss’s law, one can easily show that the electric field outside of the Earth is

equal to , where Q is the charge present on the Earth’s “conducting” surface. The scalar potential is

rrQE^

20 )4/( πε=

)4/( 0rQ πε=Φ . The capacitance is equal to

aarQC 04)(/ πε=== Φ

aQE 0max 4/(= πε

, where a is the radius of the Earth. Evaluation of the expression yields C = 709 µF.

1 22

2

∂∂

=Φ∇ rrr

1 )(r =Φ

2 )(r =Φ

ΦΦ

(b) The air typically can sustain an electric field of 3 x 106 V/m before breakdown. This would imply that the maximum charge that can be put onto the Earth is governed by

CaQ 1020

6max

62 1035.1)4(103103) ×=×=⇒×= πε 6 (a). Since there is no free charge in the dielectric regions, we have the Laplace equation to describe the electrostatic potential inside the regions. Exploiting the spherical symmetry of the problem, we have:

0sin1sin

sin1

2

2

222 =∂Φ∂

∂∂

+

∂Φ∂

∂∂

+

∂Φ∂

φφθθθ

θθ rrr

Because of the symmetry, Φ only varies in r. Thus only the radial term in the above expression remains. Thus, we have

DrC

rC

r

Cr

r

rr

r

+−=Φ

=∂Φ∂

=∂Φ∂

=

∂Φ∂

∂∂

2

2

2 0

where, C and D are constants. To account for the different dielectric constants in the two regions, we set up the following expressions:

11 D

rC

+− for rR bi R≤≤

22 D

rC

+− for rR ob R≤≤

The subscript 1 and 2 denotes the dielectric region closer to the inner shell and to the outside shell, respectively. With an assumption that the inner shell has a voltage V0 and the outer one is grounded, the boundary conditions are:

01 )( VRi = (1) 0)(2 =oR (2)

)()( 11 bb RR Φ=Φ (3)

)()( 21 bnbn RDRD = (4) The last condition is true because there is no free charge on the boundary surface between the two dielectric. Applying the conditions towards this problem, we have

011 VD

RC

i

=+−

022 =+− D

RC

o

22

11 D

RCD

RC

bb

+−=+−

22

221

1bb R

CRC

εε −=−

Expressing D1 and D2 in terms of C1 and C2 using the first and second equations, and substituting the results into the third equation yield:

2101111 CRR

CRR

Vbobi

−=

−+

Using the fourth equation, one can find

( )ibo RRR

VC111

01

2

1

2

1 1 −−+=

εε

εε

Therefore,

( )

−

−−++=++−=+−=

rRVV

RCV

rCD

rC

iRRRi ibo

111 111

00

10

11

11

2

1

2

1εε

εεΦ

( )

−

−−+=+−=+−=

rRV

RC

rCD

rC

ibo RRRo

111 0

1110

2

11

2

11

2

12

22

2

1

2

1εε

εεε

εεε

εε

Φ

Then, use the fact that ε2 = 2 ε1 = 2 εr, we get

rr rVE

boi RRRr

^

22

12

110

1

^

11

−−=−∇= Φ and rr r

VD

boi RRR

rr

^

22

12

110

^ 1−−

=ε

rr rVE

boi RRRr

^

21120

1

^

21

−−=−∇= Φ and rr r

VDboi RRR

rr

^

22

12

110

^ 1−−

=ε

(b) The surface density of the inner shell is given by

22

12

110

^

1

^ 1)(iRRR

rirsi R

VRrD

boi

rr −−==⋅=

ερ

The capacitance is then equal to

boi RRR

rsii

VR

VQC

21

211

0

2

0

44−−

===περπ

7. The problem can be analyzed using method of image, namely the ground plane can be replaced by two image wires distance h below the ground.

2

h

d 1

In general, the voltage on the surface of the wire 1 is given by

21211112111 QPQPVVV +=+= (1) V11 is the voltage on the surface of the wire 1 due to a charge Q1 in the wire 1 (and -Q1 in its image wire), and V12 is the voltage of wire 1 due to a charge Q2 in the wire 2 (and –Q2 in its image wire). To find V11, we neglect wire 2 and calculate the potential on the wire surface due to a charge Q1 inside the wire and an image charge –Q1. This potential is same as the one due to a dipole evaluated on the surface of the positive charge.

hQ

aQ

V0

1

0

111 84 πεπε

−=

or

haP

0011 8

14

1πεπε

−=

Likewise, for wire 2, we have

22122221222 QPQPVVV +=+= (2) Because of the symmetry, P11 = P22, and P12 = P21. To calculate P21 (and thus P12), we need find V21, the voltage on the surface of wire 2 due to a charge Q1 inside wire 1 (and -Q1 in its image wire). The potential is

2/1220

1

0

121 )4(44 dh

Qd

QV

+−=

πεπε

P21 and P12 are equal to

2/12200

1221 )4(41

41

dhdPP

+−==

πεπε

The equation (1) and (2) can be expressed as V = P Q, or . We

can express the Q vector in terms of V vector by inverting P matrix, i.e.

=

2

1

2221

1211

2

1

QQ

PPPP

VV

−

−

−=

−

−−

=

=

−

2

1

1112

12112

212

112

1

1121

1222

122122112

11

2221

1211

2

1 11VV

PPPP

PPVV

PPPP

PPPPVV

PPPP

QQ

In the equation form, we have

2111122

2121111

VcVcQVcVcQ

+=+=

where )(11 2

212

11

11PP

P−=c , )(12 2

212

11

12PP

P−−=c . These two equations can be expressed by self

capacitance C11 and mutual capacitance C12, such that

)()(

12122112111122

21121112121111

VVCVCVcVcQVVCVCVcVcQ−+=+=−+=+=

One can see that C11 = c11+ c12, and C12 = - c12. So

C11 = ( ) ( )24

112211

411

211

022

22

4dhdha

dhdha

+

+

−−−

+−−πε and C12 = ( ) ( )2

4112

211

411

022

22

4dhdha

dhd

+

+

−−−

−πε .

8(a). In both dielectric-filled and vacuum-filled capacitor areas, the electric field is equal

to -V0/d , where y axis is pointing up normal to the capacitor plates. But in the dielectric

region, D = ε

j^

r E = - εr V0/d and ρj^

s = - εr V0/d. The vacuum region has D = ε0 E = - ε0

V0/d and ρj^

s = - ε0 V0/d. (b). The electrostatic energy Ud stored inside the dielectric region is equal to

xwd

Vxdw

EU rr

d 22

20

2εε

== . By the same token, The electrostatic energy Ud stored

inside the vacuum region is described by wxld

Vdwxl

Ed )(

2)(

2

200

20 −=−=

εεU . The

value of x for equal amount of energy stored in each region is given by 0//)( εε rxxl =− , or )/( 00 rlx εεε += .