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., D1n = D2 n . So the scalar potential
is continuous across the boundary. The tangential component of E is also continuous, i.e.,
E1t = E 2t .
5. Using Gauss’s law, one can easily show that the electric field outside of the Earth is
^
equal to E = Q /(4πε 0 r 2 ) r , where Q is the charge present on the Earth’s “conducting”
surface. The scalar potential is Φ = Q /(4πε 0 r ) . The capacitance is equal to
C = Q / Φ (r = a) = 4πε 0 a , where a is the radius of the Earth. Evaluation of the
expression yields C = 709 µF.
(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
E = Qmax /(4πε 0 a 2 ) = 3 × 10 6 ⇒ Qmax = 3 × 10 6 (4πε 0 a 2 ) = 1.35 × 1010 C
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:
∇ 2Φ =
∂ 
∂Φ 
1 ∂  2 ∂Φ 
1
1
∂ ∂ 2Φ
r
θ
+
sin
+
=0




∂θ  r 2 sin 2 θ ∂φ ∂φ 2
r 2 ∂r  ∂r  r 2 sin θ ∂θ 
Because of the symmetry, Φ only varies in r. Thus only the radial term in the above
expression remains.
Thus, we have
∂  2 ∂Φ 
r
=0
∂r  ∂r 
∂Φ
r2
=C
∂r
∂Φ C
=
∂r r 2
C
Φ=− +D
r
where, C and D are constants. To account for the different dielectric constants in the two
regions, we set up the following expressions:
C1
+ D1
r
C
Φ 2 (r ) = − 2 + D2
r
Φ 1 (r ) = −
for Ri ≤ r ≤ Rb
for Rb ≤ r ≤ Ro
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:
Φ 1 ( Ri ) = V0
Φ 2 ( Ro ) = 0
Φ 1 ( Rb ) = Φ 1 ( Rb )
(1)
(2)
(3)
D1n ( Rb ) = D2 n ( Rb )
(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
C1
+ D1 = V0
Ri
C
− 2 + D2 = 0
Ro
C
C
− 1 + D1 = − 2 + D2
Rb
Rb
C
C
− ε 1 12 = −ε 2 22
Rb
Rb
−
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:
 1
 1
1 
1 
C 2
C1 = 
V0 +  −
−
 Ro Rb 
 Ri Rb 
Using the fourth equation, one can find
C1 =
ε1 1
ε 2 Ro
V0
+ 1 − εε12
(
)
1
Rb
−
1
Ri
Therefore,
Φ1 = −
C
C
C1
+ D1 = − 1 + V0 + 1 = V0 +
Ri
r
r
ε1 1
ε 2 Ro
V0
+ 1 − εε12
Φ2 = −
ε C ε C ε
C2
+ D2 = − 1 1 + 1 1 = 1
ε 2 r ε 2 Ro ε 2
r
ε1 1
ε 2 Ro
V0
+ 1 − εε12
(
(
)
)
1
Rb
1
Rb
−
1
Ri
 1 1
 − 
 Ri r 
1
Ri
 1 1
 − 
 R0 r 
−
Then, use the fact that ε2 = 2 ε1 = 2 εr, we get
V0
^
E1r r = −∇ Φ1 =
1
Ri
− 2 1Ro − 2 1Rb
1
r2
^
r
and Dr
^
r
=
ε r V0
1
Ri
−
1
2 Ro
−
1
2 Rb
1
r2
^
r
V0
^
E2 r r = −∇ Φ1 =
2
Ri
−
1
Ro
−
1
Rb
1
r2
^
r
ε rV0
^
and Dr r =
1
Ri
− 2 1Ro − 2 1Rb
1
r2
^
r
(b) The surface density of the inner shell is given by
^
ε r V0
^
ρ si = r ⋅ D1r (r = Ri ) r =
1
Ri
−
1
Ri
4πε r
− 2 1Ro − 2 1Rb
1
2 Ro
−
1
2 Rb
1
2
Ri
The capacitance is then equal to
Q 4πRi ρ si
C=
=
=
V0
V0
2
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.
d
1
2
h
In general, the voltage on the surface of the wire 1 is given by
V 1= V11 + V12 = P11Q1 + P12 Q2
(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.
V11 =
or
Q1
4πε 0 a
−
Q1
8πε 0 h
1
P11 =
4πε 0 a
−
1
8πε 0 h
Likewise, for wire 2, we have
V 2= V22 + V21 = P22 Q2 + P21Q2
(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
V21 =
Q1
−
4πε 0 d
Q1
4πε 0 (4h 2 + d 2 )1 / 2
P21 and P12 are equal to
P21 = P12 =
1
4πε 0 d
−
1
4πε 0 (4h 2 + d 2 )1 / 2
V1   P11 P12   Q1 
The equation (1) and (2) can be expressed as V = P Q, or   = 
   . We
V2   P21 P22  Q2 
can express the Q vector in terms of V vector by inverting P matrix, i.e.
−1
 Q1   P11
Q  =  P
 2   21
P12  V1 
1
=



P22  V2  P11 P22 − P21 P12
 P22
− P
 21
− P12  V1 
1
= 2



P11  V2  P11 − P21 2
− P12  V1 
P11  V2 
 P11
− P
 12
In the equation form, we have
Q1 = c11V1 + c12V2
Q2 = c12V1 + c11V2
where c11 =
P11
( P112 − P212 )
, c12 = − P12 ( P112 − P212 ) . These two equations can be expressed by self
capacitance C11 and mutual capacitance C12, such that
Q1 = c11V1 + c12V2 = C11V1 + C12 (V1 − V2 )
Q2 = c12V1 + c11V2 = C11V2 + C12 (V2 − V1 )
One can see that C11 = c11+ c12, and C12 = - c12. So
C11 = 4πε 0
1
a
−
1
2h
− d1 +
( 1a − 21h )2 − (d1 −
1
4h2 + d 2
4h + d
)
2
1
2
2
and C12 = 4πε 0
1
d
−
1
( a1 − 21h )2 − (d1 −
4h2 + d 2
2
4h + d
)
2
1
2
.
8(a). In both dielectric-filled and vacuum-filled capacitor areas, the electric field is equal
^
to -V0/d
j , where y axis is pointing up normal to the capacitor plates. But in the dielectric
^
region, D = εr E = - εr V0/d
^
V0/d
j
j
and ρs = - εr V0/d. The vacuum region has D = ε0 E = - ε0
and ρs = - ε0 V0/d.
(b). The electrostatic energy Ud stored inside the dielectric region is equal to
2
εr E
ε rV0 2
Ud =
xdw =
xw . By the same token, The electrostatic energy Ud stored
2
2d
2
ε0 E
ε 0V0 2
inside the vacuum region is described by U d =
(l − x)dw =
(l − x) w . The
2
2d
value of x for equal amount of energy stored in each region is given by
(l − x) / x = ε r / ε 0 , or x = lε 0 /(ε 0 + ε r ) .