ECE 222B, Winter, 2010
Solutions to homework set 1
1. Assume that you are given the parallel plate capacitor as shown in Figure
1. The region between the plates is ﬁlled with two layers of dielectric, each
with a ﬁnite (and diﬀerent) conductivity (
σ
1
and
σ
2
). In what follows
assume that
x
and
y
dependencies can be neglected, and assume that all
time variations are slow enough so that magnetic ﬁeld may be neglected
and that the electric ﬁeld within each of the dielectrics is independent of
z
. Note, however, that the electric ﬁeld can be a function of time, and
that the ﬁelds within the two regions can diﬀer.
(a) Assume that for times
t <
0 the capacitor is totally discharged and
the voltage drop across the plates is 0. At time
t
= 0 a voltage drop
of
V
0
is established across the plate and maintained. What are the
electric ﬁelds
E
1
and
E
2
in the two dielectric regions just after the
voltage drop is established? Justify your solution. You may assume
that the electric ﬁeld points in the
z
direction.
Solution:
For very early times no charge can accumulate on the
dielectric boundary. Thus,
E
1
=
E
2
. Since
V
0
= (
L/
2)(
E
1
+
E
2
), we
obtain
E
1
(0) =
E
2
(0) =
V
0
/L
.
(b) Assuming that the voltage drop has been established for a time long
compared for any transients to die out, what is the ﬁnal (asymptotic)
value of
E
1
and
E
2
. Note that this can be obtained
without
solving
for the timedependent evolution of the capacitor.
Solution:
At late times, steadystate is achieved, and thus
J
1
=
J
2
at the dielectric boundary (no charge accumulates). Thus,
σ
1
E
1
=
σ
2
E
2
and
V
0
= (
L/
2)(
E
1
+
E
2
), leading to
E
1
(
∞
) =
2
σ
2
σ
1
+
σ
2
V
0
L
E
2
(
∞
) =
2
σ
1
σ
1
+
σ
2
V
0
L
(c) Prove that, for this problem setup
σ
2
E
2
(
t
)

σ
1
E
1
(
t
) =

∂
∂t
(
εE
2
(
t
)

εE
1
(
t
))
(1)
Solution:
Starting with the continuity equation
∇ ·
~
J
=

∂
∂t
ρ
=

∂
∂t
∇ ·
~
D
and then integrating in
z
across the dielectric boundary yields
J
2

J
1
=

∂
∂t
(
D
2

D
1
). Using
J
1
=
σ
1
E
1
,
J
2
=
σ
2
E
2
,
D
1
=
εE
1
, and
D
2
=
εE
2
yields the desired result.
(d) Calculate the charge per unit area (as a function of time) that ac
cumulates at the boundary between the two dielectrics. What is
1
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 Magnetic Field, Electric charge, Ampere, E Ez

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