A capacitor with a non-uniform dielectric
Answers
Electricity and Magnetism I (P331)
M. R. Shepherd
October 17, 2008
A capacitor with a non-uniform dielectric
Consider a parallel-plate capacitor filled with non-uniform dielectric. The
dielectric is still linear, in other words,
P
=
0
χ
e
E
, but this time
χ
e
varies
as a function of position in the dielectric. Let’s say that
χ
e
varies such that
r
=
r
0
+
ax.
Here
r
0
is a dimensionless constant,
x
is the distance from one plate, and
a
is constant with units 1
/
length. Let’s assume the area of the plates is
A
and their separation is
d
. What is the capacitance?
x=0
x=d
x
Remember the strategy for computing capacitance: put a charge
Q
on the capacitor and then compute
the potential difference between the plates in terms of this charge. The capacitance can then be identified
by comparing this expression to
V
=
Q/C
. In order to compute
V
we first need to find the electric field.
Let’s charge up the plates so that we have a free surface charge density of +
σ
f
on the left plate and
−
σ
f
on the right plate. We can then compute the electric displacement
D
based on the free charge, and obtain
D
=
σ
f
ˆ
x
.
This then gives
E
=
1
D
=
σ
f
0
(
r
0
+
ax
)
ˆ
x
Now, to obtain the potential, we need to compute the line integral of this field from the negative plate
to the positive plate. We have
V
=
−
∫
E
·
d
l
=
−
σ
f
0
∫
0
d
1
r
0
+
ax
dx
=
σ
f
0
a
[ln(
r
0
+
ad
)
−
ln(
r
0
)]
Let’s rewrite the difference of logs as a log of a ratio and substitute in
σ
f
=
Q/A
. This allows us to identify
the capacitance
V
=
Q
ln(1 +
ad/
r
0
)
A
0
a
=
Q
C
=
⇒
C
=
A
0
a
ln(1 +
ad/
r
0
)
.
It is interesting to examine this expression in the limit that the dielectric is uniform. In this case we have
a
→
0 and
r
=
r
0
. This poses a little problem since, in the expression above, substituting in
a
= 0 we have
0
/
ln(1) = 0
/
0. It is helpful to remember that for small values of
y
we can write
ln(1 +
y
) =
y
−
y
2
2
+
y
3
3
+
...
This means that for
a
very small we have
C
=
A
0
a
ln(1 +
ad/
r
0
)
≈
A
0
a
(
ad/
r
0
)
−
(
ad/
r
0
)
2
/
2
=
A
0
(
d/
r
0
)
−
a
(
d/
r
0
)
2
/
2
.
We can see that in the limit
a
→
0 this expression reduces to
r
0
0
A/d
, which is exactly what we expect for
a parallel plate capacitor with dielectric constant
r
0
.
Exercise:
Compute the bound charge densities
σ
b
on
both sides of the capacitor. Which side should have the higher bound charge density? In this case,
ρ
b
6
= 0 –
compute it also. Electricity and Magnetism I (P331)
M. R. Shepherd
October 17, 2008
A capacitor with a non-uniform dielectric
Consider a parallel-plate capacitor filled with non-uniform dielectric. The
dielectric is still linear, in other words,
P
=
0
χ
e
E
, but this time
χ
e
varies
as a function of position in the dielectric. Let’s say that
χ
e
varies such that
r
=
r
0
+
ax.
Here
r
0
is a dimensionless constant,
x
is the distance from one plate, and
a
is constant with units 1
/
length. Let’s assume the area of the plates is
A
and their separation is
d
. What is the capacitance?
x=0
x=d
x
Remember the strategy for computing capacitance: put a charge
Q
on the capacitor and then compute
the potential difference between the plates in terms of this charge. The capacitance can then be identified
by comparing this expression to
V
=
Q/C
. In order to compute
V
we first need to find the electric field.
Let’s charge up the plates so that we have a free surface charge density of +
σ
f
on the left plate and
−
σ
f
on the right plate. We can then compute the electric displacement
D
based on the free charge, and obtain
D
=
σ
f
ˆ
x
.
This then gives
E
=
1
D
=
σ
f
0
(
r
0
+
ax
)
ˆ
x
Now, to obtain the potential, we need to compute the line integral of this field from the negative plate
to the positive plate. We have
V
=
−
∫
E
·
d
l
=
−
σ
f
0
∫
0
d
1
r
0
+
ax
dx
=
σ
f
0
a
[ln(
r
0
+
ad
)
−
ln(
r
0
)]
Let’s rewrite the difference of logs as a log of a ratio and substitute in
σ
f
=
Q/A
. This allows us to identify
the capacitance
V
=
Q
ln(1 +
ad/
r
0
)
A
0
a
=
Q
C
=
⇒
C
=
A
0
a
ln(1 +
ad/
r
0
)
.
It is interesting to examine this expression in the limit that the dielectric is uniform. In this case we have
a
→
0 and
r
=
r
0
. This poses a little problem since, in the expression above, substituting in
a
= 0 we have
0
/
ln(1) = 0
/
0. It is helpful to remember that for small values of
y
we can write
ln(1 +
y
) =
y
−
y
2
2
+
y
3
3
+
...
This means that for
a
very small we have
C
=
A
0
a
ln(1 +
ad/
r
0
)
≈
A
0
a
(
ad/
r
0
)
−
(
ad/
r
0
)
2
/
2
=
A
0
(
d/
r
0
)
−
a
(
d/
r
0
)
2
/
2
.
We can see that in the limit
a
→
0 this expression reduces to
r
0
0
A/d
, which is exactly what we expect for
a parallel plate capacitor with dielectric constant
r
0
.
Exercise:
Compute the bound charge densities
σ
b
on
both sides of the capacitor. Which side should have the higher bound charge density? In this case,
ρ
b
6
= 0 –
compute it also.