Solutions to Quiz 3
March 22, 2010
1
Capacitor with two Dielectrics
A parallel plate capacitor consists of two plates of width w, length L and
plate separation d. The space between is filled with slabs of two materials of
different dielectric constants κ1 and κ2 . Material 1 has a thickness x. Find
the capacitance C of this capacitor.
1.1
Solution
We can break up this capacitor into a problem of two different capacitors,
connected in parallel. These individual capacitors then have capacitances of
C1 =
xwκ1 0
d
(1)
and
(L − x)wκ2 0
(2)
d
where xw and (L − x)w are the areas of the plates of these capacitors.
Since they are connected in parallel, C = C1 +C2 , so the total capacitance
is
w0
C=
(κ1 x + κ2 (L − x))
(3)
d
When κ1 = κ2 , the x’s cancel and wL = A, giving
C2 =
Cκ1 =κ2 =κ =
Aκ0
d
(4)
so we recover the capacitance of a parallel plate capacitor filled with a single
dielectric.
1
2
Tuning the Capacitance
Imagine that material κ1 is air and that material κ2 is oil. This would then
describe an air filled capacitor partially submerged in oil. Say you start to
lift this capacitor out of the oil at a speed v. Find the rate of change of
capacitance during this process i.e. what is dC/dt ?
2.1
Solution
We can get the rate of change of capacitance by taking the derivative of
equation (3).
w0
dC
=
(κair − κoil )v
(5)
dt
d
where we have used that dx/dt = v and dL/dt = 0 since L is a constant.
The rate of change of capacitance is proportional to how fast you pull the
capacitor out of the oil. Since κoil > κair , dC/dt < 0 i.e. the capacitance
is decreasing. If κ1 = κ2 , the above equation gives no change in capacitance. This is reasonable, as in this case nothing physical is changing for the
capacitor.
2