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Dielectrics and Capacitance
I.
Introduction.
Let us consider a polarizable medium with a dielectric constant k. When placed in an
electric field the dipoles within the medium will partially align with the field producing
induced charges on the edges of the medium, qin(See fig’s 1 and 2). The larger k, the
more polarizable the material, and therefore the larger qin. We can then write down the
difference in charges as a function of k: Q-qin= Q/k.
_ _ _ _ _ _ _ _ _ _ _ -Q
+
Dipole
moments
randomly
aligned
_
+
+
_
+
_
+
_
+
_
qin
-qin
+qin on surface
of dielectric
Dipole moments
partially align in
field, producing
induced charges on
surface of dielectric
++++++++++++++Q
Polarizable medium
Fig. 1
Fig. 2
As an example, let’s put the material between the plates of a parallel plate capacitor and
calculate the resulting capacitance.
_ _ _ _ _ _ _ _ _ _ _ -Q
+
+
+
+
+
+
qin
Gaussian surface:
Qenc=(Q-qin)/A
_
_
_
_
_
-qin
++++++++++++++Q
Fig. 3
From fig. 3 we see Gauss’ law gives us:
E
Q  qin
Q



o A
k o A k o
Physics 122—Matt DePies
1
First we find V:
d


dx
d
   E  dl   

k o
k o
0
0
d
Vab
Here eta is the surface charge density on the plates which have an area A and separation
d.
Now we find C:
C
 A
Q
A

k o
d
Vab
d
k o
So the capacitance is increased by a factor of k;
C=kCo
where Co is the vacuum capacitance.
We would like to see what happens to the potential, potential energy, and charge on the
capacitor due to the insertion of the dielectric. In other words, we want to see the
usefulness of the dielectric in a circuit. For the following we will observe two different
processes: one is where the charge on the cap is fixed, and the other is where the
potential is fixed. An example of fixed potential is when it is attached to a battery.
II.
Fixed charged on a capacitor.
For this we know Q=Qo, where the “o” indicates the vacuum filled capacitor value.
Qo  C oVo  CV
V 
Qo
Q
 o
C
kCo
V 
Vo
k
The potential is reduced by a factor of 1/k! What about the potential energy?
2
1 Qo
Uo 
2 Co
2
Physics 122—Matt DePies
U 
1 Q 2 1 Qo

2 C
2 kCo
U 
Uo
k
2
The potential energy stored in the capacitor is also reduced. Where did the energy go?
On what did the field do work?
III.
Fixed potential on a capacitor.
In this case we will keep the potential across the cap constant as we insert the dielectric,
V=Vo.
V 
Q Qo

C Co
Q
Qo C
 kQo
Co
The charge increases by a factor of k. Now for the potential energy:
1
2
C oVo
2
1
1
2
U  CV 2  kCoVo
2
2
U  kU o
Uo 
So the potential energy stored in the capacitor increases. Where does the energy come
from? The simple answer is the battery. Thus we see the use of a dielectric in a circuit: it
increases the stored energy in the capacitor. Also, it keeps the plates from sticking
together due to coulomb attraction!
IV.
Conclusion.
There are complications with inserting the dielectric into the cap that we have ignored
here. One question is whether or not the dielectric is sucked into the cap or has to be
pushed in. It turns out both cases, constant V and constant Q, are the same, but the
potential energies are different. Why is this?
Physics 122—Matt DePies
3