
Capacitors
Capacitors are devices that can store electrical energy. The example we
are familiar with now is the set of parallel, charged plates. If we hook
a battery up to the plates then we get charges separating across the
divide, and by virtue of having done work to accomplish this charge
imbalance, we have stored energy in the capacitor.
We defined capacitance as the ratio of the magnitude of the charge on one
plate to the potential difference across a capacitor.
(1)
These two quantities, charge and potential difference, are of
paramount importance in determining what happens to our capacitor.
If you think of charge as "bang" and potential difference as "buck,"
then capacitance is just a measure of how much bang you get for your
buck.
We also said that the energy stored in a capacitor is proportional to
the charge and to the potential difference. We can find this relation by
integrating (see p. 410 in S&K), and, when we combine the result with
the equation above, we get:
(2)
These equations hold for capacitors in general. Now let's look at what
happens with our specific case of a parallel plate capacitor. We know that
the electric field strength between the plates is
(3)
If we substitute this into our first equation above, we get
(4)
a result that is valid only for parallel plate capacitors.
Dielectrics
When another material (besides air) is put between the plates, the
capacitance changes. Since the dipole field in the dielectric tends
to partially cancel the capacitor's field, the net field is reduced.
The dielectric constant, K, is a measure of a material's ability to alter
the field strength between the plates. A capacitor with a dielectric
has a higher capacitance than one without because of it gets more "bang
for the buck." Since breakdown does not happen as readily, you can put
more charge on a dielectricfilled capacitor. For the parallel plate
capacitor, the capacitance is thus
(5)
In general, when you're dealing with dielectrics, just stick a K in
front of every epsilonzero.
Fiddlin' with the Knobs
Now, if you want to know what happens to the energy when you increase
the distance between the plates or what happens to the energy when you
put a dielectric between the plates, you have to be very careful about
what stays constant and what does not. There are three cases to consider.
Constant capacitance
This is the easiest of the three cases. If we take a capacitor off the
shelf and put a battery across it, it charges up and we get an amount
of charge equal to CV on the plates. If we put a bigger battery across
the plates, we get more charge. The greater the potential difference is,
the more charge we get and the more energy is stored.
Constant charge
This is the case we've generally been talking about when we've talked
about charged plates. We hook a battery up to the capacitor, charge it
up so that the plates have charges of +Q and Q on them, and then remove
the battery. Since the capacitor is isolated, no
charge can be added or removed (at least not in our ideal world), so Q
is constant.
What effect does this have on the energy if we change the separation
distance of the plates? Well, as we said in class in our ConcepTest,
if we pull the plates apart, the stored energy increases. We do work
to move them apart.
We can see this by looking at the equations too. We choose the form that
has the constant quantity in it so that we know what happens when we vary
d. As the following equation shows, when Q and A are constant and d increases,
the energy increases.
(6)
We can easily see from here what happens if we stick a dielectric between
the plates too. In this case, the capacitance increases by a factor of K,
so
(7)
which indicates that sticking a dielectric into a charged, isolated
capacitor lowers the stored energy.
Where did the energy go? If all we did was to put a slab of dielectric
between the plates, how did our capacitor lose energy? If you had been
holding the slab, you would have felt where it went. The field between
the plates does work to pull the slab in. If there were no friction, the
slab would become a harmonic oscillator.
Constant potential difference
Now let's see what happens if we don't disconnect the battery. Here is
a schematic representation of the the constant potential difference case.
In this case, the potential difference is constant when we move
the plates or add a dielectric. From Q = CV, we know that if C
changes (by changing the separation distance or adding a dielectric),
then so does the charge on the plates. How does this affect the
stored energy? Let's look at the equation. This time we want the form
that has C and V in it, since V is constant.
(8)
Unlike for the constant charge case, this time the energy decreases when
you move the plates farther apart. Why does this happen? Think about how
much bang you get for your buck. If you increase d, the capacitance gets
smaller. You're spending the same bucks (V), but your bang to buck ratio
(C) is smaller, so you get less bang (Q) in return.
Now, I know what you're thinking here. You're saying to yourself, "Allison,
that's screwed up. You told us in class that the positive and negative plates
will attract each other, and we have to do work to move them apart." Yes,
it seems like a paradox, but in class we were considering the constant charge
case. Here when we move the plates apart, the amount of charge on the plates
changes. So we have two competing mechanisms: increasing distance and
decreasing charge. You can prove to yourself, using equations (1), (2), and
(4), that decreasing charge wins out. (Go ahead and do it now. I'll wait
here.)
OK, the next step is easy now. What happens when we put a dielectric between
the plates and keep the potential difference constant? As we said earlier,
all we need to do is put a K in front of our epsilonzero above.
(9)
This time the energy increases. How will you experience that? You'll feel
it when you try to push the slab between the plates. The field will resist
your efforts because you're increasing the capacitance, which means that
for a given potential difference, you get more charge.
You have to do work to build up more charge on the plates, thus adding energy
to the system.
How do I remember this!?!
This may seem like a confusing mess, but it's not really. You just have
to keep certain things clear in your mind, and then you can always figure
out what happens. Here are the important points:
 Capacitance is the ratio of charge to potential difference.
 Capacitance always decreases when you move the plates apart.
 Capacitance always increases when you put a dielectric
(instead of air) between the plates.
 You have to do work (i.e., add energy) to increase either the
charge on the plates or the potential difference between the plates.
 When you're doing things that change the capacitance, always pay attention
to whether the charge or the potential difference is constant.
 When the charge on the capacitor is constant, you have to do work
to increase the potential difference between the plates.
If you increase V while Q is constant, you're getting less bang for
your buck, so you have to do work to decrease the capacitance.
 When the potential difference across the capacitor is constant, you
have to do work to put more charge on the plates, and the only way to
get more charge is by increasing the capacitance.
Questions to ponder
 When the charge is held constant, what happens to V when you double
the separation distance between the plates?
 When the charge is held constant, what happens to V when you put a
slab of glass (K=10) between the plates?
 When V is held constant, what happens to the charge when you halve the
distance between the plates?
 When V is held constant, what happens to the charge when you put a
piece of paper (K=3.5) between the plates?
(Answers below)
