Without Equations
Relations among the Fundamental Quantitites
In this note, we're going to discuss how to
understand--without equations--what happens in a
capacitor when we start making changes. (You can check out how to do this
with equations in the other capacitance note in the section called
fiddlin' with the knobs.)
Although I'm not going to give any equations in this note, you still need
to know the basic relationships between the quantities. As shown in the
figure below, the most important quantities in the study of capacitors are:
- electric field, E
- potential difference, V = Vtop - Vbot
- magnitude of charge on one plate, Q
- distance between plates, d
Also of importance are the capacitance, the area of the plates, and
the dielectric constant. 
Figure 1. A parallel-plate capacitor
The important relations between the quantities that you need to know
are: - The electric field, E, increases linearly with charge, Q, and is constant between the plates (until we fiddle with the knobs).
- The potential difference, V, for a given capacitor is directly proportional to the electric field, E.
- The potential difference, V, for a given electric field is directly proportional to the distance between the plates, d.
- The capacitance is a measure of how much charge we get on the plates for a given amount of potential difference.
- The capacitance increases if we increase the area of the plates.
- The capacitance increases if we put the plates closer together, (i.e., decrease d).
- The capacitance increases if we put a dielectric material between the plates rather than air.
- The energy stored in a capacitor increases with increasing charge or with increasing potential difference.
What are the reasons for these relations? Let's think about what's happening
in each case. If we put more charge in our system, it makes sense that the
electric field gets stronger. Likewise, the potential difference between
two points increases as we make the field stronger. To understand the
dependence of potential difference on distance, just think of what the
potential difference is between a point and itself versus the potential
difference between two points on
opposite sides of the capacitor. In the former case, the potential difference
is zero, and in the latter it is not.
The ratio of charge to potential difference is just the definition of
capacitance. For a given capacitor, the more potential difference we
put across it, the more charge we put on it. If we use a bigger capacitor
and the same potential difference, we get more charge. If our capacitor
consists of parallel plates, we can see that making the plates
bigger allows more charge to be stored.
Why does putting them closer
together increase the capacitance though? Well, if we put them closer
together, the oppositely charged plates attract each other more and
thus can hold more charge. Remember, putting charge on the plates takes
work because we are piling up a lot of like charges on each plate. There
comes a point when the repulsion of the charges on one plate cancels the
attraction of the charges on the other plate. That limit defines the
capacitance. If we move the plates closer together, we get more attraction
so more charges can pile in there.
What effect does the dielectric have that causes capacitance to increase?
Think about what's likely to happen in the face of the dielectric near
the positive plate: The positive charge on the plate will attract negative
charge in the dielectric. On the opposite side, the negative plate attracts
positive charge in the dielectric. This leads to an electric field inside
the dielectric that is in the opposite direction to the capacitor's E-field.
Thus, some cancellation occurs and the net E-field is reduced. If our
capacitor is already charged up and disconnected from the battery, the
charge Q is constant. If E is reduced, then we know that V also
must be smaller. If we have the same charge and smaller potential difference,
then we're getting more bang for our buck, i.e., higher capacitance.
The last relation is fairly straightforward. To put more charge on the
plates we have to do work. To increase the potential difference between
the plates, we have to do work. In the former case, we're having to
overcome the repulsion between the like charges to put more charge in
there. In the latter, we are effectively putting the positive charges
at higher potential and the negative charges at lower potential, and
those are not the directions that positive and negative charges want
to change their potential. (See the section on
potential curves in the note on
electric potential.)
OK, let's put all this to use now.
Fiddlin' with the Knobs, part II
In this section, we revisit how the charge, potential difference, and
stored energy are affected when make changes in the nature of our
capacitor. The capacitance depends on the area of the plates, the
distance between the plates, and the material between the plates.
We're going to examine only the last two of these.
Changing the plate separation
First let's consider the case of constant charge. We've charged
up the capacitor so that each plate has a total charge Q (pos. or
neg.) and then removed the battery.
Figure 2. Moving the plates apart while holding the charge constant.
The figure above illustrates what happens in this case. We have
two oppositely charged objects, and we want to pull them farther apart
from each other. We are moving them away from where they want to go,
so we have to do work to move them. We're fighting against the electric
force here, so we do work and add energy to the system.
What happens to the potential difference when we move the plates apart?
As long as the charge is constant--and we don't pull the plates so far
apart that the finite size of the plates is felt--the electric field
stays the same too. If the E-field is constant and the plate separation
increases, the potential difference increases. We could also have
deduced this from the increase in energy. If the charge is constant and
the energy increases, then the potential difference had to increase.
Now, what if we charge up the plates but don't disconnect the battery,
as shown below.
Figure 3. Moving the plates apart while holding the potential difference constant.
This time the charge on the plates will change, but does it increase
or does it decrease? If we move the plates apart and keep V constant,
then the E-field has to get smaller. If the field weakens, we know
there's less charge. If there's less charge, there's less energy.
Recall that there are two things going on with the charge: It's
attracted to the opposite charge on the other plate, but it's
repelled by all of the like charge on its own plate. If we move the
plates further from each other, the attraction is weakened. That means
that they feel the repulsion of their neighbors even more. With the
battery still connected, some the charges are going to be pushed out.
Changing the material between the plates
OK, that wasn't too bad (I hope). Now let's see what happens when we
put a material between the plates that has a higher dielectric constant,
i.e., one that tends to lower the E-field.
Figure 4. Adding a dielectric while holding the charge constant.
Figure 4 above shows the addition of a dielectric for the constant
charge case. What happens? Do we have to push the slab and do work
to insert it, or does it get sucked in by the field? Remember that
the dielectric tends to lower the E-field. If the field strength
decreases and the charge and plate separation stay the same, then
the potential difference also decreases. We know that the energy will
go in the same direction that the potential difference goes, so
if V decreases so does U.
Where did the energy go? As we said in the other
fiddlin' note, the field
does work on the slab. It pulls it in and, in the absence of friction,
would cause the slab to become a harmonic oscillator.
All righty. We've got only one case left to consider: increasing the
dielectric constant while holding the potential difference constant
(Fig. 5).
Figure 5. Adding a dielectric while holding the potential difference constant.
This time the amount of charge on the plates will change in response
to our action. If we push the slab between the plates, the dielectric
tends to lower the E-field. In this case, however, we're still hooked
up to the battery. A decreasing E-field would cause the potential
difference to decrease--if it could. But it can't here, so the battery
does what it can to make the E-field correspond with the amount of
potential difference. How do you adjust the charge to increase the
E-field? You have to increase it. So the battery pumps more charge to
the plates. The charge increases and the energy increases.
Where does the energy come from? Well, that charge doesn't come from
the battery for free. As you push the slab into the capacitor, you're
doing work that pumps charge to the plates. Instead of the slab being
sucked in, as in the constant charge case, now you have to force it
in there.