Physics 102 Page
CapacitorsCapacitors 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.
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:
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
If we substitute this into our first equation above, we get
a result that is valid only for parallel plate capacitors.
DielectricsWhen 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 dielectric-filled capacitor. For the parallel plate capacitor, the capacitance is thus
In general, when you're dealing with dielectrics, just stick a K in front of every epsilon-zero.
Fiddlin' with the KnobsNow, 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.
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.
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
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
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.
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 epsilon-zero above.
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!?!
Questions to ponder
Physics 102 Page