Cracking the SAT Physics Subject Test

After the switch is closed, what will be the common voltage

across each of the parallel capacitors (once electrostatic

conditions are reestablished)?

Here’s How to Crack It

When C 1 is fully charged, the charge on (each of the plates of) C 1 has the magnitude

Q = C 1 V = (2 mF)(12 V) = 24 mC. After the switch is closed, this charge will be

redistributed in such a way that the resulting voltages across the two capacitors, V′,
are equal. This happens because the capacitors are in parallel. So if Q′ 1 is the new

charge magnitude on C 1 and Q′ 2 is the new charge magnitude on C 2 , we have

, so C 1 V′ + C 2 V′ = Q, which gives us

Now let’s examine the case in which a capacitor is first charged up and, while it’s
still connected to its voltage source, has a dielectric inserted between its plates.
First, since the capacitor is still connected to the battery, the voltage between the
plates must match the voltage of the battery. Therefore, V will not change. Next,
because the capacitance C increases, the equation Q = CV tells us that the charge Q
must increase; in fact, because V doesn’t change and C increases by a factor of k,
we see that Q will increase by a factor of k. Next, using the equation V = Ed, we