Question

An empty parallel plate capacitor is connected between the terminals of a 10.7-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

Answer 1

Here's a trick that makes this question easy:

If the plates are very close together, so that the distance, D, between them is much smaller than their width and length, then the electric field between them is uniform and perpendicular to the plates with magnitude:
E = V/D,
which is just the voltage difference divided by the distance.

Now here's the trick. If you double the distance, it is still quite small compared to the size of the plates so there is same amount of electric flux between them. It doesn't spread out over a larger area, the way it would around a point charge or a line charge, because the lines of flux are parallel to each other. The only spreading is at the edges, which is insignificant for small D. In other words, E does not change.

But D is twice as big as before, which means that V must double to keep E = V/D constant. So, the new voltage is 25 V.

Of course, the traditional way to do this is with:
C = ??A/D
where C is the capacitance between the plates, and A is their area.

So, if D doubles then C is halved. But we also know that:
Q = CV
= (??A/D)V

The charge, Q, does not change (because the plates are not connected to anything), and neither does ?? nor A, so V must double as well.