Question

The charge in a conductor tends to have a higher density in pointy regions (smaller radius of curvature). So, why is it not the case for the two parallel plates of a capacitor? In a capacitor, the charge is over the plane region of the plate.

Answer 1

In a parallel plate capacitor, as the name implies, the plates are perfectly parallel and flat like this | | . You have a charge that builds up on one side and since the plate is a conductor (and since like charges repel each other) the charges space themselves out equally to minimize their potential energy. Again since like charges repel an excess of electrons on one plate will push electrons away from the other plate, thus creating 2 plates one of + and one of - charge.

The reason why charge builds higher densities on that conductor is simply a factor of its geometry. The charges are always trying to space themselves out to minimize their potential energy, and on a parallel plate capacitor they are just able to do it more precisely!

Answer 2

But wouldn't it be reasonable to think that, since the conductor plate is not spherical, there should be a higher charge density in the vertices?

Answer 3

Do you mean the conductor plate as in the parallel plate capacitor? As i mentioned when like charges accumulate on the plate they all repel eachother to minimize their energy. This spacing (on an ideal, perfect plate conductor) is perfect in that all the charges are equidistant to eachother giving a uniform charge density across the whole plate. Parallel plate capacitors are used specifically because of this. It makes them ideal!

Answer 4

it is true that the charge density would be higher at the vertices/edges of the plate... these effects are called "edge effects" and can be difficult to deal with. the use of parallel plate capacitors in problems almost always assumes that the plates are infinite (very large in comparison to the plate separation) and that we only deal with the fields away from the edges where they are uniform