We can think the charges go to the outer surface of a conductor to minimize the electrostatic potential energy of the system. We can check this using a simple calculation using a charged sphere.
A uniformly charged sphere would have $20\,$% more energy than that of a very thin spherical shell with the same radius and the same charge as in the former case. As the potential energy would be less if we distribute charges uniformly over the surface rather than distributing the charges uniformly throughout, we can say, the conductor prefers having charges on the outer surface. Here's my question then:
Suppose I have a conducting sphere on which a charge $Q$ is to be distributed. Now, suppose I divide the charge into two equal parts, $Q/2$ each, and place them at two diametrically opposite ends of the sphere. Now the potential energy of the configuration becomes $75\,$% less than that for charges distributed uniformly over the surface. Then why doesn't the conductor prefer this kind of distribution which can minimize the potential energy further?
Best Answer
If there is any net charge inside the bulk of the conductor Gauss's law guarantees that there is some electric field lines with directions towards the outside of that region.
Any such electric field inside moves charges around inside, so it will only stop when there is no way the charges can move any. That guarantees zero electric field inside, and charges on the surface distributed so the tangential electric fields must be zero. Only fields perpendicular to the surface can survive.