A conductor is in electrostatic equilibrium when the charge distribution (the way the charge is distributed over the conductor) is fixed. Basically, when you charge a conductor the charge spreads itself out. At equilibrium, the charge and electric field follow these guidelines:
- the excess charge lies only at the surface of the conductor
- the electric field is zero within the solid part of the conductor
- the electric field at the surface of the conductor is perpendicular to the surface
- charge accumulates, and the field is strongest, on pointy parts of the conductor
Let's see if we can explain these things. Consider a negatively-charged conductor; in other words, a conductor with an excess of electrons. The excess electrons repel each other, so they want to get as far away from each other as possible. To do this they move to the surface of the conductor. They also distribute themselves so the electric field inside the conductor is zero. If the field wasn't zero, any electrons that are free to move would. There are plenty of free electrons inside the conductor (they're the ones that are canceling out the positive charge from all the protons) and they don't move, so the field must be zero.
A similar argument explains why the field at the surface of the conductor is perpendicular to the surface. If it wasn't, there would be a component of the field along the surface. A charge experiencing that field would move along the surface in response to that field, which is inconsistent with the conductor being in equilibrium.
Why does charge pile up at the pointy ends of a conductor? Consider two conductors, one in the shape of a circle and one in the shape of a line. Charges are distributed uniformly along both conductors. With the circular shape, each charge has no net force on it, because there is the same amount of charge on either side of it and it is uniformly distributed. The circular conductor is in equilibrium, as far as its charge distribution is concerned.
With the line, on the other hand, a uniform distribution does not correspond to equilbrium. If you look at the second charge from the left on the line, for example, there is just one charge to its left and several on the right. This charge would experience a force to the left, pushing it down towards the end. For charge distributed along a line, the equilibrium distribution would look more like this:
The charge accumulates at the pointy ends because that balances the forces on each charge.
Information provided by: http://physics.bu.edu