-2
$\begingroup$

In electrostatics, why in the internal surface of a hollow charged sphere there aren't charges?

$\endgroup$
1
$\begingroup$

I think it is strange that the charge on a conductor flows to the surface. Because of their mutual repulsion, the charges naturally spread out as much as possible, but for all of them to go to the surface seems like a waste of the interior space. Surely we could do better, from the point of view of making each charge as far as possible from its neighbors, to sprinkle some of them throughout the volume... Well, it simply is not so. You do best to put all the charge on the surface, and this is true regardless of the size or shape of the conductor. The problem can also be phrased in terms of energy. Like any other free dynamical system, the charge on a conductor will seek the configuration that minimizes its potential energy. The electrostatic energy of a solid object (with specified shape and total charge) is a minimum when that charge is spread over the surface. For instance, the energy of a sphere is (1/8lfEo)(q2 / R) if the charge is uniformly distributed over the surface, but it is greater, (3 /20lf EO)(q2/ R), if the charge is uniformly distributed throughout the volume.

Reference: Chapter 2- Electrostatics - Introduction to Electrodynamics by Griffith

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Same charges repel each other. A hollow sphere is a confined object. The charges repel and move as far from each other as possible. Therefore, they bubble up towards the outer surface.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.