# Why does free charge reside on the surface of a conductor?

1. Dec 2, 2008

### kaustavdgreat

1.Prove that free charge resides on the surface of a conductor.(Using Gauss's law or laws of capacitance)

2. Relevant equations

3. The attempt at a solution
I could not figure out how to prove this using Gauss's law,though i tried to solve it using the laws of capacitance.This is what i did-

Let us consider a spherical conductor of radius R having charge Q.
The potential of the conductor (V)=Q/C
If the charges are spread in such a way that it forms a spherical shell of radius r then, C=r
where $$r\leq{R}$$
Now the maximum value of r should be R
We know that every system tends to attain a state of minimum potential.
V will be min, when C is max or, C=R
So, r=R
Hence, the charge will always reside on the surface.

2. Dec 2, 2008

### Ben Niehoff

Try using a Gaussian surface that is just inside the conductor. What is the electric field on this Gaussian surface?

3. Dec 2, 2008

### nasu

This is not enough (or complete).
With the same argument you'll have the charge distributed on the surface of an insulator (or not?). Where is the conductivity in this picture?

4. Dec 2, 2008

### kaustavdgreat

The same argument can't be used in case of an insulator as you know, insulators don't have any (or any significant amount of) free charge.

5. Dec 3, 2008

### nasu

So you mean that in insulators the charge cannot take the configuration of minimum potential energy, right? (Because it's not free to do it.) Then this will make the difference, I think.
However assuming that the free charge distributes itself on a spherical shell (even of arbitrary radius) is not really proven or obvious. And what if the conductor is not spherical? The problem does not say so.

Last edited: Dec 3, 2008
6. Dec 4, 2008

### kaustavdgreat

I think $$C\propto R$$ is applicable for any other curved geometrical shape(R=radius of curvature);so, you can assume any shape.

You are right, one can prove it for any curved conductor but the problem arises when we are dealing with cubical(or such other shapes) conductors, as radius of curvature of a plane surface is infinity.I think, it can be proved more convincingly using Gauss's law,I am still working with it.

If you have placed a ball on the roof of a building, it will always tend to fall on the ground to minimize its gravitational potential;but this will happen only when you allow it to fall.The same thing happens in case of bound electrons in insulators,they cannot take the configuration of minimum potential just because they are not allowed to do it due to the strong nuclear force.The configuration they attain is that of minimum potential available for them.
Moreover,there is no point in arguing with the insulator issue since first of all charge doesn't reside on the surface in case of an insulator and secondly as I told you earlier insulators don't have any free charge.

PLZ HELP ME SOLVE IT ANYONE.THNX