Prove That Charge Distributes Uniformly on a Disk Conductor

1. Jul 7, 2010

fobos3

I heard it's quite hard to prove that electric charge is distributed uniformly on a disk conductor. Can you point me to some resources online on the subject?

2. Jul 7, 2010

zhermes

Consider any non-uniform distribution of charge. In this case there will be some charges closer to each-other than others. Therefore there will be a greater repulsive force between the nearer charges, tending to move them away. The only equilibrium position, is uniformly distributed. (With friction dissipating energy to prevent oscillation)

You could 'prove' this more rigorously by making an equation for the total potential energy of a configuration of charges (e.g. in a grid with variable x and y spacing). Its easy to show that the lowest energy (most stable) configuration is that with uniform separation.

3. Jul 8, 2010

fobos3

But how do you know that you have an electrostatics problem('most stable')? What's preventing the charges from moving in some weird way?

4. Jul 8, 2010

Eynstone

The statement should read 'At equilibrium,electric charge is distributed uniformly on a disk conductor'. Nothing keeps the charges from moving when in a non-equilibrium configuration.
A rigourous proof can be given with calculus of variations.Assume a charge density function which minimises energy & equate it's variation to zero . Using Gauss's law,you get that the function must be a constant.

5. Jul 8, 2010

fobos3

I understand and can prove the electrostatics case. What I don't understand is that if you have a non-uniform distribution of charge, it rearranges itself to an equilibrium position. Why is that?

6. Jul 8, 2010

AJ Bentley

Inside a conductor there are no fields.
The surface of a conductor is therefore an equipotential surface.

Therefore the problem reduces to a two dimensional one and the only stable solution is a uniform charge distribution over the entire surface.

This applies whatever the shape of the conductor.
EDIT - this is wrong see analysis below.

Last edited: Jul 9, 2010
7. Jul 8, 2010

zhermes

A equilibrium arrangement is one in which the potential energy is at a local minima (in general a local extrema, but in this case there aren't any relevant maximums). Away from any minimum, there will be an increased potential, and therefore a force towards the equilibrium position.

Recall:
$$F = - \nabla U$$

8. Jul 8, 2010

Ben Niehoff

Not so! The surface of the conductor must be flat.

One can show that when the surface is not flat, the charge density is a constant plus a term that is proportional to the mean curvature (i.e., the sum of the principal curvatures).

Contrary to the OP's assertion, this is not impossibly hard to prove; in fact it is a homework problem in Jackson. It follows from Gauss' Law.

9. Jul 8, 2010

AJ Bentley

Even not so-er.
Simple example, a sphere is not flat.

It's an equipotential. The potential away from the surface may have a non-uniform distribution but the surface itself is an equipotential. That fact is often the only way in which the field can be determined where conductors have a complex shape (if at all).

In practice, one rarely considers the distribution of charge, being interested only in the fields generated.

However, on such a surface I see no alternative but a uniform distribution as the solution to the reduced two dimensional problem. I'm willing to be corrected if someone shows me a contradiction.

10. Jul 8, 2010

Ben Niehoff

On a sphere, the mean curvature is constant, and so the distribution is uniform. ;)

11. Jul 9, 2010

AJ Bentley

That's true but I'm having a hard time trying to determine if the charge distribution can be anything other than uniform.

I'm not convinced by my own argument but at the moment can't see how to disprove it.

12. Jul 9, 2010

AJ Bentley

Consider two spheres of different sizes, connected by a thin wire so that we may neglect the influence of one sphere on the other.

Let the radii be a and b and the charge on each A and B respectively.

The potential of each is A/a and b/B. These are equal so A/a = b/B

But the surface charge density must be the charge of each sphere divided by the area.

For the smaller sphere that is A/4.pi.a^2 and for the other B/4.pi.b^2.
So the ratio of the charge densities is A/a^2 to B/b^2
Which means the ratio of surface charge densities must b/a.

So my statement is wrong: The charge density of an equipotential is not necessarily constant.

Well, I guess that doesn't help but at least I'm happy

PS.
If that is true, the surface charge density, like the E field, varies as the radius - in which case the surface charge of a disc conductor can't be uniform (unless it's infinite)

Last edited: Jul 9, 2010
13. Jul 9, 2010

Phrak

This is simply wrong. Electric charge is not uniformally distributed on a conducting disk in isolation. Charge is displaced toward the parimeter.

I think AJ Bentley has the idea.

Now two disks with a small gap between and oppositely charged, is another matter altogether.

14. Jul 9, 2010

AJ Bentley

It raises the interesting point that if a conductor is deformed, current must flow on the surface.

Maybe we could invent a new kind of generator, that works by pounding lumps of metal with a big hammer.