# Why is surface charge distribution uniform for a conducting sphere? Can't it be arbitrary?

If the charge $q_1$ has to repel the charge $q_2$, the electric field has to go inside the conductor which contradicts the fact that electric field inside conductors is zero. Then why do the charges distribute themselves in a particular manner? Why can't they be distributed over a small place on the conductor?

The statement "electric field inside a conductor is zero" is true only after charges have distributed themselves in the most optimal way on the surface - it is an electrostatic result. Starting with an arbitrary charge distribution, there will be forces that cause a redistribution of the charge until, for a sphere, they are distributed uniformly. At that time, there is no electric field inside the conductor, and so no force on the charges that impels them to move to another, energetically more favorable, location.

A simple proof for spherical conductor is this: if the sphere is symmetrical, then the solution must also be symmetrical (there is nothing about a sphere that would drive an asymmetrical solution, and the uniqueness theorem says that if you have "a" solution that meets the boundary conditions, it must be "the" solution. Since uniform distribution meets the boundary conditions, it must be the solution.). But if that is so, then the electric field inside the sphere must also be spherically symmetrical. And we know from Gauss's theorem that the integral $\int E\cdot dS$ must equal the $\frac{Q}{\epsilon_0}$. Since $Q=0$, it follows that $E=0$.

• Electric field inside a conductor is zero is a result of the distribution. But, why should that distribution happen? How is that distribution the most "comfortable"? – Tea is life Jan 9 '15 at 0:03
• As long as there is an electric field, there is a force on the charge that will move it. The charges stop moving when there is no longer a field that generates a force. Energetically this is like "rolling down the hill". – Floris Jan 9 '15 at 0:04
• It's kind of subtle but I want to point this out: the symmetry argument is more or less a lie; it doesn't really work when we're talking about particles. (Imagine if only two electrons existed on a sphere... would the solution still be spherically symmetric?) – user541686 Jan 9 '15 at 8:53
• @Mehrdad Gauss' Law assumes a continuous charge distribution. This is a perfectly fine approximation for the vast number of electrons floating around in a conductor of macroscopic size. Gauss' Law is a classical therom; of course it doesn't hold up if you want to get detailed down to the atomic level. – jpmc26 Jan 9 '15 at 9:23
• A uniform distribution on the surface is the most "comfortable" because it places the repelling charges as far apart as possible. – semi-extrinsic Jan 9 '15 at 12:27

The electric field due to a charge surface element $dq$ does have to go inside the conductor, but it is cancelled out by other charges elsewhere on the sphere. So the field inside the conductor is zero, and there is no contradiction.

The precise nature of this cancellation can be seen by doing a surface integral, but it seems mysterious if you do it that way. If you apply Gauss's law, the precise cancellation becomes clear.

So the constant charge distribution satisfies $E=0$ inside the conductor. To show that it is unique, one needs the uniqueness theorems that are expounded upon in, for example, Griffiths Introduction to Electrodynamics.

Good question, but normally there are more than two charges.

Note that the electric field inside conductors is zero because the charges on the outside move to an arrangement where it is zero - or as close to zero as possible.

Consider a sphere with a uniform density of charge on the outside. If the contributions of all the charges on the surface are summed up to calculate the electric field at any point inside the sphere the result is zero electric field.