# Why does excess of charge in an isolated conductor move to the surface?

A remark in my textbook goes as follows: "If an excess charge is placed on an isolated conductor, that amount of charge will move entirely to the surface of the conductor. None of the excess charge will be found within the body of the conductor."

The author explains this with the help of the above picture, which is a lump of copper hanging from an insulating thread. He says that the electrical field in the Gaussian surface must be zero, but he does not really explain it well.

Can someone explain why this must necessarily be so?

• Remember Gauss's law? $\rho = \text {div} E$ – pppqqq Jan 14 '17 at 18:50

The excess charge is repelled from itself (all charge carriers have the same sign). It can't move any further away than to the surface.

So, it is actually logical that a net charge will be present on the surfaces (in the static situation) only.

• But isn't there charge on the surface too? So these charges will repel each other aswell? For me it is much more logical that in a conductor there can be no equilibrium; all the charges just move chaotically past each other, trying to avoid each other (if they are of the same sign). But still, somehow, they all come to rest and the excess of charge moves to the surface. – Kamil Feb 15 '15 at 12:04
• @Kamil The charges within the volume are a bit like "musical chairs" (if you know the game en.wikipedia.org/wiki/Musical_chairs ). There is a transient conflict of movable charges , electrons, to fill available minimum energy holes. The electrons that lose are the ones ending on the surface, because there are no more atoms/energy-levels to offer bound states ( half of their neighborhood is empty). Excess charge means more electrons than positive nucleus charges. The gaus integral describes this situation. – anna v Feb 15 '15 at 12:52

If you are considering electrostatic then by definition all the charges do not move. A conductor is an element whose charges are not bounded, i.e. they will move if any force acts of them. In electrostatic thus we must have that the net force on any charge in the conductor must be zero. Considering only electrostatic forces this means that inside the conductor the net electric field $\mathbf{E}$ is zero as well. If you take the Gaussian surface suggested by your book you will see that the flux through that surface will be zero (since on it $\mathbf{E} = 0$); thus from Gauss' law it follows that there is no net charge inside the volume enclosed by that surface. Thus if the conductor has a net total charge, it must lies on the conductor surface.

One way to think about it is that prior to the placement of excess charge on the object, it is electrically neutral. So you can almost pretend that there is nothing inside the object at all. No atoms, no protons, no neutrons, no electrons, no nothing. Pretend that object is the full extent of our universe and its boundaries are the boundaries of our existence. If you now place all these excess charges inside this empty space, they will move as far away from each other as possible. Ergo, they will move to the surface.

• Nope, because when they're at the surface they end up being more densely packed than if they were spread out trough the whole conductor. – MaDrung Feb 28 '17 at 8:51