# 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

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.