- #1

yoran

- 118

- 0

In my physics we derived certain properties of conductors in electrostatic equilibrium using Gauss' law. One property seemed to "mathematical" to me. It says that the whole charge of a conductor in electrostatic equilibrium is located at the surface.

The proof is at follows. Since the electrical field inside the conductor is zero, then one can put a Gaussian surface inside the conductor. The net flux through the surface is given by

[tex]\Phi = \oint \bar{E}d\bar{A} = 0[/tex]

Then, according to Gauss' law,

[tex]\Phi = \frac{Q_{in}}{\epsilon_0} = 0[/tex]

thus

[tex]Q_{in} = 0[/tex]

One can make a Gaussian surface that gets bigger and bigger but stays inside the conductor. The charge inside the surface is then 0. From that follows that the charge is located at the surface.

What bothers me in this proof is that, what stops you from making a Gaussian surface that is exactly a copy of the conductor? That means that the total charge in the conductor is 0 and that is a contradiction. Why do you have to stop "growing" your Gaussian surface just before you reach the surface of the conductor? Furthermore, if that's true, then that means that the whole charge is a located in the infinitesimal thin outer layer of the conductor.

How is the charge really distributed in a conductor in electrostatic equilibrium? I ask that question because the physical world doesn't know infinitesimal thin layers. Or does it?

Thanks.