1. Nov 22, 2015

betelgeuse91

1. The problem statement, all variables and given/known data
I am having trouble understanding how a Faraday cage works.

2. Relevant equations
$$\oint \vec{E}\cdot\,d\vec{A} = \dfrac{Q_{encl}}{\epsilon_{0}}$$

3. The attempt at a solution
It says that Faraday cage is a hollow metallic conductor and hence, inside the cage, $$\vec{E} = 0$$
I am aware that it holds for a eletrostatic metallic conductor, since if there is any electric field within the conductor, then the charges will move around by the field and it would be no longer electrostatic. Is Faraday cage a eletrostatic conductor as well? Is that why this still holds for Faraday cage? If it is, why is a Faraday cage electrostatic even if there is a electric field from outer source? Furthermore, I am wondering if all metallic conductors are naturally electrostatic and if they are, why. Thank you very much for your help.

2. Nov 22, 2015

betelgeuse91

So why the charges on the surface of the cage move around to cancel the electric field from outside?

3. Nov 22, 2015

J Hann

A Faraday cage is basically a conducting shell.
Suppose you have 3 of these shells (spherical for simplicity).
One has all of its excess charge on the outside.
Another has an equal distribution of its excess charges on both its inner and outer surfaces.
The third has all of its charge on its inner surface.
Which of the three has the lowest electrical potential?
Since charge is free to move about within a conductor it will assume the lowest energy configuration.

4. Nov 22, 2015

haruspex

You seem to be making some distinction between a conductor and an electrostatic conductor. (The metallic aspect is clearly irrelevant: it's a conductor.) Electrostasis is a circumstance, not a property of the material. If you drop a charge onto one spot on a conductor it will not immediately be electrostatic, it will take some fraction of a second for the charge to redistribute. (In fact, if it is a superconductor a current will continue to flow. It only settles down because of resistance.) Once redistributed, it will be electrostatic again.
If there is a field within the material of the conductor, it will push the free electrons around. They will only stop moving when there is no field.