# Why do charges placed inside a metallic cavity have no effect on outer charges and vice versa?

Why do charges placed inside a metallic cavity have no effect on outer charges and vice versa?

My teacher explained it by electrostatic shielding. He said that the conductor does not allow the electric field to pass through it. I agree that inside a conductor the electric field is zero, because free electrons of the conductor move opposite to the external electric field to create an equal and opposite electric field. But what about inside the cavity? There is nothing like free electrons in the cavity, so an external electric field should reach there.

• But a charged placed inside a hollow conductor does cause field lines external to the conductor, this is a result easily verifiable through Gauss' law. When you say outside charges do you mean inside the conductor still or completely outside the entire conductor? Commented Sep 30, 2023 at 16:49

## 1 Answer

This is easiest to see if you look at it in terms of the electrostatic potential. In particular, we will want to use the following two facts:

1. The potential is the same everywhere inside the bulk of a conductor.
2. If we follow a field line through space, in the direction of the electric field, then the potential must decrease along this line.

With these two facts, we can construct a proof by contradiction. Consider a conductor with a cavity. The cavity is empty (no charges inside it.) Suppose that there was an electric field inside this cavity.

Pick some field line of this electric field that lies inside the cavity. Since there are no charges inside the cavity, the electric field must start at one point $$A$$ on the surface of the cavity, on some positive surface charge; and it must end at another point $$B$$ on the surface of the cavity, on some negative surface charge.

By fact #2 above, the potentials at points $$A$$ and $$B$$ must be different: we followed an electric field line to get from $$A$$ to $$B$$, so the potential at $$B$$ must be less than the potential at $$A$$. But by fact #1 above, the potentials at $$A$$ and $$B$$ must be the same. This is a contradiction; we conclude that there cannot be an electric field inside the cavity.