# Electrostatic Shielding

suppose there is a solid conducting body with arbitrarily shaped cavity inside. Now a charge is placed inside the cavity.Now the entire setup is placed in an uniform electric field. Will the redistribution of induced charges take place only on the outer surface of the conductor or inside the cavity as well?

## 2 Answers

The outer surface charge distribution will redistribute itself such that $\textbf{E} = \textbf{0}$ everywhere inside the conductor. Why? because if $\textbf{E} \neq \textbf{0}$ charges inside the conductor will move according to $\textbf{F} = q\textbf{E}$, charges will continue to move until equilibrium is reached, i.e. $\textbf{E} = \textbf{0}$ inside the conductor.

Therefore the electric conditions on the inner surface are the same regardless of whether or not an external field is present, and so the inner charge distribution will not change.

I tried addressing this question properly and logically in an article from the European Journal of Physics, from a few years back:

https://arxiv.org/abs/1609.04248