Above there is a 2D pic of this problem.

$S$ is a conducting sphere with no charge. I am considering the electrostatics case.

It is a hollow sphere: inside its cavity lies a point charge $q$, $q > 0$.

What is the electrostatic force $\vec{F}$ on the point charge $q$?

My attempt:

If $\partial S $ is border of the cavity, I know there is a total charge of $-q$ on it (because $S$ is a conductor).

Now, $\vec{F} = q \vec{E}$, where $\vec{E}$ is the electric field on the charge $q$ caused by the charge $-q$ on $\partial S$.

The problem is now about $\vec{E}$. $\vec{E} = 0$ inside the cavity if no charge is inside the cavity. What if there is $q$ inside it?

Because of symmetry, I thought that $\vec{E} = 0$ as well: there is no "main direction" the electric field should have. Neither do the force on the charge. And I also thought that the electric field on every point inside the cavity should be zero as well. Whereas it would be non-zero if charge if moved and the symmetry is lost.

However, I couldn't find a rigorous way to prove it. If I consider a Gauss surface *inside* the cavity, the flux is $> 0$ because $\frac{q}{\epsilon_0} > 0$, so why should the electric field be zero?

## Best Answer

There is a difference between the field at the location of the charge $q$ and the field at another point in the cavity. Indeed, you are correct that by symmetry $E=0$ at the charge $q$ by charges on the outside of the cavity. However, if you are looking at a Gaussian sphere centered on $q$, then you are looking at the field

caused by $q$.Since this is a homework problem I will leave it to you to apply Gauss's law inside the cavity. But you can reason that the field in the cavity must be radial centered on $q$.

You can also use superposition. You already said that $E=0$ inside of the cavity without a charge in it. So then what is the field inside the cavity with the charge if we know superposition is valid for electric fields?