0
$\begingroup$

I read that if there were a charge $+q \ $ kept inside a spherical conducting shell of thickness $t$, a charge of $-q \ $ would develop on the inner surface of the shell. Not more, not less.

Why is this so? Why doesn't the charge accumulating on the inner surface depend on the geometry of the conductor? Why wouldn't the charge on the inner surface be different if I changed the radius of the shell?

$\endgroup$
0
$\begingroup$

If you took a Gaussian surface within the conductor and the positive inducing change was not equal in magnitude to the negative induced charge there would be a net electric flux through the Gaussian surface - Gauss’s law.
If that were the case then there must be an electric fields inside the conductor which cannot be present otherwise there would be electric current within the conductor.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.