Question

Why do extra charges given to a conductor reside on the outer surface?

Answer 1

We can deduce this from the Gauss law. Let us construct a Gaussian surface which lies just inside the surface of the conductor.
$\oint E.ds=\frac{Q}{{ϵ}_o}$
Here Q is the enclosed charge.
We know that the electric field-strength inside a conductor must be zero. This implies that the left-hand side of the above equation is zero, and therefore the right-hand side must also be zero. This is possible only if there is no net charge enclosed by the Gaussian surface and the excess charges move to the surface of the conductor.