My textbook gives the following explanation on how excess charges are spread over conductors:
The excess charge on an isolated conductor moves entirely to the conductor's surface. However, unless the conductor is spherical, the charge does not distribute itself uniformly.
I was studying for a test and there was the following question, regarding the thin-wall conducting cylindrical shell below (which is coaxial to the rod inside it):
$Q_1 > 0, Q_2 < 0$
What is the charge on the interior and exterior surface of the shell?
At first I thought the excess charge on the interior surface would be zero, since all the excess charge would move to the outer surface of the shell. But according to my textbook, it is as follows:
We consider a cylindrical Gaussian surface whose radius places it within the shell itself. The electric field is zero at all points on the surface since any field within a conducting material would lead to current flow (and thus to a situation other than the electrostatic ones being considered here), so the total electric flux through the Gaussian surface is zero and the net charge within it is zero (by Gauss's law). Since the central rod has charge $Q_1$, the inner surface of the shell must have charge $Q_{in} = -Q_1 = -3.40 \times 10^{-12}\: \mathrm{C}$.
Since the shell is known to have total charge $Q_2 = -2.00\: Q_1$ it must have charge $Q_{out} = Q_2 - Q_{in} = - Q_1 = -3.40 \times 10^{-12}\: \mathrm{C}$ on its outer surface.
So there is $-Q_1$ excess charge on the inner surface because said charges are being attracted by the rod? What if there was no rod, how would the excess charge be distributed over the shell's surface? Can I tell how the excess charge will be distributed over the surface of any non-spherical conductor, or only in special cases such as the above?