A solid conducting sphere of radius R has a total charge q. Find the potential everywhere, both outside and inside the sphere.
From the previous analysis, you know that the charge will be distributed on the surface of the conducting sphere.
Exploiting the spherical symmetry with Gauss’s Law, for $r \geq R$,
$$\begin{aligned} EA &= \frac{q}{\epsilon_{0}} \\ E (4 \pi r^{2}) &= \frac{q}{\epsilon_{0}} \\ E &= \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r^{2}} \end{aligned}$$
$$V = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{r}$$
At r = R, just sub. r = R into the above equations. Note that the surface charge density can be introduced to give $E = \frac{\sigma}{\epsilon_{0}}$
For $r \leq R$,
$$E = 0$$
Since the electric field within the conductor is 0, the whole conductor must be at the same potential (equipotential). The electric potential within the conductor will be:
$$V = \frac{1}{4 \pi \epsilon_{0}} \frac{q}{R}$$
Next: Electric Field Of A Uniformly Charged Sphere
