**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