### You should know¶

#### Electric Potential¶

- For a point charge,

where

$Q$: point charge

$r$: distance

- In a conductor,

### Solution¶

Let's start from infinity, where the electric potential is assumed to be 0, and move toward to this object.

When the distance $r$ is far enough ($r > b$), the net charge can be considered as the sum of the point charge $Q$ and the net charge of the shell $q$, thus, the electric potential is

$$ \large V = \frac{Q+q}{4\pi\epsilon_0 r}. $$When the distance is equal to $b$, which is the outer radius of the shell, then the electric potential on the outer surface is

$$ \large V = \frac{Q+q}{4\pi\epsilon_0 b}. $$Since the electric potential in a conductor is constant, the electric potential inside the shell ($a < r < b$) is the same as the potential on the outer radius, therefore,

$$ \large V = \frac{Q+q}{4\pi\epsilon_0 b}. $$### Answer¶

(E) $ \large \frac{Q+q}{4\pi \epsilon_0 b} $