GRE Physics GR 1777 Problem Solution¶

038. Electrodynamics (Electric Potential)¶

You should know¶

Electric Potential¶

• For a point charge,
$$\large V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$$

where

$Q$: point charge

$r$: distance

• In a conductor,
$$\large V = \text{constant}$$

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,

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

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