GRE Physics GR 1777, Problem 038 Solution

You should know

Electric Potential

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


$Q$: point charge

$r$: distance

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


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} $

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