Answer
The correct answer is:
(b) $E = 0$, but $V$ is the same as before the charges were moved.
Work Step by Step
At point P, the magnitude of the electric field due to each point charge is $E = \frac{k~q}{r^2}$. In part (b), the electric field due to each point charge is equal in magnitude and they point in the opposite direction. Therefore, $E = 0$
We can find an expression for the electric potential at point P:
$V = \frac{k~q}{r}+ \frac{k~q}{r}$
Note that the electric potential is the same in part (a) and part (b).
The correct answer is:
(b) $E = 0$, but $V$ is the same as before the charges were moved.
