Answer
We can rank the four situations according to the magnitude of the net electric field at the central point:
$(2) \gt (4) \gt (3) \gt (1)$
Work Step by Step
We can find write the general equation for the magnitude of an electric field:
$E = \frac{\vert q \vert}{4\pi ~\epsilon_0~r^2}$
Note that an electric field points toward a negative charge and away from a positive charge.
We can find the magnitude of the net electric field at the central point in each situation:
(1) The electric fields due to the two charges on the left cancel with the electric fields due to the two charges on the right. The magnitude of the net electric field at the central point is zero.
(2) We can find the magnitude of the net electric field:
$E = 2\times \Big(\frac{\vert e \vert}{4\pi ~\epsilon_0~(2d)^2}+\frac{\vert e \vert}{4\pi ~\epsilon_0~d^2}\Big)$
$E = \frac{5}{2}\times \Big(\frac{\vert e \vert}{4\pi ~\epsilon_0~d^2}\Big)$
(3) The electric fields due to the two interior charges cancel with each other.
We can find the magnitude of the net electric field:
$E = 2\times \Big(\frac{\vert e \vert}{4\pi ~\epsilon_0~(2d)^2}\Big)$
$E = \frac{1}{2}\times \Big(\frac{\vert e \vert}{4\pi ~\epsilon_0~d^2}\Big)$
(4) The electric fields due to the two exterior charges cancel with each other.
We can find the magnitude of the net electric field:
$E = 2\times \Big(\frac{\vert e \vert}{4\pi ~\epsilon_0~d^2}\Big)$
We can rank the four situations according to the magnitude of the net electric field at the central point:
$(2) \gt (4) \gt (3) \gt (1)$
