Answer
We can rank the situations in order of electric potential energy, from highest to lowest:
$a = d \gt b = e \gt c$
Work Step by Step
We can write a general equation for the electric potential energy:
$U = \frac{k~Q_1~Q_2}{d}$
Let $q_1 = 1~\mu C$
Let $q_2 = 1~\mu C$
Let $r = 1~m$
Let $U_0 = \frac{k~q_1~q_2}{r}$
We can find an expression for the electric potential energy in each case.
(a) $U = \frac{k~(q_1)~(2q_2)}{r} = 2~U_0$
(b) $U = \frac{k~(2q_1)~(-q_2)}{r} = -2~U_0$
(c) $U = \frac{k~(2q_1)~(-4q_2)}{2r} = -4~U_0$
(d) $U = \frac{k~(-2q_1)~(-2q_2)}{2r} = 2~U_0$
(e) $U = \frac{k~(4q_1)~(-2q_2)}{4r} = -2~U_0$
We can rank the situations in order of electric potential energy, from highest to lowest:
$a = d \gt b = e \gt c$
