Answer
$U=0.225J$
Work Step by Step
The electric potential energy of the system is
$U=K\frac{q_1 q_2}{r}$
We know that $q_1=q_2=q$
So $U=K\frac{qq}{r}$ which becomes $U=K\frac{q^2}{r}$
We plug in the known values to obtain:
$U=9\times 10^9\frac{(5\times 10^{-6})^2}{1.00}$
$U=0.225J$
