Answer
$-5.9\times 10^{-14}J$
Work Step by Step
We can find the required electric potential energy as follows:
$U=\frac{Kq_1q_2}{r}+\frac{Kq_1q_3}{r}+\frac{Kq_2q_3}{r}$
We plug in the known values to obtain:
$\implies U=\frac{8.99\times 10^9N.m^2/C^2}{1.3\times 10^{-15}m}[(\frac{-e}{3})(\frac{-e}{3})+(\frac{-e}{3})(\frac{2e}{3})+(\frac{-e}{3})(\frac{2e}{3})]$
$\implies U=\frac{8.99\times 10^9N.m^2/C^2}{1.3\times 10^{-15}m}(\frac{-e^2}{3})$
$\implies U=\frac{8.99\times 10^9N.m^2/C^2(-1.6\times 10^{-19}C)^2}{1.3\times 10^{-15}m\times 3}$
$U=-5.9\times 10^{-14}J$
