College Physics (4th Edition)

The electric potential energy is $~-0.018~J$
We can find the electric potential energy: $U = \frac{k~q_1~q_2}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(5.0\times 10^{-6}~C)(-2.0\times 10^{-6}~C)}{5.0~m}$ $U = -0.018~J$ The electric potential energy is $~-0.018~J$