## College Physics (4th Edition)

(a) The electric potential energy is $~-4.4\times 10^{-18}~J$ (b) The negative sign shows that it would require external work to separate the two charges.
(a) We can find the electric potential energy: $U = \frac{k~q_1~q_2}{r}$ $U = \frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)(-1.6\times 10^{-19}~C)}{0.0529\times 10^{-9}~m}$ $U = -4.4\times 10^{-18}~J$ The electric potential energy is $~-4.4\times 10^{-18}~J$ (b) The negative sign shows that it would require external work to separate the two charges.