Answer
(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.
Work Step by Step
(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.
