Answer
$U=1.15\times10^{-19}J$
Work Step by Step
Electric potential $U$ can be calculated as
$U=K\frac{q_1q_2}{r}$
Putting the values in this formula, we get
$U=9\times10^9\times\frac{(-1.6\times10^{-19})(-1.6\times10^{-19})}{2\times10^{-9}}$
$U=1.15\times10^{-19}J$
