Answer
If they are 20 nm apart, the potential energy is 1.5 J.
Work Step by Step
First,
$E \propto \frac{1}{r}$
$E_1 \propto \frac{1}{r_1}$
$E_2 \propto \frac{1}{r_2}$
We can use the ratio of $\frac{E_2}{E_1}$ to find $E_2$;
$\frac{E_2}{E_1} = \frac{r_1}{r_2}$
$E_2 = \frac{E_1~r_1}{r_2}$
$E_2 = \frac{(1.0~J)(30~nm)}{20~nm}$
$E_2 = 1.5~J$
If they are 20 nm apart, the potential energy is 1.5 J.
