Answer
$U = -1.1\times 10^{-5}~J$
Work Step by Step
We can find the electric potential energy due to the $+10.0~nC$ charge and the $-10.0~nC$ charge:
$U = \frac{k~q_1~q_2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(10.0\times 10^{-9}~C)(-10.0\times 10^{-9}~C)}{0.080~m}$
$U = -1.125\times 10^{-5}~J$
We can find the electric potential energy due to the $+10.0~nC$ charge and the $-4.2~nC$ charge:
$U = \frac{k~q_1~q_2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(10.0\times 10^{-9}~C)(-4.2\times 10^{-9}~C)}{0.040~m}$
$U = -9.45\times 10^{-6}~J$
We can find the electric potential energy due to the $-10.0~nC$ charge and the $-4.2~nC$ charge:
$U = \frac{k~q_1~q_2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(-10.0\times 10^{-9}~C)(-4.2\times 10^{-9}~C)}{0.040~m}$
$U = 9.45\times 10^{-6}~J$
We can find the total potential energy of the system of three charges:
$U = -1.125\times 10^{-5}~J-9.45\times 10^{-6}~J+9.45\times 10^{-6}~J$
$U = -1.1\times 10^{-5}~J$
