Answer
$U = 0.00285~J$
Work Step by Step
We can find the electric potential energy due to the $+4.0~\mu C$ charge and the $+3.0~\mu C$ charge:
$U = \frac{k~q_1~q_2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(4.0\times 10^{-6}~C)(3.0\times 10^{-6}~C)}{5.0~m}$
$U = 0.0216~J$
We can find the electric potential energy due to the $+4.0~\mu C$ charge and the $-1.0~\mu C$ charge:
$U = \frac{k~q_1~q_3}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(4.0\times 10^{-6}~C)(-1.0\times 10^{-6}~C)}{3.0~m}$
$U = -0.012~J$
We can find the electric potential energy due to the $+3.0~\mu C$ charge and the $-1.0~\mu C$ charge:
$U = \frac{k~q_2~q_3}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(3.0\times 10^{-6}~C)(-1.0\times 10^{-6}~C)}{4.0~m}$
$U = -0.00675~J$
We can find the total potential energy of the system of three charges:
$U = 0.0216~J-0.012~J-0.00675~J$
$U = 0.00285~J$
