Answer
The work required by an external force to set up this system of charges is $-3.0~J$
Work Step by Step
We can find the electric potential energy due to the $5.5~\mu C$ charge and the $-6.5~\mu C$ charge:
$U = \frac{k~q_1~q_2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(5.5\times 10^{-6}~C)(-6.5\times 10^{-6}~C)}{0.12~m}$
$U = -2.68~J$
We can find the electric potential energy due to the $2.5~\mu C$ charge and the $-6.5~\mu C$ charge:
$U = \frac{k~q_1~q_2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(2.5\times 10^{-6}~C)(-6.5\times 10^{-6}~C)}{0.16~m}$
$U = -0.914~J$
We can find the electric potential energy due to the $5.5~\mu C$ charge and the $2.5~\mu C$ charge:
$U = \frac{k~q_1~q_2}{r}$
$U = \frac{(9.0\times 10^9~N~m^2/C^2)(5.5\times 10^{-6}~C)(2.5\times 10^{-6}~C)}{0.20~m}$
$U = 0.619~J$
We can find the total electric potential energy in the system:
$U = -2.68~J-0.914~J+0.619~J = -3.0~J$
The work required by an external force to set up this system of charges is $-3.0~J$
