Three point charges are placed at the corners of an equilateral triangle ABC as shown. The work done to put together these charges from infinity to their respective locations is

$-3.42 × 10^{-4}J$

The correct answer is Option (3) → $-3.42 × 10^{-4}J$

Given charges at corners of an equilateral triangle of side $r=5\ \text{cm}=0.05\ \text{m}$:

$q_1=+3\times10^{-8}\ \text{C},\; q_2=-5\times10^{-8}\ \text{C},\; q_3=+2\times10^{-8}\ \text{C}$

Electrostatic potential energy (work to assemble) :

$U = \frac{1}{4\pi\epsilon_0}\left(\frac{q_1q_2}{r}+\frac{q_1q_3}{r}+\frac{q_2q_3}{r}\right)$

Compute pair products and sum:

$q_1q_2=-1.5\times10^{-15}\ \text{C}^2,\; q_1q_3=0.6\times10^{-15}\ \text{C}^2,\; q_2q_3=-1.0\times10^{-15}\ \text{C}^2$

Sum $= -1.9\times10^{-15}\ \text{C}^2$

Using $\frac{1}{4\pi\epsilon_0}=9\times10^{9}\ \text{N·m}^2\text{·C}^{-2}$:

$U = 9\times10^{9}\times\frac{-1.9\times10^{-15}}{0.05} = -3.42\times10^{-4}\ \text{J}$

Answer: $-3.42\times10^{-4}\ \text{J}$