The electric potential energy for a system of two point charges of 6…

The correct answer is Option (1) → -0.9 J

$\text{Given: } q_1 = 6~\mu\text{C} = 6 \times 10^{-6}~\text{C},~ q_2 = -3~\mu\text{C} = -3 \times 10^{-6}~\text{C}$

$\text{Positions: } x_1 = -9~\text{cm},~ x_2 = +9~\text{cm}$

$\text{Distance between charges: } r = x_2 - x_1 = 9 - (-9) = 18~\text{cm} = 0.18~\text{m}$

$\text{Electric potential energy of the system: } U = \frac{k q_1 q_2}{r}$

$k = 9 \times 10^9~\text{Nm}^2/\text{C}^2$

$U = \frac{(9 \times 10^9) \cdot (6 \times 10^{-6}) \cdot (-3 \times 10^{-6})}{0.18}$

$U = \frac{-162 \times 10^{-3}}{0.18}$

$U = -0.9~\text{J}$

$\text{Answer: } U = -0.9~\text{J}$