A system consisting of two point charges 7 µC and -4 µC are placed at (-9, 0, 0) cm and (9, 0, 0) cm respectively. The electrostatic potential energy of the system is:

- 1.4 J

The correct answer is Option (2) → - 1.4 J

Given,

$q_1$, charge = $7μc=7×10^{-6}C$

$q_2$, charge = $4μc=-4×10^{-6}C$

d, distance = $(+9,0,0)-(-9,0,0)=18cm$

Distance between the two charges is 18cm.

$ \Rightarrow U = \frac{kq_1q_2}{r}$

$= \frac{ 9\times 10^9 \times -28\times 10^{-12}}{18\times 10^{-2}} = -1.4J$