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:
Answer & explanation
Correct answer: option 2
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}$
$= \frac{ 9\times 10^9 \times -28\times 10^{-12}}{18\times 10^{-2}} = -1.4J$