Question

Two point charges 4 microcoulomb and 10 microcoulomb are placed at 20cm distance the electrostatic potential energy of the system is
1)3.6J 2)2.4J
3)1.8J 4)18J

Answer 1

HEYA!!

$e = \frac{1}{4\pi \: e} \times \frac{q1q2}{ {r}^{2} }$

now according to question,

$e = \frac{1}{4\pi \: e} \times \frac{4 \times {10}^{ - 6} \times 10 \times {10}^{ - 6} } {20}^{2}$

on solving the above you will get your answer

note: the square is on 20 i.e. r^2

☺FOLLOW ME☺