Question

The side of the cube is ${}^{\prime }{l}^{\prime }$ and point charges are kept at each corner as shown in diagram. Interactive electrostatic potential energy of all the charges is $\frac{4k{q}^2}{\sqrt{N}l}$, then $N$ is

Answer 1

Total electrostatic potential energy is $\sum \frac{k{q}_1{q}_2}{r_{12}}$

$U=\frac{k{q}^2}{l}\left(6\right)+\left(\frac{-k{q}^2}{d}\right)\left(6\right)+\left(\frac{-k{q}^2}{l\sqrt{2}}\right)\left(6\right)+\left(\frac{k{q}^2}{l\sqrt{2}}\right)\left(6\right)+\frac{k{q}^2}{l\sqrt{3}}\left(4\right)=\frac{4k{q}^2}{\sqrt{3}l}\Rightarrow N=3$

Ans $3$