Question

Identical charges $(-q)$ are placed at each corner of a cube of side b then E.P.E. of charge $(+q)$ which is placed at the centre of the cube will be:

• A. $\frac{-4\sqrt{2}{q}^2}{\pi {ϵ}_{0}b}$
• B. $\frac{-8\sqrt{2}{q}^2}{\pi {ϵ}_{0}b}$
• C. $\frac{-4q^2}{\sqrt{3}\pi {ϵ}_{0}b}$
• D. $\frac{8\sqrt{2}{q}^2}{4\pi {ϵ}_{0}b}$

Answer 1

The correct option is C $\frac{-4q^2}{\sqrt{3}\pi {ϵ}_{0}b}$

$\begin{array}{c}Lengthofthedaigonalofacubehavingeachsidebis\sqrt{3b.}\\ So,dis\tan ceofcentreofcubefromeachvertexis\frac{\sqrt{3b}}{2}.\\ hence,potentialenergyofthegivensystemofcheis\\ U=8\times \left\{\frac{1}{4\pi {\in }_0}\cdot \frac{\left(-q\right)\left(q\right)}{\sqrt{3}b/2}\right\}=\frac{-4q^2}{\sqrt{3}\pi {\in }_{0}b}\end{array}$

Hence,

option $\left(C\right)$ is correct answer.