A) \[\frac{8\sqrt{2}{{q}^{2}}}{4\pi {{\varepsilon }_{0}}b}\]
B) \[\frac{-8\sqrt{2}{{q}^{2}}}{\pi {{\varepsilon }_{0}}b}\]
C) \[\frac{-4\sqrt{2}{{q}^{2}}}{\pi {{\varepsilon }_{0}}b}\]
D) \[\frac{-4{{q}^{2}}}{\sqrt{3}\pi {{\varepsilon }_{0}}b}\]
Correct Answer: D
Solution :
Length of the diagonal of a cube having each side b is \[\sqrt{3}\,b.\] So distance of centre of cube from each vertex is \[\frac{\sqrt{3}\,b}{2}.\] Hence potential energy of the given system of charge is \[U=8\times \left\{ \frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{(-q)\,(q)}{\sqrt{3}\,b/2} \right\}=\frac{-4{{q}^{2}}}{\sqrt{3}\pi {{\varepsilon }_{0}}b}\]
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