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Identical charges (-q) are placed at each comer of a cube of side b then electrical potential energy of charge (+q) which is placed at the centre of the cube will be :

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$\frac{- 4 \sqrt{2} q^{2}}{{πε}_{0} b}$

$\frac{- 4 q^{2}}{\sqrt{3} {πε}_{0} b}$

$\frac{- 8 \sqrt{2} q^{2}}{{πε}_{0} b}$

$\frac{8 \sqrt{2} q^{2}}{4 {πε}_{0} b}$

answer is C.

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Detailed Solution

Distance of each charge placed at comer from the centre of cube

$= \frac{\sqrt{3} b}{2}$

So, potential energy = $8 x \frac{1}{4 {πε}_{0}} \frac{q (- q)}{\frac{\sqrt{3} b}{2}} = - \frac{4 q^{2}}{\sqrt{3} {πε}_{0} b}$

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