Identical charges – q each are placed at 8 corners of a cube of each side b. Electrostatic potential energy of a charge + q which is placed at the centre of cub will be

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$\frac{- 4 \sqrt{2} q^{2}}{π ε_{0} b}$

$\frac{- 4 q^{2}}{\sqrt{3} π ε_{0} b}$

$\frac{- 8 \sqrt{2} q^{2}}{\sqrt{3} π ε_{0} b}$

$\frac{- 8 \sqrt{2} q^{2}}{π ε_{0} b}$

answer is C.

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Detailed Solution

Length of body diagnonal $= \sqrt{3} b .$

$∴$ Distance of centre of cube from each corner, $r = \frac{\sqrt{3}}{2} b$

Total P.E. of charge + q at the centre $= \frac{8 q (- q)}{4 π ε_{0} r} = \frac{- 8 q^{2}}{4 π ε_{0} (\sqrt{3} b / 2)} = \frac{- 4 q^{2}}{π ε_{0} \sqrt{3} b}$

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