A charge 'Q' is placed at each corner of a cube of side 'a'. The potential at the centre of the cube is

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$\frac{4 Q}{\sqrt{3} {πϵ}_{0} a}$

$\frac{8 Q}{{πE}_{0} a}$

$\frac{4 Q}{{πϵ}_{0} a}$

$\frac{4 Q}{4 {πϵ}_{0} a}$

answer is C.

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Detailed Solution

Distance of each charge from the centre = $\frac{\sqrt{3}}{2} a$

Net potential at the centre

$\begin{array}{l} = 8 (\frac{1}{4 {πε}_{0}} \frac{Q}{d}) \\ = 8 (\frac{1}{4 {πε}_{0} \frac{\sqrt{3}}{2} a}) = \frac{4 Q}{\sqrt{3} {πε}_{0} a} \end{array}$

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