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Eight charges, each of magnitude q are placed at the vertices of a cube placed in vacuum. Electric potential at the centre of the cube due to this system of charges is: ( $ε_{0}$ is permittivity of vacuum and a is length of each side of the cube)

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$\frac{2 q}{π ε_{0} a}$

$\frac{4 q}{\sqrt{3} π ε_{0} a}$

zero

$\frac{\sqrt{3} q}{π ε_{0} a}$

answer is B.

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Detailed Solution

$V = 8 \frac{k q}{(\frac{\sqrt{3} a}{2})} = \frac{16 k q}{\sqrt{3} a} = \frac{4 q}{\sqrt{3} π ε_{0} a}$

$Where \sqrt{3} a is the body diagonal and distance of the center from the corner is \frac{\sqrt{3} a}{2}$

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