A charge Q is placed at each corner of a cube of side r. The potential at the centre of the cube is ( ε₀ = permittivity of free space) (a) (4Q)/(πε₀ r√(3)) (b) (8Q)/(πε₀ r√(3)) (c) (16Q)/(3πε₀ r) (d) (32Q)/(πε₀ r√(3))

Answer: (a). Centre-to-corner distance = √(3)/2r; all 8 charges contribute equally. V = 8 × (Q)/(4πε₀)·(2)/(√(3)r) = (16Q)/(4πε₀√(3)r) V = (4Q)/(πε₀ r√(3)) So the answer is (a).

A charge Q is placed at each corner of a cube of side r. The… | MHT CET 2026 (20 Apr, Shift 2)

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Electric Potential & Equipotentials

A charge $Q$ is placed at each corner of a cube of side $r$ . The potential at the centre of the cube is ( $\epsilon_0$ = permittivity of free space)

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Answer: (a)

Centre-to-corner distance $= \frac{\sqrt{3}}{2}r$ ; all 8 charges contribute equally.
$V = 8 \times \frac{Q}{4\pi\epsilon_0}\cdot\frac{2}{\sqrt{3}r} = \frac{16Q}{4\pi\epsilon_0\sqrt{3}r}$
$V = \frac{4Q}{\pi\epsilon_0 r\sqrt{3}}$
So the answer is (a).

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