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MHT CET 2026 (12 May, Shift 1) · Electrostatic Potential Energy

Electrostatic Potential Energy question from MHT CET 2026 (12 May, Shift 1)

Q7MHT CET 2026 · 12 MAY · SHIFT 1Medium

Electrostatic Potential Energy

The equal charges (say 'q') are placed at a distance of $4r$ and a third charge $(-2q)$ is placed at a mid-point. The potential energy of the system is ( $\epsilon_0 =$ permittivity of free space)

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Answer: (c)

Pairs: $(q, q)$ at $4r$ ; two pairs $(q, -2q)$ at $2r$ .
$U = k\left[\dfrac{q^2}{4r} - \dfrac{2q^2}{2r} - \dfrac{2q^2}{2r}\right] = -\dfrac{7kq^2}{4r}$
$U = -\dfrac{7q^2}{16\pi\epsilon_0 r}$ . So the answer is (c).

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