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MHT CET 2026 (16 Apr, Shift 1) · Electrostatic Potential Energy

Electrostatic Potential Energy question from MHT CET 2026 (16 Apr, Shift 1)

Q12MHT CET 2026 · 16 APR · SHIFT 1Medium

Electrostatic Potential Energy

Two point charges $q_1 = 6\,\mu$ C and $q_2 = 4\,\mu$ C are kept at points A and B in air where distance $AB = 10$ cm. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ , by 2 cm? $\Bigg(\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}$ SI units $\Bigg)$

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Answer: (c)

$\Delta U = kq_1q_2\left(\dfrac{1}{r_2} - \dfrac{1}{r_1}\right)$ with $r_1 = 0.10$ m, $r_2 = 0.08$ m.
$kq_1q_2 = 9\times10^9 \times 24\times10^{-12} = 0.216$
$\Delta U = 0.216(12.5 - 10) = 0.54$ J
So the answer is (c) 0.54 J.

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