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MHT CET 2025 (23 Apr, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from MHT CET 2025 (23 Apr, Shift 2)

Q28MHT CET 2025 · 23 APR · SHIFT 2Medium

Electrostatic Potential Energy

Two point charges $+10\mu\text{C}$ and $4\mu\text{C}$ are placed 10 cm apart in air. The work required to be done to bring them 2 cm closer is

$\Bigg(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ SI units $\Bigg)$

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Answer: (b)

$W = \frac{q_1q_2}{4\pi\varepsilon_0}\left(\frac{1}{r_2} - \frac{1}{r_1}\right)$ with $r_1 = 0.10$ m, $r_2 = 0.08$ m.
$q_1q_2 \times 9\times10^9 = 10\times10^{-6}\times4\times10^{-6}\times9\times10^9 = 0.36$
$W = 0.36\,(12.5 - 10) = 0.9$ J
So the answer is (b) 0.9 J.

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