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MHT CET 2024 (10 May, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from MHT CET 2024 (10 May, Shift 2)

Q33MHT CET 2024 · 10 MAY · SHIFT 2Medium

Electrostatic Potential Energy

Two point charges $q_1 = 6\mu\text{C}$ and $q_2 = 4\mu\text{C}$ are kept at points $A$ and $B$ in air where $AB = 10\text{ cm}$ . What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by 2 cm? $\Bigg(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ SI units $\Bigg)$

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Answer: (b)

$\Delta U = kq_1q_2\left(\dfrac{1}{0.08} - \dfrac{1}{0.10}\right)$
$kq_1q_2 = 9\times10^9 \times 24\times10^{-12} = 0.216$
$\Delta U = 0.216 \times 2.5 = 0.54$ J
So the answer is (b) 0.54 J.

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