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JEE Main 2021 (26 Feb, Shift 2) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2021 (26 Feb, Shift 2)

Q46JEE Main 2021 · 26 FEB · SHIFT 2Medium

Electrostatic Potential Energy

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is __________ times that of a smaller drop.

Numerical answer type. Work it out, then open the answer.

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Answer: 243

Self-energy $U = \frac{kQ^2}{2R}$ , so $U \propto \frac{Q^2}{R}$ .
27 drops: $R = 3r$ , $Q = 27q$ .
$\frac{U_{big}}{U_{small}} = \frac{27^2}{3} = 243$
So the answer is 243.

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