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JEE Main 2023 (11 Apr, Shift 1) · Charge Sharing & Loss of Energy

Charge Sharing & Loss of Energy question from JEE Main 2023 (11 Apr, Shift 1)

Q14JEE Main 2023 · 11 APR · SHIFT 1Easy

Charge Sharing & Loss of Energy

A parallel plate capacitor of capacitance $2\text{ F}$ is charged to a potential $V$ , The energy stored in the capacitor is $E_1$ . The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$ . The ratio $E_2/E_1$ is:

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Answer: (a)

Charge is fixed and capacitance doubles: $E = \frac{Q^2}{2C}$ .
$E_2 = \frac{Q^2}{2(2C)} = \frac{E_1}{2}$
$E_2 : E_1 = 1 : 2$ , so the answer is (a).

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