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MHT CET 2026 (12 May, Shift 2) · Capacitance of a Conductor

Capacitance of a Conductor question from MHT CET 2026 (12 May, Shift 2)

Q1MHT CET 2026 · 12 MAY · SHIFT 2Easy

Capacitance of a Conductor

A capacitor of unknown capacitance is connected across a battery of $'V'$ volt. The charge stored in it is $'Q'$ C. When potential across the capacitor is reduced by $'V_1'$ volt, the charge stored in it becomes $'Q_1'$ C. The potential $'V'$ is

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Answer: (a)

Capacitance is constant: $C = \frac{Q}{V} = \frac{Q_1}{V - V_1}$ .
$Q(V - V_1) = Q_1 V \Rightarrow V(Q - Q_1) = QV_1$
So the answer is (a) $V = \frac{QV_1}{Q - Q_1}$ .

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