JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Capacitance

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Two capacitors of capacities 2C and C are joined in parallel and charge up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be:

(A) $\frac{3 V}{K}$
(B) $\frac{V}{K}$
(C) $\frac{V}{K + 2}$
(D) $\frac{3 V}{K + 2}$ ✓ Correct

Answer: (D)

Solution

Initially, the total charge is

Q = (2 C + C) V = 3 C V

. After the battery is removed and the dielectric is inserted, the charge remains constant. The capacitor of capacity C becomes KC. The new potential is

V^{'} = \frac{Q}{2 C + K C} = \frac{3 C V}{C ( K + 2 )} = \frac{3 V}{K + 2}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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