Two capacitors with capacitance 2C and C are joined in parallel and charged 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 equals:

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$\frac{V}{K}$

$\frac{3 V}{K}$

$\frac{V}{K + 2}$

$\frac{3 V}{K + 2}$

answer is A.

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Detailed Solution

$q_{1} = 2 C V, q_{2} = C V$

Now condenser of capacity C is filled with dielectric K, therefore C₂ = KC

As charge is conserved

$∴ q_{1} + q_{2} = (C_{2} + 2 C) V$

$\Rightarrow V^{'} = \frac{3 CV}{(K + 2) C} = \frac{3 V}{K + 2}$

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