Two parallel-plate capacitors of capacitance $C$ and $2 C$ are connected in parallel and charged to a potential difference $V$ . The battery is then disconnected and the region between the plates of capacitor $C$ is filled completely with a material of dielectric constant $K$ . The common potential difference across the combination becomes

\frac{2 V}{K + 2}

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\frac{2 V}{K + 2}

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\frac{3 V}{K + 3}

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\frac{3 V}{K + 2}

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Solution

The correct option is D $\frac{3 V}{K + 2}$
Initally, Case I

$C_{e q} = C_{1} + C_{2} = C + 2 C = 3 C$
$Q_{n e t} = 3 C \times V$
$Q_{n e t} = 3 C V$

Case II

$C^{'} = K C + 2 C$
$C^{'} = C (K + 2)$
Now
$V_{c o m m o n} = \frac{Q_{n e t}}{C_{n e t}} = \frac{3 C V}{C (K + 2)} = \frac{3 V}{(K + 2)}$

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