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Two parallel plate capacitors of capacitance $C_{0}$ and $2 C_{0}$ are connected in parallel and charged to a potential difference $V_{0}$ . The battery is disconnected and the region between the capacitor plates of dielectric constant K. The potential difference across the capacitors now becomes:

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$\frac{V_{0}}{(3 K + 1)}$

$\frac{3 V_{0}}{(K + 2)}$

$\frac{V_{0}}{(2 K + 1)}$

$\frac{2 V_{0}}{K}$

answer is B.

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Detailed Solution

$Q = 3 C_{0} V_{0} = (K C_{0} + 2 C_{0}) V$

$v = \frac{3}{K + 2} V_{0}$

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