Two capacitors of capacities 2C and C are joi | vedclass

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(A) Initially,the capacitors are in parallel,so the total charge $Q$ is given by $Q = (2C + C)V = 3CV$. When the battery is removed,the total charge $

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

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(A) Initially,the capacitors are in parallel,so the total charge $Q$ is given by $Q = (2C + C)V = 3CV$. When the battery is removed,the total charge $Q = 3CV$ remains conserved. After filling the capacitor of capacity $C$ with a dielectric of constant $K$,its new capacitance becomes $C' = KC$. The new equivalent capacitance of the parallel combination is $C_{eq} = 2C + KC = C(K + 2)$. Since the charge is conserved,the new potential difference $V'$ across the capacitors is given by $V' = \frac{Q}{C_{eq}} = \frac{3CV}{C(K + 2)} = \frac{3V}{K + 2}$.

Two capacitors of capacities $2C$ 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 will now be

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