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(A) $1$. Initially, the capacitors $2C$ and $C$ are in parallel and connected to a battery of potential $V$. The total charge $Q$ stored is $Q = (2C +

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

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(A) $1$. Initially, the capacitors $2C$ and $C$ are in parallel and connected to a battery of potential $V$. The total charge $Q$ stored is $Q = (2C + C)V = 3CV$.$2$. When the battery is disconnected, the total charge $Q = 3CV$ remains constant.$3$. After filling the capacitor $C$ with a dielectric $K$, its new capacitance becomes $C' = KC$. The capacitor $2C$ remains unchanged.$4$. The capacitors are still in parallel, so they share the same potential difference $V'$.$5$. The new total capacitance is $C_{eq} = 2C + KC = C(K + 2)$.$6$. Using the relation $Q = C_{eq}V'$, we get $3CV = C(K + 2)V'$.$7$. Solving for $V'$, we get $V' = \frac{3CV}{C(K + 2)} = \frac{3V}{(K + 2)}$.

Two condensers of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is then disconnected and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors in the second case is

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