Two capacitors C and 2C are connected in para | vedclass

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(D) Initially,the capacitors $C$ and $2C$ are connected in parallel to a battery of potential $V$. The total charge $Q$ stored is given by $Q = C_{eq}

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

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

Two capacitors $C$ and $2C$ are connected in parallel and charged to a potential $V$ by a battery. The battery is then removed. The capacitor $C$ is filled with a dielectric of constant $K$. What is the new potential difference across the capacitors?

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