Two capacitors of capacitance 2C and C are jo | vedclass

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Question

(A) Initially,the capacitors are connected in parallel and charged to potential $V$. The total charge stored in the system is $Q_{	ext{total}} = (2C)

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be

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