Two parallel plate capacitors of capacity C a | vedclass

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(C) $1$. Initial state: The capacitors $C$ and $3C$ are in parallel and connected to an $18V$ battery. The total charge $Q_{total}$ stored in the syst

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$6$

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(C) $1$. Initial state: The capacitors $C$ and $3C$ are in parallel and connected to an $18V$ battery. The total charge $Q_{total}$ stored in the system is:$Q_{total} = (C + 3C) 	imes 18V = 4C 	imes 18V = 72CV$.$2$. After disconnecting the battery: The total charge $Q_{total} = 72CV$ remains constant because the system is isolated.$3$. Dielectric insertion: $A$ dielectric of constant $K=9$ is inserted into the capacitor of capacity $C$. Its new capacitance becomes $C^{\prime} = K 	imes C = 9C$. The other capacitor remains $3C$.$4$. Final state: The capacitors are still in parallel. Let the new common potential difference be $V^{\prime}$. The total capacitance of the combination is $C_{eq} = C^{\prime} + 3C = 9C + 3C = 12C$.$5$. Conservation of charge: $Q_{total} = C_{eq} 	imes V^{\prime}$$72CV = 12C 	imes V^{\prime}$$V^{\prime} = \frac{72CV}{12C} = 6V$.

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