Two parallel plate capacitors C_1 and C_2 eac | vedclass

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(C) Initial state: Both capacitors have $C = 10 \mu F$ and are charged to $V_0 = 100 \, V$.For $C_1$: It remains connected to the source. When a diele

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$55$

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(C) Initial state: Both capacitors have $C = 10 \mu F$ and are charged to $V_0 = 100 \, V$.For $C_1$: It remains connected to the source. When a dielectric of $K = 10$ is inserted,the new capacitance becomes $C_1' = K \cdot C = 10 	imes 10 \mu F = 100 \mu F$. Since it is connected to the source,the potential remains $V_1 = 100 \, V$. The charge on $C_1$ is $Q_1 = C_1' \cdot V_1 = 100 \mu F 	imes 100 \, V = 10000 \mu C$.For $C_2$: It is disconnected from the source,so its charge remains constant. $Q_2 = C \cdot V_0 = 10 \mu F 	imes 100 \, V = 1000 \mu C$. When the dielectric is inserted,the new capacitance becomes $C_2' = K \cdot C = 100 \mu F$. The potential across $C_2$ becomes $V_2 = Q_2 / C_2' = 1000 \mu C / 100 \mu F = 10 \, V$.Final state: The capacitors are connected in parallel. The total charge $Q_{total} = Q_1 + Q_2 = 10000 \mu C + 1000 \mu C = 11000 \mu C$. The total capacitance $C_{eq} = C_1' + C_2' = 100 \mu F + 100 \mu F = 200 \mu F$.The common potential $V = Q_{total} / C_{eq} = 11000 \mu C / 200 \mu F = 55 \, V$.

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