Two capacitors C_1 and C_2 = 2C_1 are connect | vedclass

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(B) Initially,capacitor $C_1$ has charge $Q$ and capacitor $C_2$ is uncharged. When the switch $S$ is closed,the two capacitors are connected in paral

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$Q/3, 2Q/3$

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(B) Initially,capacitor $C_1$ has charge $Q$ and capacitor $C_2$ is uncharged. When the switch $S$ is closed,the two capacitors are connected in parallel across each other (or effectively form a closed loop where charge redistributes until potentials are equal).Since the capacitors are connected in parallel,they will have the same potential difference $V$ across them.The total charge $Q_{total} = Q$ is conserved.Let the final charges be $Q_1'$ and $Q_2'$.$Q_1' = C_1 V$ and $Q_2' = C_2 V = 2C_1 V = 2Q_1'$.Since $Q_1' + Q_2' = Q$,we have $Q_1' + 2Q_1' = Q$,which gives $3Q_1' = Q$.Therefore,$Q_1' = Q/3$ and $Q_2' = 2Q/3$.

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

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