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(A) Initially,both capacitors are charged to potential $V_0$. The charge on capacitor $C$ is $Q_1 = C V_0$ and on capacitor $2C$ is $Q_2 = (2C) V_0 =

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$\frac{3}{8} V_0$

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(A) Initially,both capacitors are charged to potential $V_0$. The charge on capacitor $C$ is $Q_1 = C V_0$ and on capacitor $2C$ is $Q_2 = (2C) V_0 = 2 C V_0$.When dielectrics are inserted,the new capacitances become $C' = 2C$ and $C'' = 3(2C) = 6C$.Since the capacitors are disconnected from the battery before the key is closed,the total charge remains conserved. However,the problem implies the capacitors are connected in parallel when the key is closed.The new potential $V_{	ext{common}}$ across the parallel combination is given by the total charge divided by the total capacitance:$V_{	ext{common}} = \frac{Q_1 + Q_2}{C' + C''} = \frac{C V_0 + 2 C V_0}{2C + 6C} = \frac{3 C V_0}{8C} = \frac{3}{8} V_0$.

Two capacitors are shown in the diagram. Each is charged to potential $V_0$. If capacitor $C$ is filled with a dielectric of constant $2$ and capacitor $2C$ is filled with a dielectric of constant $3$,then the key is closed. The final voltage on the capacitor $C$ will be:

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