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(C) Initially,the switch $S$ is closed. Both capacitors $A$ and $B$ are connected in parallel to the battery of potential $V$. The capacitance of each

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$3:5$

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(C) Initially,the switch $S$ is closed. Both capacitors $A$ and $B$ are connected in parallel to the battery of potential $V$. The capacitance of each capacitor is $C$.The initial total energy stored in the system is:$U_1 = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$ --- $(i)$When the switch $S$ is opened,capacitor $A$ remains connected to the battery,so its potential difference remains $V$. Capacitor $B$ is disconnected from the battery,so its charge $Q$ remains constant. The initial charge on capacitor $B$ was $Q = CV$.After filling the space with a dielectric of constant $K = 3$,the new capacitance of each capacitor becomes $C' = KC = 3C$.For capacitor $A$,the potential difference remains $V$. The new energy is $U_{A}' = \frac{1}{2}C'V^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$.For capacitor $B$,the charge $Q = CV$ remains constant. The new energy is $U_{B}' = \frac{Q^2}{2C'} = \frac{(CV)^2}{2(3C)} = \frac{1}{6}CV^2$.The final total energy stored in the system is:$U_2 = U_{A}' + U_{B}' = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \left(\frac{9+1}{6}\right)CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$ --- (ii)The ratio of total electrostatic energy before and after is:$\frac{U_1}{U_2} = \frac{CV^2}{(5/3)CV^2} = \frac{3}{5}$

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