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Question

(C) Let the capacitance of each capacitor be $C$ and the battery voltage be $V$.Initially,when the switch $S$ is closed,both capacitors are connected

Vedclass · Accepted Answer

$3: 5$

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(C) Let the capacitance of each capacitor be $C$ and the battery voltage be $V$.Initially,when the switch $S$ is closed,both capacitors are connected in parallel to the battery $V$.The total electrostatic energy stored initially is $U_1 = \frac{1}{2} CV^2 + \frac{1}{2} CV^2 = CV^2$.Now,the switch $S$ is opened. Capacitor $A$ remains connected to the battery,so its potential difference remains $V$. Capacitor $B$ is disconnected,so its charge $Q = CV$ remains constant.After introducing a dielectric of constant $K = 3$ into both capacitors:For capacitor $A$,the new capacitance is $C' = KC = 3C$. The energy stored is $U_A = \frac{1}{2} C' V^2 = \frac{1}{2} (3C) V^2 = \frac{3}{2} CV^2$.For capacitor $B$,the new capacitance is $C' = KC = 3C$. Since the charge $Q = CV$ is constant,the energy stored is $U_B = \frac{Q^2}{2C'} = \frac{(CV)^2}{2(3C)} = \frac{CV^2}{6}$.The total energy after introducing the dielectric is $U_2 = U_A + U_B = \frac{3}{2} CV^2 + \frac{1}{6} CV^2 = \left( \frac{9+1}{6} \right) CV^2 = \frac{10}{6} CV^2 = \frac{5}{3} CV^2$.The ratio of the total electrostatic energy before and after is $\frac{U_1}{U_2} = \frac{CV^2}{\frac{5}{3} CV^2} = \frac{3}{5}$.

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