Consider the arrangement shown in the figure. | vedclass

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(A) When the key $K$ is closed,both capacitors are in parallel with the battery of voltage $V$. The equivalent capacitance is $C_{eq} = C + C = 2C$. T

Vedclass · Accepted Answer

$\frac{2\epsilon_r}{1 + \epsilon_r^2}$

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(A) When the key $K$ is closed,both capacitors are in parallel with the battery of voltage $V$. The equivalent capacitance is $C_{eq} = C + C = 2C$. The total energy stored is $U_1 = \frac{1}{2} (2C) V^2 = CV^2$.When the key $K$ is opened,the capacitor connected to the battery remains at voltage $V$,while the other capacitor is isolated with charge $Q = CV$. For the first capacitor (connected to the battery),the new capacitance is $C' = \epsilon_r C$. The energy stored is $U_{2,1} = \frac{1}{2} C' V^2 = \frac{1}{2} \epsilon_r C V^2$.For the second capacitor (isolated),the charge $Q = CV$ remains constant. The new capacitance is $C' = \epsilon_r C$. The energy stored is $U_{2,2} = \frac{Q^2}{2C'} = \frac{(CV)^2}{2 \epsilon_r C} = \frac{CV^2}{2 \epsilon_r}$.The total energy stored is $U_2 = U_{2,1} + U_{2,2} = \frac{1}{2} \epsilon_r C V^2 + \frac{CV^2}{2 \epsilon_r} = \frac{CV^2}{2} \left( \epsilon_r + \frac{1}{\epsilon_r} \right) = \frac{CV^2}{2} \left( \frac{\epsilon_r^2 + 1}{\epsilon_r} \right)$.The ratio $U_1/U_2 = \frac{CV^2}{\frac{CV^2}{2} \left( \frac{\epsilon_r^2 + 1}{\epsilon_r} \right)} = \frac{2 \epsilon_r}{1 + \epsilon_r^2}$.

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