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(C) Case $I$: Switch $K$ is closed.Both capacitors are in parallel with the source $V$.Equivalent capacitance $C_{eq} = C + C = 2C$.Total energy $E_{1

Vedclass · Accepted Answer

$\frac{5}{13}$

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(C) Case $I$: Switch $K$ is closed.Both capacitors are in parallel with the source $V$.Equivalent capacitance $C_{eq} = C + C = 2C$.Total energy $E_{1} = \frac{1}{2} C_{eq} V^{2} = \frac{1}{2} (2C) V^{2} = CV^{2}$.Case $II$: Switch $K$ is opened.The left capacitor remains connected to the source $V$,so its potential remains $V$ and its new capacitance is $C' = K_{d}C = 5C$. Its energy is $E_{L} = \frac{1}{2} (5C) V^{2} = \frac{5}{2} CV^{2}$.The right capacitor is disconnected,so its charge $Q = CV$ remains constant. Its new capacitance is $C' = K_{d}C = 5C$. Its energy is $E_{R} = \frac{Q^{2}}{2C'} = \frac{(CV)^{2}}{2(5C)} = \frac{CV^{2}}{10}$.Total energy $E_{2} = E_{L} + E_{R} = \frac{5}{2} CV^{2} + \frac{1}{10} CV^{2} = \frac{25+1}{10} CV^{2} = \frac{26}{10} CV^{2} = \frac{13}{5} CV^{2}$.Ratio $\frac{E_{1}}{E_{2}} = \frac{CV^{2}}{\frac{13}{5} CV^{2}} = \frac{5}{13}$.

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