A battery is used to charge a parallel plate | vedclass

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(C) Let $C$ be the capacitance of the capacitor and $e$ be the electromotive force $(emf)$ of the battery.When the capacitor is fully charged,the pote

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$1/2$

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(C) Let $C$ be the capacitance of the capacitor and $e$ be the electromotive force $(emf)$ of the battery.When the capacitor is fully charged,the potential difference across the plates $V$ is equal to the $emf$ of the battery,i.e.,$V = e$.The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2 = \frac{1}{2} C e^2$.The work done by the battery in charging the capacitor is given by $W = qV = (Ce)e = C e^2$.The ratio of the energy stored in the capacitor to the work done by the battery is $\frac{U}{W} = \frac{\frac{1}{2} C e^2}{C e^2} = \frac{1}{2}$.

$A$ battery is used to charge a parallel plate capacitor until the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor to the work done by the battery is

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