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(A) When the capacitor is fully charged, the charge on it is $q = CE$.The total work done by the $DC$ source (battery) is $W = qE = (CE)E = CE^2$.The

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$\frac{1}{2} C E^{2}$

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(A) When the capacitor is fully charged, the charge on it is $q = CE$.The total work done by the $DC$ source (battery) is $W = qE = (CE)E = CE^2$.The energy stored in the capacitor is $U = \frac{1}{2} CE^2$.According to the law of conservation of energy, the work done by the battery is equal to the sum of the energy stored in the capacitor and the energy dissipated as heat in the resistor $R$.Therefore, the energy dissipated in the resistance $R$ is given by:$H = W - U$$H = CE^2 - \frac{1}{2} CE^2$$H = \frac{1}{2} CE^2$.

$A$ capacitor of capacitance $C$ is connected in series with a resistance $R$ and a $DC$ source of $emf$ $E$ through a key. The capacitor starts charging when the key is closed. By the time the capacitor has been fully charged, what amount of energy is dissipated in the resistance $R$?

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