Work done by an external agent in separating | vedclass

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Question

(C) When a capacitor is disconnected from the battery,the charge $Q$ on its plates remains constant.The capacitance of a parallel plate capacitor is g

Vedclass · Accepted Answer

$\frac{1}{2}CV^2$

Vedclass · Answer

(C) When a capacitor is disconnected from the battery,the charge $Q$ on its plates remains constant.The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.As the plates are separated,the distance $d$ increases,so the capacitance $C$ decreases.The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.Since $Q$ is constant and $C$ decreases,the energy $U$ increases.The work done by the external agent is equal to the change in the potential energy of the capacitor: $W = \Delta U = U_f - U_i$.If the capacitance changes from $C$ to $C'$,the work done is $W = \frac{Q^2}{2C'} - \frac{Q^2}{2C}$.Given the options provided,the standard expression for energy stored in a capacitor is $\frac{1}{2}CV^2$.

Work done by an external agent in separating the plates of a parallel plate capacitor (disconnected from the battery) is:

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