A parallel plate capacitor with a dielectric | vedclass

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(D) When the dielectric slab is removed from the capacitor,the capacitance changes from $KC$ to $C$. Since the battery is disconnected,the charge $Q =

Vedclass · Accepted Answer

Zero

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(D) When the dielectric slab is removed from the capacitor,the capacitance changes from $KC$ to $C$. Since the battery is disconnected,the charge $Q = KCV$ remains constant.Work done by an external agent to remove the slab is $W_{ext} = U_{final} - U_{initial} = \frac{Q^2}{2C} - \frac{Q^2}{2KC} = \frac{Q^2}{2C}(1 - \frac{1}{K}) = \frac{K^2C^2V^2}{2C}(\frac{K-1}{K}) = \frac{1}{2}KC(K-1)V^2$.When the slab is reinserted,the external agent does negative work of the same magnitude to restore the initial state.Since the process is conservative and the system returns to its initial state,the net work done by the system (or external agent) over the complete cycle is zero.

$A$ parallel plate capacitor with a dielectric slab of dielectric constant $K$ between its plates has a capacitance $C$ and is charged to a potential $V$. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system is:

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