The capacitance of a parallel plate capacitor | vedclass

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(D) Initial state (with dielectric): Capacitance $C_i = kC$,Charge $Q_i = kCE$,Energy $U_i = \frac{1}{2}kCE^2$.Final state (without dielectric): Capac

Vedclass · Accepted Answer

All of the above.

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(D) Initial state (with dielectric): Capacitance $C_i = kC$,Charge $Q_i = kCE$,Energy $U_i = \frac{1}{2}kCE^2$.Final state (without dielectric): Capacitance $C_f = C$,Charge $Q_f = CE$,Energy $U_f = \frac{1}{2}CE^2$.Charge flow through the cell: $\Delta Q = Q_i - Q_f = kCE - CE = CE(k - 1)$. Thus,option $A$ is correct.Energy absorbed by the cell: $W_{cell} = E \cdot \Delta Q = E \cdot CE(k - 1) = CE^2(k - 1)$. Thus,option $B$ is correct.Work done by external agent: By the work-energy theorem,$W_{ext} + W_{cell} = U_f - U_i$.$W_{ext} = (U_f - U_i) - W_{cell} = (\frac{1}{2}CE^2 - \frac{1}{2}kCE^2) - CE^2(k - 1)$.$W_{ext} = -\frac{1}{2}CE^2(k - 1) - CE^2(k - 1) = -\frac{3}{2}CE^2(k - 1)$.Note: The magnitude of work done by the external agent to remove the slab against the electrostatic force is $\frac{1}{2}CE^2(k - 1)$. Thus,option $C$ is correct.Since all statements are correct,the answer is $D$.

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