A parallel plate capacitor of plate area A an | vedclass

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(B) Initial capacitance $C = \frac{{\varepsilon _0}A}{d}$. Initial charge $Q = CV = \frac{{\varepsilon _0}AV}{d}$. Since the battery is disconnected,t

Vedclass · Accepted Answer

$W = \frac{{\varepsilon _0}A{V^2}}{{2kd}}$

Vedclass · Answer

(B) Initial capacitance $C = \frac{{\varepsilon _0}A}{d}$. Initial charge $Q = CV = \frac{{\varepsilon _0}AV}{d}$. Since the battery is disconnected,the charge $Q$ remains constant.After inserting the dielectric slab of constant $k$,the new capacitance is $C' = kC = \frac{k{\varepsilon _0}A}{d}$.The new potential difference is $V' = \frac{Q}{C'} = \frac{CV}{kC} = \frac{V}{k}$.The new electric field is $E = \frac{V'}{d} = \frac{V}{kd}$.The work done $W$ on the system is the change in stored energy: $W = U_{final} - U_{initial}$.$U_{initial} = \frac{1}{2}CV^2$.$U_{final} = \frac{Q^2}{2C'} = \frac{(CV)^2}{2kC} = \frac{CV^2}{2k}$.$W = \frac{CV^2}{2k} - \frac{1}{2}CV^2 = \frac{1}{2}CV^2 \left( \frac{1}{k} - 1 \right) = -\frac{{\varepsilon _0}AV^2}{2d} \left( 1 - \frac{1}{k} \right)$.Comparing the options,option $B$ is incorrect as it does not match the derived expression for work done.

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