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(D) Since the battery is disconnected,the charge $Q$ on the plates remains constant.Initial charge $Q = C V = \frac{\varepsilon_0 A}{d} V$.When a diel

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(D) Since the battery is disconnected,the charge $Q$ on the plates remains constant.Initial charge $Q = C V = \frac{\varepsilon_0 A}{d} V$.When a dielectric slab of constant $K$ is inserted,the new capacitance becomes $C' = KC = \frac{K\varepsilon_0 A}{d}$.The new potential difference is $V' = \frac{Q}{C'} = \frac{Q}{KC} = \frac{V}{K}$.The new electric field is $E = \frac{V'}{d} = \frac{V}{Kd}$.The initial potential energy is $U_i = \frac{1}{2}CV^2 = \frac{\varepsilon_0 AV^2}{2d}$.The final potential energy is $U_f = \frac{1}{2}C'V'^2 = \frac{1}{2}(KC) \left(\frac{V}{K}\right)^2 = \frac{1}{2} \frac{CV^2}{K} = \frac{\varepsilon_0 AV^2}{2Kd}$.The work done on the system is $W = U_f - U_i = \frac{\varepsilon_0 AV^2}{2Kd} - \frac{\varepsilon_0 AV^2}{2d} = -\frac{\varepsilon_0 AV^2}{2d} \left(1 - \frac{1}{K}\right)$.

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