$(A)\;Q= \frac{\in _0 AV}{d} \\ (B)\;E= \frac{V}{kd} \\ (C)\; w= \frac{\in_0 AV^2}{2 kd} \\ (D)\;w= \frac{\in _0 AV^2}{2d} (1-\frac{1}{k}) $

After inserting slab

New capacitance $C'=KC=\large\frac{K \in _0 A}{d}$

New potential difference, $V'=\large\frac{V}{K}$

New electric field $E'= \large\frac{V'}{d}=\frac{V}{kd}$

New charge $Q=C'V'= \large\frac{\in _0 AV}{d}$

Work= Final energy - Initial energy

$\qquad= \large\frac{1}{2}$$ C'V'^2 -\large\frac{1}{2} CV^2$

$\qquad=\large\frac{1}{2} $$ (KC ) \bigg ( 1-\large\frac{1}{K}\bigg)$

=> $ |w| =\large\frac{\in _0 AV^2}{2d} \bigg(1 -\large\frac{1}{k}\bigg)$

Hence C is the correct answer.

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