What is the mechanical work done in pulling the slab out of the capacitor if the battery is connected

$\frac{1}{2} CE^2(k-1)$ $ CE^2$ $ CE^2(k-1)$ Zero

$\frac{1}{2} CE^2(k-1)$

Work done  = change in potential energy = U2 – U1 $ U_1 = \frac{1}{2} kCE^2$ $U_2 = \frac{1}{2}CE^2 $ $\Rightarrow W = U_2 - U_1 = \frac{1}{2} CE^2(k-1)$