What is the mechanical work done in pulling the slab out of the capacitor after disconnecting it from the battery

$\frac{1}{2} E^2 C\left(\varepsilon_{r}-1\right)$

Work done = change in potential energy = U₂ – U₁

v₁ = (1/2) E²C

v₂ = $\frac{1}{2} \frac{(E C)^2}{C'}=\frac{1}{2} \frac{E^2 C^2}{C} \varepsilon_r=(1 / 2) E^2 C \varepsilon_r$

∴ Work done = $(1 / 2) E^2 C\left(\varepsilon_r-1\right)$

∴ (A)