JEE Main 2014 — Capacitance Question with Solution

A parallel plate capacitor is made of two plates of length 1 , width $w$ and separated by distance $d$. A dielectric slab (dielectric constant $\mathrm{K}$ ) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $\mathrm{F}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}$ where $\mathrm{U}$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is: