$F=-\frac{\delta U}{\delta x}$

The energy stored in the capacitor is

$U=\frac{1}{2} \frac{Q^{2}}{C}$

The capacitance in the case shown in the figure is

$C = {C_1} + {C_2} = \frac{{{\varepsilon _0}wxK}}{d} + \frac{{{\varepsilon _0}w(l - x)}}{d}$

$=\varepsilon_{\circ} w \frac{x(K-1)+l}{d}$

$U=\frac{1}{2} \frac{Q^{2}}{C}$

$F=-(\delta U) /(\delta x)=-\frac{Q^{2} d}{2 \varepsilon_{\mathrm{o}} w} \frac{\delta}{\delta x}\left(\frac{1}{x(K-1)+l}\right)$

$F=\frac{Q^{2} d}{2 \varepsilon_{\mathrm{o}} w} \cdot \frac{K-1}{(x(K-1)+l)^{2}}$

At $x=0$ (edge):

$F=\frac{Q^{2} d(K-1)}{2 \varepsilon_{0} w l^{2}}$