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A parallel plate capacitor is made of two plates of length $l$, width $w$ and separated by distance $d$. A dielectric slab ( dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{{\partial U}}{{\partial x}}$ where $U$ 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
$\frac{{{Q^2}d}}{{2w{l^2}{\varepsilon _0}}}K$
$\frac{{{Q^2}w}}{{2d{l^2}{\varepsilon _0}}}\left( {K - 1} \right)$
$\frac{{{Q^2}d}}{{2w{l^2}{\varepsilon _0}}}\left( {K - 1} \right)$
$\frac{{{Q^2}w}}{{2d{l^2}{\varepsilon _0}}}K$
Solution
The electric force on the slab close to the edge is
$F=-\frac{\delta U}{\delta x}$
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):
$F=\frac{Q^{2} d(K-1)}{2 \varepsilon_{0} w l^{2}}$
