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

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Answer & Solution

Correct Answer

(C) \(\frac{{{Q^2}d}}{{2w{l^2}{\varepsilon _0}}}\left( {K - 1} \right)\)

Detailed explanation

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):

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

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