A parallel-plate capacitor consists of a fixed plate and a movable plate that is allowed to slide in the direction parallel to the plates. Let $x$ be the distance of overlap, as shown in the figure. The separation between the plates is fixed. Assume that the plates are electrically isolated, so that their charges $±Q$ are constant. Force on the movable plate is proportional to
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$\mathrm{F}=\frac{\mathrm{d} \mathrm{U}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\right)=\frac{1}{2} \frac{\mathrm{Q}^{2}}{\mathrm{C}^{2}} \frac{\mathrm{d} \mathrm{C}}{\mathrm{dx}}$
$\mathrm{b} :$ length of one side of plates
$C = \frac{{{ \in _0}A}}{{{\ell _0}}} = \frac{{{ \in _0}bx}}{{{\ell _0}}}$ $\frac{{dC}}{{dx}} = \frac{{{ \in _0}b}}{{{\ell _0}}}$
$F = \frac{1}{2}\frac{{{Q^2}{\ell _0}}}{{e \in _0^2{b^2}{x^2}}}\frac{{{ \in _0}b}}{{{\ell _0}}}$
${\rm{F}} = \left( {\frac{1}{2}\frac{{{{\rm{Q}}^2}{\ell _0}}}{{{ \in _0}{\rm{b}}}}} \right)\frac{1}{{{{\rm{x}}^2}}}$
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