The force between the plates is:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{r}}{\mathbf{\left(}}{\mathbf{U}}{\mathbf{\right)}}}$

The average energy stored in a capacitor is:

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}{{\mathbf{V}}}^{{\mathbf{2}}}}$

For a parallel plate capacitor, the capacitance is given by:

$\overline{){\mathit{C}}{\mathbf{=}}\frac{{\mathit{\epsilon}}_{\mathbf{0}}\mathit{A}}{\mathit{d}}}$, where A is the cross-sectional area and d is the separation of the plates.

Consider a parallel-plate capacitor with plates of area A and with separation d. Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor. Express your answer in terms of given quantities and ϵ_{0}.

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