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(A) The force of attraction between the plates of a parallel plate capacitor is given by $F = \frac{q^2}{2 A \varepsilon_0}$.In this system,the inner

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$\frac{q^2}{2 A \varepsilon_0 k}$

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(A) The force of attraction between the plates of a parallel plate capacitor is given by $F = \frac{q^2}{2 A \varepsilon_0}$.In this system,the inner plates are attracted towards their respective fixed outer plates due to the electrostatic force of attraction.At equilibrium,the spring force $F_s = kx$ must balance the electrostatic force of attraction $F$ acting on the inner plate.Therefore,$kx = \frac{q^2}{2 A \varepsilon_0}$.Solving for the extension $x$,we get $x = \frac{q^2}{2 A \varepsilon_0 k}$.

Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant $k$. The charge on each capacitor is $q$. Find the extension in the spring at equilibrium.

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