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(D) In equilibrium,the attractive force between the plates of the capacitor is balanced by the spring force.The force between the plates of a parallel

Vedclass · Accepted Answer

$\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}$

Vedclass · Answer

(D) In equilibrium,the attractive force between the plates of the capacitor is balanced by the spring force.The force between the plates of a parallel plate capacitor is given by $F = \frac{q^2}{2 \varepsilon_0 A} = \frac{C^2 V^2}{2 \varepsilon_0 A}$.Since $C = \frac{\varepsilon_0 A}{d'}$,where $d'$ is the new distance between the plates,we have $F = \frac{(\varepsilon_0 A / d')^2 V^2}{2 \varepsilon_0 A} = \frac{\varepsilon_0 A V^2}{2 d'^2}$.Given that the initial distance is $d$ and the final distance is $d' = 0.75 d = \frac{3}{4} d$,the extension in the spring is $x = d - d' = d - \frac{3}{4} d = \frac{1}{4} d$.The spring force is $F_s = kx = k \left( \frac{1}{4} d \right)$.Equating the forces: $\frac{\varepsilon_0 A V^2}{2 (\frac{3}{4} d)^2} = k \left( \frac{1}{4} d \right)$.$\frac{\varepsilon_0 A V^2}{2 (\frac{9}{16} d^2)} = k \left( \frac{d}{4} \right)$.$\frac{\varepsilon_0 A V^2}{\frac{9}{8} d^2} = k \left( \frac{d}{4} \right)$.$\frac{8 \varepsilon_0 A V^2}{9 d^2} = k \left( \frac{d}{4} \right)$.$k = \frac{32 \varepsilon_0 A V^2}{9 d^3}$.

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