One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each…
One plate of a parallel plate capacitor is connected to a spring as shown in the figure. The area of each plate of the capacitor is $A$ and the distance between the plates is $d$, when the battery is not connected and the spring is unstretched. After connecting the battery, in the steady state the distance between the plates is $0.75 d$, then the force constant of the spring is
$\frac{3}{8} \frac{\varepsilon_0 V^2 A}{d^3}$
$\frac{8}{3} \frac{\varepsilon_0 V^2 A}{d^3}$
$\frac{9}{32} \frac{\varepsilon_0 V^2 A}{d^3}$
$\frac{32}{9} \frac{\varepsilon_0 V^2 A}{d^3}$
Solution
In equilibrium, force between plates of capacitor $=$ spring force $\Rightarrow \frac{q^2}{2 \varepsilon_0 A}=k x$ where, $x=$ extension in spring. Now before charging, when spring is unstretched $(x=0)$ distance of plates is $d$ and after charging it is $\frac{3}{4} d$.
So, $\quad x=d-\frac{3}{4} d=\frac{1}{4} d$
Hence, $\begin{aligned} k & =\frac{q^2}{2 \varepsilon_0 A \cdot x}=\frac{C^2 V^2}{2 \varepsilon_0 A x} \\ & =\left\{\frac{\frac{\varepsilon_0^2 A^2}{\left(\frac{3}{4} d\right)} \cdot V^2}{2 \varepsilon_0 A \cdot \frac{1}{4} d}\right\}=\frac{32}{9}\left(\frac{\varepsilon_0 V^2 A}{d^3}\right)\end{aligned}$
