Question

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 $0.75d$, then the force constant of the spring is

• A. $\frac{3}{8}\frac{{\in }_0{V}^{2}A}{d^3}$
• B. $\frac{8}{3}\frac{{\in }_0{V}^{2}A}{d^3}$
• C. $\frac{9}{32}\frac{{\in }_0{V}^{2}A}{d^3}$
• D. $\frac{32}{9}\frac{{\in }_0{V}^{2}A}{d^3}$

Answer 1

Answer: (D)

In equilibrium, force between plates of capacitor $=$ spring force
$\Rightarrow \frac{q^2}{2{\epsilon }_{0}A}=kx$
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, $x=d-\frac{3}{4}d=\frac{1}{4}d$
Hence, $k=\frac{q^2}{2{\epsilon }_{0}A\cdot x}=\frac{C^2{V}^2}{2{\epsilon }_{0}Ax}$
$=\left\{\frac{\frac{{\epsilon }_0^2{A}^2}{\left(\frac{3}{4}d\right)}\cdot V^2}{2{\epsilon }_{0}A\cdot \frac{1}{4}d}\right\}=\frac{32}{9}\left(\frac{{\epsilon }_0{V}^{2}A}{d^3}\right)$