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 $0.75\, d$, 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
$\Rightarrow \frac{q^{2}}{2 \varepsilon_{0} A}=k x$
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, $x=d-\frac{3}{4} d=\frac{1}{4} d$
Hence, $k =\frac{q^{2}}{2 \varepsilon_{0} A \cdot x}=\frac{C^{2} V^{2}}{2 \varepsilon_{0} A x}$
$=\left\{\frac{\frac{\varepsilon_{0}^{2} A^{2}}{\left(\frac{3}{4} d\right)} \cdot V^{2}}{2 \varepsilon_{0} A \cdot \frac{1}{4} d}\right\}=\frac{32}{9}\left(\frac{\varepsilon_{0} V^{2} A}{d^{3}}\right)$