Question

One plate of a capacitor is fixed, and the other is connected to a spring as shown in figure. The area of both the plates is $A$. In the steady state (equilibrium), separation between the plates is $0.8d$ (spring was unstretched, and the distance between the plates was $d$, when the capacitor was uncharged). The force constant of the spring is approximately.

• A. $\frac{125}{32}\frac{{\epsilon }_{0}A{E}^2}{d^3}$
• B. $\frac{{2\epsilon }_{0}A{E}^2}{d^3}$
• C. $\frac{{6\epsilon }_0{E}^2}{{Ad}^2}$
• D. $\frac{{\epsilon }_{0}A{E}^3}{{2d}^3}$

Answer 1

The correct option is A $\frac{125}{32}\frac{{\epsilon }_{0}A{E}^2}{d^3}$

Force on plates due to charged capacitor $=$ Force due to spring
$F=Kx$
$\frac{Q^2}{2{\epsilon }_{0}A}=K\left(0.2d\right)$

Now $Q=CE$, $C=\frac{{ϵ}_{0}A}{D}=\frac{{ϵ}_{0}A}{0.8d}$
$\Rightarrow Q=\frac{{ϵ}_{0}A}{0.8d}E$

$\Rightarrow \frac{{\left(\frac{{\epsilon }_{0}A}{0.8d}E\right)}^2}{{2\epsilon }_{0}A}=0.2Kd$

$\Rightarrow K=\frac{125{\epsilon }_{0}A{E}^2}{32d^3}$