Question

One plate of a capacitor is connected to a spring as shown. Area of both the plates is $A$ and separation is $d$ when capacitor is uncharged. When capacitor is charged in steady state separation between the plate is $0.8d$ the force constant of the spring is

• A. $\frac{5{\epsilon }_{0}A{E}^2}{2d^3}$
• B. $\frac{2{\epsilon }_{0}AE}{d^2}$
• C. $\frac{6{\epsilon }_0{E}^2}{A{d}^3}$
• D. $\frac{4{\epsilon }_{0}A{E}^2}{d^3}$

Answer 1

The correct option is A $\frac{5{\epsilon }_{0}A{E}^2}{2d^3}$

The attractive force between the plates since they are oppositely charged,

$F=$ Electric by one plate $\times$ charge on the another plate

$\therefore F=\frac{\sigma }{2{ϵ}_0}\times Q$

$F=\frac{Q}{2{ϵ}_{0}A}\times Q=\frac{Q^2}{2{ϵ}_{0}A}$

$F=\frac{C^2{E}^2}{2{ϵ}_{0}A}[\because Q=CE]$

$F=\frac{{ϵ}_{0}A{E}^2}{2d^2}[\because C=\frac{{ϵ}_{0}A}{d}]$

The spring force is given by,

$F^{\prime }=k\left(\Delta x\right)=k(d-0.8d)$

$F^{\prime }=k\left(0.2d\right)$

Therefore for steady state,

$F^{\prime }=F$

$0.2dk=\frac{{ϵ}_{0}A{E}^2}{2d^2}$

$k=\frac{{ϵ}_{0}A{E}^2}{0.4d^3}=\frac{10{ϵ}_{0}A{E}^2}{4d^3}$

$k=\frac{5{ϵ}_{0}A{E}^2}{2d^3}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (a) is the correct answer.