Question

One plate of a capacitor is connected to a spring as shown in figure. Area of both the plates is A. In steady state separation between the plates is 0.6 d (spring was to un stretched and the distance between the plates was d when the capacitor was uncharged. The force constant of the spring is

• A. $\frac{3.5{ϵ}_{0}A{E}^2}{d^3}$
• B. $\frac{2{ϵ}_{0}AE}{d^2}$
• C. $\frac{6{ϵ}_0{E}^2}{A{d}^3}$
• D. $\frac{{ϵ}_{0}A{E}^2}{2d^3}$

Answer 1

The correct option is A $\frac{3.5{ϵ}_{0}A{E}^2}{d^3}$

In equilibrium electro static attraction between the plates = spring force
$\because \frac{q^2}{2{ϵ}_{0}A}=kx$
$\because \frac{(CE{)}^2}{2{ϵ}_{0}A}=k(d-0.6d)$
$\because \frac{{\left(\frac{{ϵ}_{0}A}{0.6d}\right)}^2{E}^2}{2{ϵ}_{0}A}=0.4dk$
$\Rightarrow k=\frac{{ϵ}_{0}A{E}^2}{0.256d^3}\approx \frac{3.5{ϵ}_{0}A{E}^2}{d^3}$