Question

One plate of a capacitor is connected to a spring as shown in Fig. Area of both the plates is A. In steady state separation between the plates is 0.8 d (spring was un stretched and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is

• A. $\frac{4{ϵ}_{0}A{E}^2}{d^3}$
• B. $\frac{2{ϵ}_{0}A{E}^2}{d^2}$
• 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

$\frac{4{ϵ}_{0}A{E}^2}{d^3}$

In equilibrium electrostatic attraction between the plates = spring force $\because \frac{q^2}{2{ϵ}_{0}A}=k_x\because \frac{(CE{)}^2}{2{ϵ}_{0}A}=k(d-0.8d)\because \frac{{\left(\frac{{ϵ}_{0}A}{0.8d}\right)}^2{E}^2}{2{ϵ}_{0}A}=0.2dk\Rightarrow k=\frac{{ϵ}_{0}A{E}^2}{0.256d^3}\approx \frac{4{ϵ}_{0}A{E}^2}{d^3}$