Question

One plate of a capacitor is fixed and the other is connected to a spring as shown in the figure. Area of both the plates is $A$. In 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. $\frac{125}{32}\frac{{\in }_{0}A{E}^2}{d^3}$
• B. $\frac{2{\in }_{0}A{E}^2}{d^3}$
• C. $\frac{6{\in }_0{E}^2}{A{d}^2}$
• D. $\frac{{\in }_{0}A{E}^3}{2d^3}$

Answer 1

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

In steady state the forces acting on the right plate are spring force due to elongation of spring and second is force due to electric field of left plate.
As the spring gets elongated by distance $0.2d$ , the spring force can be given by
$F_s=K\left(0.2d\right)$
Force due to electric field can be given by
$F_E=Q\frac{\sigma }{2{ϵ}_0}$-------(let Q be the charge on the plates due to battery E)
$⟹F_E=\frac{Q^2}{2A{ϵ}_0}$------$(\because \sigma =\frac{Q}{A})$
We know that in equilibrium $F_E=F_s$
$\therefore K\left(0.2d\right)=\frac{Q^2}{2A{ϵ}_0}$-----(i)
Also $Q=CE=\frac{A{ϵ}_{0}E}{0.8d}$----(ii)
Put (ii) in (i) and we will get
$K=\frac{125}{32}\frac{{\in }_{0}A{E}^2}{d^3}$