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 0.8d (spring was stretchered and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is approximately

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

Answer 1

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

Given : Separation between plates, when uncharged $=d$

Separation between plates, when charged $=0.8d$

In steady state, the force by spring is balanced by the attractive force between plates.

therefore, attractive force between plates = spring force

Now, the plates are oppositely charged, so the attractive force $F$ between them will be:

$F=$electric field by one plate $\times$ charge on another plate,

or $F=\frac{\sigma }{2{\epsilon }_0}\times Q$

or $F=\frac{Q}{2A{\epsilon }_0}\times Q=\frac{Q^2}{2A{\epsilon }_0}$,

or $F=\frac{C^2{E}^2}{2A{\epsilon }_0}$, (by $Q=CE$)

or $F=\frac{{\epsilon }_{0}A{E}^2}{2d^2}$ , (by $C=\frac{{\epsilon }_{0}A}{d}$) .

Now, spring force is given by:

$F^{\prime }=k\Delta x=k(d-0.8d)=0.2dk$ , (where $k=$spring constt.) ,

Therefore, for steady state,

$F^{\prime }=F$

or $0.2dk=\frac{{\epsilon }_{0}A{E}^2}{2d^2}$

or $k=\frac{{\epsilon }_{0}A{E}^2}{0.4d^2}$

or $k=\frac{5{\epsilon }_{0}A{E}^2}{2d^2}$