Two parallel plate capacitors with area A are connected through a conducting spring of natural length l in series as shown. Plates P and S have fixed positions at separation d. Now the plates are connected by a battery of emf $\mathrm{\xi }$ as shown. If the extension in the spring in equilibrium is equal to the separation between the plates, find the spring constant k.

Sol: Let charge on capacitors be q and separation between plates P & Q and R & S be x at any time. Distance between plates P & Q and R & S is same because force acting on them is same.
 

Capacitance of capacitor $\mathrm{PQ},{\mathrm{C}}_1=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{x}}$
Capacitance of capacitor $\mathrm{RS},{\mathrm{C}}_2=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{x}}$
From KVL, we have$\frac{\mathrm{q}}{{\mathrm{C}}_1}+\frac{\mathrm{q}}{{\mathrm{C}}_2}=\mathrm{\xi } \mathrm{or}$
$\mathrm{q}=\frac{{\mathrm{\epsilon }}_0\mathrm{A\xi }}{2\mathrm{x}}$
At this moment extension is spring = d - 2x - l . Force on plate Q towards P
${\mathrm{F}}_1=\frac{{\mathrm{q}}^2}{2{\mathrm{A\epsilon }}_0}=\frac{{\mathrm{\epsilon }}_0{\mathrm{A}}^2{\mathrm{\xi }}^2}{8{\mathrm{Ax}}^2{\mathrm{\epsilon }}_0}=\frac{{\mathrm{A\epsilon }}_0{\mathrm{\xi }}^2}{8{\mathrm{x}}^2}$
Spring force on plate Q due to extension in spring. At equilibrium, separation between plates = extension is spring
Thus x = y =( d - 2x -l )
$\Rightarrow \mathrm{x}=\frac{\mathrm{d}-\mathrm{I}}{3}$……….(1)
At equilibrium,  F₁ = F₂ …… (2)
From equation (1) and (2), we have
$\frac{{\mathrm{A\epsilon }}_0{\mathrm{\xi }}^2}{8{\mathrm{x}}^2}=\mathrm{ky}=\mathrm{kx}\Rightarrow \mathrm{x}=\left(\frac{{\mathrm{A\epsilon }}_0{\mathrm{\xi }}^2}{8\mathrm{k}}\right)\dots \dots$(3)
From equation (1) and (3),
${\left(\frac{\mathrm{d}-1}{3}\right)}^3=\frac{{\mathrm{A\epsilon }}_0{\mathrm{\xi }}^2}{8\mathrm{k}}\Rightarrow \mathrm{k}=\frac{27}{8}\frac{{\mathrm{A\epsilon }}_0{\mathrm{\xi }}^2}{(\mathrm{d}-\mathrm{l}{)}^3}$