Consider an air filled parallel plate capacitor with one plate connected to a spring having a force constant $k$ and the other plate is held fixed. The system rests on a frictionless table top. The plates have change density $\sigma$ and $-\sigma$, area $A$, charges $Q$ and $-Q$. The electric field created by the plate $b$ is $E , magnitude of force acting on each plate is F_e.$ and the spring gets an extension $x$. Then

The spring force ${\overrightarrow{F}}_S$ acting on a plate $a$ is given by ${\overrightarrow{F}}_S=-kx\hat{i}$

Similarly the electrostatic froce ${\overrightarrow{F}}_e$ due to the electric field created by plate $b$ is ${\overrightarrow{F}}_e=QE\hat{i}=Q\left(\frac{\sigma }{2{\epsilon }_0}\right)\hat{i}=\frac{Q^2}{2A{\epsilon }_0}\hat{i}$

Where $A$ is the area of the plate. Notice that charge on plate $a$ cannot exert a force on itself, as required by Newtons's Third Law. Thus only the electric field due to plate $b$ is considered. At equilibrium the two forces cancel ans we have

$kx=Q\left(\frac{Q}{2A{\epsilon }_0}\right)=QE$ which gives $x=\frac{Q^2}{2kA{\epsilon }_0}=\frac{QE}{k}$