Figure shows two capacitors in series. The rigid center section of length ‘b’ is movable. The area of each plate is S. If the voltage difference between the plates is maintained constant at V₀. The change in stored energy if the center section is removed is

Initially,

On assuming the distance between the plates of C₁ be x, the distance between the plates of second capacitor will be 'a-b-x'

the capacitances will be given by 

 ${\mathrm{C}}_1=\frac{{\mathrm{\epsilon }}_0\mathrm{S}}{\mathrm{x}}{\mathrm{andC}}_2=\frac{{\mathrm{\epsilon }}_0\mathrm{S}}{(\mathrm{a}-\mathrm{b}-\mathrm{x})}$

${\mathrm{C}}_{\mathrm{eff}}=\frac{{\mathrm{ϵ}}_0\mathrm{S}}{(\mathrm{a}-\mathrm{b})}$is independent of x

So the energy stored in the capacitors is given by ${\mathrm{U}}_{\mathrm{i}}=\frac{1}{2}\left(\frac{{\mathrm{\epsilon }}_0\mathrm{S}}{\mathrm{a}-\mathrm{b}}\right){\mathrm{V}}_0^2$

When the center section is released, the plates of the capacitor will stick together and act as a capacitor with a distance between the plates as 'a'.

So the energy stored in the capacitor will be given by ${\mathrm{U}}_{\mathrm{f}}=\frac{1}{2}\left(\frac{{\mathrm{\epsilon }}_0\mathrm{S}}{\mathrm{a}}\right){\mathrm{V}}_0^2$

Change in energy stored will be $∆\mathrm{U}=\frac{1}{2}\left(\frac{{\mathrm{\epsilon }}_0\mathrm{S}}{\mathrm{a}-\mathrm{b}}\right){\mathrm{V}}_0^2-\frac{1}{2}\left(\frac{{\mathrm{\epsilon }}_0\mathrm{S}}{\mathrm{a}}\right){\mathrm{V}}_0^2$=$\frac{{\mathrm{S\epsilon }}_0{\mathrm{V}}_0^2\mathrm{b}}{2\mathrm{a}(\mathrm{a}-\mathrm{b})}$