Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Initially charge on either capacitor is QA = QB = CV.

After dielectric is introduced, new capacitance of either capacitor =KC After opening switch potential across capacitor A is V volts. Let potential across capacitor B be V1 Therefore
QB = CV = C1V1 = KCV1
${\mathrm{V}}_1=\frac{\mathrm{V}}{\mathrm{K}}$
Initial energy in both capacitors $=\frac{{\mathrm{CV}}^2}{2}+\frac{{\mathrm{CV}}^2}{2}={\mathrm{CV}}^2$
Final energy of capacitor $\text{A=}\frac{{\mathrm{KCV}}^2}{2}$
$\mathrm{B}=\frac{{\mathrm{KCV}}^2}{2{\mathrm{K}}^2}=\frac{{\mathrm{CV}}^2}{2\mathrm{K}}$
Final energy of capacitor
Total final energy of both capacitors $=\frac{{\mathrm{KCV}}^2}{2}+\frac{{\mathrm{CV}}^2}{2\mathrm{K}}=\left(\frac{{\mathrm{K}}^2+1}{2\mathrm{K}}\right){\mathrm{CV}}^2$
Ratio of the total electrostatic energy stored in both capacitors before and after the
introduction of the dielectric $=\frac{{\mathrm{CV}}^2}{\left(\frac{{\mathrm{K}}^2+1}{2\mathrm{K}}\right){\mathrm{CV}}^2}=\frac{2\mathrm{K}}{{\mathrm{K}}^2+1}$