Figure given below shows two identical parallel plate capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plate of capacitors is filled with a dielectric of dielectric constant 3. What will be ten times the ratio of total electrostatic energy stored in both capacitors before and after the introduction of the dielectric

 

Initially potential difference across both the capacitor is same hence energy of the system  is ${\mathrm{U}}_1=\frac{1}{2}{\mathrm{CV}}^2+\frac{1}{2}{\mathrm{CV}}^2={\mathrm{CV}}^2$
 
……(i) In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes 3C, while potential difference across A is V and potential difference across B is $\frac{V}{3}$   hence energy of the system now is   ${\mathrm{U}}_2=\frac{1}{2}\left(3\mathrm{C}\right){\mathrm{V}}^2+\frac{1}{2}\left(3\mathrm{C}\right){\left(\frac{\mathrm{V}}{3}\right)}^2=\frac{10}{6}{\mathrm{CV}}^2$
……(ii) So,   $\frac{{\mathrm{U}}_1}{{\mathrm{U}}_2}=\frac{3}{5}$