Figure shows two identical parallel plate capacitors connected to a switch `S.` Initially ,the switch is closed so that the capacitors are completely charged .The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

Explanation

Initial total energy is `1/2CV^(2)+1/2CV^(2)=CV^(2)`
when the switch is open and dielectric is filled.
Then capacitance of each capacitor is `C=KC=3 C`.
Then energy stored in `A` is `1/2(3C)V^(2)=3/2CV^(2)`
Since switch is open, so chang will be same in B.
Energy in `B=(q^(2)B)/(2C')=(CV)^(2)/(2(3C)//6). v^(2)`
So total final energy stored is `3/2CV^(2)+1/6CV^(2)=(9CV^(2)+CV^(2))/6 =(10)/6CV^(2)`
So, rwquired ratio `=(CV^(2))/((10)/6CV^(2)) =3/5 =3:5`.