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
` = 1 / 2 CV^(2) + 1 / 2 CV^(2) = CV^(2) `
when the switch is open dielectric induced. Then capacitance
`C = 1 /2 3 CV^(2) = 3 /2 CV^(2)`
Since switch is open so change be same in B so Energy in
` B = 1 /2 . C/ 3 . V^(2)`
So, total final energy
` = 3 / 2 CV^(2) + 1 / 6 CV^(2) `
`= 9 CV^(2) + 1 CV^(2) / 6 = 10 / 6 CV^(2) `
So, required ratio `= CV_2 / (10 / 6 CV^(2)) = 3 /5 = 3.5` .