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

(a) when switch `S` is closed.
`p.d.` across each capacitor is `V`. Initial total energy stored
`U_(i)=(1)/(2)CV^(2)+(1)/(2)CV^(2)=CV^(2)`
charge on `B:Q=CV`
(b) Now dielectric is placed in both capacitors, `p.d.` across `A` remains same because it is connected to battery and charge on `B` remais.
`U_(f)=U_(A)+U_(B)=(1)/(2).3C.V^(2)+(Q^(2))/(3C)`
`=(3CV^(2))/(2)+(C^(2)V^(2))/(3C)=CV^(2)((3)/(2)+(1)/(3))=(5)/(3)CV^(2)`
`(U_(i))/(U_(f))=(CV^(2))/((5)/(3)CV^(2))=(3)/(5)`