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.

Options

A. `2:1`
B. `5:1`
C. `4:5`
D. `5:4`

Answer

C

Explanation

When S is closed
`U_(i)=(1)/(2)CV^(2)+(1)/(2)CV^(2)=CV^(2)`
Charge on `B:Q=CV`
When `S` is open and dielectric `(K=2)` is inserted in both capacitors, potential of `A` remains same. While cahrged on `B` remains same
`U_(f)=(1)/(2).2C.V^(2)+(Q^(2))/(2.2C)`
`=CV^(2)+(C^(2)V^(2))/(4C)=(5)/(4)CV^(2)`
`(U_(i))/(U_(f))=(4)/(5)`