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Figure shows two identical parallel plat...

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.

A

`9:16`

B

`5 :9`

C

` 2: 3`

D

` 3: 5`

Text Solution

Verified by Experts

The correct Answer is:
D
Initial total energy `=(1)/(2)CV^(2)+(1)/(2)CV^(2)=CV^(2)`
When the switch is open, dielectric is introduced. Then capacitance, `C=KC=3C`
Energy stored in `C` is
`(1)/(2)3CV^(2)`
So total final energy `=(3)/(2)CV^(2)+(1)/(6)CV^(2)`
`=(9CV^(2)+1CV^(2))/(6)=(10)/(6)CV^(2)`
So required ration`=(CV^(2))/(10/(6)CV^(2))=(3)/(5)=3:5`
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