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The figure shows two identical parallel ...

The figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant(or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Text Solution

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Initally, when the swich S is closed, both the capacitors have same potential difference (V) across them.
`:.` Initial energy stored in both the capacitors
is `U_(i) = U_(A) + U_(B) = (1)/(2) CV^(2) + (1)/(2) CV^(2) = CV^(2)`
When the dielectric (K = 3) is introduced, the capacitance of each capacitor becomes 3C. The p.d. across A is still V as it is still connected across the battery, with swich S open, p.d across B attains new value but charge remains constant
`q = CV = 3C xx V' or V' = V//3`.
So final energy `U_(f) = U'_(A) + U'_(B)`
`= (1)/(2) (3C) V^(2) + (1)/(2) (3C) ((V)/(3))^(2) = (5C)/(3) V^(2)`
`(U_(i))/(U_(f)) = (CV^(2))/((5)/(3) CV^(2)) = 3 : 5`
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