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A parallel plate capacitor has plates wi...

A parallel plate capacitor has plates with area A and separation d. A battery charges the plates to a potential difference `V_(0)`. The battery is then disconnected and a dielectric slab of thikness d is introduced. The ratio of the enrgy stored in the capacitor before and after the slab is introduced, is

A

K

B

`(1)/(K)`

C

`(A)/(d^(2)K)`

D

`(d^(2)K)/(A)`

Text Solution

Verified by Experts

The correct Answer is:
a
Potential energy of charged capacitor is ,
`U=(1)/(2)(Q^(2))/(C)`
On introduction of dielectric slab `C.=KC`
Q - remains unaltered as battery is disconnected.
`thereforeU.=(1)/(2)(Q^(2))/(C.)=(1)/(2)(Q^(2))/(KC)therefore(U)/(U),=((1)/(2)(Q^(2))/(C))/((1)/(2)(Q^(2))/(KC))=K`
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