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`

5 views

1 Answers

Correct Answer - A
`EpropCandCpropk`
`therefore" "Epropk`
`E_(2)/E_(1)=k_(2)/k_(1)=k/k_(1)=k/1" "(thereforek_(2)=kandk_(1)=1)`

5 views

Related Questions