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Figure shows a parallel plate capacitor ...

Figure shows a parallel plate capacitor with plate area `A` and plate separation `d`. A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant `K` is placed in between the plates of the capacitor as shown.

The electric field in the gaps between the plates and the electric slab will be

A

`(epsilon_0AV)/d`

B

`V/d`

C

`(KV)/d`

D

`V/(d-t)`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(air)=E_0=V/d`
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