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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 dielectric constant K is placed in between the plates of the capacitor as shown.

The electric field in the dielectric slab is

A

`V/(Kd)`

B

`(KV)/d`

C

`V/d`

D

`(KV)/t`

Text Solution

Verified by Experts

The correct Answer is:
A
`E_("dielectric")=E_0/K=V/(Kd)`
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