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A parallel plate capacitor has a dielect...

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers `1//3` of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is `C_1`. When the capacitor is charged, the plate area covered by the dielectric gets charge `Q_1` and the rest of the area gets charge `Q_2`. The electric field in the dielectric is `E_1` and that in the other portion is `E_2`. Choose the correct option/options, ignoring edge effects.

A

`(E_(1))/(E_(2))=1`

B

`(E_(1))/(E_(2))=1/K`

C

`(Q_(1))/(Q_(2))=3/K`

D

`(C)/(C_(1))=(K+2)/(K)`

Text Solution

Verified by Experts

The correct Answer is:
A, D
`c=c_1+c_2`
`c_1=(Kepsilon_0A//3)/dand c_2=(epsilon_02A//3)/d implies c=((K+2)epsilon_0A)/(3d)implies c/(c_1)=(K+2)/k`
Also `E_1=E_2=v/d` where V is potential difference between the plates .
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