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in Physics by (74.8k points)
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.
image
A. `(E_(1))/(E_(2))=1`
B. `(E_(1))/(E_(2))=(1)/(K)`
C. `(Q_(1))/(Q_(2))=(3)/(K)`
D. `(C)/(C_(1))=(2+K)/(K)`

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1 Answer

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by (74.1k points)
Correct Answer - A::D

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