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)) = (2+k)/(k)`

A parallel plate capacitor has a dielectric slab of dielectric constan

1.	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)) = (2+k)/(k)`
Answer» Correct Answer - A::D

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