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,…

Question

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.

Aicharya · Accepted Answer

As E = v d E 1 = v d E 2 = v d (As both parts have some P.D.) upper capacitor have capacitance C 1 = K &epsilon; 0 A / d Lower capacitor C 2 = 2 &epsilon; 0 A / d &there4; equivalent capacitance C = k + 2 ϵ 0 A / d &there4; C C 1 = k + 2 k And E 1 : E 2 = v d : v d = 1

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3…

Solution