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JEE Advanced2014PhysicsCapacitanceActual

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 t

Options

  1. AE 1 E 2 = 1
  2. BE 1 E 2 = 1 K
  3. CQ 1 Q 2 = 3 K
  4. DC C 1 = 2 + K K

Correct answer

A. E 1 E 2 = 1

Step-by-step solution

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 ε 0 A / d Lower capacitor C 2 = 2 ε 0 A / d ∴ equivalent capacitance C = k + 2 ϵ 0 A / d ∴ C C 1 = k + 2 k And E 1 : E 2 = v d : v d = 1

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