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JEE Main20199 Jan 2019Evening ShiftPhysicsCapacitanceActual

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K 1 , K 2 , K 3 , K 4 arranged as shown in the figure. The effective dielectric constant K will be:

Options

  1. AK = K 1 K 2 K 1 + K 2 + K 3 . K 4 K 3 + K 4
  2. BK = K 1 + K 2 K 3 + K 4 2 K 1 + K 2 + K 3 + K 4
  3. CK = ( K 1 + K 4 ) ( K 2 + K 3 ) 2 K 1 + K 2 + K 3 + K 4
  4. DK = K 1 + K 2 ( K 3 + K 4 ) K 1 + K 2 + K 3 + K 4

Correct answer

A. K = K 1 K 2 K 1 + K 2 + K 3 . K 4 K 3 + K 4

Step-by-step solution

C 1 and C 2 are in series. The equivalent capacitance of C 1 and C 2 is C 12 = ∈ 0 A 2 d 2 K 1 + d 2 K 2 = ∈ 0 A d K 1 K 2 K 1 + K 2 C 3   and   C 4   are in series C 34 = ∈ 0 A 2 d 2 K 3 + d 2 K 4 = ∈ 0 A d K 3 K 4 K 3 + K 4 C e q = C 12 + C 34 = ∈ 0 A d K 1 K 2 K 1 + K 2 + K 3 K 4 K 3 + K 4 ∴ K e q = K 1 K 2 K 1 + K 2 + K 3 . K 4 K 3 + K 4

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