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A parallel plate capacitor of area A , plate separation d and capacitance C is filled with three dielectric materials having dielectric constants k 1 , k 2 a n d k 3 as shown. If a single dielectric material is to be used to have the same capacitance C in the capacitor, then its dielectric constant k is given by

Options

  1. Ak = k 1 + k 2 + 2 k 3
  2. Bk = k 1 k 2 k 1 + k 2 + 2 k 3
  3. C1 k = 1 k 1 + 1 k 2 + 1 2 k 3
  4. D1 k = 1 k 1 + k 2 + 1 2 k 3

Correct answer

D. 1 k = 1 k 1 + k 2 + 1 2 k 3

Step-by-step solution

The capacitances, C 1 = k 1 ε 0 A / 2 d / 2 = k 1 ε 0 A d C 2 = k 2 ε 0 A / 2 d / 2 = k 0 ε 0 A d And C 3 = k 3 ε 0 A / 2 d / 2 = 2 k 3 ε 0 A d The equivalent capacitance is given by ⇒ 1 C e q = 1 C 1 + C 2 + 1 C 3 = 1 ε 0 A d k 1 + k 2 + 1 ε 0 A d × 2 k 3 ⇒ 1 C e q = 1 C 1 + C 2 + 1 C 3 = 1 ε 0 A d k 1 + k 2 + 1 ε 0 A d × 2 k 3 ⇒ 1 C e q = d ε 0 A 1 k 1 + k 2 + 1 2 k 3 ⇒ C e q = 1 k 1 + k 2 + 1 2 k 3 - 1 . ε . A d ∴ 1 k = 1 k 1 + k 2 + 1 2 k 3 ∵ C = K ε A d k = k 1 k 3 k 1 + k 3 + k 2 k 3 k 2 + k 3 C e q = ε o A /

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