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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Capacitance

JEE Main 2018 · 8 April · Physics · Capacitance

A parallel plate capacitor of capacitance 90pF is connected to a...

A parallel plate capacitor of capacitance 90pF90pF is connected to a battery of EMF 20V20V. If a dielectric material of dielectric constant K=53K = \dfrac{5}{3} is inserted between the plates, the magnitude of the induced charge will be:

Options

  1. A0.9nC0 . 9nC
  2. B1.2nC1 . 2nC
  3. C0.3nC0 . 3nC
  4. D2.4nC2 . 4nC

Correct answer

B. 1.2nC1 . 2nC

Step-by-step solution

Induced charge on the dielectric,
Qinduced=KCV(1−1k)=53×90×10−12×20(1−35)=1.2nC{Q}_{induced} = KCV\left(1 - \dfrac{1}{k}\right) = \dfrac{5}{3} \times 90 \times {10}^{- 12} \times 20\left(1 - \dfrac{3}{5}\right) = 1.2nC.
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