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Dielectrics question from JEE Main 2018

Q100JEE Main 2018Medium

Dielectrics

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be:

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Answer: (b)

  1. Battery connected: new charge Q=KCV=53×90×20 pC=3 nCQ=KCV=\frac{5}{3}\times90\times20\text{ pC}=3\text{ nC}
  2. Induced charge Qi=Q(1−1K)=3×25Q_i=Q\left(1-\frac{1}{K}\right)=3\times\frac{2}{5}
  3. So the answer is (b) 1.2 nC.
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