The material filled between the plates of a parallel plate capacitor has resistivity 200   Ω &#160…

The material filled between the plates of a parallel plate capacitor has resistivity 200 Ω m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permittivity of material is 50)

  1. 0.9 mA
  2. 9.0 mA
  3. 9.0 μA
  4. 0.9 μA

Solution

ρ=200 Ωm

K=50

C=ϵ0KAd=2×10−12 F

R=ρdA

I=VR=AVρd=Vρ⋅Cϵ0K=40200×2×10−128.85×10−12×50=0.9mA

Asked in: JEE Main 2021 (26 Aug Shift 1)

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