
ρ=200Ωm
K=50
C=dϵ0KA=2×10−12F
R=Aρd
I=RV=ρdAV=ρV⋅ϵ0KC=20040×8.85×10−12×502×10−12=0.9mA
JEE Main 2021 — Physics Electromagnetism
The material filled between the plates of a parallel plate capacitor has resistivity 200Ωm. The value of capacitance of the capacitor is 2pF. If a potential difference of 40V 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)
Held on 26 Aug 2021 · Verified 6 Jul 2026.
0.9mA
9.0mA
9.0μA
0.9μA
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