JEE Main202126 Aug 2021Morning ShiftPhysicsCapacitanceActual
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 )
Options
- A0.9   mA
- B9.0   mA
- C9.0   μ A
- D0.9   μ A
Correct answer
A. 0.9   mA
Step-by-step solution
ρ = 200   Ωm K = 50 C = ϵ 0 KA d = 2 × 10 − 12   F R = ρ d A I = V R = AV ρ d = V ρ ⋅ C ϵ 0 K = 40 200 × 2 × 10 − 12 8.85 × 10 − 12 × 50 = 0.9 mA