AC voltage V(t) = 20 sinωt of frequency 50Hz is applied to a parallel plate capacitor. The separation between the plates is 2mm and area is 1m2. The amplitude of oscillating displacement current for applied AC voltage is___________ μA.(Take ∈0=8.85×10−12 F/m)(Round off to nearest integer)

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AC voltage $V (t) = 20 \sin ω t$ of frequency 50Hz is applied to a parallel plate capacitor. The separation between the plates is 2mm and area is 1m². The amplitude of oscillating displacement current for applied AC voltage is___________ $μ A . (T a k e \in_{0} = 8.85 \times 10^{- 12} F / m)$
(Round off to nearest integer)

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answer is 28.

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Detailed Solution

Displacement current $i = C \frac{d v}{d t} \Rightarrow i = C 20 w \cos w t$
Amplitude of displacement current is I_max = C20w $= (\frac{\in_{0} A}{d}) 20 (2 π f) = (\frac{8.85 \times 10^{- 12} \times 1}{2 \times 10^{- 3}}) 20 (2 π \times 50) = 27.79 μ A = 28 μ A$