AC voltage $\mathrm{V}(\mathrm{t})=20 \sin \omega \mathrm{t}$ of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is $1 \mathrm{~m}^{2}$. The amplitude of the oscillating displacement current for the applied Ac voltage is $\_\_\_\_$. [Take $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$]

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Alternating Current

AC voltage

V (t) = 20 sin ω t

of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is

1 m^{2}

. The amplitude of the oscillating displacement current for the applied Ac voltage is

____

.
[Take

ε_{0} = 8.85 \times 1 0^{- 12} F / m

]

Answer: (D)

The displacement current between the plates equals the conduction current charging them,

I_{d} = \frac{d q}{d t} = C \frac{d V}{d t}

. The capacitance is

C = \frac{ε _{0} A}{d} = \frac{8.85 \times 1 0 ^{- 12} \times 1}{2 \times 1 0 ^{- 3}} = 4.425 \times 1 0^{- 9} F

With

V = 20 sin ω t

and

ω = 2 π \times 50

, the derivative has amplitude

ω V_{0}

, so

I_{0} = ω C V_{0} = 2 π \times 50 \times 4.425 \times 1 0^{- 9} \times 20 = 2.78 \times 1 0^{- 5} A

The amplitude of the oscillating displacement current is

27.79 μ

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option