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An alternating voltage V(t) = 20 sin(omega*t) of frequency 50 Hz is applied across a parallel plate capacitor whose plates are separated by 2 mm and each have an area of 1 m². Determine the peak (amplitude) value of the displacement current set up in the capacitor. [Take epsilon₀ = 8.85 * 10⁻¹² F/m]

1. 21.14 microA
2. 83.37 microA
3. 27.79 microA
4. 55.58 microA

Correct answer: 27.79 microA

Solution

Maxwell's displacement current between capacitor plates equals C dV/dt, so its amplitude is C*V0*omega. Here omega = 2*pi*f with f = 50 Hz.

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