AC voltage V(t) = 20 sinωt of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m². The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take ε₀ = 8.85 × 10⁻¹² F/m]

Solution

Physics sir · 4 steps

The displacement current between the plates equals the charging current, so find it from C dV/dt.

1. Displacement current. Between the plates, i_d = ε₀ dΦ_E/dt. For a parallel plate capacitor this works out to CdV/dt, which is the same as the current in the wires.

   i_d = CdV/dt = C V₀ ω cos ω t ⇒ i₀ = C V₀ ω

2. Find C. C = (ε₀ A)/d = (8.85 × 10⁻¹² × 1)/(2 × 10⁻³) = 4.425 × 10⁻⁹ F.
3. Put in numbers. ω = 2π × 50 = 314 rad s⁻¹ and V₀ = 20 V. So i₀ = 4.425 × 10⁻⁹ × 20 × 314 = 2.779 × 10⁻⁵ A.
4. So. The amplitude of the displacement current is about 27.79 μA, option (c).

The idea to keep. Displacement current in a capacitor is just C dV/dt. For V = V₀ sin ω t, its amplitude is C V₀ ω, exactly like the conduction current in the circuit.

Answer from the NTA final answer key. Solution by PrepWiser.