Displacement current density at 1 kHz

In the metal, the conduction current J_c = σE, and the displacement current density J_d = ∂D/∂t = ε ∂E/∂t.

Given: f = 1 kHz, ω = 2πf = 6283 rad/s, σ = 5×10⁷ S/m, ε = ε_r ε₀ = 8.854×10⁻¹² F/m, J_c = 10⁷ sin(6283t − 444z) a_y A/m².

Step 1: E from J_c.

E = J_c / σ = (10⁷ / 5×10⁷) sin(6283t − 444z) a_y = 0.2 sin(6283t − 444z) a_y  V/m

Step 2: D and J_d.

D = ε₀ E = 8.854×10⁻¹² × 0.2 sin(...) a_y
J_d = ∂D/∂t = 8.854×10⁻¹² × 0.2 × 6283 cos(6283t − 444z) a_y
    = 1.1126×10⁻⁸ cos(6283t − 444z) a_y   A/m²

Step 3: ratio.

J_d / J_c = ωε/σ = 1.1126×10⁻⁸ / 10⁷ ≈ 1.1×10⁻¹⁵

So displacement current is negligible compared with conduction current in a metal at 1 kHz.

Answer: amplitude of J_d = 1.11×10⁻⁸ A/m² (leading J_c by 90°, since it varies as cosine while J_c varies as sine).