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Displacement current density, dE/dt, induced B

Short answer

dE/dt = J_d/ε₀ = 2.26×10¹² V m⁻¹ s⁻¹. The displacement current density is 20 A/m². Induced magnetic field at the plate edge is B = μ₀ J_d R/2 = 1.26×10⁻⁶ T.

A parallel plate capacitor with circular plates of 10cm radius is charged producing uniform displacement current of magnitude 20A/m². Calculate (i) dE/dt in the region (ii) Displacement current density and (iii) Induced magnetic field.

Given: plate radius R = 10 cm = 0.10 m, displacement current density J_d = 20 A/m² (uniform). ε₀ = 8.85×10⁻¹² F/m, μ₀ = 4π×10⁻⁷ H/m.

(ii) Displacement current density

This is given: J_d = 20 A/m². The total displacement current is I_d = J_d πR² = 20 × π × 0.01 = 0.628 A.

(i) dE/dt

Displacement current density is J_d = ε₀ dE/dt.

dE/dt = J_d / ε₀ = 20 / 8.85×10⁻¹²
      = 2.26×10¹² V/(m·s)

(iii) Induced magnetic field

Ampere-Maxwell law for a circle of radius r: B(2πr) = μ₀ J_d (πr²), so B = μ₀ J_d r/2. At the edge of the plates, r = R:

B = μ₀ J_d R / 2
  = (4π×10⁻⁷ × 20 × 0.10) / 2
  = 1.26×10⁻⁶ T

Answer: (i) dE/dt ≈ 2.26×10¹² V m⁻¹s⁻¹; (ii) J_d = 20 A/m²; (iii) B ≈ 1.26×10⁻⁶ T at the plate edge.

✓ Answer checked by Claude Opus 5.5 (high effort)

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