Skip to main content

Dielectric boundary at y=0, find fields

Short answer

The normal D_y is continuous; tangential D is scaled by ε_r2/ε_r1=0.625, giving D₂=−18.8a_x+50a_y+43.8a_z nC/m², |D₂|=69.0 nC/m². E₂=D₂/(2ε₀), |E₂|=3.90 kV/m. P₁=0.6875D₁, P₂=0.5D₂ with components as given.

Consider the region y < 0 composed of a uniform dielectric material for which the relative permittivity (ε_r) is 3.2, while the region y > 0 is characterized by ε_r = 2. Let the flux density in region 1 be D1 = −30a_x + 50a_y + 70a_z nC/m². Find: (a) the magnitude of flux density and electric field intensity at region 2; (b) polarization (P) in region 1 and region 2.

Boundary: the plane y = 0, with normal along a_y. Region 1 (y < 0): ε_r1 = 3.2. Region 2 (y > 0): ε_r2 = 2. D₁ = −30 a_x + 50 a_y + 70 a_z nC/m².

Conditions: the normal component of D is continuous; the tangential component of E is continuous, so D_t1/ε_r1 = D_t2/ε_r2.

(a) D₂ and E₂

D_N1 = 50 a_y  →  D_N2 = 50 a_y  nC/m²
D_t1 = −30 a_x + 70 a_z
D_t2 = (ε_r2/ε_r1) D_t1 = (2/3.2)(−30 a_x + 70 a_z)
     = −18.75 a_x + 43.75 a_z

D₂ = −18.75 a_x + 50 a_y + 43.75 a_z  nC/m²
|D₂| = √(18.75² + 50² + 43.75²) = √4765.6 = 69.03 nC/m²

Electric field in region 2: E₂ = D₂/(ε₀ε_r2) = D₂/(2 × 8.854×10⁻¹²):

E₂ = (−18.75 a_x + 50 a_y + 43.75 a_z)×10⁻⁹ / (17.708×10⁻¹²)
   = −1058.8 a_x + 2823.6 a_y + 2470.6 a_z  V/m
|E₂| = 3898 V/m ≈ 3.90 kV/m

(b) Polarization, P = D − ε₀E = D(1 − 1/ε_r)

Region 1: P₁ = D₁ (1 − 1/3.2) = 0.6875 D₁
        = −20.63 a_x + 34.38 a_y + 48.13 a_z  nC/m²

Region 2: P₂ = D₂ (1 − 1/2) = 0.5 D₂
        = −9.38 a_x + 25 a_y + 21.88 a_z  nC/m²

Answers: |D₂| = 69.0 nC/m², |E₂| = 3.90 kV/m; P₁ = −20.6 a_x + 34.4 a_y + 48.1 a_z nC/m², P₂ = −9.38 a_x + 25 a_y + 21.9 a_z nC/m².

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electromagnetism chapter 2 questions →