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Polarization and E in both regions

Short answer

With boundary normal a_x, D_x continuous. Region 1: E₁ = (−30a_x+50a_y+70a_z)×10⁻⁹/(3.2ε₀) and P₁=0.6875D₁. Region 2: D₂ = −30a_x+31.25a_y+43.75a_z nC/m²; E₂ from D₂/(2ε₀); P₂=0.5D₂. Components as computed.

The region x < 0 is composed of a uniform dielectric material for which ε_r1 = 3.2, while the region x > 0 is characterized by ε_r2 = 2. The electric flux density at region x < 0 is D1 = −30a_x + 50a_y + 70a_z nC/m². Find the polarization (P) and electric field intensity (E) in both regions.

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

Region 1

E₁ = D₁/(ε₀ε_r1) = D₁ ×10⁻⁹ /(3.2 × 8.854×10⁻¹²)
   = (−30 a_x + 50 a_y + 70 a_z) × 35.29
   = −1058.8 a_x + 1764.7 a_y + 2470.6 a_z  V/m

P₁ = D₁ (1 − 1/ε_r1) = 0.6875 D₁
   = −20.63 a_x + 34.38 a_y + 48.13 a_z  nC/m²

Region 2

Normal D continuous: D_x2 = D_x1 = −30. Tangential E continuous, so D_t2 = (ε_r2/ε_r1) D_t1 = 0.625 D_t1:

D_y2 = 0.625 × 50 = 31.25 ,  D_z2 = 0.625 × 70 = 43.75
D₂ = −30 a_x + 31.25 a_y + 43.75 a_z  nC/m²

E₂ = D₂/(ε₀ε_r2) = D₂ ×10⁻⁹ /(2 × 8.854×10⁻¹²)
   = −1694.1 a_x + 1764.7 a_y + 2470.6 a_z  V/m
   (E_y and E_z agree with region 1, as required)

P₂ = D₂ (1 − 1/2) = 0.5 D₂
   = −15 a_x + 15.63 a_y + 21.88 a_z  nC/m²
QuantityRegion 1 (x < 0)Region 2 (x > 0)
E (V/m)−1059 a_x + 1765 a_y + 2471 a_z−1694 a_x + 1765 a_y + 2471 a_z
P (nC/m²)−20.6 a_x + 34.4 a_y + 48.1 a_z−15 a_x + 15.6 a_y + 21.9 a_z

✓ Answer checked by Claude Opus 5.5 (high effort)

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