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Dielectric boundary at z=0, region 1 data

Short answer

Tangential D₁ = −30 a_x+50 a_y nC/m². P₁=0.6875 D₁ = −20.6 a_x+34.4 a_y+48.1 a_z nC/m². Normal E₂ = 3.95 a_z kV/m. Tangential E₂ = −1.06 a_x+1.76 a_y kV/m.

Let the region z < 0 be composed of a uniform dielectric material for which ε_R1 = 3.2, while the region z > 0 is characterized by ε_R2 = 2. Let D1 = −30a_x + 50a_y + 70a_z nC/m² and find: (a) D_t1 (tangential component of D in region 1); (b) polarization (P1); (c) E_n2 (normal component of E in region 2); (d) E_t2 (tangential component of E in region 2).

Given: Boundary at z = 0. Region 1 (z < 0): ε_R1 = 3.2. Region 2 (z > 0): ε_R2 = 2. D₁ = −30 a_x + 50 a_y + 70 a_z nC/m². The normal to the boundary is a_z.

Boundary conditions (no free surface charge on a dielectric interface):

  • Normal D is continuous: D_N1 = D_N2.
  • Tangential E is continuous: E_t1 = E_t2, i.e. D_t1/ε₁ = D_t2/ε₂.

(a) Tangential component of D in region 1

D_t1 = −30 a_x + 50 a_y   nC/m²        (|D_t1| = 58.3 nC/m²)

(b) Polarization P₁

P₁ = D₁ − ε₀E₁ = D₁ − D₁/ε_R1 = D₁ (1 − 1/3.2) = 0.6875 D₁
P₁ = −20.6 a_x + 34.4 a_y + 48.1 a_z   nC/m²

(c) Normal component of E in region 2

D_N2 = D_N1 = 70 nC/m²:

E_N2 = D_N2 / (ε_R2 ε₀) = 70 × 10⁻⁹ / (2 × 8.854 × 10⁻¹²) = 3953 V/m
E_N2 = 3.95 a_z kV/m

(d) Tangential component of E in region 2

E_t2 = E_t1 = D_t1 / (ε_R1 ε₀):

E_t2 = (−30 a_x + 50 a_y) × 10⁻⁹ / (3.2 × 8.854 × 10⁻¹²)
     = (−30 a_x + 50 a_y) × 10⁻⁹ / (2.833 × 10⁻¹¹)
     = −1059 a_x + 1765 a_y   V/m

Answers:

  • (a) D_t1 = −30 a_x + 50 a_y nC/m²
  • (b) P₁ = −20.6 a_x + 34.4 a_y + 48.1 a_z nC/m²
  • (c) E_N2 = 3.95 kV/m (along a_z)
  • (d) E_t2 = −1.06 a_x + 1.76 a_y kV/m

✓ Answer checked by Claude Opus 5.5 (high effort)

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