Dielectric boundary, E1 given, find E2 D2 P2

Electric dipole

An electric dipole is a pair of equal and opposite point charges +Q and −Q separated by a small distance d. Its dipole moment is p = Q d (vector from −Q to +Q), in C·m.

Polarization

Polarization P is the dipole moment per unit volume of a dielectric: P = lim (Σ pᵢ)/Δv, in C/m². When an external field is applied, bound charges in the dielectric shift and form dipoles. For a linear isotropic material P = ε₀χ_e E = ε₀(ε_r − 1)E, and D = ε₀E + P.

Boundary: z = 0, normal along a_z

Region 1 (z < 0): ε_r1 = 2.5. Region 2 (z > 0): ε_r2 = 4. E₁ = −30 a_x + 50 a_y + 70 a_z V/m.

Conditions: tangential E is continuous; normal D is continuous (no free surface charge), so ε_r1 E_z1 = ε_r2 E_z2.

(a) E₂

E₂ tangential = E₁ tangential : −30 a_x + 50 a_y
E_z2 = (ε_r1/ε_r2) E_z1 = (2.5/4)(70) = 43.75
E₂ = −30 a_x + 50 a_y + 43.75 a_z  V/m

(b) D₂

D₂ = ε₀ε_r2 E₂ = (8.854×10⁻¹²)(4) E₂
   = 35.42×10⁻¹² (−30 a_x + 50 a_y + 43.75 a_z)
   = −1.062 a_x + 1.771 a_y + 1.549 a_z  nC/m²

(Equivalently D₂ = ε₀(−120 a_x + 200 a_y + 175 a_z).)

(c) P₂

P₂ = ε₀(ε_r2 − 1)E₂ = 3 ε₀ E₂ = 26.56×10⁻¹² (−30, 50, 43.75)
   = −0.797 a_x + 1.328 a_y + 1.162 a_z  nC/m²

Answers: E₂ = −30 a_x + 50 a_y + 43.75 a_z V/m; D₂ = −1.06 a_x + 1.77 a_y + 1.55 a_z nC/m²; P₂ = −0.80 a_x + 1.33 a_y + 1.16 a_z nC/m².