Line and point charges, find D

Given: Line charge ρ_L = 2 nC/m along the x-axis; point charges 8 nC at (0, 0, 1) and (0, 0, −1). Find D at P(2, 3, −4). Use the superposition principle.

1. Line charge (along x-axis)

The perpendicular distance from the x-axis to P is in the yz-plane: R_L = 3 a_y − 4 a_z, |R_L| = 5.

D_L = ρ_L / (2π ρ) · a_ρ = 2 / (2π × 5) × (3 a_y − 4 a_z)/5   nC/m²
    = 0.06366 (0.6 a_y − 0.8 a_z)
    = 0.0382 a_y − 0.0509 a_z   nC/m²

2. Point charge at (0, 0, 1)

R₁ = P − (0, 0, 1) = 2 a_x + 3 a_y − 5 a_z,   |R₁| = √38 = 6.164
|R₁|³ = 234.2
D₁ = Q R₁ / (4π |R₁|³) = 8 (2, 3, −5) / (4π × 234.2)  nC/m²
   = 2.718 × 10⁻³ (2, 3, −5)  nC/m²

D₁ = (0.00544 a_x + 0.00815 a_y − 0.01359 a_z)  nC/m²

3. Point charge at (0, 0, −1)

R₂ = P − (0, 0, −1) = 2 a_x + 3 a_y − 3 a_z,   |R₂| = √22 = 4.690
|R₂|³ = 103.2
D₂ = 8 (2, 3, −3) / (4π × 103.2) = 6.17 × 10⁻³ × (2, 3, −3)  nC/m²
   = (0.01234 a_x + 0.01851 a_y − 0.01851 a_z)  nC/m²

4. Total

D = D_L + D₁ + D₂
D_x = 0 + 0.00544 + 0.01234 = 0.01777
D_y = 0.0382 + 0.00815 + 0.01851 = 0.0649
D_z = −0.0509 − 0.01359 − 0.01851 = −0.0830

Answer: D = 0.0178 a_x + 0.0649 a_y − 0.0830 a_z nC/m², i.e. D = 17.8 a_x + 64.9 a_y − 83.0 a_z pC/m². (|D| ≈ 0.107 nC/m².)