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Line charge and point charge, find E

Short answer

Total E is the vector sum of line and point charge fields. Line: using ρ_L/(2πε₀ρ) with ρ=5 m gives 4.31 a_y − 5.75 a_z V/m. Point: using Q/(4πε₀R²) with R from (0,0,1) to (2,3,−4) gives 0.614 a_x + 0.921 a_y − 1.535 a_z V/m. Result: E = 0.61 a_x + 5.23 a_y − 7.29 a_z V/m.

An infinite uniform line charge ρ_L = 2 nC/m lies along the x-axis in free space, while the point charge of 8 nC is located at (0, 0, 1). Find E at (2, 3, −4).

The total field is the vector sum of the field of the infinite line charge and the field of the point charge (superposition). Use ε₀ = 8.854×10⁻¹² F/m, so 1/(4πε₀) = 8.99×10⁹ and 1/(2πε₀) = 17.98×10⁹.

Field of the line charge (along the x-axis)

For a line charge, E = ρ_L/(2πε₀ρ) a_ρ, where ρ is the perpendicular distance from the line, measured from the axis to the field point (2, 3, −4).

ρ vector = 0 a_x + 3 a_y − 4 a_z ,  ρ = √(9+16) = 5 m
a_ρ = (3 a_y − 4 a_z)/5 = 0.6 a_y − 0.8 a_z
|E_L| = (2×10⁻⁹)(17.98×10⁹)/5 = 7.19 V/m
E_L = 7.19 (0.6 a_y − 0.8 a_z) = 4.31 a_y − 5.75 a_z V/m

Field of the point charge at (0, 0, 1)

R = (2,3,−4) − (0,0,1) = 2 a_x + 3 a_y − 5 a_z
|R| = √(4+9+25) = √38 = 6.164 m
|E_P| = Q/(4πε₀R²) = (8×10⁻⁹)(8.99×10⁹)/38 = 1.892 V/m
a_R = (2 a_x + 3 a_y − 5 a_z)/6.164
E_P = 1.892 (0.324 a_x + 0.487 a_y − 0.811 a_z)
    = 0.614 a_x + 0.921 a_y − 1.535 a_z V/m

Total field

E = E_L + E_P
  = (0 + 0.614) a_x + (4.314 + 0.921) a_y + (−5.752 − 1.535) a_z
  = 0.61 a_x + 5.23 a_y − 7.29 a_z V/m

Answer: E = 0.61 a_x + 5.23 a_y − 7.29 a_z V/m (magnitude about 9.0 V/m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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