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Line charge at y=3, z=5, E at three points

Short answer

E = ρ_L/(2πε₀ρ) a_ρ with ρ_L/(2πε₀)=539.3 V. P_A: ρ=5.83 m, E=−47.6 a_y−79.3 a_z V/m. P_B: ρ=5 m, E=64.7 a_y−86.3 a_z V/m. P_C same as P_B.

An infinitely long uniform line charge is located at y = 3, z = 5. If ρ_L = 30 nC/m, find E at (i) P_A(0, 0, 0) (ii) P_B(0, 6, 1) (iii) P_C(5, 6, 1).

The line is infinite and parallel to the x-axis at y = 3, z = 5. Its field is E = ρ_L/(2πε₀ρ) a_ρ, with ρ the perpendicular distance from the line (x does not matter) and a_ρ pointing from the line to the field point.

ρ_L/(2πε₀) = (30×10⁻⁹)(17.975×10⁹) = 539.3 V

(i) P_A(0, 0, 0)

ρ vector = (0−3) a_y + (0−5) a_z = −3 a_y − 5 a_z
ρ = √34 = 5.831 m
E = 539.3/5.831 × (−3 a_y − 5 a_z)/5.831
  = 92.48 × (−0.5145 a_y − 0.8575 a_z)
  = −47.6 a_y − 79.3 a_z  V/m

(ii) P_B(0, 6, 1)

ρ vector = (6−3) a_y + (1−5) a_z = 3 a_y − 4 a_z ,  ρ = 5 m
E = (539.3/5) × (0.6 a_y − 0.8 a_z)
  = 107.9 × (0.6 a_y − 0.8 a_z)
  = 64.7 a_y − 86.3 a_z  V/m

(iii) P_C(5, 6, 1)

Its y and z coordinates equal those of P_B, so the perpendicular distance is the same and x makes no difference for an infinite line:

E = 64.7 a_y − 86.3 a_z  V/m
Pointρ (m)E (V/m)
P_A5.83−47.6 a_y − 79.3 a_z
P_B564.7 a_y − 86.3 a_z
P_C564.7 a_y − 86.3 a_z

✓ Answer checked by Claude Opus 5.5 (high effort)

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