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Uniform line charge, find E at P

Short answer

The perpendicular distance from the line to P is 5 m, and the unit radial vector is 0.6a_y−0.8a_z. Field magnitude is (20×10⁻⁹)/(2πε₀·5)=71.9 V/m, so E = 71.9(0.6a_y−0.8a_z) = 43.1a_y − 57.5a_z V/m.

A uniform line charge density of 20 nC/m is located at y = 3 and z = 5. Find E at P(5, 6, 1).

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

ρ vector from line to P(5,6,1): R_⊥ = (6−3) a_y + (1−5) a_z = 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| = ρ_L/(2πε₀ρ)
    = (20×10⁻⁹)(17.975×10⁹)/5
    = 71.9 V/m

E = 71.9 (0.6 a_y − 0.8 a_z)
  = 43.1 a_y − 57.5 a_z  V/m

Answer: E = 43.1 a_y − 57.5 a_z V/m (magnitude 71.9 V/m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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