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Sheet and line charge, E at origin

Short answer

Superpose fields at origin. Line charge at y=3 gives E_L = (ρ_L/(2πε₀·3))(−a_y) = −17.98 a_y V/m. Sheet at x=2 gives E_S = (ρ_s/(2ε₀))(−a_x) = −11.29 a_x V/m. Total E = −11.3 a_x − 18.0 a_y V/m.

Let a uniform line charge density of 3 nC/m be at y = 3 and a uniform surface charge density of 0.2 nC/m² be at x = 2. Find E at the origin.

Assumptions: the line charge at y = 3 is an infinite line parallel to the z-axis (x = 0, y = 3), and the sheet at x = 2 is the infinite plane x = 2. Superpose the two fields at the origin. Take ε₀ = 8.854×10⁻¹² F/m.

Field of the line charge

E_L = ρ_L/(2πε₀ρ) a_ρ. The distance from the line to the origin is ρ = 3 m, and the field points away from the line, from (0,3) towards the origin, which is the −a_y direction.

|E_L| = (3×10⁻⁹)/(2π × 8.854×10⁻¹² × 3)
      = (3×10⁻⁹)(17.98×10⁹)/3
      = 17.98 V/m
E_L = −17.98 a_y V/m

Field of the sheet

E_S = (ρ_s/2ε₀) a_n. The origin lies at x < 2, so the normal pointing from the sheet to the origin is −a_x.

|E_S| = 0.2×10⁻⁹/(2 × 8.854×10⁻¹²) = 11.29 V/m
E_S = −11.29 a_x V/m

Total

E = E_S + E_L = −11.29 a_x − 17.98 a_y V/m
|E| = √(11.29² + 17.98²) = 21.2 V/m

Answer: E = −11.3 a_x − 18.0 a_y V/m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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