Skip to main content

Potential due to sheet and line charge

Short answer

Sheet field is 20 a_x V/m, contribution −20(6−2)=−80 V. Line contribution uses −(ρ_L/2πε₀) ln(ρ_P/ρ_A)=+5.85 V. Total V_P = 10 − 74.15 = −64.2 V.

A uniform sheet of charge ρ_s = 40ε0 C/m² is located in the plane x = 0 in free space. A uniform line charge ρ_L = 0.6 nC/m lies along the line x = 9, y = 4 in free space. Find the potential at point P(6, 8, −3) if V = 10 V at A(2, 9, 3).

Use ε₀ = 8.854×10⁻¹² F/m. Treat the sheet as the infinite plane x = 0 and the line as an infinite line parallel to the z-axis through (x, y) = (9, 4). Both A(2, 9, 3) and P(6, 8, −3) lie on the side x > 0 of the sheet.

Potential difference due to the sheet

E_S = (ρ_s/2ε₀) a_x = (40ε₀/2ε₀) a_x = 20 a_x V/m for x > 0.

V_P − V_A (sheet) = −∫(A→P) E·dl = −20 (x_P − x_A) = −20 (6 − 2) = −80 V

Potential difference due to the line

E_L = ρ_L/(2πε₀ρ) a_ρ, so moving from ρ_A to ρ_P changes the potential by −ρ_L/(2πε₀) ln(ρ_P/ρ_A).

Distance to line:
 A: (2−9, 9−4) = (−7, 5)   ρ_A = √74 = 8.602 m
 P: (6−9, 8−4) = (−3, 4)   ρ_P = 5 m

ρ_L/(2πε₀) = (0.6×10⁻⁹)(17.975×10⁹) = 10.785 V

V_P − V_A (line) = −10.785 ln(5/8.602)
                 = −10.785 (−0.5426)
                 = +5.852 V

Total

V_P − V_A = −80 + 5.85 = −74.15 V
V_P = V_A − 74.15 = 10 − 74.15 = −64.15 V

Answer: V_P ≈ −64.1 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electromagnetism chapter 2 questions →