- 2067 Mangsir · 6 marks
Potential from point and line charge
Short answer
Point charge contribution (Q/4πε₀)(1/R_P−1/R_ref)=−3.80 V. Line charge contribution −(ρ_L/2πε₀) ln(ρ_P/ρ_ref)=+8.24 V. Total V_P = 4.44 V.
Find the potential at a point P(2, 3, 3) due to a 1 nC charge located at Q(3, 4, 4) and a 1 nC/m uniform line charge located at x = 2, y = 1, if V = 0 V at (3, 4, 5).
Use ε₀ = 8.854×10⁻¹² F/m. The reference point is V = 0 at (3, 4, 5). The line (x = 2, y = 1) is parallel to the z-axis. Add the potentials of the two sources, each measured relative to the reference point.
Point charge Q = 1 nC at (3, 4, 4)
V = (Q/4πε₀)(1/R_P − 1/R₀).
R_P : from (3,4,4) to P(2,3,3): (−1,−1,−1) → √3 = 1.732 m
R₀ : from (3,4,4) to (3,4,5): (0,0,1) → 1 m
Q/(4πε₀) = (10⁻⁹)(8.988×10⁹) = 8.988 V·m
V₁ = 8.988 (1/1.732 − 1/1) = 8.988 (0.5774 − 1) = −3.80 V
Line charge ρ_L = 1 nC/m at x = 2, y = 1
V = −(ρ_L/2πε₀) ln(ρ_P/ρ₀) relative to the reference radius.
ρ_P : from line to P(2,3): (0, 2) → 2 m
ρ₀ : from line to (3,4): (1, 3) → √10 = 3.162 m
ρ_L/(2πε₀) = (10⁻⁹)(17.975×10⁹) = 17.975 V
V₂ = −17.975 ln(2/3.162) = −17.975 (−0.4581) = +8.24 V
Total
V_P = V₁ + V₂ = −3.80 + 8.24 = 4.44 V
Answer: V_P ≈ 4.44 V.
✓ Answer checked by Claude Opus 5.5 (high effort)
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