--- title: "E due to line charge using Gauss's law" subject: "Electromagnetism" chapter: 2 key: "semester_3/electromagnetism:line-charge-field-gauss-law" source: https://www.bctengineer.com/q/semester_3/electromagnetism/line-charge-field-gauss-law site: BCT Engineering --- # E due to line charge using Gauss's law Electromagnetism · Chapter 2: Electric field - Asked in: 2072 Chaitra, 4 marks; 2067 Mangsir, 6 marks; 2067 Shrawan, 6 marks (asked 3 times) - Video: [2.3 ELECTRIC FIELD DUE TO LINE CHARGE for IES,GATE](https://www.youtube.com/watch?v=8yYdxf0y5Bs) — EPOV CHANNEL - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electromagnetism/line-charge-field-gauss-law - Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2 ## Question Derive the expression for the electric field intensity due to an infinitely long line charge with uniform charge density ρ_L using Gauss's law. ## Short answer Using Gauss's law with a coaxial cylindrical Gaussian surface of radius ρ and length L, symmetry gives D_ρ (2πρL) = ρ_L L, so D_ρ = ρ_L/(2πρ). The electric field intensity is E = ρ_L/(2πε₀ρ) a_ρ V/m. ## Answer **Gauss's law:** The total electric flux through any closed surface equals the free charge enclosed: ∮ D · dS = Q_enc. #### Setup Consider an infinitely long straight line of charge along the z-axis with uniform linear density ρ_L (C/m) in a medium of permittivity ε₀. By symmetry (the line is infinite and uniform): - D (and E) has only a radial component, D = D_ρ a_ρ. - D_ρ depends only on the distance ρ from the line, not on φ or z. ```text z │ ▲ ┌───────────┐ │ │ │ Gaussian │ closed cylinder: │ │ │ surface │ radius ρ, length L ● ─ ─┼─ ─ ─ ─ ─ ─┼──→ ρ │ └───────────┘ │ (D points radially outward) ``` #### Gaussian surface Choose a coaxial closed cylinder of radius ρ and length L. - On the curved side, D is normal to the surface and constant in magnitude, so D · dS = D_ρ dS. - On the two end caps, D is parallel to the surface (D · dS = 0). ```text ∮ D · dS = D_ρ ∫ dS = D_ρ (2πρL) ``` Enclosed charge: ```text Q_enc = ρ_L L ``` #### Apply Gauss's law ```text D_ρ (2πρL) = ρ_L L D_ρ = ρ_L / (2πρ) E = D / ε₀ = ρ_L / (2π ε₀ ρ) a_ρ V/m ``` **Result: E = ρ_L / (2πε₀ρ) a_ρ** The field is radial, directed away from a positive line charge, and decreases as 1/ρ. For a line not on the z-axis, ρ is the perpendicular distance and a_ρ the unit vector perpendicular to the line. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does symmetry tell us about D for an infinite line charge? A. D has only a radial component depending only on distance ρ from the line B. D has only a z-component C. D depends on z D. D is zero Answer: A. The line is infinite and uniform, so D = D_ρ a_ρ and D_ρ depends only on ρ, not on φ or z. (Key point: Symmetry) 2. Which Gaussian surface is chosen, and what is the flux through the end caps? A. A sphere; flux is zero B. A coaxial cylinder of radius ρ and length L; the end caps carry no flux since D is parallel to them C. A cube; flux is equal on all faces D. A cylinder where flux flows through the caps Answer: B. D is radial, so it is normal to the curved side and parallel to the end caps, giving D·dS = 0 on the caps. (Key point: Gaussian surface) 3. What equation results from applying Gauss's law? A. D_ρ (2πρL) = ρ_L L B. D_ρ (4πρ²) = ρ_L L C. D_ρ (πρ²L) = ρ_L D. D_ρ (2πρ) = ρ_L L Answer: A. The flux is D_ρ times the curved area 2πρL and the enclosed charge is ρ_L L. (Key point: Flux and charge) 4. What is E? A. ρ_L/(4πε₀ρ²) a_ρ B. ρ_L/(2πε₀ρ) a_ρ, falling as 1/ρ C. ρ_L/(2πε₀) a_ρ, constant D. ρ_L ρ/(2πε₀) a_ρ Answer: B. D_ρ = ρ_L/(2πρ) and E = D/ε₀, giving a field that decreases as 1/ρ. 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