--- title: "Field of charged rod along its axis" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:charged-rod-field-along-axis" source: https://www.bctengineer.com/q/semester_1/engineering_physics/charged-rod-field-along-axis site: BCT Engineering --- # Field of charged rod along its axis Engineering Physics · Chapter 7: Electrostatics - Asked in: 2068 Chaitra, 5 marks - Video: [Electric field of a uniformly charged rod (electric field integral for a uniform line of charge).](https://www.youtube.com/watch?v=2_vr4gL8dls) — Zak's Lab - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/charged-rod-field-along-axis - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question Calculate the electric field due to a uniformly charged rod of length l at a point along its long axis at a distance 'a' from its nearest end. ## Short answer Using linear density λ = Q/l, integrate dE = (1/4πε₀) λ dx/x² from x=a to a+l, giving E = Q/(4πε₀ a(a+l)), directed along the rod away from it. ## Answer Rod of length l, total charge Q, uniform linear density λ = Q/l. Find the field at point P on the line of the rod, a distance *a* from the nearest end. ```text |==========================| . P 0 l <-a-> x axis along rod -> dq at distance x from P ``` Measure distance x from P along the rod, so the rod extends from x = a to x = a + l. #### Steps 1. Take an element of length dx at distance x from P, carrying dq = λ dx. 2. Its field at P is along the rod (all elements are collinear with P, so contributions add directly): dE = (1/4πε0) λ dx / x² 3. Integrate over the rod: E = (λ/4πε0) ∫ₐ^(a+l) dx/x² = (λ/4πε0) [−1/x]ₐ^(a+l) = (λ/4πε0) [1/a − 1/(a+l)] 4. Simplify: 1/a − 1/(a + l) = l / (a(a + l)) E = (λ l) / (4πε0 a (a + l)) 5. Since λ l = Q: **E = Q / (4πε0 a (a + l))**, directed along the rod, away from it for positive Q. Check: for a ≫ l, E → Q/(4πε0 a²), the field of a point charge. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the contributions of all elements be added directly? A. They have equal magnitude B. All lie along the rod, so fields are collinear with P C. Their x-components cancel D. The rod is neutral Answer: B. Every element lies on the line through P, so each dE points along the rod. (Key point: Collinear elements) 2. Which integral gives the field at P with limits a to a+l? A. (λ/4πε0)∫dx/x B. (λ/4πε0)∫x dx C. (λ/4πε0)∫dx/x^3 D. (λ/4πε0)∫dx/x^2 Answer: D. dE = λ dx/(4πε0 x^2), integrated from a to a+l. (Key point: Integral) 3. The final field is A. Q/(4πε0 a(a+l)) B. Q/(4πε0 (a+l)^2) C. Q/(4πε0 a^2) D. Q l/(4πε0 a) Answer: A. 1/a − 1/(a+l) = l/(a(a+l)), and λl = Q. (Key point: Result) 4. What does the result reduce to for a ≫ l? A. Zero B. Q/(4πε0 a l) C. Field of a point charge Q/(4πε0 a^2) D. Q/(2πε0 a) Answer: C. a(a+l) ≈ a^2, so the rod looks like a point charge. 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