--- title: "Field at center of bent charged rod" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:plastic-rod-arc-field-center" source: https://www.bctengineer.com/q/semester_1/engineering_physics/plastic-rod-arc-field-center site: BCT Engineering --- # Field at center of bent charged rod Engineering Physics · Chapter 7: Electrostatics - Asked in: 2073 Shrawan, 5 marks - Video: [Physics 36 Electric Field (11 of 18) An Arc of Charge](https://www.youtube.com/watch?v=p-W2mKDgbbs) — Michel van Biezen - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/plastic-rod-arc-field-center - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question A plastic rod contains uniformly distributed Q charge. The rod has been bent in a 120° circular arc of radius 'r' as shown in the figure [Figure: a circular arc of 120° subtending its centre, radius r, centre at the origin, arc symmetric about the horizontal axis]. Prove that the electric field intensity at the center of the bent rod is E = 0.83Q/(4πε0r²). ## Short answer Linear charge density λ = 3Q/(2πr). By symmetry only the x‑component survives: E = (λ/4πε₀r) ∫ cosθ dθ from −60° to 60° = (λ√3)/(4πε₀r). Substituting λ gives E = (3√3/(2π)) Q/(4πε₀r²) ≈ 0.83 Q/(4πε₀r²). ## Answer Let the arc of radius r subtend 120° at the centre O (take Q positive), symmetric about the x-axis, running from θ = −60° to θ = +60°. ```text y | . | . (arc, 120 deg) -------O-------)-----> x | . | . each end at 60 deg from the x-axis ``` #### Steps 1. Arc length L = r × (2π/3) = 2πr/3. Linear charge density λ = Q/L = 3Q/(2πr). 2. Take an element dl = r dθ at angle θ, carrying dq = λ r dθ. 3. Its field at O has magnitude dE = (1/4πε0) dq/r² = (1/4πε0) λ dθ / r. For positive Q it points away from the element, along the line from the element through O. 4. By symmetry the y-components of elements at +θ and −θ cancel. Only the x-component survives: dEₓ = dE cosθ = (λ/4πε0 r) cosθ dθ 5. Integrate from −60° to +60°: Eₓ = (λ/4πε0 r) ∫ cosθ dθ = (λ/4πε0 r) [sinθ]₋₆₀°⁺⁶⁰° = (λ/4πε0 r)(2 sin 60°) = (λ/4πε0 r)√3 6. Substitute λ = 3Q/(2πr): E = (1/4πε0) · (3√3 / 2π) · Q / r² 7. 3√3/(2π) = 5.196/6.283 = 0.827 ≈ 0.83 **E = 0.83 Q/(4πε0 r²)**, directed along the axis of symmetry, away from the middle of the arc (for positive Q). Proved. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the linear charge density for charge Q on a 120° arc of radius r? A. λ = Q/(2πr/3) = 3Q/(2πr) B. λ = Q/(2πr) C. λ = Q/(πr) D. λ = Q/r Answer: A. A 120° arc has length r x 2π/3. (Key point: Charge density) 2. Which field components at the centre cancel? A. The components along the symmetry axis B. Both components C. None; they all add D. The components perpendicular to the symmetry axis (y), from elements at +θ and -θ Answer: D. Only dE cos θ along the axis survives. (Key point: Symmetry) 3. What is ∫cos θ dθ from -60° to +60°? A. 2 cos 60° = 1 B. sin 120° = 0.5 C. 2 sin 60° = √3 D. 0 Answer: C. [sin θ] from -60° to +60° is 2 sin 60° = √3. (Key point: Integral) 4. What numerical factor multiplies Q/(4πε0 r^2) in the field at the centre? A. 3/(2π) ≈ 0.48 B. 3√3/(2π) ≈ 0.83 C. √3 ≈ 1.73 D. 2π/(3√3) ≈ 1.21 Answer: B. E = (λ/4πε0 r)√3 with λ = 3Q/(2πr). 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