--- title: "Field on axis of charged disc" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:disc-axial-field" source: https://www.bctengineer.com/q/semester_1/engineering_physics/disc-axial-field site: BCT Engineering --- # Field on axis of charged disc Engineering Physics · Chapter 7: Electrostatics - Asked in: 2069 Chaitra, 5 marks; 2068 Chaitra, 5 marks; 2067 Asoj, 5 marks (asked 3 times) - Video: [Physics 36 The Electric Field (9 of 18) Disc of Charge](https://www.youtube.com/watch?v=i68CGhgGkxs) — 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/disc-axial-field - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question Write an expression for electric field at any point on the axial line of a charged ring. Using this equation, calculate the electric field at any point on the axial line of a charged disc (circular disc of radius a, uniform charge). ## Short answer For a ring of radius s, axial field dE = (dq x)/[4πε₀(s²+x²)^(3/2)]. For a uniformly charged disc, integrate with dq = σ·2πs ds from 0 to a to get E = (σ/2ε₀)[1 − x/√(x²+a²)] along the axis. ## Answer #### Field on the axis of a ring For a ring of radius s carrying charge dq, at a point on its axis at distance x from the centre, E = dq · x / [4πε₀ (s² + x²)^(3/2)] (directed along the axis; the perpendicular components cancel by symmetry). #### Field on the axis of a disc A uniformly charged disc of radius *a* has surface charge density σ. ```text P | x ------+------ disc, radius a ring of radius s, width ds ``` 1. Divide the disc into thin concentric rings. A ring of radius s and width ds has area 2πs ds and charge dq = σ 2πs ds. 2. Its axial field at P (distance x from the disc): dE = σ (2πs ds) x / [4πε₀ (s² + x²)^(3/2)] = σ x s ds / [2ε₀ (s² + x²)^(3/2)] 3. Integrate from s = 0 to s = a. Let u = s² + x², so s ds = du/2: E = (σx / 2ε₀) ∫ s ds / (s² + x²)^(3/2) = (σx / 4ε₀) ∫ u^(−3/2) du = (σx / 2ε₀) [ −(s² + x²)^(−1/2) ]₀ᵃ = (σx / 2ε₀) [ 1/x − 1/√(a² + x²) ] 4. Result: **E = (σ / 2ε₀) [ 1 − x / √(x² + a²) ]** (along the axis, away from the disc for σ > 0) #### Special cases - Infinite plane sheet (a → ∞): E → σ / 2ε₀. - Far away (x ≫ a): expand x/√(x² + a²) ≈ 1 − a²/2x², so E ≈ σa² / 4ε₀x² = Q / 4πε₀x², where Q = σπa². The disc acts as a point charge. - At x → 0⁺: E → σ/2ε₀. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What charge does a thin ring of radius s and width ds on the disc carry? A. dq = σ(πs^2) B. dq = σ(2πs) C. dq = σ ds D. dq = σ(2πs ds) Answer: D. The ring's area is 2πs ds, multiplied by surface density σ. (Key point: Ring element) 2. How is the disc's axial field built from the ring formula? A. Integrate from x = 0 to x = a with s fixed B. Integrate from s = a to infinity C. Integrate the ring field from s = 0 to s = a with x fixed D. Multiply one ring's field by the number of rings Answer: C. Each ring contributes dE at the same point P, and the rings fill s from 0 to a. (Key point: Integration limits) 3. What is the field on the axis of the disc? A. E = (σ/2ε0)[1 + x/√(x^2 + a^2)] B. E = (σ/2ε0)[1 - x/√(x^2 + a^2)] C. E = σ x/(2ε0 √(x^2 + a^2)) D. E = (σ/ε0)[1 - x/(x^2 + a^2)] Answer: B. Integration with u = s^2 + x^2 gives (σx/2ε0)[1/x - 1/√(a^2 + x^2)]. (Key point: Result) 4. What does E become as a -> infinity (infinite sheet)? A. σ/2ε0, independent of x B. σ/ε0 C. Zero D. Q/(4πε0 x^2) Answer: A. x/√(x^2 + a^2) -> 0, leaving σ/2ε0. 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