--- title: "Ring on axis: potential then field" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:ring-axial-potential-then-field" source: https://www.bctengineer.com/q/semester_1/engineering_physics/ring-axial-potential-then-field site: BCT Engineering --- # Ring on axis: potential then field Engineering Physics · Chapter 7: Electrostatics - Asked in: 2073 Chaitra, 5 marks; 2069 Chaitra, 5 marks (asked 2 times) - Video: [Physics 36 The Electric Field (8 of 18) Ring of Charge](https://www.youtube.com/watch?v=Ta1q5_d-7yM) — 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/ring-axial-potential-then-field - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question Derive an expression for the electric potential at a point P at an axial distance x from the center of a ring of radius 'a' and (linear) charge density λ. Hence develop the expression for electric field intensity at the same point. ## Short answer Potential V = λa/(2ε₀√(a²+x²)) = q/(4πε₀√(a²+x²)). Electric field is E = −dV/dx = λ a x / [2ε₀ (a²+x²)^(3/2)] = q x / [4πε₀ (a²+x²)^(3/2)] along the axis. ## Answer **Setup:** a ring of radius *a* carries uniform linear charge density λ (total charge q = 2πaλ). Point P is on the axis at distance x from the centre. ```text P | r = √(a²+x²) | / ----+-/----- ring x ``` #### Potential 1. Take an element of length dl with charge dq = λ dl. 2. All elements are at the same distance r = √(a² + x²) from P. 3. Potential of an element: dV = dq / (4πε₀ r) = λ dl / (4πε₀ √(a² + x²)). 4. Integrate round the ring (total length 2πa): V = [λ / (4πε₀ √(a² + x²))] ∫ dl = λ (2πa) / [4πε₀ √(a² + x²)] **V = λ a / [2ε₀ √(a² + x²)] = q / [4πε₀ √(a² + x²)]** Potential is a scalar, so no direction considerations are needed. #### Electric field from the potential By symmetry, E has only an axial component, E = −dV/dx. ```text E = −d/dx [ (λa/2ε₀)(a² + x²)^(−1/2) ] = −(λa/2ε₀) · (−1/2)(a² + x²)^(−3/2) · 2x ``` **E = λ a x / [2ε₀ (a² + x²)^(3/2)] = q x / [4πε₀ (a² + x²)^(3/2)]** directed along +x (away from the ring) for λ > 0. This agrees with the direct vector integration. At x = 0, E = 0; at large x, E → q/(4πε₀x²). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the potential of a ring simple to calculate at an axial point? A. All elements contribute opposite signs B. Potential components cancel by symmetry C. Only the nearest element contributes D. Potential is a scalar and every element is at the same distance √(a^2 + x^2) Answer: D. V = ∫dq/(4πε0 r) with constant r gives q/(4πε0 √(a^2 + x^2)). (Key point: Distance to elements) 2. What is the potential at P? A. V = q x/[4πε0 (a^2 + x^2)^(3/2)] B. V = q/[4πε0 (a^2 + x^2)] C. V = q/[4πε0 √(a^2 + x^2)] = λa/[2ε0 √(a^2 + x^2)] D. V = q/(4πε0 x) Answer: C. q = 2πaλ and every part of the ring is at distance √(a^2 + x^2). (Key point: Potential result) 3. How is the axial field obtained from V? A. E = +dV/dx B. E = -dV/dx, giving E = qx/[4πε0 (a^2 + x^2)^(3/2)] C. E = V/x D. E = ∫V dx Answer: B. Differentiating (a^2 + x^2)^(-1/2) gives -x(a^2 + x^2)^(-3/2). (Key point: Field from potential) 4. At x = 0 the potential is maximum and E = 0. Why? A. dV/dx = 0 there, and E = -dV/dx B. The charge is zero at the centre C. E = V/x equals zero when V is maximum D. Potential is zero at the centre Answer: A. The slope of V is zero at the centre, so the axial field vanishes. 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