--- title: "Field on axis of charged ring" subject: "Engineering Physics" chapter: 7 key: "semester_1/engineering_physics:ring-axial-field-maximum" source: https://www.bctengineer.com/q/semester_1/engineering_physics/ring-axial-field-maximum site: BCT Engineering --- # Field on axis of charged ring Engineering Physics · Chapter 7: Electrostatics - Asked in: 2072 Chaitra, 5 marks; 2071 Chaitra, 5 marks; 2070 Asar, 5 marks; 2069 Asar, 5 marks (asked 4 times) - Video: [22. Electric field due to uniformly charged ring | Electrostatics | #cbse 12th Physics](https://www.youtube.com/watch?v=2ehgNmGkpLM) — PHYSICS with Umesh Rajoria - 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-field-maximum - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-7 ## Question Derive an expression for the electric field at any point on the axis of a uniformly charged ring (radius R, charge q). Find the point at which the electric field is maximum. ## Short answer On the axis of a uniformly charged ring, E = qx/[4πε₀(R²+x²)^(3/2)] directed away from the centre. The field is maximum at x = R/√2 with magnitude E_max = q/(6√3π ε₀R²). ## Answer **Setup:** a ring of radius R carries total charge q uniformly. Find E at a point P on the axis at distance x from the centre. ```text dq . | . r = √(R² + x²) R | . | . θ O-------------------P ----> E (along axis) |<-------- x ------>| | ring seen edge-on (radius R) ``` #### Derivation 1. Divide the ring into small elements of charge dq. Each element is at distance r = √(R² + x²) from P. 2. Field of one element: dE = dq / (4πε₀ r²), directed along the line from the element to P. 3. Resolve dE into a component along the axis, dE cos θ, and a perpendicular component, dE sin θ, where cos θ = x/r. 4. By symmetry, elements on opposite sides of the ring give equal and opposite perpendicular components, which cancel. Only the axial components add up. 5. E = ∫ dE cos θ = (1/4πε₀) (x / r³) ∫ dq = q x / [4πε₀ (R² + x²)^(3/2)] **E = q x / [4πε₀ (R² + x²)^(3/2)]** (along the axis, away from the ring if q > 0) Checks: at x = 0, E = 0; for x ≫ R, E ≈ q/(4πε₀x²), the field of a point charge. #### Point of maximum field Set dE/dx = 0: ```text d/dx [ x (R² + x²)^(-3/2) ] = (R² + x²)^(-3/2) − 3x² (R² + x²)^(-5/2) = 0 R² + x² = 3x² x = R / √2 ``` The second derivative is negative there, so this is a maximum. **Maximum field:** ```text E_max = q (R/√2) / [4πε₀ (3R²/2)^(3/2)] = q / (6√3 π ε₀ R²) ≈ 0.385 q / (4πε₀R²) ``` **Answer: E = qx / [4πε₀(R² + x²)^(3/2)], maximum at x = R/√2 (on either side of the ring) with E_max = q / (6√3 π ε₀ R²).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do only the axial components of dE add up for a charged ring? A. The perpendicular components are zero for each element B. The axial components cancel instead C. Perpendicular components of opposite elements are equal and opposite and cancel D. Charge is concentrated on the axis Answer: C. Opposite elements give equal and opposite perpendicular parts, so only the cos θ parts survive. (Key point: Symmetry) 2. Where does the factor x/(R^2 + x^2)^(3/2) in E = qx/[4πε0 (R^2 + x^2)^(3/2)] come from? A. dE multiplied by sin θ = R/r B. dE = dq/(4πε0 r^2) multiplied by cos θ = x/r C. dE = dq/(4πε0 r) with no angle factor D. The integral of dq over x Answer: B. Each element's field is dq/(4πε0 r^2); its axial part is multiplied by x/r, and the integral of dq is q. (Key point: Origin of the x factor) 3. At what axial distance is the field of the ring maximum? A. x = R/√2 B. x = R C. x = 0 D. x = R√2 Answer: A. Setting dE/dx = 0 gives R^2 + x^2 = 3x^2, hence x = R/√2. (Key point: Point of maximum field) 4. What does E become for x much larger than R? A. Zero B. q/(4πε0 R^2) C. Proportional to x D. q/(4πε0 x^2), the field of a point charge Answer: D. For x >> R, (R^2 + x^2)^(3/2) ≈ x^3 so E ≈ q/(4πε0 x^2). 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