A thin ring of radius R = 0.100 m carries total uniform charge Q = +4.00 nC. A field point lies on its axis x = 0.0500 m from the centre. Starting from Coulomb's law, derive the axial electric field and calculate its magnitude and the potential there, taking V = 0 at infinity. Show the x ≪ R approximation, compare it numerically here, and explain why the transverse field components cancel.

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E_x = kQx/(x²+R²)^(3/2) = 1.29 × 10³ N C⁻¹ away from the ring; V = kQ/√(x²+R²) = 322 V. The small-x approximation gives 1.80 × 10³ N C⁻¹ and is not accurate at x/R = 0.5. Every ring element is the same distance s = √(x²+R²) from the field point. An element dq contributes magnitude k dq/s². Components perpendicular to the axis cancel pairwise for diametrically opposite elements. The axial component is dE_x = (k dq/s²)(x/s) = kx dq/s³. Integrating around the ring gives E_x = kQx/(x²+R²)^(3/2). Substitution yields 1.29 × 10³ N C⁻¹ in the +x direction for positive Q and positive x. Potential adds as a scalar: V = ∫k dq/s = kQ/√(x²+R²) = 322 V. Differentiating −dV/dx reproduces the field expression. For x ≪ R, the denominator is approximately R³, so E_x ≈ kQx/R³ = 1.80 × 10³ N C⁻¹. Here x/R = 0.5 is not very small, and the approximation overestimates the exact field by about 40%.

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