Ring on axis: potential then field

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.

        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.

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).