Field on axis of charged ring

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.

          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:

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:

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).