Field on axis of charged disc

Field on the axis of a ring

For a ring of radius s carrying charge dq, at a point on its axis at distance x from the centre,

E = dq · x / [4πε₀ (s² + x²)^(3/2)]

(directed along the axis; the perpendicular components cancel by symmetry).

Field on the axis of a disc

A uniformly charged disc of radius a has surface charge density σ.

         P
         |  x
   ------+------   disc, radius a
   ring of radius s, width ds

1. Divide the disc into thin concentric rings. A ring of radius s and width ds has area 2πs ds and charge dq = σ 2πs ds.
2. Its axial field at P (distance x from the disc): dE = σ (2πs ds) x / [4πε₀ (s² + x²)^(3/2)] = σ x s ds / [2ε₀ (s² + x²)^(3/2)]
3. Integrate from s = 0 to s = a. Let u = s² + x², so s ds = du/2: E = (σx / 2ε₀) ∫ s ds / (s² + x²)^(3/2) = (σx / 4ε₀) ∫ u^(−3/2) du = (σx / 2ε₀) [ −(s² + x²)^(−1/2) ]₀ᵃ = (σx / 2ε₀) [ 1/x − 1/√(a² + x²) ]
4. Result:

E = (σ / 2ε₀) [ 1 − x / √(x² + a²) ] (along the axis, away from the disc for σ > 0)

Special cases

• Infinite plane sheet (a → ∞): E → σ / 2ε₀.
• Far away (x ≫ a): expand x/√(x² + a²) ≈ 1 − a²/2x², so E ≈ σa² / 4ε₀x² = Q / 4πε₀x², where Q = σπa². The disc acts as a point charge.
• At x → 0⁺: E → σ/2ε₀.