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Potential on axis of charged disk

Short answer

By integrating ring contributions, the axial potential is V = (σ/(2ε₀))[√(z²+R²) − z]. The electric field from E = −dV/dz is E = (σ/(2ε₀))[1 − z/√(z²+R²)] directed away from the disk.

Derive an expression for the electric potential at any point on the axis of a uniformly charged disk. Extend your result to calculate electric field.

Disc of radius R with uniform surface charge density σ. Find V and E at point P on the axis, distance z from the centre.

            P
            .
           /|
        s / | z
         /  |
   -----o---O-----   disc
      ring  R
      (radius w, width dw)

Potential

  1. Divide the disc into thin rings of radius w and width dw. Charge on a ring: dq = σ (2πw dw).
  2. Every point on the ring is at distance s = √(w² + z²) from P, so its potential is

dV = (1/4πε0) dq/s = (1/4πε0) σ 2πw dw / √(w² + z²) = (σ/2ε0) w dw / √(w² + z²)

  1. Integrate w from 0 to R:

V = (σ/2ε0) ∫₀ᴿ w dw/√(w² + z²) = (σ/2ε0) [√(w² + z²)]₀ᴿ

V = (σ/2ε0) [√(z² + R²) − z]

Electric field

Since E = −∇V and by symmetry E is along the axis:

E = −dV/dz = −(σ/2ε0) [ z/√(z² + R²) − 1 ]

E = (σ/2ε0) [1 − z/√(z² + R²)], directed along the axis away from the disc (σ > 0).

Limiting cases

  • R → ∞ (infinite sheet): E = σ/(2ε0), uniform.
  • z ≫ R: expanding gives V → Q/(4πε0 z) and E → Q/(4πε0 z²) with Q = σπR², i.e. a point charge.

✓ Answer checked by Claude Opus 5.5 (high effort)

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