As z grows large above a uniformly charged disk, what happens to the axial field E_z?

As z grows large above a uniformly charged disk, what happens to the axial field E_z?

• A. E_z → σ/(2 ε0)
• B. E_z → σ/ε0
• C. E_z → 0
• D. E_z → σ/(4π ε0)

Answer: (C)

The axial field from a finite uniformly charged disk weakens with distance and must vanish far away because the disk carries a finite total charge. If you slice the disk into coaxial rings and sum their contributions, you get the on-axis field as E_z = (σ/(2ε0)) [1 − z/√(z^2 + R^2)]. As z becomes very large, √(z^2 + R^2) ≈ z, so z/√(z^2 + R^2) ≈ 1 and the bracket goes to zero. That means E_z → 0 at large distances. Intuitively, the disk behaves like a point charge Q = σπR^2 at far distances, and the field on the axis falls off as 1/z^2, also tending to zero.

The axial field from a finite uniformly charged disk weakens with distance and must vanish far away because the disk carries a finite total charge. If you slice the disk into coaxial rings and sum their contributions, you get the on-axis field as E_z = (σ/(2ε0)) [1 − z/√(z^2 + R^2)]. As z becomes very large, √(z^2 + R^2) ≈ z, so z/√(z^2 + R^2) ≈ 1 and the bracket goes to zero. That means E_z → 0 at large distances. Intuitively, the disk behaves like a point charge Q = σπR^2 at far distances, and the field on the axis falls off as 1/z^2, also tending to zero.