The electric field along the axis of a uniformly charged disk of radius R and total charge Q is given below.

Ex = 2πkeσ(1 − x/(x2 + R2)1/2

Show that the electric field at distances x that are large compared with R approaches that of a particle with charge

Q = σπR2.

Suggestion: First show that

x/(x2 + R2)^1/2 = (1 + R2/x2)^−1/2,

and use the binomial expansion

(1 + δ)n ≈ 1 + nδ

when

δ « 1.

## Answers (1)

Your equation for the axial electric field is not correct. It should be

Ex = σ/(2*k*e0)*(1 − x/(x² + R²)^½)

x/(x² + R²)^½ divide numerator and denominator by x:

1/(1 + R²/x²)^½ = (1 + R²/x²)^-½

when R/x is small, (1 + R²/x²)^-½ ≈ 1 - ½(R/x)²

Ex = σ/(2*k*e0)*[1 - 1 + ½(R/x)²]

Ex = σ/(4*k*e0)*(R/x)²

The field from a particle of charge Q at distance x is Q/(4*π*k*e0*x²)

Q = σ*π*R²

Ex = σ

π*R²/(4π*e0*x²)Ex = σ/(4*k*e0)*(R/x)²

Reference:demonstrations.wolfram.com/AxialEl...