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 = σπ*R²/(4π*e0*x²)
Ex = σ/(4*k*e0)*(R/x)²
Reference: demonstrations.wolfram.com/AxialEl...