Coaxial cable charge density, D and E

Given: length L = 0.5 m, inner radius a = 1 mm, outer radius b = 4 mm, air dielectric, charge on the inner conductor Q = +30 nC. The outer conductor carries −30 nC (equal and opposite). Fringing at the ends is neglected.

(i) Surface charge densities

Inner:  ρ_s(a) = Q/(2πaL) = 30×10⁻⁹ /(2π × 10⁻³ × 0.5)
               = 30×10⁻⁹ / 3.1416×10⁻³
               = 9.55×10⁻⁶ C/m²   = 9.55 μC/m²

Outer:  ρ_s(b) = −Q/(2πbL) = −30×10⁻⁹ /(2π × 4×10⁻³ × 0.5)
               = −2.39×10⁻⁶ C/m²  = −2.39 μC/m²

(ii) D between the conductors (a < ρ < b)

Apply Gauss's law with a coaxial cylinder of radius ρ and length L: D(2πρL) = Q.

D = Q/(2πρL) a_ρ = 30×10⁻⁹/(2π × 0.5 × ρ) a_ρ
  = 9.55×10⁻⁹ / ρ  a_ρ   C/m²     (ρ in metres)

Check at ρ = a: D = 9.55×10⁻⁹/10⁻³ = 9.55×10⁻⁶ = ρ_s(a). D is zero for ρ < a and ρ > b (net enclosed charge is zero outside).

(iii) E between the conductors

E = D/ε₀ = 9.55×10⁻⁹ /(8.854×10⁻¹² ρ) a_ρ
  = 1078.5/ρ  a_ρ   V/m

For example, at ρ = a = 1 mm, E = 1.08×10⁶ V/m, and at ρ = b = 4 mm, E = 2.70×10⁵ V/m.

Answer: ρ_s(inner) = 9.55 μC/m², ρ_s(outer) = −2.39 μC/m²; D = 9.55×10⁻⁹/ρ a_ρ C/m²; E = 1078.5/ρ a_ρ V/m.