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Field between coaxial conductors

Short answer

Using a coaxial Gaussian cylinder of radius ρ (a<ρ<b), Gauss’s law gives D_ρ(2πρL)=ρ_s(2πaL), so E = (aρ_s/(ερ)) a_ρ. Inside the inner conductor and outside the outer conductor, E=0.

Derive the electric field intensity (E) in between the two co-axial cylindrical conductors, the inner of radius 'a' and outer of radius 'b', each infinite in extent and assuming a surface charge density ρ_s on the outer surface of the inner conductor.

Given: Two infinitely long coaxial cylindrical conductors, inner radius a, outer radius b. The outer surface of the inner conductor carries uniform surface charge density ρ_s. Find E for a < ρ < b.

       ╭────────────────╮
      ╱  ╭────────────╮  ╲    Gaussian surface:
     │  │  ●──a       │  │    coaxial cylinder radius ρ, length L
      ╲  ╰────────────╯  ╱    (a < ρ < b)
       ╰───────b────────╯

Symmetry

By symmetry, D has only a radial component that depends only on ρ: D = D_ρ(ρ) a_ρ.

Gaussian surface

Take a coaxial cylinder of radius ρ (a < ρ < b) and length L.

  • Curved surface: D ∥ dS and constant, so ∮ D · dS = D_ρ (2πρL).
  • Two end caps: D ⊥ dS, so no flux.

Enclosed charge

The charge on a length L of the inner conductor's outer surface is

Q_enc = ρ_s (2πaL)

(If the charge per unit length is ρ_L, then ρ_L = 2πaρ_s.)

Apply Gauss's law

D_ρ (2πρL) = ρ_s (2πaL)

D_ρ = a ρ_s / ρ

E = D / ε = ( a ρ_s / (ε ρ) ) a_ρ        (a < ρ < b)

In terms of line charge: E = ρ_L / (2περ) a_ρ.

Remarks:

  • Inside the inner conductor (ρ < a) the field is zero (a conductor in electrostatics). Outside the outer conductor (ρ > b) the field is zero if the two conductors carry equal and opposite charges, because the net enclosed charge is then zero.
  • Integrating E gives the potential difference V_ab = (a ρ_s/ε) ln(b/a), which is used to find the capacitance.

✓ Answer checked by Claude Opus 5.5 (high effort)

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