---
title: "Field between coaxial conductors"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:coax-cylinders-field-and-line-charge"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/coax-cylinders-field-and-line-charge
site: BCT Engineering
---

# Field between coaxial conductors

Electromagnetism · Chapter 2: Electric field

- Asked in: 2075 Asoj, 4 marks
- Video: [KTU Electromagnetics Module-2 | Capacitance of co-axial cables](https://www.youtube.com/watch?v=N5_W2XJ_kEo) — 100 Million Thoughts
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/coax-cylinders-field-and-line-charge
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

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.

## 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.

## Answer

**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.

```text
       ╭────────────────╮
      ╱  ╭────────────╮  ╲    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

```text
Q_enc = ρ_s (2πaL)
```

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

#### Apply Gauss's law

```text
D_ρ (2πρL) = ρ_s (2πaL)

D_ρ = a ρ_s / ρ
```

```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does symmetry tell us about D between the cylinders?
   A. D has only a z-component
   B. D is uniform
   C. D depends on φ
   D. D has only a radial component depending only on ρ
   Answer: D. Infinite coaxial geometry gives D = D_ρ(ρ) a_ρ. (Key point: Symmetry)

2. Why do the end caps of the Gaussian cylinder carry no flux?
   A. D is zero everywhere
   B. Their area is zero
   C. D is perpendicular to dS there (D lies radially, caps have axial normal)
   D. They are inside the conductor
   Answer: C. D is radial, so D · dS = 0 on the caps. (Key point: Gaussian surface)

3. The charge enclosed in length L of the Gaussian surface is:
   A. ρ_s (2πρL)
   B. ρ_s (2πaL)
   C. ρ_s (πa²L)
   D. ρ_s L
   Answer: B. Only the inner conductor's surface of radius a carries charge. (Key point: Enclosed charge)

4. The field between the cylinders is:
   A. E = a ρ_s / (ε ρ) a_ρ
   B. E = ρ_s / ε
   C. E = a ρ_s / (ε ρ²)
   D. E = ρ ρ_s / (ε a)
   Answer: A. D_ρ(2πρL) = ρ_s(2πaL) gives D = aρ_s/ρ, E = D/ε. (Key point: Result)

## Related in Electromagnetism

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- [Capacitance of coaxial capacitor](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-capacitance-laplace-cylindrical) — Electromagnetism, chapter 2, asked 2×
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- [∇×H in coaxial cable regions](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acurl-h-in-coax-regions) — Electromagnetism, chapter 3

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