---
title: "Coaxial cable charge density, D and E"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:coax-cable-charge-density-d-e"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/coax-cable-charge-density-d-e
site: BCT Engineering
---

# Coaxial cable charge density, D and E

Electromagnetism · Chapter 2: Electric field

- Asked in: 2067 Shrawan, 5 marks
- Video: [Finding the electric field everywhere in a charged coaxial cable using Gauss's law Ex 12716](https://www.youtube.com/watch?v=bpqZ7pGmo1A) — Marx Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/coax-cable-charge-density-d-e
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Consider a co-axial cable of length 50 cm having inner radius of 1 mm and an outer radius of 4 mm with the space between the conductors filled with air. Total charge on the inner conductor is 30 nC. Find (i) the charge density on the inner conductor and outer conductor (ii) D (iii) E.

## Short answer

Inner surface charge density 9.55 μC/m², outer −2.39 μC/m². Displacement field D = 9.55×10⁻⁹/ρ a_ρ C/m². Electric field E = 1078.5/ρ a_ρ V/m.

## Answer

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
```text
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.

```text
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
```text
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.

## Check yourself

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

1. Why is the outer conductor's |ρ_s| smaller than the inner one's?
   A. It carries less charge
   B. It has the same charge −Q spread over a larger area (2πbL > 2πaL)
   C. The dielectric reduces it
   D. The outer conductor is negative so smaller
   Answer: B. Charge is equal and opposite, but ρ_s = Q/(area); b = 4a gives a quarter of the magnitude: −2.39 μC/m² versus 9.55 μC/m². (Key point: Surface charge density of conductors)

2. Using a coaxial Gaussian cylinder between the conductors, D equals:
   A. Q/(4πρ²)
   B. Q/(2πρL) a_ρ
   C. Q/(2πL)
   D. ρ_s ρ
   Answer: B. D(2πρL) = Q gives D = Q/(2πρL) a_ρ = 9.55×10^−9/ρ C/m². (Key point: D via Gauss)

3. What is D for ρ > b?
   A. 9.55×10^−9/ρ
   B. Q/(2πbL)
   C. Zero, because net enclosed charge is zero
   D. Infinite
   Answer: C. The inner +Q and outer −Q cancel, so enclosed charge is zero. (Key point: Field outside)

4. For air between the conductors, E is:
   A. D ε0
   B. D/ε0 = 1078.5/ρ a_ρ V/m
   C. D/μ0
   D. constant between the conductors
   Answer: B. E = D/ε0 = 9.55×10^−9/(8.854×10^−12 ρ) = 1078.5/ρ V/m, falling as 1/ρ. (Key point: E from D)

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