Capacitance per unit length of a coaxial cable Two coaxial metallic cylinders have inner radius R1 and outer radius R2. The space between them is uniformly filled with a dielectric of relative permittivity εr. What is the capacitance per metre length?
-
AC′ = 2 * π * ε0 * εr / ln(R2 / R1)
-
BC′ = π * ε0 * εr / ln(R2 / R1)
-
CC′ = 2 * π * ε0 * εr * ln(R2 / R1)
-
DC′ = ε0 * εr / (R2 − R1)
Answer
Correct Answer: C′ = 2 * π * ε0 * εr / ln(R2 / R1)
Explanation
Introduction / Context:The capacitance per unit length of a coaxial line is a staple result in electromagnetics and transmission line theory. It follows from Gauss’s law under cylindrical symmetry with a homogeneous dielectric between the conductors.
Given Data / Assumptions:
- Inner conductor radius R1, outer conductor inner radius R2 (R2 > R1).
- Dielectric is linear, homogeneous, isotropic with permittivity ε = ε0 * εr.
- Length considered is 1 metre (per-unit-length capacitance C′).
Concept / Approach:
Use Gauss’s law in integral form over a cylindrical surface to find the electric field E(r). Integrate the potential difference V between R1 and R2. The per-unit-length capacitance is C′ = Q / V for a 1 m length. The result contains a logarithm due to the 1/r dependence of E in cylindrical coordinates.
Step-by-Step Solution:
Choose Gaussian surface of radius r, length 1 m: ∮ D · dA = Q_free ⇒ D_r * (2 * π * r * 1) = Q ⇒ D_r = Q / (2 * π * r).Relate fields: D = εE ⇒ E(r) = D_r / ε = Q / (2 * π * ε * r).Potential: V = ∫_{R1}^{R2} E(r) dr = Q / (2 * π * ε) * ln(R2 / R1).Capacitance per metre: C′ = Q / V = 2 * π * ε / ln(R2 / R1) = 2 * π * ε0 * εr / ln(R2 / R1).Verification / Alternative check:
Dimensional analysis: ε has units F/m; logarithm is dimensionless; result has units F/m as required. For εr = 1, this reduces to the vacuum formula.
Why Other Options Are Wrong:
- Halving the numerator (π instead of 2π) underestimates C′ by 2.
- Multiplying by ln instead of dividing reverses the dependence.
- Using (R2 − R1) ignores cylindrical field variation (not a parallel-plate capacitor).
Common Pitfalls:
Confusing natural log with log base 10; forgetting ε = ε0 * εr; using diameter rather than radius.
Final Answer:
C′ = 2 * π * ε0 * εr / ln(R2 / R1)