Learning objective: Derive the electric-field expression for a coaxial geometry and calculate capacitance per metre from material properties.
Understand the idea
A coaxial cable has a central conductor and an outer cylindrical conductor separated by a dielectric. Because the geometry is cylindrically symmetric, Gauss’s law turns a three-dimensional field problem into a one-variable radial calculation. We neglect end effects and treat the cable as long compared with its radius.
Worked example: given data
- Inner conductor radius a = 1 mm and inner radius of outer conductor b = 5 mm.
- Relative permittivity εr = 2.5, with vacuum permittivity ε₀ ≈ 8.854 × 10⁻¹² F/m.
- The potential difference between conductors is 100 V. Calculate capacitance per metre and electric field at r = 2 mm.
Detailed solution
Apply Gauss’s law
For a cylindrical Gaussian surface of radius r and length L between the conductors, E(2πrL) = λL/ε, where λ is charge per unit length. Thus E(r) = λ/(2πεr), valid for a < r < b.
Relate voltage and charge
Integrate E(r) radially: V = ∫ab E(r)dr = (λ/(2πε))ln(b/a). Rearranging gives the capacitance per unit length C′ = λ/V = 2πε/ln(b/a).
Substitute dimensions and dielectric
ε = ε₀εr = 2.2135 × 10⁻¹¹ F/m. Since ln(5/1) = ln 5 ≈ 1.609, C′ ≈ 86.4 pF/m. The logarithm uses a ratio of radii expressed in the same units.
Find the electric field at r = 2 mm
Eliminate λ using the voltage relation: E(r) = V/[r ln(b/a)]. At r = 0.002 m, E ≈ 100/[0.002 × 1.609] ≈ 31.1 kV/m.
Explain where the field is strongest
E(r) decreases as 1/r, so the largest field in the dielectric occurs at the inner conductor surface, r = a. This is important when checking dielectric breakdown limits.
What the answer means
The dielectric permittivity affects the capacitance at a fixed geometry. When voltage is specified and the dielectric is uniform, the expression E = V/[r ln(b/a)] is independent of ε; the required charge changes instead. This distinction often helps eliminate incorrect multiple-choice answers.
Three mistakes to avoid
- Using a spherical Gauss surface for a cylindrical cable.
- Confusing the radius of the outer conductor with the thickness of the dielectric.
- Treating the ratio b/a as having units or forgetting to convert millimetres when calculating E.
Try a related question
Keeping the same cable and dielectric, what happens to capacitance per metre if b becomes 10 mm?
Show the worked-answer check
C′ becomes 2πε/ln(10) ≈ 60.4 pF/m. Increasing the separation reduces the capacitance.
How to revise this concept
Practise parallel-plate and spherical capacitor calculations and compare the appropriate symmetry. First attempt the example on paper without opening the answer, then compare the intermediate steps. Explain each assumption before memorising a formula.
Content note: This is a newly prepared practice example, not an official GATE previous-year question. It was drafted with AI assistance for this educational website. Students should verify calculations and exam requirements against standard textbooks and official sources; educator review is recommended before publication.