UY1: Capacitance Of A Cylindrical Capacitor
Key idea: Derive the capacitance of a coaxial cylindrical capacitor and its per-unit-length form using Gauss's law.
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The core idea
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Learning objectives
- Analyse capacitance, resistance, energy transfer, and transient circuit behaviour.
This page gives the UY1 working model/result for Capacitance Of A Cylindrical Capacitor. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.
- Module path: Electromagnetism (UY1)
- Practice: UY1 Electromagnetism Quiz
- Full routing: UY1 Assessment Map
- Math toolkit: Mathematics for Undergraduate Physics
A coaxial capacitor is formed by two long concentric conductors, common in cables and transmission systems.
1) At a glance
- Prerequisites: cylindrical Gauss symmetry (Common Charge Distributions) and log integrals (Integration Techniques)
- Outcomes: derive the capacitance of a long coaxial capacitor and understand why the answer depends on ln(b/a)
- Key result: C = 2πε₀L/(ln(b/a)), C'≡C/L = 2πε₀/(ln(b/a))
- Common trap: sign confusion between Vₐ-V_b and V_b-Vₐ
Motivation / intuition
In cylindrical symmetry, the field falls like 1/r, so the potential difference involves ∫ dr/r, which is why logarithms appear. The per-unit-length form C' is especially useful because real coaxial capacitors are often “long” compared to their radii.
2) Setup
Inner cylinder radius a has linear charge density + λ, outer cylinder inner radius b has -λ, with a < b.
Assume long geometry so edge effects are negligible.
3) Core derivation/explanation
For a < r < b, use cylindrical Gaussian surface (radius r, length L): E(2π rL) = (λ L)/ε₀ E(r) = λ/2πε₀r
Potential difference magnitude: V_ab = Vₐ-V_b = ∫ₐ^b E dr V_ab = λ/2πε₀∫ₐ^bdr/r = λ/2πε₀ ln(b/a)
Total charge on length L is Q = λ L, so C = Q/V_ab = 2πε₀L/(ln(b/a))
Per unit length: C' = C/L = 2πε₀/(ln(b/a))
- Cylinders are long enough that end effects/fringing are negligible.
- Charge resides on conductor surfaces and the field exists mainly in a < r < b.
- Units: [C'] = F/m and [C] = F; ln(b/a) is dimensionless.
- Limits/signs: b → a⁺ ⇒ ln(b/a) → 0⁺ so C' → ∞; scaling both radii by the same factor keeps b/a fixed, so ideal C' is unchanged.
4) Worked example(s)
Given a = 1.0 mm, b = 5.0 mm, L = 2.0 m.
C' = 2πε₀/(ln(5)) ≈ 3.45 × 10⁻¹¹ F/m C = C'L ≈ 6.90 × 10⁻¹¹ F = 69.0 pF
5) Practice set (with hints + answers)
- If b/a increases, what happens to C'?
- If both radii scale up by same factor, does C' change?
- Why do we report both C and C' for cylindrical capacitors?
Hints
- Look at denominator ln(b/a).
- Ratio b/a is unchanged.
- Practical devices often differ mainly in length.
Answers
- C' decreases.
- No, ideal C' stays the same.
- C' is geometry-intrinsic for long coaxial structures; C then scales with L.
6) Summary + next steps
- Gauss’s law gives the 1/r field in coaxial geometry.
- Integrating field gives logarithmic voltage dependence.
- Capacitance depends strongly on radius ratio through ln(b/a).
Next: Transferring Charge And Energy Between Capacitors Previous: Capacitance Of Spherical Capacitor Back To Electromagnetism (UY1)
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Course and syllabus information
- Course
- University Physics Year 1
- Edition
- University Physics Year 1