UY1: Capacitance Of Spherical Capacitor
Key idea: Derive the capacitance of concentric spherical conductors and interpret useful limits of the formula.
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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 Spherical 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 spherical capacitor consists of two concentric conducting shells separated by an insulating region.
1) At a glance
- Prerequisites: Gauss’s law (Gauss’s Law (Simple Version)) and potential from field (Electric Potential)
- Outcomes: derive the capacitance of two concentric spherical conductors and interpret useful limits
- Key result: C = 4πε₀ab/(b-a) (a < b)
- Common trap: integrating potential with wrong limits/signs
- Physical check: as gap shrinks, capacitance increases
Motivation / intuition
Spherical geometry is where Gauss’s law is at its cleanest: symmetry gives E(r) immediately. Once you can integrate E to get Δ V, the capacitance drops out as a pure geometry constant.
2) Setup
Inner conductor radius a carries + Q, outer conductor inner radius b carries -Q, with a < b.
Field exists only for a < r < b (ideal electrostatic case).
3) Core derivation/explanation
For a < r < b, spherical Gaussian surface gives E(4π r²) = Q/ε₀ E(r) = (1/4πε₀)Q/r²
Potential difference magnitude between shells: V_ab = V(a)-V(b) = ∫ₐ^b E dr V_ab = ∫ₐ^b(Q/4πε₀r²)dr = (Q/4πε₀)(1/a-1/b)
Therefore, C = Q/V_ab = 4πε₀ab/(b-a)
- Units: [4πε₀ab/(b-a)] = (F/m)m = F.
- Limits/signs: b → ∞ ⇒ C → 4πε₀a (isolated sphere); b → a⁺ ⇒ C → ∞; C is always positive for a < b.
4) Worked example(s)
Let a = 0.040 m and b = 0.060 m.
C = 4πε₀ab/(b-a) = 4π(8.854 × 10⁻¹²)(0.040)(0.060)/0.020 C ≈ 1.34 × 10⁻¹¹ F = 13.4 pF
5) Practice set (with hints + answers)
- If b → ∞, what does spherical-capacitor formula reduce to?
- If b-a decreases, does C increase or decrease?
- Why is field outside outer conductor ideally zero?
Hints
- Take the limit in 4πε₀ab/(b-a).
- Inspect denominator.
- Use spherical symmetry plus zero net enclosed charge outside both shells.
Answers
- C → 4πε₀a (isolated sphere).
- Increases.
- For r > b, spherical symmetry makes E uniform on a Gaussian sphere and Q_encl = 0, so E = 0.
6) Summary + next steps
- Gauss’s law gives E(r) in the gap, then integration gives Δ V.
- The concentric spherical capacitor has C = 4πε₀ab/(b-a)
- Limiting cases provide good checks against algebra errors.
Next: Capacitance Of A Cylindrical Capacitor Previous: Energy Stored In Capacitors Back To Electromagnetism (UY1)
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Course and syllabus information
- Course
- University Physics Year 1
- Edition
- University Physics Year 1