UY1: Capacitance Of Spherical Capacitor

Key idea: Derive the capacitance of concentric spherical conductors and interpret useful limits of the formula.

  • University Physics Year 1
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Learning objectives

  • Analyse capacitance, resistance, energy transfer, and transient circuit behaviour.
Why this matters + quick links

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.

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)

Quick checks (units + limits/sign)
  • 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)

  1. If b → ∞, what does spherical-capacitor formula reduce to?
  2. If b-a decreases, does C increase or decrease?
  3. Why is field outside outer conductor ideally zero?

Hints

  1. Take the limit in 4πε₀ab/(b-a).
  2. Inspect denominator.
  3. Use spherical symmetry plus zero net enclosed charge outside both shells.

Answers

  1. C → 4πε₀a (isolated sphere).
  2. Increases.
  3. 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)

Continue with the next resource in this course.

Course and syllabus information
Course
University Physics Year 1
Edition
University Physics Year 1