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Capacitance

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Capacitance measures charge stored per volt

A capacitor is any pair of conductors carrying equal and opposite charges ±Q. The potential difference V between them is proportional to Q — double the charge and the field everywhere doubles, so the line integral giving V doubles too. The ratio is therefore a constant of the hardware: the capacitance C = Q/V, measured in farads (1 F = 1 C/V). C depends only on the sizes, shapes, and separation of the conductors (and any dielectric between them) — never on Q or V individually. A farad is enormous; real capacitors live in the pF–mF range.

Definition of capacitance
C = Q / V (1 F = 1 C/V)
Q is the magnitude of the charge on either plate; V is the potential difference between the plates.
Worked example

Derive the capacitance of a parallel-plate capacitor (plate area A, separation d, vacuum gap) from Gauss’s law, then evaluate for A = 0.10 m², d = 1.0 mm, and find the charge stored at 100 V.

  1. 1.Put ±Q on the plates, giving surface densities ±σ with σ = Q/A. For d small compared to the plate size, the field between the plates is uniform.
  2. 2.Apply Gauss’s law with a pillbox through the positive plate’s inner surface (conductor boundary condition): E = σ/ε₀ = Q/(ε₀A), directed from + plate to − plate. Outside the gap the fields cancel.
  3. 3.Integrate for the potential difference across the uniform field: V = ∫→E·d→l = E·d = Qd/(ε₀A).
  4. 4.Divide: C = Q/V = ε₀A/d. Q cancels, confirming capacitance is pure geometry — bigger plates raise C, wider gaps lower it.
  5. 5.Numerically: C = (8.85 × 10⁻¹²)(0.10)/(1.0 × 10⁻³) = 8.9 × 10⁻¹⁰ F ≈ 0.89 nF. At 100 V: Q = CV = 8.9 × 10⁻⁸ C ≈ 89 nC.
Answer: C = ε₀A/d ≈ 0.89 nF; Q ≈ 89 nC at 100 V
Parallel-plate capacitor
C = ε₀A / d, with ε₀ = 8.85 × 10⁻¹² F/m
Valid when d is much smaller than the plate dimensions, so fringing fields at the edges are negligible.

A recipe that works for any geometry

The parallel-plate derivation is one instance of a universal four-step recipe: (1) assume charges ±Q on the conductors; (2) find E between them, usually via Gauss’s law; (3) integrate V = ∫→E·d→l from one conductor to the other; (4) form C = Q/V — the Q always cancels. The same steps give C = 4πε₀·ab/(b − a) for concentric spheres and C = 2πε₀L/ln(b/a) for a length-L coaxial cable. Even a single conductor has capacitance, defined relative to V = 0 at infinity.

Worked example

Find the capacitance of an isolated conducting sphere of radius R, then estimate the capacitance of the Earth (R = 6.4 × 10⁶ m).

  1. 1.Step 1–2: put charge Q on the sphere; outside, it behaves as a point charge, so the potential of the surface (relative to infinity) is V = kQ/R.
  2. 2.Step 3–4: C = Q/V = Q/(kQ/R) = R/k = 4πε₀R. Again Q cancels — capacitance is set by the radius alone.
  3. 3.For the Earth: C = (6.4 × 10⁶)/(8.99 × 10⁹) = 7.1 × 10⁻⁴ F.
  4. 4.Perspective: the entire planet is only ≈ 710 μF — a supercapacitor from an electronics catalog beats it. This is why the farad is such a huge unit.
Answer: C = 4πε₀R; Earth ≈ 7.1 × 10⁻⁴ F ≈ 710 μF
On the exam

The capacitance derivation is a guaranteed FRQ pattern — commit the recipe: (1) assume ±Q, (2) Gauss’s law for E, (3) V = ∫→E·d→l between the conductors, (4) C = Q/V. Show the Q canceling; that line is what proves C is geometric. Key results: C = ε₀A/d (plates), C = 4πε₀R (sphere).

Checkpoint

The plate separation of an air-filled parallel-plate capacitor is doubled (nothing else changes). Its capacitance:

Checkpoint

A capacitor holds 20 μC of charge when the potential difference across it is 5.0 V. Its capacitance is:

Answer the 2 checkpoints as you read.

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