Capacitance from Field Theory
Deriving C = Q/V from geometry, not from a datasheet
You have used capacitors as circuit elements before—two-terminal devices with a value printed on the side, measured in farads. But where does that number come from? It is not a property of the material alone; it is a property of the geometry. In this lesson we derive capacitance from the electric field, starting from Gauss's law and ending with exact formulas for three canonical shapes. By the end, you will see every capacitor datasheet value as a consequence of shape and size.
Capacitance: A Geometric Quantity
Take two conductors, place charge +Q on one and −Q on the other, and measure the potential difference V between them. The ratio C = Q/V is the capacitance. What makes this remarkable is that C depends only on the geometry—the shapes, sizes, and separation of the conductors—and on the permittivity of whatever sits between them. It does not depend on Q or V individually. Double the charge and the voltage doubles too; the ratio stays fixed.
This is why a capacitor's value is stamped on its case rather than measured at runtime: the geometry was fixed at manufacture. In your circuits course you treated capacitance as a given constant. Here we open the box and derive it. The strategy is always the same three-step recipe:
Step 1: Assume charge ±Q on the conductors and use Gauss's law to find the electric field E in the region between them.
Step 2: Integrate E along a path from the negative conductor to the positive conductor to obtain the potential difference V = −∫E·dl.
Step 3: Form the ratio C = Q/V. The charge Q cancels, leaving an expression in terms of geometry and ε₀ only.
We now apply this recipe to three canonical geometries, all in vacuum (or air, which is nearly the same). Dielectrics—materials with permittivity greater than ε₀—come in the next lesson.
Parallel-Plate Capacitor
Two flat conducting plates of area A, separated by a small gap d, carry charges +Q and −Q. If d is small compared to the plate dimensions, edge effects are negligible and the field between the plates is uniform. Gauss's law with a pillbox straddling the positive plate gives E = σ/ε₀ = Q/(ε₀A), directed from the positive to the negative plate. The potential difference is simply V = Ed = Qd/(ε₀A), and the capacitance follows immediately.
The result is the most recognizable capacitor formula in engineering. It says: bigger plates store more charge per volt, and a smaller gap strengthens the field for the same charge, also raising capacitance. Every parallel-plate capacitor you have ever used in a circuit lab traces back to this single line of field theory.
- Use C = ε₀A/d with ε₀ = 8.854 × 10⁻¹² F/m.
- Substitute: C = (8.854 × 10⁻¹²)(0.02) / (0.001).
- C = (8.854 × 10⁻¹²)(20) = 1.7708 × 10⁻¹⁰ F.
- Convert to picofarads: C = 177.08 pF.
Coaxial Cable
A coaxial cable has an inner conductor of radius a and an outer conductor of radius b, length L, with charge +Q on the inner and −Q on the outer. By cylindrical symmetry, Gauss's law gives the field in the region a < r < b as E = Q/(2πε₀Lr), pointing radially outward. Integrating from a to b gives V = (Q/(2πε₀L)) ln(b/a), and the capacitance is the clean expression below.
As a quick worked example: take a = 1 mm, b = 5 mm, L = 2 m. Then C = 2π(8.854 × 10⁻¹²)(2) / ln(5) = (1.113 × 10⁻¹⁰) / 1.6094 = 6.915 × 10⁻¹¹ F = 69.15 pF. Notice that most of the cable's capacitance comes from the small inner radius—the closer the conductors, the larger the field and the more charge per volt.
Concentric Spheres
Our final geometry is two concentric spherical shells: an inner sphere of radius a carrying +Q and an outer sphere of radius b carrying −Q. The field between them, by spherical symmetry and Gauss's law, is E = Q/(4πε₀r²) for a < r < b. Integrating from a to b gives V = Q/(4πε₀) · (1/a − 1/b) = Q(b − a)/(4πε₀ab), and the capacitance emerges as the compact expression below. Note that as b → ∞, this reduces to C = 4πε₀a—the capacitance of an isolated sphere, a useful sanity check.
Every capacitance derivation in this lesson followed the same path: (1) assume ±Q, find E with Gauss's law; (2) integrate E to get V; (3) compute C = Q/V and watch Q cancel. The geometry enters through the Gaussian surface (planar, cylindrical, or spherical), and the permittivity ε₀ enters through Gauss's law. When we introduce dielectrics in the next lesson, the only change will be replacing ε₀ with ε = ε₀εr—a single substitution that multiplies every capacitance by the relative permittivity εr.
- Substitute a = 0.02 m, b = 0.03 m into C = 4πε₀ab/(b − a).
- Numerator: 4π × 8.854 × 10⁻¹² × (0.02 × 0.03) = 4π × 8.854 × 10⁻¹² × 6 × 10⁻⁴.
- Denominator: b − a = 0.03 − 0.02 = 0.01 m.
- C = (4π × 8.854 × 10⁻¹² × 6 × 10⁻⁴) / 0.01 = 6.676 × 10⁻¹² F = 6.676 pF.
Check your understanding
- Capacitance C = Q/V is a geometric quantity: it depends only on conductor shapes, sizes, separation, and the permittivity of the medium—not on Q or V individually.
- Parallel-plate: C = ε₀A/d. Larger plates or smaller gaps increase capacitance.
- Coaxial cable: C = 2πε₀L/ln(b/a). The logarithmic dependence on the radius ratio means small inner conductors dominate the capacitance per unit length.
- Concentric spheres: C = 4πε₀ab/(b−a). As the outer sphere grows to infinity, this reduces to C = 4πε₀a, the capacitance of an isolated sphere.