Two conductors separated by an insulating region can carry equal and opposite charges while maintaining a potential difference between them. A device designed around this charge separation is a capacitor.

The capacitance is defined by

$$\boxed{C=\frac{Q}{|\Delta V|}},$$

where $Q$ is the magnitude of the charge on either conductor and $|\Delta V|$ is the magnitude of the potential difference between them.

The SI unit is the farad:

$$1,\mathrm F=1,\mathrm{C/V}.$$

For an ideal linear electrostatic system, capacitance depends on the conductor geometry and the material between the conductors, not on the particular values of $Q$ and $\Delta V$ used to measure it.

Parallel-plate capacitor

Consider two large parallel conducting plates of area $A$, separated by distance $d$ in vacuum. Let the plates carry surface charge densities $+\sigma$ and $-\sigma$.

Ignoring edge effects, each ideal infinite sheet produces field magnitude

$$\frac{\sigma}{2\varepsilon_0}.$$

Between the oppositely charged plates the two fields point in the same direction and add, giving

$$\boxed{E=\frac{\sigma}{\varepsilon_0}}.$$

Outside the plates, the ideal sheet fields cancel.

Because

$$\sigma=\frac{Q}{A},$$

we have

$$E=\frac{Q}{\varepsilon_0A}.$$

The field between ideal large plates is uniform, so the magnitude of the potential difference is

$$|\Delta V|=Ed =\frac{Qd}{\varepsilon_0A}.$$

Therefore

$$C=\frac{Q}{|\Delta V|} =\boxed{\varepsilon_0\frac{A}{d}}.$$

This result explains the geometry directly:

• increasing plate area gives more surface over which charge can be distributed for the same voltage, so $C$ increases;
• increasing the separation requires a larger potential difference for the same field and charge, so $C$ decreases.

Worked example

A vacuum parallel-plate capacitor has

$$A=0.020,\mathrm{m^2}$$

and

$$d=1.0\times10^{-3},\mathrm m.$$

Its capacitance is

$$C=(8.85\times10^{-12})\frac{0.020}{1.0\times10^{-3}} \approx1.77\times10^{-10},\mathrm F.$$

Thus

$$\boxed{C\approx177,\mathrm{pF}}.$$

If the voltage magnitude is $100,\mathrm V$, the charge magnitude on each plate is

$$Q=C|\Delta V| \approx(1.77\times10^{-10})(100) =1.77\times10^{-8},\mathrm C.$$

Beyond ideal parallel plates

Different conductor geometries have different capacitances. Spherical and cylindrical capacitors, for example, have nonuniform fields and require the corresponding field-potential calculation.

Real finite plates also have fringing fields near their edges, so $C=\varepsilon_0A/d$ is most accurate when the plate dimensions are much larger than the separation.

Capacitance describes how strongly a geometry relates separated charge to voltage. How capacitors combine, how much energy their fields store, and how dielectrics change the capacitance are separate concepts built on this definition.