Capacitance and Its Geometries

Charge and potential grow together

In the chapter Electric Potential, Energy and the Dipole, the potential of a charge distribution was a sum of \(kq/r\) over its pieces. Double every piece of charge and every term doubles, so the potential doubles too. Put a charge \(Q\) on an isolated metal body and it spreads over the surface in a pattern fixed by the shape. Give it \(2Q\) and the same pattern holds twice the charge everywhere.

So for any isolated conductor the potential \(V\), measured from zero at infinity, is proportional to its charge \(Q\). The ratio \(Q/V\) does not depend on how much charge is there. It depends only on the size and shape of the conductor. This ratio is its capacitance, \(C\).

The same holds for two conductors that carry equal and opposite charges, \(+Q\) and \(-Q\). Such a pair is a capacitor, and each conductor is called a plate. Now \(V\) means the potential difference between the plates. Doubling \(Q\) doubles the field everywhere between them, and so doubles \(V\). The ratio \(C = Q/V\) is again fixed by the geometry.

The unit of capacitance is the coulomb per volt, called the farad (F) after Michael Faraday. One farad is a very large capacitance, as you will see in a moment. So capacitors are marked in microfarads (1 μF \(= 10^{-6}\) F), nanofarads (1 nF \(= 10^{-9}\) F) and picofarads (1 pF \(= 10^{-12}\) F).

An isolated sphere

A metal sphere of radius \(R\) holding charge \(Q\) spreads it evenly over its surface. In the chapter Electric Potential, Energy and the Dipole you found its potential: \(V = kQ/R\), where \(k = 1/4\pi\varepsilon_0\). So its capacitance is \(C = Q/V = R/k\), or \(4\pi\varepsilon_0R\). It grows in proportion to the radius and has nothing to do with the charge.

A sphere of radius \(9\,\text{cm}\) has \(C\) \(= 0.09/(9 \times 10^9)\) \(= 10^{-11}\,\text{F}\) \(= 10\,\text{pF}\). Even the whole Earth, radius \(6400\,\text{km}\), has only \(6.4 \times 10^6/(9 \times 10^9)\) \(\approx 7.1 \times 10^{-4}\,\text{F}\), about \(711\,\mu\text{F}\). A sphere of one farad would need \(R = kC\) \(= 9 \times 10^9\,\text{m}\). That is about 1400 times the radius of the Earth, which shows how large a farad is.

The parallel-plate capacitor

The simplest capacitor is two flat metal plates of area \(A\), facing each other a small distance \(d\) apart. Connect them to a battery and one plate takes \(+Q\) while the other takes \(-Q\). These charges attract each other, so they sit on the two facing surfaces, as the chapter Conductors in Electrostatics shows. Each facing surface then carries a surface charge density \(\sigma = Q/A\).

In the chapter Electric Flux and Gauss's Law you met two large sheets with \(+\sigma\) and \(-\sigma\). Between them their fields add to \(E = \sigma/\varepsilon_0\), pointing from the positive plate to the negative one. Outside the pair the two fields cancel. So the whole field of the capacitor lives in the gap, and it is uniform there, as Fig 1.1 shows.

The capacitance grows with the area and falls with the gap. A larger plate holds more charge at the same field. A narrower gap needs a smaller potential difference for the same field. So doubling \(d\) halves \(C\), and doubling \(A\) doubles it.

Run the same capacitor the other way. Give its plates \(\pm 8.85\,\text{nC}\). Then \(\sigma\) \(= 8.85 \times 10^{-9}/10^{-2}\) \(= 8.85 \times 10^{-7}\,\text{C/m}^2\), and \(E\) \(= \sigma/\varepsilon_0\) \(= 10^5\,\text{V/m}\) in the gap and zero outside. The potential difference is \(V = Ed\) \(= 10^5 \times 10^{-3}\) \(= 100\,\text{V}\). As a check, \(Q/V\) \(= 8.85\,\text{nC}/100\,\text{V}\) \(= 88.5\,\text{pF}\) again.

How big would a one-farad capacitor of this kind be? With a \(1\,\text{mm}\) gap, \(A = Cd/\varepsilon_0\) \(= 10^{-3}/(8.85 \times 10^{-12})\) \(\approx 1.13 \times 10^8\,\text{m}^2\). That is a square more than \(10\,\text{km}\) on a side, bigger than many towns. Real capacitors of large value need much thinner gaps and something better than air between the plates.

Fringing at the edges

Near the ends of the plates the field lines bow outward instead of running straight across. This is fringing. The field there is weaker and not uniform, and a little of it spreads outside the gap. When the gap is much smaller than the plates, the edges hold a tiny share of the field and \(C = \varepsilon_0A/d\) is accurate. The bowed lines store a little extra field, so the true capacitance is slightly larger than the formula.

The spherical capacitor

Put a metal sphere of radius \(a\) inside a concentric metal shell of inner radius \(b\), as on the left of Fig 1.2. Give the sphere \(+Q\) and the shell \(-Q\), which gathers on its inner face. Between them the field is that of the sphere alone, \(kQ/r^2\), pointing outward. By Gauss's law there is no field inside the sphere, and none outside the shell, where the enclosed charge is zero.

Take \(a = 9\,\text{cm}\) and \(b = 10\,\text{cm}\). Then \(C = ab/k(b - a)\) \(= 0.09 \times 0.1/(9 \times 10^9 \times 0.01)\) \(= 10^{-10}\,\text{F}\) \(= 100\,\text{pF}\). The same sphere alone had \(10\,\text{pF}\). An earthed shell only \(1\,\text{cm}\) away makes it ten times larger. The shell's \(-Q\) pulls the potential of the sphere down, so the sphere holds more charge per volt.

Now let the shell grow without limit, \(b \to \infty\). Then \(ab/(b - a) \to a\), and \(C \to 4\pi\varepsilon_0 a\): the formula of the isolated sphere. An isolated sphere is a spherical capacitor whose second plate is at infinity. When the gap is narrow, \(b - a = d\) is small and \(ab \approx a^2\), so \(C \approx 4\pi\varepsilon_0 a^2/d\) \(= \varepsilon_0 A/d\), with \(A\) the area of the sphere. The parallel-plate result comes back.

The cylindrical capacitor

A long rod of radius \(a\) inside a coaxial tube of radius \(b\), length \(L\), is a cylindrical capacitor. Give the rod a charge \(\lambda\) per metre. From the chapter Electric Flux and Gauss's Law, the field between them is \(E = \lambda/2\pi\varepsilon_0 r\). Adding \(E\,dr\) from \(a\) to \(b\) gives \(V\) \(= (\lambda/2\pi\varepsilon_0)\ln(b/a)\). With \(Q = \lambda L\), the capacitance is \(C = 2\pi\varepsilon_0 L/\ln(b/a)\).

Choose \(b = 2.72\,a\), so that \(\ln(b/a) = 1\). Then each metre has \(C\) \(= 2\pi \times 8.85 \times 10^{-12}\) \(\approx 55.6\,\text{pF}\). The cable that brings a television signal into many homes is built like this. It has a central copper wire, an insulating layer and a braided metal tube. Every metre of such a cable has a capacitance of some tens of picofarads, which the engineers must allow for.

The fan capacitor of the opening can now be read. Its \(2.5\,\mu\text{F}\) means \(2.5\,\mu\text{C}\) of charge for every volt. At a steady \(220\,\text{V}\) it would hold \(Q = CV\) \(= 550\,\mu\text{C}\); on the mains the voltage swings to and fro, as the chapter on alternating current shows. With air plates \(1\,\text{mm}\) apart, \(A = Cd/\varepsilon_0\) \(\approx 282\,\text{m}^2\), the floor of a large house. The real box holds two long strips of metal film separated by a plastic film a few micrometres thick, rolled up tight. The thin gap does most of the work; Lesson 5 shows why the plastic does the rest.