Energy and the Force Between the Plates

The work of charging

Charging a capacitor means carrying charge from one plate to the other. At first the plates are neutral and the first bit of charge moves for free. But once the plates hold \(q\) and \(-q\), there is a potential difference \(v = q/C\) between them. Moving a further small charge \(dq\) against it takes work \(dW = v\,dq = (q/C)\,dq\).

Each new bit of charge costs more than the one before, because the potential difference keeps rising. In Fig 2.1 the potential difference grows along a straight line as the charge builds up. The work for one small \(dq\) is a thin strip under the line. The strips fill the triangle under the line, whose area is \(\tfrac12 QV\).

The factor one half has a plain reason. During charging the potential difference is not \(V\) all the time. It rises from \(0\) to \(V\), so on average each coulomb crosses only \(V/2\). Use \(U = \tfrac12 CV^2\) when the voltage is known, and \(Q^2/2C\) when the charge is fixed.

Half the battery's work is lost

Now look at the battery. It moves the whole charge \(Q\) through its full voltage \(V\), from the start. So it does work \(QV\), the whole rectangle of Fig 2.1. For the \(20\,\text{nF}\) capacitor at \(200\,\text{V}\) that is \(4 \times 10^{-6} \times 200\) \(= 8 \times 10^{-4}\,\text{J}\). Only half of it, \(4 \times 10^{-4}\,\text{J}\), is stored.

The other half becomes heat in the connecting wires, with a little lost as radiation. The surprise is that the resistance of the wires does not matter. A thin wire lets the charge flow slowly with a small current for a long time. A thick wire lets it flow fast with a large current for a short time. The heat comes out the same, half of \(QV\); the chapter on circuits and RC transients proves it.

A charged conductor

A single charged conductor stores energy in the same way, \(U = \tfrac12QV\), with \(V\) its potential. A sphere of radius \(9\,\text{cm}\) holding \(3\,\mu\text{C}\) is at \(V = kQ/R\) \(= 9 \times 10^9 \times 3 \times 10^{-6}/0.09\) \(= 3 \times 10^5\,\text{V}\). It stores \(\tfrac12QV\) \(= 0.45\,\text{J}\). This is the self-energy \(kQ^2/2R\) of a charged shell from the chapter on potential, found now as a capacitor.

The energy lives in the field

Where in a capacitor is the energy kept? Write \(U = \tfrac12CV^2\) for parallel plates, with \(C = \varepsilon_0A/d\) and \(V = Ed\). Then \(U\) \(= \tfrac12(\varepsilon_0A/d)(Ed)^2\) \(= \tfrac12\varepsilon_0E^2 \times Ad\). The product \(Ad\) is the volume of the gap, the only place where the field exists. So each cubic metre of field holds \(\tfrac12\varepsilon_0E^2\).

The result is more general than the parallel plates that gave it. Every electric field carries energy \(\tfrac12\varepsilon_0E^2\) per cubic metre, wherever it is. That is the energy density of the field. So the energy of a capacitor is not on its plates. It is spread through the space between them, where the field is.

The plates pull each other

The positive plate and the negative plate attract. How hard? The field in the gap is \(E = \sigma/\varepsilon_0\), but a plate cannot push or pull on itself. Its own charge makes half of that field, \(\sigma/2\varepsilon_0\). Only the other half, \(E/2\), comes from the other plate and acts on it.

So the force on a plate is its charge times \(E/2\): \(F = \tfrac12QE\). With \(E = Q/\varepsilon_0A\), this is \(F = Q^2/2\varepsilon_0A\). For the plates carrying \(\pm8.85\,\text{nC}\), \(F\) \(= \tfrac12 \times 8.85 \times 10^{-9} \times 10^5\) \(\approx 4.4 \times 10^{-4}\,\text{N}\). That is the weight of a mass of about 44 mg, a couple of grains of rice.

The energy gives the same answer. Keep the charge fixed and let the gap be \(x\). Then \(U = Q^2/2C\) \(= Q^2x/2\varepsilon_0A\), which grows as the gap grows. From the chapter Potential Energy and Conservation of Energy, \(F = -dU/dx\) \(= -Q^2/2\varepsilon_0A\). The minus sign says the force points towards smaller \(x\): the plates attract.

Divide by the area and you get the force on each square metre of plate, \(\sigma^2/2\varepsilon_0\). For the same plates it is \(4.4 \times 10^{-4}/10^{-2}\) \(= 0.044\,\text{N/m}^2\). This is the electrostatic pressure that the chapter Conductors in Electrostatics finds on any charged metal surface. It equals the energy density, \(0.044\,\text{J/m}^3\), because pulling a plate out by \(dx\) creates \(A\,dx\) of new field.

Pulling the plates apart

Charge the plates and then change the gap. What happens depends on one question: is the battery still connected? If it has been removed, the charge has nowhere to go, so \(Q\) stays fixed. If it is connected, the battery holds the potential difference, so \(V\) stays fixed. Every other quantity follows from the one that stays fixed.

Remove the battery and double the gap. The same \(8.85\,\text{nC}\) sits on the plates, so the field stays \(10^5\,\text{V/m}\). The same field across twice the gap gives twice the voltage, \(200\,\text{V}\), and twice the energy, \(0.89\,\mu\text{J}\). The extra energy came from your hands: the plates attract, and you did work pulling them apart.

Keep the battery and double the gap instead. The voltage stays \(100\,\text{V}\), so the field \(V/d\) halves, and so does the charge, to \(4.43\,\text{nC}\). The energy \(\tfrac12CV^2\) halves too, to \(0.22\,\mu\text{J}\). Half the charge flowed back into the battery through its voltage, recharging it.