A parallel plate capacitor.

Any pair of conductors has a capacitance that can be calculated in the
same way we found the capacitance of the two shells in the previous
chapter. The most famous two-piece capacitor consists of two
identical sheets of metal, each with area \(A\), that are separated by
a distance \(d\). This is called a parallel-plate
capacitor. The plates are usually drawn as squares, but the
shape doesn't matter so long as they are identical and overlap one another.
Find the capacitance of a parallel-plate capacitor. (Hint: the
electric field between the plates is roughly the same as the field of
two infinite sheets.)

Just as before, we suppose that we move charge from one plate to the other, so that the positive plate has charge \(+Q\) and the negative plate has charge \(-Q\). Now we need to find the potential difference between the plates, and we do that by finding the electric field due to each plate in the space between them.

Field of two infinite plates

If the plates are close together, then the electric field in the gap between the plates is roughly the same as if the plates were infinitely large, especially towards the center of the plates. In an earlier chapter we found that the electric field of an infinite sheet with charge density \(\sigma\) is \(2\pi k\sigma\). The top, positive plate has a charge density of \(Q/A\), and so it creates an electric field of \(2\pi kQ/A\) downward (i.e. away from the plate, because it is positive). The bottom, negative plate has a charge density of \(-Q/A\), and it creates an electric field of \(2\pi kQ/A\), also downward because the field points towards the negative plate. Therefore the net electric field between the two plates is \(4\pi kQ/A\) pointing downward, and it is uniform.

Integrating to find the potential difference

To find the potential difference between the two plates, we integrate the electric field from the negative plate to the positive plate. To do that, we define a coordinate axis \(z\) which runs from \(z=0\) at the bottom plate to \(z=d\) at the top plate. In this coordinate system \(\vec E=4\pi kQ/A(-\hat z)\), and so \(E_z=-4\pi kQ/A\). We can then write $$\Dl V=-\int_0^d (-4\pi kQ/A)\,dz =4\pi kQ/A \int_0^d\,dz = 4\pi kQd/A$$ and the capacitance is $$C={Q\over \Dl V}=\frac{Q}{4\pi kQd/A}=\frc{4\pi k}{A\over d}$$

$$ C=\epsilon_0\frac{A}{d} $$

where
$$\epsilon_0=\frc{4\pi k}=8.85\ten{-12}\u{F/m}$$
is a fundamental constant.
Theoretically, we can make the capacitance of this configuration as big as we want, just by moving the plates really close together: just as with the two spherical shells in the previous section , the capacitance is inversely proportional to the gap between the pieces. In practice, however, there is a limit to how close we can put the plates together. The atoms in a metal plate are separated by a few angstroms. If I can place another metal plate within this distance from the first, then for practical purposes the two plates become a single conductor, and the charges will equilibrate: positive charge will flow from the positive plate to the negative plate. Electrons may be able to jump the gap between the plates even before this happens. And this assumes that the two plates are perfectly flat and smooth; a typical metal surface has bumps on the order of microns which will come in contact with each other even before the bulk of the plates touch.

Suppose we want to create a 1 Farad capacitor out of two square metal plates
that are a nanometer apart (which is probably unrealistically small).
How long are the plates on each side?

We want a \(C=1\u{F}\) capacitor with \(d=10^{-9}\u{m}\) separation.
The area of the plates must be
$$C=\epsilon_0\frac{A}{d} \implies A=\frac{dC}{\epsilon_0}
=\frac{(10^{-9}\u{m})(1\u{F})}{8.85\ten{-12}\u{F/m}}=113\u{m^2}$$
A square with this area must have a side \(\sqrt{113\u{m^2}}=\boxed{10.6\u{m}}\)
long. That's the size of a small classroom: not astronomical, but not
terribly realistic for consumer electronics!