FIGURE 27.5 shows a capacitor charged to $\pm 16\text{nC}$ as it is being discharged by a 2.0-mm-diameter, 20-cm-long copper wire. How long does it take to discharge the capacitor? We’ve noted that a fairly typical drift speed of the electron current through a wire is ${10}^{-4}\text{m}\text{/}\text{s}$. At this rate, it would take 2000 s, or about a half hour, for an electron to travel 20 cm.

FIGURE 27.5 How long does it take to discharge this capacitor?

But this isn’t what happens. As far as our senses are concerned, the discharge of a capacitor is instantaneous. So what’s wrong with our simple calculation?

The important point we overlooked is that the wire is already full of electrons. As an analogy, think of water in a hose. If the hose is already full of water, adding a drop to one end immediately (or very nearly so) pushes a drop out the other end. Likewise with the wire. As soon as the excess electrons move from the negative capacitor plate into the wire, they immediately (or very nearly so) push an equal number of electrons out the other end of the wire and onto the positive plate, thus neutralizing it. We don’t have to wait for electrons to move all the way through the wire from one plate to the other. Instead, we just need to slightly rearrange the charges on the plates and in the wire.

Let’s do a rough estimate of how much rearrangement is needed and how long the discharge takes. Using the conduction-electron density of copper in Table 27.1, we can calculate that there are $5\times {10}^{22}$ conduction electrons in the wire. The negative plate in FIGURE 27.6, with $Q=-16\text{nC,}$ has ${10}^{11}$ excess electrons, far fewer than in the wire. In fact, the length of copper wire needed to hold ${10}^{11}$ electrons is a mere $4\times {10}^{-13}\text{m}$.

FIGURE 27.6 The sea of electrons needs only a minuscule rearrangement.

The instant the wire joins the capacitor plates together, the repulsive forces between the excess ${10}^{11}$ electrons on the negative plate cause them to push their way into the wire. As they do, ${10}^{11}$ electrons are squeezed out of the final $4\times {10}^{-13}\text{m}$ of the wire and onto the positive plate. If the electrons all move together, and if they move at the typical drift speed of ${10}^{-4}\text{m}\text{/}\text{s}$—both less than perfect assumptions but fine for making an estimate—it takes $4\times {10}^{-9}\text{s,}$ or 4 ns, to move $4\times {10}^{-13}\text{m}$ and discharge the capacitor. And, indeed, this is the right order of magnitude for how long the electrons take to rearrange themselves so that the capacitor plates are neutral.