FIGURE 27.8a shows two metal wires attached to the plates of a charged capacitor. The wires are conductors, so some of the charges on the capacitor plates become spread out along the wires as a surface charge. (Remember that all excess charge on a conductor is located on the surface.)

FIGURE 27.8 The surface charge on the wires before and after they are connected.

Figure 27-8 Full Alternative Text

This is an electrostatic situation, with no current and no charges in motion. Consequently—because this is always true in electrostatic equilibrium—the electric field inside the wire is zero. Symmetry requires there to be equal amounts of charge to either side of each point to make $\vec{E}=\vec{0}$ at that point; hence the surface charge density must be uniform along each wire except near the ends (where the details need not concern us). We implied this uniform density in Figure 27.8a by drawing equally spaced $+$ and $-$ symbols along the wire. Remember that a positively charged surface is a surface that is missing electrons.

Now we connect the ends of the wires together. What happens? The excess electrons on the negative wire suddenly have an opportunity to move onto the positive wire that is missing electrons. Within a very brief interval of time $(\approx {10}^{-9}\text{s})$, the sea of electrons shifts slightly and the surface charge is rearranged into a nonuniform distribution like that shown in FIGURE 27.8b. The surface charge near the positive and negative plates remains strongly positive and negative because of the large amount of charge on the capacitor plates, but the midpoint of the wire, halfway between the positive and negative plates, is now electrically neutral. The new surface charge density on the wire varies from positive at the positive capacitor plate through zero at the midpoint to negative at the negative plate.

This nonuniform distribution of surface charge has an extremely important consequence. FIGURE 27.9 shows a section from a wire on which the surface charge density becomes more positive toward the left and more negative toward the right. Calculating the exact electric field is complicated, but we can understand the basic idea if we model this section of wire with four circular rings of charge.

FIGURE 27.9 A varying surface charge distribution creates an internal electric field inside the wire.

In Chapter 23, we found that the on-axis field of a ring of charge

• Points away from a positive ring, toward a negative ring;

• Is proportional to the amount of charge on the ring; and

• Decreases with distance away from the ring.

The field midway between rings A and B is well approximated as $\vec{E}{}_{\text{net}}\approx \vec{E}{}_{\text{A}}+\vec{E}{}_{\text{B}}$. Ring A has more charge than ring B, so $\vec{E}{}_{\text{net}}$ points away from A.

The analysis of Figure 27.9 leads to a very important conclusion:

A nonuniform distribution of surface charges along a wire creates a net electric field inside the wire that points from the more positive end of the wire toward the more negative end of the wire. This is the internal electric field $\overrightarrow{\mathit{\text{E}}}$ that pushes the electron current through the wire.

Note that the surface charges are not the moving charges of the current. Further, the current—the moving charges—is inside the wire, not on the surface. In fact, as the next example shows, the electric field inside a current-carrying wire can be established with an extremely small amount of surface charge.

EXAMPLE 27.2 The surface charge on a current-carrying wire

Table 23.1 in Chapter 23 gave a typical electric field strength in a current-carrying wire as 0.01 N/C or, as we would now say, 0.01 V/m. (We’ll verify this value later in this chapter.) Two 2.0-mm-diameter rings are 2.0 mm apart. They are charged to $\pm Q$. What value of $Q$ causes the electric field at the midpoint to be 0.010 V/m?

MODEL Use the on-axis electric field of a ring of charge from Chapter 23.

VISUALIZE FIGURE 27.10 shows the two rings. Both contribute equally to the field strength, so the electric field strength of the positive ring is $E_{+}=0.0050\text{V/m}$. The distance $z=1.0\text{mm}$ is half the ring spacing.

FIGURE 27.10 The electric field of two charged rings.

FIGURE 27.10 Full Alternative Text

SOLVE Chapter 23 found the on-axis electric field of a ring of charge $Q$ to be

$$E_{+}={\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{zQ}{{(z^2+R^2)}^{3\text{/}2}}}$$

Thus the charge needed to produce the desired field is

$$\begin{array}{lll}\hfill Q& =\hfill & {\displaystyle \frac{4\pi {ϵ}_0{(z^2+R^2)}^{3\text{/}2}}{z}}{E}_{+}\hfill \\ \hfill & =\hfill & {\displaystyle \frac{{\left({(0.0010\text{ m})}^2+{(0.0010\text{ m})}^2\right)}^{3\text{/}2}}{(9.0\times {10}^9{\text{N m}}^2{\text{/C}}^2)(0.0010\text{ m})}}(0.0050\text{ V/m})\hfill \\ \hfill & =\hfill & 1.6\times {10}^{-18}\text{C}\hfill \end{array}$$

REVIEW The electric field of a ring of charge is largest at $z\approx R$, so these two rings are a simple but reasonable model for estimating the electric field inside a 2.0-mm-diameter wire. We find that the surface charge needed to establish the electric field is very small. A mere 10 electrons have to be moved from one ring to the other to charge them to $\pm 1.6\times {10}^{-18}\text{C}$. The resulting electric field is sufficient to drive a sizable electron current through the wire.