23.4 The Electric Fields of Some Common Charge Distributions

In this section we’ll derive the electric fields for several important charge distributions.

EXAMPLE 23.4 The electric field of a ring of charge

A thin ring of radius R is uniformly charged with total charge Q. Find the electric field at a point on the axis of the ring (perpendicular to the ring).

MODEL Because the ring is thin, we’ll assume the charge lies along a circle of radius R. You can think of this as a line of charge of length $2 π R$ wrapped into a circle. The linear charge density along the ring is $λ = Q / 2 π R$ .

VISUALIZE FIGURE 23.14 shows the ring and illustrates the steps of the problem-solving strategy. We’ve chosen a coordinate system in which the ring lies in the xy-plane and point P is on the z-axis. We’ve then divided the ring into N small segments of charge $Δ Q$ , each of which can be modeled as a point charge. As you can see from the figure, the component of the field perpendicular to the axis cancels for two diametrically opposite segments. Thus we need to calculate only the z-component $E_{z}$ .

The figure illustrates the electric field of a thin charged ring placed on the x y plane of an x y z coordinate system.

FIGURE 23.14 Full Alternative Text

SOLVE The z-component of the electric field due to segment i is

{(E_{i})}_{z} = E_{i} cos θ_{i} ​ = \frac{1}{4 π ϵ_{0}} \frac{Δ Q}{r_{i}^{2}} cos θ_{i}

You can see from the figure that every segment of the ring, independent of i, has

\begin{array}{l} r_{i} = \sqrt{z^{2} + R^{2}} \\ cos θ_{i} = \frac{z}{r_{i}} = \frac{z}{\sqrt{z^{2} + R^{2}}} \end{array}

Consequently, the field of segment i is

{(E_{i})}_{z} = \frac{1}{4 π ϵ_{0}} \frac{Δ Q}{z^{2} + R^{2}} \frac{z}{\sqrt{z^{2} + R^{2}}} = \frac{1}{4 π ϵ_{0}} \frac{z}{{(z^{2} + R^{2})}^{3 / 2}} Δ Q

The net electric field is found by summing ${(E_{i})}_{z}$ due to all N segments:

E_{z} = \sum_{i = 1}^{N} {(E_{i})}_{z} = \frac{1}{4 π ϵ_{0}} \frac{z}{{(z^{2} + R^{2})}^{3 / 2}} \sum_{i = 1}^{N} Δ Q

We were able to bring all terms involving z to the front because z is a constant as far as the summation is concerned. Surprisingly, we don’t need to convert the sum to an integral to complete this calculation. The sum of all the $Δ Q$ around the ring is simply the ring’s total charge, $\sum Δ Q = Q$ , hence the field on the axis is

(E_{ring})_{z} = \frac{1}{4 π ϵ_{0}} \frac{z Q}{{(z^{2} + R^{2})}^{3 / 2}}

This expression is valid for both positive and negative z (i.e., on either side of the ring) and for both positive and negative charge.

REVIEW It will be left as a homework problem to show that this result gives the expected limit when $z ≫ R$ .

FIGURE 23.15 shows two representations of the on-axis electric field of a positively charged ring. Figure 23.15a shows that the electric field vectors point away from the ring, increasing in length until reaching a maximum when $| z | \approx R$ , then decreasing. The graph of $(E_{ring})_{z}$ in Figure 23.15b confirms that the field strength has a maximum on either side of the ring. Notice that the electric field at the center of the ring is zero, even though this point is surrounded by charge. You might want to spend a minute thinking about why this has to be the case.

The figure labeled ("a") illustrates the on-axis electric field of a positively charged ring.

The figure labeled (b) illustrates the graph of the on-axis electric field of a positively charged ring.

Full Alternative Text

23.4 The Electric Fields of Some Common Charge Distributions

EXAMPLE 23.4 The electric field of a ring of charge

FIGURE 23.14 Calculating the on-axis electric field of a ring of charge.

FIGURE 23.15 The on-axis electric field of a ring of charge.