23.3.1 Integration Is Summation

Calculating the electric field of a continuous charge distribution usually requires setting up and evaluating integrals—a skill you’ve been learning in calculus. It is common to think that “an integral is the area under a curve,” an idea we used in our study of kinematics.

But integration is much more than a tool for finding areas. More generally, integration is summation. That is, an integral is a sophisticated way to add an infinite number of infinitesimally small pieces. The area under a curve happens to be a special case in which you’re summing the small areas $y (x) d x$ of an infinite number of tall, narrow boxes, but the idea of integration as summation has many other applications.

Suppose, for example, that a charged object is divided into a large number of small pieces numbered $i = 1, 2, 3, \dots, N$ having small quantities of charge $Δ Q_{1}, Δ Q_{2}, Δ Q_{3}, \dots, Δ Q_{N}$ . Figure 23.9 showed small pieces of charge for a charged rod and a charged sheet, but the object could have any shape. The total charge on the object is found by summing all the small charges:

Q = \sum_{i = 1}^{N} Δ Q_{i}

(23.14)

If we now let $Δ Q_{i} \to 0$ and $N \to \infty$ , then we define the integral:

Q = lim_{Δ Q \to 0} \sum_{i = 1}^{N} Δ Q_{i} = \int_{object} d Q

(23.15)

That is, integration is the summing of an infinite number of infinitesimally small pieces of charge. This use of integration has nothing to do with the area under a curve.

Although Equation 23.15 is a formal statement of “add up all the little pieces,” it’s not yet an expression that can actually be integrated with the tools of calculus. Integrals are carried out over coordinates, such as $d x$ or $d y$ , and we also need coordinates to specify what is meant by “integrate over the object.” This is where densities come in.

Suppose we want to find the total charge of a thin, charged rod of length L. First, we establish an x-axis with the origin at one end of the rod, as shown in FIGURE 23.10. Then we divide the rod into lots of tiny segments of length $d x$ . Each of these little segments has a charge $d Q$ , and the total charge on the rod is the sum of all the $d Q$ values—that’s what Equation 23.15 is saying. Now the critical step: The rod has some linear charge density $λ$ . Consequently, the charge of a small segment of the rod is $d Q = λ d x$ . Densities are the link between quantities and coordinates. Finally, “integrate over the rod” means to integrate from $x = 0$ to $x = L$ . Thus the total charge on the rod is

Q = \int_{rod} d Q = \int_{0}^{L} λ d x

(23.16)

The figure illustrates a rod of length L with a uniform positive charge.

FIGURE 23.10 Full Alternative Text

Now we have an expression that can be integrated. If $λ$ is constant, as it is for a uniformly charged rod, we can take it outside the integral to find $Q = λ L$ . But we could also use Equation 23.16 for a nonuniformly charged rod where $λ$ is some function of x.

This discussion reveals two key ideas that will be needed for calculating electric fields:

Integration is the tool for summing a vast number of small pieces.
Density is the connection between quantities and coordinates.