In the previous chapter we saw how Gauss' Law was able to help us calculate the electric field of certain very symmetric charge distribution. For most charge distributions, we need to integrate. If you've taken integral calculus already, you've already learned how to *evaluate* integrals, but in this chapter we will focus on *constructing* integrals. Some of them we'll evaluate, some of them we won't, some of them we can't — but in physics, the integral itself can be interesting regardless of what it evaluates to.

In case you haven't had much practice in constructing integrals, let's begin with a more basic example before tackling electric field integration.

Suppose we have a line segment which is 1 meter long, and has a uniform line charge density of \(\lambda=3\u{mC/m}\). What is the total charge on that line? Well, we know from a previous section that

$$q=\lambda L=(3\u{mC/m})(1\u{m})=3\u{mC}$$

Suppose we took the line and chopped it up into a bunch of tiny pieces. Each of *these* is a line segment as well, and if the pieces are small enough, the charge density λ is not going to vary much from one side to the other. Thus we can make the approximation that λ is constant on each segment, and calculate the segment's charge using the formula *q*=*λL*. The total charge of the entire line is the sum of all of these little charges. The following animation shows how this works.

Interactive 6.1.1

This is an approximation, but one which gets better and better as we make the pieces smaller and smaller, and we can get an exact result by taking the limit as the segment size goes to zero. This is, of course, what calculus does. In calculus notation, we put the letter *d* in front of every tiny quantity (e.g. the width *dx* and charge *dq* of each little piece) to indicate that they are infinitesimal. The total charge on the line is the sum of all the little charges *dq*: in calculus we use the integral symbol ∫ when we add up tiny quantities, so we write

$$q=\int\,dq$$

This integral may look trivial, but it's not complete. In particular, we don't know what the limits of integration are so we can't come up with a number: generally speaking, we know how to find the limits of integration if we're integrating over space (i.e.
$$dq=\lambda\,dx$$

and if we substitute this into the integral above, we will change the variable of integration from
$$q=\int\lambda\,dx=\int\multiplier x\,dx$$

Now that the integration variable is $$q=\int_0^1 \multiplier x\,dx=\multiplier\left[\frac12x^2\right]_0^1=\halfmultiplier\u{mC}$$