Our initial definition of the electric flux assumed that the electric field $\vec{E}$ was constant over the surface. How should we calculate the electric flux if $\vec{E}$ varies from point to point on the surface? We can answer this question by returning to the analogy of air flowing through a loop. Suppose the airflow varies from point to point. We can still find the total volume of air passing through the loop each second by dividing the loop into many small areas, finding the flow through each small area, then adding them. Similarly, the electric flux through a surface can be calculated as the sum of the fluxes through smaller pieces of the surface. Because flux is a scalar, adding fluxes is easier than adding electric fields.

FIGURE 24.14 shows a surface in a nonuniform electric field. Imagine dividing the surface into many small pieces of area $\delta A$. Each little area has an area vector $\delta \vec{A}$ perpendicular to the surface. Two of the little pieces are shown in the figure. The electric fluxes through these two pieces differ because the electric fields are different.

FIGURE 24.14 A surface in a nonuniform electric field.

Consider the small piece i where the electric field is $\vec{E}{}_i$. The small electric flux $\delta {\text{Φ}}_i$ through area ${\left(\delta \vec{A}\right)}_i$ is

The flux through every other little piece of the surface is found the same way. The total electric flux through the entire surface is then the sum of the fluxes through each of the small areas:

Now let’s go to the limit $\delta \vec{A}\to d\vec{A}$. That is, the little areas become infinitesimally small, and there are infinitely many of them. Then the sum becomes an integral, and the electric flux through the surface is

The integral in Equation 24.6 is called a surface integral.

Equation 24.6 may look rather frightening if you haven’t seen surface integrals before. Despite its appearance, a surface integral is no more complicated than integrals you know from calculus. After all, what does $\int f\left(x\right)dx$ really mean? This expression is a shorthand way to say “Divide the x-axis into many little segments of length $\delta x$, evaluate the function $f\left(x\right)$ in each of them, then add up $f\left(x\right)\delta x$ for all the segments along the line.” The integral in Equation 24.6 differs only in that we’re dividing a surface into little pieces instead of a line into little segments. In particular, we’re summing the fluxes through a vast number of very tiny pieces.

You may be thinking, “OK, I understand the idea, but I don’t know what to do. In calculus, I learned formulas for evaluating integrals such as $\int x^{2}dx$. How do I evaluate a surface integral?” This is a good question. We’ll deal with evaluation shortly, and it will turn out that the surface integrals in electrostatics are quite easy to evaluate. But don’t confuse evaluating the integral with understanding what the integral means. The surface integral in Equation 24.6 is simply a shorthand notation for the summation of the electric fluxes through a vast number of very tiny pieces of a surface.

The electric field might be different at every point on the surface, but suppose it isn’t. That is, suppose a flat surface is in a uniform electric field $\vec{E}$. A field that is the same at every single point on a surface is a constant as far as the integration of Equation 24.6 is concerned, so we can take it outside the integral. In that case,

The integral that remains in Equation 24.7 tells us to add up all the little areas into which the full surface was subdivided. But the sum of all the little areas is simply the area of the surface:

This idea—that the surface integral of dA is the area of the surface—is one we’ll use to evaluate most of the surface integrals of electrostatics. If we substitute Equation 24.8 into Equation 24.7, we find that the electric flux in a uniform electric field is ${\text{Φ}}_{\text{e}}=EA\text{cos}\theta$. We already knew this, from Equation 24.2, but it was important to see that the surface integral of Equation 24.6 gives the correct result for the case of a uniform electric field.