Most of the Gaussian surfaces we considered in the last section were curved surfaces. FIGURE 24.15 shows an electric field passing through a curved surface. How do we find the electric flux through this surface? Just as we did for a flat surface!

FIGURE 24.15 A curved surface in an electric field.

Divide the surface into many small pieces of area $\delta A$. For each, define the area vector $\delta \vec{A}$ perpendicular to the surface at that point. Compared to Figure 24.14, the only difference that the curvature of the surface makes is that the $\delta \vec{A}$ are no longer parallel to each other. Find the small electric flux $\delta {\text{Φ}}_i=\vec{E}{}_i\cdot {\left(\delta \vec{A}\right)}_i$ through each little area, then add them all up. The result, once again, is

We assumed, in deriving this expression the first time, that the surface was flat and that all the $\delta \vec{A}$ were parallel to each other. But that assumption wasn’t necessary. The meaning of Equation 24.9—a summation of the fluxes through a vast number of very tiny pieces—is unchanged if the pieces lie on a curved surface.

We seem to be getting more and more complex, using surface integrals first for nonuniform fields and now for curved surfaces. But consider the two situations shown in FIGURE 24.16. The electric field $\vec{E}$ in Figure 24.16a is everywhere tangent, or parallel, to the curved surface. We don’t need to know the magnitude of $\vec{E}$ to recognize that $\vec{E}\cdot d\vec{A}$ is zero at every point on the surface because $\vec{E}$ is perpendicular to $d\vec{A}$ at every point. Thus ${\text{Φ}}_{\text{e}}=0$. A tangent electric field never pokes through the surface, so it has no flux through the surface.

FIGURE 24.16 Electric fields that are everywhere tangent to or everywhere perpendicular to a curved surface.

Full Alternative Text

The electric field in Figure 24.16b is everywhere perpendicular to the surface and has the same magnitude E at every point. $\vec{E}$ differs in direction at different points on a curved surface, but at any particular point $\vec{E}$ is parallel to $d\vec{A}$ and $\vec{E}\cdot d\vec{A}$ is simply E dA. In this case,

As we evaluated the integral, the fact that E has the same magnitude at every point on the surface allowed us to bring the constant value outside the integral. We then used the fact that the integral of dA over the surface is the surface area A.

We can summarize these two situations with a Tactics Box.

These two results will be of immeasurable value for using Gauss’s law because every flux we’ll need to calculate will be one of these situations. This is the basis for our earlier claim that the evaluation of surface integrals is not going to be difficult.