24.3.4 The Electric Flux Through a Closed Surface

Our final step, to calculate the electric flux through a closed surface such as a box, a cylinder, or a sphere, requires nothing new. We’ve already learned how to calculate the electric flux through flat and curved surfaces, and a closed surface is nothing more than a surface that happens to be closed.

However, the mathematical notation for the surface integral over a closed surface differs slightly from what we’ve been using. It is customary to use a little circle on the integral sign to indicate that the surface integral is to be performed over a closed surface. With this notation, the electric flux through a closed surface is

Φ_{e} = \oint \vec{E} \cdot d \vec{A}

(24.11)

Only the notation has changed. The electric flux is still the summation of the fluxes through a vast number of tiny pieces, pieces that now cover a closed surface.

NOTE A closed surface has a distinct inside and outside. The area vector $d \vec{A}$ is defined to always point toward the outside. This removes an ambiguity that was present for a single surface, where $d \vec{A}$ could point to either side.

Now we’re ready to calculate the flux through a closed surface.

TACTICS BOX 24.2 Finding the flux through a closed surface

1 Choose a Gaussian surface made up of pieces that are everywhere tangent to the electric field or everywhere perpendicular to the electric field.

2 Use Tactics Box 24.1 to evaluate the surface integrals over these surfaces, then add the results.

Workbook Exercise 10

EXAMPLE 24.2 Calculating the electric flux through a closed cylinder

A charge distribution with cylindrical symmetry has created the electric field $\vec{E} = E_{0} (r^{2} / r_{0}^{2}) \hat{r}$ , where $E_{0}$ and $r_{0}$ are constants and where unit vector $\hat{r}$ lies in the xy-plane. Calculate the electric flux through a closed cylinder of length L and radius R that is centered along the z-axis.

MODEL The electric field extends radially outward from the z-axis with cylindrical symmetry. The z-component is $E_{z} = 0$ . The cylinder is a Gaussian surface.

VISUALIZE FIGURE 24.17a is a view of the electric field looking along the z-axis. The field strength increases with increasing radial distance, and it’s symmetric about the z-axis. FIGURE 24.17b is the closed Gaussian surface for which we need to calculate the electric flux. We can place the cylinder anywhere along the z-axis because the electric field extends forever in that direction.

The figure consists of two parts labeled (“a”) and (b) that illustrate the electric field and electric flux through a closed cylinder.

FIGURE 24.17 Full Alternative Text

SOLVE To calculate the flux, we divide the closed cylinder into three surfaces: the top, the bottom, and the cylindrical wall. The electric field is tangent to the surface at every point on the top and bottom surfaces. Hence, according to step 1 in Tactics Box 24.1, the flux through those two surfaces is zero. For the cylindrical wall, the electric field is perpendicular to the surface at every point and has the constant magnitude $E = E_{0} (R^{2} / r_{0}^{2})$ at every point on the surface. Thus, from step 2 in Tactics Box 24.1,

Φ_{wall} = E A_{wall}

If we add the three pieces, the net flux through the closed surface is

\begin{matrix} Φ_{e} & = & \oint \vec{E} \cdot d \vec{A} = Φ_{top} + Φ_{bottom} + Φ_{wall} = 0 + 0 + E A_{wall} \\ = & E A_{wall} \end{matrix}

We’ve evaluated the surface integral, using the two steps in Tactics Box 24.1, and there was nothing to it! To finish, all we need to recall is that the surface area of a cylindrical wall is $circumference \times height$ , or $A_{wall} = 2 π R L$ . Thus

Φ_{e} = (E_{0} \frac{R^{2}}{r_{0}^{2}}) (2 π R L) = \frac{2 π L R^{3}}{r_{0}^{2}} E_{0}

REVIEW $L R^{3} / r_{0}^{2}$ has units of $m^{2}$ , an area, so this expression for $Φ_{e}$ has units of ${N m}^{2} / C$ . These are the correct units for electric flux, giving us confidence in our answer. Notice the important role played by symmetry. The electric field was perpendicular to the wall and of constant value at every point on the wall because the Gaussian surface had the same symmetry as the charge distribution. We would not have been able to evaluate the surface integral in such an easy way for a surface of any other shape. Symmetry is the key.