Electric Flux and Gauss’s Law

In electrostatics, calculating the electric field “from scratch” can become very costly when the geometry of the charge distribution is not trivial. Electric flux and Gauss’s law offer a more efficient route: instead of struggling with endless integrals, you choose an appropriate closed surface and exploit the symmetry of the system to obtain clean and verifiable results. In practice, this translates into fewer steps, fewer errors, and greater conceptual control over what you are doing. If you want to move from “I know the recipe” to “I understand the method,” here you will see how Gauss transforms problems that appear cumbersome into direct solutions, and when it is actually convenient to use it.

Learning Objectives

1. Explain the operation of Gauss’s law for the electric field.
2. Use Gauss’s law to calculate electric fields by exploiting the symmetries of Cartesian, cylindrical, and spherical coordinates.
3. Relate the integral and differential forms through the Divergence Theorem, identifying what each term represents.
4. Contrast the Gauss approach with direct calculation via Coulomb’s integral, explaining when it reduces complexity and when it does not yield a closed-form solution.

CONTENT INDEX:
Resolution of electrostatics
Electric field lines
Note on the density of field lines and their representation
Electric Field Flux
Gauss’s Law
Problems with spherical symmetry
More Symmetries
Problems with cylindrical and planar symmetry

Resolution of Electrostatics

From what has been reviewed so far, we have that it is sufficient to know the form of the electric field element and its distribution in space in order to determine the total electric field. If we have a volumetric distribution, then

\displaystyle \vec{E}(\vec{r}) = \int_V d\vec{E}(\vec{r})= \int_V \frac{\rho(\vec{r}^\prime)}{4\pi\epsilon_0}\frac{\vec{r}-\vec{r}^\prime}{\|\vec{r}-\vec{r}^\prime\|^3}dV

where \rho(\vec{r}^\prime) is the volumetric charge density. In the case of a surface or linear charge density, we will replace \rho with \sigma or \lambda, respectively. From this point onward, what determines whether or not we can find the electric field is whether or not we are able to evaluate the integral.

Although formulating the problem is usually straightforward, sooner or later we will discover that it is not always easy to evaluate it. In fact, a large part of the study of electrostatics consists of developing strategies that allow us to avoid the calculation of unnecessarily complicated integrals. Many of these simplifications arise from vector analysis, in particular from the use of the divergence.

Electric Field Lines

Before introducing vector analysis into our study of electrostatics, we will present some ideas that will help make the topic somewhat more intuitive. I am referring to electric field lines.

Let us begin with the simplest case: the electric field of a point charge located at the origin of coordinates. This has the form

\displaystyle \vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0}\frac{q}{\|\vec{r}\|^2}\hat{r}

This allows us to represent the electric field in space as a set of “arrows” whose direction and magnitude describe the direction and intensity of the electric field at each point.

Since the intensity of the electric field decreases with the square of the distance from the origin, the vectors become progressively smaller as we move away. In addition, they point radially outward from the charge.

This representation is useful, but there is another that is even more informative: “connecting the continuum of arrows” to form a field of lines. In this way, it is no longer the length of the arrows that indicates the intensity of the electric field, but rather the “density of field lines” in the diagram.

Note on the Density of Field Lines and Their Representation

Before continuing, it is worth noting a detail regarding the diagram of electric field lines. This type of representation is not entirely faithful when drawn on a plane (2D). In a 2D drawing, if we consider a circle of radius r, the total number of lines is distributed along the circumference, so that the linear density is

\displaystyle \frac{n}{2\pi r}

This decreases with respect to r and not with respect to r^2, as would be expected for the intensity of the electric field. However, if we interpret the model in three dimensions (like a hedgehog), then the total number of lines would be divided by the surface of a sphere

\displaystyle \frac{n}{4\pi r^2}

and this does decrease with respect to r^2. In other words, although the representation of field lines is usually carried out in two dimensions, what is actually intended is to synthesize a three-dimensional situation. We simply do not have three-dimensional paper to draw it: we represent in 2D what we wish to communicate in 3D.

Electric Field Flux

When we ask about the number of electric field lines that cross a given surface, the answer is given by the flux of the electric field through that surface. Thus, the electric flux of a field \vec{E} through a surface S is defined as

\Phi_{\vec{E},S} =\displaystyle \int_S \vec{E}\cdot d\vec{S}

We should not be misled by the intuitive notion of the “number of electric field lines crossing a surface.” Recall that this number of lines (or line density) is a way of representing the intensity of the electric field. Therefore, the electric flux that we calculate is a scalar quantity associated with the intensity of the electric field passing through the surface S.

Gauss’s Law

Since the intensity of the electric field is proportional to the electric charge, we should be able to express the electric flux through a surface that encloses a certain charge as a quantity proportional to the enclosed charge. In fact, it is not difficult to show that this is the case. Consider the following figure:

From this, we obtain that:

\begin{array}{rl} \displaystyle \oint_S \vec{E}\cdot d\vec{S} &= \displaystyle \oint_S \left(\frac{1}{4\pi\epsilon_0} \frac{q_{enc}}{\|\vec{r}\|^2}\hat{r} \right)\cdot d\vec{S} \\ \\ & = \displaystyle \frac{q_{enc}}{4\pi\epsilon_0} \oint_S \frac{\hat{r}}{\|\vec{r}\|^2}\cdot d\vec{S} \\ \\ & = \displaystyle \frac{q_{enc}}{4\pi\epsilon_0} \underbrace{\oint_S d{\Omega}}_{= 4\pi} = \frac{q_{enc}}{\epsilon_0} \end{array}

In summary, we obtain:

\displaystyle\color{blue}{\oint_S \vec{E}\cdot d\vec{S} = \frac{q_{enc}}{\epsilon_0}}

This is Gauss’s Law for the electric field in its integral form, and it shows a proportionality relationship between the electric flux through a closed surface and the enclosed charge. Note that it has been presented in its “integral form” to emphasize that a differential form also exists, which is obtained by using Gauss’s Divergence Theorem in the context of vector analysis.

 

Applying the divergence theorem to the flux of the electric field over the closed surface S, we obtain

\displaystyle \oint_S\vec{E}\cdot d\vec{S} = \int_V (\vec{\nabla}\cdot\vec{E})dV = \frac{q_{enc}}{\epsilon_0}

On the other hand, we also have

\displaystyle \frac{q_{enc}}{\epsilon_0} = \int_V \frac{\rho}{\epsilon_0} dV

From these last two equations, we finally obtain

\displaystyle \color{blue}{\vec{\nabla}\cdot\vec{E} = \frac{\rho}{\epsilon_0}}

This is Gauss’s Law for the electric field in its differential form.

We can now make use of Gauss’s Law to better exploit the geometric symmetries of certain problems and greatly simplify the calculation of the integrals that lead to the electric field.

Problems with Spherical Symmetry

1. Find the electric field at a distance z from the center of a spherical surface of radius R that has a uniform surface charge density \sigma. Analyze both cases: when z\lt R, and when z\geq R.
2. Carry out the same analysis as in the previous exercise, but now considering a solid sphere uniformly charged with a volumetric charge density \rho. Then construct a plot of \|\vec{E}\| as a function of z.
3. Suppose that the electric field, at a distance r from the origin of coordinates, is \vec{E}=kr^2\hat{r}, with k constant. Find the charge density \rho associated with this field.

More Symmetries

Gauss’s law is always true, but it is not always useful. In the previous examples, if \rho were not uniform, if we did not have spherical symmetry, or if a different shape were chosen for the Gaussian surface, it would still be true that the electric flux is q_{enc}/\epsilon_0, but the electric field would not necessarily be constant nor oriented in the same direction as the element d\vec{S}; and without these conditions we cannot extract \|\vec{E}\| from the integral.

Symmetry is crucial in the application of Gauss’s Law to problem solving.

There are many types of symmetries that we can exploit. Among all of them, the following three are the most frequent:

1. Spherical symmetry: The Gaussian surface is a concentric sphere.
2. Cylindrical symmetry: The Gaussian surface is a coaxial cylinder.
3. Planar symmetry: The Gaussian surface is a rectangular box.

Problems with cylindrical and planar symmetry

1. Consider an infinitely long straight cylindrical wire of radius R, charged with a charge density \rho of the form

   \rho(r) = \left\{\begin{array}{lll} kr & ; & r\lt R \\ \\ 0 & ; & R\lt r \\ \\ \end{array}\right.

   where k is a constant. Compute the electric field inside the cylinder.

2. Find the electric field produced by an infinite plane endowed with a uniform surface charge density \sigma.