Gauss’s Law for electrostatics (no time-dependent charge distributions) can be written in differential form in vacuum as:
\vec{\nabla} \cdot \vec{E} = \rho/\epsilon_0
where \rho represents the volume charge density. Our goal is to determine the \vec{E} that this charge distribution produces.
It turns out that the integral form, under the right conditions, might be the more useful form for this purpose:
\int_S \vec{E} \cdot d\vec{a} = Q_\mathrm{enclosed}/\epsilon_0
where S is a closed surface and Q_\mathrm{enclosed} is the total charge enclosed within the volume created by the closed surface S.
Gauss’s Law in both forms is always true, but it is not always useful for calculating the E-field. What does this statement mean? The first part indicates that the electric flux (the integral) is proportional to the total charge enclosed within a surface. If you imagine the magnitude of the electric field as being the number of field lines per unit area, then the interpretation of the electric flux is the total number of electric field lines passing through a closed surface. The density of field lines depends on both the amount of charge and the distance from the charge. The direction of the field line arrows points away (towards) from (to) the charge if it is positive (negative). The total net number of field lines passing through a closed surface only depends on the total enclosed charge. The sign convention is that field lines going into (out of) the volume contribute a negative (positive) amount.
The second part of the statement above simply indicates that the field lines can point in all sorts of complicated ways/directions depending on the exact shape of the charge distribution and the exact shape of the closed surface used to perform the integration. This surface is hypothetical and is an artifice simply used to perform the integral. It is conventionally referred to as a Gaussian surface.
Before we begin, let’s write out explicitly what we are trying to do. Without loss of generality, let’s inspect the integral form of Gauss’s Law in Cartesian coordinates. We’ll choose a Gaussian surface composed of six flat plates that form a box with one corner at the origin. The length, width, and height of this box are x, y, and z. Putting all this together, we find for Gauss’s Law:
\int_S \vec{E} \cdot d\vec{a} = \Phi_\mathrm{bottom} + \Phi_\mathrm{top} + \Phi_\mathrm{front} + \Phi_\mathrm{back} + \Phi_\mathrm{left} + \Phi_\mathrm{right} = Q_\mathrm{enclosed}/\epsilon_0
\Phi_\mathrm{bottom} = \int_0^x \int_0^y \vec{E}(x',y',0) \cdot \hat{z} dy' dx'
\Phi_\mathrm{top} = \int_0^x \int_0^y \vec{E}(x',y',z) \cdot \hat{z} dy' dx'
\Phi_\mathrm{front} = \int_0^z \int_0^y \vec{E}(x,y',z') \cdot \hat{x} dy' dz'
\Phi_\mathrm{back} = \int_0^z \int_0^y \vec{E}(0,y',z') \cdot \hat{x} dy' dz'
\Phi_\mathrm{left} = \int_0^z \int_0^x \vec{E}(x',0,z') \cdot \hat{y} dx' dz'
\Phi_\mathrm{right} = \int_0^z \int_0^x \vec{E}(x',y,z') \cdot \hat{y} dx' dz'
Q_\mathrm{enclosed}/\epsilon_0 = \frac{1}{\epsilon_0} \int_0^z \int_0^y \int_0^x \rho(x',y',z')\ dx'\ dy'\ dz'
Note that the integral is over the volume enclosed by the surface over which the flux \Phi is being calculated. This volume might not be completely filled by charge and, in that case, the integration limits could be modified and reduced to just the volume that (1) contains charge and (2) is completely surrounded by the Gaussian surface.
We want to know the magnitude of the E-field which means that it must be a constant that we pull out of the integrals above. The only way that can happen is the E-magnitude is a constant on the surface with respect to the flux integration variables. Furthermore, the dot product must be maximal so that the whole E-field magnitude contributes to the flux as opposed to some fraction of it. This is only true if the E-field direction is perpendicular to the surface that is being integrated over. In other words, the E-field direction must be parallel to the unit vector of the third coordinate variable where the first two coordinate variables are being integrated over. The E-field magnitude may vary over one of the other surfaces. The contribution of this region of the E-field does not contribute if the E-field is perfectly parallel to the surface being integrated over. All of these conditions are rarely every true. If they are true, then there is a high degree of symmetry in the charge distribution.
If there is a high degree of symmetry in the problem, then we can use Gauss’s Law to determine the E-field. What is meant by high symmetry? Let’s try to walk through the general argument.:
- First, the E-field lines must be completely perpendicular to the Gaussian surface at all points on the Gaussian surface where we want to know the magnitude of the E-field.
- At locations where we are not interested in knowing the magnitude of the E-field, the E-field must be completely parallel to the Gaussian surface so that it does not contribute to the integral.
If both of these statements are true, then the dot product in the integral becomes super simple and we don’t have to worry about any \cos(\mathrm{some\ angle}) factors that would only provide information about just the component of the E-field perpendicular to the surface. Stop and think about this. An E-field partially parallel to the Gaussian surface does not contribute fully to the integral and therefore we have no way of determining anything about its parallel component and its total magnitude. Therefore, we have to construct a Gaussian surface that is completely perpendicular or completely parallel to the E-field lines everywhere on the surface so that we can ``sense’’ the full contribution of the entire E-field. This is a key part of the argument, so you need to really stop here and think about this and you cannot proceed until your convince yourself of this argument. Remember, the goal is to determine when you can use the integral form of Gauss’s Law to determine the entire E-field and not just one component of the E-field.
How does one choose a Gaussian surface that is completely perpendicular or completely parallel to the E-field on all points on its surface? Note that the surface may be composed of two or more parts. The E-field just needs to be completely parallel or completely perpendicular to each of these surfaces individually and you allowed to mix and match. This depends crucially on the shape of the charge distribution being very highly symmetrical. To make this concrete, image a spherical charge distribution centered on the origin. This charge distribution is represented by the volume charge density \rho(r, \theta, \phi) \neq 0 out to some sphere radius r= a and \rho = 0 for r >a.
Because of the spherical shape of the charge distribution, it is tempting to choose a concentric spherical shell as the Gaussian surface. Concentric just means that this Gaussian spherical shell is also centered at the origin. The normal vector for this Gaussian surface is +\hat{r} namely it must point directly out and away from the origin at all points on the Gaussian surface.
It is possible for a spherical charge distribution to create an E-field which also points only in the \pm \hat{r} direction on all points on the Gaussian surface. To determine a specific point on sphere, you must specify the sphere radius as well as the polar \theta and azimuthal \phi angles. In order to visualize this situation, just imagine that you are sitting on the surface of the Earth. Your location is given by the longitude (azimuthal) and latitude (polar). What properties does the charge distribution must have in order for the E-field to always have a radial direction regardless of location on the sphere; in other words, the direction of the E-field is independent of \theta and \phi? Don’t try to overthink this. The answer, like Beach, is that the volume charge density must be independent of the polar and azimuthal angles altogether. Convince yourself of this before moving on.
\rho = \rho(r) \rightarrow \partial \rho/\partial \theta = \partial \rho/\partial \phi = 0
Note that there is no restriction on the r dependence of the volume charge density! Now, let’s go back to our Gaussian surface. It is a spherical surface of radius r and the integral is over the surface, namely for all values of \theta and \phi. But we just painstakingly argued that the E-field cannot have a \theta or \phi dependence! This makes evaluating the integral relatively straightforward:
\int_S \vec{E} \cdot d\vec{a} = \int_{0}^{\pi} \int_0^{2\pi} E(r) \hat{r} \cdot \hat{r} r^2 \sin(\theta)\ d\phi\ d\theta
Stop here to think about how many of the properties that we had to guess/impose about the E-field that we are seeking to determine. The logic leading up to this point forced the E-field to point in the \pm\hat{r} direction and to only have an r dependence.
At this point, we can finally use the integral form of Gauss’s Law to determine only just the magnitude of the E-field:
\int_{0}^{\pi} \int_0^{2\pi} E(r) \hat{r} \cdot \hat{r} r^2 \sin(\theta)\ d\phi\ d\theta = E(r) r^2 \int_{0}^{\pi} \int_0^{2\pi} \sin(\theta)\ d\phi\ d\theta
= E(r) r^2 4 \pi = Q_\mathrm{enclosed}/\epsilon_0 \rightarrow E(r) = Q_\mathrm{enclosed}/(4 \pi \epsilon_o r^2)
Putting this altogether, we find:
\vec{E} = \frac{\hat{r}}{4 \pi \epsilon r^2} \int_{V} \rho(r')\ d\tau'
where V' is the volume enclosed by S, \rho can only have an r' dependence, and d\tau' is the differential volume element in Griffiths notation.
How can we generalize this to other geometries or coordinate systems? Here are the rules:
- The E-field must be completely perpendicular or completely parallel to the Gaussian surface at all points on this surface. The surface can be composed of different parts and you can mix and match the perpendicular/parallel with respect to each individual surface. For example, image a Gaussian cylinder composed of a top disk, a bottom disk, and a curved side. The E-field could be in the \hat{s} direction making it completely perpendicular to the curved side but completely parallel to the top and bottom disks.
- The charge distribution must be highly symmetric and there needs to be a “match” between the shape of the charge distribution and the shape of the Gaussian surface.
- The E-field magnitude must be the same at all points on the Gaussian surface for which the E-field is perpendicular to the surface. This implies that the charge distribution must be independent of at least two of the coordinate variables that would be necessary to specify the location on the Gaussian surface. The magnitude of the E-field at locations where the E-field is completely parallel to the surface does not matter since the dot product (being zero) makes those contributions zero.
These conditions restrict the utility of Gauss’s Law for the calculating the magnitude of the E-field to just a small handful of forms for the charge distribution:
- a flat infinite uniform sheet of charge (crucial to determining boundary conditions for E-fields)
- a spherical charge distribution which does not depend on \theta and \phi, but could depend on r and only r (very useful since we can make these in real life without any approximations)
- an infinitely long charged cylinder with a charge distribution that does not depend on z and \phi, but could depend on s and only s (useful as a model for very long transmission cables such as coaxial cables)
where we have used Griffiths notations and definitions for \theta, \phi, r, and s.