Unit 1 - Gauss's Law

As I indicated at the end of the page on field lines, since electric field lines can begin or end only on charges, we can learn something about the number of charges by simply counting field lines. Specifically, the net number of field lines leaving a region must be proportional to the net charge inside that region. A line can enter the region, pass through, and come out; it does not correspond to a charge inside the region, so it must count negatively when entering as well as positively when leaving; thus, the net number of lines leaving is what counts. Also, if there are both positive and negative charges inside, then many lines can originate on the positive(s) and end on the negative(s) without reaching the surface of the region where we count lines, but the number of such line ends must equal the number of such line beginnings, so the corresponding negative and positive charge amounts are equal; thus, the net charge inside is what counts.

The next step in developing this relation to the form called Gauss's Law is to relate field lines to electric field. (Number of lines) equals (density of lines) times area, and (density of lines) is proportional to E. Hence (number of lines) is proportional to E times area. (Number of lines) is a scalar, so to get a statement connecting E to a scalar we should be doing a scalar product. That means area should be a vector, which to physicists it is. Besides making the vector-scalar relationships work out, this also takes care of orientation questions: the same field line density and the same size area will give different numbers of lines crossing the area if the area is oriented differently. (If E is in the plane of the surface, for instance, no lines at all cross.) Since frequently the field is different at different places, and certainly (for the surface of a region) the area vector changes direction, the product (density of lines) times area must become an integral of E dot da.

The last step in finalizing the integral form of Gauss's Law is determining the proportionality constant. For this step, we only need to consider a single case where we know both charge and field; a point charge, and a spherical region around it, will do nicely.

Once we have Gauss's Law, what can we do with it? In general, there are two ways to use it. You may be given an electric field function and be asked "Where's the charge?" (That is, how much charge is at what location(s)?) Or you may be given the charge distribution, and asked for the field.

Using the integral form to find the charge is fairly straightforward, provided you know enough to identify the possible locations. You just have to pick a surface that encloses the location you want to know about, but which does not enclose any other unknown charges that would confuse the issue.

Using the integral form of Gauss's Law to find the field is trickier for a beginner. When it works, it's much faster than Coulomb's Law, but it doesn't always work for you. The reason why it doesn't always work is simple: E enters the equation as a function, not as a value, and that function is being integrated over. Hence every instance of a Gauss's Law integral involves an infinite number of distinguishable values of E. To illustrate, suppose you are dealing with the integral of E over the Earth's surface. Then you must consider E at Kingsville, E at Canyon, E at Moscow (in Idaho and Russia both), E at Timbuktu, .... So you have one equation in an infinite number of unknowns, and the only way to succeed is to have enough information to connect all those distinct E's to a single unknown magnitude. That requires symmetry, so there is only a limited number of problem types where you use the integral form of Gauss's Law to solve problems.

The problem types where Gauss's Law in integral form works to find E are these: spherical problems; cylindrical problems; flat problems; and problems that are approximately one of those three types. A point charge or a solid sphere and any number of concentric spherical shells can be combined in a spherical problem; a long line charge or a long solid cylinder and any number of concentric long cylindrical shells can be combined in a cylindrical problem; any number of large flat sheets (thin or thick) can make up a flat problem. Note the words "long" and "large"; unless those could be replaced by "infinite", the problem is actually only approximately symmetric and the Gauss's Law solution is only an approximate field, though it is often quite good enough.

For those cases where the integral form of Gauss's Law doesn't help enough, there is also the differential form. If you really have an infinite number of distinct unknowns, then you need an infinite number of distinct equations to solve for them in, and that is one aspect of a differential equation: it is a distinguishable equation for each value of the independent variable.

To get the differential equation form, we apply the 'divergence theorem', also called Gauss's Theorem (different from Gauss's Law, but named for the same man) or Green's Theorem (one of several theorems with that name). This theorem relates the flux of any vector field, how much it crosses a surface, to how much it increases inside (if it doesn't increase inside, then it had to enter in the same strength as it left and the net flux is zero). The amount of increase inside is the divergence of the vector, or del dot the vector. Hence the flux part of Gauss's Law equals the volume integral of del dot E. The charge enclosed, of course, by definition of density, equals the volume integral of charge density. Since these integrals are proportional (proportionality constant e_o) for every volume, the integrands must be in the same proportion (just consider infinitesimal volumes). Now that you have this equation, you can use it on other problem configurations, if your skills with differential equations are adequate; probably, much of what's new to you in this course will be aimed at developing such skills.

E-M Home Page/Unit 1/

Format for this page adapted with permission from pages
constructed by Dr. Lionel D. Hewett for his course
Modern Physics 1.