## Gauss' Law

1-21-98

Relevant sections in the book : 18.9

Please note that although these notes deal primarily with Gauss' Law, we're going to downplay the derivations using Gauss' Law. They're useful to see, but from this class we'll really expect you to be able to apply basic ideas about electric field rather than use Gauss' Law to derive electric fields.

### Electric flux

Gauss' Law is a powerful method for calculating the electric field from a single charge, or a distribution of charge. It can appear complicated, but it's straightforward as long as you have a good understanding of electric flux. Electric flux is a measure of the number of electric field lines passing through an area. To calculate the flux through a particular surface, multiply the surface area by the component of the electric field perpendicular to the surface. If the electric field is parallel to the surface, no field lines pass through the surface and the flux will be zero. The maximum flux occurs when the field is perpendicular to the surface.

### Calculating the electric field

We'll take this idea of electric flux, and, by applying Gauss' Law, we'll use it to figure out the electric field. Gauss' Law looks a little complicated, but when the geometry of a situation is straightforward, which it will be for any case we consider, the equation reduces to a simpler form. We'll state the general form, and then apply it to two examples to see how it gets used.

Gauss' Law - the sum of the electric flux through a surface is equal to the charge enclosed by a surface divided by a constant , the permittivity of free space.

What is the permittivity of free space? It's a constant related to the constant k that appears in Coulomb's law. The relationship between the two is this:

### Calculating the field from a point charge

Let's use Gauss' law to calculate the electric field from a point charge of size Q, at a distance r away from the charge. We already have an equation for this:

This is the answer we should get if we apply Gauss' law. To apply Gauss' law, all we need to do is to surround the point charge with an imaginary sphere. To determine the field a distance r away from the charge, the sphere should have a radius of r, and it should be centered on the charge.

If the charge is positive, the field lines will radiate out from the charge and pass through the sphere perpendicular to the surface of the sphere. With a negative charge, the field lines will be directed toward the charge; they will still pass through the sphere perpendicular to the surface, however. The other important thing to realize is that the magnitude of the electric field will be the same at each point on the sphere, because each point is the same distance away from the charge.

The fact that the field lines are perpendicular to the surface simplifies Gauss' law. The angle in Gauss' law is the angle between the electric field and the line perpendicular to the surface (also known as the normal to the surface), pointing out from the surface. For a positive charge, the electric field is in the same direction as the normal, the angle between them is 0, and the cosine is 1. With a positive charge, E and the normal are in opposite directions, giving an angle of 180° and a -1 for the cosine. Because E is constant everywhere over the surface, it can be brought out of the summation sign in Gauss' law. In this case, then, Gauss' law becomes:

That's for a positive charge. With a negative charge there'd be a minus sign in front of the E. So, what happens with the sum of the areas? Basically what we're doing is this. We split the sphere up into pieces, work out for each piece, and add them all up. If we've reduced that sum to a sum of the area of all the pieces, that's just the total area of the sphere, . We can put that in place of the sum of the areas to get:

Solving this for the electric field gives:

So, using Gauss' law we've derived the equation for the field from a point charge.

### The field from a large flat plate

Let's apply Gauss' law to figure out the electric field from a large flat conductor that has a charge Q uniformly distributed over it. In reality, some of the charge will pile up at the edges of the conductor, but we'll assume that the conductor is large enough that these edge effects are negligible.

To apply Gauss' law, we need to come up with a surface (known as a Gaussian surface) that we can use to apply the law to. With a point charge, the appropriate Gaussian surface was a sphere. In this case, let's work with a cylinder. A surface in the form of a box would work, too, but it's got six sides while a cylinder only has three, which simplifies matters a little. A diagram of the plate, with the Gaussian surface passing through the plate, is shown here:

To apply Gauss' law, we need to have some idea of what the field looks like. If the plate has a positive charge, the field lines will emerge perpendicular to the plate. No field lines pass through the side of the cylinder. The field lines do, however, pass through the two ends of the cylinder, perpendicular to the ends. As long as the cylinder is far away from an edge of the plate the charge will be uniformly distributed, and the field will be constant, as shown in the figure.

Applying Gauss' law means adding up the electric flux passing through each part of the cylinder. The amount through one end is simply EA, where E is the electric field and A is the area of an end. The amount through the side is zero. Doing the sum in Gauss' law, then, gives us EA + 0 + EA = 2EA. As with the point charge, if the plate had a negative charge rather than a positive charge then the sum of the flux have a minus sign, -2EA.

The sum of the flux is related to the charge enclosed by the cylinder. Let's call that Q, so:

If the plate itself has an area a and a total charge q, and we assume the charge is uniformly distributed over the surface of the plate, then q / a = Q / A. We'll call this ratio the surface charge density, the charge per unit area:

Writing the electric field in terms of the surface charge density gives:

This field is perpendicular to the plate on both sides, and points away from the plate for a positive charge, towards the plate for a negative charge. Note that the field does not depend on the distance from the plate! As long as the distance is less than the length of the plate, the field is constant.

### The field between two plates

After calculating the field from one plate, the field from two plates placed close together can be found simply by adding the two fields. If the two plates are identical, but one has a charge of +Q and the other has a charge of -Q, the field in the region between them will point from the positive plate to the negative plate and will have a value of :

In the region outside the plates, the two fields will exactly cancel. This arrangement is known as a parallel-plate capacitor. We'll look at that again as a device used to store charge.