Gauss' Law

Gauss' Law, òE•ndA = Q_inside/ε₀, is useful for determining the electric field at a given point where the field has a great deal of symmetry. To be specific we use Gauss' Law in three cases; infinitely large plates where E(-x) = -E(x), infinitely long cylinders where the charge density ρ (charge per volume) is only a function of the radial distance from the centre of the cylinder, and spheres where the charge density ρ (charge per volume) is only a function of the radial distance from the centre of the sphere.

The Right Hand Side

If the symmetry exists, it can be shown that the nasty integral on the left hand side of Gauss' Law, òE•ndA, reduces to the following:

The above results can be derived quite simply as shown below. The trick in each case is to find a gaussian surface for which E•n is either zero or a constant.

The Parallel Plate

In Figure 1 below, a cross-section of a parallel plate is shown. The centre of the plate is at x = 0. Provided that the distribution of charge in the plate depends only on x, i.e. that ρ(x,y,z) = ρ(x), and that the distribution is symmetric about x = 0, i.e. ρ(-x) = ρ(x), then symmetry implies that the electric field produced by the charge distribution must also be symmetric about x = 0, i.e. E(-x) = E(x), and that the field must point in opposite direction, i.e. E(-x) = -E(x). What is meant by symmetry here is that one cannot distinguish between right and left. Note that if the charge on the parallel plate shown in Figure 1 was negative, the field lines would point in the opposite direction.

The blue box is a side view of the gaussian "pillbox" in this case. Note that the centre of the pillbox is also at x = 0. The simplest pillbox is a cylinder but it could be a box. The only requirement is that the ends are perpendicular to the field lines and that the sides be parallel to the field lines. The surface integral then reduces to integrals over the ends and the side

Now the middle integral vanishes since we have cleverly chosen the pillbox such that E is parallel to the sides so that E•n_side = 0. As for the right and left sides, which we will assume are at x = ±a, the electric field is constant everywhere along the right or left faces of the pillbox. Thus E•n_left simplifies to E(-a) and likewise E•n_right becomes E(a). E(-a) and E(a) are constants and may be brought outside the integrals. We now have

The surface integrals that we are left with are simply the surface areas of the ends of the pillbox. We will call this area A. So we have the further simplification that

and, since E(x) is symmetric, we have the standard result

where the direction of the field depend on the sign of the charge on the plate. Note that x = a could be either inside or outside the plate.

The Radial Cylinder

In Figure 2, we have a picture of the an end view and a side view of an infinitely long wire. As long as the charge distribution in the wire depends only on r, i.e. ρ(r,θ,z) = ρ(r), the electric field will be radially outwards or inwards depending on the sign of the charge on the wire, i.e. E(r) = E(r)r. The symmetry here is that a cylinder has no "sides"; any one point on the side of a cylinder looks like any other.

The gaussian surface in this case is a cylinder of length L and radius a centred on the wire. The ends of the gaussian cylinder are flat so that the cylinder looks like a box in the side view, Figure 2(b). Again the surface integral can be broken into parts

However, here it is the contribution from the ends which vanish since E•n_left = E•n_right = 0. Since the gaussian cylinder is centred on the wire, the electric field is constant E = E(a)r everywhere on the gaussian cylinder's side. Thus E•n_radial = E(a)r•r = E(a), which is a constant and can be taken outside the integral, and we have

The integral is the surface area of the side of the cylinder, thus

the standard result for wires. Note that the radius of the gaussian cylinder, r = a, can be smaller than the radius of the wire.

The Radial Sphere

In Figure 3 we have a sphere of charge and a gaussian sphere surrounding it. As long as the charge distribution in the sphere depends only on r, i.e. ρ(r,θ,φ) = ρ(r), the electric field will be radially outwards or inwards depending on the sign of the charge on the sphere, i.e. E(r) = E(r)r. The symmetry here is that a sphere has no "sides"; any one point on the side of a sphere looks like any other.

A sphere only has one side so

Note that since the spheres are concentric, the electric field has the same magnitude E(a) everywhere on the surface of the gaussian sphere. Note that the radius of the gaussian sphere, r = a, can be either inside or outside the sphere.

The Left Hand Side

Consider the left-hand side of Gauss' Law, Q_inside/ε₀. Q_inside refers to the total charge contained by the gaussian surface. If the gaussian surface is bigger than the charged object this is simply the total charge on the object. However when the gaussian surface is smaller than the object, Q_inside will be less than the total charge. To calculate Q_inside in such cases, we introduce the charge distribution or charge density ρ. The charge density may be a constant or a complicated function. For an object with a uniform charge distribution, the charge density is a constant given by

For example, if a solid sphere of radius R had a charge Q smeared uniformly through it, then

Note that this is only true for object with a uniform charge density. If the charge distribution is not uniform you will be given ρ.

In general, for each of the case discussed above, Q_inside can be expressed as an integral involving ρ, Q_inside = òr(r)dV. This nasty volume integral reduces to a much simpler linear integral for the highly symmetric cases we encounter. The standard results are:

Occasionally the charge on a object, particularly with conductor, resides in a very thin shell. It is customary to introduce a surface charge density Σ defined by

where A_surface is the surface area of the shell. In problems you are given Σ and calculate Q_inside = ΣA_surface.

Summary

The standard results are:

Memorize these results.

Note that the integral depends on the size of the gaussian surface, a, but that ρ = 0 outside the object.

In problems you will be asked to find the electric field for all values of a (or more usually all values of x or r).

Be sure to always indicate the direction of E since it is a vector.

A sketch of the electric field as a function of the distance from the centre of the object should always be included even if it isn't explicitly asked for.

EXAMPLE: A very long charged object, as shown in the following diagram, consists of an inner non-conducting wire separated by a vacuum from a non-conducting shell. The inner wire has inner radius r = R unit and a charge distribution ρ(r) = Br². The gap is has a thickness of R unit. The outer shell has a constant surface charge density Σ = C. Use Gauss' Law to find the electric field as a function of the distance from the centre of the inner wire.

Solution: There are three regions that we have to solve Gauss' Law for; (a) 0 < a < R which is for a gaussian cylinder inside the solid wire, (b) R < a < 2R units which is for a gaussian cylinder in the vacuum gap, and (c) a > 2R units outside the object completely.

(a) Note that a is inside the inner wire. We know that in this region ρ_wire(r) = Br². So we have

Thus we find

(b) Note that a is bigger than the inner wire. We must break our integral into two pieces. We have

Since ρ_gap(r) = 0, as there is nothing in the gap, the second integral above vanishes. We know, as we've already seen, ρ_wire(r), so we have

Note that the right-hand side no longer depends on a, and we find

It is typical of these problems when all the charge is in a volume smaller than the gaussian surface that the electric field strength varies as 1/a.

(c) We have to take the charge on the shell into account as well as the radial distribution. The outer shell has a constant surface charge density Σ = C.

We have

The second and fourth integrals vanish since ρ_gap(r) = ρ_outside(r) = 0. We have already evaluated the first integral and Q_surface = ΣA_shell = Σ2πrL = 4πRLC, where the radius of the shell is r = 2R. Hence,

Our final result, therefore, is

As expected, the electric field drops as 1/R since the total charge inside is now a constant for a > 2R. Since we haven't been told the sign or magnitude of B and C, we cannot determine the direction of the electric field. We will assume that it is positive for convenience. In summary the electric field is

Note that r has been substituted for a since r is the customary symbol for radial distance. We have been using a to avoid confusion with the variable of integration r but since we are finished that is no longer a concern. A sketch of the electric field as a function of r would look like the following.

Note that there are discontinuities in the electric field at the edge of the inner wire r = R and at the outer shell r = 2R. This is typical of objects with a non-continuous charge distribution (see Tipler Chapter 19-3). Note the jump at r = 2R, this is typical of surface charges.

Questions?mike.coombes@kwantlen.ca