# PHYS 2212 Module 2 Class Activities

## Gauss’s Law

### Goals:

• To understand and explain in words the physical meaning of Gauss’s law.
• To employ symmetry arguments to determine the direction of the electric field for simple charge distributions.
• To use Gauss’s law to calculate the the electric field produced by an appropriate charge distribution.

### Introduction

In mathematical form, Gauss’s Law can be written as

### Part 1

Describe a procedure for applying Gauss’s Law of electromagnetism in your own words, without using equations.

1. Define the following terms carefully: surface element dA, closed surface S, enclosed charge qenc.
2. Describe the dot product in Gauss’s Law in terms of vector components.
3. Describe the process of integration in terms of a sum over many small surface elements.

### Part 2: Using symmetry to determine the direction of the electric field

Gauss’s Law can be used to determine the magnitude of the electric field in several important geometries. Gauss’s Law is always true, but it doesn’t help you determine the magnitude of the electric field unless you know the direction of the field. In the special cases where Gauss’s Law can be used to calculate the electric field, the direction of the field can be determined from the symmetry of the charge distribution.

Figure 1 shows a uniformly charged, square plate in the x-y plane with its center at the origin. Rotating the charge distribution by 180° around the z-axis does
not change the distribution of charge. Every charge element that is moved by rotation is replaced by another, identical charge element after rotation. Therefore, we say that the charge distribution is symmetric with respect to a 180° rotation about the z-axis. Figure 1: A thin, uniformly charged square plate in the x-y plane, with its center at the origin ofan x-y-z coordinate system before and after rotation by 180°. The gray vectors at Point P show thethree components of E at Point P. The charge distributions before and after rotation are identical.Therefore we expect identical electric fields. Although the z-component of E is the same beforeand after rotation, the x- and y-components change sign. The only way these observations can bereconciled is if Ex = Ey = 0.

Symmetry arguments rely on two principles. First, changing the orientation of the source of an electric field (the charge) changes the orientation of its electric field in the same way. Second, identical charge distributions produce identical fields. Figure 1 shows that Ex and Ey at Point P change sign after a 180° rotation about the z-axis. They rotate along with the plate. But this rotation does not change the charge distribution, so it cannot change the electric field. The only way these two facts can both be true is if the x- and y-components of the electric field at Point P are
zero. In contrast, Ez is not changed by this rotation. Therefore the z-component of the electric field does not have to be zero. The electric field at Point P must point in the +z or −z directions.

This procedure yields the same result for all points on the z-axis. The procedure fails for points with nonzero x- and y-components, unless the square is infinite in extent. We cannot use this procedure to find the direction of the electric field for points that are not on the z-axis, except for the special case of a plate that extends to infinity in the x- and y-directions.

Three useful groups of symmetry operations are described below.

#### Translations (straight-line displacements)

Charge distributions with translational symmetry in the x-direction are not changed when the object is moved in the x-direction. A infinite, charged plane perpendicular to the z-axis has translational symmetry in the x- and y-directions. Movements in the x- and y-directions do not change the charge distribution, because every patch of moved charged is replaced by an identical patch of charge from elsewhere on the plane. Since the charge distribution is not changed if you move the plane in the
x- or y-directions, the electric field must not be changed by this motion. That is, the electric field vector at any Point P above a charged, infinite plane cannot depend on the x- or y-components of that point.

#### Rotations

Charge distributions with rotational symmetry are not changed by a rotation about some axis. The axis and the angle of rotation must be specified, although some shapes are symmetric about a special axis for all angles. The square in Figure 1 is symmetric under rotations of 90°, 180°, and 270° about the z-axis.

#### Reflections

The difference between a vector and its reflection is similar to the difference between a vector and its reflection in a mirror. Reflection through the x-y plane changes the sign of vector z-components without changing their magnitude. The x- and y-components are not changed. You will not need to use reflections in this exercise, but they can be useful. Figure 2: Sketch representing an infinitely long cylindrical shell (like a pipe) of radius R and thickness ∆R, centered on the z-axis. The capital R indicates the position of a point in the charge distribution. The lower case r indicates the position of a point at which the electric field is to be determined.

Figure 2 represents an infinitely long cylindrical shell of radius R and thickness ∆R, centered on the z-axis. Assume that it is made of a material with a uniform, positive volume charge density .

Sketch “before” and “after” pictures of this charge distribution and the electric field components at the point of interest for each symmetry operation, as in the example above.

1. Use symmetry arguments to show that the electric field outside the cylinder at Point P on the x-axis is directed in the +x-direction. (Rotate the cylinder 180° about the x-axis.)
2. Use symmetry arguments to show that the electric field outside the cylinder at Point W on the y-axis is directed in the +y-direction.
3. Use symmetry arguments to show that the electric field is directed along a cylinder radius— that is, pointing directly toward or away from the z-axis. Further, show that the magnitude of the electric field is constant on circles of radius , the distance from the z-axis to Point P.
4. Use additional symmetry arguments to show that the electric field outside the cylinder does not depend on the z-component of position.

### Part 3: Constructing a Gaussian surface

To perform the integral in Gauss’s Law, one must be able to compute the dot product inside the integral where and θ is the angle between E and dA. The circle on the integral sign indicates that the surface must be closed. Like cubes or spheres, closed surfaces are composed of pieces whose orientations vary.

Since the direction of dA is along the surface normal direction, it changes depending where you are on the closed surface. From Part 2, we know the direction of E due to an infinitely long cylindrical shell. We get to choose the surface, and so have some control over the direction of dA.

The best we can hope for is to make constant for each part of the surface. This requires that θ be constant for each part of the surface. For closed surfaces, the only workable angles are 0°, 90°, 180° and 270°. Then cos θ = 0 or ±1.

Sketch a Gaussian surface for the infinitely long cylindrical shell in Figure 2, making sure that the angle between E and dA is one of the approved angles for each side of the closed surface. Since the surface must be closed, your Gaussian surface cannot be infinitely long. Show the dimensions of the Gaussian surface on your sketch. Write the Gauss’s Law integral as the sum of the integrals over each part of your surface, using the fact that cos θ = 0 or ±1.

### Part 4: Calculating the flux through the Gaussian surface

Even though we have determined θ for each of the required flux integral(s), we still must be careful of the functional dependence of the remaining scalars E and dA. You can’t calculate an integral unless you know the function. This is not a problem where cos θ = 0, as the integral for those parts must equal zero. But for surfaces where cos θ = ±1, the integral can be a problem. Practically, the only hope for a solution is if the magnitude E is constant.

Fortunately, you showed in Part 2 that for a fixed r, the magnitude of the electric field E is indeed constant. If (and only if) r is constant for the parts of your Gaussian surface where cos θ = ±1, you can perform the integral. After factoring out the constant factor of E, the remaining integral should have the form: where A is the area of a part of your Gaussian surface where cos θ = +1 or −1.

Calculate the total flux through your Gaussian surface S.

### Part 5: Calculating the charge enclosed by the Gaussian surface

Since S is a closed surface, with a definite inside and outside, it encloses a well defined volume. If all the charges in the system are simple point charges, one can simply identify which point charges are inside the volume and sum their values. Another simple case is when the charge density in the volume is uniform, or constant. Then the enclosed charge is given by the product of the volume V
inside S and the charge density ; that is, qEnclosed = V. Care must be taken to include only the charge inside S. If part of a charge distribution is not inside S (that is, some parts poke through the surface), only the part inside S contributes to qEnclosed.

If the charge density is a function of R only, it can still have rotational symmetry. (In this case, the shape is not changed by any rotation about the axis of symmetry.) Then the enclosed charge may be found by integration. To minimize confusion, we will use the variable R to refer to the radial coordinate of a position in the charge distribution (the cylinder). We will use the variable r to refer to the radius of our Gaussian surface.

In your calculus class, you used the method of cylindrical shells to determine the volume of shapes with rotational symmetry. You can use the same method to determine the total charge in such an object by introducing a factor of , the volume charge density. In the shell method, the volume of a thin cylindrical shell is given by where h is the (constant) length of the shell. To see that this must be true, consider a solid cylinder with radius RCyl whose charge density might be a function of R. In this case one can approximate the volume of the cylinder as the sum of the volumes of a series of N thin cylindrical shells of radii R1, R2, R3 … RN. If we take the thickness of each shell to be ∆R = RCyl/N, we can construct a series of shells with radii RJ = J∆R (where J = 1, 2, 3 … N). As N goes to infinity the sum of the
shell volumes VJ becomes an integral, and the integral yields the exact value of VCyl. The progression from thin shells to integrals can be written:

To find the charge enclosed in the entire cylinder, qCyl, one need only add a factor of to the integral:

You can find qCyl for almost any charge distribution (R) that depends only on R. If the radius of your Gaussian surface is greater than the radius of the cylinder, qEnclosed = qCyl; the upper limit of integration is then RCyl. If the radius of your Gaussian surface is less than the radius of the cylinder, you must include only the charge inside the Gaussian surface. To get qEnclosed, you reduce the upper limit of the integral from RCyl to r, the radius of your Gaussian surface.

Compute the total charge inside in a cylinder of length h and radius RCyl when (R) = αR. Use the result to compute the electric field produced by the cylinder at points outside the cylinder (r > RCyl). Note that since r > RCyl, the Gaussian surface (with radius r) encloses all the charge in the cylinder. State the direction of the electric field inside and outside the cylinder when α > 0, that is, when the cylinder carries positive charge.

Find the charge enclosed by a Gaussian surface as a function of its radius, r, when (R) = αR, for the case of r < RCyl. Since r < RCyl, a Gaussian surface with radius r encloses only part of the cylinder’s charge. Use the result with the rest of Gauss’s Law to compute the magnitude of the electric field inside the cylinder as a function r for r < RCyl.