Module 2 Class Activities
Electric Field, Flux, and Gauss’s Law
I. Area as a vector
A. Hold a small piece of paper (e.g., an index card) flat in front of you. The paper can be thought of as a part of a larger plane surface.
- What single line could you use to specify the orientation of the plane of the paper (i.e., so that someone else could hold the paper in the same, or in a parallel, plane)?
B. The area of a flat surface can be represented by a single vector, called the area vector .
- What does the direction of the vector represent?
- What would you expect the magnitude of the vector to represent?
C. Place a large piece of graph paper flat on the table .

- Describe the direction and magnitude of the area vector,
, for the entire sheet of paper.
- Describe the direction and magnitude of the area vector,
, for each of the individual squares that make up the sheet.
D. Fold the graph paper twice so that it forms a hollow triangular tube.

- Can the entire sheet be represented by a single vector with the characteristics you defined above? If not, what is the minimum number of area vectors required?
E. Form the graph paper into a tube as shown.

- Can the orientation of each of the individual squares that make up the sheet of graph paper still be represented by
vectors as in part C above? Explain.
F. What must be true about a surface or a portion of a surface in order to be able to associate a single area vector A with that surface?
II. Electric Field
A. Imagine a plastic rod that is electrically charged. You are going to explore the region surrounding the rod by using a pith ball that has a charge of the same sign as the rod.

- Sketch vectors at each of the marked points to represent the electric force exerted on the ball at that location.
- How does the magnitude of the force exerted on the ball at point A compare to the magnitude of the force on the ball at point B?
B. Suppose that the charge, qtest, on the pith ball were halved.
- Would the electric force exerted on the ball at each location change? If so, how? If not, explain why not.
- Would the ratio F/qtest, change? If so, how? If not, explain why not.
C. The quantity evaluated at any point is called the electric field
at that point.
- How does the magnitude of the electric field at point A compare to the magnitude of the electric field at point B? Explain.
D. Sketch vectors at each of the marked points to represent the electric field at that location.

- Would the magnitude or the direction of the electric field at point A change if:
- the charge on the rod were increased? Explain.
- the magnitude of the test charge were increased? Explain.
- the sign of the test charge were changed? Explain.
The electric field is typically represented in two ways: by vectors or by electric field lines. In the vector representation, vectors are drawn at various points to indicate the direction and magnitude of the electric field at those points. In the field line representation, straight or curved lines are drawn so that the tangent to each point on the line is along the direction of the electric field at that point. Below , we explore how the field line representation can also reflect the magnitude of the electric field.
E. The diagram at right shows a two-dimensional top view of the electric field lines representing the electric field for a positively charged rod.
- You determined previously that the magnitude of the electric field at point A was larger than the field at point B. What feature of the electric field lines reflects this information about the magnitude of the field?

III. Flux
The nails in this bed of nails will represent uniform electric field lines.

At right is a two-dimensional representation of the same electric field as viewed from the side.

A. Compare the magnitude of the electric field at points P and Q. Explain your reasoning.
Suppose you were given another bed of nails with nails representing a weaker uniform electric field than the one above. How would the two beds of nails differ? Explain.
B. Imagine a wire loop. The loop represents the boundary of an imaginary flat surface of area A. (In order to allow the nails that represent the field to pass through the surface, you have only been given the boundary of the surface.)
Draw a diagram to show the relative orientation of the loop and the electric field so that the number of field lines that pass through the surface of the loop is:
- the maximum possible.
- the minimum possible.
For a given surface, the electric flux, , is proportional to the number of field lines through the surface. For a uniform electric field, the maximum electric flux is equal to the product of electric field at the surface and the surface area (i.e., EA). The electric flux is defined to be positive when the electric field E has a component in the same direction as the area vector A and is negative when the electric field has a component in the direction opposite to the area vector.
C. Sketch vectors and
such that the electric flux is:
- positive
- negative
- zero
D. You will now examine the relationship between the number of field lines through a surface and the angle between and
.
Imagine placing the loop over the nails so that the number of field lines through it is a maximum. You can use this image for reference:

- Determine the angle between
and
. Record both that angle (
) and the number of field lines (n) that pass through the loop.
- Rotate the loop until there is one fewer row of nails passing through it. Determine the angle between
and
and record your measurement. Continue in this way until
= 180°.
- On graph paper, plot a graph of n versus
. (Let the number of field lines through the surface be a negative number for angles between 90° and 180°.)
E. When and
were parallel, we called the quantity EA the electric flux through the surface. For the parallel case, we found that EA is proportional to the number of field lines through the surface.
- By what trigonometric function of
must you multiply EA so that the product is proportional to the number of field lines through the area for any orientation of the surface?
- Rewrite the quantity described above as a product of just the vectors
and
.
Consult with an LA or instructor before proceeding.
Gauss’s Law
I. Electric Flux through closed surfaces
In the previous part, we found that the electric flux through a set of imaginary surfaces, ; each with a uniform electric field,
, can be written as:

The area vectors at each point on a closed surface (i.e., a surface that surrounds a region so that the only way out of the region is through the surface) are chosen by convention to point out of the enclosed region. A closed imaginary surface is called a Gaussian surface.
In the following questions, a Gaussian cylinder with radius a and length L is placed in various electric fields. The end caps are labeled A and C and the side surface is labeled B. In each case, base your answer about the net flux only on qualitative arguments about the magnitude of the flux through the end caps and side surface.
A. The Gaussian cylinder is in a uniform electric field of magnitude E, aligned with the cylinder axis.
- Find the sign and magnitude of the flux through:
- surface A
- surface B
- surface C
- Is the net flux through the Gaussian surface positive, negative, or zero?

B. The Gaussian cylinder encloses a negative charge. (The field from part A is removed.)
- Find the sign of the flux through:
- surface A
- surface B
- surface C
- Is the net flux through the Gaussian surface positive, negative, or zero?

C. The Gaussian cylinder encloses opposite charges of equal magnitude. (The charges are on the axis of the cylinder and equidistant from the center.)
- Find the sign of the flux through:
- surface A
- surface B
- surface C
- Is the net flux through the Gaussian surface positive, negative, or zero?

D. A positive charge is located above the Gaussian cylinder.
- Find the sign of the flux through:
- surface A
- surface B
- surface C
- Is the net flux through the Gaussian surface positive, negative, or zero?

Consult with an LA or instructor before proceeding.
II. Gauss’s Law
Gauss’ law states that the electric flux through a Gaussian surface is directly proportional to the net charge enclosed by the surface:

A. Are your answers to parts A-C of the previous section consistent with Gauss’ law? Explain.
B. In part D, you tried to determine the sign of the flux through the Gaussian cylinder shown.
- If you have not done so already, use Gauss’ law to determine whether the net flux through the Gaussian surface is positive, negative, or zero. Explain.
- If
A = -10 Nm2/C and
C = 2 Nm2/C, what is
B?

C. Find the net flux through each of the Gaussian surfaces below.

D. The three spherical Gaussian surfaces at right each enclose a charge +Q0. In case C there is another charge -6Q0 outside the surface.

Consider the following conversation:
Since each Gaussian surface encloses the same charge, the net flux through each must be the same.
Student 1
Gauss’s Law doesn’t apply here. The electric field at the Gaussian surface in case B is weaker than in case A because the surface is farther from the charge. Since the flux is proportional to the electric field strength, the flux must also be smaller in case B.
Student 2
I was comparing A and C. In C, the charge outside changes the field over the whole surface. The areas are the same so the flux must be different.
Student 3
Do you agree with any of the students? Explain.
Consult with an LA or instructor before proceeding.
III. Application of Gauss’s Law
A. A large sheet has charge density . A cylindrical Gaussian surface encloses a portion of the sheet and extends a distance L0 on either side of the sheet. A1, A2, and A3 are the areas of the ends and curved side, respectively. Only a small portion of the sheet is shown.

- On the diagram, indicate the location of the charge enclosed by the Gaussian cylinder.
- In terms of
, and other relevant quantities, what is the net charge enclosed by the Gaussian cylinder?
- In terms of
- Sketch the electric field lines on both sides of the sheet.
- Does the Gaussian cylinder affect the field lines or the charge distribution? Explain.
- Let EL and ER represent the magnitude of the electric field on the left and right ends of the Gaussian surface.
- How do the magnitudes of EL and ER compare? Explain.
- How do the magnitudes of the areas of the ends of the Gaussian surface compare?
- Through which of the surfaces (A1, A2, A3) is there a net flux? Explain using a sketch showing the relative orientation of the electric field vector and the area vectors.
- Write an expression for the net electric flux
, through the cylinder in terms of the three areas (A1, A2, A3), EL, and ER.
- Use the relationships between the electric fields EL and ER and between the areas A1 and A2 to simplify your equation for the net flux.
- Write an expression for the net electric flux
- Gauss’ law (
) relates the net electric flux through a Gaussian surface (which you found in step 4) to the net charge enclosed (which you found in step 1 ). Use this relationship to find the direction and magnitude of the electric field at the right end of the cylinder in terms of
.
- What is the electric field at the left end of the cylinder?
- Does the electric field near a large sheet of charge depend on the distance from the sheet? Use your results above to justify your answer.
- Is your answer consistent with the electric field lines you sketched in step 2 ? Explain.
Consult with an LA or instructor before proceeding.
B. This Gaussian cylinder encloses a portion of two identical large sheets. The charge density of the sheet on the left is ; the charge density of the sheet on the right is
.

- Find the net charge enclosed by the Gaussian cylinder in terms of
and any relevant dimensions.
- Let EL and ER be the magnitudes of the electric fields at the left and right end caps of the Gaussian cylinder respectively.
- Is EL greater than, less than, or equal to ER? Explain.
- Find the net flux through the Gaussian cylinder in terms of EL, ER, and any relevant dimensions.
- Use Gauss’ law to find the electric field a distance L0, to the right of the rightmost sheet.
- Are your results consistent with the results you would obtain using superposition? Explain.