Activity 19 – Electric Fields
Studio Physics I
This activity will be carried out using a simulation module on the computer. You will need to
install and run a “share-ware” program on your computer. (You can uninstall it later.) It is in
your Physics I folder as efield12.exe or go to the Activities page of the web site and download it.
Run the installation program and accept the default locations. After installation, start the
program with Start – Program – Electric Field – Electric Field. When the program starts, select
File / New and accept the defaults. The program will encourage you to register, but you should
just exit the registration window. We will only use this program today. You will probably want
to maximize the program and plotting windows on your computer.
Part I – Electric Field Vectors and Electric Field Lines
1. Click on the Add Charge button (third to the right after the Printer button) and then click
approximately in the center of the plotting grid. Specify 10 units. (The default unit of charge is
10 nano Coulombs.) You should see a red circle where you clicked with a “10” inside. On your
write-up, put “1. Done.”
2. Next, click on the Tester button (looks like a little volt meter). As you move the cursor
around on the graph, you will see an arrow representing the electric field vector. The length of
the vector is roughly proportional to the magnitude of the vector. You should see that the vector
always points away from the positive charge. On your write-up, put “2. Done.”
3. Click on the Field Lines button (first to the right after the Printer button). The program will
draw electric field lines. A field line is a continuous line that follows the direction of the field
vector at each point on the line. If you put
the cursor back in the data area and move it
around, you can see that the electric field
lines are in the same direction as the electric
field vectors at every point. Your plot
should look something like the picture
below:
Answer the following questions:
3A. If an electric field line or vector is not
drawn at a point, does that mean that the
electric field is zero there?
3B. Is there a fundamental law of physics
that gives the length in meters for drawing
an electric field vector of a given magnitude of N/C, or is it just a matter of scaling to make a
nice visual representation?
3C. Where the field lines are closer together, is the magnitude of the electric field vector less
than, equal to, or greater than locations where they are further apart. (Check your answer.)
3D. Do the field vectors and field lines point in the direction you expect for a positive charge?
4. Click on the Add Charge button again and put a charge of –10 units to the right of the positive
charge Erase the field lines with the Cleaning Cloth button and then redraw the lines with the
Rev. 23-Mar-07 Bedrosian
Field Lines button. The field lines should curve from the positive charge to the negative charge.
Click on the Tester button again and check the directions of the field vectors at various points
versus the directions of the field lines. Also check the magnitude of the electric field to verify
the relationship with the distance between field lines that you should have found for question 3C.
Do your results make sense in terms of what you know about electric field vectors and field
lines? If not, check with your TA or instructor.
Part II – Connecting Electric Field to Electric Force
5. The direction of the electric field at a given point is defined to be the direction the
electrostatic force would be on a positive charge if one were placed at the point. The magnitude
of the electric field at a given point is the magnitude of the electrostatic force that would be
present on a 1.0 C charge if it were placed at the point. However, there does not actually have
to be a charge present at that point.
Using this information, your knowledge of the electrostatic force and the simulation of the
positive and negative charges, describe at least two ways that what you observe with electric field
vectors is consistent with this definition of the direction and magnitude of the electric field.
Part III – More Complex Configurations
6. Add 4-6 additional charges between –10 and 10 units in random locations. Erase the old field
lines and draw new ones. Can you make a general statement about where electric field lines
begin and end?
7. Erase all charges and start over. Create a horizontal chain of about eight +1 charges and
below it a chain of the same number of –1 charges. Draw the electric field lines (click the
button). (This will look something line the picture shown below.) How would you describe the
electric field between the charges (as compared to the field you saw in step 6) – approximately
constant in magnitude and direction or changing rapidly from spot to spot? If you had to
calculate the force on a charged particle due to a distribution of charges like those shown below,
in the region where the electric field is nearly constant, would it be easier to use the electric field
approach or go back to Coulomb’s Law? Explain.
Rev. 23-Mar-07 Bedrosian
Part IV – Exercise
8. In this exercise, we will calculate the total electric field at a specified location due to two
point charges as shown in the figure below. We will use two different methods: the one
recommended in the book and then the one that uses unit vectors. Obviously, they should give
the same answer. Copy the figure below and sketch the electric field vectors at the location
marked with an "×" from each point charge separately, and the total field, without calculating
numbers. Use your knowledge of electric fields and
Y
symmetry.
The locations of the points are as follows:
Point A (where we calculate the field) is (0,0).
The –1 nC (–1.0 × 10–9 C) charge is at (–0.25,–0.25√3).
The +1 nC (+1.0 × 10–9 C) charge is at (+0.25,–0.25√3).
Your sketch of the figure at the right with the electric field
vectors should help you check your calculations.

E 
1 q
4  0 r 2
X
point A
0.50 m
-1 nC
0.50 m
0.50 m
+1 nC
9. Calculate the magnitude of the electric field at Point A created by the –1 nC charge. What is
the angle of the electric field from this charge with respect to the horizontal axis? Use
trigonometry to determine the X and Y components of the electric field from the –1 nC charge.
Remember to include – signs as needed.
10. Repeat step 9 for the +1 nC charge. You should be able to reuse several of your calculations.
11. Add the X and Y components of the electric fields from each charge to get the total electric
field. Does it point in the direction you predicted in step 8? If not, check your work.
12. We will now use the alternative method to calculate the electric field:

1 q
 r̂ 
E
4  0 r 2
First, we will calculate "r-hat" for the –1 nC charge. Find the displacement vector from Point A
to the location of the –1 nC charge. Call the X component of that vector dx and the Y
component dy. The distance between the points is marked on the diagram. However, you should
check it using r = √(dx2+dy2). Also, be careful with + and – signs for dx and dy.
r̂ 
dy
dx
î 
ĵ
r
r
Finally, since we actually use the negative of r-hat in the formula, switch the signs of the
components:
 r̂ 
 dy
 dx
î 
ĵ
r
r
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13. The second part of the calculation is the multiplier:
E0 
1 q
4  0 r 2
Note that this is a scalar and it could be + or – depending on the sign of q. In this case, it will be
negative. Calculate E0 and then multiply by the result of step 12 to get the electric field:
  E0 dx
 E0 d y
E
î 
ĵ
r
r
You should have gotten the same result as step 9. If not, check your work. Watch the – signs!
14. Repeat steps 12 and 13 for the +1 nC charge. Again, you should be able to reuse several of
your calculations. You should have gotten the same result as step 10. If not, check your work.
15. Add the X and Y components of the electric fields from steps 13 and 14 to get the total
electric field. Does it agree with your answer from step 11? If not, check your work.
16. Which method seemed most understandable to you, meaning most likely for you to get the
correct answer on a test? This is the method you should use on Exam 3 and the Final Exam.
(There is no right or wrong answer here.)
Important: There is a common student error when calculating the formula for the electric field,
when you need to divide by r2, to divide instead by dx2 for the X component of the field and by
dy2 for the Y component of the field. This is incorrect! Always divide by r2 = dx2+dy2.
Rev. 23-Mar-07 Bedrosian