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Magnitude of Electric Field Question

  1. Aug 21, 2007 #1
    1. The problem statement:
    Four point charges are placed at the corners of a square. Starting from the top left corner and working around in the clockwise direction, their values are as follows; 4 micro coulombs, 1 micro coulomb, 4 micro coulombs, and negative 3 micro coulombs. Each side of the square has length 2.0 m. Determine the magnitude of the electric field at point "P" at the center of the square.

    2. Relevant equations
    Electric Field: E= F/q
    Force: F= q1 x q2/r^2

    3. The attempt at a solution
    I not sure where to begin with this problem. I believe that the first and third point charges would cancel each other out. Based on what I know, I need a test charge at point "P" and I need to know the magnitude of that test charge. Then I can apply coulombs law by using the force equation above. Once I know force, I can plug that value into the electric field equation which should answer the question. Is there another way to solve this problem without a test charge?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Aug 21, 2007 #2

    Doc Al

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    Staff: Mentor

    That's true.

    You don't need an actual value for the test charge. Just call the test charge "q"; you'll find that it cancels out when you solve for the field.
    You should know the formula for the E field from a point charge. (Essentially you'll be deriving that formula from Coulomb's law if you follow the above approach.) The field at point P is the vector sum of the contributions from each of the four point charges.
  4. Aug 21, 2007 #3
    Thanks for the reply.

    1. I know that the formula for electric field of a point charge is: (k)(q)/r^2

    2. I see that the point charge has been cancelled algebraically.

    3. I think I don't need to consider the 1st and 3rd point charges since they cancel out.

    4. I think the vectors should both point in the same direction.

    5. Applying the formula for the electric field of a test charge using the magnitude of the second point charge then doing the same for the fourth point charge will give me two values for "E".

    6. Then I should add those values to get the vector sum...is this correct?
  5. Aug 21, 2007 #4

    Doc Al

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    Staff: Mentor

    Good. Realize that "q" is the charge creating the field, not a "test charge". You don't need a test charge, since you already have the formula for the field.
    Yes, the fields of those two charges cancel out at point P. So you can forget them.

    Yes, the field contributions from the 2nd and 4th charge will point in the same direction. (What direction?)

    No need for any "test charges". Just find the field from each point charge (the 2nd and 4th charge).

  6. Aug 21, 2007 #5
    Thanks for the additional help; however I'm still not getting the correct answer. According to my answer key, I should get 1.8 x 10^4.

    1. Both vectors point toward the fourth charge.

    2. My formula for the electric field created by the second charge is:
    (8.99 x 10^9)(1 x 10^-6)/1.41^2. When I plug it into my calculator I get 4.522 x 10^3.

    3. My formula for the electric field created by the fourth charge is:
    (8.99 x 10^9)(3 x 10^-6)/1.41^2. When I plug it into my calculator I get 1.357 x 10^4.

    4. Adding these two results together don't give me the answer I'm looking for.
  7. Aug 21, 2007 #6

    Doc Al

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    Staff: Mentor

    Your answers look good to me. And if you add them up, you get an answer that matches the answer key (when you round off to two digits). Check your addition!
  8. Aug 21, 2007 #7
    Thanks for the help....I finally got it! Hope you are around in the future because I'm sure I'll be posting again pretty soon. Physics is interesting but it doesn't come easy for me.
  9. Aug 21, 2007 #8


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    Homework Helper

    You worked through the problem very well. I'm sure it will get easier as you do more and more problems.
  10. Aug 21, 2007 #9

    Good job solving that problem. I'm currently going through calculus based electricity and magnetism, so if I come by one of your posts. I'll try to help.

    Best of luck,

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