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Homework Help: Determine the magnitude and direction of the force on each charge

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Two Q = +5.86 mC charges are placed at opposite corners of a square d = 0.9 m on each side, and -Q = -5.86 mC charges are placed at the remaining corners.

    Determine the magnitude and direction of the force on each charge.

    2. Relevant equations

    Coulomb's Law

    3. The attempt at a solution

    My teacher just gave this to us without any explanation of how to go about solving such a problem. I have a basic understanding of electric fields and I know coulomb's law, can someone please help me work through this? I'm wondering if this is at all similar to the problems where the charges are simply placed on a horizontal plane with a set distance between two charges. I know how to solve those problems, but I'm not sure if you use the same strategy to solve this problem.

    Thanks in advance
  2. jcsd
  3. Apr 1, 2009 #2


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    In this problem you are using the F = q*E

    You need to take the ∑ of the Forces. But as before in your other problem you are dealing with vectors, and that is important since you have charges separated in both x and y.

    Happily though because of symmetry and the uniform charges and distances, once you determine the Force on one from the other 3, you should see that the other answers are easily inferred.
  4. Apr 1, 2009 #3
    So for each charge F = q* E
    And E is the sum of the forces?
    And where are the X and Y planes drawn for this situation?
  5. Apr 1, 2009 #4


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    The charges themselves occupy the x-y plane. They are in a square of .9m.

    At any charge there is an equal charge of opposite sign at .9m in x direction and the same in y direction. Then there is the 3rd charge (same sign) along the diagonal that will contribute an x and a y component.
  6. Apr 1, 2009 #5
    I'm still a bit confused by what your telling me, I talked to a friend and here is what he said:

    To calculate the forces you really only need to do it once and then apply the direction as if you were doing it in each quadrant.
    . K= 9x109. r = .9m for doing the adj and r = √.92+.92 for the ones across the square. This gives you the chance to break them into x and y components and add them to get the ΣF.

    Does that seem right to you? I couldn't really decipher what he was saying so if you can could you help me understand it?
  7. Apr 1, 2009 #6


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    Since all the charges are the same (except for sign) it might be easier to calculate it symbolically.

    For the 1q you have an attractive Force in vector notation that looks like - q2/d2 i - q2/d2 j and the one on the diagonal has a repulsive force with components of +q2/2d2*cosθ i + q2/2d2*sinθ j

    In this case sinθ and cosθ are equal to √2/2 which should help a little to calculate the magnitudes.
  8. Apr 1, 2009 #7
    You say
    "for the 1q"
    What is the 1q? Is it the one quadrant? You gave me two equations for the vectors for attractive and repulsive forces, do I need to add these two vectors? I don't know what I need to do. I can plug the numbers in for these equations:
    q2/d2 i - q2/d2 j
    q2/2d2*cosθ i + q2/2d2*sinθ j
    But once I plug the numbers in to both what do I do?
    Sorry, but I'm becoming fairly confused.
  9. Apr 1, 2009 #8


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    In your problem opposite charges are repelling and adjacent charges are attracting. Let F be the forces between adjacent charges, Since charges are same except sign, forces between opposite charges will be F/(2). At any corner resultant of Fs is in the opposite direction to F/2. Resultant of two adjacent forces = (2)^1/2*F
    Therefore Fr = (2)^1/2*F - F/2
    Here F = k* (5.86*10*-3)^2/(0.9)^2
    Last edited: Apr 2, 2009
  10. Apr 1, 2009 #9


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    I meant to say "for 1 charge q". For that charge you will have 3 forces to be added by superposition. You must treat these forces as vectors.

    From 1 corner it will be -q2/d2 and it will be along the x axis say. Give that force a value

    F = -q2/d2 i

    From the other adjacent corner you have the same magnitude force but it is directed along the y axis.

    F = -q2/d2 j

    From the opposite corner the magnitude of the force |Force| will be + q2/2d2 because it is a distance of d*√2 away. But it will have both an x and a y component relative to the other charges.

    F = q2/2d2cosθ i + q2/2d2sinθ j

    Substitute √2/2 for the sinθ and cosθ and add the 3 F's all together.

    This vector sum is your answer. Pay careful attention to the sign, and note that I left out a factor of k for simplification. Since the i and j components are equal then you know your angle direction is along the diagonal. The sign of your answer should help you to understand whether it is attractive or repulsive.
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