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Electric Field Forces

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A point charge q1 = -3.00 µC is situated at the origin. Another point charge q2 = 5.00 µC is located at the location x = 1.00 m. If another charge Q1 = 10.0 µC is placed at x = -0.100 m, what is the magnitude of the force on it?


    2. Relevant equations
    F=[k(q1)(q2)]/r^2
    E=F/Q (I'm not sure if this one is relevant, considering how I have no idea how to do this).



    3. The attempt at a solution

    For two particles, I did the same thing and it worked. However, when I tried with 3 particles, obviously it didn't give me the right answer or else I wouldn't be here?

    F1= [(9x10^9)(10x10^-6)(-3x10^-6)]/(.1^2) = -27 i-hat
    F2= [(9x10^9)(10x10^-6)(5x10^-6)]/(1.1^2) = -.37 i-hat
    Fnet = -27 i-hat + -.37 i-hat = -27.37 i-hat.

    Meaning the magnitude would be 27.37 N, and it would be going in the negative i-hat direction. Of course, that was not the right answer, and I was really hoping for some help because I have no clue what to do.

    Thank you!
     
  2. jcsd
  3. Feb 6, 2009 #2

    LowlyPion

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    Welcome to PF.

    Just curious how both forces ended up with the same sign?
     
  4. Feb 6, 2009 #3
    That's pretty interesting actually. I think what I did was that the F1 ended up being negative because of the -3x10^-6 force and because the i-hat is positive for that, so that one stayed the same.

    As for F2, I wasn't exactly sure about this one because I have a problem with figuring out when to use - or + but. I figured the i-hat was negative because of a method our teacher taught us. But I'm not really sure if the i-hats even change based on direction because of a previous example he showed us. Having three charges just really throws me off. I feel it shouldn't be much different than two...
     
  5. Feb 6, 2009 #4

    LowlyPion

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    If you have a + & - sign then the charges are attractive and that means the force is toward the charge that you are figuring the force for.

    In this case both q1 and q2 charges affecting the force are located in the + X direction, so a + Force toward those charges (if -) on the Q1 will be positive X.

    The opposite is true for the same sign charges. Repulsive force at Q1 will be - X.
     
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