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Coulomb's Law and microcoulombs

  1. Jul 2, 2008 #1
    There are four charges, each with a magnitude of 2.5 microcoulombs. Two are positive and two are negative. The charges are fixed to the corners of a 0.29-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.
     
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  3. Jul 2, 2008 #2

    Hootenanny

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    Hi ramalik,

    You are expected to show your efforts are solving a problem, before asking for help. What have you tried thus far?
     
  4. Jul 2, 2008 #3
    Oh sorry!
    I tried adding the 3 forces acting on each charge, so its Fx + Fy + Fdiagonal. Im just confused about what to do with the signs since the way I'm doing it does not incorporate them but I know I have to since the direction of the force is important in determining the net force acting on each charge.
     
  5. Jul 2, 2008 #4

    Hootenanny

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    How have you been doing it?
     
  6. Jul 2, 2008 #5
    For Fx and Fy I used the formula F = (k qq)/r^2 and used the data given. For Fdiagonal, I found the radius of the square which was sq root of( .29^2 + .29^2) and used the same formula. Then I added the three together. I know thats not right because the three forces aren't positive I just don't know how to determine the signs.
     
  7. Jul 2, 2008 #6
    and i did make sure to convert from microcoulombs to coulombs
     
  8. Jul 2, 2008 #7

    Hootenanny

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    If you correctly substituted in the charges, then two of your forces should be negative (i.e. attractive) and one should be positive (i.e repulsive). Don't forget to include the sign of the charges in Coulomb's formula.
     
  9. Jul 2, 2008 #8
    i thought you have to use the absolute value of the charge so that the sign doesn't matter
     
  10. Jul 2, 2008 #9

    Hootenanny

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    You can do that, but then you have to figure out in which direction the force acts. If you simply input the charges into the formula then the direction is given: positive for repulsive and negative for attractive.
     
  11. Jul 2, 2008 #10

    Hootenanny

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    I should also mention that you cannot simply add the magnitude of the diagonal force to the other two. Instead you need to split the diagonal force into horizontal and vertical components and then add these components to the other corresponding component (Fx and Fy respectively). You should then find the magnitude of this net force.

    Do you follow?
     
  12. Jul 2, 2008 #11
    Aren't the other two the components of the diagonal force? I found the diagonal force using the pythagorean theorem so isn't that the same thing?
     
  13. Jul 2, 2008 #12

    Hootenanny

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    You have three forces acting on the charge, yes? How do you normally find the net force acting on an object?
     
  14. Jul 2, 2008 #13

    rbj

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    ya know, given the subsequent posts, i might come to a conclusion that it is never useful to think of Coulombs Law in terms of magnitude only. even in a single dimensional problem (so it's the scaler version of Coulomb's Law), there is no good reason to throw away sign information. what Coulomb's Law is in 3D vector form can be expressed in 3 identical scaler equations. but sign needs to be included in these 3 scaler equations.
     
  15. Jul 4, 2008 #14
    I'm still not getting this problem. I found the force of the diagonal and since the angle it makes is 45 degrees, I found the x and y components of this force. Then I found the force on 1 in the Y direction and since the charges are +, +, -, - in the clockwise direction, there are no other forces in the x direction. Thus the net force should be Fnet = Fx + F(all forces in y direction). However, this answer is incorrect. I think I'm still doing something wrong with the signs...
     
  16. Jul 4, 2008 #15

    Hootenanny

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    This is your problem, your charge configuration is wrong. Notice that the question states that:
    This is not the case with the charge configuration you have been using. You need to determine the charge configuration such that the electric field of the array is directed towards the centre.

    Do you follow?
     
  17. Jul 4, 2008 #16
    ohhh I see, I got it right! Thank you for your help!
     
  18. Jul 5, 2008 #17

    Hootenanny

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    No problem :smile:
     
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