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Electrostatics (four equal point charges)

  1. Sep 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Four equal point charges q = +22.4 μC sit at each corner of a square. What charge Q, in μC, must be placed in the center so all the charges are in equilibrium? Your answer must include the sign of the charge.

    2. Relevant equations

    F = q1*q2 * k / r^2

    3. The attempt at a solution


    My attempt was to choose one corner (let's say q3), try to find the electrostatic force between q3 and other charges (q1,q2,q4), then add all the y and x components of forces acting to q3, then find the magnitude. Then make the magnitude equal to force acting from the center to q3.

    I don't need someone solving this problem for me, I just need a hint to start with. Please check my logic and see if there is any flow. Because I solved it this way once and I had an equation with both q and x (distance) in one equation...
     

    Attached Files:

  2. jcsd
  3. Sep 6, 2015 #2

    Doc Al

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    Sounds like a good plan to me.
     
  4. Sep 6, 2015 #3
    But I am ending up with two unknowns in one equation...
     
  5. Sep 6, 2015 #4

    Doc Al

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    What are your unknowns?
     
  6. Sep 6, 2015 #5
    I am taking the side of the square as "r" [so half is r/2] and 4 of the charges are all the same, q, and last (fifth, in the center) charge is Q.

    After doing what I've said in the original post, my equation (after all the simplification) is 4 * r^2 * q + √(2) * q = 8 * r^2 * Q

    :/ It's given that q is + 22. 4 C, but, r and Q are not given... my original task is to find Q so that everything will be balanced (equilibrium.)
     
  7. Sep 6, 2015 #6
    ..
     
    Last edited: Sep 6, 2015
  8. Sep 6, 2015 #7

    Doc Al

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    Rethink this equation. All terms must have the same units for an equation to make sense. (When you get it right, the r will cancel.)
     
  9. Sep 6, 2015 #8
    I can't get rid off that one "r^2" ...

    Checked my algebra thrice. Should be alright. Do you really think the logic was fine? Where else the problem might be... :/
     
  10. Sep 6, 2015 #9

    Doc Al

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    Show each step. List the force you get from each charge.
     
  11. Sep 6, 2015 #10
    Before I do that. Can you please check these equations (whether they are set up correctly or no):

    ( I attached the photo of the image)

    F 1 due 2 = (k q^2 / r^2) i
    F 1 due 3 = (k q^2 / 4 r^4) i

    F 1 due 4 = (k q^2 / r^2) j
    F 1 due 3 = (k q^2 / 4 r^4) j

    F 1 due 5 = 4 k q Q / r^4

    My logic: Add all the i's and add all the j's. Square both i and j and take the square root and equal the answer to F 1 due 5.

    :/
     

    Attached Files:

  12. Sep 6, 2015 #11

    Doc Al

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    How did you get an r^4? That's where you are going wrong.
     
  13. Sep 6, 2015 #12
    Well, q3 lies on the opposite (diagonally) direction from q1. So I needed "r" [Electrostatic force between two charged particles: F = k q1 q 2 / r^2] So to find it I used pythagorean theorem [ two sides are equal since it's a square ] so the diagonal ( the distance between q1 and q3) = r^2 + r^2 = 2 r^2 BUT when I plug it in the formula I need to raise it to another square so it's (2r^2)^2 = 4 r^4 ... Am I wrong?
     
  14. Sep 6, 2015 #13
    What am I talking about. D:

    Forget it.

    So silly.

    BRB
     
  15. Sep 6, 2015 #14

    Doc Al

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    The diagonal squared is 2 r^2.

    Yes, you are wrong. :smile:
     
  16. Sep 6, 2015 #15
    Aha! Such a careless mistake!

    I got Q = √(q^2/2 = - 15.839 μC

    And the negative answer is the one I want. :)

    Hopefully it's correct.

    Thank you very much, Doc Al!
     
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