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Electric Charge problem

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    http://snag.gy/Zg3SO.jpg

    At the moment is it question A


    3. The attempt at a solution
    A)
    I've realized that Force applied by F_31 + F_41 = F_21

    SO i've writte this equation up but Mathmathica, says there is an errror, and gives me an incorrect result.

    http://snag.gy/rQ2kz.jpg


    Mathmatica says it should {{D -> -0.225008}, {D -> 0.185008}}

    The correct answer should be 2,27cm.
     
  2. jcsd
  3. Feb 9, 2013 #2

    TSny

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    Did you take into account the directions of the forces F_31 and F_41? They have both x and y components. :smile:
     
  4. Feb 9, 2013 #3
    No..
    y -component must be zero since it is zero for q2
    and the sum of q3 and q4 would negate the y component.
     
  5. Feb 9, 2013 #4

    TSny

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    Correct. So, you only need to work with the x components of the forces.
     
  6. Feb 9, 2013 #5
    But aren't I doing it already?
     
  7. Feb 9, 2013 #6

    TSny

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    I don't think so. It appears to me that on the left hand side of your equation you are using the full magnitude of the forces F_31 and F_41, rather than their x-components.
     
  8. Feb 9, 2013 #7
    Ahh.. so you want me to skip
    d/cos([itex]\theta[/itex])
    and just use d.. ?
     
    Last edited: Feb 9, 2013
  9. Feb 9, 2013 #8

    TSny

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    No, that part is correct, r = d/cosθ is the correct distance between charges 4 and 1. Draw a sketch showing the direction of the force F41. How would you express the x-component of F41 in terms of F41 and θ?
     
  10. Feb 9, 2013 #9
    well the distance it moves is d.. so I have to write an expression for d..

    So it would be rcosθ = d??
     
  11. Feb 9, 2013 #10

    TSny

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    Sorry, but I don't understand. What is moving a distance d? I thought you were trying to find the distance D so that the net force on q1 is zero.

    Your first equation which you derived looks good except it doesn't take into account that you want the x-components of the forces to add to zero. (You already know the y-components add to zero by symmetry.)
     
  12. Feb 9, 2013 #11
    well q1 lies on a distance d from q3 and q4...

    and isn't that distance given by r cos(θ)= d

    where r is the Hypotenuse of the triangle`..
     
  13. Feb 9, 2013 #12

    TSny

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    Yes. All of that is correct. So, your expression (d/cosθ)2 in the denominator of your equation is correct. :smile:

    Force is a vector quantity. The equation F = q1q2/(4∏εor2) gives the magnitude of the force. You need to determine the x-component of the force vector. Again, a sketch of the forces will help.
     
  14. Feb 9, 2013 #13
    ... hmm would it then be
    F_x = q1q2/(4∏εo(d))

    Since both lies in a distance d?... or how..
     
  15. Feb 9, 2013 #14

    TSny

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    No. Consider the x-component of F31. Can you use trigonometry or similar triangles to find Fx in the attached picture?
     

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  16. Feb 9, 2013 #15
    hmmm.. is it then cos(\theta)=r/x <=> x = cos(theta)/r
     
  17. Feb 9, 2013 #16

    TSny

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    Not sure what you are writing here. You need to find Fx, rather than x.
     
  18. Feb 9, 2013 #17
    Ahh.. I see what you mean now..
    My drawings weren't correct..
    But by using youre drawing I see that Cos(theta) = F_x/F <=> F_x = Fcos(theta)

    So the expression for F_X = (q1q2/(4∏εo(d^2))) cos(theta)
     
  19. Feb 9, 2013 #18

    TSny

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    Yes. Good.

    [EDIT: oops, you did mean r instead of d in the denominator, right? And the charges are q1 and q3 ]
     
    Last edited: Feb 9, 2013
  20. Feb 9, 2013 #19
    Hmm.. but i still get an incorrect answer.. :(

    http://snag.gy/LDwAe.jpg

    When i Tries to calculate D i get complex number..
     
  21. Feb 9, 2013 #20

    TSny

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    It seems to work out ok. Note that you don't want to plug in a negative value for q3 since you have set up your equation to state that the magnitudes of the forces balance out.
     
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