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Homework Help: Yet another electric charge problem

  1. May 13, 2007 #1
    I got this problem wrong in my text's student companion website, but I don't understand why. Could someone let me know what went wrong?

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

    What are the horizontal and the vertical components of the resultant electrostatic force on the charge in the lower left corner of the square if q = 0.10 micro Coulombs (Or uC as I cannot remember how to type the Greek letter equivalent.) and a = 5.0cm?


    2. Relevant equations

    Fh = F-q + F-2q(cos(45))

    Fv = -F+q + F-q(sin(45))

    3. The attempt at a solution

    I get .17E-6 N for Fh and -2.1E-7 N for Fv. I think I may have mixed up some decimal places on Fh, but the solution presented is .17 N for Fh and -0.046 for Fv. I just don't understand what I did wrong though.
  2. jcsd
  3. May 13, 2007 #2
    for Fh: your first term should be [tex]F_{-2q}[/tex], and your second term should be [tex]F_{-q}\cos(45^o)[/tex]

    for both: Please show more work. What is the general formula for the electrostatic force? What values did you use for each variable in the equation?
  4. May 13, 2007 #3
    My apologies. I'll try out your suggestion for Fh. Otherwise...

    Fh = ((8.99E9)(2E-7)(1E-7)/.05^2) + (((8.99E9)(2E-7)^2)/.05^2)cos(45)

    Fv = -((8.99E9)(2E-6)(1E-6)/.05^2) + ((8.99E9)(2E-6)(1E-6)/.05^2)sin(45)
  5. May 13, 2007 #4
    For Fv: as in Fh, the charges should be multiplied by 10-7
  6. May 13, 2007 #5
    After your suggestions I get .19E-8 for FH and -.02E-8. Why do my answers still differ from what the website presents? I don't get it.
  7. May 13, 2007 #6
    for the charge on the far corner: the distance is not 0.05m. It is along the diagonal of the square, not along one of the sides.
    I'm checking now to see if that clears things up...

    edit: I got the correct answer for both Fh and Fv when I took both of my suggestions (and the rest of your work :wink:).
    Last edited: May 13, 2007
  8. May 13, 2007 #7
    So for the charge in the far corner, it's not 0.05m? If not then how can I get the right measurement? The diagram doesn't give any indication of what the diagonal is. Or do you mean just the two suggestions you gave and the diagonal is a red herring so to speak?
  9. May 13, 2007 #8
    Cheese, frank, I wouldn't trick you like that!

    You don't know the formula for the diagonal of a square? It's the same as the hypotenuse of a 45-45-90 triangle.
  10. May 13, 2007 #9
    Ah okay, that helped. Thanks!
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