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Coulomb's Law and Electrostatic Forces Problem

  1. Aug 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Three charges are arranged as shown in the picture I attached. Find the magnitude and direction of the electrostatic force on the charge at the origin.


    2. Relevant equations
    I know that Coulomb's Law must be applied.


    3. The attempt at a solution
    I've drawn a force diagram with FCB pointing to the left and FAB pointing down. I've also broke the vectors into components using Coulomb's Law and added them together. By doing this, I got 1.42 x 10-5 [itex]\hat{x}[/itex] + 8.54 [itex]\hat{y}[/itex]. From here what do I do to get an answer?

    I know the answer is supposed to be 1.38 x 10-5 at 77.5 degrees so I need help with how to get there.
     

    Attached Files:

  2. jcsd
  3. Aug 26, 2011 #2

    tiny-tim

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    hi cheerspens! :smile:
    i don't :redface:

    how did you get that?​
     
  4. Aug 27, 2011 #3
    I solved for these vectors then added them:

    FCB=[(9x109|(6x10-9)(5x10-9)|) / (.32)][cos 180][itex]\widehat{x}[/itex] + [(9x109|(6x10-9)(5x10-9)|) / (.32)][sin 180][itex]\widehat{y}[/itex]

    FAB=[(9x109|(-3x10-9)(5x10-9)|) / (.12)][cos 198.43][itex]\widehat{x}[/itex] +[(9x109|(-3x10-9)(5x10-9)|) / (.12)][sin 198.43][itex]\widehat{y}[/itex]
     
  5. Aug 27, 2011 #4

    tiny-tim

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    hi cheerspens! :smile:

    (just got up :zzz: …)

    where does 198.43° come from? :confused:
     
  6. Aug 27, 2011 #5
    Good morning! :smile:

    So I plugged the 198.43 degrees in for some reason but that was the angle at which the ball is being pushed by the other balls.

    So then in the vectors I had before, I fixed the degree in the second one to 270 degrees but then I get 2.7x10-5[itex]\widehat{x}[/itex] + 0 [itex]\widehat{y}[/itex] and that still does not match the correct answer.
     
  7. Aug 27, 2011 #6

    tiny-tim

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    let's see …

    it's |-9*10-9 (6*5/.09y + 3*5/.01x)|

    = |-9*10-7 (2*5/3y + 3*5/1x)|

    = 9*10-7 √(100/9 + 225)

    = 9*10-7 √(111.11 + 225)

    … hmm, that's not the answer in the book :redface:

    (though the angle is correct, isn't it?)
     
  8. Aug 27, 2011 #7
    The angle I got isn't even correct. According to the book it should be 77.5 degrees below the -x axis.
     
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