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Electric Force

  1. Feb 9, 2015 #1
    1. Given that q=+12uc distance=19cm
    Find Magnitude and direction exerted on point q2.
    What happens to the force if the distance is tripled?


    q____________ q2 ____________q3
    q1=+12
    q2=-2.0q
    q3=+3.0q
    distances between point are equidistant.


    3. The attempt at a solution
    F=k Q Q/ d^2
    F=F+F
    F=k/d^2(QQ+QQ)

    F=k(q)(q)/d^2

    (9*10^9)(12*10^-6)(2*10^-6)
    (.19)^2

    =5.98

    (9*10^9)(-2*10^-6)(3*10^-6)

    (.19)^2

    =-1.50

    thus 5.98-(-1.50)=7.48





     
    Last edited: Feb 9, 2015
  2. jcsd
  3. Feb 9, 2015 #2

    BiGyElLoWhAt

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    You're sure you don't have a value for q1?
     
  4. Feb 9, 2015 #3
    I think q1 is 12uc
     
  5. Feb 9, 2015 #4

    BiGyElLoWhAt

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    Ok, so how do you treat separate forces? You have coloumbs law correct. I guess I don't see where you're stuck, unless you don't understand the formula. If that's the case, please clarify.
     
  6. Feb 9, 2015 #5
    I think I'm just confused because the book does it completely differently and my answer isn't the same. I think there Is error rounding. and I didn't initially understand the formula. The more I looked at it I had a urikea moment.
     
  7. Feb 9, 2015 #6
    How do you remove this from the feed?
     
  8. Feb 9, 2015 #7

    BiGyElLoWhAt

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    Well I hope you did. As to removing it, this thread will stay here in case someone else can get some use out of it at a later time.

    Would you humor me? Walk me through this, so I know what's going on.
     
  9. Feb 9, 2015 #8

    BiGyElLoWhAt

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    By the way, welcome to Physics Forums. :)
     
  10. Feb 9, 2015 #9
    So I used Coulombs equation.
    F=kQQ/d^2
    I knew that F=F+F
    So I just wrote the equation
    F=k/d^2(QQ+QQ)
    I plugged the constant in for k. converted 19cm to meters so it changed to .19m.
    Then plugged in ((-12)(2)+(2)(3))
    but then I didn't change them to coulombs.
    So that through me off.

    So instead I plugged everything into the original format of the equation and did it in parts because it just was easier on the eyes.

    I hope your humored.
     
  11. Feb 9, 2015 #10

    BiGyElLoWhAt

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    =]
    I am.
     
  12. Feb 9, 2015 #11
    but are you sure I did it right?
     
  13. Feb 9, 2015 #12

    BiGyElLoWhAt

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  14. Feb 9, 2015 #13

    BiGyElLoWhAt

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    Assuming there are subscripts below your q's, and also assuming that (QQ+QQ) is in the numerator, then yes.
     
  15. Feb 9, 2015 #14

    BiGyElLoWhAt

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    Also assuming that you ultimately converted your charges correctly. Neglected that initially, because you, yourself, mentioned it. I figured I might as well include it.
     
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