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Electric force on a charge

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Q1-----Q2-----Q3; Q1=1.78E-6C;Q2=Q3=-Q1, r1=r2=1.82m
    Calculate the total force on Q2. Give your answer with a positive number for a force directed to the right.



    2. Relevant equations

    F= kq1q2/r^2

    3. The attempt at a solution
    F2 = F12+F23
    F12 = k(1.78E-6)(-1.78E-6)/(1.82^2)
    F23 = k(-1.78E-6)(-1.78E-6)/(1.82^2)
    The problem is these forces cancel and I know that they shouldn't, so I am sure is some thing I am just not seeing here.
     
  2. jcsd
  3. Jan 11, 2012 #2

    SammyS

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    F12 is attractive, so it's in what direction?

    F32 is repulsive, so it's in what direction?
     
  4. Jan 11, 2012 #3
    F12 is negative so to the right and F32 is positive so it is to the right because Q2 is negative
     
  5. Jan 12, 2012 #4

    SammyS

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    Therefore, they don't cancel !

    Usually right is positive.
     
  6. Jan 12, 2012 #5
    okay so then I am not sure where I am going wrong since I know that it shouldn't cancel the charges and the distances are the same for F12 and F32 so it should be 2*F12 = F2, I am not sure what I am doing wrong here this shouuld be a fairly simple problem.
     
  7. Jan 12, 2012 #6

    SammyS

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    The sign given by these two equations really only is useful in showing whether you have attraction (negative sign) or you have repulsion (positive sign).
    F1,2 = k(1.78E-6)(-1.78E-6)/(1.822)

    F2,3 = k(-1.78E-6)(-1.78E-6)/(1.822)​
    You can't blindly add the results together.

    Since each force is in the positive x direction, they are both positive and should be added together.
     
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