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Another Charge question

  1. Aug 26, 2007 #1
    1. The problem statement, all variables and given/known data
    The charges Q1= 1.90·10-6 C and Q2= -3.03·10-6 C are fixed at their positions, distance 0.279 m apart, and the charge Q3= 3.33·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.

    2. Relevant equations
    [tex] F=\frac{KQq_{1}q_{2}}{r^2}[/tex]



    3. The attempt at a solution
    I get 0.6483m
     
  2. jcsd
  3. Aug 26, 2007 #2

    Doc Al

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    Please give the coordinate of each particle along an axis.
     
  4. Aug 26, 2007 #3
    is the answer 1.06?
     
  5. Aug 26, 2007 #4
    Q(0,0), Q2(.279,0), Q3(x,0). Where X>.279m
     
  6. Aug 26, 2007 #5

    Doc Al

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    That's what I thought you were saying. Note that X > .279 is not a requirement of the problem, but something that you added. In fact, it can't be true!
     
  7. Aug 26, 2007 #6
    True, i did add that in.
    I started by finding the force between q1 and q2:
    [tex]F=\frac{9.0e^9 1.90e^-6 -3.03e^-6}{.279^2}=-6.65e^-1N[/tex]
    Then I can set that equal to [tex] F=\frac{KQq_{1}q_{2}}{(x-r)^2}[/tex], right?
     
  8. Aug 26, 2007 #7
    just making it way too complex,
    find where E is 0, and that would be the position of Q3
    and you would get something like
    q/r^2 = q1/(d+r)^2
     
  9. Aug 26, 2007 #8

    Doc Al

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    The force between Q1 and Q2 is irrelevant. You want the force on Q3! (Hint: Where does the electric field of Q1 + Q2 equal zero?)

    The first thing I would do is figure out which region Q3 must be in:
    (a) to the left of Q1
    (b) between Q1 and Q2
    (c) to the right of Q2
     
    Last edited: Aug 26, 2007
  10. Aug 26, 2007 #9
    So: [tex]E_{1}+E_{2}=0[/tex]
    [tex]\frac{kq_{1}}{x^2}-\frac{kq_{2}}{(r+x)^2}=0[/tex]
     
  11. Aug 26, 2007 #10

    Doc Al

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    What region are you analyzing? What's the definition of x?

    If it's not obvious which region Q3 must be in, study each region systematically.
     
  12. Aug 26, 2007 #11
    the region right of q2. X is distance from q1 to q3
     
  13. Aug 26, 2007 #12

    Doc Al

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    For some reason you are choosing the region to the right of q2 (even though I told you that it was impossible! :wink:). No problem! If x is the distance from q1, what must be the distance from q2 in that region?
     
  14. Aug 26, 2007 #13
    ok, so I the region where E=0, q3 would have to inbetween q1 & q2. getting confused
     
  15. Aug 26, 2007 #14

    Doc Al

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    Don't give up on studying any region. If you pick the wrong region, the equation you get will not give you an answer in the region.

    To study the region in between q1 & q2, draw yourself a picture. In that region, which way would the field from q1 point? The field from q2? Could they cancel?
     
  16. Aug 26, 2007 #15
    I know the q1 field is outward and q2 field in inward. So the field(right side) from q1 to q2 will interect with each other.
     
  17. Aug 26, 2007 #16

    Doc Al

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    Right! So, in the region between q1 and q2, which way does each field point: to the right or to the left? If we let positive stand for "to the right", are the fields positive or negative in that region?
     
  18. Aug 26, 2007 #17
    positve to the right, negitive to the left
     
  19. Aug 26, 2007 #18

    Doc Al

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    That's the sign convention. But which way do the fields point?
     
  20. Aug 26, 2007 #19
    The combination of the q1 and q2 field points to the right.
     
  21. Aug 26, 2007 #20

    Doc Al

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    In that middle region, the field from q1 points to the right (and thus is positive) and the field from q2 also points to the right (and is positive). So is it possible for the fields to cancel out in that region?
     
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