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Where to place a third charge so it experiences no force

  1. Jul 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A point charge of 5.6 microcoulombs is placed at the origin (x=0) of a coordinate system, and another charge of -1.9 microcouloumbs is placed on the x axis at .29m. Where on the axis can a third charge be placed so that it experiences no charge?

    q1= 5.6 E-6
    q2= -1.9 E-6
    d= .29m

    2. Relevant equations
    F= k*(q1*q2)/r^2

    3. The attempt at a solution
    Fnet = F1,3 + F2,3
    Fnet = 0
    F1,3 = -(F2,3)
    k*(q1*q3)/r^2 =k*(q2*q3)/(r-distance)^2
    i know the k and q3 variables cancel out so i get
    q1/r^2 = q2/(r-distance)^2

    after i get this far I can't seem to get a correct answer. any help would be greatly appreciated!
     
  2. jcsd
  3. Jul 17, 2016 #2
    Shouldn't the distance from the second charge be (.29 - r) where r is the distance from the origin?
     
  4. Jul 17, 2016 #3
    Yes, I accidentally reversed it in my post. Thanks for pointing that out.
     
  5. Jul 17, 2016 #4
    Also, remember that the third (test) charge cannot be between the first two charges because one charge will
    attract the test charge and one charge will repel the test charge.
     
  6. Jul 17, 2016 #5
    so should the charge on q2 be q2/(.29 PLUS r)?
     
  7. Jul 17, 2016 #6

    SammyS

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    You have typo which I have indicated above.

    Your equation, q1/r2 = q2/(r - d)2 is missing a negative sign. (Remember, q2 is itself negative.)

    That makes it q1/r2 = -q2/(r - d)2 .

    Take the square root of both sides remembering to use a ± on one side or the other. Solve for r .

    That does not make sense. Did you mean "Should q3 be q2/(.29 + r) " ?
     
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