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Homework Help: Point Charges

  1. Sep 17, 2008 #1
    Charge 1 is located along the x axis and has a magnitude of +1.6E-6C and Charge 2 is located 1m away from charge 1 with a magnitude of -4.8E-6C. A third point charge of magnitude Q is brought from far apart and placed on the line joining the two charges and at a point where the charge q experiences no net force at all. What is the distance of that point from the origin and what should be the sign of this unknown charge q?

    This is what I came up with so far...

    F1 = K (Q1)(Q2)/X^2
    F2 = K (Q2)(Q3)/(1m-x)^2
    You set them equal to eachother and get
    K(Q1)(Q2)/X^2 = K(Q2)(Q3)/(1.00-x)^2
    Cancel out K and Q3 leaving;
    (1.00-x^2)/X^2 = (Q2)/(Q1)
    1.00-x/x = square root (4.8E-6C/1.6E-6C)
    1.00-1x = 1.73x
    1.00=1.73x +1x
    1.00=2.73x
    x = .366

    Does anybody know if this is correct?

    Also I do not know how to figure out what the sign of this charge should be so if anybody could help me I would appreciate it!
     
  2. jcsd
  3. Sep 17, 2008 #2

    LowlyPion

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    You have chosen a point between the positive and negative charges. There is an E-field between those two charges at all points. What you found is the point at which the magnitudes are equal, but the direction is the same direction for both.

    Choose a point outside the 2 charges, away from the larger one and closer to the smaller to fgure your Null point.
     
  4. Sep 18, 2008 #3
    So if I choose a point to the right of the -4.8 charge then the vectors are in the opposite direction. So are you saying in that equation I should add x
    so it would be

    1.00+x = 1.73x
    1.00=1.73x - 1x
    1.00 = .73x

    x = 1.4m

    If that is right then how do I figure out the sign of the charge?

    Thanks for your help!
     
  5. Sep 18, 2008 #4

    LowlyPion

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    Refresh my memory, does an e-field of 0 exert a force on any charge?
     
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