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Homework Help: Charge Placement

  1. Jul 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A charge of +2q is placed at the origin and a second charge of –q is placed at
    x = 3.0 cm. Where can a third charge +Q be placed so that it experiences a zero
    force?


    I think I've found the solution, I just need someone to verify why the math is the way it is at the end.

    1) The overall force acting on the the new charge must equal zero, therefore I set them equal to eachother:

    k (2q x Q)/(r^2) = k (q x Q)/(r-0.03)^2

    (r - 0.03)^2 = r^2 (2q x Q)/(q x Q)

    (r - 0.03)^2 = 2r^2

    r - 0.03 = +/- 1.414r

    -0.03 = -2.414r
    r = .0124

    OR

    -0.03 = 0.414r
    r = -.072


    So would this mean that the charge would be .012 metres to the right of the negative charge? (Assuming that -q is placed 3 cm to the right of +2q)
     
  2. jcsd
  3. Jul 12, 2010 #2
    Sorry, one more thing.

    I'm just confused because my textbook shows you how to do a similar problem, and I don't understand why they don't use one of the variables to determine the answer.

    For example:

    the "r" in (r - 0.03) doesn't come into play in their answer because they just divide 1.414 by 0.03 giving a different answer than what I found. Why don't they use the "r" from (r - 0.03) like I do in my answer?
     
  4. Jul 13, 2010 #3
    bump*
     
  5. Jul 14, 2010 #4
    Note that force (or electric field) is a vector. For two forces to cancel out, not only should they have the same magnitude but they should point in opposite directions.

    So, you cannot have the point in between the two charges.

    You have made an error in your second step.
    [tex]\frac{2q}{{r}^{2}} = \frac{q}{(r-3)}^{2}[/tex]
    [tex]2{(r-3)}^{2} = {r}^{2}[/tex]

    That'll give you two values of r, of which only one can be possible.
     
  6. Jul 15, 2010 #5

    lightgrav

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    Homework Helper

    Conceptually, it should make sense that the new charge should be farther from the double charge ...
    but how may TIMES farther? 1.414x farther, right?
    So, the distance from the far charge is 1.414 * near , but far - near = 3cm,
    so : far - 1.414 * far = 0.03m = far (1 - 1.414) .

    that is, they isolated all the "r" terms in a ratio on one side, with their relationship factor on the other. [and most folks would rather add than subtract : (r+3)/r = 1.414 ]
     
  7. Jul 24, 2010 #6
    Hmm, I'm wondering if I went about this all wrong.

    My original answer was following the steps of a similar question in my text, but I have re-thought it out and maybe I'm making it more difficult then it should be. Here's a diffeernt solution:

    k (2Qq)/(r+3)^2 = k (qq)/r^2
    2Qq(r^2) = qq (r+3)^2
    2r^2 = r^2 + 9
    r^2 = 9
    r = 3

    Does that make sense? So essentially, the charge would be 6 cm from +2q (3 cm to the right of the -q charge)
     
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