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Magnitude of q ~ What am I doing wrong?

  1. May 7, 2009 #1
    Magnitude of "q" ~ What am I doing wrong?

    1. The problem statement, all variables and given/known data
    This is my homework problem:
    "Two equal charges are separated by 3.7 x 10^-10 m.
    The force between the charges has a magnitude of 2.37 x 10^-3 N. What is the magnitude of q on the charges?"



    2. Relevant equations
    F(electricity) = k(Coulomb)(q1 • q2)/r^2
    E = F(e)/q0
    E = k(C) • q/r^2
    ...I can't really seem to think of any other relevant equations for this particular problem, though I have a lot of others that go along with this electricity unit in Physics. If you can think of some that apply that I missed, please let me know. :)


    3. The attempt at a solution
    Here's my insanity work below, hahaha. I tried two approaches and got the same answer, but I still don't think it's correct:

    (a)
    F = kc(q1q2/r^2)

    2.37 x 10^-3 = 8.99 x 10^9 • q1q2 / (3.7 x 10^-10)^2

    F • r^2 = kc • q1q2
    F • r^2 / kc = q1q2
    2.37 x 10^-3 • (3.7 x 10^-10)^2 / 8.99 x 10^9 = q1q2

    q1q2 = 3.61 x 10^-32

    (b)
    E = kc • q/r^2

    2.37 x 10^-3 = 8.99 x 10^9 • q/(3.7 x 10^-10)^2
    2.37 x 10^-3 / 8.99 x 10^9 = q/(3.7 x 10^-10)^2
    2.37 x 10^-3 / 8.99 x 10^9 • (3.7 x 10^-10)^2 = q

    q = 3.61 x 10^-32

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Please help me. I don't really know what I'm doing wrong, or if I'm doing anything wrong. I have a feeling I got this incorrect, however. Also, I don't exactly understand the concept of "q". I know what q1 and q2 are, but if anyone could explain it just one more way (I've already heard it three different approaches, but I still don't comprehend), I'd be much obliged for any input as soon as possible. Thank you! :)
     
  2. jcsd
  3. May 7, 2009 #2

    Cyosis

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    Re: Magnitude of "q" ~ What am I doing wrong?

    The problem statement gives you one very important hint, two EQUAL charges. Therefore q1=q2=q and the equation for the force between two charges becomes, [tex]F=k \frac{q^2}{r^2}[/tex]. What they mean with the magnitude of q on the charges is, how much charge do q1 and q2 have,ignoring the sign.
     
  4. May 7, 2009 #3
    Re: Magnitude of "q" ~ What am I doing wrong?

    Thanks!

    So basically, I calculated for q^2? I just take the square-root of my answer for q?
     
  5. May 7, 2009 #4

    Cyosis

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    Homework Helper

    Re: Magnitude of "q" ~ What am I doing wrong?

    Yep, you can then check your answer by calculating what the force would be between q1 and q2 given q=q1=q2.
     
  6. May 7, 2009 #5
    Re: Magnitude of "q" ~ What am I doing wrong?

    Yay! The sun came up and suddenly Physics isn't half as evil as I thought! ;)
     
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