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Columb's law

  1. Jan 8, 2008 #1
    1. The problem statement, all variables and given/known data

    In a vacuum, two particles have charges of q1 and q2, where q1 = +3.5 µC. They are separated by a distance of 0.25 m, and particle 1 experiences an attractive force of 2.3 N. What is q2 (magnitude and sign)?

    2. Relevant equations



    3. The attempt at a solution

    not sure why this wouldn't work 2.3 N = (8.99 x 10^9 * Q2)/ .25^2

    should that work? and would the sign be + or -?

    thanks.
     
  2. jcsd
  3. Jan 8, 2008 #2

    olgranpappy

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    Homework Helper

    you want the force to be attractive, as stated in the problem. do like charges attract or repel?
     
  4. Jan 8, 2008 #3
    like charges repel. however, is my formula equation correct?
     
  5. Jan 8, 2008 #4
    You are missing a charge, remember

    [tex] F = k \frac{qQ}{r^2}[/tex]
     
  6. Jan 8, 2008 #5
    Okay, I see, so then I solve for Q2, correct?

    which shows now 2.3 N = ((8.99 x 10^9) * (2.3) * (Q2) )/ .25^2

    i get something like: -4.57 x 10^-12 C
     
  7. Jan 8, 2008 #6
    You should solve for what you want algebraically before you put in numbers. It is a lot easier that way. So you are right, you want the value of the other charge.

    [tex]q_2 = \frac{r^2 F}{kq_1}[/tex]

    in terms of the numbers that would be... (you do it)

    *note: you made a mistake for the value of the first charge. Also note that I just gave you the magnitude of the charge, you will have to figure out the sign.
     
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