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Net Force of Point Charges, Coulomb's law

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +8.02 C; the other two have identical magnitudes, but opposite signs: q2 = -4.73 C and q3 = +4.73 C. (a) Determine the net force exerted on q1 by the other two charges. (b) If q1 had a mass of 1.50 g and it were free to move, what would be its acceleration?


    2. Relevant equations

    F= kq1q2/r^2

    3. The attempt at a solution

    F = k(8.02E-6C)(4.73E-6C)/(1.3^2)
    F= 0.20179N

    The horizontal componenets of the Force vectors cancel out and the vertical components are both pointing straight up. I think I want to find the net vertical force so I did:

    sin23 = x/0.2017
    x = -0.1707N <--I thought this was the Force on q1 from one of the charges so in order to get the net force I doubled it = -0.3415N

    This answer didn't look right and it wasn't but i'm really confused how to get the net force.
    For part b I'm pretty sure I understand how to figure it out I just need the answer from part a to solve it.
    F = ma --> a=F/m
    m=1.5g -->0.0015kg

    a= (?F?)/0.0015kg
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 2, 2009 #2
    "sin23 = x/0.2017
    x = -0.1707N"

    If you are using a calculatro, check the settings, I get something different from this one. Also pay attention to the sign, now it's contradicting with the picture.
  4. Mar 2, 2009 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Rethink the direction of the vertical components.

    The angle is 23 degrees, not radians. (You have your calculator set to radian mode.)
  5. Mar 2, 2009 #4
    Thanks! I got it right, and because I changed my calculator back to degrees I also was able to figure out why another problem wasn't working - Thanks for your help!
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