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Point charges magnitude and direction

  1. Sep 2, 2010 #1

    J89

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    1. The problem statement, all variables and given/known data
    Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 0.40 µC charge. (A = 0.20 µC, B = 6.60 µC, and C = -3.80 µC.). Diagram below...



    2. Relevant equations

    F= Ke |q1||q2|/r^2, cos and sin symbols



    3. The attempt at a solution

    a) .20*10^-6*(6.60*10^-6)*8.99*10^9/(.500)^2 = .0474672
    b) .20*10^-6*(3.80*10^-6)*8.99*10^9/(.500)^2 =.0273296

    .0474672*cos(240) = .015463926386
    .0474672*sin(240) = .044877634258

    = .015463926386 + .0273296 = .042793526386

    .042793526386^2 + .044877634258^2 = .003845287957 = sqrt(.003845287957 ) = 0.062 N. Direction is 275 degrees. However, according the real solution, the answer is not 0.062! Can someone find what I did wrong here?
     

    Attached Files:

  2. jcsd
  3. Sep 2, 2010 #2
    Where is the 0.40 µC charge located in the picture?

    Edit: At any rate, I found the problem. You had your calculator in radians mode. In degrees mode, 0.0474672sin(240) = -0.0411.
     
    Last edited: Sep 2, 2010
  4. Sep 2, 2010 #3

    J89

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    it is supposed to be .20, sorry, but thanks :)
     
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