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Coulomb's Law Problem and net force

  1. Jan 26, 2012 #1
    1. The problem statement, all variables and given/known data
    In a region of two-dimensional space, there are three fixed charges: +1 mC at (0, 0), -2 mC at (16 mm, -6 mm), and +3 mC at (-6 mm, 20 mm). What is the net force on the -2-mC charge?
    -magnitude
    -direction (° counterclockwise from the +x-axis)

    2. Relevant equations
    F = k*q1*q2 / d^2

    3. The attempt at a solution
    I got the magnitude correct. Fnetx: 8.77e07 N and Fnety: 5.71e07 N ; magnitude: 1.05e08 N. When I try to get the angle however my answer of 145 degrees is wrong. I don't understand why. I used arctan(Fnety/Fnetx).
     
  2. jcsd
  3. Jan 26, 2012 #2

    gneill

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    Staff: Mentor

    Could be an accuracy issue. Your Fx and Fy values look okay, so try calculating your angle again. Show your work.
     
  4. Jan 26, 2012 #3
    @=theta
    F12 = k*2mC*1mC/(17.1mm)^2 = 6.16e7
    F32 = k*2mC*3mC/(34.1mm)^2 = 4.65e7

    Fnetx = F12 cos(@) + F32cos(@)
    Fnety = F12 sin(@) + F32sin(@)

    Fnetx = 5.77e7 + 3e7 = 8.77e7
    Fnety = 2.16e7 + 3.55e7 = 5.71e7

    Answer for the magnitude of the force.
    sqrt [ Fnetx^2 + Fnety^2 ] = 1.05e8 N
    Answer for the angle of the force counterclockwise from the x-axis.
    180 - arctan(Fnety/Fnetx) = 147 degrees

    I've tried values of 33, 57, 123, 147, 145 and I still get the answer wrong (the reason why I tried 33, 57 etc. was to see if the computer was mistaken and wanted the angle from a different reference axis).
     
  5. Jan 26, 2012 #4

    gneill

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    Staff: Mentor

    Your work looks good. When I calculate the angle, keeping several extra decimal places for all intermediate results, the result is 146.933 degrees. So I think that your 147° answer should have been acceptable. I suggest that you contact your lecturer and present the issue.
     
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