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Magnitude of force on a charge

  1. Aug 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Trying to find the magnitude of the resulting force on the -7 nC charge at the origin and my answers arent matching up. Is there a step I am missing? I used the pythagorean to find r for both ΔOrigin-Q1 and ΔOrigin-Q2. Here is the graph: http://imgur.com/90TQmQU.


    2. Relevant equations
    I've been using
    Fy = kQ1Qy/r^2
    Fx = kQ1Qx/r^2
    Fnet = √(Fx)^2 + (Fy)^2


    3. The attempt at a solution
    The current attempts have landed me at
    Fy = 1.736 * 10^-2
    Fx = 1.481 * 10^-2
    Fnet = .23 N
     
  2. jcsd
  3. Aug 31, 2013 #2

    gneill

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    Staff: Mentor

    You'll have to show how you arrived at your results. They don't look right. Consider that the charges have values in the nanocoulomb range, and the distances are in meters. The resulting forces should be much smaller than you've found.

    If you show your work, we can see how to help.
     
  4. Sep 1, 2013 #3
    Thanks for you reply. Here is what gave me those results:

    Fy = ((8.99*10^9)(-7*10^-6)(8*10^-6))/5.385^2
    Fx = ((8.99*10^9)(-7*10^-6)(4*10^-6))/4.123^2
    Fnet = sqrt((1.736*10^-2)^2 + (1.481*10^-2)^2 ) = 0.23 N
     
  5. Sep 1, 2013 #4

    gneill

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    Staff: Mentor

    Ah. You've taken the individual forces and called them Fx and Fy, but that is not the case. Each of those two forces has X and Y components which contribute to the total X and Y force components.

    You would have been correct if the 8nC charge was situated on the Y-axis and the 4nC charge on the X-axis. But both are located off-axis and must therefore their locations are vectors with both x and y components. You'll need to sort out the force components of each with a bit of trig.

    Also, nano is smaller than micro... micro is 10-6, nano 10-9. So adjust your charge magnitudes.
     
  6. Sep 1, 2013 #5
    Thank you for the reply. So to find the X and Y vectors of the individual charges I need to find theta and multiply each magnitude by its respective sin/cos?
     
  7. Sep 2, 2013 #6

    gneill

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    Staff: Mentor

    That's the idea unless you are familiar with vector algebra, in which case you might forgo the trig and work directly with vector operations.
     
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