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Finding Magnitude and Direction of Force on a Charge

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Three charges are arranged as shown in the picture I attached. Find the magnitude and direction of the electrostatic force on the charge at the origin.


    2. Relevant equations
    I know that Coulomb's Law must be applied.


    3. The attempt at a solution
    I've drawn a force diagram with FCB pointing to the left and FAB pointing down. I've also broke the vectors into components using Coulomb's Law and added them together. By doing this, I got 2.7 x 10-5 [itex]\hat{x}[/itex] + 0 [itex]\hat{y}[/itex]. From here what do I do to get an answer?

    I know the answer is supposed to be 1.38 x 10-5 at 77.5 degrees so I need help with how to get there.
     

    Attached Files:

  2. jcsd
  3. Aug 27, 2011 #2

    Doc Al

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    Staff: Mentor

    You must have made a mistake, since the resultant must have a component in the y direction. Show how you computed FCB and FAB.
     
  4. Aug 27, 2011 #3
    FCB=[(9x109|(6x10-9)(5x10-9)|) / (.32)][cos 180][itex]\widehat{x}[/itex] + [(9x109|(6x10-9)(5x10-9)|) / (.32)][sin 180][itex]\widehat{y}[/itex]

    FAB=[(9x109|(-3x10-9)(5x10-9)|) / (.12)][cos 270][itex]\widehat{x}[/itex] +[(9x109|(-3x10-9)(5x10-9)|) / (.12)][sin 270][itex]\widehat{y}[/itex]
     
  5. Aug 27, 2011 #4

    Doc Al

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    OK, now simplify these results by evaluating those trig functions.
     
  6. Aug 27, 2011 #5
    I did and thats how I got 2.7x10-9[itex]\hat{x}[/itex] + 0[itex]\hat{y}[/itex]
    Was my set up right and I went wrong somewhere in the math? I was always thinking I set it up wrong or wasn't using the right values.
     
  7. Aug 27, 2011 #6

    Doc Al

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    Give me your values for sin(180), cos(180), sin(270), and cos(270).
     
  8. Aug 27, 2011 #7
    sin(180) = 0
    cos(180) = -1
    sin(270) = -1
    cos(270)=0
     
  9. Aug 27, 2011 #8

    Doc Al

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    Good. Now simplify your equations in post #3 accordingly.
     
  10. Aug 27, 2011 #9
    What's the point of incorporating Trig functions? If you compute them individually then you get two separate forces, one along the y axis and one along the x axis. Remember also that the force between charges A and B is repulsive, so the force you drew downwards is meant to be upwards.
     
    Last edited by a moderator: Aug 27, 2011
  11. Aug 27, 2011 #10

    Doc Al

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    While you don't need to use trig functions, there's nothing wrong with using them.
    Charges A and B are oppositely charged, thus the force is attractive.
     
  12. Aug 28, 2011 #11
    C and B, sorry. They'll produce an opposite charge.
     
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