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Particle Physics Problem

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  1. Aug 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Three point charges are located on the positive x-axis of a coordinate system. Charge q1 = 2.0 nC is 2.0 cm from the origin, charge q2 = -3.5 nC is 4.0 cm from the origin and charge q3 = 5.0 nC located at the origin.

    What is the net force ((a)magnitude and (b) direction) on charge q1 = 2.0 nC exerted by the other two charges?.



    2. Relevant equations
    F= k (q1q2/r^2)

    3. The attempt at a solution
    I know that the direction (b) is 0 degrees as all the points are all located on the x-axis, I just can't figure out part (a). I keep mixing up my diagram so that when the equation is applied, my answer comes out incorrectly. Any help would be great!
     
  2. jcsd
  3. Aug 22, 2015 #2

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    Superposition of two forces. Do them one at a time.
     
  4. Aug 22, 2015 #3
    Okay, so would it be...
    F of 2 on 1 = 9*10^-6((-3.5*10^-6)*(-2*10^-6)/(0.04)^2 = 39.375 N (+ve therefore forces repel)
    F of 3 on 1 = 9*10^-6((2*10^-6)*(-5*10^-6)/(0.02)^2 = -225N (-ve therefore forces attract)
    F total = 39.375 - (-225) = 264.375 N
     
  5. Aug 22, 2015 #4

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    "n" means what? Like charges attract or repel?
     
  6. Aug 22, 2015 #5
    'N' in the final result was for Newtons
     
  7. Aug 22, 2015 #6

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    Lower case "n," as in "nC."
     
  8. Aug 22, 2015 #7
    nC = nanocoulomb
     
  9. Aug 22, 2015 #8

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    "Nano" means what?
     
  10. Aug 22, 2015 #9
    ohhh.... 1*10^-9 not 1*10^-6 thank you I didn't even see that !
    so the new solutions come out at
    F of 2 on 1 = -0.000225 N
    F of 3 on 1 =0.000039 N
    So the final answer would be -0.000225 N - 0.000039 N = -2.46*10^-4


    I just tried to enter that in and it still came out incorrect... I'm just not sure what I'm doing wrong
     
    Last edited: Aug 22, 2015
  11. Aug 22, 2015 #10

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    4-2=?
     
  12. Aug 22, 2015 #11
    4-2=2 but I'm not sure where that fits in..
     
  13. Aug 22, 2015 #12

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  14. Aug 22, 2015 #13
    Okay so it should be
    F of 2 on 1 = 9*10^9((-3.5*10^-9)*(-2*10^-9)/(0.02)^2 = 0.000158 N
    F of 3 on 1 = 9*10^9((2*10^-9)*(-5*10^-9)/(0.02)^2 = -0.00025N
    Therefore, Ft= 0.000158-(-0.00025) = 3.8*10^-4 N

    Which I just put in and it's correct! Thank you so much for your help!!! :) :)
     
  15. Aug 22, 2015 #14

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    You are welcome. Things to remember: one step at a time; and, pay attention to details.
     
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