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Force on a point from electric charges

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    knight_Figure_25_46.jpg
    What is the net force on the bottom charge?
    Please give the force in Fx and Fy
    2. Relevant equations

    F=qE
    E=kq/r^2
    3. The attempt at a solution
    Fx=2.88x10^-9N
    Fy=-1.01x10^-4N

    These aren't correct. I used the equation above and used

    r=0.035m for both the side charges.

    For the top charge,

    r=0.05m

    Why is the formula not working?

    Thanks!
     
  2. jcsd
  3. Apr 3, 2013 #2

    ehild

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    Why do you use r=3.5 cm for the side charges when r=5 cm is indicated?

    ehild
     
  4. Apr 4, 2013 #3
    I have to separate the forces into their x and y components, so the distance on the x axis should be 3.5 cm. is this not the right approach?
     
  5. Apr 4, 2013 #4

    ehild

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    How do you use the distance on the x axis at all? Show your working in detail.

    ehild
     
  6. Apr 4, 2013 #5

    tiny-tim

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    hi rocapp! :smile:
    nooo :redface:

    you want Q/r2 times cosθ,

    but you're using Q/(rcosθ)2 :wink:
     
  7. Apr 4, 2013 #6
    Ohh! Thanks so much!
     
  8. Apr 4, 2013 #7
    Does the 1nC charge affect itself in terms of force?

    I keep getting wrong answers still.

    -2.124×10^-23 N is the y-component I get from the -6nC charge.
    1.00x10^-23 N is the y-component I get from both 2 nC charges on the 1 nC charge.
    Fy = -1.124x10^-24 N

    But that's not correct.

    Fx=1.00x10^-23 N

    Not sure if either of these are correct.
     
    Last edited: Apr 4, 2013
  9. Apr 4, 2013 #8

    ehild

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    How did you get these data? Show your work in detail, please.

    ehild
     
  10. Apr 4, 2013 #9
    In the images attached are the exact equations I used for Fx and Fy.
     

    Attached Files:

  11. Apr 4, 2013 #10

    ehild

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    You use wrong value for k in Coulomb's Law.

    You need the resultant force on the bottom charge. Fx should be the horizontal component of the resultant force, Fy has to be the vertical component. What are the directions of the force components exerted bas the charges?

    ehild
     
    Last edited: Apr 5, 2013
  12. Apr 5, 2013 #11
    Ah. Thanks for that!

    The horizontal x component of the -6 nC charge is zero because it is directly above the 1 nC charge.

    The horizontal x component of one of the 2 nC charges is:

    Fx = (k*q*q/r^2)*cos(theta)
    Fx = ((8.99x10^9)*(2x10^-9)*(1.0x10^-9)/(.05^2))*cos(45)
    Fx = 5.09x10^-6 N

    Since there are two, the total Fx = 2*5.09x10^-6 N = 1.02x10^-5 N

    The y-components of the two 2 nC charges are the same as the x components because sin(45)=cos(45).

    The y-component of the -6 nC charge is

    Fy=((8.99x10^9)*(1.0x10^-9)*(-6x10^-9)/(0.05^2))
    Fy= -2.16x10^-5 N

    So the total Fy= Fy + Fy = (-2.16x10^-9) + (1.02x10^-5) = 1.02x10^-5 N

    This is still not correct, though.
     
  13. Apr 5, 2013 #12

    ehild

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    Check the direction of these forces. Are not the x components opposite to each other?

    Show what you mean on positive y direction, up of down?

    Do not note the different force components with the same Fy. And you made a mistake, it is not 2.16x10^-9 but 2.16x10^-5.

    ehild
     
  14. Apr 5, 2013 #13
    Thanks!
     
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