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Electrostatic force between more than 2 charges

  1. Jan 20, 2008 #1
    hi
    i dont know if this has been posted or not but i am sort fo stuck on this one

    What is the force F_vec on the 1 nC charge at the bottom? View Figure
    Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures.

    the figure looks like this knight_Figure_25_46.jpg

    attempt to this problem:

    i found out the forces individually and added them together .. but it didnt work

    x-component:
    6nC = 0
    2nC = 5.09 x 10[tex]^{-5}[/tex] each

    y-component:
    6nC = 2.16 x 10[tex]^{-5}[/tex]
    2nc = 5.09 x 10[tex]^{-5}[/tex] each

    when i add them: 1.02 x 10 [tex]^{-5}[/tex] [tex]\hat{x}[/tex] , 7.76 x 10[tex]^{-5}[/tex] [tex]\hat{y}[/tex]

    but that was wrong.. i also tried the x-comp to be 0 .. but that was wrong too

    thanx in advance for the help
     
  2. jcsd
  3. Jan 20, 2008 #2

    Doc Al

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    Staff: Mentor

    How does the x-component of the force exerted by one 2 nC charge compare to that of the other?
     
  4. Jan 20, 2008 #3
    they are opposite of each other .. which means there is no net x- component .. but thats not the right answer tho..
     
  5. Jan 20, 2008 #4
    I see 2 things you might do differently to help.
    Make sure you're doing vectors (right is +, up is +). Keep the vector notation throughout.
    Check your use of scientific notation. If you're unsure, have your calculator switch it to regular decimal.
     
  6. Jan 20, 2008 #5
    i tried a similar question, with 1 + and 1 - 2nC charge .. and that worked out
    but somehow i cant get this working ..

    i am not too sure if the y - component it right as well or not .. even tho it seems right
     
  7. Jan 20, 2008 #6

    Doc Al

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    Show how you calculated the vertical components of the forces from the 2 nC charges.
     
  8. Jan 20, 2008 #7

    Doc Al

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    What are the directions (signs) of these components?
     
  9. Jan 20, 2008 #8
    y-comp = Esin[tex]\vartheta[/tex]
    r= 5x10^-2
    E = Kq[tex]_{1}[/tex]q[tex]_{2}[/tex] / r[tex]^{2}[/tex]
    = (9.0x10^9)(2x10^-9)(1x10^-9)/ ( 2.5x10^-3)
    E= 7.2x10^-6

    y-comp = E sin(45) = 5.09x10^-6
     
  10. Jan 20, 2008 #9
    6nC = +y, no x
    2nC = +y, +x (on the right)
    2nC = +y, -x (on the left)

    and thats why my understand was that the x components would cancel out .. but i am not sure whats going on
     
  11. Jan 20, 2008 #10

    Doc Al

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    Realize that the 2 nC charges are positive and the 6 nC charge is negative. Which way do their forces on the + 1 nC charge act?
     
  12. Jan 20, 2008 #11
    so the 2 +y .. should be -y ..?
     
  13. Jan 20, 2008 #12
    i did the calc again and i get

    x = 0
    y = 1.14x10^-5 ..

    doe sthat seem right?
     
  14. Jan 20, 2008 #13

    Doc Al

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    Correct. Since the charges repel, the y component of force is downward and thus negative.
     
  15. Jan 20, 2008 #14
    thank you ., thank you ... thank you..
     
  16. Jan 20, 2008 #15

    Doc Al

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    What's the distance between the charges?
     
  17. Jan 20, 2008 #16
    what you said abt the charges repelling .. this answer is right ..

    thank you so much once again
     
  18. Jan 20, 2008 #17

    Doc Al

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    You are most welcome.

    Never mind. You had written the distance squared. :smile:
     
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