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Point charges on square - find magnitude of force

  1. Feb 3, 2004 #1
    Four point charges, each of magnitude 12.92 x 10^-6 C, are placed at the corners of a square 65.5 cm on a side. Given K_e = 8.98755 x 10^9 Nm^2/C^2. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. Answer in units of N.

    can someone help me out with this? thanks
     
  2. jcsd
  3. Feb 3, 2004 #2

    Doc Al

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    Can you find the force that each positive charge exerts on the negative charge? (Hint: Coulomb's law) Draw yourself a picture to keep the direction of each force straight. Now add the force vectors up. (Another hint: take advantage of fact that the charges are symmetrically distributed; you should be able to tell which direction the net force points.)
     
  4. Feb 3, 2004 #3
    im not sure what u mean
    this is what i tried

    i found the magnitude of each vector using kqq/r^2
    then i squared all 3 magnitudes and added, and then took the square root. but its wrong.
    what can i do?
     
  5. Feb 3, 2004 #4

    Doc Al

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    You found each separate force vector. Excellent. Now describe the direction of each vector. You must first add these vectors before you can find the magnitude of the resultant force. I have attached a picture of the four charges that may prove helpful. Why don't you use the picture to describe the forces.
     
  6. Feb 3, 2004 #5

    Doc Al

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    Oops... here it is.
     

    Attached Files:

    Last edited: Feb 3, 2004
  7. Feb 3, 2004 #6
    how do i recieve the attachment?
     
  8. Feb 3, 2004 #7

    Doc Al

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    Oh well, the attachment doesn't seem to be working. Why not draw one of your own?
     
  9. Feb 3, 2004 #8

    i think i have the directions correct, how do i add the vectors here?
     
  10. Feb 3, 2004 #9

    Doc Al

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    To add the vectors, I suggest finding their components. Add the components, then you can find the magnitude of the resultant.
     
  11. Feb 4, 2004 #10
    i keep getting it incorrect
    i would use cos45 for the x compoenent and sin45 for the y right?
    then i add those components to the other two magnitudes?
     
  12. Feb 4, 2004 #11

    Doc Al

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    Not quite. Let's do it step by step. It looks like my diagram did finally make it into the system; go to my previous posts and take a look.

    Charges at A, B, and C exert forces on the negative charge at D. Tell me the x and y components of each of those forces.
     
  13. Feb 4, 2004 #12
    F_db*cos45 and F_db*sin45 ?
     
  14. Feb 4, 2004 #13

    Doc Al

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    Which force is this? Do it systematically for each force:

    What force does A exert on D: magnitude and x & y components.
    What force does B exert on D: magnitude and x & y components.
    What force does C exert on D: magnitude and x & y components.
     
  15. Feb 4, 2004 #14
    this is my view of it.

    A+ B+


    C- D+


    F_ad = kqq/R^2
    F_dc = kqq/R^2
    F_bc = kqq/(sqr(x))^2 [x = distance of side of square]
    F_bc_x = kqq/(sqr(x))^2 * cos45
    F_bc_y = kqq/(sqr(x))^2 * sin45
     
  16. Feb 5, 2004 #15

    Doc Al

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    I just realized that you were drawing your own diagram. D'oh!
    Here's what I get (using your diagram) for the forces on C:
    [tex]F_{AC} = \frac{kq^2}{x^2} [/tex] (magnitude)
    [tex]F_{BC} = \frac{kq^2}{2x^2} [/tex] (magnitude)
    [tex]F_{DC} = \frac{kq^2}{x^2} [/tex] (magnitude)

    Now the components:
    [tex]F_{AC-x} = 0; F_{AC-y} = \frac{kq^2}{x^2} [/tex]
    [tex]F_{BC-x} = \frac{kq^2}{2x^2}cos(45); F_{BC-y} = \frac{kq^2}{2x^2}cos(45)[/tex]
    [tex]F_{DC-x} = \frac{kq^2}{x^2}; F_{DC-y} = 0 [/tex]

    Add the components:
    [tex]F_{x} = \frac{kq^2}{x^2}(1 + \frac{cos(45)}{2}) [/tex]
    [tex]F_{y} = \frac{kq^2}{x^2}(1 + \frac{cos(45)}{2}) [/tex]

    Now the magnitude of the resultant:
    [tex]F_{net} = \sqrt{2}\frac{kq^2}{x^2}(1 + \frac{cos(45)}{2}) [/tex]

    Plug in the numbers and you'll get your answer. (Unless I goofed.)
     
  17. Feb 5, 2004 #16
    for the part where you showed the y component do you mean sin45 or is it really cos45?
     
  18. Feb 5, 2004 #17

    Doc Al

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    sin(45)=cos(45), so it doesn't matter.
     
  19. Feb 5, 2004 #18
    hah oh right..forgot

    thanks for ur patience and help doc, ill reply back to tell you if the answer is correct
     
  20. Feb 5, 2004 #19
    It's correct. Thanks a bunch Doc.
     
  21. Feb 5, 2004 #20

    Doc Al

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    You are welcome. :smile:
     
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