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Coulomb's Law and forces

  1. Jan 22, 2008 #1
    [SOLVED] Coulomb's Law

    The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +3.5 µC, x1 = 3.0 cm, y1 = 0.50 cm, and q2 = -4.0 µC, x2 = -2.0 cm, y2 = 1.5 cm.
    At what coordinates should a third charge q3 = +5.5 µC be placed such that the net electrostatic force on particle 3 due to particles 1 and 2 is zero?

    I set both Forces together, F13= F23. When i simplified i i get the equation (q1/x^2)=(q2/(x-l)^2). I used the distance formula to find l and i came up with the answer .026.
    When i solve for x i get .0126m or 1.26 cm. I understand i have to use sin and cosine to get the answers. I used the formula x= x2-x1cos -11.3 and the same for y using different numbers. X1 is the value i just solved for. by doing this i get an answer of -3.57 for x and -1.686 which are incorrect. I was able to find the angle by using the equation theta=Tan-1((y2-y1)/(x2-x1). i know that this value is correct because the system i used called webassign checks the answers immediately. Can someone please tell me what i did wrong.
     
  2. jcsd
  3. Jan 22, 2008 #2

    Doc Al

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    Are you sure that that is the distance between the charges? (Double check.)

    Also, where must q3 be placed? Between the other charges? Or to one side?
     
  4. Jan 22, 2008 #3

    Astronuc

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    Why x-l, as opposed to x+l?

    What is the significance of the fact the magnitude of q2 > q1?
     
  5. Jan 22, 2008 #4
    i believe that the charge should be to the left of q1 since q2 has a greater charge than q1 and and sorry the actual calculation was 5.099 cm which is equal to .05099 m
     
  6. Jan 22, 2008 #5
    i assumed it was x-l but i learned that i am wrong by figuring out where the charge will be located
     
  7. Jan 22, 2008 #6

    Doc Al

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    When you say "to the left of q1" do you mean between the charges?
     
  8. Jan 22, 2008 #7
    i placed the charges in a cordinate plane and q2 was to the left of q1 considering it had a coordinate of -2 and to answer your question no it is outside the charges
     
  9. Jan 22, 2008 #8

    Doc Al

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    Right. Which side?
     
  10. Jan 22, 2008 #9
    it would have to be on the positive side since the charge on q2 is greater that the charge of q2
     
  11. Jan 22, 2008 #10

    Doc Al

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    Right. So redo your force equation accordingly.
     
  12. Jan 22, 2008 #11
    so as Astronuc said it would be x+l not x-l if that is the case the x value would be -2.4 cm or 73.9 cm
     
  13. Jan 22, 2008 #12

    Doc Al

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    Right. And only one of those answers makes sense.
     
  14. Jan 22, 2008 #13
    It would have to 73.9 because it is positive and i am looking for a positive value
     
  15. Jan 22, 2008 #14

    Doc Al

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    Exactly. (Assuming your arithmetic is correct.) A negative answer would put the third charge between the other two--which is no good.
     
  16. Jan 22, 2008 #15
    Thank you Doc Al for your quick responses i was able to figure out the rest of the problem on my own using the angle. Once again thank you for your help!!
     
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