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Homework Help: Two point charges along the y-axis.

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A charge of q1 = -8.5 μC is at y = 6.0 m, and a charge of q2 = -6 μC is at y = -4.0 m. Locate the point (other than infinity) at which the total electric field is zero.


    2. Relevant equations
    E=k(q/r^2)
    E1=E2
    Quadratic Equation

    3. The attempt at a solution

    9x10^9 (8.5x 10^-6)/(10-r)^2 = 9x10^9(6x10^-6)/r^2

    8.5r^2 = 6(100-20r + r^2)

    8.5r^2= 600-120r+6r^2

    2.5r^2+120r-600=0

    so using the quadratic equation

    -120+-SqRt [120^2-(4)(2.5)(-600)] / 2(2.5)

    -120+-SqRt [(14400-(-6000)] / 5

    -120+-SqRt 20400 / 5

    -120 +-142.8/ 5

    R= 4.56, -148.56

    with realistic value being 4.56

    Im using webassign and I submitted 8 previous different answers before submitting 4.5. I know the answer was not 4.5 but 4.56, but is it that sensitive. Since it is my last allowed attempt, I want to be sure it is 4.56 before submitting.

    Thank you

    sorry if its hard to understand, i dont know how to do the symbols and stuff
     
    Last edited: Feb 10, 2010
  2. jcsd
  3. Feb 10, 2010 #2

    rl.bhat

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    Homework Helper

    since the two charges are opposite in nature, the point of zero electric field will not be between the charges. It will be out side the charges and it will be nearer to the small charge. So you rewrite the equations and solve for r.
     
  4. Feb 10, 2010 #3
    Ive tried like a hundred different set ups could you write out the equation its supposed to be its due tonight?
     
  5. Feb 10, 2010 #4

    rl.bhat

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    If d is the distance between the two charges, at the neutral point
    Kq1/r^2 = kq2/(d+r)^2. Here q2>q1.
    Now substitute the values and find r.
     
  6. Feb 10, 2010 #5
    Charges are the same sign (q1 = -8.5 μC, q2 = -6 μC) so the point of zero electric field is between them. Draw a picture and pick an arbitrary point between the two and write an expression for the electric field at that point (rl.bhat's equation except d-r not d+r) and solve for r.
     
  7. Feb 10, 2010 #6

    rl.bhat

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    Yes. You are right. I missed that point.
     
  8. Feb 10, 2010 #7
    Kq1/r^2 = kq2/(d-r)^2

    so

    kq2/(d-r)^2= Kq1/r^2 correct?

    So my original 9x10^9 (8.5x 10^-6)/(10-r)^2 = 9x10^9(6x10^-6)/r^2 should be ok correct?

    k=9x10^9
    q1=8.5^10-6
    q2=6x10^-6
    d=10

    So according to your responses I did set it up correctly??

    I made r2 to r. Doesnt change anything.

    meaning the answer should be 4.56?
     
    Last edited: Feb 10, 2010
  9. Feb 10, 2010 #8
    Yeah, the electric field is 0 4.56 meters from the charge with the smaller magnitude, which is around y=+.56m.
     
  10. Feb 10, 2010 #9
    DOH!!!!! This whole time I was putting 4.56 in the answer instead of -4y + 4.56=.56.

    Now I can understand the movie A Beautiful Mind so much more, its addicting to get the right answer. I was up untill 3:30 Am last night.

    Thank you so much...

    but Ill be back lol!
     
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