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Uniform Distributed Electric Charge

  1. Jan 25, 2007 #1
    1. The problem statement, all variables and given/known data
    A ring-shaped conductor with radius a = 2.20 cm has a total positive charge Q = 0.123 nC uniformly distributed around it.

    (Question relevant to main problem)
    (a)What is the magnitude of the electric field at point P, which is on the positive x-axis at x = 35.0 cm?

    [**Actual Question**]
    A particle with a charge of - 2.70 mC is placed at the point P described in part (a). What is the magnitude of the force exerted by the particle on the ring?

    2. Relevant equations

    F = 1/4*pi*e_o * |q| / r^2 for single point charge

    e_o = 8.85 * 10^-12


    3. The attempt at a solution

    1/(4*pi*8.85 * 10^-12) * ( 0.123*10^-9 * 2.70*10^-9 / .35^2 )

    The online system says I'm off by an additive constant,

    I also tried

    1/(4*pi*8.85 * 10^-12) * ( 0.123*10^-9 * 2.70*10^-9 / .35^2 + 0.0220^2)

    Since that equation was also in the book, but still no luck.
     
  2. jcsd
  3. Jan 25, 2007 #2

    Kurdt

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    The electric fields at a point due a to a charge distribution is given by:

    [tex] \mathbf{E} = \frac{1}{4\pi \epsilon_0} \int \frac{\sigma}{r^2} \mathbf{\hat{r}} dl[/tex]

    [itex]\sigma[/itex] is the charge density of the ring of charge.
     
  4. Jan 25, 2007 #3
    Given your equation:

    e_o = 8.85 * 10^-12

    rho = 0.0220m

    r = .35m

    I'm not sure what r_hat is in this situation?
     
  5. Jan 25, 2007 #4

    Kurdt

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    The r-hat is the unit vector pointing in the direction of the field. You will have to perform the integral which will be different depending on what coordinate system you use.

    Do you have a diagram? The geometry is important aswell. I assume the axis of rotation of the circle is the x-axis.
     
  6. Jan 25, 2007 #5
    Here is the diagram:

    [​IMG]
    http://www.uploadyourimages.com/view/505148yf.figure.21.21.jpg
     
  7. Jan 25, 2007 #6

    Kurdt

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  8. Jan 25, 2007 #7

    Kurdt

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    Oh hang on problem diagnosed. You've put 2.7 nC in the equation instead of 2.7mC (2.7x10-3). I jumped the gun and started answering the first part.
     
  9. Jan 25, 2007 #8
    Wait, so what am I suppose to do?
     
  10. Jan 25, 2007 #9

    Kurdt

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    Your original method should work you just mistook the units of one of the charges. When I said I jumped the gun I wasn't kidding I had you on the path of deriving the equation you already had. I just didn't recognise it at first because you'd put all the numbers in instead of symbols.
     
  11. Jan 25, 2007 #10
    yea haha, okay thanks. I see what you mean.
     
  12. Jan 25, 2007 #11

    Kurdt

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    I must be cursed with you. I keep sending you down the wrong path :eek: many apologies.
     
  13. Jan 26, 2007 #12
    it's okay. Thankfully homework isn't worth much in my overall grade, least this will help me prepare for the exam which is worth a lot more.
     
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