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E field from half ring of charge.

  1. Feb 19, 2007 #1
    1. The problem statement, all variables and given/known data
    How do I find the electric field from a semi-ring of charge at the center of the would-be circle? Does that make sense? Or another way: If the semi-circle of charge is centered at the origin, what is the electric field at the origin? I hope you know what I mean now.

    the total charge of the ring is Q, the radius is R.

    Oh yeah, the semi circle is in the left hand plane

    2. Relevant equations
    E = k\int_{}^{l}\frac{(r-r')\rho _l(r')}{|r-r'|^3} dl'

    3. The attempt at a solution

    E = k\int_{\pi/2}^{-\pi/2}\frac{(-r')\rho _l(r')}{|-r'|^3} d\theta

    Is converting to cylindrical coordinates the right move? Is this even the equation I should be using? What should I use as the charge density?
    Last edited: Feb 19, 2007
  2. jcsd
  3. Feb 20, 2007 #2
    Try applying Gauss's Theorem. Assume a gaussian surface enclosing the ring and [tex]\int E.ds=\frac{q}{\epsilon}[/tex]
  4. Feb 20, 2007 #3

    Meir Achuz

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    Gauss won't work here because there is not full circular symmetry
    There is enough symmetry to see the direction of the E field.
    Just put cos\theta into the numerator of your integral.
    rho is really a linear density lambda=Q/2pi R.
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