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Charge density - exact answer?

  1. Sep 10, 2006 #1
    Question: The electric field on the surface of an irregularly shaped conductor varies from 56.0 kN/c to 28.0kN/c. calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is (a) greatest and (b) smallest.

    I am stuck as far as how to get an exact number. The only thing I concluded was that at the largest radius of curvature, charge density is smallest, and I called this lamda. Because electric field ranges to double the smallest value, I concluded at the smallest radius of curvature the charge density is 2 lamda. Is this totally incorrect?
     
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  3. Sep 10, 2006 #2

    Kurdt

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    The charge density is proportional to 1/(radius of curvature at that particular point).
     
  4. Sep 10, 2006 #3
    I understand that, but I'm just given the information above, which is simply an electric field, so I couldn't find the exact charge density any more than what I have above, could I?
     
  5. Sep 10, 2006 #4

    Kurdt

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    Ok correct my mistake I totally read that wrong. The equation relating the potential and the electric field just outside the surface is:

    [tex] v=aE=\frac{a\sigma}{\epsilon_0} [/tex]

    where a is the radius of curvature at a particular point and sigma is the charge density.
     
  6. Sep 10, 2006 #5
    thanks man, I was overlooking that for some reason, appreciate it.
     
  7. Sep 10, 2006 #6

    Kurdt

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    sorry about the original post just didn't have my head screwed on.
     
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