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Surface charge density of conducting disc

  1. Mar 11, 2016 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    This is problem 3.4 from Prucell and Morin if you have the book.
    Capture.PNG

    2. Relevant equations
    None

    3. The attempt at a solution
    Electric field inside a conducting sphere is zero. Let P be a point on one of its equatorial plane. The field along the plane is zero. So I know the charge distribution that can produce zero electric field along a disc.

    Let the charge density of the shell be ##\sigma##. It is a constant function.

    Consider a patch like the red line in the below diagram. The yellow surface is the disc.

    Let the thickness of the patch be ##dr##. Let its area be ##A##.
    ##A## is a projection of the circle on the shell having the red line as diameter.
    2.PNG
    If the area of that circle is ##A_1##, then area of the patch is ##A_1\cos\theta=A##
    Hence ##A=\frac{A_1}{\cos\theta}##

    Here, $$A_1=2\pi R\cos\theta\cdot\frac{dr}{\cos\theta}$$

    Now, $$Q=\int_0^R{\sigma A}=\sigma\int\frac{2\pi R^2}{\sqrt{R^2-r^2}}dr$$.

    Is my understanding correct? I am not getting the correct answer.
     
    Last edited: Mar 11, 2016
  2. jcsd
  3. Mar 12, 2016 #2

    haruspex

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    The area of your band A does not depend on theta. (Ref Archimedes.)
    But anyway, the band A is specific to the location of P. Clearly, if there is a field from a rotationally symmetric field it will be in the radial direction, and the band A will not have a component in that direction at P.
    You need to consider the field at P due to the entire shell.
     
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