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Continuous Surface Charge Source - Integral Confusion

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the total charge on a circular disc of radius ρ = a if the charge density is given by
    ρs = ρs0 (e^−ρ) sin2 φ C/m2 where ρs0 is a constant.
    Are the two limits of integration from 0 -> a for ρ and 0->2∏ for φ? In the example given in the notes, ρ varies, instead of being a constant value. Does this mean that I cannot integrate ρ in this problem?


    2. Relevant equations

    Q = ∫ρdS

    3. The attempt at a solution

    Q = ∫∫ρs0 (e^−ρ) sin^2 (φ) dφdρ...
     
  2. jcsd
  3. Jan 28, 2012 #2

    SammyS

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    You have the Greek letter rho, ρ, used in two very different ways here.

    ρ as a coordinate: This is the role usually player by the letter, r, in polar coordinates.

    ρ as (surface) charge density: Here ρ seems to be accompanied by what looks like should be a subscript. ρs and ρs0.

    Yes, the limits of integration are as you indicate.

    The charge density, ρs, is a function of the coordinate, ρ. (somewhat confusing).

    For polar coordinates, the element (differential) for area, dS, is given by: [itex]\displaystyle dS=\rho\,d\varphi\,d\rho\,.[/itex]

    Other than that, you seem to be on the right track.
     
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