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Surface charge density

  1. Jul 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A surface is defined by a hemisphere of radius b, centered on the x-y plane. The surface charged density is given by [tex]\rho_s(z) = z (\frac{Coul}{m^3})[/tex].

    2. Relevant equations
    [tex]\rho_s(z) = z = Rcos(\theta) = bcos(\theta) (\frac{Coul}{m^3})[/tex].

    3. Question
    My question is how can the surface charge density equal to [tex]Rcos(\theta) = bcos(\theta)[/tex]? That is a measure of the distance from the origin to the surface [element], and thus only [to my knowledge] have units of radius b, or meters [tex]\neq (\frac{Coul}{m^3})[/tex].


  2. jcsd
  3. Jul 13, 2009 #2


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    z has no units, it's just a coordinate in the coordinate system, and what they are saying is that the charge density at points with coordinates (x,y,z) is equal to z C/m2. (You wrote C/m3 which I'm assuming is a typo since we're talking about a surface, not a 3D region.)
  4. Jul 13, 2009 #3
    Sorry about the typo. But what about the variable b which has units of meters? How does that fit into the interpretation of surface charges?

    Thank you.
  5. Jul 13, 2009 #4


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    No typo. One unit of length in z cancels one unit of length in the denom. of C/m^3.

    This is misleading (i.e. wrong). The first inequality should be confusing. I would suggest:
    \rho_s(z)=\rho_0z\rightarrow{}\rho_0Rcos(\theta)=bcos(\theta) (C/m^3)
    where [itex]\rho_0[/itex] is some unknown constant that has units of charge-per-volume and R and b have units of length.
    Or, better yet,
    where [itex]\rho_0[/itex] is some unknown constant with units of C/m^2.
  6. Jul 13, 2009 #5
    That makes much more sense, thank you.
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