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Calculating the charge densities

  1. Nov 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Lets say you have a infinitely long surface with one side of length ##L## and a surface charge density ##ρ_s## and you need to transform that into a linear charge density ##Q'## so that you can represent the surface along some axis ##y## so the the surface is placed normal to the axis and goes from ##y=0## to ##y=L## how would you make that transition? Take the infinitely long dimension to be ##h##.
    2. Relevant equations
    3. The attempt at a solution

    I tried this line of thinking:
    $$Q'=\frac{dQ}{dh}=\frac{ρ_sdS}{dh}=\frac{ρ_sdhdl}{dh}=ρ_sdl$$
    Could this be correct and how would it be in a more complicated case? Is there a pattern here?
     
  2. jcsd
  3. Nov 8, 2016 #2

    haruspex

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    I don't understand the arrangement. It is an infinite strip width L, and the y axis is normal to the strip? So say it is L in the x axis and infinite in z. But now you have y from 0 to L? And in what sense are you wanting to "represent" the surface?
     
  4. Nov 9, 2016 #3
    Sorry, i think i was a little confusing in choosing the words. Lets imagine it like this. You see the wall, the ##y-axis## is to the right and left of this wall, the ##x-axis## is up and down and the ##z-axis## is behind and in front of you. The surface is placed so that it is infinite up and down and of length ##L## from ##y=0## to ##y=L##. Is now clearer?
     
  5. Nov 9, 2016 #4

    haruspex

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    Ok, but then what do you mean by the required transformation into a linear density? Where is this in the picture, and in what sense does it relresent the original charge? The fields are obviously not the same.
     
  6. Nov 9, 2016 #5
    So with the set up coordinate system you rotate it so that the x-axis looks to you and you see only the z ad the y. The surface looks like a line on the y axis you right? So given the surface charge density can you transform it into a linear charge density along the y axis bu cutting the surface into small elements of ##dy## (lines along the surface). This is the problem.
     
  7. Nov 9, 2016 #6

    haruspex

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    Ok. I think you are asking whether the field in the YZ plane can be simulated by a uniform charge along the Y axis.
    The answer is no.
    Close to the plane, the field is approximately constant (as for an infinite plane sheet). For a line of charge, the field gets very much stronger as you approach the line.
     
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