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Expressing surface charge density as volume charge density

  1. Mar 5, 2017 #1
    I'm working on this: When I consider a disc with radius ##a## and total charge ##Q## uniformly distributed (placed in the XY plane and centered at the origin) and determine the volume charge density in cylindrical coordinates, I have assumed is of the form ##\rho=A \delta (z) U(R-r)##, (##U## is the unit step function) and obtained just what I expected $$\rho=\frac{Q}{\pi R^2} \delta (z) U(R-r)$$
    The problem arise when I try to solve this problem in spherical coordinates because my first assumption was ##\rho=A \delta (\theta-\pi/2) U(R-r)## (note that here ##r## is not the same as in the previous paragraph) but some people include the scale factor corresponding to each variable appearing in each Dirac delta; in this case $$\rho=A \frac {\delta (\theta-\pi/2)}{r} U(R-r)$$
    What's the reason for including such factor? Is there any mathematical result that support this?
     
  2. jcsd
  3. Mar 5, 2017 #2

    Orodruin

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    If you do not include the r you do not get a uniform charge distribution. Also note that ##\delta(\theta-\pi/2)## would not have the correct physical dimension for ##A## to be a surface charge density.
     
  4. Mar 5, 2017 #3
    That's what makes me confused. If I include the scale factor ##r## I get ##\rho\propto 1/r ##, how could it be a uniform charge distribution if density decreases as radius increases?
     
  5. Mar 5, 2017 #4

    Orodruin

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    No you don't. If you do not include it you get a density that increases with radius and has the wrong dimensions. I suggest you check it by integrating over a small volume containing an area element.
     
  6. Mar 5, 2017 #5

    Orodruin

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    Note that ##\delta(z) = \delta(r\cos(\theta)) = \delta(\cos(\theta))/r = \delta(\theta-\pi/2)/r##.
     
  7. Mar 5, 2017 #6
    Finally I can see it, the last relation is so clear... Thanks a lot.
     
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