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Electrostatics: Charge distribution and Dirac's delta, lots of questions

  1. Apr 14, 2010 #1

    fluidistic

    User Avatar
    Gold Member

    The problem was to give the charge distribution of a uniformly charged disk of radius R centered at the origin and perpendicular to the z-axis.
    The professor explained us how to do so (though many parts were unclear to me and I couldn't copy everything that was on the blackboard because we were too much students and I couldn't see the bottom part of the blackboard).
    For a punctual charge, the general form of the charge distribution is [tex]\rho (\vec x) = q \delta (x) \delta (y) \delta (z)[/tex].
    For a charged plane: [tex]\rho (\vec x) = q \delta (z)[/tex] and for a charged line: [tex]\rho (x,y,z)=q \delta (x) \delta (y)[/tex].

    However for a finite (in extension) plane perpendicular to the z-axis, [tex]\rho (\vec x) = f(x,y)\delta (z)[/tex]. The professor told us that the Dirac's delta "reduces" the dimension (though I don't understand why/how. If you have any comment on his, this would be nice). That's why for a point-like charge we need 3 of these deltas, 1 for a plane and 2 for a line.
    The purpose of the function f in the above example is to make the final value coincide with what it should be; that's what I understood.
    Using cylindrical coordinates [tex](\rho, z, \phi)[/tex], for the uniformly charged disk of radius R we have that [tex]\rho (\vec x)=\delta (z) \theta (\rho -R) f(\rho)[/tex] where [tex]\theta (\rho - R) = 1[/tex] for [tex]\rho \leq R[/tex] and 0 for [tex]\rho \geq R[/tex]. Thus the theta function is a constraint that delimit the disk of radius R.
    Then I don't really know what the professor has done, some integration I believe and he reached a final result I couldn't copy.
    In spherical coordinates [tex](r, \theta, \phi )[/tex] and for the same problem, he reached that [tex]\rho (\vec x) =\frac{Q}{\pi R^2 r} \delta (\theta - \frac{\pi}{2}) \hat \theta (r-R)[/tex] which looks different from the result he reached in cylindrical coordinates but it must be the same.
    If I understand well the final result, it has no sense to ask for the charge distribution in a particular point in [tex]\mathbb{R}^3[/tex]. However it makes sense to ask for the charge distribution in any surface element; though I don't know how to proceed. I think I should do an integration or so, but I'm not sure and an example would be welcome.
    So the final answer to the question is a function that is ill defined because for example in the origin it is not defined nor at the boundaries although there are charges there.
    And if I'm not wrong, if I do integrate the final answer over an area (or volume?!) A, then I should get [tex]\sigma _0 A[/tex] ? Namely the amount of charges enclosed into this small area? Is this right?
    Why is the function "ill defined" as I call it? Is it because we DO NOT assume charges like points but like a property of a non zero space extension?

    If I have the problem "What is the charge distribution of a uniformly charged cube?", I'd rather answer [tex]\rho _0[/tex] for [tex]-a \leq x \leq a[/tex], [tex]-a \leq y \leq a[/tex], [tex]-a \leq y \leq a[/tex]. 0 everywhere else.
    My answer wouldn't be ill defined and I could give you the charge distribution in any particular point in space, unlike the method used in more complicated problems.
     
  2. jcsd
  3. Apr 15, 2010 #2

    Dale

    Staff: Mentor

    A disk is a cylinder whose height has been "flattened". So, if you can come up with the expression for a cube, then you can probably come up with the expression for a cylinder. Then, simply take the part of the expression of the cylinder that sets the height, and replace it with the delta function.
     
  4. Apr 15, 2010 #3

    fluidistic

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    Gold Member

    Thanks for this insight.
    I use cylindrical coordinates (I can't remember the variables so I take [tex](\rho , \theta , z)[/tex] ).
    [tex]0\leq \rho \leq a[/tex] , [tex]0 \leq \theta \leq 2 \pi[/tex] , [tex]0\leq z \leq b[/tex] for a cylinder. I replace z by [tex]\delta (z-z_0)[/tex] and I get a disk perpendicular to the z-axis which crosses it at [tex]z=z_0[/tex] and with radius [tex]a[/tex].
    Is this good? How would you personally write it? I feel my description of the cylinder is kind of "the ugly way".
     
  5. Apr 16, 2010 #4

    Dale

    Staff: Mentor

    Well, your description is fine, but if you want to make it prettier then you can define a couple of useful functions.

    1) The unit step function:
    [tex]u(x)=\int_{-\infty}^x{\delta(X)dX}[/tex]

    2) The boxcar function:
    [tex]B_{a,b}(x)=u(x-a)-u(x-b)[/tex]

    Then your charge distribution becomes
    [tex]\frac{Q}{A}B_{0,a}(\rho) B_{0,2\pi}(\theta)\delta(z-z_0)[/tex]
     
    Last edited: Apr 16, 2010
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