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Charge density of a disk with radius a in cylindrical coordinates

  1. Mar 5, 2014 #1
    To write the uniform charge density of a disk with radius a in cylindrical coordinates, If we do this form:
    [itex]\rho (x)=\frac{A\delta(z)\Theta (a-\rho)}{\rho}[/itex] (A is constant that sholud be determined and [itex]\theta [/itex]is step function), we get [itex]A=\frac{Q}{2\pi a} [/itex] and so:
    [itex]\rho (x)=\frac{\frac{Q}{2\pi a}\delta(z)\Theta (a-\rho)}{\rho}[/itex]
    But we know that the correct one is:
    [itex]\rho (x)=Q\frac{\delta(z)\Theta (a-\rho)}{2\pi a^2}[/itex].
    Could anyone please tell me what is my mistake?
  2. jcsd
  3. Mar 5, 2014 #2
    While I'm no expert on the subject, I DO notice that your "ro" term suddenly vanishes (term in the denominator).
  4. Mar 5, 2014 #3
    Try starting with [itex] \rho (x)=A\delta(z)\Theta (a-\rho)[/itex]
  5. Mar 5, 2014 #4
    Ok, in this special case it gives the correct answer, but in the spherical coordinate system it doesn't give the correct answer. My principle question is what is the routine way of solving such problems?
  6. Mar 6, 2014 #5


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    Take a cylindrical disk of finite height filled uniformly with a total charge [itex]Q[/itex]. Obviously the charge density is given by
    [tex]\rho=\frac{Q}{\pi a^2 h} \Theta(a-r_{\perp}) \Theta(-h/2<z<h/2).[/tex]
    I write [itex]r_{\perp}[/itex] for the radial coordinate in order to avoid conflicts with [itex]\rho[/itex] as the symbol for the charge density. In the limit [itex]h \rightarrow 0^+[/itex] this gives
    [tex]\rho=\frac{Q}{\pi a^2} \Theta(a-r_{\perp}) \delta(z).[/tex]
    You can easily check that this gives you the correct total charge,
    [tex]\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(\vec{x})=\int_{0}^{a} \mathrm{d} r_{\perp} \int_0^{2 \pi} \mathrm{d} \varphi \int_{\mathbb{R}} \mathrm{d}z r_{\perp} \frac{Q}{\pi a^2} \delta(z) = \frac{Q}{\pi a^2} 2 \pi \frac{a^2}{2}=Q.[/tex]
  7. Mar 6, 2014 #6
    What do you mean it doesn't work in spherical coordinates? If you post your calculation I might be able to comment.
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