If we have a straight wire with linear charge density $\lambda$ and $l$ lenght. Is it correct to define the volumetric charge density as:

$ \rho(r,\phi,z) = \lambda \frac{1}{\rho} \delta(\rho) H(l-z) $

Where $\delta$ stands for the delta dirac distribution expressed in cylindrical coordinates and $H$ for the heaviside step function?

  • $\begingroup$ No. If you plot that step function, you'll see that that describes a semi-infinite wire. $\endgroup$
    – Chris
    Jan 24, 2018 at 20:05

1 Answer 1


I think one thing you have to correct is the fact that that charge density you wrote there is non-vanishing for all $z<l, \, \rho=0$. And you want it to be finite right? One way is this $\lambda \frac{1}{2 \pi \rho} \delta(\rho) \left[ H(z+\frac{l}{2})-H(z-\frac{l}{2})\right]$ which is a bit more symmetric. The $2 \pi$ is for the angular integration.

  • $\begingroup$ Thanks @secavara. The heaviside modification makes a lot of sense (I was not thinking about z being negative which needs to be included as well!). I still do not understand the $2\pi$ factor. I know we need it to get the total charge but what is the (previous to integration) reason? Is it related with the delta dirac in cylindrical coordinates? $\endgroup$
    – JoeSwap
    Jan 24, 2018 at 18:31
  • $\begingroup$ The integration argument is fundamental to me, in the sense that the least that we could demand is that $Q = \int \rho \, dV$. There are not many alternatives... something like $\delta(\phi)$ which in principle could work for $Q$ is something I've never seen (given the fact that the $z$ axis is a sick region for the angular coordinate) whereas you find this $2 \pi$ in, for instance, eq. 3.132 of Jackson's book, Classical Electrodynamics. $\endgroup$
    – secavara
    Jan 24, 2018 at 18:44
  • $\begingroup$ The factor of $2\pi$ comes from the dirac distribution naturally: fen.bilkent.edu.tr/~ercelebi/mp03.pdf $\endgroup$
    – JoeSwap
    Feb 16, 2018 at 19:56

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