# Charge density definition in Cylindrical Coordinates

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?

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

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.
• 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? Jan 24, 2018 at 18:31
• 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. Jan 24, 2018 at 18:44
• The factor of $2\pi$ comes from the dirac distribution naturally: fen.bilkent.edu.tr/~ercelebi/mp03.pdf Feb 16, 2018 at 19:56