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