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Feb 16, 2018 at 19:56 comment added JoeSwap The factor of $2\pi$ comes from the dirac distribution naturally: fen.bilkent.edu.tr/~ercelebi/mp03.pdf
Jan 24, 2018 at 18:45 vote accept JoeSwap
Jan 24, 2018 at 18:44 comment added secavara 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 18:31 comment added JoeSwap 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 18:27 history edited secavara CC BY-SA 3.0
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Jan 24, 2018 at 18:21 history answered secavara CC BY-SA 3.0