AxiomOfChoice
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Can someone tell me how to express a line of charge of charge per unit length [tex]\lambda[/tex] as a delta function volume charge density in cylindrical coordinates?
The discussion centers around expressing a line of charge with charge per unit length \(\lambda\) as a delta function volume charge density in cylindrical coordinates. Participants explore the formulation of this expression and its implications in both cylindrical and rectangular coordinate systems.
Participants express varying degrees of confidence in their formulations, with some agreeing on the correctness of the cylindrical coordinate expression while others question the rectangular coordinate version. The discussion remains unresolved regarding the optimal representation in rectangular coordinates.
Participants discuss the implications of using delta functions in different coordinate systems, but there are no explicit resolutions to the assumptions or limitations of their formulations.
[itex] <br /> This is correct. Why? There are two things: first the 2-D delta function is right, and second, the charge density is right. First the delta function. You are integrating over an area in the x-y plane, so <br /> [tex] dA = r\, d\phi \, dr.[/tex]<br /> You require your 2-D delta function [tex]\delta(\mathbf{r}-\mathbf{\rho})[/tex] to satisfy<br /> [tex] f(a,\phi_0) = \int \, dA \, \delta(\mathbf{r}-\mathbf{\rho}) \, f(r, \phi).[/tex]<br /> Above, [tex]\mathbf{\rho}[/tex] is your source location in the x-y plane, and [tex]\delta(\mathbf{r}-\mathbf{\rho})[/tex] is simply a symbol to represent the delta function that you are looking for. Clearly your choice above satisfies this:<br /> [tex] \int dr \, \int r\, d\phi \, \frac{1}{a} \delta(r - a) \delta (\phi - \phi_0) \, f(r, \phi)<br /> = \int dr \, r\, \frac{1}{a} \delta(r - a) \, f(r, \phi_0)<br /> = a \frac{1}{a} f(a,\phi_0) = f(a,\phi_0).[/tex]<br /> Note that your cartesian coordinate version you posted is also okay.<br /> <br /> Now for the charge density. Beyond the delta function you basically just need to make sure you have the right units. your "r" delta function has units of one over length, so your charge density has the right units.[/itex]AxiomOfChoice said:Ok, here's my guess:
[tex] \rho(\vec{r}') = \frac{1}{a}\lambda \delta(r' - a) \delta (\phi' - \phi_0).[/tex]
Assuming that the line of charge is infinite, parallel to the [itex]z[/itex]-axis, a distance [itex]a[/itex] from the origin, and at an angle [itex]\phi = \phi_0[/tex] w.r.t. the [itex]x[/itex]-axis. Sound good?[/itex]