A line of charge as a delta function

Click For Summary

Discussion Overview

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.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to express a line of charge as a delta function volume charge density in cylindrical coordinates.
  • Another participant proposes a formulation: \(\rho(\vec{r}') = \frac{1}{a}\lambda \delta(r' - a) \delta (\phi' - \phi_0)\), assuming the line of charge is infinite, parallel to the z-axis, at a distance \(a\) from the origin, and at an angle \(\phi = \phi_0\) with respect to the x-axis.
  • A third participant questions whether a corresponding charge density in rectangular coordinates would be \(\rho(\mathbf{r}') = \lambda \delta(x' - a) \delta (y' - a)\).
  • A later reply reiterates the initial cylindrical coordinate formulation and discusses the correctness of the delta function and charge density, emphasizing the need for proper units and integration over the area in the x-y plane.

Areas of Agreement / Disagreement

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.

Contextual Notes

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.

AxiomOfChoice
Messages
531
Reaction score
1
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?
 
Physics news on Phys.org
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]
 
...anyone? Please?

Would this be the correct charge density in rectangular coordinates:

[tex] \rho(\mathbf{r}') = \lambda \delta(x' - a) \delta (y' - a)[/tex]
 
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]
[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]
 
Last edited:

Similar threads

Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
507