A line of charge as a delta function

[AxiomOfChoice]

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?

 

[AxiomOfChoice]

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?

 

[AxiomOfChoice]

...anyone? Please?

Would this be the correct charge density in rectangular coordinates:

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

 

[jasonRF]

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?

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
[tex]
dA = r\, d\phi \, dr.
[/tex]
You require your 2-D delta function [tex]\delta(\mathbf{r}-\mathbf{\rho})[/tex] to satisfy
[tex]
f(a,\phi_0) = \int \, dA \, \delta(\mathbf{r}-\mathbf{\rho}) \, f(r, \phi).
[/tex]
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:
[tex]
\int dr \, \int r\, d\phi \, \frac{1}{a} \delta(r - a) \delta (\phi - \phi_0) \, f(r, \phi)
= \int dr \, r\, \frac{1}{a} \delta(r - a) \, f(r, \phi_0)
= a \frac{1}{a} f(a,\phi_0) = f(a,\phi_0).
[/tex]
Note that your cartesian coordinate version you posted is also okay.

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 teh right units.

 

[sardar]

A line of charge can be represented as a delta function in cylindrical coordinates by using the formula for volume charge density, which is given by ρ = λδ(ρ)δ(φ)δ(z). Here, ρ represents the radial distance from the origin, φ represents the angle in the cylindrical coordinate system, and z represents the distance along the axis of the line of charge. By multiplying the charge per unit length, λ, with the delta function in the cylindrical coordinates, we can express the line of charge as a delta function volume charge density. This representation allows us to simplify calculations and analyze the electric field and potential due to the line of charge in cylindrical coordinates. I hope this helps.