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- Thread starter AxiomOfChoice
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- #2

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[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?

- #3

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Would this be the correct charge density in rectangular coordinates:

[tex]

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

[/tex]

- #4

jasonRF

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[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.

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