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 \lambda as a delta function volume charge density in cylindrical coordinates?

 

[AxiomOfChoice]

Ok, here's my guess:

<br /> \rho(\vec{r}&#039;) = \frac{1}{a}\lambda \delta(r&#039; - a) \delta (\phi&#039; - \phi_0).<br />

Assuming that the line of charge is infinite, parallel to the z-axis, a distance a from the origin, and at an angle \phi = \phi_0[/tex] w.r.t. the x-axis. Sound good?

 

[AxiomOfChoice]

...anyone? Please?

Would this be the correct charge density in rectangular coordinates:

<br /> \rho(\mathbf{r}&#039;) = \lambda \delta(x&#039; - a) \delta (y&#039; - a)<br />

 

[jasonRF]

Ok, here's my guess:

<br /> \rho(\vec{r}&#039;) = \frac{1}{a}\lambda \delta(r&#039; - a) \delta (\phi&#039; - \phi_0).<br />

Assuming that the line of charge is infinite, parallel to the z-axis, a distance a from the origin, and at an angle \phi = \phi_0[/tex] w.r.t. the x-axis. Sound good?

<br /> <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 /> &lt;br /&gt; dA = r\, d\phi \, dr.&lt;br /&gt;<br /> You require your 2-D delta function \delta(\mathbf{r}-\mathbf{\rho}) to satisfy<br /> &lt;br /&gt; f(a,\phi_0) = \int \, dA \, \delta(\mathbf{r}-\mathbf{\rho}) \, f(r, \phi).&lt;br /&gt;<br /> Above, \mathbf{\rho} is your source location in the x-y plane, and \delta(\mathbf{r}-\mathbf{\rho}) is simply a symbol to represent the delta function that you are looking for. Clearly your choice above satisfies this:<br /> &lt;br /&gt; \int dr \, \int r\, d\phi \, \frac{1}{a} \delta(r - a) \delta (\phi - \phi_0) \, f(r, \phi)&lt;br /&gt; = \int dr \, r\, \frac{1}{a} \delta(r - a) \, f(r, \phi_0)&lt;br /&gt; = a \frac{1}{a} f(a,\phi_0) = f(a,\phi_0).&lt;br /&gt;<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 &quot;r&quot; delta function has units of one over length, so your charge density has teh right units.