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## Homework Statement

Hi! This is not really a problem. I'm just confused on how to express the charge distribution of a set of point charges in spherical coordinates. From our discussion,

[itex]ρ(\vec{r})=\sum\limits_{i=1}^N q_i δ(\vec{r}-\vec{r}')[/itex]

where [itex]\vec{r}[/itex] is the position of the point where charge density is being evaluated and [itex]\vec{r}'[/itex] is the position of the point charge.

Three-dimensional Dirac delta function in general is

[itex]δ(\vec{r})= \frac{1}{h_1 h_2 h_3} δ(u_1 - u_1 ')δ(u_2 -u_2 ') δ(u_3 - u_3 ')[/itex]

So in spherical coordinates, it is

[itex]δ(\vec{r})= \frac{1}{r^2 sin(θ)} δ(r - r')δ(θ -θ') δ(\phi - \phi')[/itex]

The continuous charge distribution of these point charges is therefore

[itex]ρ(\vec{r})=\sum\limits_{i=1}^N \frac{1}{r^2 sin(θ)} δ(r - r'_i)δ(θ -θ'_i) δ(\phi - \phi'_i)[/itex]

So here's the part where I'm confused. For the factor [itex]\frac{1}{r^2 sin(θ)}[/itex]. do I retain it as is (i.e. r and θ are coordinates of position vector [itex]\vec{r}[/itex] or do I subsitute the coordinates of my point charge (i.e. the coordinates of [itex]\vec{r}'[/itex])? Thank you very much.

## Homework Equations

See above.

## The Attempt at a Solution

I've seen a problem and an online solution and it seems that he substituted the coordinates of [itex]\vec{r}'[/itex] but I can't reconcile it from the definition of 3-D Dirac delta function I've read where the coordinates of [itex]\vec{r}[/itex] are substituted instead. Thank you very much.