# Homework Help: Expressing surface charge density as volume charge density

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1. Mar 5, 2017

### Mictlantecuhtli

I'm working on this: When I consider a disc with radius $a$ and total charge $Q$ uniformly distributed (placed in the XY plane and centered at the origin) and determine the volume charge density in cylindrical coordinates, I have assumed is of the form $\rho=A \delta (z) U(R-r)$, ($U$ is the unit step function) and obtained just what I expected $$\rho=\frac{Q}{\pi R^2} \delta (z) U(R-r)$$
The problem arise when I try to solve this problem in spherical coordinates because my first assumption was $\rho=A \delta (\theta-\pi/2) U(R-r)$ (note that here $r$ is not the same as in the previous paragraph) but some people include the scale factor corresponding to each variable appearing in each Dirac delta; in this case $$\rho=A \frac {\delta (\theta-\pi/2)}{r} U(R-r)$$
What's the reason for including such factor? Is there any mathematical result that support this?

2. Mar 5, 2017

### Orodruin

Staff Emeritus
If you do not include the r you do not get a uniform charge distribution. Also note that $\delta(\theta-\pi/2)$ would not have the correct physical dimension for $A$ to be a surface charge density.

3. Mar 5, 2017

### Mictlantecuhtli

That's what makes me confused. If I include the scale factor $r$ I get $\rho\propto 1/r$, how could it be a uniform charge distribution if density decreases as radius increases?

4. Mar 5, 2017

### Orodruin

Staff Emeritus
No you don't. If you do not include it you get a density that increases with radius and has the wrong dimensions. I suggest you check it by integrating over a small volume containing an area element.

5. Mar 5, 2017

### Orodruin

Staff Emeritus
Note that $\delta(z) = \delta(r\cos(\theta)) = \delta(\cos(\theta))/r = \delta(\theta-\pi/2)/r$.

6. Mar 5, 2017

### Mictlantecuhtli

Finally I can see it, the last relation is so clear... Thanks a lot.