# Surface charge density

1. Jul 13, 2009

### jeff1evesque

1. The problem statement, all variables and given/known data

A surface is defined by a hemisphere of radius b, centered on the x-y plane. The surface charged density is given by $$\rho_s(z) = z (\frac{Coul}{m^3})$$.

2. Relevant equations
$$\rho_s(z) = z = Rcos(\theta) = bcos(\theta) (\frac{Coul}{m^3})$$.

3. Question
My question is how can the surface charge density equal to $$Rcos(\theta) = bcos(\theta)$$? That is a measure of the distance from the origin to the surface [element], and thus only [to my knowledge] have units of radius b, or meters $$\neq (\frac{Coul}{m^3})$$.

Thanks,

JL

2. Jul 13, 2009

### dx

z has no units, it's just a coordinate in the coordinate system, and what they are saying is that the charge density at points with coordinates (x,y,z) is equal to z C/m2. (You wrote C/m3 which I'm assuming is a typo since we're talking about a surface, not a 3D region.)

3. Jul 13, 2009

### jeff1evesque

Sorry about the typo. But what about the variable b which has units of meters? How does that fit into the interpretation of surface charges?

Thank you.

4. Jul 13, 2009

### turin

No typo. One unit of length in z cancels one unit of length in the denom. of C/m^3.

This is misleading (i.e. wrong). The first inequality should be confusing. I would suggest:
$$\rho_s(z)=\rho_0z\rightarrow{}\rho_0Rcos(\theta)=bcos(\theta) (C/m^3)$$
where $\rho_0$ is some unknown constant that has units of charge-per-volume and R and b have units of length.
Or, better yet,
$$\rho_s(z)\rightarrow\rho_0cos(\theta)$$
where $\rho_0$ is some unknown constant with units of C/m^2.

5. Jul 13, 2009

### jeff1evesque

That makes much more sense, thank you.

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