Relationship between volume and surface density

I was not sure if it fits better here, or in the math forum, but this is the problem I am trying to solve:

Say you have a full cylinder of radius $$R$$, height $$L$$, and its uniformly charged with volume density $$\rho$$. And now I draw a hollow cylinder inside (with the same height, and radius $$r < R\,$$), is there any numerical relationship between its surface density $$\sigma$$(r) and $$\rho$$ ?

• What is the thickness of the hollow cylinder? – Bio Apr 13 at 14:37
• I don't understand the question.. how can a hollow cylinder have thickness? – Frogfire Apr 13 at 14:44
• Bio means the thickness of the two-dimensional wall of the cylinder, which is zero if the cylinder is two-dimensional. – descheleschilder Apr 14 at 3:33

Now let's see.

The total charge, $$C_t$$, in the cylinder can be written as:

$$C_t=\pi R^2 L\rho$$

The total charge on the hollow cylinder, $$C_{thc}$$, with radius $$r$$ is

$$C_{thc}=2\pi rL\sigma,$$ where $$\sigma$$ is independent of $$r$$, just like $$\rho$$.

The point though is that because the hollow cylinder has a two-dimensional surface (zero thickness), we can't relate $$\rho$$ to $$\sigma$$. If $$\sigma \gt 0$$ then $$C_t$$ would be infinite. $$\rho$$ is only defined for three-dimensional cases, while $$\sigma$$ only for two-dimensional cases.

Maybe you could think that

$$C_t=\int _{r=0} ^R 2\pi r L \sigma dr,$$

but this gives for the dimension of $$C_t$$ $$(Cm)$$.

You cán get the volume though of the cylinder by considering the surface of the hollow cylinder and integrate from $$r=0$$ to r=R.

So maybe it is possible if we consider the inverse of $$\sigma=\frac 1{\sigma}$$, a constant, which has as its unit $$(\frac{m^2}{C})$$, and the inverse of $$\rho=\frac 1 {\rho}$$, also a constant in a uniform medium, with unit $$(\frac{m^3}{C})$$.

Now let's (for simplicity) consider a cube of which the sides L are $$1(m)$$, and proceed in the way we calculate the volume from the integral of surfaces:

$$\frac 1{\rho}=\int _{l=o}^{L}\frac 1 {\sigma}dl=\frac 1 {\sigma}\int _{l=0}^{L} 1 dl=\frac 1 {\sigma}|_{l=0}^L l=\frac 1 {\sigma}L ,$$ so

$$\rho=\frac{\sigma}{L}.$$

Off course, when $$L=1$$, this reduces to $$\sigma=\rho$$, but keep in mind the units are the same in this case.

You can try the same procedure for the cylinder and find out yourself the relation between $$\rho$$ and $$\sigma$$ (keep an eye on the units!).