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 $\rho$ ?

Thanks in advance.

  • $\begingroup$ What is the thickness of the hollow cylinder? $\endgroup$ – Bio Apr 13 '19 at 14:37
  • $\begingroup$ I don't understand the question.. how can a hollow cylinder have thickness? $\endgroup$ – Frogfire Apr 13 '19 at 14:44
  • $\begingroup$ Bio means the thickness of the two-dimensional wall of the cylinder, which is zero if the cylinder is two-dimensional. $\endgroup$ – descheleschilder Apr 14 '19 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


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!).

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