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Let us say we have a volume $V$ having a volumetric mass(charge) density $\rho(x,y,z)$.

Suppose we have a smooth surface $S$ within the volume with a surface mass(charge) density $\sigma(x,y,z)$, if possible, how can one obtain $\sigma (x,y,z)$ from $\rho (x,y,z)$?

If it is an integral, it must be of the form: $$\sigma = \int \rho dx$$

where $x$ is some spatial coordinate. But a simple example made me doubt of the veracity of the expression above: Consider a solid sphere with mass $M$ and radius $R$, supposing that the formula above is correct and using the spherical symmetry of the problem, the surface density for a surface sphere within the solid sphere will be:

$$\sigma =\int \rho dr$$ since $\rho$ is constant, we can extract it from the integral:$$\sigma =\rho\int dr$$

Also, since we are considering a spherical sheet, the limits for $r$ are $r_{0}-\delta r$ and $r_{0} +\delta r$, where $r_{0}$ is the radius of our sphere, thus:

$$\sigma(r_{0})= \rho \int_{r_{0}-\delta r}^{r_{0}+\delta r} dr=\rho \times 2\delta r$$

taking the limit $\delta r \rightarrow 0$ we finally get:

$$\sigma(r_{0}) =0 ~C/m^2$$

this last equality must certainly be incorrect, for it states that there are no charges on $S$, which is not true.

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  • $\begingroup$ $$\int_{r_{0}-\delta r}^{r_{0}+\delta r} dr=(r_{0}^2-\delta r^2)$$... Are you sure of this? $\endgroup$ – acarturk Sep 1 '19 at 18:55
  • $\begingroup$ @acarturk Thank you, I'll correct it. There's still a problem with $\sigma (r_{0})$ though. $\endgroup$ – Hilbert Sep 1 '19 at 19:29
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    $\begingroup$ Look up Dirac delta function. $\endgroup$ – Razor Sep 1 '19 at 19:38
  • $\begingroup$ I' m aware that delta functions can be used to express volume densities in terms of surface or linear ones, here I am looking for the other way around, from volume densities to surface densities. The question above is different from this one. $\endgroup$ – Hilbert Sep 1 '19 at 19:44
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Modelling a true volumetric distribution of charge by a surfacic distribution usually only makes sense if one of the dimensions of the distribution is small (yet finite) compared to the others. It would be the case for a shell with a small thickness (difference between outer and inner radii) compared to the surface of the shell. In this case, you could integrate your distribution along the thickness, as you do, but without letting your thickness tend to zero. The implicit assumption while performing this is that you are able to compute the total charge as $ Q= \int \rho(r) S(r)dr \approx S \int \rho(r) dr$, that is $S$ remains approximately constant constant and $\int \rho(r) dr$ becomes your surfacic charge. For a sphere, the only thing that could make sense is to compute the total charge $Q$ and divide it by the surface of the sphere to obtain the surfacic charge $\sigma = Q/ S$.

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  • $\begingroup$ "compute the total charge Q and divide it by the surface of the sphere to obtain the surfacic charge σ=Q/S." Actually, the $\sigma$ obtained would be meaningless and certainly not the surface charge present on the the outermost surface said sphere. $\endgroup$ – Hilbert Sep 2 '19 at 17:39
  • $\begingroup$ It's not completely meaningless since a sphere with this surfacic charge would yield the same outer electric field as the original sphere. I understand your point, though, and what you were trying to get at. $\endgroup$ – user8736288 Sep 2 '19 at 19:32

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