What is the physical analogy between the flux of a flow of fluid through a closed surface and the flux of an electric field through a closed surface?

Question

In Griffiths' Introduction to Electrodynamics, in writing about electric field lines, he says

...the flux of $\vec{E}$ through a surface $S$ is a measure of the "number of field lines" passing through $S$. I put this in quotes because of course we can only draw a representative sample of the field lines - the total number would be infinite. But for a given sampling rate the flux is proportional to the number of lines drawn, because the field strength, remember, is proportional to the density of field lines (the number per unit area), and hence $\vec{E} \cdot d\vec{a}$ is proportional to the number of lines passing through the infinitesimal area $d\vec{a}$. This suggests that the flux through any closed surface is a measure of the total charge inside.

What does "a given sampling rate" mean here?

I suppose this means, for a given number of field lines that you choose to draw (if you could draw in 3D), assuming that you are consistent in how you draw them, following a set of basic rules

a) if $q$ gets 8 lines, then $2q$ gets 16 lines

b) when they emanate from the charge, you draw them "fairly" spaced, meaning symmetrically in all directions

then you will see that for any area you choose, the number of your drawn field lines that pass through the area is proportional to the flux in that area.

Is this interpretation correct?

The dot product $\vec{E}\cdot d\vec{a}$ is thus some measure of the strength of the electric field in this infinitesimal area.

I remember a similar calculation when flux was introduced to me in multivariable calculus. There, the physical setup was that of a flow of a fluid described by a velocity vector field and a fluid density vector field.

For a small area on a closed surface, the dot product of a velocity vector and an outer unit vector normal to the surface gave a height, multiplied by a $dA$ element we obtained a volume, and multiplied by the fluid density we obtained a mass.

Integrating over the surface we obtained a mass of fluid leaving the closed surface.

There seems to be somewhat of a direct analogy here, but I don't see what the analogue of fluid mass is in the case of an electric field.

After all, $\vec{E}$ is measured in $Newton/Coulomb$, and a surface integral of $\vec{E} d\vec{a}$ would have units $\frac{Newton \times m^2}{Coulomb}$. Is there an intuitive interpretation of these units, and what electric flux represents?

Answer 1

I think the most intuitive way of looking at this is using the integral form of Gauss' law, which in integral form says: $$ \unicode{x222F} \vec{D} \cdot d\vec{a} = \iiint\rho dV$$ Which says that the flux through a closed surface is equal to the charge enclosed by the surface, with $\rho$ the charge density.

$\vec{D}$, which has units coulombs per square meter, is usually called the electric flux density, and is related to the electric field by the permittivity of the medium, $\vec{D} = \epsilon \vec{E}$. The units for the electric field becomes intuitive when you keep in mind that the lorentz force due to the electric field is given by $\vec{F} = q\vec{E}$.

As for the usage of the words 'for a given sampling rate', your explanation a) is not correct because double the charge would mean double the flux, so double the field lines. If you would draw 10 lines coming from a charge $q$, then you would draw 20 lines coming from a charge $2q$.

I don't think there's an analogue as in your fluid flow example. At best you could say that since the integral over a closed surface of the flux equals the charge, that in the absence of other charges a surface integral would 'capture' part of this charge.