# Why is electric flux defined as $\Phi = E \cdot S$?

Flux, as I understand it, is the amount of substance passing through a particular surface over some time. So, from a simple perspective, considering photons that go through some virtual surface $A$ (or $S$, doesn't matter). They have a fixed speed in vacuum, $v=299,792,458$ $\text m/\text s$. To simplify even further, they're all hitting the surface head-on. So, if we wanted to figure out how many photons go through the surface, we conclude that at a constant velocity they will only pass through the surface if they are in the volume bounded by sweeping the surface area along the velocity vector (perpendicular to the surface, the opposite of its normal) a distance $d$ in the alloted time $t$: $d = vt$

So the flux volume is well-defined as $V(t) = Avt$. We could just look for a period of unit time and "drop" the dependency on time. But even then, it's useless if we cannot sample photon volume density $\rho_p$ to determine how many photons occupy a unit volume in order to determine the actual flux. That makes sense:

$$\Phi = \rho_pV(t)$$

And now comes along the electric flux and thwarts my understanding of the whole notion completely. An electric field is generated when a charge is dropped somewhere in space. Any other charge, especially idealized point charges, placed in its vicinity would experience a force exerted on them by the source charge, its magnitude modulated by the amount of charge. So, the electric field maps points in space with force vectors (ie. a vector field) whose direction and magnitude is parametrized by the interaction between the source charge and the point charge.

And this electric flux is defined as $\Phi = E \cdot S$ (I'll use $\Phi = E \cdot A$) and I just cannot interpret the semantics of this dot product, the product's dimensionality is not what I've come to expect from the notion of flux ($\text{Vm}^{-1}$ or any other). How does this in any way show how much electric field flow goes through a surface? Furthermore, what is this vague thing called electric field flow? It seems like it is completely disconnected.

I've tried expressing it in different ways from the derivation of the expression of an electric field, which makes sense (non-vector form, dropped unit vector):

$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$

I've intentionally separated the inverse square of the distance and the $4\pi$ which is, I presume, a part of the normalization factor (steradians of sphere) -- but I noticed that together they forge the area of a sphere $A_r = 4\pi r^2$. This way, I could see the expression as the uniform charge density distribution on the surface of a sphere, scaled by the vacuum permittivity.

$$E = \frac{Q}{A_s\varepsilon_0}$$

And then, due to presumed uniformity, by multiplying by an arbitrary area, I could get the flux, the amount of charge(?) flowing through a particular surface in unit time(?):

$$\Phi = E \cdot A = \frac{Q}{A_s\varepsilon_0} A$$

I could see that as flux, but I'm really not sure can I really reinterpret parts of the normalization factor and the inverse-square of the distance into the area of a sphere. From the perspective of voltage over distance it makes absolutely no sense to me.

Any help would be appreciated.

-
Because why not? x) It's a definition. Suggestion: Consider a velocity field when talking about fluids. – Schlomo Jul 11 '13 at 18:16
So, there's absolutely no way to derive it in a way where it makes semantical sense? How is it quantized? What exactly does it mean for it to flow through? What flows through? For example, flux in light transport makes perfect sense and can be derived. – Delgado Jul 11 '13 at 18:21
I think Jerry Schirmer answered very well how it could be explained to make some semantical sense. As I mentioned, fluid mechanics is the key ^^ – Schlomo Jul 11 '13 at 18:26
P.S. There's also a mathematical definition of "flux", check this if you're interested, ocw.mit.edu/courses/mathematics/… – Schlomo Jul 11 '13 at 18:28

When the notions of electric and magnetic fields were conceptualized, they imagined that there was an invisible fluid being pushed around by charges, and they leveraged some of the equations and terminology of fluid mechanics.

Modern understanding of field have largely gotten rid of this picture, but some colorful langauge like "electric flux" remains. If you want to picture positive charge as "amount of fluid added to region per unit time" and negative charge as "amount of fluid removed from region per unit time", you can, but this thinking only gets you so far. Safer to just think of it as an abstract mathematical definition.

-
Ah, precisely! That's why it bothered me so much. Thank you! – Delgado Jul 11 '13 at 18:22
As Feynman said, if you use the bullet analogy, you can grab the sense but this analogy can't be used anywhere. So, it is apt enough to consider it as absurd concept. +1. – MAFIA36790 Jun 18 '15 at 14:33

A very good example to understand flux, any kind of flux:-

Imagine that it is raining.

Let the rain fall in a direction denoted by unit vector $\vec v$, with a rain intensity of $I\frac { litres}{m^2}$. In this rain, you are holding a vessel with a plane open area on one side. Take $\vec A$ as a vector perpendicular to the plane of the open area of the vessel, having the same magnitude as the area.

Now the flux of rain through that open area is defined as the total volume of water which goes through the open area at any moment.
If the plane of the area was perpendicular, flux would be given by $$\Phi = I|\vec A|$$ But if the area is at any general angle, we would say $$\Phi = I|\vec A|\cos\theta$$ where $|\vec A|\cos\theta$ is the projection of the area perpendicular to the direction of rainfall. Thus $\theta$ is the angle $\vec A$ makes with $\vec v$. Using the definition of Dot Product, I can generalize this as $$\Phi = I\vec v\cdot\vec A$$ I can also write $I\vec v$ as the intensity vector $\vec I$(note that $|\vec v|=1$ as it is a unit vector), giving me finally $$\Phi = \vec I\cdot\vec A$$

This same analogy can be used to understand electric flux, because the electric field $\vec E$ is nothing but the electric field INTENSITY. Thus for electrostatics you get $$\Phi = \vec E\cdot\vec A$$

-
you could similarly define the flux of photons, but I would rather consider photon power flux rather than photon number flux. Maybe that's what got you confused – udiboy1209 Jul 11 '13 at 18:26
+1 for an intuitive answer. – centralcharge Sep 14 '13 at 8:02

## protected by Qmechanic♦Jul 11 '13 at 19:36

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).