B Understanding Flux: The Measure of Field Lines Passing Through a Surface

[rudransh verma]

It's of course a metaphor to say $\vec{E} \cdot \mathrm{d}^2 \vec{f}$ is a "flow through the surface element". For a static electric field nothing is really flowing in the literal sense.

The metaphor of course has its origin from some real flow. The most simple case is that you have some conserved quantity flowing (in the literal sense) with the matter. In electromagnetism you have this case concerning the electric charge, which is a conserved quantity carried with the electrically charged matter, on a microscopic level it's carried by the atomic nuclei (positive charge) and electrons (negative charge), but in classical electrodynamics as in all of classical physics we don't consider this microscopic resolution at the atomic level but take averages over macroscopically small but microscopically large regions of space.

Then you can describe the charge in a macrocsopic continuum mechanical way. Here you describe it by the charge density $\rho(t,\vec{x})$. The meaning is that if you measure at time $t$ the charge in a small volume element $\mathrm{d}^3 x$ around the position $\vec{x}$ the (net) charge contained in this volume element is $\mathrm{d} Q=\mathrm{d}^3 x \rho(t,\vec{x})$.

Now we also know that charge is conserved, which means that (in this macroscopic picture) the charge contained in the volume element $\mathrm{d}^3 x$ can only change, because there is some charge flowing through the boundary of this volume, which is a closed surface. Let now $\vec{v}(t,\vec{x})$ be the flow field of the charged medium, i.e., if you look at time $t$ at the position $\vec{x}$ the matter just being there has the velocity $\vec{v}(t,\vec{x})$.

Now we consider the charge flowing through the surface. For that purpose at any point $\vec{x}$ on the surface you define the surface normal element $\mathrm{d}^2 \vec{f}$, which is a vector being directed perpendicular to the surface pointing out of the volume (by convention) with the magnitude being the area of this surface element. Then in an infinitesimal time $\mathrm{d} t$ the particles just at the surface element going out of the volume is given by $\mathrm{d}^2 \vec{f} \cdot \vec{v}(t,\vec{x})$. By choice of the direction of $\mathrm{d}^2 \vec{f}$ if the particles go out of the volume that's counting as positive, and if the particles go into the volume it's counted negative. So the particles sweep out a volume of $\mathrm{d} t \vec{v} \cdot \mathrm{d}^2 \vec{f}$, and the charge being carried through the surface is given by $$\mathrm{d} Q=\mathrm{d} t \mathrm{d}^2 \vec{f} \cdot \vec{v}(t,\vec{x} \rho(t,\vec{x}).$$
Now we can make the charge balance to express the charge-conservation law. The change of the charge contained in the volume during the infinitesimal time interval $\mathrm{d} t$ is
$$\mathrm{d} Q_V=\mathrm{d} t \partial_t \rho(t,\vec{x}) \mathrm{d}^3 x).$$
On the other hand we know that this change can only be due to charge flowing through the surface. Now with our convention of directing the surface elements $\mathrm{d}^2 \vec{f}$ out of the volume this means
$$\mathrm{d} Q_V = -\mathrm{d} Q=-\mathrm{d} t \mathrm{d}^2 \vec{f} \cdot \vec{v}(t,\vec{x}) \rho(t,\vec{x})$$
and thus you get
$$\partial_t \rho(t,\vec{x}) \mathrm{d}^3 \vec{x} = -\mathrm{d}^2 \vec{f} \cdot \vec{v}(t,\vec{x}) \rho(t,\vec{x})$$
or integrating over a finite volume $V$ with its boundary $\partial V$ you get
$$\dot{Q}_V=\int_V \mathrm{d}^3 x \partial_t \rho(t,\vec{x})=-\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{v}(t,\vec{x}) \rho(t,\vec{x}).$$
Usually you introduce the current density,
$$\vec{j}(t,\vec{x})=\rho(t,\vec{x}) \vec{v}(t,\vec{x})$$
and write
$$\dot{Q}_V=\int_V \mathrm{d}^3 x \partial_t \rho(t,\vec{x})=-\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{j}(t,\vec{x}).$$
This shows that the current density is a vector field telling you how much charge is flowing through a surface element $\mathrm{d}^2 \vec{f}$, $\mathrm{d} Q=\mathrm{d}^2 \vec{f} \cdot \vec{j}(t,\vec{x})$.

An important mathematical theorem now is Gauss's Theorem, according to which
$$\int_V \mathrm{d}^3 x \vec{\nabla} \cdot \vec{j}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{j},$$
where the surface normal elements of the closed boundary surface $\partial V$ of the volume $V$ are defined to be pointing outward of the volume. Here the divergence of the current (in Cartesian coordinates) is
$$\vec{\nabla} \cdot \vec{j}=\frac{\partial j_1}{\partial x_1} + \frac{\partial j_2}{\partial x_2} +\frac{\partial j_3}{\partial x_3},$$
and the above derived charge-conservation law reads
$$\int_V \mathrm{d}^3 x \partial_t \rho(t,\vec{x})=-\int_V \mathrm{d}^3 x \vec{\nabla} \cdot \vec{j}(t,\vec{x}).$$
Now you can make the volume infinitesimally small again and then consider the fields you integrate over as constant within this small region. Then the volume element $\mathrm{d}^3 x$ cancels on both sides and you get the local form of the charge-conservation law,
$$\partial_t \rho(t,\vec{x})=-\vec{\nabla} \cdot \vec{j}.$$
Now back to the original idea to calculate the "flow" of the electric field through a surface. This comes from taking the volume integral
$$\int_V \mathrm{d}^3 x \vec{\nabla} \cdot \vec{E}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E},$$
where we have used again Gauss's integral theorem. Using Coulomb's law you can derive that
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
This is named Gauss's Law for the electric field. The factor $1/\epsilon_0$ is due to our choice of SI units with the Coulomb as the unit of the electric charge.

Using this in the above integral you get
$$Q_V=\int_V \mathrm{d}^3 x \rho(t,\vec{x})=\epsilon_0 \int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E},$$
i.e., (up to the unit-conversion factor $\epsilon_0$) the "flow of the electric field" through any closed boundary surface $\partial V$ of the volume $V$ gives the amount of charge contained in this volume.

It's clear that "flow" is here to be understood in the sense of being a mathematical analog to the ideas about something really flowing like the electric charge considered above.

You almost said everything but not why E is there in the formula of flux. If you have please point out.

 

[weirdoguy]

You almost said everything but not why E is there in the formula of flux.

Because without it formula for flux would not be useful. In most cases the reason that definitions look the way they look is just this - they are useful. You can define whatever you want, you can try to define a flux without $E$, or with $E^3$ or with $E^\pi$. Do what you want, but those definitions would be useless. I think you are wasting your time trying to dig in this issue too much. You should focus on where this definition of flux is used - and it's used in Gauss' theorem.

 

[vanhees71]

You almost said everything but not why E is there in the formula of flux. If you have please point out.

I don't understand that question. What is needed in electromagnetic theory are equations to describe, how the fields are related to the charge and current distributions and how the charge and current distributions are affected by the fields. This is at the end given by the four Maxwell equations and the Lorentz force. These are fundamental laws of Nature which can only be deduced from experimental facts. Historically it took quite some time to get this complete system of equations to describe all phenomena (except those related to quantum theory) in terms of Maxwell's equations and Lorentz force law.

One of the fundamental Maxwell equations is Gauss's Law for the electromagnetic field, which is the here discussed law about the flux of the electric field,
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} Q_V,$$
where $V$ is an arbitrary volume and $\partial V$ is its boundary, $Q_V$ the charge within the volume.

One way to make this plausible is to consider first charges at rest (electrostatics) and use the empirically found Coulomb force law to deduce that a charge at rest in the origin of our coordinate system leads to the electric field
$$\vec{E}=\frac{Q}{4 \pi \epsilon_0 r^3} \vec{r}.$$
Then you can show, using vector calculus, that indeed Gauss's Law holds for the point charge and arbitrary volums $V$ with their boundaries $\partial V$.

Further, another empirical fact is that the electric fields due to more than 1 point charge simply add up to the total electric field (superposition principle). Since the surface integral over $\vec{E}$ is a linear operator, from this we conclude immediately that Gauss's Law holds also for any distribution of point charges and, finally, using the continuum description in terms of the charge density,
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} \int_{V} \mathrm{d}^3 r \rho.$$
Now using Gauss's integral law you can write the left-hand side as a volume integral,
$$\int_V \mathrm{d}^3 r \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \int_{V} \mathrm{d}^3 r \rho.$$
Since this holds for all volums $V$, you can make the volume arbitrarily small and deduce the local form of Gauss's Law,
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho,$$
and that's the most general form of Gauss's Law and one of Maxwell's equations. That it holds not only for electrostatics but generally is justified by the fact that, together with the other Maxwell equations, it describes the electromagnetic phenomena correctly.

 

[rudransh verma]

I don't understand that question.

My question is very simple. In the formula of flux ie flux=integration of(E.dA) why is there E in it?
Or in the formula flux=E.dA=EAcos theta we know flux depends on A and theta between E and A. But why on E? Why is E vector in the equation. That’s it!
By the way you liked my post# 24. I thought you understood what I said.
I understand now “why”.

 

[robphy]

My question is very simple. In the formula of flux ie flux=integration of(E.dA) why is there E in it?
Or in the formula flux=E.dA=EAcos theta we know flux depends on A and theta between E and A. But why on E? Why is E vector in the equation. That’s it!
By the way you liked my post# 24. I thought you understood what I said.
I understand now “why”.

The full name of the "integration of(E.dA)" (symbolically, \iint \vec E \cdot d\vec A)
is the "flux of the electric field [through a given area]"... often shortened to "electric flux".
(See Griffiths Eq. (2.11) on the page before your quoted passage. I'm making reference to Griffiths since you mentioned it in the OP.)

Similarly,
\iint \vec B \cdot d\vec A is called the magnetic flux. (See Griffiths Eq. (5.71) and onward, (7.12) and onward. See also Eq. (8.9) to (8.11) for the flux of the Poynting vector field. )

\iint \vec g \cdot d\vec A is called the gravitational flux.

\iint \vec Q \cdot d\vec A is called the flux of the vector field \vec Q(\vec r) (see Griffiths 1.3.1 (b) "Surface integrals"), see also 1.3.5 for "flux of a curl" and 1.6.2 for "potentials"].

The flux of a vector field is a measure of the
"component of the vector field that perpendicular to a surface",
where \vec E\cdot d\vec A is evaluated on each tile of a surface,
and those flux-of-E contributions from these tiles are added together.

If the surface is closed (so it encloses a volume), symbolized as \unicode{x222F}\vec Q \cdot d\vec A or \iint_{\partial V}\vec Q \cdot d\vec A where \partial V is the "boundary of a volume" ( as in the posts above by @vanhees71 ),
the flux of the vector field is usually the "outward" component of the vector field.

(To play around with the electric flux, consult this post I wrote elsewhere
https://physics.stackexchange.com/a/468365 .)

 

[rudransh verma]

Because without it formula for flux would not be useful. In most cases the reason that definitions look the way they look is just this - they are useful. You can define whatever you want, you can try to define a flux without E, or with E3 or with Eπ. Do what you want, but those definitions would be useless. I think you are wasting your time trying to dig in this issue too much. You should focus on where this definition of flux is used - and it's used in Gauss' theorem.

I missed your post. Only now I am looking