Concept of flux

  • #1
rudransh verma
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Phi= int(E.da) through a surface S
is a measure of “number of field lines” passing through S. I put this in quotes because of course we can only draw a representative sample of field lines. The total number would be infinite. But for a given sampling rate the flux is proportional to the number of field lines drawn, because the field strength, remember is proportional to the density of field lines(the number per unit area), and hence E.da is proportional to the number of field lines passing through the infinitesimal area da.
I didn’t get it?
It’s from chapter 2 Electrostatics from Introduction to electrodynamics by Griffiths.

I have read in a book that field lines number crossing is proportional to Edelta S.
I can understand that something is proportional to area but to product of field and area makes no sense.
 

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  • #2
PeroK
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I can understand that something is proportional to area but to product of field and area makes no sense.
You mean ##\mathbf E \cdot \mathbf {da}##?
 
  • #3
rudransh verma
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You mean ##\mathbf E \cdot \mathbf {da}##?
Yes!
 
  • #4
PeroK
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Yes!
A surface can be given a direction in terms of a normal to the surface at each point. That allows the dot product to be carried out between a field and a surface. It's not too hard to see that the dot product represents the amount of the field going through the surface. Consider the two extreme cases where the field and surface are parallel (maximum flux) and where they are perpendicular (zero flux).
 
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  • #5
rudransh verma
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It's not too hard to see that the dor product represents the amount of the field going through the surface.
Well I can’t see it. Can you elaborate?
Consider the two extreme cases where the field and surface are parallel (maximum flux) and where they are perpendicular (zero flux).
I think you are saying the opposite of what is true.
 
  • #6
PeroK
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I think you are saying the opposite of what is true.
Good point. I was talking about parallel and perpendicular to the normal of the surface. That is the opposite in terms of the tangent to the surface. It's the normal that is used, by convention, in the dot product.

Imagine the field as a stream of particles. Take the surface just to be a plane. If the stream is parallel to the normal to the surface (perpendicular to the tangent) then the max number of particles go through the surface per unit time. This reduces if the stream is directed at an angle to the surface and becomes zero when the stream is perpendicular to the normal (parallel to the tangent).
 
  • #7
rudransh verma
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@PeroK i still don’t get it why it’s E.da. What you told is the explanation of why we should consider area as a vector.
(It says in the book that it’s analogous to the rate of flow of water which is the volume and nothing flows unlike water.)

When we see it’s unit E.da is not a volume ie Nm^2/C. I don’t know what this means or called?
 
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  • #8
PeroK
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@PeroK i still don’t get it why it’s E.da. What you told is the explanation of why we should consider area as a vector.
You need either a diagram and/or a numerical calculation. Remember that ##\mathbf{E} \cdot \mathbf{da} = Eda \cos \theta##, where ##\theta## is the angle between the field and the normal to the surface.

The justification is given here:

https://www.ux1.eiu.edu/~cfadd/1360/24Gauss/Flux.html

There's a Khan Academy video:

https://www.khanacademy.org/math/mu...tal-understanding-of-flux-in-three-dimensions

Any textbook on EM must cover this.
 
  • #9
rudransh verma
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Remember that E⋅da=Edacos⁡θ, where θ is the angle between the field and the normal to the surface.
I know it’s Edacostheta because that is a better way to account for the fact that it’s not always perpendicular. So it’s E.da
In the link you provided it directly writes EA as flux in the very beginning. That is what I don’t understand. The product of E and A?
If I think it this way then again it depends on what you consider F as. Here very easily he said it’s kg/s but that depends on taking F=rhov. Here there is E.da which is Nm^2/C. What meaning do I derive from this. How is this the amount of field lines flowing through a surface?
 
  • #10
PeroK
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I know it’s Edacostheta because that is a better way to account for the fact that it’s not always perpendicular. So it’s E.da

In the link you provided it directly writes EA as flux in the very beginning. That is what I don’t understand. The product of E and A?

If I think it this way then again it depends on what you consider F as. Here very easily he said it’s kg/s but that depends on taking F=rhov. Here there is E.da which is Nm^2/C. What meaning do I derive from this. How is this the amount of field lines flowing through a surface?
Seriously, I can't write anything here that could possibly be better than the graphics on those links. You have to work through the geometry for yourself.
 
  • #11
vanhees71
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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.
 
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  • #12
rudransh verma
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@PeroK sorry but your link doesn’t explain it why it’s product of E and A called flux. I was reading my NCERT Book and Maybe I got it. If you see the picture you can well see that at points R and S field is stronger at R because there are more number of field of lines crossing the surface. And The number of field of lines crossing a surface is directly proportional to the magnitude of field at that point. So if we place a surface delta S in a field E perpendicular to it then the number of field lines crossing it is proportional to EdeltaS and that is where E comes in the equation of flux.
 

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  • #13
PeroK
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It's really about the cross-sectional area of a surface in a given direction. You can see in the second diagram that a surface that is titled at an angle to a field presents less surface area in the direction of the field.

Again, if you think about the field as a stream of particles, then a titled surface of a fixed size intersects less of the stream. A bit of geometry shows you that it depends on the cosine of the angle between the stream and the normal to the surface - and that's the dot product.
 
  • #14
rudransh verma
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It's really about the cross-sectional area of a surface in a given direction. You can see in the second diagram that a surface that is titled at an angle to a field presents less surface area in the direction of the field.
Sorry but I don’t agree with you , flux or amount of field lines also depends upon electric field magnitude E and not just on area or angle between them.
 
  • #15
vanhees71
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You should not think in terms of field lines, because that's a bit confusing. It's more a tool to picture the direction of the electric field than a quantitative tool.

You should get an intuition for what flow (current densities as the local concept) and flux/current (global concept) is using quantities that are really flows in the literal sense. I tried to explain it as simple as possible but not simpler using the example of electric charge and current densities. This you need anyway in your further study of electromagnetism.
 
  • #16
PeroK
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flux or amount of field lines also depends upon electric field magnitude E and not just on area or angle between them.
That goes without saying. For the same field and the same surface area, the flux depends on the cosine of the angle between them.
 
  • #17
rudransh verma
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That goes without saying.
How is this so obvious? Apart from my explanation how do you see this so obvious thing?
 
  • #18
PeroK
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How is this so obvious? Apart from my explanation how do you see this so obvious thing?
Learning is partly about a willingness to engage with those who try to help you. I'm picking up some bad vibes here, so I'll let this be.
 
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  • #19
256bits
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Sorry but I don’t agree with you , flux or amount of field lines also depends upon electric field magnitude E and not just on area or angle between them.
Don't mix up flux with flux density.
For a magnetic field, the flux is the Weber, and the flux density is the Tesla.

For your picture is post 12 , P1 and P2 have the same flux, but P1 has a greater flux density.
 
  • #20
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(It says in the book that it’s analogous to the rate of flow of water which is the volume and nothing flows unlike water.)
Could be anything - mass, number of H20 molecules, liters, crossing the surface.
 
  • #21
rudransh verma
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Learning is partly about a willingness to engage with those who try to help you. I'm picking up some bad vibes here, so I'll let this be.
No man! I think you are presuming what is obvious to you is also obvious to me. I don’t see why E is there in the eqn in a very obvious way.
And I was not offensive to you but I am angry on myself. 😆
 
  • #22
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@rudransh verma
Because the electric field can be understood as the density of field lines passing through space, the ##\int \vec E \cdot d \vec A~## is obviously used to calculate the number of field lines passing through an area, and the number of field lines passing through an area is equal to the electric flux passing through this area.
 
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  • #23
vanhees71
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Don't mix up flux with flux density.
For a magnetic field, the flux is the Weber, and the flux density is the Tesla.

For your picture is post 12 , P1 and P2 have the same flux, but P1 has a greater flux density.
It's rather flux versus flow or current density. The flux is a scalar quantity and a surface integral over the corresponding current density. The former is a global formulation, i.e., referring to some extended region like here a surface and the latter a local formulation referring to properties at a certain point in space. It's clear that in general the local formulation is simpler than the global one.

To understand the idea of a surface-normal vector and its relation to a current, maybe this figure from my E&M lecture noes helps:

flow.png

That's referring to my above long posting about the subject using the charge flow of a fluid as an example. I hope with this is becomes clear, why you need the dot product between the flow velocity and the direction of the surface normal (together with ##\mathrm{d}^2 f##, the magnitude of the area leading to the surface-normal vector element ##\mathrm{d}^2 \vec{f}=\vec{n} \mathrm{d}^2 f##.
 
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  • #24
rudransh verma
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@vanhees71
In griffiths it is explained like this. “You draw field vectors at points around the charge. The vector lengths decreases as we move away from the charge. Then it says join the vectors and we get field lines.

But doing so have we lost the information about the field strength. No , the strength of field is determined by how close the lines are . Near the charge it’s closer so more strength and away the field lines are apart so field is weaker. So the density is what tells us about the strength and that is defined as flux. This flux is directly proportional to charge enclosed by the surface and that is gauss law. And the amount of lines passing through a surface is directly proportional to E vector at a point and on the area vector at that point. And hence the formula we know of flux.

Have I understood it correctly?
 
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  • #25
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well the idea of electric field lines is supposed to be analogous to that of streamlines in fluid dynamics

i.e. there is a vector field ##\mathbf{u}(\mathbf{r}, t)## representing the velocity of the local fluid flow, then the curve defined by ##\mathbf{r} = \mathbf{R}(\lambda, \mathbf{r}_0, t)## such that ##\dfrac{\partial \mathbf{R}}{\partial \lambda}(\lambda, \mathbf{r}_0, t) = \mathbf{u}(\mathbf{R}(\lambda, \mathbf{r}_0, t), t)## and where ##\mathbf{r}_0## and ##t## are fixed (with ##\mathbf{R} = \mathbf{r}_0## at ##\lambda=0##), is the streamline starting at ##\mathbf{r}_0## in the snapshot at time ##t##.

if you take two arbitrary streamlines at the same time but at differential initial ##\mathbf{r}_0##'s and follow them through the flow, then loosely the average flow velocity between them is greater when they are closer together and smaller when they are further apart. that's where the idea of "higher density of field lines ##\equiv## faster flow" comes from.

but you mustn't take the idea of "density of field lines" too seriously, because of course you can always keep drawing more and more streamlines between those two and you won't be affecting the flow velocity in the slightest :smile:
 
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  • #26
rudransh verma
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but you mustn't take the idea of "density of field lines" too seriously,
Ya! I want a rough idea. So am I correct?
 
  • #27
alan123hk
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Electric field lines is a graphical method of representing electric fields It can intuitively express the strength and direction of the electric field in different places in space. We can choose to generate 1, 10, 100 or any number of field lines per unit charge. Therefore, we can set the field line density of the unit electric field as needed, and this field line density will change with the change of the electric field strength.
 
  • #28
robphy
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It might be helpful to think of conducting surfaces, with charge distributed on them according to electrostatics. (Each conducting surface is an equipotential.)

Now consider the local electric field near the surfaces.
Consider the distributed charges in bunches of some small fixed unit of charge. To each bunch, there is an associated electric field vector that is perpendicular to the surface and is proportional to the surface charge density (the small fixed unit of charge). Places with high surface charge density will have more representative electric field vectors per unit surface area of the conductor. Generate field lines from those representative electric field vectors.

For electric field lines far from a conducting surface,
use a uniqueness theorem to pretend there is a conducting surface there with some appropriate pretend surface charge density.
 
  • #29
rudransh verma
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@alan123hk @robphy I am not in the state of engulfing more information. I am still trying to process the existing ones. So am I correct?
( I have a need of listening yes) 🤤
 
  • #30
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Have I understood it correctly?
I basically agree with what you said except for the following.
So the density is what tells us about the strength and that is defined as flux.
More accurate it should be electric flux density ##~ \mathbf{D} =\epsilon \mathbf{E}~##, and the flux ##~ \Phi_e = \mathbf{D} \cdot {d\mathbf{a}} = \epsilon \mathbf{E}\cdot {d\mathbf{a}} = Eda \cos \theta~##, where ##~\mathbf{E}~ ## is the electric field strength, ##~\theta~## is the angle between the field and the normal to the surface
 
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  • #31
rudransh verma
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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.
 
  • #32
weirdoguy
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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.
 
  • #33
vanhees71
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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.
 
  • #34
rudransh verma
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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”.
 
  • #35
robphy
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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, [itex]\iint \vec E \cdot d\vec A[/itex])
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,
[itex]\iint \vec B \cdot d\vec A[/itex] 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. )

[itex]\iint \vec g \cdot d\vec A[/itex] is called the gravitational flux.

[itex]\iint \vec Q \cdot d\vec A[/itex] is called the flux of the vector field [itex]\vec Q(\vec r)[/itex] (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 [itex]\vec E\cdot d\vec A[/itex] 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 [itex]\unicode{x222F}\vec Q \cdot d\vec A[/itex] or [itex]\iint_{\partial V}\vec Q \cdot d\vec A[/itex] where [itex]\partial V [/itex] 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 .)
 
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