The Concept of Flux in Electromagnetism: Exploring Gauss' Law

[rockyshephear]

Thanks. I believe I need to perfect this analogy so it doesn't stir debate in the engineering community, first. So I need the best description possible relating this analogy to the true nature of a point of charge.
Anyone care to help refine this analogy to as close a level of perfection possible. With it, I am sure I could easily teach an 8 year old this concept in a few hours.

So the previous poster refined my analogy as such----------------

We have a totally smooth spherical world upon which stand an infinite tribe of indians all over the planet. They are only allowed to shoot normal to the infinitely small space they stand on. They are surrounded by a bubble located somewhere in the ionosphere. A shape is cut out of the bubble allowing whatever arrows are pointed at this opening to come out. We assume only one charge exists inside this bubble which is the perfect center of the planet. And it all happens in an otherwise empty universe. Now...

Charge would equal how far back the indians pull on their bows (seems to conflict with the revised definition of electric field below)

Electric field would be the pressure exerted on the ionospheric bubble, which is related to how far the indians pull back their bows and depicted by the magnitude of the arrows.

Electric Flux would be a scalar quantity depicting how many arrows escape a closed surface cut into the ionospheric bubble, irrespective of direction since it's a scalar quantity. (Ironically it seems it would always be infinite since if you take the infinite field and divide it up, you still get infinity)

A is the total area of the ionospheric bubble

dA is the size of the area cut out of the bubble, which is summed up to give the total flux, and which approaches zero as we approach the most accurate total flux.

permittivity of free space (since it is a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless).


Are we getting closer or farther away?

 

[rockyshephear]

Something that bothers me about my analogy is that having the ionospheric bubble with an opening in it, every point on the surface that is cut out is normal to every arrow coming thru at that point. That's an idealized scenario since in reality many surfaces are at different angles to the incoming arrows.

If the closed surface and arrows have 0 angle between them, then there is 0 flux thru that dA surface. My analogy does not take this into account. But there's no way around having every point on the surface normal to each arrow if the ionspheric is a concentric sphere around the planet. I guess offsetting the planet inside the bubble would give arrows impinging on the cut away surface at DIFFERING angles other than normal. Is this correct?
Then the flux at an infinitely small location on the surface would then be Phi= EA cos q (it is even when normal).
So here's my new analogy based on the above consideration.
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We have a totally smooth spherical world upon which stand an infinite number of indians, completely covering the planet. They are only allowed to shoot normal to the infinitely small space they stand on, on the surface of the planet. They are surrounded by a concentric bubble located somewhere in the ionosphere. A closed shape is drawn on the bubble and this closed shape is extruded away from the center of the bubble for infinity. The ideal goal of the indians is to have their arrow go thru the extruded shape and never hit the sides but go as far as they can travel thru the extruded shape. Only those in the correct position can do so, or even get their arrow in the drawn shape at all. We are only concerned with arrows entering and leaving this closed surface. We assume only one charge exists inside the planet, which is the perfect center of the planet but not neccessarily colinear with the center point of the bubble. Assume that the surface of the planet is no where near the inside of the bubble. And it all happens in an otherwise empty universe. Now...

Charge would equal how far back the indians pull on their bows. If they are lucky, they get the highest flux score by making the arrow go really far and not touching the inside of the extruded shape by having the perfect entry angle. If they do not go as far or do not have the perfect entry angle, they lose points. (Phi=EA cos theta)

Electric field would be tantamount to, all the arrows shot at once, over and over, and hitting the inside of the bubble, and the force with which they hit, based on how far back the bowstring is pulled. It would be evenly distributed if the center of the bubble and the center of the planet are colinear.

Electric Flux would be a scalar quantity depicting how many arrows escape the drawn closed surface on the ionospheric bubble. Flux is stronger (they get more points), the more the tendency for the arrow to travel thru an infinite extruded shape of the closed surface, projected along the arrows trajectory away from the bubble center, and not hit the inside of the extruded shape. The ideal is the arrow goes on for infinity and never hits the sides.

A is the total area of the ionospheric bubble

dA is the size of the area drawn on the bubble, which is summed up to give the total flux on the entire bubble, and when dA approaches zero, we approach the most accurate measurement of the total flux for the entire charge inside the bubble.

permittivity of free space (a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless). The higher this number of permittivity (a constant), the harder it is for the indians to make their arrows go very far. To go further you need more charge (pulling back harder on the bowstrings).

I think I'm almost there with this analogy.

 

[rockyshephear]

Correction:

We have a totally smooth spherical world upon which stand an infinite number of indians, completely covering the planet. They are only allowed to shoot normal to the infinitely small space they stand on, on the surface of the planet. They are surrounded by a bubble located somewhere in the ionosphere but not necessarily concentric to the center of the planet. A closed shape is drawn on the bubble and this closed shape is extruded away from the center of the bubble for infinity. The ideal goal of the indians is to have their arrow go thru the extruded shape and never hit the sides but go as far as they can travel thru the extruded shape. Only those in the correct position can do so, or even get their arrow in the drawn shape at all. We are only concerned with arrows entering and leaving this closed surface. We assume only one charge exists inside the planet, which is the perfect center of the planet but not neccessarily colinear with the center point of the bubble. Assume that the surface of the planet is no where near the inside of the bubble. And it all happens in an otherwise empty universe. Now...

Charge would equal how far back the indians pull on their bows. If they are lucky, they get the highest flux score by making the arrow go really far and not touching the inside of the extruded shape by having the perfect entry angle. If they do not go as far or do not have the perfect entry angle, they lose points. (Phi=EA cos theta)

Electric field would be tantamount to, all the arrows shot at once, over and over, and hitting the inside of the bubble, and the force with which they hit, based on how far back the bowstring is pulled. It would be evenly distributed if the center of the bubble and the center of the planet are colinear.

Electric Flux would be a scalar quantity depicting how many arrows escape the drawn closed surface on the ionospheric bubble. Flux is stronger (they get more points), the more the tendency for the arrow to travel thru an infinite extruded shape of the closed surface, projected along the arrows trajectory away from the bubble center, and not hit the inside of the extruded shape. The ideal is the arrow goes on for infinity and never hits the sides.

A is the total area of the ionospheric bubble

dA is the size of the area drawn on the bubble, which is summed up to give the total flux on the entire bubble, and when dA approaches zero, we approach the most accurate measurement of the total flux for the entire charge inside the bubble.

permittivity of free space (a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless). The higher this number of permittivity (a constant), the harder it is for the indians to make their arrows go very far. To go further you need more charge (pulling back harder on the bowstrings).

 

[cepheid]

Just a minor point. I am Indian (racially or ethnically), because my ancestors once lived in India. This is the only type of Indian I tend to generally accept in modern parlance. Here in Canada, I believe that the people who used to live in the region before the arrival of Europeans are officially referred to as either First Nations or Aboriginals, and are sometimes called Natives. In the U.S. I believe they are known as Native Americans. I'm not claiming that the term has disappeared entirely. I believe that our federal government still has a Minister of Indian Affairs (and no, his job is not to deal with the influx of brown people immigrating from the subcontinent, LOL), but my point is that these are generally relics of the 19th century. I'm not taking any offence to your statement, but I feel obliged to warn you that some people might.

I certainly can't claim to have read all of this, but let's examine some parts of your analogy.

Now...

Charge would equal how far back the indians pull on their bows (seems to conflict with the revised definition of electric field below)

An electric field is a vector field (a function that assigns a vector to every point in space). In other words, the electric field is a function of position: E = E(x,y,z).

For a static field, the vectors don't move, so your analogy of firing speed doesn't really have any bearing here.

What the charge determines is the strength of the electric field at any given point in space. The strength of the electric field is typically visualized in terms of the length of the electric field vector at that point in space. However, this also has no bearing to your analogy.

As an alternative to mapping out the electric field vectors in space, we can visualize the electric field using field lines, which are curves to which the electric field vectors are tangent at every point. Now, sometimes, the density of the electric field lines (how closely spaced they are) is used to indicate the strength of the field (so the field line density would be higher for a higher charge). In this case, the charge would be analogous to the number of people firing arrows. However, I want to emphasize that this is an arbitrary convention. The field lines are not actually more or less dense with more or less charge. Since the field has a value and direction at EVERY point in space, we would actually have to draw an infinite number of field lines (in every scenario) in order to be "accurate." This, of course, is useless. Therefore, the number of field lines drawn in practice is entirely a matter of convenience.

Electric field would be the pressure exerted on the ionospheric bubble, which is related to how far the indians pull back their bows and depicted by the magnitude of the arrows.

The direction of the electric field would be the direction of the arrows. The magnitude of the electric field, as I have stated already, would not be analogous to anything at all (at least, not anything obvious). Again, using the field line analogy, the closest thing we might say is that the the stronger the field, the more closely-spaced the arrows are.

Electric Flux would be a scalar quantity depicting how many arrows escape a closed surface cut into the ionospheric bubble, irrespective of direction since it's a scalar quantity. (Ironically it seems it would always be infinite since if you take the infinite field and divide it up, you still get infinity)

It does depend on direction: a scalar can be positive or negative, and as I explained several posts back, the dot product of the electric field vector with the unit normal vector ensures that what is calculated is the NET flux. Field vectors whose normal component points inward from the surface contribute to the flux in a negative way, and field vectors whose normal component point outward from the surface contribute in a positive way. In this simplified scenario (a single positive charge at the origin), obviously all of the field vectors point radially outward (radially because you have no sources that are offset from the centre of the sphere, and outward because all the sources are inside the sphere and none are outside).

I would say that a reasonable analogy for flux in this scenario is the number of arrows passing through the surface per unit area, per unit time. EDIT: No. Scratch that. That's a different use of the word "flux" that I was thinking of. As diazona pointed out, there is no good analogy for the flux here.

permittivity of free space (since it is a constant) would be equivalent to the mass of the arrows. (Not sure I buy this since electro-magnetic propogation is at the speed of light, precipitated by photons which are massless).

You are confusing two different issues. It is true that if I have some charges "over here", and I shake them around a bit, then I ask myself, how long will it take before the people "over there" a distance x away will realize that the electric field at their location has changed due to the time variation of the source charge distribution, the answer is that they will not realize that anything has happened until a time t = x/c later. This is because c is the speed at which information about the change propagates.

HOWEVER, in this case, the field we are talking about is static. It is not propagating anywhere. It merely permeates space. This is just one example of where the arrow analogy is leading you astray. What permittivity measures is the tendency of a medium to respond to an applied electric field by becoming "polarized" (meaning that the applied field induces a spatial separation of + and - charges in the atoms or molecules of that medium, which is known as an electric dipole). If this medium is polarized (has a bunch of electric dipoles in it), then those dipoles will set up a field of their own, one that tends to oppose the applied field. The net result is a shielding effect -- the net field within the medium is weakened. That's what permittivity is, period.

Are we getting closer or farther away?

To be honest, I do not think that this analogy is a useful intellectual exercise that will lead you to greater insights about electrostatics. Analogies by their very nature are imprecise, so trying to "perfect" this one makes little sense. Please please take the advice of diazona in his/her last post (#35 -- the very last sentence) to heart. Your intellectual energies would be better spent trying to get a handle on vector calculus. Physical theories are formulated mathematically, and vector calculus is the most natural language for formulating electromagnetism (at least at this level of understanding).

 

[rockyshephear]

Well, thanks for that. I don't feel like I'm making progress with this analogy thing. I am sensing engineering antipathy for all but using math and vector calc to describe this. I feel that's somewhat narrow thinking since people all learn in differing ways. But your critique leaves me not knowing how to proceed. I can't be satisfied with an analogy full of holes.
I can't believe the concept is so complex it defies a concise analogy that could be memorized as a narrative that would enable a higher degree of understanding to the young and those who are far more visually minded that analytical.

 

[diazona]

This "engineering antipathy" you think you're sensing is - well, first of all, not engineering, since I at least am a physicist, not an engineer But seriously, it's not that we have some ignorant disrespect for every learning method that isn't math. I know I've tried to use many analogies for these sorts of concepts, and inevitably they all fall apart at some point - the only description that I know to be completely accurate and not misleading is the mathematical one. So by telling you to forget about analogies and learn to use the math, we're trying to give you the benefit of experience.

(P.S. @cepheid: I'm a "him" )

 

[rockyshephear]

Thanks but I'm a dyed in the wool optomist and I KNOW that an amazingly accurat analogy can be devised. I just know it. And no, I don't believe in Santa anymore and I'm agnostic. So for me to say this indicates a true sincerity on my part. There is nothing that cannot be explained with analogy. Sorry, physicist, not engineer. You've been wrongly categorized twice today. lol
So, I challenge all on this forum to come up with an excellent analogy for electrostatics.

So without this, my next question would be...

You said..."The strength of the electric field is typically visualized in terms of the length of the electric field vector at that point in space."

So considering a sphere with a point charge at it's center, around the center will have the greatest magnitude vector and they will radially fall off as the inverse of R^2.

So I see two things alone that can be differentiated at the surface dA. One of them is not how many lines of flux pass thru, since it is infinite. The only two things are the angle that the vector makes at the surface dA and the magnitude of that vector. These, to me are the only two pieces of information to work with in describing flux.
So this agrees with finding the total flux by Phi(total)=q/epsilon(0).
Uses:
If you know Phi you can find q

I

 

[rockyshephear]

I only see two pieces of data that are useful at the dA surface. The angle that the vector makes to the surface and the magnitude of the vector. Since flux vectors are infinite, they are infinite thru dA.

So are these equations correct?

Phi (total)= Surf Integral of E dot producted with dA = Surf Integral |E| times |dA | times cos theta (scalar)

Phi at dA is
|E| times |dA | time cos (scalar)

Phi (total)=q/epsilon (0) (scalar)

Oh, and there is not such thing as an electric field that is not radially spherical. Right?

Thx

 

[cepheid]

Hi rockyshephear,

I'm sorry if I sounded overly critical. I flipped through a few pages of Electrodynamics by David J. Griffiths. It was my undergraduate text for both of my third-year E & M (Electricity and Magnetism) courses. It's a pretty standard text for undergraduate physics, although I am sure that some people here feel that it is no good, preferring a more rigourous text like Jackson (which I'm told is the de facto standard for Classical E & M). In any case, I don't think that Griffiths is used in Electrical Engineering, in which the course that introduces Maxwell's equations tends to be called "Electromagnetics", and has a slightly different emphasis (the goal being to get right to applications like transmission line theory and waveguides, from what I gather). Nevertheless, because my education was in Engineering Physics, I got the best of both worlds. In particular, I had the good fortune (in my opinion) of being able to learn from Griffiths, which I felt to be really excellently written, and highly entertaining to boot. One of my friends said that an acquaintance of his once commented that the book, "reads like a novel," and was among his favourite books! I would highly recommend borrowing a copy of this book from the library if you get the chance. I took vector calculus before E & M, and I didn't really have much of an intuitive feel for divergence, gradient, and curl (just the definitions). Seeing them applied in Griffiths really helped. I think it was the right order in which to do things, however, because knowing the math and then seeing it applied meant that I had no impediments to learning the physics. I would not recommend doing things the other way around (tackling E & M first, and learning vector calculus as you go along).

So, like I said, I was flipping through the section that introduces Gauss' Law, and I realized that I may have been too quick to pooh-pooh the density of field lines as an indicator of electric field strength. Even if you haven't drawn the radial vectors decreasing in length, and have only drawn the field lines themselves, you have still not thrown out the information about the strength of the electric field. The reason is because for these field lines for a single point charge, which extend radially outward like spokes on a wheel, the lines (spokes) really are denser (more closely-spaced) near the centre of the wheel than they are farther out. Now, what I said is true. The number of lines you draw is arbitrary. However, so long as you choose a consistent way of sampling the infinite number of field lines (i.e you decide that a charge of q will have x number of lines emanating from it, and a charge of 2q will have 2x lines), then the density of field lines is indeed an indicator of the electric field strength.

Based on these notions, Griffiths goes on to state that the flux through a surface can simply be thought of as the number of lines passing through it. From this idea, he immediately concludes that the flux through a closed surface is proportional to the amount of electric charge enclosed by it. His reasoning is entirely intuitive (not mathematical). He points out that there are only two possible fates for the field lines of any positive charge enclosed by the surface: these lines must either pass through the surface, or terminate on another negative charge within the surface. Therefore, the total number of lines passing through the surface (i.e. the flux of the electric field) is proportional to the net charge within that surface. From this argument, two conclusions also immediately follow:

1. Charges outside the surface do NOT contribute to the NET flux, because their field lines will pass into the surface through one side, and out of it through the other side. Hence it is only the charge *enclosed* that matters.

2. The shape and size of the closed surface doesn't matter, because any closed surface will capture exactly the same number of field lines as any other closed surface. (This is supported by the math, for although the strength of the electric field and resulting density of lines falls off as 1/r², the surface area available to catch them increases as r², and the two dependencies cancel out).

I think that these ideas might help you out in understanding some questions in another thread you posted here in General Physics. Griffiths does not go through a formal mathematical derivation of Gauss' Law until after he has outlined these intuitive ideas. I sincerely hope that this helps you in your understanding!

diazona: Sorry for my gender uncertainty. As you can see, I've been around PF for quite some time, but I wouldn't say I really know too many of the members or interact with them too much. I definitely don't know too much about people's backgrounds or personal info. Also, I am just a lowly grad student, so if I say anything that is wrong, be sure to step in and put me in my place.

 

[diazona]

diazona: Sorry for my gender uncertainty. As you can see, I've been around PF for quite some time, but I wouldn't say I really know too many of the members or interact with them too much. I definitely don't know too much about people's backgrounds or personal info. Also, I am just a lowly grad student, so if I say anything that is wrong, be sure to step in and put me in my place.

No apology necessary I certainly didn't take any offense at your uncertainty, I just figured it couldn't hurt to clarify. And actually I'm just a lowly grad student myself! (Sorry if I suggested otherwise... I'm either embarrassed or flattered ) So it's certainly not my place to put you in your place (though if I saw you posting something that seemed egregiously wrong, I wouldn't hesitate to bring it up).

I do agree with you about Griffiths, by the way. I also used it as the primary textbook for my 3rd-year E&M course and I thought it was great as an introductory textbook, at least compared to the (admittedly few) other E&M books I've looked at in any detail.

rockyshephear, if you'd like a recommendation for a book at a somewhat less technical level, a standard is http://he-cda.wiley.com/WileyCDA/HigherEdTitle/productCd-0471758019.html by Halliday, Resnick, and Walker (unfortunately Griffiths hasn't written a book at this level ) Part 3 covers electromagnetism, including electric flux and Gauss's law. I think that if you're not already familiar with those concepts, the HRW book may be more useful to you than Griffiths right now. Just a suggestion, of course.