# Do you wonder why people get confused?

1. Jul 28, 2009

### rockyshephear

Electric flux is a measure of the number of electric field lines passing through an area. To calculate the flux through a particular surface, multiply the surface area by the component of the electric field perpendicular to the surface. If the electric field is parallel to the surface, no field lines pass through the surface and the flux will be zero. The maximum flux occurs when the field is perpendicular to the surface.

Sorry! WRONG!! For any surface there are an infinite number of electric field lines passing thru a surface. Maybe the hand drawn kind you see in illustrations but not in reality. So this is sheer and utter bogus.

The only thing that makes sense is 'not the number of field lines' but the angle that the infinite amount of flux lines makes to the surface. Then there is no issue with 'count' of whatever. Just angle. So the flux is always the same if whatever coming in is normal to the surface. There's nothing that can vary the flux since there are not discrete flux lines floating around in space with a few nanometers between them.

Can anyone finally clear this up for me. And don't use equations please.

2. Jul 28, 2009

### diazona

False. The intensity of the electric field contributes to the flux. A strong electric field produces more flux than a weak electric field, even if both are normal to the same surface.

The idea that the number of electric field lines is proportional to flux refers to electric field lines in a drawing, where only a finite number are drawn.

With that attitude towards math, it's no wonder you're confused...

3. Jul 29, 2009

### rockyshephear

Ok. So what then is flux? We know it's not the count of number of lines. What makes one flux different than another? And please don't say the intensity of the electric field. That's not an answer. While it may be true it is not what I'm driving at.
If you were to take some kind of as of yet undiscovered scope and watch perpendicuar flux A of intensity 100 coming thru a surface and compared it to perpendicuar flux B of intensity 900 coming thru the same surface at a different time of course, what difference could you see? More photons driving thru the surface at faster rates?

4. Jul 29, 2009

### rockyshephear

I love math, but you can't use math to understand math.

5. Jul 29, 2009

### Born2bwire

I think you need to brush up on your vector calculus. If I have a finite continuous function, f(x), then there are an infinite number of points between a<x<b. Yet, if I take the integral of f(x) from a to b, the result will be a finite number despite the infinite amount of contributions. This is because each contribution is scaled by an infinitesimal line element.

Your definition of electric flux is wrong. We are taking the integral of the electric field components perpendicular to a given surface over that surface.

Only electromagnetic waves have real photons. If you were to have a static electric field, there would be no observable photons.

6. Jul 29, 2009

### LeonhardEuler

rockyshephear, you are right that the statement you quoted is not literally true, and I can understand why it confuses you. To make it literally true, this part:

"Electric flux is a measure of the number of electric field lines passing through an area."

should really be :

"Electric flux is a measure of the number of electric field lines passing through an area in a pictoral representation of the system."

because, as you rightly point out there is no limit to the number of lines you could use to represent the field. You're right that that statement seems to imply a greater amount of physical reality for electric field lines than is justified. I think they are just trying to get a picture in your head. It will make perfect sense if you think of it as applying to diagrams, rather than physical charges in space.

7. Jul 29, 2009

### rockyshephear

Thanks Mr Euler :) but that leads to the next question. If it's is as Born2BWire stated

the integral of the electric field components perpendicular to a given surface over that surface

Then what fundamentally gives you differing answers for one integral calculation vs any other? Do you see where I'm going with this?

8. Jul 29, 2009

### Born2bwire

The difference is the amplitude and the angle of the electric field over the surface varies from problem to problem (and surface to surface for the same situation).

9. Jul 29, 2009

### LeonhardEuler

You get different answers for different calculations because the strength of the electric field, its angle with respect to the surface, and the size of the surface can all vary.

As far as your second post goes, you seem to want to equate the flux with the flow of something. Actually, nothing is really flowing. If the E field represented a fluid velocity rather than a force, the flux would be the amount of fluid flowing across the surface. But it does not represent any sort of velocity, so there is no such interpretation.

10. Jul 29, 2009

### maverick_starstrider

Yes, you have no idea what you're talking about and are going to get banned soon. What makes a different integral have a different value? The vector function f(x,y,z)...

11. Jul 29, 2009

### rockyshephear

Thanks Leonhard_Euler.

12. Jul 29, 2009

### maverick_starstrider

I'm sorry but I feel you are abusing the spirit of this forum. I wish I'd known about this forum when I was in undergrad because it really is an amazing way to get basic questions you have answered. Learning physics is a difficult endeavour and it's a total boon to have someone with a greater understanding on hand to bounce ideas off of. And some questions posed by people are easy to answer and others aren't and it is very much the essence of physics to determine between them. And I'm sure in your mind these questions you are filling the board with are real duzies (spelling?), however, by my summation, without exception, they all result from you being unwilling to put in the effort to understand these concepts. If I felt that you had even put in a token effort to resolve these questions yourself I would be a lot more accomodating. However, it is clear that you just want to understand classical EM without learning vector calc. Unfortunately, classical EM IS VECTOR CALC. You have literally posted about a dozen posts on basic classical EM within a couple hours that could have been easily resolved by taking the time and effort to actually learn EM. If you REALLY want to understand basic EM then get yourself a copy of griffiths (and possibly a copy of boas) and just work through that book. It may be difficult (did you actually expect physics to be easy) but trust that classical EM was resolved over ONE HUNDRED YEARS AGO and the fact that it hasn't been replaced with a better classical theory and use that as a reason to KEEP READING/LOOKING when, practically every sentence, you think you've found a flaw. Then if there are actually things that you've looked at multiple sources and you can't resolve, come here. However, until then, this is not a free comprehensive tutoring service and don't monopolize the forum's time because you can't be bothered to concentrate for 10 minutes.

13. Jul 29, 2009

Hello rocky shepear are you saying that the concept of magnetic field lines has its limitations?If so you are right but I think people have been aware of these limitations since the time of Faraday.Field lines are imaginary, not real but the concept does have some limited uses for example when some phenomena are described qualitatively.There is no harm in sometimes using the concept ,for example you might like to imagine flux cutting when describing electromagnetic induction.If, however, you are to progress further you need to become aware of the limitations of the concept(which I suspect you partly are already) and be prepared to move on to and use the more encompassing mathematical descriptions.You can ,for example, describe that if a coil in a generator turns more quickly a bigger voltage will be generated but without maths you cannot calculate what that voltage is.If you bear in mind that field lines are not real and that the concept of field lines is severely flawed then I think you can make progress.

Last edited: Jul 29, 2009
14. Jul 29, 2009

### rockyshephear

Maverick: You do misunderstand the work I've put into understanding this topic and some of the questions I post. I have spent many hours reading Jerrold Franklin and hundreds of websites which each describe EM in uniquely different ways. I've also watched scores of hours of EM lectures. Yet, I tend to be able to regurgitate it yet I know I don't have the understanding I seek. And yes, I do sometimes post things I know not to be true for dramtic effect to emphasize my frustration at not getting answers I can comprehends, like flux density does not exist, but I'm attempting, and likely in an inconsiderate way, to emphasize my confusion and to stimulate more and clearer responses because the responder will tend to get down to brass tacks.
Sorry if I have taken up valuable forum processes but I have done preliminary work before posing these questions. I may be totally ignorant of the topic but I doubt it. I just have lots of gaps between what I know and am trying to fill them.

15. Jul 29, 2009

### negitron

I don't get what it is you're trying to understand. It's only necessary to understand how electric and magnetic fields behave; the math does that just fine. If you're trying to understand what they are on a fundamental level, you won't find an answer here because nobody knows what they are exactly.

16. Jul 29, 2009

Staff Emeritus
Inconsiderate? It's positively hostile.

Why do you feel that your lack of understanding is somehow our fault? Particularly if you haven't taken our advice - have you read the book by Schey yet?

17. Jul 29, 2009

### rockyshephear

I have to order it first.

18. Jul 29, 2009

### Tac-Tics

Where did you get this snippet from?

The snippet you provide is technically not correct, but it's not entirely false either. It serves to create an intuition of what flux is. That paragraph gives you the flavor of what flux is. What is wrong with that?

It sounds like you are looking for a halfway point between that paragraph's handwaving and the dense mathematical formulation. Mind yourself, though. Physics IS math. If you don't understand the basics of how vectors work, you'll never get it. If you do, then you need only be polite, but persistent, and you'll figure it out in time.

Let me give it a try at a mid-way explanation.

First, forget the field lines. The "number of field lines" business is a euphemism for the STRENGTH of the electric field at a point. If you draw field lines in a natural way, they spread out the further you get from the electric charge, and more importantly, they spread out in the same "inverse square" fashion, which is why the analogy is used.

But, really, we're talking about fields. At each point in space, we associate a direction and a strength (a "vector" or an "arrow"). When you have a single, positive point charge, the arrows always point away from the charge. When you have a single, negative point charge, the arrows always point in towards it. The strength, of course, is determined by the charge of the point charge and its distance.

One math technique that is absolutely essential to understand here is the dot product (or inner product). I'm not going to go into detail on how it works, but it's hugely important for lots of things in geometry. The basic property here is that if you take the normal to a plane and a second vector and take their dot product, you get the amount the second vector "pushes" against the surface. If the vector is pointed head on at the surface, the dot product is the length of the vector. If the vector is at a 90ยบ angle, it skims the surface, and the dot product is 0. If the vector is OPPOSITE the surface, it pulls instead of pushes, and the dot product is the NEGATIVE of the length of the vector.

If you have a very tiny square piece of paper floating in space, we define the flux on it as the dot product of the normal of the sheet of paper with the electric field at the center of the paper, multiplied by the area of the paper.

Read that last part very carefully. You really have to know what each of the terms mean: dot product, normal, the value of the field at the center.

Once you have that down, the rest is easy. Say you want to know the electric flux of a cube. You can approximate it by cutting the surface of the cube up into many tiny squares, finding the flux of each square, and adding them together.

When the squares are infinitesimally small and infinitely many, we change from a sum of very many small things to an integral.

There. No equations to be seen. But if you look at the formula and interpret it how I've described it here, the formula for electric flux says the same thing.

19. Jul 29, 2009

### vin300

You must try to know better before posting the next time rocky

20. Jul 29, 2009

### rockyshephear

Tac-Tics: Thank you for that attempt at explaining. I do understand the field and the dot product as you've described it. Tell me if I'm wrong.

The orientation of the normal of the piece of paper to the incoming vectors is dot producted. If the result of the cos of theta is zero then the two arrows are perpendicular and the incoming vector pushes minimally on the piece of paper. If the cos of theta is 1 then your vector length is the abs value of the magnitudes of the two vectors.

So the electric field is the sum of infinite vectors as described above. Yet they are not randomly pointing in any old direction. If you have a point charge, the vectors eminate out of the charge in a spherical shape.

Each point in space in the field will have it's own possibly unique vector angle and magnitude but the angle is always normal to the point charge and the magnitude of the vector is always falling off as a function of the inverse square law.

Am I missing anything here?

So if you will give me that I understand this much, here are a few questions.

So let's see if the explanation you gave and my interpretation of it now match the equations. If so, it's been a successful Q&A.

1.Since E=F/q then F (the force experienced by a test charge placed in the field) is equal to the electric field times the test charge itself.
2.The electric field is the limit as q approaches 0 of F/q
3.The electric field is also q times [-Nabla theta - partial deriviative of A wrt t]

A test charge will experience the same scenario of force the piece of paper does but only by a single vector since the charge is the limit as q goes to 0. Is this right?

How do you determine if the incoming vector interacting with the test charge is coming in at 0 degrees to the charge or 90 degrees to the charge since a charge is not flat and has no normal vector? Or is it that the two point charges only find the one single field vector that is perfectly in the line between the two charge points? Is the dot product useful in this case?

How do equations 2 and 3 relate to each other. Or more precisely what is the meaning of the E term in 3?

Thank you!

21. Jul 29, 2009

### diazona

The electric field is the set of vectors associated with every point in space. They are not infinite vectors (well, there are an infinite number of them, but if you pick one, there is nothing infinite about it), and there is no sum of the vectors.

Where you say "angle," I would say "direction." Each point in space will have its own direction and magnitude of the electric field. And when you have a point charge, the direction is always away from the charge (if it's positive) or toward the charge (if it's negative).

Seems reasonable so far.

#1 and #2 sound just fine. (In #2, you don't even need to take a limit, really - it's true even for nonzero charge)

For #3, you're describing something like this?
$$\vec{E} = q\left(\nabla\theta - \frac{\partial \vec{A}}{\partial t}\right)$$
I'm not sure what you mean by that, but it doesn't look like any equation I've ever seen. (Well, actually it does look suspiciously like the equation for a dynamic electric field in terms of scalar and vector potentials, but that's not what you meant, is it?) What do those variables stand for?

The paper was just an analogy for area, I don't think it was really supposed to experience a force. But you are right that a test charge will experience a force determined by a single electric field vector, since it exists only at a single point.

When you're dealing with a test charge, there is no concept of angle. An angle only exists when you have two different things that are pointing in (possibly different) directions. For instance, the electric field vector and the area normal vector - those two have an angle between them. But with a test charge, there is no angle. The test charge experiences a force either in the direction of the electric field vector (if it's positive) or in the opposite direction (if it's negative). So if you combine this idea with the fact that electric field vectors from a point charge point either toward the charge or away from it, you can see that two point charges either directly attract or directly repel, along the line between the two charges. (There's no dot product involved with two point charges)

22. Jul 29, 2009

### rockyshephear

23. Jul 29, 2009

### diazona

I'm using LaTeX... the next nifty thing you should learn after math Well... maybe. It is very handy. I'm pretty sure that somewhere on these forums there is a guide to LaTeX but someone else will have to point it out to you; I'm not sure where to find it.

Anyway, the equation you found was
$$\vec{F} = q\left[-\vec{\nabla}\phi - \frac{\partial \vec{A}}{\partial t}\right]$$
That is indeed the equation for the electric force due to scalar and vector potentials. The $\phi$ is phi, not theta, and that $\vec{A}$ was vector potential, not area. So I suspect that that equation is completely irrelevant to the discussion we're having here.

24. Jul 29, 2009

### rockyshephear

Irrelevant but interesting. How does is relate to an electric field of a single point charge in space?

25. Jul 29, 2009

### diazona

What, the equation? Well... I'm not sure how much sense this will make to you, but here goes:

The general equation for electric field (rather than electric force) is
$$\vec{E} = -\vec{\nabla}\phi - \frac{\partial\vec{A}}{\partial t}$$
In this equation $\phi[/tex] is the "scalar potential" or sometimes called "electric potential." Potential is the potential energy per unit charge, which is more or less the same thing you may know as voltage. This [itex]\phi$ is a field, like $\vec{E}$; that means that is has a value at every point in space, but unlike the electric field $\vec{E}$, that value is a scalar - it has only a magnitude, no direction.

The formula for the scalar potential produced by a single point charge in otherwise empty space is
$$\phi = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$$
Note that its magnitude is proportional to 1/r, not 1/r^2 as with the electric field. But when you take the gradient ($\vec{\nabla}$) of this formula - that means finding the rate of change of the potential over space - you do get a formula with 1/r^2. Specifically, it's
$$\vec{E} = -\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$
which is precisely the formula for the electric field of a point charge. So for a single point charge sitting still in empty space, you have
$$\phi = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$$
and
$$\vec{A} = 0$$

I suspect you're curious as to what $\vec{A}$ is? It was originally developed as the equivalent of $\phi$, but for magnetism. But magnetism is slightly different from electricity, so there was no way to describe magnetism with a scalar potential - they had to use a vector. So $\vec{A}$ has a magnitude and a direction at every point in space. (It is, of course, a field, like $\phi$ and $\vec{E}$)

If you have just a single point charge not moving, there is no magnetism. Hopefully that should be intuitive, but the fact that $\vec{A}=0$ for a stationary point charge confirms that. But if the point charge were moving around, there would be a magnetic field, and $\vec{A}$ would not be zero. The details depend on how the charge is moving, though.