Do you wonder why people get confused?

[rockyshephear]

Read this.

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.

 

[diazona]

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.

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.

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

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

 

[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?

 

[rockyshephear]

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

 

[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.

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?

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

 

[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.

 

[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?

 

[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).

 

[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.

 

[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)...

 

[rockyshephear]

Thanks Leonhard_Euler.

 

[maverick_starstrider]

Please don't respond to any more questions mr maverick. You don't have the temperment for it.

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.

 

[Dadface]

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.

 

[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.

 

[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.

 

[Vanadium 50]

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.

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?

 

[rockyshephear]

I have to order it first.

 

[Tac-Tics]

Read this.

Where did you get this snippet from?

Electric flux is a measure of the number of electric field lines passing through an area. [...] Sorry! WRONG!

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 90º 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.

 

[vin300]

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

 

[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!

 

[diazona]

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.

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.

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.

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).

Am I missing anything here?

Seems reasonable so far.

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]

#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?

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?

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.

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?

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)

 

[rockyshephear]

I found that equation here
http://en.wikipedia.org/wiki/Electric_field
It seems to be defining a simple electric field but then it talks about magnetism also.
(how are you putting equations in your posts?)

 

[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.

 

[rockyshephear]

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

 

[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 [itex]\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 $\phi$ 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.

 

[Tac-Tics]

A few points of correction on your last post.

If the cos of theta is 1 then your vector length is the abs value of the magnitudes of the two vectors.

What you have here isn't incorrect, but you can think of it even more simply. Forget the angles entirely. It's usually an unnecessary hassle to compute the angles involved, so forget the "cosine theta" stuff. Try to get a good feeling of what the dot product does, then think of the flux entirely in terms of the dot product.

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.

Make sure to distinguish between the electric field and the electric flux. The field assigns a force-per-charge to each point in space. The field doesn't have a "shape" in the conventional sense. It exists throughout all of space. Mathematically, it's a function from R^3 to R^3.

The flux is a measure of a field's behavior on a surface. The flux is the "tiny sheets of paper" idea in my last post. Given a field and a surface, the flux of that field through that surface is a scalar that tells you how much "flow" there is through the surface. (Flux means "flow" in Latin).

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.

You are mixing up concepts again here. (That's ok, you'll get it soon).

There is no such thing as "vector angle". Angles aren't so useful in 3D space as they are in 2D. The field assigns to each point in space a *vector* -- which has both a magnitude and a direction. (Not an angle, a direction!)

In the special case where you have a single point charge, the direction of the field at any point either points at or away from the point charge. If you have two or more charges in space, the field can get pretty wonky, and certainly isn't governed by such a simple rule.

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.

This is true.

2.The electric field is the limit as q approaches 0 of F/q

This is false. In fact, it contradicts the last question you just asked. The electric field is the force when q=1.

The "approaches 0" thing you have in your head is probably stemming from the use of the integral in the formal definition of flux. (Again, I think you mixed up "field" and "flux"). When the tiny pieces of paper, you just have a good approximation of the flux. It's only in the limit as the area of the pieces of paper approaches 0 that you get the exact flux.

3.The electric field is also q times [-Nabla theta - partial deriviative of A wrt t]

I'm not sure what you mean here. What's A? (Nevermind, I see it's on Wikipedia. They don't define it there either... Sometimes Wikipedia can suck for science articles).

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?

The piece of paper does not experience a force. It's not charged. The pieces of paper (or the surface as a whole) are not real objects.

The whole idea behind the flux of a surface is that it is proportional to the charge contained by the surface. If I have a gift-wrapped box, you can use Gauss's law to figure out the total charge inside of the box without counting the electrons inside of it. You simply need to measure the electric potential around the box, and you can figure out how much charge the stuff inside has.

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?

The idea of a test charge is simply to help you visualize the field. Test charges don't really exist. The dot product is between the normal of the surface and the field's value at that point on the surface.I hope that helps.

 

[rockyshephear]

A bit rough for sure. I've read that...

"The electric field at a point is equal to the negative gradient of the electric potential there. In symbols,"

E = negative Nabla times phi.

DEF A: I suppose meaning that the vector in the electric field at this point is specified by the gradient of the scalar field at that point. So the vector points to the greatest rate of increase of voltage in the immediate surrounds of that point?

--------

You've stated

E = negative Nabla times phi (or the gradient of the electric potential) - dA/dt

Could you modify DEF A: such that it agrees with the longer equation of the two?

 

[maverick_starstrider]

A bit rough for sure. I've read that...

"The electric field at a point is equal to the negative gradient of the electric potential there. In symbols,"

E = negative Nabla times phi.

DEF A: I suppose meaning that the vector in the electric field at this point is specified by the gradient of the scalar field at that point. So the vector points to the greatest rate of increase of voltage in the immediate surrounds of that point?

--------

You've stated

E = negative Nabla times phi (or the gradient of the electric potential) - dA/dt

Could you modify DEF A: such that it agrees with the longer equation of the two?

You DEF A is from the electro-static case (i.e. no magnetic field). Your later version (the actual maxwell equation) is the correct relation. It is true even in the presence of a magnetic field (the A). I should point out that A is not uniquely determined and is only defined up to a gauge transformation. But that's advanced stuff.

P.S. Wikipedia is not designed to be a learning tool but a refernce tool. You really won't get anywhere by just wiki'ing this stuff (of course now that I recognize your name and realize who I'm responding to I realize I've already told you this before).

 

[rockyshephear]

Tac-Tics: But I thought the definition of the dot product IS |A||B| cos theta times some unit vector. How can we ignore the angle in the light of this and still consider the dot product?

So electric field is what it is without any charges dropped in and the flux is the interaction after a charge is dropped in? So without test charge there is a field but no flux. Once you throw in a charge, the you have flux?

Ok. Can I claim semantics on this one? A vector has direction but no angle. Ok. But a vector in 3D space makes angles to each of x hat, y hat and z hat. So can we say it has both? Or it has angle only in the context of a coordinate system? This one is losing me entirely. Ok so a vector has direction. Which is either away from or into a charge? There seem to be many variants of the away from when I look at the vector arrows. The all point in differing directions...but its still only away from?

My brain is fried again. Thanks for the help. I hope you are correct that it will be clear as glass soon. Do you see what I mean though. If all we've been discussing were made into a 3D movie, it would be easier to then look at the equations and comprehend them. All our banter is analogous to a movie of sorts where visualization plays a large part in understanding the math. I wonder do all you guys have visualizations of each concept or do you think only in term of the equations?

 

[rockyshephear]

Taking a break for a few days. Thanks for the help.

 

[maverick_starstrider]

Taking a break for a few days. Thanks for the help.

May I suggest you relax with an EM TEXTBOOK? Just a thought.

 

[maverick_starstrider]

Tac-Tics: But I thought the definition of the dot product IS |A||B| cos theta times some unit vector. How can we ignore the angle in the light of this and still consider the dot product?

So electric field is what it is without any charges dropped in and the flux is the interaction after a charge is dropped in? So without test charge there is a field but no flux. Once you throw in a charge, the you have flux?

Ok. Can I claim semantics on this one? A vector has direction but no angle. Ok. But a vector in 3D space makes angles to each of x hat, y hat and z hat. So can we say it has both? Or it has angle only in the context of a coordinate system? This one is losing me entirely. Ok so a vector has direction. Which is either away from or into a charge? There seem to be many variants of the away from when I look at the vector arrows. The all point in differing directions...but its still only away from?

My brain is fried again. Thanks for the help. I hope you are correct that it will be clear as glass soon. Do you see what I mean though. If all we've been discussing were made into a 3D movie, it would be easier to then look at the equations and comprehend them. All our banter is analogous to a movie of sorts where visualization plays a large part in understanding the math. I wonder do all you guys have visualizations of each concept or do you think only in term of the equations?

We're the sorry the universe does not obey laws that are sufficiently accessible to you. Take it up with your local deity of choice.

 

[diazona]

For when you get back:

Tac-Tics: But I thought the definition of the dot product IS |A||B| cos theta times some unit vector. How can we ignore the angle in the light of this and still consider the dot product?

The definition of the dot product is that you multiply corresponding components of the vectors and add them up.
$$\vec{A}\cdot\vec{B} = (A_x \hat{x} + A_y \hat{y} + A_z \hat{z})\cdot(B_x \hat{x} + B_y \hat{y} + B_z \hat{z}) = A_x B_x + A_y B_y + A_z B_z$$
That is equal to multiplying the magnitudes times the cosine of the angle between the vectors (not times a unit vector):
$$\vec{A}\cdot\vec{B} = |\vec{A}|\,|\vec{B}|\cos\theta$$

So electric field is what it is without any charges dropped in

Yes

and the flux is the interaction after a charge is dropped in?

NO. You're thinking of force.

So without test charge there is a field but no flux. Once you throw in a charge, the you have flux?

Without a test charge there is a field but no FORCE. Once you throw in a charge, then you have force.

Ok. Can I claim semantics on this one? A vector has direction but no angle. Ok. But a vector in 3D space makes angles to each of x hat, y hat and z hat. So can we say it has both? Or it has angle only in the context of a coordinate system?

Sure, there are angles only in the context of a coordinate system. Remember what I said before, that you only have an angle between two vectors. Any of $\hat{x}$, $\hat{y}$, or $\hat{z}$ can be one of those two. So for a vector in 3D space, there is an angle between it and each coordinate axis, once you have defined the axes. But it doesn't have an angle by itself.

Ok so a vector has direction. Which is either away from or into a charge?

Generally a vector can have any direction. But the electric field of a single point charge at a particular point (which is an example of a vector) does have a direction either away from or into the charge.

There seem to be many variants of the away from when I look at the vector arrows. The all point in differing directions...but its still only away from?

Sure. At any given point, the field points away from the charge (if it's a positive charge). Look at some of the pictures http://www.lightandmatter.com/html_books/0sn/ch10/ch10.html .

My brain is fried again. Thanks for the help. I hope you are correct that it will be clear as glass soon.

Gee, I hope so, but I'm not holding my breath...

Do you see what I mean though. If all we've been discussing were made into a 3D movie, it would be easier to then look at the equations and comprehend them. All our banter is analogous to a movie of sorts where visualization plays a large part in understanding the math. I wonder do all you guys have visualizations of each concept or do you think only in term of the equations?

Sure, we have visualizations. Just hunt around online and you can find a bunch http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/estatics/u8l4c.html http://qbx6.ltu.edu/s_schneider/physlets/main/efield.shtml . (The visuals come from the math and aid understanding of the math, though, not the other way around.)

 

[Tac-Tics]

We're the sorry the universe does not obey laws that are sufficiently accessible to you. Take it up with your local deity of choice.

If you've got an issue with the poster, use the report button.

But I thought the definition of the dot product IS |A||B| cos theta times some unit vector.

Math often presents us with several related definitions. This definition is given to most students first, because we all knows what an angle is, and it's easy to see that the dot product is related to the angle two vectors meet.

However, the inner product can also be defined in terms of orthogonal projection. It's something like one vector casting a "shadow" over another (to use fluffy, visual terms). Inner products can also be defined on other kinds of "vectors". Later on in physics, you often deal with function spaces -- sets where each "vector" is actually a complex-valued function. Unlike vectors in R^3, functions don't have a direction and don't meet at angles. But we can still define an inner product on them which preserves many useful properties. (The inner product between two functions is usually defined as the integral of the product of the two functions).

So electric field is what it is without any charges dropped in and the flux is the interaction after a charge is dropped in? So without test charge there is a field but no flux. Once you throw in a charge, the you have flux?

The electric field is what it is. The test charges are only there to show you the "shape" of the field. After you understand how the field looks, forget everything you know about test charges. Just understand that the field can point one way, and the particle might even move backwards! (If the charge on the particle is negative)

One way to think of test charges is like this. Imagine the field is invisible (because... it actually is...) Now, your test charge is like a microscope. Wherever you hold the test charge, you are allowed to see the field at that point in space. Take the test charge away, and the field is invisible, but it's still there.


There is no "flux" until you pick a surface. You have to have a field AND a surface, and then you talk about flux of the field through the surface.

Ok. Can I claim semantics on this one?

Take a silly example. I have the vector <3, 4, 0>. It's length is 5. What is it's angle? There's no good answer. You have to take the angle with respect to another vector... and with respect to a given plane (... and orientation on top of that!) Granted the notion of "direction" isn't great either. But intuitively, "angle" is more misleading.

My brain is fried again.

Your brain becoming fried is probably an indication you have learned something. I remember struggling with these ideas when I started too. Give your mind some rest, and when you're showering or ready to go to bed, it will just sort of click. Replay the definitions of everything and the visual examples in your head, and you'll get it eventually.

I wonder do all you guys have visualizations of each concept or do you think only in term of the equations?

Visualizations are a huge part of a lot of math. The mind works a great deal better in two dimensional images than it does in symbols on a page. The hard work you have to do is be able to translate back and forth between them. It's almost like reading and writing sheet music or tabs. A tab doesn't sound at all like the music. And the music doesn't look like a tab. But the two are in some way the same. Part of the job of a musician is learning to translate between the two.

Whenever you see an integral or a direction derivative or a vector, think back to the simplest examples. There's a joke that goes like this:

A physicist and a mathematician go to a string theory conference. After the conference, the physicist exclaims "This nonsense about 9-dimensional spacetime is impossible to visualize!" The mathematician turns to him and says "Oh, it's not so hard." The physicist asks him, "Well, how do you do it then?" The mathematicians explains, "It's simple, I just imagine n-dimensional space, then set n equal to 9".

Of course, no one can really "see" 9-dimensional space, much less "n-"dimensional space. But the basic ideas in n-dimensional space are the same as they are in 3. There are vectors. They have lengths. You can take the dot product between any two. You can rotate any vector in any plane by any angle. You can add vectors in the same way. You can scale them in the same way. They are very similar.

 

[maverick_starstrider]

Of course, no one can really "see" 9-dimensional space, much less "n-"dimensional space. But the basic ideas in n-dimensional space are the same as they are in 3. There are vectors. They have lengths. You can take the dot product between any two. You can rotate any vector in any plane by any angle. You can add vectors in the same way. You can scale them in the same way. They are very similar.

Cross product is actually only defined for 3 and 7 dimensions ;)