Moving electrical charges and Maxwell's equations

[Phrak]

elect_eng: Which state of a coaxial cable are you discussing? I've lost track of the thread on this.

Either a simple resistor or an imperfect conductor with a DC current has a radial Poynting vector. The Poynting vector is constant. E and B are constant. The usual treatment is to say that the surface integral of S round a portion of the conductor is equal to the energy lost to heat in that portion. The heat is the result of lattice collisions. I don't see any mystery.

The Poynting vector is directed inward.
http://www.cheniere.org/images/EMfndns1/LorentzInt%20sm.jpg"

Maybe I'm being a bit dense. So I considered one electron or charged object at a time. You accelerate some charged body in an electric field and crash it into something. Repeating this process should give you an average inwardly pointing Poynting vector. It would even happen if the charge carrier were subject to a continuous resistant to acceleration such as friction.

To make it look more like a bunch of charge carriers in a wire, let the charge carrier be a uniformly charged rod of material subject to sliding friction which is immersed in a uniform co parallel electric field.

If you are inclined to presume the mathematics is correct, then at this point it may seem uninteresting. But why should the local actions of a charged body resisting acceleration by an electric field appear as the cross product of the applied electric field and the magnetic field induced by the body?

 

[elect_eng]

elect_eng: Which state of a coaxial cable are you discussing? I've lost track of the thread on this.

...

The Poynting vector is directed inward.

Perhaps the issue is the definition of a coax cable. A coax cable has a wire conductor on the axis and a cylindrical shell conductor as a shield. This is a very effective waveguide, and a good example of one is cable TV coax that transmits hundreds of TV channels with low loss over a wide bandwidth range in the RF spectrum. I showed the E and H fields for the coax in a previous thread. The E field is in the radial direction and the H field is in the angular direction (using cylindrical coordinates, of course). You can see that they are orthogonal and the Poynting vector is in the axial z-direction, not inward.

I agree there is no real mystery with the coax. I was just trying to give a physical interpretation directed at a specific question from kcdodd.

If you are inclined to presume the mathematics is correct, then at this point it may seem uninteresting. But why should the local actions of a charged body resisting acceleration by an electric field appear as the cross product of the applied electric field and the magnetic field induced by the body?

I'm not sure I understand this question, but I interpret you to mean, "why did kcdodd even want to ask the question". I think it is because one can apply Poynting's theorem over a closed surface that encloses the power dissipation end of the coax cable and the middle of the coax cable, but excludes the power generation end of the coax cable. One then sees power being dissipated without a generator. The only interpretation left is to note that the cross product of E and H looks like a power flow into that volume over a portion of the closed surface. Whatever interpretation we want to give to this, I don't think it should include photons.

 

[kcdodd]

Phrak is talking about a resistive coax. Since power is being dissipated throughout the cable, you don't get as much poynting out one side as goes in the other. The only way that can happen is if the divergence of the poynting vector field is negative within the cable, which is caused by the energy flowing into heat. In a perfect coax you don't get this, and so the vector points along the axis all the way through. My point is the energy is "going from one place to another". This is a classical problem, and so doesn't really require photons to begin with. I am sure (quantum) field theory has a more correct explanation, I just don't know it.

 

[Dunnis]

The solution inside a long open coax cable (perhaps with a voltage source at one end) is essentially that of a capacitor with fixed charge generating the electric field. A DC field is present only if there is a charge on the capacitor. The field is in the radial direction and proportional to the stored charge, and inversely proportional to radial distance.

E_r={{q}\over{2\pi \epsilon r}}

Where did you get that information and formula from? cables are electrically neutral, there will be no measurable net electric field outside the cable. And, in the case cable happens to have some charge, then the magnitude of electric field potential will be inversely proportional to radial distance squared. Why are you comparing a wire with a capacitor? For electric fields it is irrelevant whether the wire is finite or not.

The solution inside a long shorted coax cable (perhaps with a current source at one end) is essentially that of an inductor. The DC magnetic field is in the angular direction and proportional to the current and inversely proportional to radial distance.

H_{\phi}={{I}\over{2\pi r}}

Magnetic field potential drops radially from the wire with "1/r^2" as well.

Where did you get your equations from?

 

[Phrak]

I think we've may have swapped positions on this mystery thing about coaxial cables. Signs of a good discussion. I have a thought experiment.

Take a nice room where all the walls are turned into a coaxial shield with perfectly conductive mesh. Run a perfectly conductive wire from one wall the opposite wall. At one end of the room insert a voltage source in the wire. At the other end of the room insert a resistor into the wire. Change the shape of the room into a cylinder surrounding the wire at it's center. Now we're inside this coaxial cable arrangement that was a room.

The magnetic fields form concentric hoops, uniform from one end of the room to the other because the current is uniform through the wire, battery and resistor. The radial electric field is the same as well, whether about the battery, reististor or conductor. The battery and resistor have a voltage across them so the axial electric field within the volume of each is nonzero and opposite. The axial electric field within the wire is zero valued.

What does the Poynting field look like in the cylinder?

 

[elect_eng]

Where did you get that information and formula from? cables are electrically neutral, there will be no measurable net electric field outside the cable. And, in the case cable happens to have some charge, then the magnitude of electric field potential will be inversely proportional to radial distance squared. Why are you comparing a wire with a capacitor? For electric fields it is irrelevant whether the wire is finite or not.

Magnetic field potential drops radially from the wire with "1/r^2" as well.

Where did you get your equations from?

Wow, talk about misunderstandings. Clarifications as follows ...

1. Where did I get the information from?

I got it from ... Memory, ... Derived from Maxwells Equations, ... Verified in every EM book I own. Take your pick of the answer.

2. Cables are electrically nuetral?

A coax cable with open ends is also a capacitor, it can be charged up with no problem. Every coax cable as a capacitance per unit length.

3. There will be no measureable electric field outside the cable?

Did you notice that I said "INSIDE" the cable? Yes, the shield has the effect of making the magnetic and electric field zero outside the cable.

4. The electric field will be inversely proportional to radius squared?

No, it's inversely proportional to radius. You can look up the solution or derive it if you don't believe me.

5. Why am I comparing a wire with a capacitor?

I'm not, I'm comparing a waveguide with a capacitor. So it's actually two wires. However, even a wire by itself has capacitance and one could make the comparison if it served a purpose.

6. For electric fields length is irrelevent?

Not sure why you mentioned this, but it's not true. A finite length object has end effects and a distorted electric field when compared to the solution of a similar, but infinitely long object.

7. Magnetic field drops as 1/r^2 as well?

Nope. Look it up. Solution for a coax is proportional to 1/r, and again I specified this is inside the coax, not outside.

Clearly, you have not read the post and the rest of thread carefully (or at all). It's important to do this before you jump to conclusions too hastily. It wastes time and distracts from the thread. I dont' mind some misuderstandings because I know I don't explain things in a perfectly clear way, but you need to make at least the smallest of efforts.

 

[elect_eng]

I think we've may have swapped positions on this mystery thing about coaxial cables. Signs of a good discussion. I have a thought experiment.

Take a nice room where all the walls are turned into a coaxial shield with perfectly conductive mesh. Run a perfectly conductive wire from one wall the opposite wall. At one end of the room insert a voltage source in the wire. At the other end of the room insert a resistor into the wire. Change the shape of the room into a cylinder surrounding the wire at it's center. Now we're inside this coaxial cable arrangement that was a room.

The magnetic fields form concentric hoops, uniform from one end of the room to the other because the current is uniform through the wire, battery and resistor. The radial electric field is the same as well, whether about the battery, reististor or conductor. The battery and resistor have a voltage across them so the axial electric field within the volume of each is nonzero and opposite. The axial electric field within the wire is zero valued.

What does the Poynting field look like in the cylinder?

I'm confused by what you are describing. You said to make the walls ( I assume this means only wall and not ceiling and floor) a conductor. But, then you said to run a wire from one wall to the other. This would short the wire, but then it seems you say to insert a resistor and battery. I guess this is in series so that the wire is no longer shorted. It's not clear to me how the magnetic field forms concentric hoops uniformly. This strikes me as a very asymmetrical geometry that would need a numerical solution.

Please clarify if I'm completely misinterpreting. Maybe a diagram would help.

 

[Born2bwire]

I think we've may have swapped positions on this mystery thing about coaxial cables. Signs of a good discussion. I have a thought experiment.

Take a nice room where all the walls are turned into a coaxial shield with perfectly conductive mesh. Run a perfectly conductive wire from one wall the opposite wall. At one end of the room insert a voltage source in the wire. At the other end of the room insert a resistor into the wire. Change the shape of the room into a cylinder surrounding the wire at it's center. Now we're inside this coaxial cable arrangement that was a room.

The magnetic fields form concentric hoops, uniform from one end of the room to the other because the current is uniform through the wire, battery and resistor. The radial electric field is the same as well, whether about the battery, reististor or conductor. The battery and resistor have a voltage across them so the axial electric field within the volume of each is nonzero and opposite. The axial electric field within the wire is zero valued.

What does the Poynting field look like in the cylinder?

I don't know this off-hand and I'm just jumping into the end of this discussion but I'm pretty sure I could find the answer in Balanis' electromagnetics text as he does analyze the coaxial cable transmission line. Suffice to say though, it would probably depend on your mode and frequency specifically. My gut intuition is that the Poynting vector will be normal to the \phi direction and will "bounce" back and forth between the shield and conductor. That is, the there is a cylindrical wavefront the is a standing wave in the radial direction and propagating along the axial direction.

I could be misinterpreting what you are asking, but I assume you are just treating the voltage and resistor as discrete elements simply providing an excitation and load respectively. Now there will also be reflection off of the load and the source and so forth but I assume you are familiar with the transient behavior of transmission lines.

But you are asking about in the DC case though right?

With an electromagnetic wave, the power is propagated by the waves being guided between the shield and conductor. But in the DC case... Hmmm... I'd have to think about it. If we naively ignore the return path on the shield, then the Poynting vector points along the surface of the conductor only. But with a coax we now have a static electric field applied between the shield and conductor. Question then would be what would be the magnetic field from the return current on the shield I guess. I'll take a look at Balanis tomorrow morning. The first mode of the coax problem extends down to DC and so I might be able to gleam what the answer is from that solution.

EDIT: elect_eng has mentioned some stuff I was also wondering about. I am just assuming we have a cylindrical conducting shield, a wire conductor and at one end of the coax we put a resistor between the shield and conductor and at the other end we place a voltage source across it.

 

[Phrak]

But you are asking about in the DC case though right?

That's correct. The underlying question that I'm asking is: "Does the Poynting vector correspond to energy flux in a physically meaningful way?"

With an electromagnetic wave, the power is propagated by the waves being guided between the shield and conductor. But in the DC case... Hmmm... I'd have to think about it. If we naively ignore the return path on the shield, then the Poynting vector points along the surface of the conductor only. But with a coax we now have a static electric field applied between the shield and conductor. Question then would be what would be the magnetic field from the return current on the shield I guess. I'll take a look at Balanis tomorrow morning. The first mode of the coax problem extends down to DC and so I might be able to gleam what the answer is from that solution.

EDIT: elect_eng has mentioned some stuff I was also wondering about. I am just assuming we have a cylindrical conducting shield, a wire conductor and at one end of the coax we put a resistor between the shield and conductor and at the other end we place a voltage source across it.

I made a mistake in my little DC thought experiment. There cannot be a radial component of electric field in any length of the conductor or out of the components without a net electric charge present.

Is it necessary to have a charge imbalance to have a current flow?

We are allowed to have both positive and negative charge carriers at once, so than any net charge might be canceled. If no imbalance, there is the peculiar circumstance that the Poynting vector points outward in a disk about the battery and inward in a disk about the resistor with both terminating on the shield. So there would be no interconnecting Poynting vector field to transport energy from battery to resistor.

 

[Born2bwire]

That's correct. The underlying question that I'm asking is: "Does the Poynting vector correspond to energy flux in a physically meaningful way?"

With an electromagnetic wave, the power is propagated by the waves being guided between the shield and conductor. But in the DC case... Hmmm... I'd have to think about it. If we naively ignore the return path on the shield, then the Poynting vector points along the surface of the conductor only. But with a coax we now have a static electric field applied between the shield and conductor. Question then would be what would be the magnetic field from the return current on the shield I guess. I'll take a look at Balanis tomorrow morning. The first mode of the coax problem extends down to DC and so I might be able to gleam what the answer is from that solution.

EDIT: elect_eng has mentioned some stuff I was also wondering about. I am just assuming we have a cylindrical conducting shield, a wire conductor and at one end of the coax we put a resistor between the shield and conductor and at the other end we place a voltage source across it.

I made a mistake in my little DC thought experiment. There cannot be a radial component of electric field in any length of the conductor or out of the components without a net electric charge present.

Is it necessary to have a charge imbalance to have a current flow?

We are allowed to have both positive and negative charge carriers at once, so than any net charge might be canceled. If no imbalance, there is the peculiar circumstance that the Poynting vector points outward in a disk about the battery and inward in a disk about the resistor with both terminating on the shield. So there would be no interconnecting Poynting vector field to transport energy from battery to resistor.

Bah, wouldn't you know it, Balanis doesn't have the coaxial waveguide, he does do circular cross-section though. But here we go: http://books.google.com/books?id=Ao...Q#v=onepage&q=coaxial waveguide modes&f=false

That text gives the fields for the DC mode and we can see that the Poynting vector is in the z-direction, which would point from your source to your load.

 

[elect_eng]

The underlying question that I'm asking is: "Does the Poynting vector correspond to energy flux in a physically meaningful way?"

This was exactly the question that I took kcdodd to be asking - whether of not the power flow was photons (or waves) in the DC case. My long winded answer to his question was looking exactly at the case of a long coax with a voltage source at one end and a resistor at the other end. My opinion was that the proper choice of a closed surface (one that slices a cross section in the middle of the coax dielectric far from the ends) would force us to make an interpretation that the Pointing vector is representing power flow even in the DC case. It was also my opinion that this is not physically meaningful, but that the DC fields in the dielectric of the coax represent "encoded information" about physically meaningful mechanisms of power flow in other parts of the system. Note that if one uses a closed surface that encloses the entire coax including both ends and the resistor and voltage source, then there will be no Poynting vector on the surface and only power dissipation and power generation will show up as canceling (net zero) power in the volume integral portion of Poynting's Theorem for DC case.

 

[kcdodd]

If you look at the resistor, the electrons are dropping in potential, and depositing that energy into the resistor. It is the electric field which is doing the work on the electrons, not some other mechanism. If the electric field is doing work, then the energy in the field would drop. That is poyntings theorem.

\nabla \cdot \vec{S} + \partial_t u + \vec{J}\cdot\vec{E} = 0

If you balance mechanical work and poynting flow

\nabla \cdot \vec{S} + \vec{J}\cdot\vec{E} = 0

then you reach steady field strength:

\partial_t u = 0

Same for the generator side. So, to me, without poyting's theorem, you do not construct a self-consistent scenario. That energy must be getting to the field somehow. So, how is it not physically meaningful? How else does the energy get there?

 

[elect_eng]

So, to me, without poyting's theorem, you do not construct a self-consistent scenario. That energy must be getting to the field somehow. So, how is it not physically meaningful? How else does the energy get there?

Keep in mind that we are talking about interpretations which are like opinions in that they are hard to prove. However, the way I'm thinking about it is that a DC electric field and a DC magnetic field each represent a stored energy, not an energy flow or transfer. That stored energy got there during a transient event that happened long ago in the past before the DC steady state conditions were created. Once we try to interpret the DC state, the physical meaning of power flow in the dielectric over a portion of a surface due to a Poynting vector becomes questionable.

If you choose to interpret the DC field as being an energy flow in and an energy flow out with perfect balance, I won't try to say you are wrong. It's just an interpretation that does not appeal to me personally. I like the idea of the fields representing information about power flow, rather than the power flow itself.

 

[kcdodd]

In your interpretation, how is the energy getting from the generator to the resistor?

 

[elect_eng]

In your interpretation, how is the energy getting from the generator to the resistor?

I prefer to think of the energy as force pushing charges over a distance. One can view this as an analogy with a train. Here that last car of the train gets its energy through a force linkage through all of the intermediate cars. Essentially a tension force through the chain is the linkage. Similarly the electric field around the conduction path is a force per unit charge. This force pushes charges over distance. The analogy is not perfect but is helpful. A water pipe analogy might be preferred, with a pump pushing water through a wide hose with a small outlet at the end. But, the idea of a linkage appeals to me. In mathematical terms this is similar to applying Poynting's theorem over the entire volume of the system, rather than slicing through the center. Here the integral of E.J is the only part of the equation that matters since the Poynting vector is zero everywhere on the surface. Basically, I prefer to think of the E.J as more physically meaningful than the ExH in the DC case.

Again, this is just an interpretations to help in understanding. I can't prove one interpretation is better than another. It is interesting that Poynting's theorem works for any surface. This tells us that more than one interpretation is possible. Clearly there are surfaces that show a power flow with ExH. Are these fields actually doing anything at those points in space? I think not, since they represent the ability to apply a force and there is nothing to apply a force on in the dielectric. Hence, I prefer to think of those fields as information about what is happening in other places in the system. These other places actually have charges that are moving under the influence of force pushing through a resistance.

 

[Phrak]

I'm confused by what you are describing. You said to make the walls ( I assume this means only wall and not ceiling and floor) a conductor. But, then you said to run a wire from one wall to the other. This would short the wire, but then it seems you say to insert a resistor and battery. I guess this is in series so that the wire is no longer shorted. It's not clear to me how the magnetic field forms concentric hoops uniformly. This strikes me as a very asymmetrical geometry that would need a numerical solution.

Please clarify if I'm completely misinterpreting. Maybe a diagram would help.

Sorry. Take a large conducting cylindrical surface with conducting disks at each end. Run the wire, resistor and voltage source down the axis of the cylinder. The symmetry makes the analysis easier. To make things even better, take the radius and length to infinity.

I don't see anything taking energy from the voltage source to the resistor.

-------------------------------------------------------------------------------

I'd like to see the Lorentz invariant energy continuity equation which should apparently include the Poynting vector as three out of 6 components. Maybe kcdodd already posted it. I can't tell.

 

[kcdodd]

I think you're looking for the stress tensor. This will give both energy and momentum conservation between field and matter.

 

[Dunnis]

Wow, talk about misunderstandings. Clarifications as follows ...

7. Magnetic field drops as 1/r^2 as well?

Nope. Look it up. Solution for a coax is proportional to 1/r, and again I specified this is inside the coax, not outside.

Look it up where, in your dreams? H_{\phi}={{I}\over{2\pi r}}

No, you did not specify "inside", you said: -"inversely proportional to radial distance", which now tells us that you do not even understand the meaning of what you said yourself previously. Look at the diagram below, that's what radial distance is. There, now even you know what was that which you were talking about.

                  B(r) = Km* I*L*sin(90)/r^2
                   |
                   |
                   | r
                   | 
          alpha=90 |
------------------------------------WIRE--------->>

We do not measure any magnetic fields INSIDE wires, nor electric fields for that matter - what we measure inside wires is called voltage and el. potential difference. Your equations talk about OUTSIDE too, it's just that they are not completely correct. This is correct relation of magnetic field, current and radial (and even non-radial) distance:

http://en.wikipedia.org/wiki/Biot–Savart_law -The Biot–Savart law is used to compute the magnetic field generated by a steady current, for example through a wire... The equation in SI units is:

<br /> B = \int \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{|r|^2}

1. Where did I get the information from?

I got it from ... Memory, ... Derived from Maxwells Equations, ... Verified in every EM book I own. Take your pick of the answer.

Then wake up, or sober up, you are dreaming.

Can you show me your equations anywhere in Wikipedia?

2. Cables are electrically nuetral?

A coax cable with open ends is also a capacitor, it can be charged up with no problem. Every coax cable as a capacitance per unit length.

What kind of questions are those? Have you not any cables around your house to confirm for yourself? Yes, in relation to radial distance cables are electrically neutral, of course. Wires are made of atoms and atoms are made of about the same number of positive and negative charges, so due to principle of superposition these fields combine to neutral net charge around the wire, regardless of any drift velocity, voltage or whatever is going on inside. This is macroscopic approximation so we do need some insulator to prevent direct contact and this distance does have a minimum other than zero, which is where "outside" becomes the "inside".

3. There will be no measureable electric field outside the cable?

Did you notice that I said "INSIDE" the cable? Yes, the shield has the effect of making the magnetic and electric field zero outside the cable.

Your equations and you said: "inversely proportional to RADIAL DISTANCE". What do you imagine is this "radial distance" you were talking about? Yes, as explained above, there will be no measurable electric field with any macroscopic radial distance from the cable, i.e. outside the cable.

4. The electric field will be inversely proportional to radius squared?

No, it's inversely proportional to radius. You can look up the solution or derive it if you don't believe me.

What planet are you from? I don't think you should be allowed to make any more statements without providing some source to back it up. There is nothing to derive, it is experimentally established relation and the 1st chapter in electromagnetism on this planet, it's called Coulomb's law:

E = {1 \over 4\pi\varepsilon_0}\frac{q}{r^2}

http://en.wikipedia.org/wiki/Coulomb's_lawNow, can you provide some reference for this nonsense: E_r={{q}\over{2\pi \epsilon r}}

5. Why am I comparing a wire with a capacitor?

I'm not, I'm comparing a waveguide with a capacitor. So it's actually two wires. However, even a wire by itself has capacitance and one could make the comparison if it served a purpose.

And I'm asking you about this "purpose". Why make comparison, what is the purpose and where are you getting at with it? What is the application, what is conclusion or prediction? What is it you are talking about and what is the point you are trying to make?

6. For electric fields length is irrelevent?

Not sure why you mentioned this, but it's not true. A finite length object has end effects and a distorted electric field when compared to the solution of a similar, but infinitely long object.

I mentioned it because I saw your confusion. You need to decide what kind of "objects" you are talking about. There is not many objects in this world that are electrically charged, and even batteries and el. conductors do not have any net electric charge at any macroscopic distances away from them. You need objects made exclusively, or largely, of one type of charges for that object to have some eclectic potential AROUND it, say electron beams are electrically charged even AROUND them.

+ positive charge (nucleus)
- negative charge (electron)                  E(r) = Ke* (-Q+Q)/r^2 = ~zero 
                   |
                   |
                   | r
                   | 
                   |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+WIRE ZOOM+-+-+-+-+->>
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->> 
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->>
                  E(r) = Ke* -Q/r^2 > 0
                   |
                   |
                   | r
                   | 
                   |
- - - - - - - - - - - - - - - - - ELECTRON BEAM ZOOM- - - - - ->>
 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ->> 
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ->>

You are mixing "electric field" with electric current and voltage. The equations you gave and the term "radial distance" very specifically describe field potentials in relation to their sources and/or center of their density distribution, which means "OUTSIDE". - Now, do you really think electric field potentials around cables, at some radial distance, depend on the length of the wire?

 

[Phrak]

I think you're looking for the stress tensor. This will give both energy and momentum conservation between field and matter.

You have gotten me curious about this. I'm looking for something similar to the charge continuity equation, that might look something like

\frac{\partial W}{\partial t} = -\nabla \cdot p

where W and p are energy and momentum densities in terms of the fields.

 

[Phrak]

Look it up where, in your dreams?

The radial electric field strength of a coaxial cable operating in the fundamental mode has 1/r dependence. Note this is about alternating currents.

In the limt of DC current, it is also 1/r dependent, but the radial component is zero valued; 0*(1/r)=0.

The currents in different phases along the length of the cable are not equal. Charge continuity requires that the charge be not always zero valued. It is not much different than a wave guide, where there are oscillations in all of B, E, phi and J.

 

[kcdodd]

Energy and momentum won't be invariant. Just think about the same quantities of a particle. However, you can write it in covariant form. It is interesting that the lagrangian has the form of energy and yet is invariant. But then it must be or it wouldn't work.

This is handwavy, but E^2+B^2 and ExB is a way to look at the energy and momentum of the field itself. If you start with a pure electric field, say in the x direction, and then go to a frame moving in the -z direction, you will see an ExB in the +z direction. You might say the field now has momentum in the +z direction, as well as increase in total energy. In analogy to a particle, it might be called kinetic energy. That is why you can consider ExB as energy flow, when technically energy is only a scalar. With particles, the kinetic energy can be related to the momentum. If momentum is zero, then the total energy is just the rest energy.

 

[Phrak]

Energy and momentum won't be invariant. Just think about the same quantities of a particle. However, you can write it in covariant form.

Yes, I realize all that, which is why I used the toy equation that I did. Unfortunately it works for energy and momentum, but not energy density and momentum density.

So, with your inspiration to this thread, I got out the crayons and have been looking for a manifestly covariant energy transport equation in terms of electromagnetic fields. The direct (and skew) product of F and G,

F_{[\mu \nu} G_{\sigma \rho]}

has units of energy. F is the covariant Maxwell tensor, G is the Maxwell tensor multiplied by the mixed index 4 dimensional Levi-Civita symbol in the Minkowski metric. The seemingly imbalanced brackets are deliberate and denote skew multipliction.

There are 24 nonzero terms in the product. The elements with evenly permutated indeces equal -2E². The odd permutations have entries of 2B².

BTW, I don't know that this is going in the right direction. I'm still hunting. And if all this sounds like complete jibberish, you're not alone. I'm just hoping some of it is understandable.

 

[Dunnis]

The radial electric field strength of a coaxial cable operating in the fundamental mode has 1/r dependence. Note this is about alternating currents.

In the limt of DC current, it is also 1/r dependent, but the radial component is zero valued; 0*(1/r)=0.

The currents in different phases along the length of the cable are not equal. Charge continuity requires that the charge be not always zero valued. It is not much different than a wave guide, where there are oscillations in all of B, E, phi and J.

What fields and what distance are you talking about exactly, the same ones I draw on my diagrams, and for which I provided equations and links to Wikipedia? -- So, you are just saying everything I said is wrong, including my reference, and what you say is correct, but please, if you mean to disagree, provide some link, just show me some article, paper or some other web-site where I can see what in the world you are talking about, ok?- "coaxial cable" or not, has nothing to do with this, wires are electrically neutral at any macroscopic distance away from them, which means "around them".

- "operating mode", whatever that is supposed to mean, makes no difference to the magnitude of electric field potential around any cables.

- "alternating currents" has nothing to do with 'radial distance and electric field potential around any cables'.

Wires would NOT be SAFE if they were electrically charged, i.e. if they had much more positive or negative charges per any given point over length, but cables are uniformly made both of positive and negative charges, regardless of any changes in the direction or magnitude of their drift velocity along that length. You may disagree, and if so please be specific this time, but most importantly I just want to see that information from the "1st hand", some valid source, as you seem to be misinterpreting something.

 

[Phrak]

The solution inside a long open coax cable (perhaps with a voltage source at one end) is essentially that of a capacitor with fixed charge generating the electric field. A DC field is present only if there is a charge on the capacitor. The field is in the radial direction and proportional to the stored charge, and inversely proportional to radial distance.

Where did you get that information and formula from? cables are electrically neutral, there will be no measurable net electric field outside the cable. And, in the case cable happens to have some charge, then the magnitude of electric field potential will be inversely proportional to radial distance squared. Why are you comparing a wire with a capacitor? For electric fields it is irrelevant whether the wire is finite or not.

In reply to myself, you said:

You may disagree, and if so please be specific this time, but most importantly I just want to see that information from the "1st hand", some valid source, as you seem to be misinterpreting something.

Let's start from the beginning, with your first critical statement. You might read very carefully the conditions set up by elect_eng again.

Being both insistent and wrong about the most elementary physics doesn't play well on this forum. Take this as a word to the wise.

 

[kcdodd]

Phrak, is that not the em field lagrangian?

Dunnis, I am not even sure what you are taking issue with. If you have a difference in voltage, then there is an electric field. A point's field drops off as 1/r^2, an infinite line as 1/r, and a infinite plane as a constant. This goes for both electric and magnetic sources.

 

[Dunnis]

Being both insistent and wrong about the most elementary physics doesn't play

I'm not interested in your opinions and advices given out ignorance. Basic physics are Coulomb's and Biot-Savart law, and you are the one here that wishes to refute them, so I'm asking for the 3rd time now:

1.) Show me Wikipedia article where I can see those "other" formulas for E and B field.

2.) Google whatever other web-site, paper or article where I can see those two equations.


I'm not insistent any more than you're refusing to support your claims. Whatever your status on this forum is, this is, I hope, still scientific discussion, so provide EVIDENCE and don't expect me to believe a complete stranger that uses threats and intimidation instead of valid sources and references.

 

[Phrak]

I'm not interested in your opinions and advices given out ignorance. Basic physics are Coulomb's and Biot-Savart law, and you are the one here that wishes to refute them, so I'm asking for the 3rd time now:...

I'm not interested in your discourse.

 

[Dunnis]

I'm not interested in your discourse.

Ok, I just thought you would like to know those equations are incorrect, and also that pretty much anything you attempted to disagree with me is wrong too, but I do not mean to be bothersome, so I'm sorry, and please, go on, as you were...

 

[Dunnis]

Dunnis, I am not even sure what you are taking issue with.

These two equations are wrong:

<br /> H_{\phi}={{I}\over{2\pi r}}<br />

<br /> E_r={{q}\over{2\pi \epsilon r}}<br />


No mater if "inside" or "outside" the cable, the Coulomb's law and Biot-Savart law should hold. Now, I may be mistaken of course, and all I'm asking is to see some reference where I can see exactly where and how did those formulas come to be.

If you have a difference in voltage, then there is an electric field.

You mean if you change voltage over time? Otherwise it is called "difference in electric potential" OR "voltage". It is not "difference in voltage", because the "voltage" IS "difference in electric potential". There is no such thing as "difference in voltage" when talking about a single wire, but only if you change voltage during some time period.


What "electric field" are you talking about?

ELECTRIC FIELD: E(r)=Ke*(-Q+Q)/r^2 = zero 
                 |
                 |        MAGNETIC FIELD: B(r)=Km*I*L*sin(90)/r^2
                 | r                       |
                 |                         | r
                 |                         |
MORE+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->>LESS
ELECTRONS-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+->>ELECTRONS  
LESS PROTONS+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+--+-+-+-+>>MORE PROTONS
|                                                       |
|<-------------POTENTIAL DIFFERENCE (VOLTAGE)---------->|

Is there anything here you disagree with?

Do you think that E(r>0) can be anything else but zero?

A point's field drops off as 1/r^2, an infinite line as 1/r, and a infinite plane as a constant. This goes for both electric and magnetic sources.

No, that is impossible. Please show me where do you draw your conclusions from. Can you please Google some link or show me Wikipedia article so I can see for myself what is it you are talking about?

 

[Born2bwire]

No, that is impossible. Please show me where do you draw your conclusions from. Can you please Google some link or show me Wikipedia article so I can see for myself what is it you are talking about?

Coulomb's law only deals with a single point charge. You need to do the appropriate integrals for multiple charges. Though with line and planar sources you can use Gauss' law to derive the appropriate electric fields.

 

[kcdodd]

For an infinite line, those equations are correct. You can do a google search for youself, and I suggest actually doing the calculus as an excersise to see where they come from.

Electric field is defined as the spatial derivative of voltage. In si it has units of volts/meter. So whe I say a change in voltage, that's literally what I mean. If you have two conductors at different potentials, you know there has to be an electric field between them.

 

[Dunnis]

Coulomb's law only deals with a single point charge. You need to do the appropriate integrals for multiple charges. Though with line and planar sources you can use Gauss' law to derive the appropriate electric fields.

Why is it so hard to Google some link where I can see what you telling me is not just your imagination, again? It is physically and mathematically impossible for a single electric field potential to change its distribution and gradient, perhaps in special relativity, but in any case, if what you say is true, then surely there must be some internet article to mention it. -- Show me some evidence that can support your opinion, some links and papers, please.

 

[Dunnis]

For an infinite line, those equations are correct. You can do a google search for youself, and I suggest actually doing the calculus as an excersise to see where they come from.

Electric field is defined as the spatial derivative of voltage. In si it has units of volts/meter. So whe I say a change in voltage, that's literally what I mean. If you have two conductors at different potentials, you know there has to be an electric field between them.

No, that is nonsense and is impossible, there is nothing on the internet that agrees with your assumptions. I do this calculus every day, it's my job, and if you want to support your assertions than you should be able to Google at least one of those links.

Why I can not I find anything about it in Wikipedia, and would it not be easier for you to just copy/paste the damn link? -- Seriously now, CAN YOU provide any reference to support YOUR CLAIMS, or not?

 

[kcdodd]

Don't say I never did anything for you. Second link in google.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

 

[elect_eng]

Look it up where, in your dreams? H_{\phi}={{I}\over{2\pi r}}

No, you did not specify "inside", you said: -"inversely proportional to radial distance", which now tells us that you do not even understand the meaning of what you said yourself previously. Look at the diagram below, that's what radial distance is. There, now even you know what was that which you were talking about.

We do not measure any magnetic fields INSIDE wires, nor electric fields for that matter - what we measure inside wires is called voltage and el. potential difference. Your equations talk about OUTSIDE too, it's just that they are not completely correct. This is correct relation of magnetic field, current and radial (and even non-radial) distance:

...

Then wake up, or sober up, you are dreaming.

Can you show me your equations anywhere in Wikipedia?

...

What planet are you from? I don't think you should be allowed to make any more statements without providing some source to back it up. There is nothing to derive, it is experimentally established relation and the 1st chapter in electromagnetism on this planet, it's called Coulomb's law:

E = {1 \over 4\pi\varepsilon_0}\frac{q}{r^2}

http://en.wikipedia.org/wiki/Coulomb's_lawNow, can you provide some reference for this nonsense: E_r={{q}\over{2\pi \epsilon r}}

Good God man. What is wrong with you?

I did say "inside". You can check my post and read it for yourself. Unfortunately, because you have not bothered to read the thread carefully, you don't even know what "inside" means in the context we are talking about. I'm talking about a coaxial cable that has a central thin wire and a cylindrical shell outer conductive shield. "Inside" means the volume with the dielectric between the two conductors. Outside the shield, the fields are zero. Inside the shield the fields are what I said. One of a thousand references I could site is as follows.

"Fields and Waves in Communications Electronics", by Ramo, Whinnery and van Duzer. Pages 76 and 77 show the magnetic field.

On page 10, there is example 1.4b entitled "Field about a line Charge, or between coaxial cylinders". The solution is:

E_r ={{q}\over{2 \pi \epsilon r}}

Since we are talking about two conductors and a coaxial transmission line, the concept of capacitance and charge applies here. A coax line can act as a capacitor. If you are not aware of this, don't criticize me. Go back and study what you should know. This can be found under chapter 8 of the above reference. The chapter title is "Waveguides with cylindrical conducting boundaries".

By the way, you keep talking about Wikipedia. This is not a reliable source of information. You can pick up ANY electromagnetics book and find this all worked out. Besides, don't you even know how to apply Maxwell's equations to a basic problem like this?

From Maxwell's expression of Ampere's Law in the static case \ointop \vec H \cdot d\vec l = 2\pi r H_{\phi}=I

therefore, H_{\phi}={{I}\over{2\pi r}}}

Was that so hard?

EDIT: OOPS, I forgot that I need to provide a reference for Maxwell's Equations, otherwise I will be viewed as an alien from another planet. Please see the following. "A Treatise on Electricity and Magnetism" by James Clerk Maxwell, 1873.

And, let's not forget the Wikipedia link: http://en.wikipedia.org/wiki/Maxwell's_equations
http://en.wikipedia.org/wiki/Maxwell's_equations

 

[elect_eng]

What planet are you from? I don't think you should be allowed to make any more statements without providing some source to back it up. There is nothing to derive, it is experimentally established relation and the 1st chapter in electromagnetism on this planet, it's called Coulomb's law:

E = {1 \over 4\pi\varepsilon_0}\frac{q}{r^2}

http://en.wikipedia.org/wiki/Coulomb's_law


Now, can you provide some reference for this nonsense: E_r={{q}\over{2\pi \epsilon r}}

In the previous post, I gave a reference for what you call "nonsense", but your stubbornness convinces me this will not be enough for you. Let's ignore your large number of misconceptions and focus on this one area. This is an easy one to handle because you claim to know about Coulomb's law from the first chapter of your EM fields books.

So let's proceed to chapter two and apply Coulombs law to a line of charges. I've attached a derivation as a jpg file. So, now you have a reference and a clear derivation in terms of a law you can understand and a basic application of calculus. Contrary to two of your statements above. There is something to derive, and the final equation is not nonsense.

I'd like to state for the record that I don't appreciate someone who is unwilling to read past the first chapter of his book saying I'm "dreaming", "stating nonsense" and "from another planet".

 

[Dunnis]

I did say "inside". You can check my post and read it for yourself. Unfortunately, because you have not bothered to read the thread carefully, you don't even know what "inside" means in the context we are talking about. I'm talking about a coaxial cable that has a central thin wire and a cylindrical shell outer conductive shield. "Inside" means the volume with the dielectric between the two conductors. Outside the shield, the fields are zero. Inside the shield the fields are what I said.

You "shield" magnetic fields, electric fields you "NEUTRALIZE" with superposition, simply by having conductors be made of regular material, all of them are electrically neutral, there is nothing to "shield" with electric fields, but only to prevent "sparking" or direct contact, i.e. current flow. There are no any electric fields around the conductor at any macroscopic distance, as I said. The dielectric between the two conductors is to prevent microscopic contact and keep the distance, but you are wrong to refer to it as INSIDE, as that is just another layer of OUTSIDE, the only difference is that it is not air between those two but some plastic.

                        E(r2),B(r2)
                             | 
                             | 
                             | r2
AIR OUTSIDE 2                | 
                             |
========================================RUBBER==============
+-+-+-SHIELD INSIDE-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-   |
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+   | C
. . . . . . . . . . . . .                              | O
. . . . . . . . . . . . .        E(r1),B(r1)           | A
. . PLASTIC OUTSIDE 1 . .             |                | X
. . . . . . . . . . . . .             | r1             | I
. . . . . . . . . . . . .             |                | A
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->>   | L
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->>    |
+-+-CORE INSIDE-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->>   | C
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->>    | A
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+- ->>   | B
. . . . . . . . . . . . . . . . . . . . . . . . . .    | L
. . . . . . . . . . . . . . . . . . . . . . . . . .    | E
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+   |
============================================================

Any objection to the diagram? We are talking about the same fields and different materials do not make difference to what is "outside" and what the "inside" of the CONDUCTOR ITSELF, which means - ONLY THAT WHICH IS MADE OF METAL, any point that is not located in this substance is OUTSIDE of that substance, ok? There ere TWO conductors here and that case simplifies to the case of two parallel wires, and then it boil down to the line integration against a single charge - coaxial cable or not. Having a complex combination of insulators and conductors changes nothing as the problem need to be split to its basic components to be solved, and only then to be combined (integrated) and superimposed macroscopically. -- Nothing is wrong with me, but with your equations, understanding and mathematics, as I will demonstrate it step by step, hold on...

 

[Dunnis]

Don't say I never did anything for you. Second link in google.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html

\lambda= 1C/1m = ?

E = \frac{\lambda}{2\pi\epsilon_0*r} =&gt; \frac{1C/1m}{2\pi\epsilon_0*1m} = \frac{1C}{2\pi\epsilon_0*1m^2} \ N/CCoulomb's law: E = \frac{Q}{4\pi\epsilon_0*r^2} =&gt; \frac{1C}{4\pi\epsilon_0*1m^2} \ N/CSo, all that mathematics to change 1/4Pi to 1/2Pi? -- That is not right, it's some kind of butchering and symbolic derivation, while that integral was never supposed to be derived, but numerically integrated at the time of application, with given numerical input and REAL NUMBERS, not symbols. Symbols and derivation can not generalize arbitrary scenarios, and so that equation must stay in its integral form so it actually can be applied to wires of different and arbitrary lengths in relation to different and arbitrary geometry, where given input is defined by the REAL-WORLD numerical values.

E = \int_a^b \frac{Q*dl}{4\pi\epsilon_0*r^2} =&gt; \int_a^b \frac{1C*1m}{4\pi\epsilon_0*1m^2} \ Nm/C = 1 Volt = N/(m/C)================== LET'S DO SOME REAL WORD PHYSICS, shall we? ==================

CASE A:                   CASE B:          CASE C:       E(r),B(r)=?           E(r),B(r)=?          E(r),B(r)=?
          |                     |                    |
          |r=2m                 |r=1m                | r=0.5m
          |                     |                    |
|===================|     |==========|     + ==================infinite >> -
| I1= 1A            |     | I1= 1A   |        I1= 1A
|<------- 9m ------>|     |<-- 1m -->| 
+                   -     +          -

E = \frac{\lambda}{2\pi\epsilon_0*r} \ N/C\lambda= 1C/1m = ? How do you get lambda, and what result your equation predicts for the above three scenarios: with 9 meters, 1 meter and infinite wire?

 

[elect_eng]

Any objection to the diagram?

I object to your diagram.

I object to every word you have said in this thread.

I object to you having no consideration to try to understand an existing thread.

I object to you interrupting this thread with nonsense.

I object to your harassing tone.

I object to anyone who doesn't know that Coulomb's law and Biot Savart's Law apply to differential elements and integration is required to get the answer for real objects, yet purports himself to an expert "who does this for a living".

I object to needing to waste my time trying to educate an ignoramus who will not admit when he is wrong even in the face of overwhelming evidence.

Mostly, I object to all of the stupid comments you will be spouting in response to this post.

I won't be responding to any more of your nonsense. However, I will be reporting your posts as inappropriate.

 

[Dunnis]

What is wrong with you?
...
One of a thousand references I could site is as follows.

"Fields and Waves in Communications Electronics", by Ramo, Whinnery and van Duzer. Pages 76 and 77 show the magnetic field. On page 10, there is example 1.4b entitled "Field about a line Charge, or between coaxial cylinders". The solution is:
E_r ={{q}\over{2 \pi \epsilon r}}

That equation is wrong and demonstrates nicely what you know about it.

What do you imagine 'q' stands for?

       E(r)
        |
        | r=2.5m
        |
-----------------I= 23A -------------------->

q= ??! How do you solve this problem?q= 1.602176487(40)×10^-19 coulombs

Do you realize that equation predicts the same result regardless of any wire length? What that equation does is to evaluate electric field of a SINGLE electron, regardless of any voltage, amperes or oscillation or wire length. It is the same thing as Coulomb's law for POINT CHARGES, only it has 1/2pi*r, where it should be 1/4pi*r^2. -- It is not "q", but it should be Q(net), which is zero to start with, since wires are made of atoms and atoms are electrically neutral when measured at any macroscopic distance, either from a single atom or volume of atoms, and even a line of atoms. -- Why is it surprising all the wires in your house are electrically neutral? Is that some news?

Q(net) = sum(-q,+q) --- made of atoms ---> electrically neutral (mostly and usually)THREE DIMENSION, three integrals: C/m, Cm^2, C/m^3

1.) Q(net Line) ; 2.) Q(net Surface) ; 3.) Q(net Volume)

http://en.wikipedia.org/wiki/Charge_densityYou can not "derive" integrals, integral equations must stay in their integral form so you can INTEGRATE over given distance segments and certain geometry defined by the NUMERICAL VALUES given by the specific given problem - i.e. integrals and not "derived", but numerically integrated. And to realize this you only need to actually try to USE YOUR EQUATION on some REAL WORLD SCENARIO, where the actuality is different than the one in Wonderland.

From Maxwell's expression of Ampere's Law in the static case \ointop \vec H \cdot d\vec l = 2\pi r H_{\phi}=I

therefore, H_{\phi}={{I}\over{2\pi r}}}

And, let's not forget the Wikipedia link: http://en.wikipedia.org/wiki/Maxwell's_equations

What the... ?!? There is no such equation in that Wikipedia article or on the whole internet. Nowhere in Wikipedia there is any mention of any of your equations, is that why you're angry(?) -- This is what Wikipedia says:

Free charge and current:
\oint_{\partial S} \mathbf{H} \cdot \mathrm{d}\mathbf{l} = I_{f,S} + \frac {\partial \Phi_{D,S}}{\partial t}

Total charge and current:
\oint_{\partial S} \mathbf{B} \cdot \mathrm{d}\mathbf{l} = \mu_0 I_S + \mu_0 \varepsilon_0 \frac {\partial \Phi_{E,S}}{\partial t}

So let's proceed to chapter two and apply Coulombs law to a line of charges. I've attached a derivation as a jpg file. So, now you have a reference and a clear derivation in terms of a law you can understand and a basic application of calculus. Contrary to two of your statements above. There is something to derive, and the final equation is not nonsense.

I'd like to state for the record that I don't appreciate someone who is unwilling to read past the first chapter of his book saying I'm "dreaming", "stating nonsense" and "from another planet".
Attached Images File Type: jpg Efield.jpg

You mean this below? Is that supposed to be the same?

-----------

You got it wrong it seems, can you please decide which one:

E = \frac{\lambda}{2\pi\epsilon_0*r}

-OR-

E = \frac{q}{2\pi\epsilon_0*r}
...since your equation always give the same result regardless of any specific input, I'm really curios how do you solve for DIFFERENT scenarios we just happen to have in REAL WORLD, so can you show me how do you apply that equation to the following setups and what result do you get:

CASE A:                   CASE B:          CASE C:       E(r),B(r)=?           E(r),B(r)=?          E(r),B(r)=?
          |                     |                    |
          |r=2m                 |r=1m                | r=0.5m
          |                     |                    |
|===================|     |==========|     + ==================infinite >> -
| I1= 1A            |     | I1= 1A   |        I1= 1A
|<------- 9m ------>|     |<-- 1m -->| 
+                   -     +          -

E = \frac{q = ?}{2\pi\epsilon_0*r}E = \frac{\lambda = 1C/1m = ?}{2\pi\epsilon_0*r}

 

[kcdodd]

Can we please get a moderator?

 

[Dunnis]

Can we please get a moderator?

Huh? What for? What Wikipedia articles and what equation do you believe is wrong? What kind of discussion is this where I provide all the reference and get called "stupid", yet those who can not support their claims just scream around and wave hands without actually saying anything? Just say it already, what exactly are you so nervous about?

 

[elect_eng]

Can we please get a moderator?

I've reported his crazy posts. I recommend others do the same. He can't even tell the difference between charge and charge density, yet he wants to claim every EM book ever written is wrong. This is yet another clear example of how a little bit of knowledge can be a dangerous thing.

 

[Dunnis]

I've reported his crazy posts.

He can't even tell the difference between charge and charge density..

Interesting reaction, and I only wanted to see some evidence for your assertions.YOUR FALSE EQUATION: E = \frac{q}{2\pi\epsilon_0*r}

It is you who has a single charge of a single electron in your equation, and moderators will tell you that, whatever the reason you want to call them. You may as well call your mum too, that will not change the reality and what has come to past. Next time be careful about the equations and try to use them before you make your conclusion, so to not embarrass yourself like this. Ok? -- At least you realized there is a TOTAL amount of charges here, now learn about it:

http://en.wikipedia.org/wiki/Coulomb

http://en.wikipedia.org/wiki/Charge_density

http://en.wikipedia.org/wiki/Elementary_charge

 

[elect_eng]

It is you who has a single charge of a single electron in your equation, and moderators will tell you that, whatever the reason you want to call them. You may as well call your mum too, that will not change the reality and what has come to past. Next time be careful about the equations and try to use them before you make your conclusion, so to not embarrass yourself like this. Ok? -- At least you realized there is a TOTAL amount of charges here, now learn about it:

No, it is you that has ignored all of the explanations given to you. I provided you a derivation in an attached jpg file. Did you even look at it after you asked for references and explanations? No you did not. I quoted a book. Did you bother to consult it? No you did not. The variable q is linear charge density with units of Coulombs per meter. It is not the single electron charge as you say. Anybody can go back into this thread and read proof that I explained this. Besides, it needs no explanation. This is such an obvious fact from the context of the equations. You are asking for explanations that are like asking what 2+2 is. Who do you think you are kidding? You are the one who is embarrassing himself.

 

[elect_eng]

YOUR FALSE EQUATION: E = \frac{q}{2\pi\epsilon_0*r}

Why do you call this "my false equation"? This equation can be found in every electromagnetics book. EVERY SINGLE ONE! I gave reference to one book above. The well known book by Krauss is another, and Jackson can be consulted too. I gave a jpg file with my own derivation and kcdoddd gave a link to another derivation. You keep asking for references and we give them, yet you ignore them. What is your problem? Are you just doing this as a prank? Are you just unwilling to admit when you are wrong? Whatever the issue is you'd better come to terms with it. By the way, at one point you mentioned you do this for a living. I dare you to show this thread to your boss, or your customers if you are self-employed.