Moving electrical charges and Maxwell's equations

[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:

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

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


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

$$\lambda= 1C/1m = ?$$

$$E = \frac{\lambda}{2\pi\epsilon_0*r} => \frac{1C/1m}{2\pi\epsilon_0*1m} = \frac{1C}{2\pi\epsilon_0*1m^2} \ N/C$$Coulomb's law: $$E = \frac{Q}{4\pi\epsilon_0*r^2} => \frac{1C}{4\pi\epsilon_0*1m^2} \ N/C$$So, 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.

$$E = \int_a^b \frac{Q*dl}{4\pi\epsilon_0*r^2} => \int_a^b \frac{1C*1m}{4\pi\epsilon_0*1m^2} \ Nm/C = 1 Volt = N/(m/C)$$================== 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.