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What makes you think that the electromagnetic disturbance created when you connect the terminals of a battery by cabling is a wave?
aaaa202 said:I did an experiment today with a long coaxial cable where you send a square pulse down the wire. I just want to know what this disturbance is. Is it the field disturbance or what is it.
aaaa202 said:I don't really understand how electromagnetic waves in wires are created …
aaaa202 said:… we are conserned about how the field from our battery drives the current around right? And that field has existed always so I don't understand how it should take time for the field to propagate information around. Or what field are we really looking it when we study this apparent wavel like behaviour?
Cannot make sense of this statement, but anyway just wondering if you actually bothered to look at especially pdf article as per 2nd link given in #32. If you had done so, why keep asking questions that are perfectly well answered there? Agrees with the broad thrust of some other postings, but provides all the math detail together with graphical plots of time evolution. There is no basic difference in the underlying phenomenon - initial establishment of current - whether voltage source is a capacitor or battery. The much longer time-scale behavior differs, but that is irrelevant wrt what was asked in #1.aaaa202 said:okay so with a battery we wouldn't have any waves while with for instance an oscilloscope we would have?
OK fine, but at least give this one a shot.aaaa202 said:oh sorry yes. I completely forgot about it. I will read it now and hope to get a better understanding - although I have read a lot of articles and am still not sure if I understand it.
There will be wave motion of the field itself. But it should be understood this wave is a bound surface wave and is intimately tied to and coupled with disturbance of the conduction charge distribution in the surface layer of the conductor. As discussed in previous postings - and that pdf article.What I meant with the statement was:
So far I have learned that what travels down with the speed of light is the information of an electric field originating from your voltage source. So if you plug a battery to a coaxial transmission line should there be a wave motion down it bringing information of the existence of an electric field?
Don't have great experience with oscilloscopes, but ones I have used act as detection/recording devices - not as signal source. Maybe some higher end models can act as source too. I guess you meant scope as a signal source.Because unlike an oscilloscope where you actually change the field suddenly there is a constant field in space in the case of a battery.
.So basically I ask this: Does the electrons in a conductor, wire or whatever feel the existence of a battery even when it is not directly connected to it? I feel it should since I don't see why the electric field from the battery should vanish in the space between them
They are intimately connected! Always keep this in mind - the source of any EM field, whether static or a propagating wave, is always charge, whether static or in motion. So when a guided wave moves down a line, it is always a coupled system of field and charge/current distribution that constitutes that wave.aaaa202 said:I tried to read the articles but was a little bit confused of this idea of surface charges. I think I should get a more basic understanding of what waves on a conductor are before I dive into that subject.
That article provides a decent basic overview of a step-wave propagating in a transmission-line. What transmits is a coupled disturbance - field initiates change in charge distribution which then generates additional field which then...and we have a wave. Note that disturbance of charge distribution is *not* equivalent to motion of individual charges. Think of sound wave in water - speed is much higher than in air even though speed of water molecules is much lower than air molecules. What is transmitted is a pressure/displacement disturbance, not gross motion of water or air!So instead I read a more (for me) illuminating article: https://www.speedingedge.com/PDF-Fil..._Impedance.pdf
But seeing how many questions I asked I am still confuzzled about the basic question: What do these waves transmit? Is it disturbances in the electric field? So that when you quickly switch on the voltage these waves transmit information that there has been created an electric field back at the voltage source. If this is the correct understanding please tell me so.
As per that earlier pdf article, the extremely brief setup phase of surface fields involves a coupled disturbance of field and surface charges. they act in unison. It is never just one or the other. And rapidly settles to a steady distribution whose details depends on the relative contributions to steady-state impedance of both load and connecting leads - how much potential drop occurs where under steady conditions.My increased confusion comes from the fact that I overheard two classmates discussing what this wavemotion represents. They said it was the electrons' movement causing it. So when you have too much negative charge at the - pole your excess electrons tries to get away by bouncing into the electrons in front of them and they in turn bounce into the next electrons in the line thus creating some kind of longitidiunal disturbance propagating through the wire. Is this what the wave equation describes? I doesn't really make sense for me that it should be since it would then seem that electromagnetic waves use electrons as their propagation medium...
The EM portion yes - but as I have been at pains to point out, this is intimately coupled with an accompanying charge/current wave in the conductor surface region. And the latter is not the speed at which a charge carrier can move - have you already forgotten the analogy I gave with sound waves in water/air?aaaa202 said:I hadn't realized that the actual wave motion takes place in the dielectric between our conductors.
Because one is relating the progression of a wavefront with the characteristics of the media through which it must advance. That media is assumed a continuum, with capacitance-per-unit-length and inductance-per-unit-length. Classically there is no lower limit to how finely one may sub-divide capacitance and inductance (and therefore the incremental advance in propagation distance), and indeed it is customary to take the limit of infinitely small values for C and L - to arrive at the wave equation for a continuum media/waveguide, as per e.g. "www.ece.msstate.edu/~donohoe/ece4333notes2.pdf" Given you have indicated previously having already studied something equivalent, I'm not sure what else to add.But it made me think. In the model of the telegraph equation you only consider the capacitance at a specific point to depend on the capacitance between that point and the same point on the return path of the circuit. Why is that?
Yes - as per previous comments. But note that in the case of leads connected to a battery, in general such leads will form a very non-uniform waveguide and there will be horrifically complex distributed reflections going on owing to highly non-uniform impedance along such a 'transmission-line'. That's where those computer simulated surface charge simulations come in handy - all that is accounted for.And is all these considerations of wave motion in the dielectric caused by capacitance and inductance between our two conductors in agreement with your surface charge picture?
See above comments! What moves at wave speed is a disturbance in the distribution of charge carriers - having no relation to individual motion of such charges.Also it occurred to me: Since the wave motion in a dielectric basically also is connected with the movement of electrons shouldn't the speed of the wave somehow be limited by how fast an electron can move? Or is the speed of an electron to be assumed infinite?
For reflection to occur there must be a change in line impedance - the wave equation for reflectionless propagation is based on uniform impedance of line/media. when such a line is terminated in a load having that same impedance, it is the same as if that line continued on indefinitely. When the terminating impedance is something else, a changed voltage is set up there that acts as a secondary source of signal that must then propagate a wave back down towards the original source.On a last note: The article discusses when no reflection occurs and says that it is when the characteristic impedance of the cable is equal to the resistor sitting at the end of the cable. Can you briefly explain what the idea of this is.