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What exactly do electric waves transfer from one point of the wire to other?

  1. Jul 6, 2012 #1
    so basically i was considering the speed of the charges inside wire defined by the drift velocity and speed of electric waves that's equal to speed of light..so my point is if electric waves don't carry any charge (as mostly light that we encounter from sun's radiation is em and since they don't carry any charge we don't get shock)so what do they actually carry..i was trying to figure out its resemblance to mechanical waves ..as mechanical wave don't itself carry any matter with it they only transfer energy..then so do the electric waves they can't carry charge but they must transfer energy in the form of you know electric waves..or could i say now that these electric waves are source of electric field..then thing that's confusing me is that if electric field propagates at speed of light(speed of electric waves)..then do the voltage that we define as eL or voltage or potential at a point is=electric field intensity * L(distance) then should the voltage drop or potential also varies accordingly or i could say that the potential also following the electric field intensity at the speed of light..but this is contradictory to our general observation in which we define particular time period and phasors which define variation of voltage that..so not at the speed of light..
  2. jcsd
  3. Jul 6, 2012 #2


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    welcome to pf!

    hi shivaniits! welcome to pf! :smile:
    i'm not sure i understand :confused:

    if you're asking why the speed of AC electricity depends on the frequency (which is fixed by the power supply), the answer is it doesn't, the speed has nothing to do with the frequency

    (except that speed = frequency x wavelength)
  4. Jul 6, 2012 #3
    I don't quite get your last part of the post. First don't confuse the electron drift velocity with the EM wave propagation. Electron drift velocity is so slow it's not funny!!! It is the EM wave that propagate down the line. Even it is a line in the air, there got to be a return path either through earth ground or a return wire. The first case form something like a microstrip where a line is on top of a ground plane ( earth ground or soil ground) or a parallel line of the second case. These structure form a wave guide that EM wave propagate through.

    The current that people referred to all the time is not due to electrons moving. It is due to the boundary condition of EM wave. Remember

    [tex] \nabla \times \vec B = \mu \vec J[/tex]

    The boundary condition give the surface current when integrated.

    The voltage is from

    [tex]V=\int_C \vec E \cdot d\vec l [/tex]

    Maxwell's equations give rise to the wave equation where the solution of the differential equation give you the phasor that you referred to. Go on the web and read p429 to 430 of Field and Wave Electromagnetics 2nd edition by David K. Cheng. It really explains the parallel plate tx line and the boundary condition.

    Bottom line, It's the EM wave that travel down the line. The voltage is from the E field. Then current is from V with the characteristic impedance of the tx line structure. Straightly speaking, it is all EM wave that running around the electronic circuits. You are just measuring the voltage and current due to the consequence of the EM wave. If you are waiting for the electrons to physically move from one end of a 3' wire, you are going to wait for a long time. You can inject an electron into one end, then put a potential across the line, then go get yourself a cup of coffee, then come back and wait for the same electron to come out from the other end!!!! That's drift velocity and the better the conductor, the slower it moves.
    Last edited: Jul 6, 2012
  5. Jul 6, 2012 #4
    Re: welcome to pf!

    amm.. the thing i am concerned is about travelling of electric waves through wires ..and as the electric waves travel considering a sinusoidal electric wave in which the electric field is varying continuously at speed of light ..then does it have any effect on voltage drop across wire ..i mean if electric field across a source charge changes then so do the the potential across it also changes ..so if electric field is changing at the speed of light then can i say the potential across the wire is also changing at the speed of light..as both are related through expression :- voltage drop or potential drop=electric field intensity * distance ....
    if the electric field is varying at the the speed of light (as the speed of the em waves are taken to be of that of light)then does the volatge drop or potential drop also changing at the speed of light..??
    if it is that so then there would be no meaning for ac frequency.....??
  6. Jul 6, 2012 #5
    so that means whatever i am measuring either the voltage drop or current across the wire i am measuring its average value i mean i could never follow up with instantaneous values as the electric field intensity is varying at speed of light i.e c and so the potential also varying at speed of light as both are related quantities..but what i am measuring is only the average..???..:confused:
  7. Jul 6, 2012 #6
    The EM wave propagate at high speed, not necessary at the speed of light as

    [tex]U=\sqrt{\frac 1 {\mu \epsilon}}[/tex]

    Only if the dielectric medium is air then it travel at speed of light.

    The electric field at any point varies with frequency of the EM wave, nothing to do with the speed of propagation. It is like a floating ball in the ocean. It does not matter how fast the wave travel on the surface of the ocean. It is the number of times per second the ball bobble that you referred to.

    One important thing to clarify, it is likely the EM wave traveling in wire, coax, pcb trace and most common conductor is TEM wave where both the E and H are normal to the direction of travel and they are normal to each other. So if direction of propagation is z, E is in x and H in y direction. So you can think of the E is bobbling in x direction while the wave is propagating in z direction.

    When you measure the voltage at any point on the wire, you see the waveform of the signal. If it is a sine wave of 1KHz, you see a 1KHz wave form...........even though the wave is traveling at close to light speed down the wire.
    Last edited: Jul 6, 2012
  8. Jul 6, 2012 #7
    Hello yungman,

    You need to distinguish between wave propagation in dielectrics (as you have done) and wires. Copper wire is a conductor and conduction current dominates over displacement current.

    The velocity in a conductive medium is given by

    [tex]v = \sqrt {\frac{{2\omega }}{{\mu \sigma }}} [/tex]

    Some figures

    A 50/60Hz power wave in free air (strictly in vacuum) has a velocity of 3x108m/s and consequently a wavelength of 5x106m (yes 5 million metres).

    In copper wire that same wave has a velocity of 3m/s and a wavelength of .05m.

    That is the wavelength of a power signal is longer than most if not all current power supply cables.

    We consider the power wave to be a surface effect guided by the copper wire travelling at the faster (phase) velocity.
  9. Jul 6, 2012 #8
    The wire with the return form a guided structure that EM wave travel in. It has nothing to do with the speed of electrons travel inside the conductor as you brought up. It is the EM wave that propagates in the electronics, not the current or the voltage. In this case, I assume the wire is in the air. The EM wave is traveling in the air GUIDED by the wire and the return.

    If electronics depend on conduction of wire, we don't have much electronics to work with as the speed is so slow. Everything is EM wave propagation even though we measure voltage and current.
  10. Jul 6, 2012 #9
    That is largely what I said mate.
  11. Jul 6, 2012 #10
    Yes, I looked back, you formula is for EM wave in good conductor, not electrons velocity.
  12. Jul 6, 2012 #11
    The important two points are the relationship between the TEM speed in air and copper, which is one (simple) way of understanding why the power is a surface wave,
    and the wavelength in air which is why we don't get standing wave issues between power station and consumer.
  13. Jul 7, 2012 #12

    but in accordance to relation speed of propagation=c=wavelength*time period or i could say speed=wavelength(λ)/frequency..and frequency here is no of bobbles but yes if we refer here amplitude or the distance the ball bobbles in x direction while the wave propagation is in z direction..then both are independent....i hope so
  14. Jul 7, 2012 #13
    but if electric field is varying at speed of light(not exactly at the speed of light but is actually at very higher speed)then don't you think that the the electric potential associated with it should also vary.....as both are related terms ..
  15. Jul 7, 2012 #14
    Speed is frequency times wave length!!!!

    Yes the two are independent. The bobble is only depend on frequency, propagation speed is constant for lossless material regardless of frequency. Speed does change a little with frequency if the dielectric is lossy as the [itex]\epsilon_c[/itex] is frequency dependent.

    [tex] \epsilon_c=\epsilon -j\frac σ ω[/tex]

    Where σ is the conductance of the dielectric material.
  16. Jul 7, 2012 #15

    yes it is..i typed wrong..
    but if u say that propagation speed = wavelength* frequency then how could they be independent..but yes
    the speed of propagation and the amplitude are independent to each other but the speed and the frequency are dependent things...
  17. Jul 7, 2012 #16
    I just edit the post.
  18. Jul 7, 2012 #17
    ok thanks i am sorry if i offended u i just meant that yes the frequency and speed are somehow related to each other that's it...so they can't be both independent..
  19. Jul 7, 2012 #18
    I was not offended at all, what is to offend?!!!

    For the most part, it is almost independent. You can see from the equation that any dielectric material that is not perfectly lossless will have an imaginary term, that change the speed ever so slightly with frequency. I believe that is what people talked about dispersion where different frequency travel at slightly different speed in a transmission line. So if you pump a RF square wave into one side of the tx line, the square wave is distorted coming out from the other side as the harmonics arrived the end at slightly different time. There is NO lossless dielectric. They all have different degree of loss.
  20. Jul 7, 2012 #19
    Frequency and wavelength are not both independent in a given medium.

    The speed of transmission for a given medium is constant and equal to the product of frequency and wavelength.

    v = fλ

    So if we set either frequency or wavelength we have determined the other.
  21. Jul 7, 2012 #20


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    If this were true then a simple prism wouldn't work!
    c (in a vacuum) is the only 'constant' speed.
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