"The field due to the battery sets up a surface charge in the wire. The surface charge is negative near the negative pole of the battery, and positive near the positive terminal, and varies more or less linearly along the wire. The surface charge in turn sets up inside the wire an electric field which is constant across the diameter of the wire, and along the length. This field accelerates electrons"

If there are charges being set up in the wire, the charges (electrons) have mass and cannot move at $c$. How then do they not affect the speed of electricity and gives us its speed to be equal to $c$?

Do electrons in a wire (load) affect the speed of electricity due to their own velocity?

Think about a circuit with a 5 V battery and a very long resistor with two switches at its end. So, when both switches were open there was no charge setup inside resistors. When we close both switches then, the charge will setup and the signal travel at the speed of $c$ inside that resistor. How?

Doesn't the charge being set up and their own electric fields being created slow down the signal's speed?

  • 15
    $\begingroup$ The propagation speed of an electric signal down a wire is not c. It is more like 0.6c and what it is exactly depends on the construction of the wire. $\endgroup$
    – DKNguyen
    Jun 6 at 3:06
  • 4
    $\begingroup$ It's the electromagnetic wave rippling through the electrons that propagates at close to the speed of light. The dimensions of the wire and electrical properties like its inductance affect the exact propagation speed, but usually it will be around 90 per cent of the speed of light – about 270,000 km/s. How did you get to 0.6 C? And does electrons speed affects the speed of electricity? $\endgroup$ Jun 6 at 3:59
  • 8
    $\begingroup$ 0.6C is just a rough typical number for the wires and traces used in circuit board design when you need a speed but it isn't too critical that it is absolutely correct. $\endgroup$
    – DKNguyen
    Jun 6 at 4:13
  • 2
    $\begingroup$ It doesn't. Why do you think this? Also, the field is set up everywhere in space at the speed of light. All that needs to happen to create a current is that a field moves a charge. The charges do not have to push each other (which would be very slow). $\endgroup$
    – user196418
    Jun 6 at 20:44
  • 1
    $\begingroup$ Define "speed of electricity". $\endgroup$ Jun 7 at 9:12

If there are charges being setup in the wire, the charges(electrons) have mass and cannot move at C. How then they don't affect the speed of electricity and gives us its speed to be equal to C.

They do affect the speed.

The speed of the signal in a cable or other communication channel is given by the velocity factor which is the speed of the signal expressed as a percentage of c. This is 100% for a radio wave in vacuum, but it is about 75% for CAT 7 twisted pair cable. The velocity factor primarily depends on the insulator rather than the conductor. So a coaxial cable with air as the insulator might propagate signals at 93% of c, while one using polyethylene might have a velocity factor of around 80%.

The velocity factor is sometimes considered to be the speed of light in the material as opposed to c which is the speed of light in vacuum. However, it is a little complicated since the structure is not just a simple homogenous material (the dielectric) but a complicated waveguide.

  • 1
    $\begingroup$ Does the speed of electricity changes with changing voltage across the wire? Because drift velocity will also get changed $\endgroup$ Jun 6 at 4:10
  • 3
    $\begingroup$ I don’t know about that. I suppose it could if the dielectric were nonlinear. But I haven’t seen any experimental confirmation. The drift velocity in the conductor is not as important as the dielectric properties of the insulator $\endgroup$
    – Dale
    Jun 6 at 4:15
  • $\begingroup$ It might be 66%, not 80%, for polyethylene. $\endgroup$ Jun 7 at 8:06
  • $\begingroup$ @PredakingAskboss AFAIK the "speed of electricity" is not the drift velocity, but the speed at which the drift velocity changes $\endgroup$
    – user253751
    Jun 7 at 14:17
  • $\begingroup$ @user253751 "...but the speed at which the drift velocity changes" I actually don't know what do you mean by this. $\endgroup$ Jun 7 at 15:03

The information about the disturbance of static charge equilibrium (toggling the switches) travels at the group velocity of light in matter, which is in turn related to the refractive index in matter, or more precisely to its dispersion (rate of change of refractive index with frequency/wave number). Group velocity is sometimes close to the speed of light, but sometimes significantly smaller (but never greater, which would be a violation of causality/special relativity; for causality in material electrodynamics see also Kramers-Kronig relations).

Models of dielectric permittivity (and hence, refractive index) in matter, namely in conductors, are the Lorentz and Drude models (see this reference for a compound model). These models also show how the mass of electrons enter the dispersion relations. In other words, the fact that the electron mass is non-zero, is encoded in a more or less convoluted form into the dispersion relations (and hence the speed of light) of electromagnetic waves in matter. They determine the slackness of the electrons to wiggling around in the Lorentz oscillators, so to say, which determines the effective speed of light in matter.

Needless to say that even if the speed of information is somewhere around (but, probably well, below) c, the relaxation time to regain static charge equilibrium can be considerably longer than the time to travel the whole circuit. This is because the disturbance causes oscillations, which are damped depending on the properties of the individual circuit. There is resistive damping, and there is radiative damping, all of which determine how long it takes the stirred up charges to calm down.

  • $\begingroup$ So, is there any effect of charge distribution to the speed of signal in wire? Because the signal itself is travelling due to the charges setup in wire and electrons have mass too and a very low drift velocity. In this way one electrons moves a little and it electric field propagates and then the next charge moves a little and and so on. This will affect the speed of electricity even the electrons have to cover a very less distance. $\endgroup$ Jun 6 at 4:08
  • $\begingroup$ These effects will be part of the low frequency limit of the Drude-Lorentz model (the conductivity part). It is determined mainly by the configuration of the atom bodies and the bulk of the electrons, not the small amount of surface charges. However, it is easy to imagine that, if charge distribution gets extreme enough, it will cause the model to break down, and hence, change the whole dispersion relation in an arbitrary way... $\endgroup$
    – oliver
    Jun 6 at 4:18
  • $\begingroup$ ... in other words, figuratively speaking, if you expose the circuit to a nuclear explosion, it will disintegrate, and hence, the charge distribution will certainly have an effect on the speed of the signal in the wire. $\endgroup$
    – oliver
    Jun 6 at 4:18

Electrons don't need to bump into each other to transfer movement, like billiard balls. They are charged particles which interact with each other at a large distance through the electromagnetic field.

Imagine a room packed with balloons: if you push on balloons on one side of the room, the wave travels several meters to the other end of the room in a matter of seconds, even if none of the balloons traveled more than a few centimeters. In this analogy, balloons represent electrons together with the field they produce.

This is why the electrical signals travel at e.g. 0.7c, while individual electrons move at less than 1% of that speed (depending on the temperature), even at microscopic level. At macroscopic level, the charge drift happens even slower, with typical speeds in mm/s or cm/s, depending on the charge and current densities.

  • $\begingroup$ This description reminds me of nerve axons and the jumping across nodes on the myelin sheath. $\endgroup$
    – DKNguyen
    Jun 6 at 20:00
  • 1
    $\begingroup$ “at a large distance” compared to their own size. That distance is not larger than the distance of the billiard balls when they interact with each other. You could replace your room packed with balloons with a room packed with billiard balls and the wave would travel to the end of the room in less than a second. $\endgroup$
    – Holger
    Jun 7 at 7:39
  • $\begingroup$ Something could be said about the dielectric constant and the geometry. $\endgroup$ Jun 7 at 8:10
  • $\begingroup$ @Holger Yes, but a room filled with billiard balls would contain mostly billiard ball material, while a room filled with balloons contains >99% air. Of course there's no such thing as perfect analogies. $\endgroup$ Jun 8 at 6:55
  • $\begingroup$ @PeterMortensen I think the crux of the question is the discrepancy between signal speed and electron speed, not how far away from c the signal speed is. $\endgroup$ Jun 8 at 6:57

To answer your title question, the speed of electricity does not become the speed of light. But many "messages" or, more correctly, transient effects that set up the final steady state flow do indeed travel at the speed of light in the medium your circuit is steeped in

Think of your circuit at the very instant you close the switch. The switch changes a boundary condition of the electromagnetic field, suddenly (let's say for simplcity, instantaneously) imparting a potential difference between the two conductors at the battery end, whereas the potential differnce just before the switch close was nought.

Let $+z$ be the direction along the conductors. For simplicity we think of your conductors as a waveguide whose cross section is translationally invariant in the $z$ direction.

If you solve this boundary value problem given that the initial Potential Difference (PD) was nought but is now suddenly +V at one end of the conductor, the solution is that the potential difference now travels down the conductors as a freespace electromagnetic wave, progressively raising the PD from 0 to +V at points of increasing $z$ as the wave travels.

The speed of this wave is not simple: the waveguide comprising the conductors with freespace between them have a system of eigenmodes with different speeds that depend on the cross sectional geometry. But the lowest order mode, the TEM mode (see footnote) for a two conductor system, indeed runs at speed $c$ if the conductors are steeped in free space. The disturbance comprises eigenmodes that travel at speeds $c$ and less than $c$.

As this disturbance travels, the EM field acts on the conduction electrons in the wires. They begin to move. They can't move at $c$ as you rightly point out because they have nonzero rest mass. As they accelerate, they generate their own radiation: new electromagnetic waves that travel down the conductors as a system of modes with speeds $c$ and less.

Meanwhile, the freespace wave reaches the resistor. If the resistor is not matched to the waveguide characteristic impedance, a reflected wave begins to come back, comprising eigenmodes at speeds travelling $c$ and less.

And all these waves bouncing back and forth move the electrons which radiate their own reaction fields. We get, very fleetingly, a hugely complicated system of waves bouncing back and forth,coming from both the original closing of the switch as well as radiation from the accelerated electrons.

Eventually these transients settle down with the electrons moving uniformly through the conductor at their drift velocity set by the potential difference $V$. This is MUCH slower than $c$. There are no more bouncing waves: the electromagnetic field is now static and the electrons drift in their boring old constant speed pattern around the circuit.

But, because the initial transient includes the TEM eigenmode that travels at $c$, the first motion of electrons at the resistor end of the system indeed begin to move a time $L/c$ after the switch is closed, where $L$ is the $z$-direction distance along the conductors from the battery to the resistor.

Footnote: TEM ("Transverse Electromagnetic") modes arise in two conductor waveguides, i.e. those with a froward and backward "return" path, i.e. a system that can be thought of as a closed circuit. For conductors in freespace, they always have speed $c$ and, curiously, the cross sectional EM field configuration is the same as an electrostatic / magneto static configuration, but this cross sectional static variation is subject to an aplitude change that moves as a wave, as i explain in my answer here and also here.

  • $\begingroup$ So in your explanation one thing that I note is that electrons close to terminal are accelerating in +Z direction(when we close switch) and their EM waves are pushing the electrons next to it with with a force. But it seems a different idea from charge being setup on surface to force the electrons in +Z direction. But the quickest way signal that can reach is by your way, electrons pushing electrons in front of it(+Z direction). charge being setup vertically from flow of wave seems to be a bit slower for large thickness of wires (charges that setup on surface)...... $\endgroup$ Jun 9 at 15:33
  • $\begingroup$ The electrons do not need to push to propagate waves. Wave radiate in freespace. But it is the very swift motion of the electrons to a configuration that cancels the tangential $E$ and $H$ fields that give the waveguide its characteristic guided wave wave motion, including the TEM wave modes. In practice, there is a delay, but in a thought experiment it can be made arbitrarily small by using a small diameter conductor with an arbitrarily high conductivity. This means that the electrons hardly have to move at all to establish the zero tangential field boundary condition. In practice .... $\endgroup$ Jun 10 at 10:00
  • $\begingroup$ ... there is a small delay, and it shows itself in the propagation constant for the modes: it includes the finite conductivity of the conductor. It's very involved, but one can play around with the full expressions including this delay to see that it can in practice be very close to the $c$ ideal for the TEM mode except at the very highest frequencies for even practicable materials. Also, in these idealized cases, the reaction radiation from the moving electrons is actually very small compared to the primary TEM mode field that they establish. $\endgroup$ Jun 10 at 10:10
  • $\begingroup$ @PredakingAskboss See my comments above $\endgroup$ Jun 10 at 12:56

The electrons themselves aren't moving that fast, there is an electromagnetic field that makes electrons move in a certain direction within a conductor. It can be helpful to think of it as a pressure.

It's not equivalent, but think if you hook up a hose and fill it with water. Then you pump more water into one end. The pressure will make water come out the other end very quickly, but he water molecules that come out certainly aren't the same ones you pumped into the other end.

  • $\begingroup$ one question for you! What is the comparative flow of water when two similar pumps are added in parallel than one pump? $\endgroup$ Jun 8 at 2:04
  • $\begingroup$ I'm not sure, it's not a perfect analogy by any means, but I think batteries are similar to water pumps in that parallel pumps increase the flow rate (current) and pumps in series increase the head (like added pressure, or voltage). $\endgroup$ Jun 8 at 2:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.