Consider two circuits containing a battery, bulb, switch and conducting wires but of length 1 meter in one case and 1000KM in other. When switched on at the same time both the bulbs glow instantaneously. I think it is because, electric filed travels from positive terminal of battery to negative terminal through wire almost instantaneously.

My question is there any speed for this electric field? If yes, what is it? By any modification in this circuit is it possible to induce any delay in glowing the bulb?

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    $\begingroup$ possible duplicate of Is electricity instantaneous? $\endgroup$ – ptomato Dec 8 '10 at 9:51
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    $\begingroup$ I like to imagine your wire as a river. Or a tube of fuzzy cloth balls. Your battery is the force that pushes the water, or the balls. When you start pushing it, they push on each other and eventually flow out the other side. But this isn't instantaneous -- one ball pushes on the next which pushes on the next...it seems instantaneous at short distances, but once you have a long one, the delay is noticeable. $\endgroup$ – Justin L. Dec 8 '10 at 22:59

There are two different questions at work here, that you've kind of mashed together. The first question is "What is the speed at which a change in the electric field propagates?" The answer to that is the speed of light. In QED terms, the electromagnetic interaction that we see as the electric field is mediated by photons, so any change in an established field (say, due to shifting the position of the charge creating the field) won't be felt by a distant object until enough time has passed for a photon from the source to make it to the observation point.

The second question is "What is the speed of propagation of electric current?" This speed is slower than the speed of light, but still on about that order of magnitude-- the exact value depends a little on the arrangement of wires and so on, but you won't be far off if you assume that electrical signals propagate down a cable at the speed of light.

This relates to electric field in that the charge moving through a circuit to light a light bulb has to be driven by some electric field, so you can reasonably ask how that field is established, and how much time it takes. Qualitatively, the necessary field is established by excess charge on the surface of the wires, with the surface charge being generally positive near the positive terminal of a battery and generally negative near the negative terminal, and dropping off smoothly from one to the other so that the electric field is more or less piecewise constant (that is, the field is the same everywhere inside a wire, and the field is the same everywhere inside a resistor, but the two field values are not the same).

When the circuit is first connected, there is a rapid redistribution of the charge on the surface of the wires which establishes the surface charge gradients that drive the steady-state current that will eventually do whatever it is you want it to do. The time required to establish the gradients and settle in to the steady-state condition is very fast, most likely on the order of nanoseconds for a normal circuit.

There's a good discussion of the business of how, exactly, charges get moved around to drive a current in the textbook that we use for our introductory classes, Matter and Interactions, by Chabay and Sherwood. It doesn't go into enough detail to let you calculate the relevant times directly, but it lays out the basic science pretty well.

(It's a textbook for a first-year introductory physics class, so it sweeps a lot of condensed matter physics under the metaphorical rug-- there's no discussion of band structure or surface modes, or any of that. It's fairly solid conceptually, though, at least according to colleagues who know more about those fields than I do.)


Yes, the relativity tell you that no information can travel faster than speed of light. That is the disturbance of electric field can only travel slower than the speed of light. The typically speed is around few tens percentage of speed of light for a copper wire.

You find the response almost instantaneous because you use only a short wire, say 1 m. In this case, it responses with a time delay of the order $10^{-8}$ s in which you certainly cannot feel any time delay because you cannot tell difference for events shorter than $10^{-1}$ s as a human.

If you want to introduce delay, the easiest way is to use a longer wire. By wrapping a long wire, say, 10 km, but you still cannot feel any different since the delay is still around $10^{-4}$ s. However, you may use device to measure this delay. You cannot use a very long wire, say 1000 km, because the electric field will be vanishing small for you to do the experiment.


Electric field is mediated by photons, photons are light, information of a change in the electric field thus travel at lightspeed.

This doesnt mean your bulb is gonna light that fast. For that to happend there has to be electrons moving through it, these travel much slower through the wire, of the order 0.01 cm/s IIRC. However since the photons travel at lightspeed with information that the electric field has changed the electrons near the bulb start to move once the photons has reached them.

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    $\begingroup$ Huh? So even the one-meter wire would cause the bulb to light 2 hours and 47 minutes after you flip the switch? $\endgroup$ – ptomato Dec 8 '10 at 9:31
  • $\begingroup$ Are you sure that photons travel through the wire? If the current is DC, then the frequency of these photons is 0, and so according to $E=\hbar\omega$ they have no energy? I always thought that it was just a disturbance in the electric field. $\endgroup$ – ptomato Dec 8 '10 at 9:51
  • $\begingroup$ @ptomato: There is no contradiction. A static E-field has no energy flow. $\endgroup$ – unsym Dec 8 '10 at 9:59
  • $\begingroup$ @ptomato, the electrons (or ions) travel at the speed of a few mm/sec. The electric field's change is felt at the speed of light. $\endgroup$ – CMR Jun 25 '11 at 2:06
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    $\begingroup$ @mmc: Yes you are right, the average velocity is the drift velocity, but the average is over many zeros and a few much-largers. But I have a knee-jerk reaction to this sort of thing, because it is repeated endlessly, and it is so easy to correct. $\endgroup$ – Ron Maimon Oct 14 '11 at 1:23

The electric field in a wire is residual (only strong enough for electrons to overcome the wire resistivity and its energy is tranformed in heat creating Joule´s loss- this meaning that the better a conductor is the smaller this electric field is). The flow of "useful" energy happens outside the wire (as strange as this may seem, but easily verifiable by means of a Poynting diagram) with a rotational magnetic field around the wire and an electric field normal to the wire - thus an EM field (the wire acting merely as an waveguide or a duct). I have no concrete values but i would say that the propagation speed of this EM field depends on the physical properties of the medium, varying with them.

  • We must keep in mind that the speed of an EM field depends on the medium in which it flows - if we have the wire submersed in water its propagation speed is slower than the speed of light in the vacuum.

We have been taught as drawing an electron in the center of the page and straight lines emerging out of it known as electric field..so we have a common idea that they don't travel, and they exist throughout it's length. But it isn't.

Electro-Magnetic waves are propagated due to oscillating electrons.when electron is static(assume it to be) the electric field will be a straight line. When it oscillates it behaves like a string that is oscillating like a sin wave. Which inturns produce magnetic field due to induction and electromagnetic wave that is light is born. So the change in electron position is not reflected in it's electric field instantaneously throughout the space. As we know that em waves travel with a speed of light, that means the change in electric field travels with light speed, so is the electric field(as well as the magnetic field). So the ans is electric field travel at the speed of light.

  • $\begingroup$ In a vacuum, you're right. Inside a metal wire, it is different, as the other answers and comments are discussing. $\endgroup$ – Mike Dunlavey Dec 26 '13 at 17:08

protected by Qmechanic Dec 26 '13 at 20:42


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