Keeping it simple, what is the physics approach to a current in a wire connected to a battery? How does the electric field behave?

I've read somewhere that, contrary to the common conception, electrons don't really flow as water in a hose so, what is really going on there? How is energy transported at the speed of light if electrons don't flow like very fast water?


4 Answers 4


How is energy transported at the speed of light if electrons don't flow like very fast water?

Let me reply with the hydraulic analogy, i.e. with replacing electrical current by water flow.

Open the water tap in your kitchen. Then water comes out instantaneously, although the waterworks feeding the water pipes might be many miles away from your house.

Of course, this is not surprising. Before you opened the water tap, the water was already present in the pipes all the way from the waterworks to the water tap in your kitchen. It is the pressure, not the water, which propagates so fast (theoretically with the speed of sound) through the pipes.

This water scenario above is very much analogous to the electrical scenario.

When you switch on the light in your room, the electrical current through your lamp begins to flow instantaneously, although the electrical power station might be hundreds of miles away from your house.

This is not surprising here, too. The electrons were already present in the wires all the way from from the electrical power station to the switch and the lamp in your house. It is the voltage in the wires and the electromagnetic field around the wires, not the electrons, which propagate so fast (theoretically with the speed of light).

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    $\begingroup$ "theoretically with the speed of light" <- The speed of light in whichever (likely metal or semiconductor) medium it's propagating through, obviously, not the in vacuo $c$, so there's a frequency-dependent refractive index. However, it won't make much difference. $\endgroup$
    – J.G.
    Oct 23, 2019 at 9:35
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    $\begingroup$ In telecommunications, we commonly take it as a zeroth-order approximation that signals propagate over copper wire at $2/3\,c$. This is enough of a difference to be obvious over distances of a few hundred km, but over the distance between a wall switch and a ceiling light, you'd have to set up precision instruments to measure it. $\endgroup$
    – zwol
    Oct 23, 2019 at 12:56

Below is an illustration I made of this, based on a reasonably realistic simulation of a simple DC circuit with a rectangular shape.

DC circuit with Poynting vectors

The energy does not flow in the same direction as the current -- if it did, then just as much energy would flow out of the resistor as into it. Actually, the battery is using up its chemical energy, and the resistor is putting out energy as heat, so the flow of energy from the battery to the resistor is always to the right.

In electromagnetism, we measure the flow of energy using something called the Poynting vector. For example, inside a laser beam, the Poynting vector points in the direction of the beam. The Poynting vector is proportional to $\textbf{E}\times\textbf{B}$, where $\textbf{E}$ is the electric field, $\textbf{B}$ is the magnetic field, and $\times$ is a thing called a vector cross product. The vector cross product is perpendicular to the two vectors being multiplied.

For a wire that's a perfect conductor, the electric field inside the wire is zero, so the Poynting vector vanishes. Therefore no energy flows inside the wire. The flow of energy is outside the wire. The white arrows in the figure show the direction of the Poynting vectors.

How is energy transported at the speed of light if electrons don't flow like very fast water?

The electrons flow more like very slow water, and this is another way of seeing that energy is not transported along with the electrons. If you open or close a switch in a circuit like this, there are changes that ripple outward in the electromagnetic field. Those electromagnetic waves carry the energy and information at speeds that are usually close to the speed of light.

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    $\begingroup$ Science Asylum has a good video on this: youtu.be/C7tQJ42nGno $\endgroup$
    – Wossname
    Oct 22, 2019 at 23:08
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    $\begingroup$ In a coaxial cable, one can calculate the Poynting vector as the fields are only in the cylindrical space between the conductors. If that space is narrow compared to the radius, you easily get that power is $IV$. $\endgroup$
    – user137289
    Oct 23, 2019 at 7:48
  • $\begingroup$ The density of e-field lines should be denser inside the resistor - or else it is a better conductor than the wires... $\endgroup$
    – Stian
    Oct 23, 2019 at 14:33
  • $\begingroup$ Wait, so if you can block the electromagetic fields outside the wire, then no power would be transferred? $\endgroup$
    – Michael
    Oct 23, 2019 at 22:38
  • $\begingroup$ Re "The electrons flow more like very slow water, and this is another way of seeing that energy is not transported along with the electrons.": This seems to contradict Thomas Fritsch's answer. $\endgroup$ Oct 24, 2019 at 11:42

The best way I ever had it explained to me was that if you imagine a swimming pool full to the top with ping pong balls, and you push one extra ball in at the shallow end, another ball will pop out at the deep end (well, it's a simplified logic-problem view). This is how the electron flow works: one electron will leave the negative terminal of the battery (due to the voltage, or potential difference along the length of the circuit) and enter the wiring. This wire is made up of molecules of copper; the electron will strike a molecule and displace another, which will then strike another molecule, following the flow around the circuit. The original electron itself may take a very long time to propagate around the circuit (or that specific ping pong ball you put into the shallow end may take a long time to leave the deep end). But the total effect of one-in-one-out is preserved.

Put another way, the wire is already filled to capacity with electrons (or the pool with balls) before the switch is closed. Adding balls/electrons at one end (almost) instantly pushes one out the other, but the actual electrons themselves can move very slowly from one end to the other.

Going back to the hydraulic analogy, your house is likely connected to a municipal supply (or dam, bore, whatever) through tens, or hundreds of kilometers of pipeline. Inside this pipe is water. With the tap in your house turned off, the water is under pressure (voltage) within the pipe due to the pumps at the supply/elevation of the dam. When you turn the tap on, water instantly begins to flow. However, this is the water (electrons) which were already in the pipe. If the dam/pump added a red dye to the water, it may take hours for this dye to appear out of your tap, as all of the 'clean' water (electrons) just first be removed from the pipe.

That is an extremely generalised explanation and doesn't cover things like valence bands (electrical conductivity of the molecules) or semiconductor effects (molecules becoming positively charged as they lack an electron) but it serves the purpose as far as explanations go. If you'd prefer a more detailed explanation, I'd recommend obtaining a copy of Neamen, DA 2010, Microelectronics: circuit analysis and design, 4th edn, McGraw Hill, London. Which was the undergraduate text I used, and it goes into all the detail you're likely to need about the 'how'.

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    $\begingroup$ When I read the ping pong analogy I immediately thought of the analogy of those coin balance games at carnivals where you aim a coin into the device which already has a bunch of coins lying flat across a surface up to the edge where several coins are already stacked teetering on the edge apparently ready to fall over and into your lap the moment your coin lands and indirectly pushes on them. Of course they rarely if ever do! $\endgroup$
    – Michael
    Oct 23, 2019 at 22:40

The energy transported by electrons that flow through a conducting wire in a DC circuit, is electric potential energy. The wire is attached to a battery pole, and therefore the wire acquires a surface charge which gives the wire an equipotential equal to the battery pole potential. The wire surface gets the same (monopole) voltage 'V' as the battery pole. When the wire forms a circuit (a load and a return wire to the other battery pole), then electrons flow through the wire, perpendicular to the gradient of the wire potential (the radial electric field), such that the electric energy represented by the moving electron (eV) does not change, but is transported only from battery pole to the load. An analogue is the gravitational potential of a mass M at a height 'h' with respect to the Earth surface; the mass moves parallel to the Earth surface from A to B without changing its height: the gravitational potential energy E = M.h.g is transported from A to B. For example, a 9Volt battery has two poles with monopole voltages -4.5V and +4.5V with respect to infinity. If the circuit attached to the battery carries an electric current of 0.5 Ampere, then both wires transport 2.25 Watt from battery to the load, and the load dissipates 4.5 Watt. What happens if one of the battery poles is connected to ground (which is a zero Volt monopole)? Without knowledge about the dynamic properties of surface charge on electric circuits/electronics, one cannot truly understand the physics of Electrical Engineering applications, so it is very strange nothing is taught about surface charge dynamics in EE education. How fast flows surface charge with respect to the drift current in the bulk of conductors (or semi-conductors)?


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