As I understand (I'm new), voltage is always relative. From the negative side of metal to the positive side, there is some voltage- but the negative side does not have a voltage alone. Why is it then that people say electrons flow from the higher voltage to the lower voltage?

  • \$\begingroup\$ because the higher is higher than the lower, relatively and absolute. \$\endgroup\$
    – PlasmaHH
    Dec 1, 2016 at 21:38
  • \$\begingroup\$ Yes. In fact, voltage is sometimes referred to as "potential difference", giving a much stronger indication that it's relative. At the same time, it's pretty routine to designate some part of a circuit as "ground", and everything else is measured relative to ground. \$\endgroup\$ Dec 1, 2016 at 22:02
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    \$\begingroup\$ Since electrons have negative charge, they'll actually tend to flow from the lower voltage to the higher voltage. \$\endgroup\$
    – The Photon
    Dec 1, 2016 at 23:54
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    \$\begingroup\$ Voltage is like altitude in a g-field. Rocks roll from higher altitudes to lower, yet length is relative, and one point on a hill cannot "have an altitude" (one point has many altitudes at the same time: altitude measured from sea level, measured from local countryside, measured from bottom of ocean trench (which our rolling rock could conceivably reach!)) We must measure altitude between two points. Same with voltage. It's because e-fields are Volts-per-Meter. Your battery terminal has many voltages at the same time, so one terminal can't "have voltage." Voltage needs two points. \$\endgroup\$
    – wbeaty
    Dec 4, 2016 at 23:12

5 Answers 5


wbeaty's response brings up the fact that reality is complex and that we need methods to simplify that reality and reduce complexity in order to get down to the everyday business of engineering things that manage what's important to us and will just work relatively consistently. Despite the fact that reality may be very much more complex than the workable tools we apply to think about it.

I'm going to take a stab at your question from a completely different perspective than wbeaty's, though. One that is based upon ideas that are a little closer to reality (though we cannot ever say that any of our theories actually represent reality -- that is forever beyond our reach -- consider Plato's cave allegory.)

But I'll also try and keep it understandable, too, and not just go off on a litany of terms and ideas with no explanation of them.

Matter is essentially neutral once you look at it from a distance away of more than a few atomic diameters. This fact is a testament to the incredible forces that begin to be applied when you separate charges from each other.

Now add the idea of conduction band electrons. These are electrons that would normally just orbit their neutral atom. But because of thermal agitation and the exact kind of atom and nearby atoms, can be kicked "up" in such a way that they are more easily mobile, but where they are also still near the atom and nearby atoms who "lost" them. So the matter stays neutral, as the conduction band electrons don't actually leave and go somewhere. It's just that there is a kind of thermally agitated "dust cloud" of electrons very close to their atoms but just far enough away that they can be accelerated by even a very small electric field gradient.

That's the setup. Matter is essentially neutral. But some of the electrons can be thermally agitated and because of this can respond to very modest gradients in the electric field. (Lightning is one of those tremendous results that can happen when charges get separated.)

In the case of copper for example, and at room temperatures we usually experience, the number of these conduction band electrons is about \$8\times 10^{28}\$ electrons per cubic meter, or about \$10^{23}\$ electrons per milliliter of copper. And because all these charges repel each other and can move so freely as well, they will spread out fairly uniformly throughout the material. The density of conduction electrons in copper will be uniform, in other words.

Imagine a very simple circuit that includes a battery and wires that connect it to a resistor. It's a simple circuit:


simulate this circuit – Schematic created using CircuitLab

However, let's assume for purposes here that the resistor is also mounted in such a way that it is initially parallel in orientation to the battery (as shown in the above diagram.) I want to point out that this schematic is both behavioral as well as physical, for the moment.

Electrons will flow here. From the negative terminal to the positive terminal (and we might also assume inside the battery as well.) It's a circuit. How are electrons motivated to move, though? (They must be, since they are moving.)

For conductors, it must from differences in charge balance in matter leading to a gradient. But somehow this also means that something must not be as neutral as something else.

The Coulomb force law has a very simple formula:

$$\vert\vec{F}\vert = \frac{1}{4 \pi\cdot \varepsilon_0}\cdot\frac{\vert Q_1 Q_2\vert}{r^2}$$

\$Q_1\$ and \$Q_2\$ would be the net charge balance of matter at two locations in space separated by a distance of \$r\$.

Look at the above circuit. I've placed the resistor "far away" from the battery. If we assume that the charge differences exist at the battery ends, then it would logically follow that if we brought the resistor closer to the battery then more current would flow because clearly the distance \$r\$ is now much less. But we don't observe this. It doesn't happen that way.

We might also consider rotating the resistor so that one end is closer and one end further away. We might expect a change in current from this. But that also doesn't happen.

At least some of the charge balance differences must not be at the battery, itself, we can conclude. But where?

As a current flows through a wire, each electron must move in exactly the right direction to flow along the wire. But the above equation is, without the absolute value signs shown there, is a vector. So the direction an electron takes will be pointing in some fashion. But wires bend, make loops, etc. How do the electrons follow around all these changes?

Earlier, I said that the electrons were spread out uniformly in copper. But it turns out that in the above circuit there will be a slight difference in charges at the surface of the wire or conductor. Just enough, in fact, to motivate nearby electrons to move (which means a current.) If you move along a wire from one point to another, there will be a tiny difference in the charge balance between them ... just enough in fact to cause the observed current. Here's an example using a very tiny segment of wire to show the slight differences on the surface of the wire:

wire charges

That answers the where question asked above.

How much charge is enough? Well, to get the idea across of just how small the charge difference is, imagine bending a wire so that it forms a bend as in the following image:

bent wire

For such bends, there is an accelerating force required to cause all those electrons in the current to also make that same turn. The force is \$\frac{m_e\cdot v^2}{r}\$, where \$r\$ here is the radius of the bend and \$m_e\$ is the mass of an electron and \$v\$ is the mean velocity of the electrons. To achieve this force, a SINGLE EXTRA ELECTRON at the surface of the bend can be enough!! Seriously! It can be as little as that.

This again emphasizes the sheer magnitude of the forces we are talking about when bringing in electric charge repulsion or attraction!

There is a video worth watching here that will demonstrate these slight charge balance differences that set themselves up on a wire in order to motivate a current. It takes a very large voltage between two ends in order to create enough charge to be easily demonstrated. So the Weismann Institute has to use a high voltage for their demonstration. Smaller voltages yield such tiny charge balance differences that detection is very difficult.

Reality is complex. And a real circuit is likewise going to mean a very complex arrangement of these surface charges in order that the current will seem to mysteriously follow all the twists and curves. The details are sufficient so that currents of electrons entering a node will also divide up correctly at a branching point in a circuit. If you get into the tiny details, it all still works. But then it is more physics and less electronics.

The reality is that the charge balance differences do set themselves up, and quite quickly (order of nanoseconds) in order to cause all the right motions and direction changes.

Now you realize that there are actual classical physical forces operating to make circuits work. But the details are myriad and complex and electric and electronics engineers don't go around worrying about any of that (most of the time.) Somehow, they just design.

Luckily, it turns out that we can avoid knowing about all those details almost all the time. Sure, varying charge balances in matter do form at the surface of conductors. But a designer is usually dealing with such small voltages in a circuit, and/or such large number of electrons, that they can completely ignore those details. Only very high voltage engineers worry about it (insulators are exposed on their insides to these charges and are, in fact, the weak point to be watched.) Also, these tiny differences automatically set themselves up very quickly, and do so in just the right way, that none of it matters at the emergent scale we live in and you can simply and only examine things from a node perspective without getting mired in such unhelpful details.

At a node, the important things to know are the assignment of some quantity called voltage (energy per unit Coulomb of charge) and that all the current entering a node (in equilibrium, which is the usual case for conductors) must equal the current exiting that node (nodes don't accumulate charge.)

At this heightened viewpoint, eliminating a world of unhelpful detail just works right.

Now, a potential difference is actually an integral along the wire of all those infinitesimal charge balance differences between two nodes. And one could just write down all the individual potential differences between all the nodes. But this would be a rat's nest of differences and the circuit would, once again, still be rather complex-looking. Lots and lots of potential differences along myriad point-to-point "edges" or connections.

It turns out that it is always easier to get rid of all that mess (our minds don't like it) and to turn it into a clean "node voltage" given at each node. You get the same information from that. No difference. But it's a LOT easier to read and understand.

But there is one remaining problem here. Before, you had potential differences between each node pair. And those are all relative values. But when you assign voltages to nodes, those numbers are absolute, not relative. So you need to make one node special in your mind. By convention, the convenient value of 0 is assigned to this node and often referred to by ground or the phrase reference node. Now, it's just fine to give all the other nodes definite, time-varying magnitudes relative to that reference node.

It just makes things a lot easier to handle.


If we fall back to the original physics, we see that Electrical Potentials (voltage) are one way to describe an e-field.

The second method is to use flux lines rather than voltage. So, in the language of field lines, the flux accelerates the charged particles. In the language of Potentials, the charged particles only experience a force if they're positioned among a set of differing Potentials (differing voltages) spread across space. The stacked voltage-planes are the field, just as the flux lines are the field.

It's just two ways to describe the same invisible thing: a dense bunch of flux lines, versus a dense parallel stack of Equipotential planes (the voltage in space.) And, the flux lines are always perpendicular to the voltage-planes.

For example, a single electron has a radial pattern of flux lines spreading out from the particle, but it's also surrounded by concentric onion layers made of voltage-spheres; "equipotential balloons," with the electron in the center of the onion. The e-fields' perpendicular flux-lines stab through the balloons. (Heh, and the actual voltage at the electron's 'surface' is 511 kilovolts, measured relative to a distant "balloon" at infinity. Same value in KeV as the energy released if we destroy an electron.)

Also note that voltages (potentials) are like altitudes: to have force we need a slope, and to have a slope, we need a set of differing altitudes, not just one altitude. Without a voltage-slope (a potential difference, distributed across a certain distance,) the electrons won't feel any force. Note that e-fields are measured in terms of volts/meter. So, the charged rock is rolling down the hill made of differing "voltage-altitudes" or "Potentials," spread out across a certain length of space. The strength of an e-field comes from a pair of relative things: a voltage-difference, divided by location-difference.

Yet the field itself, the thing describe by voltage ...that's just a thing itself, a "field" in empty space. The lines-of-flux and the stack-of-equipotentials aren't real. They're an array of numbers hanging in space. But they certainly help us imagine what e-fields look like and how they behave.

  • \$\begingroup\$ "the actual voltage at the electron's 'surface' is 511 kilovolts" - is this potential from the Classical electron radius [en.wikipedia.org/wiki/Classical_electron_radius] or somewhere else? I rarely hear modern ideas about an electron shape, radius, or surface. \$\endgroup\$ Dec 4, 2016 at 1:09
  • \$\begingroup\$ @Tom Anderson, yep, iirc I got that from Feynman Lectures; assume electron rest-mass is entirely due to field-energy outside the particle. 511KeV is the annihilation signature. Of course voltage itself (e-field) is a Classical concept. Yep, there it is, 28-1 and 28-3 feynmanlectures.caltech.edu/II_28.html \$\endgroup\$
    – wbeaty
    Dec 4, 2016 at 1:44

While voltage is actually as complex as they make it out to be, for a new learner, it doesn't need to be that complex. I'll try to be a little more concise and clear:

I like to think of it metaphorically as pressure. Specifically as air pressure. Pressure is all relative. It just so happens that we usually measure pressure relative to a vacuum (for atmospheric purposes, at least).

Think about an air compressor, if you empty it, so there is no difference in pressure between the air inside the compressor and outside the compressor, the pressure valve on the compressor will read '0'. When there is no pressure difference between the atmosphere and the inside of the tank, no air will flow.

If you were to take a tank pressurized at 100 psi (~690 kPa) into space (a vacuum) the gauge on the tank would then read approximately 114 psi (786 kPa), because there was a change in the relative pressure.

Similarly, if you were to take a tank pressurized to 100 psi (~690 kPa) to 1000 ft (~300 m) below the surface of the ocean, the pressure difference would be 341 psi (~3310 kPa).

In either case, if you allow a path for the pressure to equalize, air molecules will flow to equalize the pressure.

With a voltage source, there is simply something keeping that 'pressure' up so there can be a constant flow.

  • \$\begingroup\$ Would it sound right for me to say that "Your answer is ugly short winded and compressing" ? If you consider your answer is superior you can say so at the start if you must (it would probably not be well received" - but up front criticising other answers is probably unwise. IMHO of course :-) \$\endgroup\$
    – Russell McMahon
    Dec 2, 2016 at 21:58
  • \$\begingroup\$ @RussellMcMahon, you know what, you're right, when I first wrote that line in my head, it sounded a lot different. Edited to remove. \$\endgroup\$ Dec 2, 2016 at 22:13
  • \$\begingroup\$ Yep, voltage is a lot like pressure. But it's more like "dam head" in hydraulics: a water-height rather than pressure. Even better: we can pretend that voltage is like pressure, but then show all the ways that voltage is not pressure. electronics.stackexchange.com/questions/29116/… \$\endgroup\$
    – wbeaty
    Dec 4, 2016 at 2:12

By definition you can never have a single isolated point at a specified voltage. This is because voltage is by definition "potential difference".
Because it is defined as a difference it IS a difference.

By analogy -

(1) Dates are always relative to some essentially arbitrary starting point. The Gregorian & Julian calendars measures duration from 4 years after the birth of Jesus Christ. Notionally it's from the year of Christ's bit\rth BUT Dionysius got it wrong by 4 years (or is that 3 ? :-) ). We don't have much trouble understanding what is meant when people specify that eg Queen Elizabeh II was crowned in 1953. We assume the "relative to 4 years after the birth of Christ" part.
We tend to get a bit stymied if we find the years are "out" by +3226 or +3102 , -284, any of -80, -640, -691, -545, -78 or -638 unless you are Hindu (Krisna ascencion/ incarnation), Diocletian historian some flavour of Buddhist, or ... .

(2) Imagine that we eg decided on some measure of the height of mountains by measuring them relative to mean sea level (as we in fact do).
If we say that Sagarmāthā is 29,029 feet high we are actually ALWAYS including the unspoken phrase "above means sea level". We COULD express the "height" as the distance from the height above the Greenwich meridian at Greenwich or above the level of the surface of Yamhamelah (aka Lake Asphaltitas) (which would not be a bad idea apart from that level being rather variable). If we did the latter then the height of Chomolungma would be about 30,436 feet - but would then always include the unstated qualifier "... above the mean surface level of Bahr al-Mayyit".

And, just as I've used 5 terms for 2 actual locations above (Everest, Dead Sea), the value can be other than expected if the relative reference is not clearly understood.
A "9V battery" measures 9V from the +ve to -ve terminal. And/but a microwave oven's magnetron filament voltage is typically ~= 3.3 VAC. BUT failing to account for the voltage doubled 2.8 kVAC that it 'stands on top of' when measuring the filament voltage "may cause shocking problems".


You MAY be asking re DIRECTION of flow.
Conventional flow was described before the electron was discovered.
They guessed there were +ve "particles" flowing from + to -.
They guessed wrong :-).
So electrons are considered -ve "things" flowing from - to +.
The result is the same.


So, whether Voltage magnitude or polarity or electron flow direction makes sense is all relative :-).


Voltage represents electric potential energy. It's relative in the same way that gravitational potential energy is. Consider the example of a ball on a hill:

Ball on a hill

What's the gravitational potential energy of the ball? It depends on which way you push it. To define the potential energy, you have to define where the bottom is. But no matter how you define it, the ball always rolls downhill. And the ball will always gain more energy if you push it to the right than if you push it to the left.

Voltage is the same way. Instead of a gravitational field, there's an electric field. Instead of gravitational potential, we use voltage. Instead of a ball, there's an electron. You get to pick which point in the circuit counts as ground, but the voltage difference between any two points will always be the same.

  • \$\begingroup\$ But voltage is still there even when the ball (the charge) isn't present. In pre-Faraday/Maxwell era physicists believed that forces were "action-at-a-distance," and refused to believe in fields or field-mediated forces. Many textbooks still promote this discredited theory. They describe voltage as part of potential energy. In fact, potential energy is a part of charge, voltage, and distance. Remove the charge, and the potential energy is gone, yet the voltage and the distance still remain. A proper definition of voltage needs say nothing of PE. But to define PE we need voltage! \$\endgroup\$
    – wbeaty
    Dec 4, 2016 at 1:45
  • \$\begingroup\$ True, but I'm trying to keep this simple. We could define a gravitational equivalent of voltage with units of Joules/kilogram that would work the same way, but the questioner was already thinking in terms of electron flow. I figured using a test mass (analagous to a test charge) would make things easier to understand. It was an analogy, not an exact equivalent. \$\endgroup\$
    – Adam Haun
    Dec 4, 2016 at 2:00
  • \$\begingroup\$ Sorry, I was unclear. Wrong statement: "Instead of gravitational potential energy, we use voltage" it should read "Instead of gravitational potential, we use voltage." What's the nature of voltage? In your diagram voltage is like feet. Altitude is not potential energy, in the same way that voltage is not anything like potential energy. Avoid spreading the common misconception that Potential and Potential Energy are similar: with a heavier ball the PE increases, but the "Voltage" stays the same. Remove the ball, and the PE is zero, but the hill is still there: voltage is the hill. \$\endgroup\$
    – wbeaty
    Dec 4, 2016 at 3:03
  • \$\begingroup\$ What you're saying is correct, but at the questioner's level of understanding, I think it's splitting hairs. In the context of circuit theory, it is perfectly reasonable to treat voltage as normalized electric potential energy. Making voltage out to be hugely different is confusing. It's joules per coulomb. And a static E field and a static G field are very similar -- conservative, inverse square law, etc. Gravitational potential energy is familiar to anyone who's taken any intro to physics. I think it's a useful analogy. I changed the wording on that sentence anyway just to be clear. \$\endgroup\$
    – Adam Haun
    Dec 4, 2016 at 6:43
  • \$\begingroup\$ Not "splitting hairs" at all. The height of a hill is not potential energy. The height of the hill is not"normalized potential energy." Altitude is not "normalized potential energy" (since, repeating again and again, the hill is still there when we remove the ball, and the voltage is still there when we remove the test-charge.) Why do I insist? Simple: because I had no idea what voltage was until I figured out that I had a misconception, and here you are, teaching that misconception to others. Very simple: Voltages are e-fields. Complex: move a test-charge in an e-field to create energy. \$\endgroup\$
    – wbeaty
    Dec 4, 2016 at 22:53

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