I've always considered voltage to be absolute, i.e. something that is there or is not there. However, the more I think about it, it seems to be more like a delta.

For example, let's say we have a supply that lists its pins as (A) -50V and (B) 0V. If we treat pin (A) as "ground", i.e. as if it were 0V, can we treat pin (B) as +50V?

Another example might be that pin (A) is +10V, and pin B is +25V, so the potential difference is +15V. Can we treat this the same way as if it were 0V and +15V?


4 Answers 4


You are always using/measuring the difference in potential between two points.

There is no absolute zero in voltage (like there is with temperature), although it is common practice to define earth as 0V. This is not absolutely necessary, you can use any potential as reference.

  • 1
    \$\begingroup\$ Of course, Earth is not an equipotential surface... \$\endgroup\$
    – Phil Frost
    Jun 19, 2013 at 19:36
  • 2
    \$\begingroup\$ @PhilFrost Are you saying that in all the earth there is no common ground? :) \$\endgroup\$
    – user3624
    Jun 19, 2013 at 20:07
  • 1
    \$\begingroup\$ It causes many a war sadly. But seriously, if you stand on glass blocks and touch a van der graaf generator although you feel no "tension", what hair you (or I in the main) does start to stand on end... this does appear to be a phenomenon related to non-zero charge and how do we know the moon isn't a million volts higher than the earth in potential - I guess because astronaut's hair didn't stand on end so, does this mean there is a theoretical zero volts? \$\endgroup\$
    – Andy aka
    Jun 19, 2013 at 21:44
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    \$\begingroup\$ @Andyaka charge (coulomb) is not voltage (volt), nor is it electric potential energy (joule). Normally, you have on your body about as much negative charge (electrons) as positive charge (protons). Thus, the electric field around you is basically flat. When you touch the generator, it pumps you full of positive charge. These positive charges make your hair stand up because they want to be as far away from each other as possible. Your voltage relative to Earth doesn't make your hair stand up, except that if you touch Earth, you get your electrons back, and again have no net charge. \$\endgroup\$
    – Phil Frost
    Jun 20, 2013 at 3:03

Practically, voltage is a measurement of a difference between two points. You can think of it only this way, and be a very good engineer. Measuring the difference between two points is easy with a voltmeter, as you no doubt know. This thing you are measuring is usually called voltage but is more properly called electrical potential difference.

But, there is a thing that can be measured in volts that is defined at only one point, and that is the electric field potential. To understand it, you must exit the field of engineering, and enter the field of physics (no pun intended).

Say you have an electron (negative charge) and a proton (positive charge). Naturally, these two will attract, and (so far as I understand it; I'm not a physicist!) this is what keeps electrons stuck to their atomic nuclei.

But, if you can pull these two apart, you get a field between them. You might visualize it like this:

enter image description here

(image source)

These lines represent the force (in our case, the electromotive force) that would be experienced by a charge, were it to be in this field. That is, if you were an infinitesimally small charge in that picture, you would be feel a force pushing you in the direction of the arrows. You can think of the proton as spewing out an invisible fluid, and the electron sucking it in. This invisible fluid acts on other charges like wind.

Here's another way to visualize the same field. The proton is a mountain, and the electron is a valley:

3d field

(image source)

If you are a ball on this field, gravity will do work on you, and you will roll downhill. Except, this isn't a gravity field, so our "ball" is made of "charge", not mass. Of course, if you add any charge to this picture, the field changes. This is also true of gravity fields, except the Earth is so much more massive than the ball you imagine that its effect is negligible. So, imagine that your ball of charge rolling around in this field is infinitesimal.

Now one thing you will notice about this field: as we extend it out to infinity, it becomes flat. The electric field potential at this infinitely distant place is \$0V\$, by definition.

If we want to put a ball on the mountain from infinitely far away, we will have to do work. How much? Well, it depends on two things: how high we want to push it, and how big the ball is. A big ball takes more work. Pushing it higher takes more work.

One way to define the volt is joules (energy, work) per coulomb (charge):

$$ V = \frac{J}{C} $$

So you can think of it this way: if you had a ball of charge that was 1 coulomb big, and you did 1 joule of work pushing it uphill, you are one volt high. Or, if you have a 1 coulomb ball of charge, and you let it roll downhill into the electron, and stop it after 1 joule of work has been done, you are at -1 volt. If your ball was 2 coulombs big, then the work is doubled, but it's still just 1 volt.

Thus, you can pick any point in this field, and get its electric potential. It's how much work could be done, or has been done, per unit of charge, getting to there from infinitely far away. With our hill and valley analogy, electric potential is analogous to the elevation.

When you stick your probes on two points, you are asking the question:

If I let a ball of charge that is 1 coulomb big roll between these points, how many joules of work will be done on it?

Of course, we can't get infinitely far away from all charge in the universe, so we can't actually measure electric field potential directly with a multimeter. We can only measure electric potential difference. But, we can calculate the electric field potential, if we know where the charges in a system are.

Since we aren't infinitely far away from all charge in the universe, there is necessarily some electric field potential everywhere. But, we can't do work with just potential; we need a difference. You can't do any work with a ball on a mountain unless you can roll it off.

  • \$\begingroup\$ "These lines represent the force (in our case, the electromotive force)" Wrong. Not EMF; that isn't even a force. Those lines represent the direction of electric force that a possitive charge would experience if placed inside that field. And yes, it's just a convention; we could say the lines will represent the electric force experienced by negative charges, although we'd have to reverse the arrows. \$\endgroup\$
    – alejnavab
    Feb 16, 2020 at 18:10
  • \$\begingroup\$ @AlejandroNava I don't understand what you are complaining about. You call me wrong for saying the arrows represent force, and then say "those lines represent the direction of electric force". Seems you just said the same thing? \$\endgroup\$
    – Phil Frost
    Feb 17, 2020 at 23:15
  • \$\begingroup\$ I complained because you said the lines represent EMF, which is not true; I didn't complain because you said the lines represent force (which you actually never said).Then I said EMF is not a force (not according to the definition of force used in physics, i.e. a vector.) And just to clarify, I then said what the lines really represent. The lines are used for the electric field, from which you can know the direction of electric force given the sign of a charged object. \$\endgroup\$
    – alejnavab
    Feb 19, 2020 at 1:28
  • \$\begingroup\$ By EMF do you mean electromagnetic field or electromagnetic force? It can be either. Not that it matters much, since the electromagnetic field is a field of force, right? \$\endgroup\$
    – Phil Frost
    Feb 19, 2020 at 17:31
  • \$\begingroup\$ By EMF I mean electromotive force, hehe. Yes, in the end the electromacnetic field exerts electric + magnetic forces. \$\endgroup\$
    – alejnavab
    Feb 21, 2020 at 5:41

According to the first three sentences in the Wikipedia entry for Volt:

A single volt is defined as the difference in electric potential across a wire when an electric current of one ampere dissipates one watt of power. It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it.

3 sentences. 3 times the word "difference" is used.

  • \$\begingroup\$ I didn't find that quote clear at all. The first sentence makes sense to me, but didn't seem to answer my question. It's the difference in EP, but is EP the same thing as voltage? In my experience the difference between the differences is usually a derivative. If I don't know how EP relates to my supply pins, I can't apply this. The rest just seemed like alternative definitions from a physics perspective, which doesn't have any meaningful bearing on my question. When was the last time you bodged together some old bits of kit and created two parallel infinite planes spaced one meter apart? \$\endgroup\$
    – Polynomial
    Jun 19, 2013 at 19:39
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    \$\begingroup\$ @Polynomial What we colloquially call voltage is electric potential difference. It is the difference in electric potential between two given points. Both are measured in volts. \$\endgroup\$
    – Phil Frost
    Jun 20, 2013 at 3:10

Spot on. Voltage is a measurement of the electric potential difference between two points.


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