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I was in a question in Electronics SE, and a lot of people said something that I'm sure is wrong. They said, "The voltage between two points that are not part of the same circuit is undefined".

Or in other words, "if I have a circuit that is fully isolated ("floating"), then the voltage between any point in it and the ground, is undefined".

Now, I'm 100% sure that the voltage between any two points in the universe is well defined. It doesn't matter if they are in different circuits, same circuits, or even the same galaxies.

It may be hard or impossible to calculate, but it is not undefined!

This was driving me crazy so I ask, am I wrong? Are they wrong?

A follow up question. If two pieces of the same metal, are under the same external electric fields (which may be zero), and carry the same excess charge (which may be zero), then the voltage between them must be zero, even if one piece is in Jupiter and the other one in an electric circuit in my back yard. Is this wrong?

Juan Perez
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    I take two pieces of metal, no charge, zero applied electric field. One I leave on my work bench. One I place in a conducting sphere at the top of a Van de Graaff accelerator. Each piece continues to have no net charge and zero applied electric field. I now turn on the VdG machine. Without me telling you the voltage on the VdG machine, what is the potential difference between the two pieces of metal? – Jon Custer Nov 04 '21 at 17:59
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    Also remember that the whole concept of the potential difference between two points becomes undefined as soon as you introduce unconfined time-varying magnetic fields to your system. – The Photon Nov 04 '21 at 18:22
  • @JonCuster If the VdG is on then the metal on it will gain charges and feel electric fields, so it will probably have a different potential than the one in your bench. I guess that with enough data one could calculate the exact values. I don't see your point. – Juan Perez Nov 04 '21 at 19:20
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    Would you be happier with the statement "Lacking any additional information, we cannot say what the voltage is between two points that are not part of the same circuit—the potential difference could be any number at all without introducing any inconsistencies."? – Chemomechanics Nov 04 '21 at 19:21
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    @JuanPerez - no, the conducting metal sphere will acquire charge (on the outside), the objects inside it won't. – Jon Custer Nov 04 '21 at 19:23
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    Potential difference can defined everywhere for electrostatics – jensen paull Nov 04 '21 at 19:35
  • @JonCuster Very well. I still don't understand what that says about the potential difference being defined or undefined. – Juan Perez Nov 04 '21 at 19:48
  • Neither bar of metal knows the potential difference between them. It could be zero, it could be a million volts. It could be anything in between and none would be wiser. You can even replace the metals bars with people. Barring additional information, the people just don't know. – Jon Custer Nov 04 '21 at 20:15
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    If you define voltage as the work to be done per unit charge to move a charged particle from one point in space to another point in space, then voltage is never undefined, because work is never undefined. However, it will in general depend on the path the charged particle would take, as in the case of non-negligible time-varying electromagnetic fields. – alejnavab Nov 04 '21 at 21:19
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    As far as us electrical engineers are concerned, the voltage between two unconnected points is undefined, because there's no way you could know it, and it could change at any moment for any number of reasons. The real world is messy, not like the physicist's world of spherical cows. – Hearth Nov 05 '21 at 02:43
  • @jensenpaull we are not talking about PD, we are talking about voltage. Although the units of PD are Volts, they are not the same thing, just like we often talk about weight rather than mass, voltage is the phenomena you can measure using physical systems rather than the theoretical construct. – Pete Kirkham Nov 05 '21 at 13:24
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    What I find confusing about the question, most comments, and most answers is: What is the definition of "undefined"? These comments and answers seem to be referring to things that are "unknowable" or "indeterminate." But that doesn't mean the same thing as "undefined." We may define the meaning of terms (e.g., "potential difference") or the meaning of units (e.g., "volts") but we don't "define" physical quantities. They are what they are, even if we have no way to measure or calculate them. Do engineers define "undefined" in a way that would make these answers/comments make sense? – Syntax Junkie Nov 06 '21 at 02:09

7 Answers7

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The voltage between to points in the universe is always defined. But the voltage between two circuits is undefined because (at least from electronics point of view) circuits are abstract models of closed systems.

Suppose we have two devices that consist of a battery connected a resistor - one resting on the ground - the other suspended from a balloon.

Schematic showing two independent battery-resistor circuits side by side

It would be possible to measure a real potential difference between any two points in those devices. And yet if we model these devices as different circuits any voltage between them cannot be calculated and is undefined because we haven't included any information about how those circuits are related.

What we can do is refine these circuits by including parameters such as resistance, capacitance and inductance of air between the devices. But when we include such information they cease to be two distinct circuits and become a single more complex circuit where voltage between any two points can be calcuated.

Schematic showing the same circuits as before connected by parallel R L C elements

Note: Here I used simple lumped circuit approximation to illustrate the point, but the principle remains the same even for more complex models.

SilentAxe
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  • Can you help me understand this part of your answer: "circuits are abstract models of closed systems." It would make sense to me if it read, "circuit diagrams are abstract models. Is that what you meant? But isn't the original question about actual circuits? – Syntax Junkie Nov 08 '21 at 17:58
  • Circuits themselves are abstractions because everything can and does conduct electricity in one way or another. Air gaps are just weak capacitors or spark with high enough voltage, vacuum is part of a circuit in vacuum tubes and when your cell phone is handshaking it induces current in your speakers. The only way to tell if something is or is not part of particular circuit is whether it is convenient for us to do so, usually because the effects are expected to lower than some arbitrary threshold that we care about. But in nature $\mu A$ and $nA$ are conducted whether we like it or not. – SilentAxe Nov 08 '21 at 21:14
  • @SyntaxJunkie Also related: Is every open circuit a capacitor? and Does alternating current (AC) require a complete circuit? – SilentAxe Nov 08 '21 at 21:26
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Is the voltage ever undefined?

The voltage, or potential difference, between any two points is always defined. It is defined as the work per unit charge required to move the charge between the two points.

"The voltage between two points that are not part of the same circuit is undefined".

It is not defined in the sense that one cannot determine what that voltage is by applying Kirchhoff's voltage laws to two independent circuits. But any voltage that may exist between them is still defined as above.

Or in other words, "if I have a circuit that is fully isolated ("floating"), then the voltage between any point in it and the ground, is undefined".

I look at the question of voltage between a point isolated from ground and ground as a separate one. There is a known electric field between an isolated point in the atmosphere and earth ground. In areas of fair weather, the atmospheric electric field near the surface of the earth's surface typically is about 100 V/m directed vertically in such a sense as to drive positive charges downward to the earth. See:

https://glossary.ametsoc.org/wiki/Atmospheric_electric_field#:~:text=In%20areas%20of%20fair%20weather,charges%20downward%20to%20the%20earth

So if a point in the circuit is 1 meter above the ground, the voltage between it and the ground with air in between would be 100 V. However, due to air being a good insulator, the current density is only about 10 $^{-12}$ amperes per square meter parallel to the earth, so its influence on the circuit would be appear to be negligible. Its effect on you would also be negligible since you are a relative good conductor (relative to the air) so that when you stand on the ground you and the ground become equipotential surfaces. See.

https://www.feynmanlectures.caltech.edu/II_09.html

Regarding your follow up question if the two metal pieces have the same positive or negative charge, the voltage would be zero. If you connected the pieces with a conductor there would be no current in the conductor.

Hope this helps.

Bob D
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    This sparked my curiosity. Can this electric field be used to generate usable electricity? I'm guessing that not, or else things would be a lot easier in our world. But if there is such a potential difference for just 1 meter of height, why can't it be used with large enough antennae or something? – Vilx- Nov 05 '21 at 10:50
  • Read the Feynman link I provided – Bob D Nov 05 '21 at 12:02
  • "The voltage, or potential difference" in electrical engineering these are not the same thing, the answer defines the potential difference then assume that the voltage is defined (the voltage being what is measurable or calculable from Kirchoff's laws) " voltage between a point isolated from ground and ground" there is no such thing, there is a PD, not a measurable voltage. – Pete Kirkham Nov 05 '21 at 13:20
  • @PeteKirkham Re your statement ""The voltage, or potential difference in electrical engineering these are not the same thing" I refer you to the Reference Handbook for the Fundamentals Exam in Electrical and Computer Engineering, published by NCEES (National Council of Examiners for Engineering and Surveying) in the US, I quote: "The potential difference $V$ between two points is the work required per unit charge required to move the charge between the two points". – Bob D Nov 05 '21 at 14:17
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    @BobD Nice answer, but I disagree with the statement "The voltage, or potential difference, between any two points is always defined. It is defined as the work per unit charge required to move the charge between the two points." This is only strictly true in the absence of time varying magnetic fields, because $\nabla \times \mathbf E = -\partial \mathbf B/\partial t$ and conservative fields are irrotational. – ummg Nov 05 '21 at 14:18
  • @ummg Good point. I was answering in terms of the electrostatic potential difference or voltage associated with electric fields , since the OP was asking about voltages associated with electric fields. – Bob D Nov 05 '21 at 17:19
  • @BobD yes, that is the definition of 'potential difference'. What is the definition they give of 'voltage'? That is a different word, which is used almost always used to refer to a PD which can be measured/do work (work is required for a PD to be measured in practice so these amount to the same thing), which requires a circuit. – Pete Kirkham Nov 05 '21 at 17:51
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    @PeteKirkham What I quoted to you was under the heading "Voltage". So it was a description of voltage. Now it's true they didn't use the word "defined", while they used the word defined for the next item under the heading "Current" which read "Electric current $i(t)$ through a surface is defined as the rate of charge transport through that surface". Whether or not they deliberately did not use the term "defined" for voltage is a matter of conjecture. – Bob D Nov 05 '21 at 18:06
  • @BobD well that is different to how it is used in praxis. Big surprise that, real life and textbooks don't always agree! (any quoted definition is not going to change my experience in the industry of how people use the term in conversation) – Pete Kirkham Nov 05 '21 at 20:38
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    @Pete Kirkham For what it’s worth, during the course of 35yrs before retiring I both authored and contributed to many standards (UL, IEC, IEEE) and one of the hardest damn things to do was getting folks to agree on definitions. Even on a definition of the device covered by the standard! – Bob D Nov 05 '21 at 21:10
  • @PeteKirkham And BTW, thanks for your input. I do appreciate it. Bob – Bob D Nov 05 '21 at 22:21
  • @Vilx- Yes, in principle it can be used to generate usable electricity. But in the process of generating that electricity the voltage difference will be reduced. Unless either side is able to hold a lot of charge, or is connecting to something like that, the generation capacity will run out very quickly when it's used. – bdsl Nov 07 '21 at 18:15
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I interpret that statement as "the voltage difference between those two disconnected circuits cannot be solved for using Kirchhoff's laws".

niels nielsen
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I want to elaborate a little bit on what several comments and answers have stated: Undefined from an engineering perspective is different from undefined from the physics perspective.

An electrical engineer wants to control voltages and currents in a circuit to do something useful. What matters is whether they have control over the voltage between two points in the circuit. Another way to phrase this is: can they predict what a voltmeter would read if they connect two points in a circuit? If they cannot, then the voltage is undefined from an engineering perspective because an engineer cannot rely on that voltage to be any particular value. It's irrelevant whether electric potential is actually defined from a physics perspective.

As a concrete example, consider a digital NOT gate. Let's say we work with 5V logic, so passing a 5V signal into the input of a NOT gate gives 0V and passing 0V to the input of the gate gives 5V. Now, what happens to the output if the input of the digital NOT gate was not connected to any part of the circuit? Physically, the input will be at some voltage relative to the circuit ground, but you cannot predict what this voltage is as it will vary depending on the environment. Thus, for an engineer, the input voltage is indeterminate and of no engineering use, hence "undefined". The consequence is that sometimes the output will be 5V, sometimes it will be 0V, and the circuit designer has no control over when this happens, which is highly problematic if you're trying to make the circuit do something useful. Circuit designers take great care to avoid these indeterminate voltages. For example, this is why you have pull-up and pull-down resistors to make sure that the inputs to digital gates are never "floating". Hence, there is need to have vocabulary to talk about whether you have practical control over the potential difference between two parts of a circuit.

user318039
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  • Very good explanation, except for the last part. Electric potential and voltage = difference of potentials is well defined even for inductor with time-varying current, via the Coulomb potential formula. It's just that this potential by itself isn't sufficient information to determine total electric field (because of the induced electric field component). Voltage on perfect inductor is $LdI/dt$, both from physics and engineering perspective. There is no difference between the two perspectives. However maybe electrical engineers are more likely to well understand this than physicists. – Ján Lalinský Nov 05 '21 at 10:55
  • What we have is that $E=−\nabla V−\frac{\partial A}{\partial t}$ and now we have gauge freedom to choose what $V$ means and, importantly, affect what $\Delta V$ can be. We can choose the Coulomb gauge, but we don't have to. Consequently, $\Delta V$ is no longer uniquely defined. But agreed that the explanation is a bit confusing. – user318039 Nov 05 '21 at 14:24
  • In practice we do, it's always the Coulomb potential in AC circuits, even when they include inductors. – Ján Lalinský Nov 05 '21 at 17:37
  • A digital NOT gate implemented in TTL is not really floating in that sense (the effective pull-up resistance is ~5 kΩ). I suppose you mean a MOSFET NOT gate or similar. The time constants may be long (due to capacitances and high resistances), but if it was isolated from external influences/fields, wouldn't it eventually settle (the insulation resistance is not infinite)? Perhaps add an approximate settling time (backed up by a simple calculation) to demonstrate that it is impractical? – Peter Mortensen Nov 06 '21 at 15:14
  • @JánLalinský no. When a variable magnetic field is present voltage is NOT equal to potential difference. In the same way we can decompose the total electric field in the contributions of the coloumbian electric field and the induced electric field, voltage becomes the contribution of the path integral of the coloumbian part (which can be expressed as potential difference) and the path integral of the induced part (which is the contribute of the emf). See for example Popovic "Introductory Electromagnetics", sec 14.4 Potential difference and voltage in a time-varying electric and magnetic field. – Peltio Nov 06 '21 at 16:27
  • @Peltio that book's definition of voltage in 14.7 is a common belief, but it is unfortunate and inconsistent with standard methods of solving AC circuits. Voltage in AC circuits is not a path dependent quantity. In case AC circuit contains a perfect inductor, voltage on this inductor is not given by 14.7. Instead, it is given just by integral of the Coulomb field (difference of potentials). – Ján Lalinský Nov 06 '21 at 22:34
  • @JánLalinský that definition is the IEC definition of voltage. And it is perfectly consistent with the standard methods of solving AC circuits. As The Photon tried to tell you in a post above, you need to limit yourself to lumped circuit theory, where the components can be thought as pointlike so that there is no distance between the terminals. This zero distance implies zero contribution for the emf part in the gap and leaves the electrostatic contribuition only. But if your circuit is not lumped, though luck, you have to use the general definition. See Ramo Whinnery vanDuzer 3e, ch 4 p. 179 – Peltio Nov 06 '21 at 23:40
  • @Peltio it is not consistent with AC circuit theory, because there, voltage on ideal inductor is $LdI/dt$, which you can't obtain as a result of this "IEC definition of voltage", because total electric field in perfect conductor vanishes. You would need to find a special strange path going outside the inductor body for which this IEC definition would give you that result. – Ján Lalinský Nov 07 '21 at 11:17
  • @Peltio this is what that book by Ramo, Whinnery, van Duzer on p. 174 does, but in so doing, they are adding conditions to the IEC definition (use the right special path) so they get back the correct result - the result becomes integral of electrostatic field = potential difference, a definition which we could and should have used right from the start. – Ján Lalinský Nov 07 '21 at 12:04
  • @JánLalinský "total electric field in perfect conductor vanishes". Yes, that is the whole point: in an inductor voltage depends on path (exactly because of the variable magnetic field). Voltage in the conductor is zero, voltage across the terminal's jump is nonzero L di/dt (that is where the L di/dt formula come from: from applying Faraday's law to a closed loop consisting of the coil and the jump at its terminals). No inconsistencies whatsoever: we pretend voltage behaves as a potential difference to save KVL, but this is possible only for lumped circuit elements. – Peltio Nov 07 '21 at 14:49
  • Very good explanation of what an engineer would call undefined, it really clarifies the language for me. I still prefer the answer I selected, but this is very helpful still. – Juan Perez Nov 07 '21 at 15:53
  • @Peltio only "IEC voltage" depends on path, but "voltage on inductor" has a specific meaning in AC circuit theory, which does not depend on any path, it only depends on the position of the two points in space. It has unambiguous value $LdI/dt$, and its magnitude is equal, exactly, to potential difference between its terminals. Also, KVL does not need any saving here, as it is valid exactly in lumped element circuits, whether in original Kirchhoff's formulation that refers to sum of all emfs acting in the closed conductive path, or in the modern formulation using sum of potential drops. – Ján Lalinský Nov 07 '21 at 18:17
  • @JánLalinský the "IEC voltage" is THE voltage and it does depend on path in general, but does NOT depend on path for all paths in a region of space that does not contain the variable magnetic flux region. I suggest you re-read Ramo Whinnery VanDuzer. Or compute the fields and the path integrals yourself. You will see that the path integrals that, together with the gap at the inductor's terminals, do not form a loop around the variable flux, will all give the same result. Therefore, the multivalued voltage can be treated as a single-valued potential difference. But it is not the scalar PD. – Peltio Nov 07 '21 at 23:19
  • to conclude: voltage and (scalar) potential difference are different things when variable magnetic fields are present (IEC, Popovic & Popovic, and also Faria "Electromagnetic Foundations of Electrical Engineering" p.208). In electrostatics they become the same, in magnetoquasistatics you can find region of space where voltage behaves like (but is not) a potential difference but you need to limit the points to regions of space that do not enclose the variable flux (RWvD, Feynman ch 22), and this works for lumped circuits. But in unlumpable circuits you see path dependency (inside inductor). – Peltio Nov 07 '21 at 23:27
  • @Peltio you are completely ignoring my point, let's stop this discussion. – Ján Lalinský Nov 08 '21 at 16:05
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The concept of potential difference, as it is used in circuit theory, is only defined in what is called the lumped circuit approximation. This approximation requires that the extent of the circuit is physically small enough that light-speed delays between the elements of the circuit are insignificant. It also requires that there are no time-varying magnetic fields passing through the wires interconnecting the circuit.

With some effort and hand-waving circuit theory can be extended to accommodate transmission lines (where the light speed delay is not insignificant), but we generally still must pretend (even if it isn't true), that the ground potential is constant between the two ends of the transmission line.

Both of the components of the lumped circuit approximation are invalidated in the case where one part of your circuit is on Earth and one on Jupiter, so you can not expect the potential difference between two points on the two parts of the circuit to be well defined.

The Photon
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  • I suppose we can always calculate the energy required to bring a charge along a path from point A to point B. Right?Then the question becomes, under what circumstances is the energy path-dependent? – Gilbert Nov 04 '21 at 18:35
  • @Gilbert, in the presence of changing magnetic fields, the energy required will depend on the path chosen. Then you can't define a potential difference between the two points. – The Photon Nov 04 '21 at 19:09
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    So, there is an approximation to the value of potential difference, used in circuitry, that is only valid under certain circumstances. But: the actual physical non-approximated value of potential difference, is that also undefined under those circumstances? – Juan Perez Nov 04 '21 at 19:25
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    @JuanPerez, see my reply to Gilbert. The approximation is to assume the potential difference exists at all, even when there are actual differences in the energy required to get from A to B depending on the path. – The Photon Nov 04 '21 at 20:38
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    Electric potential can be always defined via the Coulomb potential formula, whether magnetic fields are present or not. It just isn't always sufficient for determining other things, such as total electric field. – Ján Lalinský Nov 05 '21 at 10:49
  • @JánLalinský, The context of the question was why OP got the answer they did to a question on EE.SE. As an EE, we're certainly hand-wavy about exactly what we mean by the potential, but we surely don't intend the word "potential" to mean the Coulomb potential, because then we'd have to say that important electrical devices like inductors, transformers, and generators produce no potential differences. – The Photon Nov 05 '21 at 14:45
  • @ThePhoton Electric potential in EE (with some exceptions in radiation theory) is always a field that is equivalent to the Coulomb potential field $\int\frac{\rho}{r}dV + C$. There is arbitrariness in the constant $C$ but difference of potential between two points of space is unambiguous. And transformers and generators certainly do produce this kind of potential difference between their terminals, as a natural response to induced/motional EMF in their winding. That's part of the practical reasons of making and using these devices - to produce convenient potential difference. – Ján Lalinský Nov 05 '21 at 17:59
  • @JánLalinský, can you share any references that explain that? The first two pages of google hits I get for "Coulomb potential" don't say anything like that. – The Photon Nov 06 '21 at 00:39
  • @ThePhoton these are basic EM concepts. Any introductory textbook on EM will discuss electric potential, e.p. difference, and books teaching AC circuits will explain its use there (Kirchhoff's voltage law). – Ján Lalinský Nov 06 '21 at 11:56
  • @JánLalinský that the Coulomb potential gives a formula for the potential around an isolated point charge is basic physics. How that relates to the potential generated by a transformer secondary or a generator is not something I've ever seen. – The Photon Nov 06 '21 at 15:17
  • "Coulomb potential" does not always mean only "potential of a point charge", it can also mean "Coulomb potential-integral of any charge distribution in space", the meaning which I used above. In the ideal case of transformer with zero ohmic resistance in wires, total electric field in the wire vanishes, so the induced electric field in the wire has to be completely cancelled by the Coulombic field due to charge distribution on the wires and other bodies nearby. That's why potential difference between two terminals has the same magnitude and opposite direction than induced EMF. – Ján Lalinský Nov 06 '21 at 22:43
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Or in other words, "if I have a circuit that is fully isolated ("floating"), then the voltage between any point in it and the ground, is undefined

Undefined in the mathematical sense? No.
Undefined in the a real world physical measurement sense? Most likely Yes.

But it depends on who you are talking to. An engineer will probably use the term undefined meaning unmeasurable or unknowable.

Floating voltages are in most cases very hard to measure because as soon as you attach a meter to a floating voltage, the resistance of the thing that you are using to measure the voltage changes the 'circuit' (air with it's really high resistance can be part of a circuit) and so a floating voltage could be considered unknowable because there isn't a good way to know what the voltage is.

From a mathematical standpoint, 'no' all the equations and values are defined.

Voltage Spike
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People seem to be overlooking something that I consider pretty important. The voltage is not always well-defined, even in principle. By definition, voltage is defined by $$ V(x) = - \int_P^x \vec{E} \cdot d\vec{x} $$ where $P$ is some chosen reference point. This definition only makes sense if the integral does not depend on the path you take to get between $P$ and $x$. For smooth enough $\vec{E}$, this is equivalent to demanding $\nabla \times \vec{E} = 0$. But Maxwell's equations tell us that $$ \nabla \times \vec{E} = -\frac{\partial B}{\partial t} $$ So whenever there are time varying magnetic fields, we cannot expect to have well-defined voltages, at least not according to the usual definition.

Charles Hudgins
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    Actually, voltage is well defined - if you specify the path. You just can no longer define a potential function to associate to the total electric field. One can decompose the total field into its conservative ans solenoidal components - following Helmoltz - and then associate a potential function to the conservative part alone. Of course that alone is not sufficient to completely characterize the field, and you need to supply the vector potential A as well. – Peltio Nov 06 '21 at 23:45
  • Incidentally, if you just change voltage into potential (and potential difference) into the above answer my objection ceases to exists. Also see https://www.electropedia.org/iev/iev.nsf/display?openform&ievref=121-11-27 and https://www.electropedia.org/iev/iev.nsf/display?openform&ievref=121-11-26 – Peltio Nov 07 '21 at 00:01
  • I think its a safe assumption that there are non-zero time varying magnetic fields everywhere (even if extremely minuscule), and thus by your answer, voltage is undefined everywhere. Or is defined-ness a gradient, from well to ill, as the magnetic fields get more significant? – Juan Perez Nov 07 '21 at 13:36
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    How much the voltage depends on the path of integration can vary. $\nabla \times \vec{E}$ in some sense measures this path-dependence of voltage. Though, as Peltio points out, there is nothing stopping us from defining voltage by $V(x) = - \int_P^x \vec{E} + \frac{\partial \vec{A}}{\partial t}$, which will always be well-defined (up to some choice of gauge), if hard to calculate. I'm not sure what the interpretation or significance of $V(x)$ defined this way would be though. – Charles Hudgins Nov 07 '21 at 14:23