Is voltage and electric potential actually the same thing? If not, why?

Question

Before I get roasted in the comments, let me say I recently got a BS in electrical engineering, so I’m not a newbie in these concepts. I’ve studied electric potential energy, electric potential, voltage (a.k.a. electric potential difference, electric tension, electric pressure), time-varying/non-conservative electromagnetic fields, Maxwell’s equations, etc. I’ve read textbooks on university physics, on electromagnetic theory, on circuit analysis/theory. (But I haven’t studied quantum mechanics or relativity.)

Before explaining why I think voltage and electric potential actually describe the same physical phenomena/process, first I’ll briefly recall certain facts about the two, so that you can see I have some standard understanding on these concepts.

Electric potential

• Electric potential is the electric potential energy, per unit charge, at a point in space. Recall that electric potential energy is a type of potential energy (i.e. energy that a particle has by virtue of its position in space or in a field, it is energy that could be used to do work), associated with the position of a charged particle.

• Electric potential is defined at a point in space, provided we’ve previously defined the zero electric potential point (usually Earth ground or a point infinitely far from the region of space under study). Since the potential is defined at any point in space (at least for conservative electric fields), and since it is a scalar value at each point, then we say the electric potential is a scalar field. It makes perfect sense. I can say “the electric potential at point $P_0$ is $\phi_0$”.

• We sometimes use the electric potential (a scalar field) to calculate a static electric field (a vector field) as the negative of the gradient of the electric potential, because the former is easier to calculate than the latter.

• Electric potential is the quantity in which Laplace's equation and Poisson's equation are described.

• We can plot the electric potential as a 2D or 3D scalar field.

• We can write Maxwell’s equations in terms of the electric potential and the magnetic vector potential.

• Physicists use the term electric potential more often than voltage.

• Etc.

Voltage or electric potential difference

• Voltage is defined as the work to be done (or energy to be transferred), per unit charge, to move a charged particle with unit charge, from one point in space to another point in space, along some path or trajectory. So voltage is a quantity between two points, while electric potential is a quantity at a single point.

• In the presence of conservative electric fields only, voltage can also be calculated as the difference of the electric potential at the two points, thus the name electric potential difference.

• Voltage is usually thought of as a scalar, without being a (scalar) field like electric potential. It makes perfect sense. I can say “the voltage between point (or node in the context of circuits) $a$ and point $b$ is $V_{ab}$”. But it wouldn’t make sense to say “the voltage at point $c$ is $V_c$”, because we’re not specifying with respect to which other point we’re measuring it, unless it is implicitly obvious.

• Voltage (and current), not electric potential, is the quantity in which Ohm’s law, the voltage-current relationship for inductors, capacitors, and diodes, are written.

• Voltage, not electric potential, is the quantity in which Kirchhoff’s voltage law is described.

• In the presence of conservative electromagnetic fields only (zero currents [electrostatics] or constant currents [magnetostatics]), the work to be done between two points is independent of any of the possible paths connecting the two points, and so the work to be done in moving a charge around a closed loop is zero, and since voltage is work per charge, then voltage is also independent of path in such case. But in the presence of non-conservative electromagnetic fields (time-varying currents [electrodynamics], for example AC circuits, with non-negligible leakage electromagnetic fields outside circuit elements/devices such as inductors), the work to be done in moving a charge from one point to another does depend on the path, and so the work to be done in moving a charge around a closed loop is in general not zero, and so voltage also depends on the path.

• Electrical/electronics engineers use the term voltage more often than electric potential.

• Etc.

Why I think voltage and electric potential are the same

Okay, now to my question. I recently had an online discussion with someone, where I said voltage and electric potential were not the same, while the person said they were. I explained them in a similar manner I just did above. But after talking and me thinking, I think those two quantities are really describing the same thing. Below I’ll try to convince you, or at least explain you why to me those quantities seem the same.

Suppose I choose one point in space, which I’ll call the reference point, with respect to which I measure the voltage at all other points in space. Isn’t that, then, the same as electric potential? For example, I could choose the reference point to be the Earth ground or a point infinitely far from us, thus the voltage would seem to have the same meaning as electric potential.

And since I’ve chosen one reference point for voltage measuring, then at each point in space the voltage has a certain scalar value. Thus voltage is now a scalar field, like electric potential. So, we can also compute a conservative electric field from this voltage “field”, and we can write Maxwell’s equations in terms of this voltage “field” (and the magnetic vector potential), just like we could with the electric potential.

And this idea of choosing a reference point for voltage measuring is not uncommon. It is widely used in a method of circuit analysis known as nodal analysis; all or most circuit simulators use that method; it is also used in power systems analysis to obtain the admittance matrix that mathematically describes the electrical behavior of a power system; even using oscilloscopes in real-life electronic circuits is the same, because we attach the negative probe to one node (usually called ground in electronics) of the circuit and then only move the positive prove for measuring the voltage.

So, as you can see, the electric potential at a point is the voltage measured at that point with respect to the zero-potential reference point, so potential is the same as voltage. Or equivalently, the voltage measured between any two points $a$ and $b$ is the electric potential at point $a$ by choosing the zero-potential reference point to be point $b$ (in other words, to obtain voltage from potential, the zero-potential point is not fixed for measuring all potentials), so voltage is the same as potential.

Please note I’m suggesting that voltage (electric tension, electric pressure, electric potential difference) and electric potential are the same thing, not that any of them is the same as electric potential energy. I know the former two are not the same as the latter.

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I searched if this question had been already asked, but didn’t find any. I found the following which ask different questions:

• Difference between voltage, electrical potential and potential difference

• Electric potential and voltage

• Why is voltage described as potential energy per charge?

• What is the difference between electric potential, potential difference (PD), voltage and electromotive force (EMF)?

• Voltage and potential

• Potential difference and voltage

• Is voltage electric potential or electric potential difference?

• Voltage about potential difference

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Edit after accepted answer: further clarification on why I say voltage and potential are the same

So far some people have answered/commented that the difference between electric potential and voltage is that the latter is a difference in potentials (which is true), in other words, that voltage is between two points (which is true) while potential is only at one point (which is not exactly true). Let me explain. You may say potential is a quantity only a point, right? Well, if you please look at equations (2.21) (for electric potential) and (2.22) (for electric potential difference, a.k.a. voltage) of David Griffiths' Introduction to Electrodynamics (3rd edition), both on page 78, you will see Griffiths defines electric potential and electric potential difference (a.k.a. voltage) as follows. (In the same manner as Roger's answer, and unlike Griffith, I'll denote voltage as $V$ and electric potential as $\phi$.)

$\phi = - \displaystyle\int_{\mathcal O}^{\mathbf r} {\mathbf E} \cdot {\mathrm d} {\mathbf l} \tag {2.21}$

$V_{ab} = \phi(\mathbf a) - \phi(\mathbf b) = \displaystyle\int_{\mathbf a}^{\mathbf b} {\mathbf E} \cdot {\mathrm d} {\mathbf l} \tag {2.22}$

Now look at the rightmost-hand side of both equations. They both are the line integral of the electric field vector from one point to another point along some path or trajectory. Aha! So electric potential is in general a quantity between two points, just like voltage!

Yes, I know that for electric potential (eq. 2.21), we usually choose the zero-potential reference point $\mathcal O$ to be at infinity, so that the potential then is a quantity at a point. But we can also choose the reference point $\mathbf b$ in the voltage (eq. 2.22) to be also at infinity (if you disagree, please explain), so that voltage is now also a quantity at a point. And as I said earlier, it is actually very common in electrical engineering to talk about the voltage at a point in electric circuits, because we choose one node as the reference node (a.k.a. ground), with respect to which we measure all voltages at the other non-reference nodes.

After discussing with Roger in the comments of his answer, it looks like indeed voltage and electric potential are the same thing, but simply different names for the same thing depending on context (engineering or physics, respectively), with the only tiny difference being that all potentials are measured with respect to a fixed reference point, while voltage can be measured with respect to an arbitrary point (although in nodal analysis we also fix such reference point to be a node, similar to potential).

Edit 2: Why my question is not this question

I've got two suggestions saying that my question is the same as the linked question in the subtitle, so I'll explain why it's not the same question. I'll refer to user11266's answer.

The first and third paragraph of user11266's answer only talk about the reason for confusion in the two terms, and what should be done instead, so those paragraphs are irrelevant to my question.

The second paragraph of user11266's answer talks about potential and potential difference (a.k.a. voltage). He/she says "Each point in space has assigned to it a value for electric potential", which is true because from Griffiths' equation (2.21) it is clear that, for a fixed reference point $\mathcal O$, the potential depends on $\mathbf r$. But guess what? You can also fix $\mathbf b$ in Griffiths' equation (2.22) (potential difference a.k.a. voltage), and now you have the exact same thing as potential: a scalar value at each point in space. And in case you say "no-one fixes a point for measuring voltage", then let me tell you that's actually extremely common: 1) we use it for nodal analysis, the most used method for solving circuits; 2) all circuit simulators use it; and 3) oscilloscopes read voltages by fixing the reference point.

And yes, I know that in the potential equation (2.21) the reference point $\mathcal O$ is usually at infinity, while in my example of circuits the reference point $\mathbf b$ for the voltage equation (2.22) is not at infinity but instead a node of the circuit. But as far as I know, there's nothing prohibiting us from choosing $\mathbf b$ to be also at infinity, in which case equations (2.21) and (2.22) become the same, and thus voltage and potential are the same thing.

And in case you say "potential is more general than voltage because voltage only applies to circuits, so your circuit example is not valid", my reply would be that simply ignore my examples, and instead focus on equation (2.22) from Griffiths and set $\mathbf b = \mathcal O$ and $\mathbf a = \mathbf r$, then swap the integral bounds/limits by adding a negative sign in front of the integral, and now you get $V_{ab} = \phi$. Et voilà, voltage is the same as potential, or potential is the same as voltage, so they're the same thing. Thus my statements in the previous paragraph hold.

Answer 1

In physics voltage is usually defined as potential difference between two points of a circuit: $$V_{ab}=\phi_a-\phi_b.$$ In other words:

• potential is a potential energy per unit charge measured in respect to an agreed reference point (e.g., an infinitely remote point), i.e. work done when bringing the charge from this point
• voltage is the work done when moving a charge between the two specified points.

One typically has the same reference point for all the potentials, but voltages can be defined between any two points.

Finally, potential is an extremely general concept, used to describe electric and electromagnetic phenomena well beyond circuit theory, whereas voltage is a term mainly limited to specific applications.

Update
The question has been further discussed in the comments, and it was agreed that much of my answer indeed reiterates the information already given in the OP. I therefore state below the main point, as it came out of the discussion:

One can say that potential is a voltage between the point of interest and the reference point. I think the real difference is indeed the usage: one will rarely speak of voltage in atomic or nuclear physics, but in case of semicodnuctor nanostructures the term is rather used. Some may say that potential is "more scientific term", as opposed to engineering.

Answer 2

No, when they are properly defined (see the IEC definitions below) voltage and potential difference are in general not the same, even though they can be considered the same in the majority of situations in electronics and quite often even in physics. The root of their difference resides in the presence (and relevance) of changing magnetic fields within the considered system.

Let's start from the physical quantity they are both intended to represent: the work per unit charge that is done against the field E (hence the minus sign) in order to go from point A to point B. This quantity is computed as a path integral and is in general path dependent, therefore we will have to specify the particular path $\gamma_{A->B}$ along which the integration takes place. $$ - \int_{\gamma_{A->B}}\vec{E}.\vec{dl} $$ Following the IEC (if we ignore the sign convention) we will call this general path dependent quantity "voltage" and we will denote it by the symbol V with a subscript that specifies the path:

$$ V_{\gamma_{A->B}}= - \int_{\gamma_{A->B}}\vec{E}.\vec{dl} $$ Note that the electric field $\vec{E}$ that appears inside the integral is the one and only electric field a charge experience along a specific point along the path. We could call this the 'total electric field' (the reason for this specifier will be made clear shortly.)

Conservative fields: voltage and PD are the same

If our physical system resides in a region of space where there are no changing magnetic fields, the electric field is conservative and we can introduce a potential function $\phi$ such that $$ \vec{E}=\vec{E}_{c}=-\nabla\phi $$

and the above path integral will be a function of the endpoints alone and not the path joining them. $$ V_{\gamma_{A->B}}= - \int_{\gamma_{A->B}}\vec{E}_{c}.\vec{dl}=- \int_{A}^{B}-\nabla\phi.\vec{dl}=\int_{A}^{B}\nabla\phi.\vec{dl}=\phi_B-\phi_A $$

That is, in the conservative case, voltage (the path integral of E along a path joining two points) can be expressed as a potential difference (the difference in the values assumed by the electric scalar potential $\phi$ at the endpoints.) In this sense, the scalar electric potential is a specialization of the more general concept of voltage.

In lumped circuit theory, where we deal with conservative electric fields (and we hide the nonconservative part inside the lumped inductive and magnetic components) there is no difference between V and phi and the (scalar) potential difference is the same thing as the voltage along any path joining A and B: since it does not depend on the path, we can use the term 'voltage difference' to denote the work per unit charge done on a charge that goes from point A to point B. It is also customary to (ab)use the same symbol V (which denotes voltage) for the scalar potential $\phi$, so that the above scalar potential difference can be written $V_B-V_A$ or, in short, $V_{BA}$.

The path independence implies that the circulation of a conservative field $\vec{E_c}$ along any closed path be zero: $$ \oint_{any \ closed \ path}\vec{E}_{c}.\vec{dl}=\int_{A}^{A}\nabla\phi.\vec{dl}=\phi_A-\phi_A=0 $$

Nonconservative fields: PD is not enough

But the above relation is not true in general. In fact, for a generic nonconservative electric field $\vec{E}$ in a non-motional context, the relevant Maxwell equation in integral form is $$ \oint_{\Gamma} {\vec{E}.\vec{dl}}=-\frac{\partial}{\partial t}\iint_{\Sigma}{\vec{B}.{dS}} $$

where the closed path $\Gamma$ is the boundary of the surface $\Sigma$. Note that here $\vec{E}$ is the total electric field that can be seen, according to Helmoltz's decomposition theorem, as the composition of an irrotational (conservative) and a solenoidal (non conservative) part $$ \vec{E} = \vec{E}_{irr} + \vec{E}_{sol} = \vec{E}_c + \vec{E}_{nc} $$ If we substitute this generic decomposition of E into the above equation, since the conservative part has a circulation that is zero, we get $$ 0 +\oint_{\Gamma} {\vec{E}_{nc}.\vec{dl}} = -\frac{\partial}{\partial t}\iint{\vec{B}.{dS}} $$ or, by expressing the magnetic field B in terms of its vector potential A and invoking Stokes' theorem

$$ \oint_{\Gamma} {\vec{E_{nc}}.\vec{dl}} = -\frac{\partial}{\partial t}\iint{\nabla\times \vec{A}.{dS}}=-\frac{\partial}{\partial t}\oint{\vec{A}.{dl}} $$ In a non-motional context where the closed path $\Gamma$ does not change its shape the above relation tells us that $\vec{E}_{nc} = -\partial{\vec{A}}/\partial{t}$.

With this expression for the non-conservative component of the total electric field, we have $$ \vec{E}=\vec{E_c}+\vec{E_{nc}}=-\nabla\phi-\partial{\vec{A}/\partial{t}} $$ Substituting this in the general expression of voltage as the path integral of the total field along a path gamma (in a non-motional context) we get:

$$ V_{\gamma_{A->B}}= - \int_{\gamma_{A->B}}\vec{E}.\vec{dl} =- \int_{\gamma_{A->B}}\vec{E_c}.\vec{dl}- \int_{\gamma_{A->B}}\vec{E_{nc}}.\vec{dl} \\ =\int_{A}^{B}\nabla\phi.\vec{dl}+\int_{\gamma_{A->B}}\partial{\vec{A}}/\partial{t}.\vec{dl} $$

The first integral is the difference in scalar potential, so we conclude that, in presence of a variable magnetic field:

$$ V_{\gamma_{A->B}}=(\phi_B-\phi_A)+\int_{\gamma_{A->B}}\partial{\vec{A}}/\partial{t}.\vec{dl} $$ That is, voltage (the path integral of the total electric field along a specific path) is the sum of the electric scalar potential difference (which is the path integral of the conservative part ) and the contribute to the emf along that path (which is path-dependent integral of the non-conservative part).

If one wants to recover the emf, they need to specify a closed path - the potential difference will go to zero and the result is the emf associated with that particular closed path.

If one is interested in the scalar potential difference alone, they need to remove the non-conservative component $\vec{E}_{nc}$ from the total electric field $\vec{E}$ and compute the path integral of

$$ \vec{E_c}=\vec{E}-\vec{E_{nc}}=\vec{E}-(-\partial{\vec{A}}/\partial{t}) \\ =\vec{E}+\partial{\vec{A}}/\partial{t} $$ That plus sign before the time derivative of A makes it seem that the Ec field is somewhat some sort of resultant field, while in reality it is just a component of the total field E.

Conclusion

The confusion around the concepts of voltage and scalar potential stems from the fact that so many engineering books use the same symbol (usually V) to denote the path integral in the conservative and nonconservative cases. Defining voltage as above is a simple way to make the two terms coexist with each other in a way that is consistent with Helmoltz's decomposition theorem. Apart from a sign convention, the above definitions of voltage, potential difference and emf are those given by the International Electrotechnical Commission (IEC):

IEC definition of voltage
IEC definition of (scalar) potential difference
IEC definition of induced voltage

(in the IEC definition of induced voltage - the path integral of the nonconservative part - it is contemplated the possibility of motional induction).

To summarize: voltage as the path integral of the total electric field is the more general quantity (and it comes from the definition of work done on a charge), while the (scalar) potential difference is a specialization that is valid only where the electric field is conservative. In lumped circuit theory we can afford the luxury of confusing these two quantities (there is a way to treat the emf contribute of a coil as if it were a voltage drop), but there are unlumpable circuits - like the Lewin ring - where this is no longer the case.

Answer 3

I think there is a semantic distinction to be made between potential and voltage/potential difference. The former is a property associated to a single point, while the latter is a property associated to an ordered pair of points. You argue that these are the same because the former still requires some zero point to be specified, but that's not quite true.

For example, consider the electric potential due to a point charge at the coordinate origin, $\phi(\mathbf r) = \frac{q}{4\pi\epsilon_0 |\mathbf r|}$, where I've chosen $\lim_{|\mathbf r|\rightarrow \infty} \phi(\mathbf r) = 0$. Note that there are in fact no points in space where $\phi(\mathbf r)=0$. On the other hand, the voltage between any two points at the same distance from the origin is equal to zero.

You might say, "well, $\phi(\mathbf r)=0$ at infinity, which I consider to be a point." However, it is generally not possible to do this, e.g. if the potential doesn't approach the same limit in every direction as $|\mathbf r|\rightarrow \infty$, or if the potential doesn't approach any limit at all, as is the case for an infinite wire. Even if you can do this, there's no reason that you should have to choose the potential to vanish there; for example, the choice $\phi(\mathbf r) = \frac{q}{4\pi\epsilon_0 |\mathbf r |}+V_0$ (with $V_0>0$) is also a perfectly valid choice for the electric potential due to a point charge at the origin, but now it's not zero anywhere even if we do include the point at infinity.

Now, you could certainly say that it's always possible to choose some $\phi(\mathbf r)$ to vanish somewhere (say, $r_0$), and that the potential at a point $\mathbf r$ is equal to the voltage between the points $\mathbf r$ and $\mathbf r_0$. However, (i) you certainly don't need to do this, and (ii) even if you do, the fact remains that potential is a property of a single point while voltage is a property of two points, which (at least to me) is a perfectly clear reason to say that potential and voltage are two distinct things which are very, very intimately related to one another.