Ambiguity between electric potential and voltage?

Question

I understand that electric potential is a location based measure of electric potential energy per unit charge in an electric field, and that voltage is then the difference between two electric potentials at different locations in a field.

I'm having a hard time reconciling this with wording in textbooks and articles, such as equating electric potential to voltage even though the latter is a difference in the former. They say: "electric potential, otherwise known as voltage…" even though I don't see how they are equal unless the electric potential at the first location is $0 \textrm{ J/C}$.

Another thing I notice is that some textbooks might use $\Delta V$, i.e. a second order difference:

$$\Delta U_E = q\Delta V$$

Why not use $qV$ since $V$ is already a change in potential? Is there some semantic reason to treat voltage as a synonym for electric potential or am I just being pedantic?

Answer 1

It is a bit of a circular argument.

For two positions $A$ and $B$ you might say that the potential difference (difference in potential) between positions $A$ and $B$ is $V_{\rm AB}$ which is to be interpreted as the potential of position $A$ relative to position $B$ , $V_{\rm A}-V_{\rm B}$.

The potential at a position is the name given to the potential difference when one of the “reference” positions, say $C$, is assigned a potential value of zero.

So the potential difference between position $A$ and position $C$ is $V_{\rm A}-V_{\rm C} = V_{\rm A}-0= V_{\rm A}$ which is called the potential at position $A$.

This means that $V_{\rm AB}$, the potential difference between position $A$ and position $B$ or the potential of position $A$ relative to position $B$, is $V_{\rm AC}-V_{\rm BC} = V_{\rm A}-0 - (V_{\rm B}-0 ) =V_{\rm A}-V_{\rm B}$.

The voltage across a circuit element is to be interpreted as the potential difference across the circuit element.

The voltage at a node (position) in an electrical circuit is to be interpreted as the potential difference between the node and another node in the circuit, the reference node, which has been assign a potential of zero ie the potential at the node.

$\Delta U_E = q\Delta V$ Why not use $qV$ since $V$ is already a change in potential?

$\Delta V$ is to be interpreted as $V_{\rm A}-V_{\rm B}$ where $V_{\rm B}$ could be $0$ whereas if the word voltage is used $V$ could interpreted as $V_{\rm A}-V_{\rm B}$, a potential difference, or $V_{\rm A}(-0)$, a potential.

This last statement highlights the problem of using the word voltage as it has two possible meanings.

Answer 2

Let's imagine moving a positive test charge from a point $i$ towards a positive source charge at a point $f$ that is far away.

The electrostatic force of the source charge is going to be fighting us the entire way, so we will have to do positive work $W$ on the test charge, giving the test charge energy, while the field does negative work $-W$, taking that energy and storing it as electric potential energy.

The total change in electric potential energy is \begin{align*} \Delta U = U_{f} - U_{i} = W \end{align*}

We can take $U_{i} = 0$ as a reference point, and say that the point $f$ has a potential energy of $U_{f} = W$.

Now, this is a little confusing because only changes in energy have physical meaning. How can a single point have a potential energy? Well let's see what happens when we don't make $U_{i} = 0$, or give it any value in particular at all.

We'll measure the change in potential energy between two points, $a$ and $b$, both of which have been measured from a common reference point $i$.

\begin{align*} \Delta U &= (U_{b} - U_{i}) - (U_{a} - U_{i})\\ &= U_{b} - U_{a} - U_{i} + U_{i}\\ &= U_{b} - U_{a} \end{align*}

The potential energy at the reference point $i$ was cancelled out! I have tried to visualise why this happens with a one dimensional example (and if you are familiar with vectors this should be intuitive):

The length $b-a$ does not depend on the position of the point $i$. You can move the point $i$ anywhere, and while that will change the length of $a$ and $b$ individually, the length of $b-a$ is constant.

Therefore, we can just set $U_{i}$ equal to anything, and the difference in potential between two points will be the same no matter what. For ease of use, we set $U_{i} = 0$, and we call $U_{f}$ the potential energy at the point, as in, "when the charge is at this point, it will have this potential energy". We then drop the $f$ and just call it $U$ as shorthand.

Now, this all applies exactly the same for electric potential differences, because they're just the same values divided by the charge. Instead of saying $\Delta V = V_{f} - V_{i}$, we recognise that

\begin{align*} V_{i} &= \frac{U_{i}}{q} \\ &= \frac{0}{q} \\ &= 0 \end{align*}

and so we say that the potential at a point is $V_{f}$, and then drop the $f$ and call it $V$ as shorthand.

For emphasis: The voltage at any point is specified as $V$, and is measured compared to a common reference point.

Why not use $qV$ since $V$ is already a change in potential? Is there some semantic reason to treat voltage as a synonym for electric potential or am I just being pedantic?

They absolutely could have just used $qV$, and some books do! Especially when you start talking about capacitors. The author is using $\Delta V$ specifically, to make sure you realise we're talking about a difference in potential between two points. Those points could be points like $a$ and $b$ (in which case $\Delta V$ is appropriate), or they could be points like $a$ and $i$, the reference point (in which case it would be appropriate to just use $V$ as shorthand for $\Delta V$ measured from a reference point of $0$ potential).

Again: When we see $V$ or $U$ by itself, we're talking about $\Delta V$ or $\Delta U$ measured from a reference point of $0$.

Answer 3

The potential and the potential difference are measured in volts, hence engineers use "voltage" or "voltage difference" . Mostly the word "difference" is dropped as it is assumed understood.

Potential energy is the contribution to energy that depends on position as opposed to kinetic or motion energy. The potential is the potential energy per unit charge. From the context it should always be clear what is meant.

Answer 4

I'm having a hard time reconciling this with wording in textbooks and articles, such as equating electric potential to voltage even though the latter is a difference in the former. They say: "electric potential, otherwise known as voltage…"

To be clear, it is the potential difference between two points that is equivalent to the voltage between the same two points. The electrical engineering definition of voltage used by the NCEE in the FE exam handbook is

The potential difference V between two points is the work per unit charge required to move the charge between the points.

The change in potential energy, $\Delta U$, when total charge $q$ moves between the points is the work done per unit charge times the amount of charge moved, or $\Delta U=qV$.

I'm not sure why one would write $\Delta V$ since, as you say, $V$ is already the difference in potential. $\Delta V$ would then be the change in the difference in potential (?)

Hope this helps.