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I ask some help in understanding better the concept of voltage. The voltage is a difference in electric potential between two points $a$ and $b$. It is defined as $$V_{ab}=-\int^a_b\mathbf{E}\cdot d\mathbf{l}$$

However, I'm a bit confused with the use of notation:

  • Is $V_{ab}$ the same as $\Delta{V}$, or rather $-\Delta{V}$? In fact, $V_{ab}$ is also written as $V_a-V_b$, while $\Delta{V}$ should be a difference between a final and an initial position.
  • What do $a$ and $b$ represent? They are extrema of integration, but how do we select them in a problem, one as the starting position and the other as the arrival? What does the integration from one to the other (and not vice versa) mean?

Eventually, I would like to add another question, this one about the integrand:

  • What is concretely $\text d\mathbf l$, and what is its direction?

Thanks very much!

Edit: My problem is somehow similar to the one expressed here: is the voltage defined as $\Delta{V}$ or $-\Delta{V}$? Sometimes I've found it is the work per unit charge done by the electric field, other times the work by an external force against the electric field. What is the correct definition?

Mabe I'm confusing the difference in potential and the voltage drop?

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  • $\begingroup$ 𝑑𝐥 is a small element of path length along which you are integrating and the integrand is the component of the electric field in the direction of 𝑑𝐥 times the magnitude of 𝑑𝐥. This example may help: youtube.com/watch?v=gnyIUqroZX4 He has an electric field with a particular form and he chooses 𝑑𝐥 to most conveniently evaluate the integral. $\endgroup$ – Not_Einstein Jan 15 at 0:29
  • $\begingroup$ @thanks Not_Einstein, i watched the video, could he go also from point (a,b) to the origin? How did he choose the initial point and the final point of integration? $\endgroup$ – Shootforthemoon Jan 15 at 6:35
  • $\begingroup$ If he started at (a,b) and ended at the origin, by whatever path, he would be calculating the negative of what he calculated by going form the origin to (a,b). The equation in Philip's post below shows the definition of the potential difference between two points. If you do the integral from a to b, then you would be calculating 𝑉𝑏 - 𝑉𝑎, which is - (𝑉𝑎−𝑉𝑏). $\endgroup$ – Not_Einstein Jan 15 at 15:14
  • $\begingroup$ @Not_Einstein But in fact I don't get what I should calculate, because my book calls (𝑉𝑎−𝑉𝑏) the potential of the field, while in general we compute the potential difference starting from the work of an external force opposed to the field (if I'm not wrong?) $\endgroup$ – Shootforthemoon Jan 15 at 15:20
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I understand your struggle with the notation. Unfortunately many texts and teachers us the letter $V$ for both voltage and electrical potential in the same context: $$V_{ab}=V_a-V_b.$$ If we ask someone to put this equation in words one might say "the voltage across the space from location $b$ to location $a$ is the difference in the electrical potentials at those locations." There you have that letter $V$ being called voltage and potential.

As others have said here, voltage is the difference in electrical potential between two finite locations (for most near-electrostatic situations - time-varying magnetic fields are more complicated).

To improve readability and understanding, I have found it useful to use the letter $\phi$ to represent the electrical potential at a point in space, with the zero reference being an infinite distance from the charge distribution producing the $\vec{E}$ field: $$\phi_s = -\int_\infty^s \vec{E}\cdot\mathrm{d}\vec{r}$$

Now if we say that the voltage $V_{ab}$ specifically refers to "starting" at point $b$ (as a local reference) and finding how much "higher" (although a negative value would say it's actually lower) the potential is at $a$, then $$V_{ab}=\phi_a-\phi_b $$ $$V_{ab} = -\int_\infty^a \vec{E}\cdot\mathrm{d}\vec{r} - \left( -\int_\infty^b \vec{E}\cdot\mathrm{d}\vec{r}\right) $$

I'll let you complete the formality of the math to arrive at your first equation.

Your question about how to choose $a$ and $b$: They are two points in space, both referenced to a common coordinate system. The integral tells you how much "higher" the potential at point $a$ is than $b$. If you get a negative value, that's okay. It simply means that $a$ actually has a lower potential than $b$. One can see $$V_{ab}=-V_{ba}.$$

Finally, regarding $\Delta \phi$ or $\Delta V$ (or even $\Delta \xi$, where $\xi$ is anything ): A delta quantity is almost always the "later" minus "earlier", or "now" minus "previous" or "this" relative to "that": $$\Delta \vec{r} = \vec{r}_{now}-\vec{r}_{previous}$$

About the order of subscripts in $V_{ab}$. There is no universal standard for the ordering of subscripts in a symbol like this. One must always look for the definitions in the book or article which uses it.

  1. $V_{ab}$ might mean the potential difference of $a$ relative to $b$. The writer may be thinking "the $a$ location is my primary interest, and $b$ is the secondary", may be looking it in the order of a $\Delta$-type calculation, $\phi_a - \phi_b$. I believe that most writers use this convention, but it's not universal. Warning Don't think of the alphabetical order of $a$ and $b$ as having any significance. It doesn't!
  2. $V_{ab}$ might mean the potential difference of $b$ relative to $a$. The writer may be thinking "I'll start at $a$ as my reference point and move toward $b$ and see how the potential changes." This is a valid definition of the symbol, but not as common.

Bottom Line Writers should always define subscripted quantities to avoid ambiguity, and readers must always look for such definitions and not make undue assumptions. We all have our preferences.

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  • $\begingroup$ Thanks a lot! Your answers is the closest to what I was looking for! just a question about the delta. Is the voltage $V_{ab}$ the Delta itself, or minus delta? Here, how do I see which between a and b is the "earlier" or the "later"? Thanks again $\endgroup$ – Shootforthemoon Jan 15 at 15:51
  • $\begingroup$ @Shootforthemoon I've added a section for the subscripts at the end. cheers! $\endgroup$ – Bill N Jan 15 at 16:44
  • $\begingroup$ Great, thanks!. $\endgroup$ – Shootforthemoon Jan 15 at 16:52
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(a) There's no ambiguity in this form of the equation $$V_a -V_b= -\int_b ^a \mathbf E\cdot \text d \mathbf l$$ Notes

(i) We could swap a and b (on both sides of the equation). In fact I'd prefer them swapped, but that's only a matter of taste. We'll leave them as they are in what follows.

(ii) I think it's usually understood in this context that $V_{ab}$ means $V_a -V_b$, but I think it's unsafe simply to assume this; it needs stating.

(iii) Likewise, I think that $\Delta V$ is usually understood to mean $V_a -V_b$, but with the same provisos.

(b) The integral is a line integral. We evaluate it along a route between points A and B. $\text d \mathbf l$ is an element (small part) of that route. $\mathbf E$ is the electric field strength vector local to that element. The electric field due to static charges is a conservative field, meaning that we get the same answer for $V_a -V_b$ whatever route we choose between A and B.

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  • $\begingroup$ Omitted minus sign in original. Now corrected. Sorry if this caused more confusion! $\endgroup$ – Philip Wood Jan 15 at 0:08
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    $\begingroup$ Good answer. The only thing I'd change is that $\vec{E}$ is the field vector. The field strength would be $|\vec{E}|$. $\endgroup$ – The Photon Jan 15 at 0:41
  • $\begingroup$ Thanks for the answer! How do we choose A and B? If they say "calculate the voltage between the two charged conductors", where do we place A and B? $\endgroup$ – Shootforthemoon Jan 15 at 6:39
  • $\begingroup$ Provided that the charges on the conductors are stationary, every point on one conductor is at the same potential ($V_1$, say) and every point on the other conductor is at the same potential ($V_2$, say). So A can be any point on one conductor and B can be any point on the other conductor. $\endgroup$ – Philip Wood Jan 15 at 9:01
  • $\begingroup$ @The Photon I think it depends on which side of the Atlantic you're on. An authoritative UK textbook (admittedly quite old now) writes about "Electric field strength, $\mathbf E$", whereas Purcell and Panofsky-and-Phillips both write about "Electric field, $\mathbf E$". I've amended my answer to use the compromise "Electric field strength vector". $\endgroup$ – Philip Wood Jan 15 at 9:22
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The integral

$$\int_\gamma \mathbf{E} \cdot d\mathbf{r}$$

represents the work done in moving a unit of charge through the electric field along the path $\gamma$. Note this has just the same form as the work integral

$$W_\gamma = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$

only we're using electric field (in effect, force per unit of electric charge emplaced), as opposed to electric force. As a result, it is the work per unit of charge.

What "voltage", $V$, is, is potential energy per unit of charge. Potential energy refers to the work that could be done were the charge to move from one point to another, or perhaps better, the stored energy in having a bit of charge resting at one point in a field, that it can give up by moving to another point. Hence it always requires two points, just as the work integral always needs a start and end to its path, and in fact the potential energy is defined as the negative of the work, that is

$$U_\gamma = -\int_{\gamma} \mathbf{F} \cdot d\mathbf{r}$$

The reason for this is because, viewing potential energy as "stored energy", it should be that as the particle moves in the force field, this energy should get used up. In particular, as the particle commences to move along $\gamma$, it starts accruing this energy, and so the store of energy should (by conservation) then start dropping. But the work integral (total work done) increases (assuming we don't have any "humps" along the way for simplicity), which is the opposite behavior. In effect, to define the potential energy sensibly, we want the work remaining to do, not the work done, which ideally we would define by taking the integral along $\gamma$ from the present position of the particle during the motion to its final place, but this definition would now technically require three points: one of these is the initial point, the other is the final point, and the third is the point we are at during a certain phase of the motion, which we may not always have. Hence, we just negate the integral to get the desired behavior, giving the negative sign in $U$.

And thus $V$ is just this per unit of charge, hence

$$V_\gamma = -\int_\gamma \mathbf{E} \cdot d\mathbf{r}$$

. :)

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