# Understanding the concept of voltage

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## Main Question or Discussion Point

Hi everyone!

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

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 $dl$, and what is its direction?

Thanks very much!

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Dale
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Looks like you are copy-pasting from stack exchange. For inline LaTeX you have to replace the SE \$ with the PF $I reverted and updated for you Ah thanks a lot! Yes, I've posted the question also there, but it hasn't been directly answered, and my doubts are still there vanhees71 Science Advisor Gold Member 2019 Award This is only true for electrostatics, because only then the electric field has a potential, independent of the integration path in your formula, i.e., you can use any path connecting the points with the position vectors$\vec{x}_a$and$\vec{x}_b$. The potential is defined as V(\vec{x})=-\int_{C(\vec{x}_0,\vec{x})} \mathrm{d} \vec{r} \cdot \vec{E}(\vec{r}), where$C(\vec{x}_0,\vec{x})$is an arbitrary curve connecting an arbitrary fixed point$\vec{x}_0$with the variable point$\vec{x}$. Then you have \vec{E}(\vec{x})=-\vec{\nabla} V(\vec{x}). The voltage is simply the difference of the potentials between the two points in question, \Delta V=V(\vec{x}_b)-V(\vec{x}_a). Since the line integral defining$V$only depends on the initial an final points of the path, you get \Delta V=-\int_{C'(\vec{x}_a,\vec{x}_b)} \mathrm{d} \vec{r} \cdot \vec{E}(\vec{r}), where$C'(\vec{x}_a,\vec{x}_b)$is an arbitrary path connecting the points at$\vec{x}_a$and$\vec{x}_b$. This is only true for electrostatics, because only then the electric field has a potential, independent of the integration path in your formula, i.e., you can use any path connecting the points with the position vectors$\vec{x}_a$and$\vec{x}_b$. The potential is defined as V(\vec{x})=-\int_{C(\vec{x}_0,\vec{x})} \mathrm{d} \vec{r} \cdot \vec{E}(\vec{r}), where$C(\vec{x}_0,\vec{x})$is an arbitrary curve connecting an arbitrary fixed point$\vec{x}_0$with the variable point$\vec{x}$. Then you have \vec{E}(\vec{x})=-\vec{\nabla} V(\vec{x}). The voltage is simply the difference of the potentials between the two points in question, \Delta V=V(\vec{x}_b)-V(\vec{x}_a). Since the line integral defining$V$only depends on the initial an final points of the path, you get \Delta V=-\int_{C'(\vec{x}_a,\vec{x}_b)} \mathrm{d} \vec{r} \cdot \vec{E}(\vec{r}), where$C'(\vec{x}_a,\vec{x}_b)$is an arbitrary path connecting the points at$\vec{x}_a$and$\vec{x}_b##.
Thanks very much for the answer!

Dale
Mentor
What does the integration from one to the other (and not vice versa) mean?
Since you have responses about the other portions, I thought I would address this. If you have a typical voltmeter then A will be your red lead wire and B will be your black lead wire. So integrating from A to B or from B to A is just a matter of switching your lead wires.