Understanding the concept of voltage

  • #1
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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

1579042968372.png


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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  • #2
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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
 
  • #3
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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 :confused::frown:
 
  • #5
vanhees71
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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##.
 
  • #7
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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!
 
  • #8
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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.
 

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