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I know that voltage is just a difference in electrical pressure, but if voltage doesn't increase the velocity, or charge of electrons, then it must just increase their energy, right? How else would an increase in voltage result in more power?

Imagine there's two conducting wires exactly the same in every way.

The first has 1 volt times 1 amp of electrical power travelling through it.

The second has 5,000,000 volts times 1 amp of electrical power travelling through it.

In both cases:

The velocity of each electron is the same.

The charge of each electron is the same.

The quantity of electrons passing a given point in the wire in one second is the same.

Does this mean that the only difference between an electron in the first wire, compared to an electron in the second wire, is that the latter is 5,000,000 times more energetic?

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  • $\begingroup$ BowlOfRed's answer talked about getting energy from a falling weight. To extend that idea a bit further, imagine a table-top experiment with a falling weight. Well, when you talk about electrons in the wire that is "at" one volt, that is like doing the experiment in a building at sea level, and when you talk about electrons in a wire that is "at" five million volts, that is like doing the same experiment in a building on top of a high mountain. The fact that you are on the mountain isn't going to matter if the weight can't fall any further than the floor. $\endgroup$ Nov 20, 2017 at 18:38

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The potential energy between two charges $Q$ and $q$ is $$ V(r)~=~\frac{1}{4\pi\epsilon}\frac{Qq}{r}. $$ The electric potential due to the charge $Q$ is $$ \Phi(r)~=~\frac{1}{4\pi\epsilon}\frac{Q}{r}, $$ which has the units of $joules/coulomb$. This one way to see the role of energy in Ohm's law, in particular $P~=~IV$. $P$ is in units of watts, or $joules/sec$, and $I$ has units $amps~=~coul/sec$. We can then see that $volts$ has units of $joules/coulomb$.

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  • $\begingroup$ Ok, well joules measures potential energy, and a coulomb is basically just a quantity of electrons (charge). So, then are you telling me yes? Volts = energy/electron. $\endgroup$ Nov 20, 2017 at 18:04
  • $\begingroup$ An electron has $1.6\times 10^{-19}$ coulombs of charge. Across one volt this charge defines an electron volt. $\endgroup$ Nov 21, 2017 at 1:54
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When you move a specific amount of charge across a particular voltage difference, then a specific amount of work is done.

Since electrons have a constant charge, then moving an electron across an increasing voltage difference requires (or produces) more work.

Imagine there's two conducting wires exactly the same in every way.

The first has 1 volt times 1 amp of electrical power travelling through it.

The second has 5,000,000 volts times 1 amp of electrical power travelling through it.

The important part about the above is that the charge has to move across a specific voltage. "Wires" are normally conductors, and they usually have very little voltage difference from one end to the other. So a wire being at "1 volt" or any other amount isn't the big deal. It's the voltage difference between that wire and some other location. The energy is released or expended as the charges move to the new potential.

Think of it like trying to get energy from a falling weight. You can hook up a weight to a rope and get energy from it. But for the same rock and the same harness, you can get more energy if you can let it fall farther.

The charges in the wires are the same, but you can get more energy from them if you have a larger potential difference with some other location.

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  • $\begingroup$ Okay, I see now. It's like an object being pulled down, or fighting against gravity. It gains energy while falling, loses energy while rising. $\endgroup$ Nov 20, 2017 at 18:15
  • $\begingroup$ Although I don't understand how you could push an electron against electrical pressure. Wouldn't you need greater voltage pushing from the other direction in the conductor? Can you have voltages just pushing against each other like that in a single wire? I guess that would be like negative and positive voltage, and if they're equal they cancel out to zero voltage. I think I get it. $\endgroup$ Nov 20, 2017 at 18:21
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    $\begingroup$ "Pushing" charges is what creates electrical pressure. You can do that with moving wires in magnetic fields (generator) or with molecular rearrangements (chemical reaction/battery) $\endgroup$
    – BowlOfRed
    Nov 20, 2017 at 18:23
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Imagine there's two conducting wires exactly the same in every way.

OK

The first has 1 volt times 1 amp of electrical power travelling through it.

The second has 5,000,000 volts times 1 amp of electrical power travelling through it.

It isn't clear what you mean by this.

I assume you do not mean that there is a $1\, \mathrm{V}$ potential difference between the ends of the first wire and a $5\, \mathrm{MV}$ potential difference between the ends of the second wire.

Rather, I assume you mean that the first wire transports $1\,\mathrm{J}$ of electrical energy per second and the second wire transports $5\,\mathrm{MJ}$ of electrical energy per second.

But it's important to understand that the flow of electrical energy for a current carrying wire is through the space in the vicinity of the wire. How?

Associated with the current through an isolated wire is a magnetic field with concentric, non-uniform field lines centered on the wire axis.

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Associated with the charge density on that wire is an electric field with radial field lines.

enter image description here

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The electric and magnetic fields are perpendicular and so their cross product is non-zero and yields a power density vector, the Poynting vector.

The power density in the fields surrounding the wire is then proportional to the product of the electric and magnetic field strengths.

It follows that the difference in the two cases you give is the charge density which is 5 million times greater on the second wire.

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