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As far as I understood, conductors connected to some DC voltage source terminals will try to "gain" corresponding terminal potential (though capacitance of typical conductor is pretty low, so amount of the required charge is tiny). And this process is incredibly fast.

If this true for AC, there must be some curious situation. Especially I'm interested in the case of a usual electrical wall outlet. Below is a description of how I see it.

At some instance of time (when power source output voltage is $0V$), "Live" and "Neutral" wires in the outlet are not charged and there is no electric field and potential difference between them. Then a small source voltage appears, the wires start to charge, there is build up of weak electric field between them and associated with the field PD. As source voltage increases, there is new charging process, the electric field becomes stronger... So in case of 240V RMS voltage, during peak +311V "Live" and "Neutral" wires are charged to maximum and magnitude of electric field between them reaches maximum as well. Then it starts decreasing to $0V$ again, then electric field flips etc. In other words, both wires keep recharging constantly. Is this correct and really happens ?

Surprisingly but I can't find any explicit confirmation or disproof to this simple question.

One more reason that confuses me - some people say that "Neutral wire is not energized". Actually I don't really understand what they mean but this and it disturbs me. I would prefer to see familiar surface charges on both wires and electric field / PD due to them.

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  • $\begingroup$ Some of the specifics of your question depends on your location. In the US, household 120V circuits have a "live" and a "neutral". Household 240V circuits have two "live" lines which are both energized (with respect to ground). $\endgroup$ – BowlOfRed Jun 23 '18 at 19:29
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In other words, both wires keep recharging constantly. Is this correct and really happens?

Yes, it is really happening exactly as you've described. The wires act as a capacitor (which they actually present) and, in order to change the voltage on this capacitor, it has to be charged and discharged accordingly.

people say that "Neutral wire is not energized". Actually I don't really understand what they mean but this and it disturbs me. I would prefer to see familiar surface charges on both wires and electric field / PD due to them.

One statement does not contradict the other. The neutral wire is not energized in a sense that its potential is always zero relative to ground (it is actually connected to ground). At the same time, as the live wire gets (notional) positive charges and becomes positively charged relative to the neutral wire and ground, the neutral wire will lose the same number of positive charges, while its potential will stay at zero.

This, again, is the same process as happens in a discrete capacitor, even if one of its plates is grounded. So, as the live and neutral wires are charged, discharged and then charged again with the reverse polarity, they both acquire and lose positive or negative charges and the field between them flips accordingly.

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  • $\begingroup$ If the neutral wire loses charges, why does the potential stay the same? $\endgroup$ – BowlOfRed Jun 23 '18 at 18:54
  • $\begingroup$ @BowlOfRed The charges are moving out because that is how a capacitor is charged. The potential stays the same because it is grounded. $\endgroup$ – V.F. Jun 23 '18 at 19:17
  • $\begingroup$ Household wires have a (minimal) capacitance, but it is not due to interactions with the other wire. The changing potential in the live wire has no effect on the neutral. You could remove the neutral wire entirely and the live wire would act in the same way. $\endgroup$ – BowlOfRed Jun 23 '18 at 19:20
  • $\begingroup$ @BowlOfRed I have to disagree on all your points. 1) Yes, a standalone wire has its own capacitance, but its capacitance is going to significantly increase if it runs in parallel with another wire. Just check any capacitance calculator for wire pairs. 2) If you don't accept the fact that live and neutral wires make up a capacitor, you can just think of charged live wire inducing charge into a nearby grounded neutral. 3) The voltage on a live wire won't depend much on the presence of neutral, but the charge will - for the reason mentioned in (1). $\endgroup$ – V.F. Jun 23 '18 at 20:15
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In this question, you are referring to an AC voltage signal that is, by definition, a sinusoidal oscillating signal. This means that yes, the wires are carrying a signal that is actually oscillating between a maximum (positive) and a minimum (negative) value of the voltage with a frequency and a root mean square value that are precisely stated.

As a reference, I think that you might find useful this link and the links therein.

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