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I've been told that in a DC circuit with a battery the negative terminal of the battery "pushes" electrons in nearby atom away of it and those by themselves cause electrons in other atoms to move, but this convention but this way of thinking brought me some problems in understanding voltage.

I mean voltage is defined as $$v = \left(\frac{Q}{4\pi\epsilon_0r}\right)$$ so it seems like the source of that voltage on an electron is the electron behind it, so my question is Is that true or no? if no why and if it is true then How something like the voltage drop can be thought of(it doesn't seem like moving in a resistor can have any effect on the equation above)?

P.S: I'm sorry if the title seemed misleading that's what I couldn't think of a better title.

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  • $\begingroup$ Does this answer your question? I don't understand what we really mean by voltage drop $\endgroup$
    – zonksoft
    May 9 '20 at 20:38
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    $\begingroup$ Does this answer your question? What maintains constant voltage in a battery? $\endgroup$
    – Sandejo
    May 9 '20 at 20:46
  • $\begingroup$ It isn't true that voltage is defined in the way that you state that it is. The formula you give is the electrostatic potential of an isolated point charge Q. $\endgroup$ May 9 '20 at 21:03
  • $\begingroup$ W=QV is the fundamental eqn. Voltage is the energy per coulomb. In a battery the voltage comes from the Gibbs Free Energy of the reaction this is RELATED to the HOMO and LUMO but these are not the true potential. Another voltage source is through Faraday's Law where a changing magnetic field produces a voltage $\endgroup$
    – ChemEng
    May 9 '20 at 21:27
  • $\begingroup$ You can also have a voltage across a simple electric field caused by charge separation $\endgroup$
    – ChemEng
    Jul 25 '20 at 1:20
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The formula you cite is not the definition of voltage. One definition of voltage, or more precisely, the electrostatic potential difference between two points is

The energy required per unit charge to move a charge from one point to the other

Your formula gives the voltage at points around a single point charge, relative to points at an infinite distance away.

like the source of that voltage on an electron is the electron behind it, so my question is Is that true or no?

No, not really. Remember that in addition to the electrons surrounding the first electron, there are also positive charges associated with protons in the nuclei of the atoms making up the wire.

How something like the voltage drop can be thought of(it doesn't seem like moving in a resistor can have any effect on the equation above)?

Consider an electron in a vacuum tube, with a positive charged plate on one side of it and a negative charged plate on the other. It will be repelled from the positively charged plate and attracted toward the negatively charged plate. As it moves toward the negative plate, it accelerates, as its electrical potential energy is converted into kinetic energy.

Now if the electron were travelling in a resistive material instead of through vacuum, instead of picking up speed as it travels, every few nanometers it would interact with the material around it (crudely, you could say it "bounces off" the atoms in the material, or in a more sophisticated model you could say it interacts with the phonon field of the material), randomizing the electrons velocity and transferring its gains in kinetic energy into thermal vibrations of the material.

Basically the electron in the resistive material can lose electrical potential energy as it travels from a lower potential to a higher one without itself gaining substantial kinetic energy like it would if it were travelling through vacuum.

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  • $\begingroup$ I'm sorry but I don't seen to understand what you mean by "there are also positive charges associated with protons in the nuclei of the atoms making up the wire." wouldn't be there net force because of the negative terminal? $\endgroup$ May 10 '20 at 12:11
  • $\begingroup$ also if the source of the voltage on two points in the wire is the battery then won't that mean that it will get smaller as we go further away from the battery? $\endgroup$ May 10 '20 at 14:02
  • $\begingroup$ @KhaledOqab, yes there will be a net force on the electrons. But you can't calculate it just from the fields produced by the nearby electrons, because there are also positive charges nearby. $\endgroup$
    – The Photon
    May 10 '20 at 14:24
  • $\begingroup$ @KhaledOqab, The voltage won't drop as 1/r as you move away from the terminals because the presence of the wire changes the field distribution. In a sense you're correct that the reason the wire changes the field distribution is because the electrons in the wire move around. But as far as I know there's no easy way to calculate the effect by considering all the electrons in the wire. It's easier to understand by just using the rule that the wire, being a very good conductive material, creates an equipotential region. $\endgroup$
    – The Photon
    May 10 '20 at 14:27
  • $\begingroup$ so you mean that moving electrons "contribute" in maintaining constant electric field through the wire? $\endgroup$ May 10 '20 at 23:41

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