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Suppose I have a cell/battery , both the terminals are at non-zero potentials.If I connect a piece of wire to say the negative terminal will there be some sort of electron redistribution in the wire. The wire was initially at zero potential , when it's attached to the terminal which is at nonzero potential , shouldn't electrons redistribute themselves to make the wire equipotential ? what will be the potential of the wire if there is such a thing , I guess its the same as that of the terminal be for the wire was connected , because the way I'm seeing things , the wire appears to be just an extension of the terminal.

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    $\begingroup$ See physics.stackexchange.com/questions/421473/…. You can model the end of the piece of wire as creating a capacitor with a very very small capacitance between the end of the wire and the other terminal of the battery. In this model a small amount of charge does transiently flow to charge the capacitor and to equilize the potential along the wire with the terminal of the cell. There is also an accompanying redistribution of electric field lines around the battery and wire. $\endgroup$ – jgerber May 17 at 20:45
  • $\begingroup$ I think the model of "capacitance with the other terminal of the battery" is not actually complete, because you are not including what I think the questioner is truly asking about, which is: en.wikipedia.org/wiki/Capacitance#Self_capacitance You can tell the difference because the further you position the wire from the other terminal of the battery, the lower their mutual capacitance will be, but the self-capacitance of the wire will remain roughly constant no matter how far away it is from the battery. $\endgroup$ – Glenn Willen May 17 at 22:26
  • $\begingroup$ I think the answer is complete. It allows you to calculate how the wire charges up in time using the mutual capacitance between the end of the wire and the other terminal of the battery. I think the relationship to self capacitance that you bring up has to do with the fact that if the battery is made very big (or the wire is made very small) so that the other terminal of the battery can be considered to be "at infinity" then the value of the mutual capacitance between the end of the wire and the other terminal will approach the value of the self capacitance of the wire itself. $\endgroup$ – jgerber May 17 at 23:57
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Unless the metal of the wire is the same as the metal of the battery terminal, there will be redistribution of charge, because of diffusion of charge carriers. The side with the higher concentration of electrons will lose some, the side with the lower concentration will gain some, in the vicinity of the connecting surfaces. This diffusion makes a small internal voltage gradient.

Because this phenomenon is temperature-dependent, it is useful in measurement (thermocouples). As for a battery, its 'negative' or 'positive' terminals aren't absolute-voltage references, those values only come into play through infinitesimal stray capacitance, stray conductance, and other minor effects, unless more connections than just one battery terminal and the wire are made.

A complete circuit can be sketched with stray capacitances, but one usually ignores the static electric effects (until there's enough energy to give you a jolt after shuffling shoes on carpet). Capacitance is a useful concept for a one-terminal device, if high voltages are present.

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