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Since $V=Ed$, why doesn't the electric potential increase when length of the wire increases. $d$ increases right? Is it because the battery only pushes the electrons from one terminal to the other and not from one end of wire to the other end?

Or is it because that electric field is a conservative field and that work done does not depend on path taken?

EDIT(copied from one of the comments):can we use the following explanation: In an electric field, the electric potential(in the case of positive charge moving away from a positive charge) is line integral of $E.dr $ . Since line integral does not change with the path taken, every path will give you the same potential

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Think about it from the definition. Why does the current flow? Due to difference in potential of the electrons caused by the battery. This potential difference is caused by a specific chemical reaction occurring inside the battery, dictated by the Nerst equation. You can not change this. This completely and totally depends upon the configuration of the cell/battery itself( type of compounds, their respective concentrations e.t.c). The wire is an external factor that can in no way influence the battery's configuration

Thus, V will remain constant. Hence, the equation V=Ed means that on increasing length of wire, electric field decreases because the battery simply can not afford any more.

Update- "can not afford more" means that the battery can not enforce a stronger electric field. As I explained before, the voltage of the battery is due to a specific internal electrochemical reaction. This is the strength or maximum potential difference it can afford. To put it simply, V will always remain constant and keeping this in mind, you change the other quantities like field.

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  • $\begingroup$ please explain what did you mean by 'cannot afford anymore' $\endgroup$
    – Allen
    Jan 31, 2017 at 6:53
  • $\begingroup$ So that means even if it was an ideal wire with zero resistance, there would be a limit to how much length the battery would be able to produce a potential difference, right? $\endgroup$
    – Allen
    Jan 31, 2017 at 7:37
  • $\begingroup$ Battery will always produce a potential difference. If you increase the length of the wire, the electric field will keep on decreasing and approach 0 as you approach infinite length. Actually, it does not matter whether the wire is real or ideal with 0 resistance as this will only affect the current according to the relation E=ρj where ρ is resitivity (which becomes o) and j is current density( which becomes very,very,very large). $\endgroup$
    – TheFool
    Jan 31, 2017 at 8:04
  • $\begingroup$ ok, can we use the following explanation too: In an electric field, the electric potential(in the case of positive charge moving away from a positive charge) is line integral of E.dr . Since line integral does not change with the path taken, every path will give you the same potential. $\endgroup$
    – Allen
    Jan 31, 2017 at 10:03
  • $\begingroup$ Yeah, that explanation is always true and is actually one of the bases of my answer. The electric field and the motion of the electrons in the circuit are always in the same direction in the circuit. The field is always along the path and hence a longer path requires a weaker field to bring about the same potential difference(which remains same because of the specific reaction in the cell) $\endgroup$
    – TheFool
    Jan 31, 2017 at 10:30
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A battery provides a constant voltage, not a constant field. That means that electric field will decrease if you increase the length of wire.

On the other hand, if you have a situation where there is a constant field (for example, between two charged parallel plates) then increasing the distance increases the voltage (you have to do work by pulling the plates apart, so energy is conserved).

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  • $\begingroup$ A voltage does mean there is a field somewhere, then where is the field present? $\endgroup$
    – Allen
    Jan 30, 2017 at 17:09
  • $\begingroup$ I've only ever used, and taught, this equations applied to the field between charged plates. And only in the portion where the field is uniform. Can this be applied across wires? And if so, is it really the length of a wire that matters or the separation between the charges? $\endgroup$
    – bpedit
    Jan 30, 2017 at 17:54
  • $\begingroup$ @bpedit - when you apply a voltage across a (resistive) wire, a current will flow. The result of that current is an E field along the wire - and the integral of that electric field over distance is the potential difference. But you should never just "grab an equation" - think about the physics of what is going on. $\endgroup$
    – Floris
    Jan 30, 2017 at 18:41
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In a circuit each charge carrier (electron) carry a certain amount of energy which is equal to the emf of the battery and the electron loses all this energy in the outer circuit and by the time the electron gets to the +ve terminal of the battery it has lost all this energy to the outer circuit.

If the circuit has short wires then the energy or potential dropped by the electron per unit length will be more and if the length of the wire were to be greater then the potential drop per unit length will be less. But the energy lost in the outer circuit is the same for both the cases.

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