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
Best Answer
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.