My basic idea of what a DC current is that, when a potential difference is applied to the ends of wire it creates a force which propagates through the wire all the way to the other end. It propagates in this way - Since the only mobile charges are electrons therefore positive part of the battery applies an EMF which pulls the nearby electrons from the wire connected to positive terminal and which subsequently leaves the blank positively charged site which is then filled by the nearby atom and so on. This happens throughout the wire, and at the end, near the negative terminal fresh electrons are supplied constantly. So in a current, blank positive sites are constantly created which then pulls the nearby electrons.

Now, my query is that - Is it necessary that the blank positive site created has the EMF equal to that of the battery at the end ? If so, why ? Will it not depend upon the amount of blank positive sites created at one point (in other words "concentration") which will eventually pull the electrons with more force ? What if the wire is very thin ?

a poor diagram showing the inside is a simple circuit

  • $\begingroup$ What is missing from this picture is the current in the loop and the corresponding drops in EMF due to the resistance of the wires, assuming circuit loop analysis holds here. Please let me know if this suggestion is helpful. Check out "Firgure 2: KVL Analysis of a Circuit" on page 3 of the PDF for the picture with the resistances of the wires shown as part of the figure. $\endgroup$ Aug 14, 2023 at 13:01

1 Answer 1


Is it necessary that the blank positive site created has the EMF equal to that of the battery at the end ?

This question doesn't make sense.

EMF is not a property of a certain position in a circuit. It is a property of a path between two locations. In the electrostatic limit, the EMF won't depend on the path between the two locations, but only on the choice of those locations.

We simply can't talk about "the EMF of point A". We have to talk about "The EMF between point A and point B (along path S)", or "The EMF around a loop from A back to itself".

We can however talk about the potential at certain point (with reference to some previously chosen reference point). So let me rephrase your question,

Is it necessary that the blank positive site created has the EMF potential equal to that of the battery at the end ?

This depends on whether you want to consider the wire as ideal (0 resistance) or real (non-zero resistance).

If the wire is ideal, then (in the lumped circuit approximation) it has the same potential at all points.

In this case, however, you seem to be modelling just a wire connected between two terminals of a battery. For this circuit, the ideal wire model is not sufficient, since it implies the same potential at the two terminals of the battery.

So you need to treat the wire as a distributed resistor. The potential will vary along the path of the wire from the value at one terminal of the battery to the potential at the other terminal. If the wire has a uniform properties and cross-section, then the variation will be linear.

If you want to consider effects outside the lumped circuit approximation, you need to consider the actual path of the wire through space, and the magnetic flux through the surface surrounded by the wire. In this case the electrostatic approximation is invalid, and you can't talk about "the potential at point B". But you can talk about the EMF along the path of the wire from point B back to itself.

  • $\begingroup$ What I meant to ask is that suppose we have an ideal wire of 1 atom thickness (a chain of atoms), and we applied the EMF at the ends. Now the electrons starts shifting from one atom from one terminal to another. They do so because when their successors electron moves it leaves an empty positive space, creating potential difference and hence attracting the electron. So my question is - will the potential difference created by the blank space of only ONE atom necessarily be equal to the EMF at the end ? You can replace one atom thickness to many atom thickness. $\endgroup$
    Jul 23, 2018 at 20:28
  • $\begingroup$ One atom thick wire means you will have quantum behavior due to confining the electron wavefunction to such a narrow path. The behavior will not be the same as for everyday wires. $\endgroup$
    – The Photon
    Jul 23, 2018 at 20:38
  • $\begingroup$ With everyday wires, the net rate of electron flow will be given by Ohm's law: proportional to the applied potential difference. The movement of individual atoms will be mostly random with only a small bias to move in the direction implied by the applied voltage. You can't really consider the movements of single electrons, only the statistical ensemble of all the free electrons in the wire. There's no reason to expect that the voltage you applied happens to be equivalent to displacing exactly one electron from the battery's electrode. $\endgroup$
    – The Photon
    Jul 23, 2018 at 20:40
  • $\begingroup$ For further reading, everyday currents are pretty well modeled by the Drude model. $\endgroup$
    – The Photon
    Jul 23, 2018 at 20:42
  • $\begingroup$ For further reading on what happens when the wire is narrow enough for quantum effects to appear: Nanowire. $\endgroup$
    – The Photon
    Jul 23, 2018 at 20:46

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