Electrical – How to an electron have 0 electric potential after exiting a resistor but have current

electronvoltage

Unfortunately, I am not able to respond and ask a question on this specific post so I am going to ask a question based on the answers of the user "Transistor" in this post because I am still confused.

https://electronics.stackexchange.com/a/217000/211088

All my statements are based on the assumption that electrons have 0 electric potential after they go through the last load or is at the end of the circuit. (Statement # 2).

Please tell me if these statements are correct:

  1. Electric Potential is the force created from all the electrons in the negative terminal of the battery. The force is created from the electrons pushing away from each other trying to go to the positive terminal which also pulls the electrons.

So does this create 2 times the force because of an electron's push and the positive terminal's pull? Or does the push and pull just the electric potential? I'm confused about this part of statement 1.

  1. If I have one battery and one resistor only, from what I understand is that after electrons exit the resistor, they have 0 electric potentials.

  2. My understanding based on the first answer of "Transistor":

It's getting pushed by the potential difference in other parts of the circuit.

This statement is referring to electrons in between resistors.

After the electrons exit the resistor, even though they have an electric potential of 0, they still flow to the positive terminal because of the fact that the electrons currently flowing through the resistor has an electric potential, therefore, it needs to move forward and exit the resistor which pushes the electrons already out of the resistor towards the positive terminal. Doesn't this mean that the electrons out of the resistor already has an electric potential because the electrons in the resistor provide a force for them? Aren't they have 0 electric potentials at the end of the circuit?

  1. In figure 2, in the post that I linked above, when the user "Transistor" replies

What's driving the current is the potential difference between the top of the tank (battery +) and the open end of the pipe (battery -).

Does this mean that after the electrons exit the resistor, the negative terminal is no longer apply a force or push but now the positive terminal is applying a pull force? So if I place a voltmeter on each side of the resistor, it reads the voltage of the negative electric potential. Is this why electrons are still able to flow to the positive terminal even though it has 0 electric potential? Basically the same question as statement 1. Can someone explain the quote above.

Sorry for bad format, this is my first post.

Best Answer

If you want to think about the force electrons experience when they are moving in a current carrying conductor, you should not think in terms of voltage, instead electric field is what is forcing the electrons to move (and what is creating the potential difference). In terms of electric field, the Ohm's law can be expressed as $$j = \sigma E,$$ where, \$j\$ is the current density (per unit area), E is the electric field and \$\sigma\$ is the conductivity. The resistors have a finite conductivity, hence a finite current (through resistor) requires it to have an electric field to force the electrons move in one direction. Due to this field, each electron will experience a force qE and will drift slightly in the direction of force (which is what is called current).
Where does this field come from?
Could come from a battery connected across the terminals of resistor.

Now, assuming the resistor is connected to the battery by an ideal wire having infinite conductivity, from Ohm's law you can see that even if the electric field is zero, you can still have a finite current density. A zero electric field means there is no potential drop across the wire but it still can have finite current.
In reality, the wire does have a finite but very high conductivity so a small electric field (and consequently potential difference) is sufficient to drive current through it. This potential difference can be neglected for practical purposes.