I understand what the voltage is and I realize that the battery makes an electric field due to the accumulation of the charges in the anode and cathode this electric field causes electric potential difference between the anode and the cathode causing the electrons or the positive charges to move in the conductor. What I can't understand is why the voltage difference between two points in the circuit is zero if there is no resistance between these two points. I mean the field exists and it exerts work on the charge moving it from point to another so there should be voltage difference between any two points.
You can also think of the battery as say one atom that has an electron and one atom that really wants another one, if you connect them the electron travels and there is no more potential. When the original atoms were apart you could feel the force pulling them together, afterwards there is no more force. In a real battery there are zillions of atoms trying to transfer electrons but the potential is the same as of one atom was trying to transfer (li ion ~3.6v,lead acid 2.2v) in practice shorting the battery takes a while and al lot of heat is generated in the conductors and battery.
Actually there is. But it is so small that it is meaningless.
All wire has resistance, so as current flows through a wire, there will be a loss of potential (voltage). The greater the current, the greater the loss.
It is how the electrical system is designed. If the maximum % voltage drop to the wire is 3%, the maximum loss at 120V would be 3.6V to the wire or feeder.
This means the voltage at the load will be less than the source voltage. 120V at the panel and the voltage at the load will be >116.4V and <=120V. Actual voltage depends on current (0A means 120V), wire distance load is from panel, how close current is to full-load ampacity, temperature.
As can be seen from the drawing, at rated current, voltage at socket depends on distance from panel.
So now scale. #10 AWG copper has a resistance of 1Ω / 1000ft @ 20°C. Maximum current 52A - maximum ampacity 41.6A. If 41.6A flowed to a load, the potential loss would be: 41.6mV/foot.
Now at 120V source, 41.6mV/foot is insignificant, but it does add up so there is a maximum distance a full-load load can be located and operate correctly. To work properly at a further distance a larger wire size is required.
Ultimately, the answer to your question is all about scale. There is a loss between any two points in a circuit, but if we are talking about V, the mV and μV are meaningless. Similarly, V are meaningless in discussions on kV or MV.