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I know that as voltage increases, current increases by $V=IR$, but I really find some difficulty in understanding this at the atomic level (what happens with the electrons inside that wire). I understand this well by the water analogy, so I don't need any explanation using it. So here are my questions:

1) If electrons move inside the wire, surely there is an electric force that cause this movement. Does this force change in various points of the simple electric circuit as the electrons move from the negative terminal to the positive, or does it stays the same all of the time??

2) If the force changes, why doesn't the current in various points change in series circuits??

3)what I am still confused about, why the greater the difference in electric potential the greater the current?? I need this be explained using electrons without the water analogy.

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If electrons move inside the wire, surely there is an electric force that cause this movement. Does this force change in various points of the simple electric circuit as the electrons move from the negative terminal to the positive, or does it stays the same all of the time??

Yes, there is an electric field within the resistor material, and this causes a force on the free carriers in the material.

For a deeper explanation, you should investigate the Drude model of electrical conduction in metals.

Does this force change in various points of the simple electric circuit as the electrons move from the negative terminal to the positive, or does it stays the same all of the time??

If the material is uniform, the field will be uniform.

If you had, say one piece of copper wire ($\rho=1.68\times{}10^{-8}\ \mathrm{\Omega{}\cdot{}m}$) and one equal-sized piece of steel wire ($\rho\approx{}6.9\times{}10^{-7}\ \mathrm{\Omega{}\cdot{}m}$) connected in series between your battery terminals, then the field in the copper wire will be lower than in the steel wire.

If the force changes, why doesn't the current in various points change in series circuits??

If the current were different in different parts of the circuit, then charge would have to build up somewhere in the circuit. This accumulated charge would repel other carriers of the same polarity, produce an opposing field, until the current equalized. This would be a dynamic effect, and when we talk about the dc steady-state solution to a circuit, we mean the solution after all such dynamic effects have settled down.

In my example with two different types of conductors, the materials have different resistivity, so the field (and resulting force) must be different exactly to produce equal currents in the two materials.

why the greater the difference in electric potential the greater the current??

Potential difference is the integral of field

$$V_{ba} = \int^b_a \vec{E}\cdot\mathrm{d}\vec{l}$$

So in order to increase the potential between two points ($a$ and $b$) you must increase the field strength along the path between them. The stronger field produces a stronger force on the free carriers, and so you get a larger current.

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  • $\begingroup$ may u please explain to me by logic or any other way why electric field in copper wire is lower than that of a steel wire?? $\endgroup$ – Omar Ali Dec 28 '15 at 20:57
  • $\begingroup$ Because if it wasn't there'd be more current in the copper wire. Negative charge would build up at the junction between the copper segment and the steel segment, and this would cause the field to drop until the currents equalized. $\endgroup$ – The Photon Dec 28 '15 at 20:58
  • $\begingroup$ i have another question regarding electric field. If we attached a resistor how the electric field will be in order to make the resistor result to a voltage drop as the electron passes through it?? thanks in advance, i really appreciate your help $\endgroup$ – Omar Ali Dec 28 '15 at 22:03

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