Why is the current in an electric circuit constant? Considering electrostatics, suppose we have two charges of equal value and opposite sign and we put electrons along the road between them: We know that the electrical force or field is different from point to point between them because of distance according to Coulomb's law and since the force is different then every electron has a different speed.
Similarly, in an electrical circuit we have a battery in which positive charges accumulate at the positive terminal and negative charges accumulate at the negative terminal. Suppose we have a wire across which has regular matter distribution and regular cross section: 
The force applied to every electron must be different because the distance is different. Then, the velocity of every electron is different - then, the current is different from point to point because current definition is that number of electrons that cross through a point per second and the point that has the most force applied has most current and least force has the least current .
Doesn't that contradict the fact that electrical current is constant at all points of a circuit? How can we explain the fields and currents in circuits mathematically, and not by assumptions?
In university books and references of field theory, they always suppose that the electrical field in the wire I described is constant and they start in their calculations with
$$ \vec E=\sigma \vec J$$
where $\sigma$ is selective conductivity and $\vec J$ is the current density. And that's it - without any proof or explanation. They didn't begin with Coulomb's law which is the most important and basic law in electricity .
I will be very thankful if someone can explain this matter in detail and give the proof of the constancy of the current starting from Coulomb's law.
 A: 
Suppose we have in static electricity two charges negative and
  positive equal in absolute value and we put electrons along the road
  between them ,we know that the electrical force or field is different
  from point to point between them because of distance according to
  columb law and since the force is different then every electron have
  different speed

If there were no charges on this road (I will call it a wire), then you would be correct.  But you have already said that there are electrons there.  Each of them also contributes to the electric field present, so you cannot say that it is simply based on distance from the external charges (or the battery).  
In a steady state, the electrons will move in such a way that the field inside the wire is zero.  This does not depend on distance from the battery.  You cannot use Coulomb's law unless you know where the charges are located, but that becomes difficult when they are all moving around.
A: Most wires are made up positively charged parts (e.g. protons) and negatively charged parts (e.g. electrons).  each generates it's own electric field, and the field gets very large as you get very close to any single electron, enough to be larger than those due to all the 10^23 or more electrons.  So the true electric field is going to jump around.  But what we usually talk about is not this microscopic electric field, but an average one.  You pick a region large enough to have many electrons, but small enough that if you translate the region a bit the average (the integral divided by the volume) doesn't change much.  It's like a population density for a city, you don't want it to spike up on each person's bed or even their bedroom, you want it to smoothly change from one part of the city to easily see where it is relatively dense.  This means the divergence of the macroscopically averaged field isn't the literal microscopic charge density but again the charge density as averaged over a similarly large region that is large enough that it has many electrons but small enough that if you translate it a bit it doesn't change much.
OK, so those are the fields and charge densities we are talking about.  So let's look at that circuit element.  If it is ohmic, then it is a resister and generally we want resisters that keep the current inside it, or at least close to it.  If there is a work function that requires some big energy to knock an electron out of the metal, and the thermal environment is cool enough so that energy if much large than a thermal jostle would likely provide then the charges will flow in, on, or near the element, but not escape away.  So we aren't seeing an acceleration of charge in the direction normal to the wire, so $\vec{E} \cdot \vec{n}=0$ where $\vec{n}$ is normal to the boundary of the circuit element, or in terms of potential $\partial V / \partial n = 0$.  How about the parts of the circuit element not on the boundary, the parts hooked up to the battery?  One of them is at potential one constant potential the other at a possibly different constant potential.  For simplicity and concreteness, take one end to be $(0,r\sin(\theta),r\cos(\theta))$ at potential zero and $(L,r\sin(\theta),r\cos(\theta))$ to be at potential $V_0$, so we have a short straight circular wire of length $L$ and radius $r$, though the length and radius are not important.  The final physical insight we need is that the macroscopic charge density is zero.  If it were nonzero and the macroscopic electric field were nonzero then there would be macroscopic acceleration and not a steady state.  Non steady states might be important and useful, but they are often transitory as a system approaches a macroscopically steady state, so looks look for a macroscopically steady state, if we find one, then it might be an accurate description of this setup.  So now we will look for a macroscopically steady state, so one where the macroscopic electric field has no divergence, and we have the boundary conditions specified on the entire boundary.
Now we can find one solution with for the electric potential $V_1$ inside the resistor, namely $V(x,y,z)=xV_0/L$.  Suppose there were another solution $V_2$ that also has $V=0$ at one end, $V=V_0$ at  the other and $\partial V /\partial n = 0$ on the edges of the resistor, and then consider the two electric fields $\vec{E}_1=-\nabla V_1$ and $\vec{E}_2=-\nabla V_1$, then consider the differences $\vec{E}_3=\vec{E}_1-\vec{E}_2$ and $V_3=V_1-V_2$, and keep in mind these are fields, not just numbers.  We note that the divergence of $E_3$ is zero because the divergence of $E_1$ and $E_2$ are each zero because there is no net macroscopic charge density.  So the identity $\nabla \cdot (V_3 \vec{E}_3)=V_3(\nabla \cdot \vec{E}_3)+\vec{E_3}\cdot(\nabla V_3))$ (which is a general vector calculus identity) reduces to $\nabla \cdot (V_3 \vec{E}_3)=\vec{E_3}\cdot(\nabla V_3))=\vec{E_3}\cdot(-\vec{E_3})=-(E_3)^2$ and that the right hand side is never positive (it is zero at most).  These two equal things $\nabla \cdot (V_3 \vec{E}_3)$ and $-(E_3)^2$are still fields, so we integrate both over the whole wire to get
$\int_W \nabla \cdot (V_3 \vec{E}_3) dxdydz=\int_W-(E_3)^2 dxdydz$, let $\partial W$ be the surface of the wire, and apply the divergence theorem to get:
$\int_{\partial W} (V_3 \vec{E}_3)\cdot d\vec{a}=\int_W-(E_3)^2 dxdydz$.  Now since the two solutions have the same potential at the two ends of the wire (so $V_3$ is zero there), we only need to worry about the sides of the wire $S$ so we get:
$\int_S (V_3 \vec{E}_3)\cdot d\vec{a}=\int_W-(E_3)^2 dxdydz$.
But $\vec{E}_3\cdot d\vec{a}=\vec{E}_3\cdot \vec{n} da=-\partial V_3 /\partial n da $ and  also $\partial V_3 /\partial n = 0$ on the sides since $\partial V_1 /\partial n = 0$ and $\partial V_2 /\partial n = 0$ there.  So the whole surface integral vanishes.  So we get:
$0=\int_W-(E_3)^2 dxdydz$.  Now if for the sake of argument, $\vec{E}_1$ didn't equal $\vec{E}_2$ then there would have to be a point where they were different with a difference in magnitude $d$ at that point.  Then, since they are continuous there would have to be a whole little ball about that point that stays inside the wire where this difference in magnitude was between $d/2$ and $d$, so the integral of $-(E_3)^2$ over that little ball would be between $d/2$ and $d$ times the negative of the volume of that little ball, but the integral over the whole rest of that wire can give at most zero and maybe even be negative, so there is no way the sum of a strictly negative number and a nonpositive number could give you zero, so it must not be the case that $\vec{E}_1$ and $\vec{E}_2$ are different since $\vec{E}_3$ is zero inside the wire.  And we know what $\vec{E}_1$ is, it is $(V/L,0,0)$.  So that's the electric field.  It is uniform.  It is the only macroscopic field that is macroscopically steady.
Does the macroscopic field have to be macroscopically steady?  Well, you have a battery, and the purpose of the battery is to provide a steady voltage, and voltage is a gauge dependant concept in full time dependant electrodynamics, so if you are saying that you have a battery and all you want to say about it is that it maintains a steady voltage and you don't want to talk about how it tries or how it dynamically interacts with nonsteady macroscopic field, then I think this is all we can say.
