Loss of energy of an electric field, what does it imply? Suppose we have a positive electrically charged particle with infinite mass, so it is fixed at one point. We have also a negatively charged particle with finite mass. This second particle will be continually attracted to the first one, so it continually gains kinetic energy. To generate this increase in the kinetic energy, the electric field generated by the first one must be doing work, so it seems like the field is continually losing energy.
My question is, what does this loss in the field energy means? The electric field gets weaker? Does that imply a decrease of the electric charge that is generating the field?
I apologize for my writing, my English is not the best.
 A: 
Suppose we have a positive electrically charged particle with infinite mass, so it is fixed at one point. We have also a negatively charged particle with finite mass.

This is - if one leaves out the assumption about infinite mass - a good model for the attraction of an electron to an ionised atom.

This second particle will be continually attracted to the first one, so it continually gains kinetic energy.

This is again what happens to the electron at the beginning. At the end, the approach ends and a kind of barrier stops the electron.

To generate this increase in the kinetic energy, the electric field generated by the first one must be doing work, so it seems like the field is continually losing energy.

This is not what you will read in books. But it is a fascinating assumption that the attraction of the nucleus and the electron weakens their electric fields. This would have the following implications:
The lack of a negative shell around each atom, which is not self-evident because all electrons surround the nucleus (it is our observation that tells us atoms are largely neutral).
However, this is not an accepted assumption because the view of the free electron as the elementary unit of charge (measured by Robert Millikan) outweighs the assumption that the interaction of an electron and a nucleus could weaken their fields.
A: You need to be careful with your statements. “Continually attracted” in no way implies “continually gains kinetic energy”.
For example, consider a satellite in a circular orbit. It is continually attracted to the planet but its KE is constant. And for a satellite in an elliptical orbit it is continually attracted, but it actually loses KE during half of the orbit.
That said, there definitely are cases where you have an increase in KE (just not continually). So it is reasonable to ask where that energy comes from. It turns out that the electromagnetic field has an energy density equal to $$U = \frac{1}{2} \left(\epsilon_0 E^2+\frac{1}{\mu_0} B^2 \right)$$
As the charges come closer, if you integrate this expression over all space you will find that it decreases in exactly the same amount as the work done on the charges. The theorem that describes this is called Poynting's theorem.
As the charges become closer no charge is lost, but the fields from the two charges add together in a way that reduces the overall field strength and the overall field energy. It is precisely this behavior that makes "opposites attract" in electromagnetism.
A: First of all, there is nothing with an infinite mass in a finite space. But we can assume that one of the charges is fixed at the origin. There is another charge (was at a finite distance to the first one at $t=0$) which is attracted towards this one due to electrical force.
$$\Delta U=-\frac{qQ}{4\pi \epsilon_0}\left( \frac{1}{r_i}-\frac{1}{r_f}\right)=\frac{1}{2}mv^2$$
That's How energy conservation will work?
