Energy-work theorem and dissipation of energy by an accelerating charge By the work energy theorem we have that the total energy of a nonrelativistic point charge, $q_0$ of mass $m$, moving in an electric field $\mathbf{E}$ is
$ E = E_k + U_e =  \frac{1}{2}mv^2 + q_0V \quad Eq.1 $
As I read in my text:

The electrical charge acquires potential energy. If the charge is
released, work is done by the field and the charge accelerates. It
means that its potential energy is converted into kinetic energy.

But if an accelerating charge emits electromagnetic radiation some electric potential energy, should be dissipated and could not be converted into kinetic energy. How can eq. 1 hold? The text mentions the dissipation of energy by accelerating charges only later in the course, and says nothing when treating energy conservation of moving charges in the electrostatic field.
 A: The conservation of energy in this problem is of course an approximation - like many things in physics (and one could even say that physics is only an approximation to the real world.) The energy dissipated by an accelerated charge can be calculated using the Larmor formula:
$$
P=\frac{2}{3}\frac{q^2a^2}{c^3},$$
where $a$ is the charge acceleration. The speed-of-light entering the numerator hints that this is a relativistic corrections - something one is likely to ignore in problems, where the kinetic energy can be described by non-relativistic expression $\frac{mv^2}{2}$.
A: 
But if an accelerating charge emits electromagnetic radiation some electric potential energy, should be dissipated and could not be converted into kinetic energy. How can eq. 1 hold?

That equation holds for quasi-static rearrangements of charges in electrostatic field (no radiation, no magnetic field). As soon as charges move, one has to take into account magnetic energy as well, which eq. 1 does not do. And if charges accelerate, radiation may carry away EM energy away at the expense of local EM energy or other kind of energy, which is not taken into account by eq.1.
More general formulation of conservation of energy is the Law of local conservation of energy. In EM theory this is based on work-energy interpretation of the Poynting theorem: work of electric forces in a region plus increase of EM energy in that region equals influx of EM energy through boundary of the region from outside. This is strictly speaking not conservation in the region - energy in the region may change due to energy coming in, or leaving out, through the boundary.
A: The standard Electric potential is not defined for electrodynamics, this is because in order for a potential to exist, the electric field needs to be able to be written in the form
$$\vec{E} = -\nabla V$$
This is the only form of the Electric field for its line integral to be path independant  and thus uniquely defining a potential V at a point in space, and is called a Conservative Vector field
In electrodynamics:
$$\vec{E} = -\nabla V- \frac{\partial \vec{A}}{\partial t}$$
This is a non Conservative vector field,Meaning your text books form of energy conservation is invalid
Poyntings theorem is the theorem for energy conservation in electrodyamics:
https://en.m.wikipedia.org/wiki/Poynting%27s_theorem#:~:text=8%20External%20links-,Definition,energy%20flux%20leaving%20that%20region.
