Energy conservation in electrodynamic system Consider two charged particles initially at rest in the configuration below.

Let us assume the following:


*

*Starting at time $t=0$, we apply a constant force $f$ to the the bottom particle so that it has a constant acceleration $a=f/m$.

*The top particle has a large mass $M$.

*The distance $r$ is large enough so that the Coulomb repulsion between the particles, which is inversely proportional to $r^2$, is negligible.


Under these conditions the Lienard-Wiechert retarded radiative electric field due to the bottom particle produces a force $F$ on the top particle given by:
$$F(t)=\frac{qQa(t-r/c)}{4\pi\epsilon_0c^2r}.$$
For simplicity we assume that the mass $M$ of the top particle is so large that its acceleration due to force $F$ is negligible. Thus it does not produce a significant radiative electric field back at the position of the bottom particle.
Now let us calculate the energy $E_{in}$ that we supply to the system.
Let us assume that we apply a force $f$ to the bottom particle for a time interval $\delta t$.
During time interval $\delta t$ the bottom particle travels a distance $d$ so that the energy supplied $E_{in}$ is given by:
$$E_{in} = f \times d$$
The bottom particle has a constant acceleration $a$ so the distance it travels in time interval $\delta t$ is given by:
$$d = \frac{1}{2}a \delta t^2$$
Using the expression for the acceleration of the bottom particle, $a=f/m$, we find from the above two relations that the energy supplied to the system during time interval $\delta t$ is given by:
$$E_{in} = \frac{f^2\delta t^2}{2m}$$
Where has this energy gone?
The kinetic energy, $KE$, of the bottom particle after the time interval $\delta t$ is:
$$KE = \frac{1}{2} m v^2$$
The velocity of the bottom particle after a time interval $\delta t$ is given by:
$$v = a \delta t$$
Since the acceleration $a=f/m$ the above two equations imply that the kinetic energy of the bottom particle is given by:
$$KE = \frac{f^2 \delta t^2}{2 m^2}$$
Therefore, as expected, all the energy $E_{in}$ that we supplied during time $\delta t$ has gone into the kinetic energy $KE$ of the bottom particle.
But, as stated above, since the bottom particle is accelerating, after a slight time delay $t=r/c$, there is a force $F$ acting on the top particle. During the time interval $\delta t$ an energy $E_{top}$ is supplied to the top particle given by:
$$E_{top} = \frac{F^2\delta t^2}{2M}$$
My question is where has this energy come from given that all the energy we supplied, $E_{in}$, is fully accounted for in the kinetic energy of the bottom particle alone?
 A: To say that

we assume that the mass $M$ of the top particle is so large that its acceleration due to force $F$ is negligible.

means to take the limit $M \to \infty$ (since $a = \frac{F}{M}$ shall produce $a = 0$), since only infinitely heavy things do not accelerate when a force is applied. Under this assumption,
$$ \lim_{M\to\infty} E_\text{top} = 0$$
hence there is no violation of energy conservation in the case considered.
A: 
My question is where has this energy come from given that all the energy we supplied, $E_{in}$, is fully accounted for in the kinetic energy of the bottom particle alone?

The work done by external force accelerating the particle 1 (bottom) goes into kinetic energy of the particle 1.
The work done by electric force of the particle 1 on the particle 2 (top) goes to kinetic energy of the particle 2. 
Work-energy theorem is valid because the kinetic energy of particle 2 increases due to work of electric force of particle 1 on particle 2. In the picture of continuous transfer of energy, energy is being accumulated in the particle 2 while there is flow of EM energy in the region surrounding it and while some EM energy is being sucked up into the particle.
The rate of increase of kinetic energy equals income of EM energy through boundary surface of the region (arbitrary closed surface that encloses the particle 2) minus the rate of increase of EM energy in the region.
