A charge q moving in vacuum with a constant velocity v has conserved overall energy because the Electric field density on the direction of v increases while the field density in the other direction decreases.
For sake of argument let's consider the following example:
particle position is [x, y, z] = [0, 0, 0] at time t1 = 0
v = [vx, vy, vz] = [1, 0, 0] = 1 m/s (moving along x axis)
Electric field along x-axis at time t1
at E(-1, 0, 0) = q / (4π ε x2) = q / (4π ε)
at E( 1, 0, 0) = q / (4π ε x2) = q / (4π ε)
At time time t2 = 0.5s the particle position will become [0.5, 0, 0]
the electric field alonx x-axis is now
at E(-1, 0, 0) = q / (4π ε (-1 - 0.5)2) =q / (4πε 2.25)
at E( 1, 0, 0) = q / (4π ε (1 - 0.5)2) = q / (4πε 0.25)
dE(-1, 0, 0) = (q / (4πε 2.25) - q / (4π ε) ) is negative
dE( 1, 0, 0) = (q / (4πε 0.25) - q / (4π ε) ) is positive
Moving particle field is conserved. In front of the particle the field increases at every point, while in the back it decreases at every point (with respect to previous position).
The potential energy of the field in front of the particle is increasing and vice verse for the back.
Epot = (1/2) ε * dE2 * dVolume
If one will integrate over all volume the potential field energy is constant, but the energy density will increase along the direction of movement while it will decrease in the opposite direction.
PS: we have a non relativistic velocity v << c, but the change in field will propagate with the speed of light that means that the field density on the direction of the movement in greater than in the back. This is like red/blue shifting. If you are in front of an ray emitting object the light will come blue-ished while if you are in the opposite direction of speed it will come red-ish.
Note: to make my explanation shorter I've only considered electric field along one axis; but you can deduce by your self what happens in all other directions.