Electric field associated with moving charge I have recently started to learn about the electric field generated by a moving charge.  I know that the electric field has two components; a velocity term and an acceeleration term. The following image is of the electric field generated by a charge that was moving at a constant velocity, and then suddenly stopped at x=0:

I don't understand what exactly is going on here.  In other words, what is happening really close to the charge, in the region before the transition, and after the transition.  How does this image relate to the velocity and acceleration compnents of the electric field?
 A: When the charge is moving at constant velocity, it emits a non-uniform electric field that can be calculated from the Lienard-Wiechert potentials. This field has a velocity component but no acceleration component, as the charge is not accelerating.
When the charge is not moving, it emits a spherically symmetric electric field that can be calculated from Coulomb's Law. The field, once again, has a velocity component and no acceleration component, as the charge is not accelerating (and the velocity component reduces to the Coulomb field when the velocity is zero).
In order to transition from moving to not moving, the charge must accelerate. Here, again, the charge's fields may be calculated from the Lienard-Wiechert potentials, but now there is a nonzero acceleration component to the field, which corresponds to radiation. The shape of this field can be reasonably approximated, for short accelerations, by requiring that the electric field lines be continuous through the transition, and this approximation appears to be used in your diagram.
When the charge suddenly stops, its field does not change instantaneously across all space. Rather, the change from non-uniform velocity field to Coulomb field propagates outward at the speed of light. If the charge suddenly stopped at $x=0,t=0$, and we examined the field at time $t$, we would find that the Coulomb field was present in the region $r < ct $, while the non-uniform velocity field was present in the region $r>ct $. This is exactly what your diagram depicts.
A: According to Special Relativity, information travels at the speed of light and this case is no different. The information here refers to the position of the particle at a certain time. 
Let me explain. When the charge was at x=1, its field lines were radially outward. When the charge reaches x=0, the information that the charge has reached that point hasn't been conveyed to the region outside the circle in the figure. Hence if the field lines outside the circular region is extrapolated, it intersects at x=1
A: You can understand rather simply by first considering an electric force between two charged particles. Let us say that we can "turn on and off" one of the particles, so that when it is off, it has no charge and will not interact with the other charge, and when it is on, it will have charge and will interact with the other charge. At the instant we turn on the charge, does the other particle feel any force? The answer is no. The force between the charges will only begin a finite time after we turn on our charge, as the electric force (like anything else) is limited by the speed of light. 
This means the instant our charge is turned on, its electric field is zero at all points in space. As time progresses, the electric field spreads out, reaching farther and farther away, "traveling" at the speed of light. This is not travel, however, it is merely delayed effects of the electric field. 
Considering the charge in the diagram you have given, when its velocity is constant, its electric field contribution is steady, no change. When it suddenly stops, we have rather extreme acceleration, and the electric field as observed by a given point will initially be exactly the same as the field before acceleration, but the changes will gradually take place, giving rise to the "ripple" shown in the picture. The arcs of the field lines are from the time when the particle was accelerating down. 
Both the relative motion of the charge initially (due to special relativity, observed as a magnetic field) and the deceleration of the charge contribute to the resulting electric field around the charge. As an accelerating field can generate far field radiation, we see that the charge radiates when it accelerates. 
This is a fairly hand-wavy explanation of the radiation of a moving charge, but it should help guide you, I hope, to more thorough and interesting treatments of the topic. 
A: No such exist in reality... Electrons or charged atoms never move in constant velocity and to stop one, one must apply electric o magnetic field...
Also electric field of an electron never changes regardless of how it is moving...
Isolated hypothetical cases are useless because one cannot prove it right or wrong... Reality is experimental physics and new properties are discovered and not invented...
A: The curved path that you see looks like the electron reacting to the Lorentz force. Here's how that works. The key insight is that a moving charge induces a magnetic field. This magnetic field, combined with the present electric field, gives you the full form of the Lorentz force:
$$\mathbf{\vec{F}}=q(\mathbf{\vec{v}} \times \mathbf{\vec{B}})+q\mathbf{\vec{E}}$$
Here you immediately see that there is both a velocity $\mathbf{v}$ of the particle and an acceleration hiding away in the force. Perhaps this illustration would be helpful:

So the full electromagnetic field influences a particle to move in a curved trajectory and the curve is dependent on the charge of the particle. The electron moves in a curve due to the cross product between the velocity and the magnetic field. From here, you could calculate the velocity and the particle from electric field and the force.
Hope this helps.
