A particle with charge q has been moving in a straight line at constant speed v0 for a long time. It runs into something, let us imagine, and in a short period of constant deceleration, of duration τ, the particle is brought to rest. The graph of velocity versus time in Fig. B.l describes its motion. What must the electric field of this particle look like after that? Figure B.2 shows how to derive it.

We shall assume that v0 is small compared with c. Let t = 0 be the instant the deceleration began, and let x = 0 be the position of the particle at that instant. By the time the particle has completely stopped it will have moved a little farther on, to x = 1/2;v0τ. That distance, although we tried to indicate it on our diagram, is small compared with the other distances that will be involved.

We now examine the electric field at a time t = T ≫ τ. Observers farther away from the origin than R = cT cannot have learned that the particle was decelerated. Throughout that region, region I in Fig. B.2, the field must be that of a charge which has been moving and is still moving at the constant speed v0. That field, as we discovered in Section 5.7, appears to emanate from the present position of the charge, which for an observer anywhere in region I is the point x = v0T on the x axis.