Alternating magnetic field, electromagnetic radiation, electromagnetic induction
An electric field moves with the velocity that the particle had while emitting that field.
Electric field of a charge that stops
Suppose a charge q has been moving at constant speed v0 in the x direction for a long time. Suddenly it stops after a short period of constant deceleration. Electric field of a charge travels radially outward and moves with the velocity that the charge had while emitting that field.
The velocity versus time graph
The electric field
Assuming that the v0 << c, we can neglect the relativistic compression of the field lines. Time t=0 it was the moment when deceleration began, and position x=0 it was the position of the particle at that moment. The particle has been moved a little farther on before coming to a stop, Δx=1/2*v0Δt1. That distance is very small compared with the other distances in the picture.
We now examine the electric field at a time t=Δt2>>Δt1. The electric field reaches a distance R=cΔt2. Thus, the field at a distance greater than R=cΔt2, in region I must be a field of a charge which has been moving and is still moving at the constant speed v0. That field appears to come from the point x=v0Δt2 on the x axis. That is where the particle would be now if it hadn't stopped. On the other hand, the field at a distance less than c(Δt2-Δt1), in region II must be a field of a charge at rest close to the position x=0 (exactly at x=1/2*v0Δt1).
What must the field be like in the transition region, the spherical shell of thickness cΔt1 between region I and region II? A field line segment AB lies on a cone around the x axis which includes a certain
amount of flux from the charge q. Due to the Gauss' law, if CD makes the same angle θ with the axis, the cone on which it lies includes that same amount of flux. (Because the v0 << c, we can neglect the
relativistic compression of the field lines.) That's why AB and CD must be parts of the same field line, connected by a segment BC. The line segment BC shows us the direction of the filed E within the shell. This field E within the
shell has both a radial component Er and a transverse component Eθ. From the geometry of the figure their ratio is easily found.
A remarkable fact is here revealed: Eθ is proportional to 1/R, not to 1/R2! As time goes on and R increases, the transverse field Eθ will eventually become very much stronger than Er. Accompanying this transverse (that is, perpendicular to R) electric field will be a magnetic field of equal strength perpendicular to both R and E. This is a general property of an electromagnetic wave.
Now we calculate the energy stored in the transverse electric field above, in the whole spherical shell. The energy density is
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