**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 v_{0} 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 v_{0} << 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*v_{0}Δt_{1}. That distance is very small compared with the other distances in the
picture.

We now examine the electric field at a time t=Δt_{2}>>Δt_{1}. The electric field reaches a distance R=cΔt_{2}. Thus, the field at a distance greater than
R=cΔt_{2}, in region I must be a field of a charge which has been moving and is still moving at the constant speed v_{0}. That field appears to come from the point
x=v_{0}Δt_{2} 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(Δt_{2}-Δt_{1}), in
region II must be a field of a charge at rest close to the position x=0 (exactly at x=1/2*v_{0}Δt_{1}).

What must the field be like in the transition region, the spherical shell of thickness cΔt_{1} 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 v_{0} << 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 E_{r} and a transverse component E_{θ}. From the geometry of the figure their ratio is easily found.

A remarkable fact is here revealed: E_{θ} is proportional to 1/R, not to 1/R^{2}! As time goes on and R increases, the transverse field E_{θ} will eventually become very much stronger
than E_{r}. 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

Total energy in transverse electromagnetic field is

**Radiating Charge simulation - PhET INTERACTIVE SIMULATIONS**

**Purcell appendix B in the CGS system - iSites**

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