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Lets consider it case by case:

Case 1: Charge particle is at rest. It has electric field around it. No problem. That is its property.

Case 2: Charge particle started moving (its accelerating). We were told that it starts radiating EM radiation? Why? What happened to it? What made it do this?

Follow up question: Suppose a charge particle is placed in uniform electric field, it accelerates because of electric force it experiences. Then work done by the electric field should not be equal to change in its kinetic energy right? It should be equal to change in K.E + energy it has radiated in the form of EM Waves. right? But then, why don't we take energy radiated into consideration when solving problem (I'm tutoring grade 12 students. I never encountered a problem in which energy radiated is considered.)

How does moving charges produce magnetic field?

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A question that comes to my mind right now is, given an acceleration of magnitude $a$, what is the resulting frequency of the EM wave? I am almost a physicist now, I did not continue because the amazing world of programming has captured me so I am working right now! But someday I will go back and continue studying physics, and it also was mentioned to us but I haven't seen any mathematical derivation. – iharob Nov 11 '15 at 0:08

A diagram may help:

enter image description here

Here, the charged particle was initially stationary, uniformly accelerated for a short period of time, and then stopped accelerating.

The electric field outside the imaginary outer ring is still in the configuration of the stationary charge.

The electric field inside the imaginary inner ring is in the configuration of the uniformly moving charge.

Within the inner and outer ring, the electric field lines, which cannot break, must transition from the inner configuration to the outer configuration.

This transition region propagates outward at the speed of light and, as you can see from the diagram, the electric field lines in the transition region are (more or less) transverse to the direction of propagation.

Also, see this Wolfram demonstration: Radiation Pulse from an Accelerated Point Charge

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Could you kindly comment on my "Follow up question" – claws May 21 '13 at 12:55
@claws, my only comment for now (I'll save more for later when I have time) is to refer you to this on the question of "Does a uniformly accelerated charge radiate?": – Alfred Centauri May 21 '13 at 13:13
@Alfred Centauri: Purcell also followed the same procedure. Check this. – MAFIA36790 Jan 11 at 3:14

The second problem is quite tough. J. D. Jackson comments, in the introductory remarks of his chapter on 'Radiation Damping, Classical Models of Charged Particles', that we know how to solve classical electrodynamics problems in two ideal conditions - a) given charge and current densities, how to compute the fields and b) given the fields, how to find the motion of charged particles in their presence. When charged particles accelerate, they do produce radiation which in turn affects the motion of all other charged particles. However, that problem is, Jackson says, still unsolved.

Coming to the first problem, if you calculate $\vec{E}$ and $\vec{B}$ for a moving charged particle, you will see that they depend on the acceleration $\vec{a}$ of the charged particle. Now calculate the Poynting vector $\vec{S}$. You will observe that $\vec{S}$, depends on acceleration but not velocity. Integrating it to get power radiated gives the famous Larmor formula. You may want to refer to Griffiths' chapter on 'Electromagnetic Radiation'.

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For Jackson, the radiation also affects the particle itself. For Griffiths, the Poynting vector does have terms that contain velocity, they just fall off faster than 1/r and so don't contribute a finite energy to an infinitely distant surface. But they are carrying power away, just not to super far away. So back to Jackson, that energy loss also affects the motion of the charge. – Timaeus Nov 12 '15 at 1:40

protected by Qmechanic Jan 27 '14 at 19:23

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