Physics Asked on October 25, 2020
My textbook states the following: " If a charge’s velocity is constant, the rate-of-change of the $E$-field is steady, and the resulting $B$-field is constant.". To my understanding, this is clearly just an application of the Ampere-Maxwell equation/Law.
Now my problem in understanding is this: Suppose we have a charge that is moving at some non-zero constant velocity relative to the earth reference frame and towards an observer standing on the earth. The charge starts off very far away from an observer and so the electric field near the observer is very weak. As the charge gets closer to the observer, the field near the observer gets stronger. Clearly the electric field detected by the observer is time dependent and this rate of change with respect to time cannot be constant or ‘steady’ due to Coulombs inverse square dependence. That is, the closer the charge to the observer the more rapidly the change in field strength near the observer. Despite this, my textbook seems to state that the electric field detected by the observer must actually have a constant rate of change. Not only that but clearly no electromagnetic wave is created because the charge has constant velocity. But my understanding leads me to think that even charges moving at constant velocity should generate changing electric fields with respect to earth reference frames and thus an EM wave should be created.
There must be something I am overlooking or under-thinking? Where am a I wrong in my understanding? Any help would be greatly appreciated!
When the charge moves at a constant velocity, the electric field registered instantaneously by an observer located at $r$ away from the charge is similar to the field registered by the same observer when the charge is at rest WRT him at $r$. Remember that the field vector is supposed to be along the relative velocity at the location of the observer. The relevant proof is slightly complicated, however, I think you can find it in the related Electrodynamics textbooks like this one.
But my understanding leads me to think that even charges moving at constant velocity should generate changing electric fields with respect to earth reference frames and thus an EM wave should be created.
The changing electric field is due to the fact that the charge approaches the observer, and according to Coulomb's law, the magnitude of the electrostatic force of attraction or repulsion between two point charges is inversely proportional to the square of the distance between them. Therefore, as the charge approaches the observer the electric field increases.
Answered by Mohammad Javanshiry on October 25, 2020
A good way to understand that situation of charge moving at constant velocity is to use special relativity.
The physical situation must be equivalent to a charge at rest, and a moving approaching observer. Clearly there is no reason for a EM wave, because the field is static in this frame.
The observer will measure greater magnitudes of the static field as getting closer. And as you said, the relation is quadratic, so the increment is not constant.
Answered by Claudio Saspinski on October 25, 2020
May be I am somewhat late too your question but I have tried to answer :
Since the charge is approaching the observer with constant velocity , the electric field at all the points between the observer and the initial position of the charge and even where the observer is standing increases which gives rise to magnetic field at all those points.
But for em waves to be produced the electric as well as the magnetic field should oscillate at some interval
!(https://i.stack.imgur.com/wu5No.jpg)
but in your case since the charge is approaching the observer, the electric as well as the magnetic field only increases to a certain amount and when the charge passes the observer, both the fields only decrease. They never oscillated . So this could be the reason why em waves are not produced.
Hope it helps.
Answered by Ankit on October 25, 2020
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