How Does the Electric Field Behave with Moving Charges?

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SUMMARY

The discussion centers on the behavior of electric fields generated by moving charges, emphasizing the limitations of Coulomb's Law. It is established that electric fields propagate at the speed of light 'c' and that a correction factor is necessary when charges are in motion. The conversation references Richard Feynman's insights on electrodynamics, highlighting that the electric field at a point is influenced by the historical behavior of charges. Participants agree that while Coulomb's Law applies to stationary charges, moving charges require additional considerations to accurately describe the electric field.

PREREQUISITES
  • Understanding of Coulomb's Law and its limitations
  • Familiarity with the concept of electric fields and their propagation
  • Basic knowledge of special relativity and reference frames
  • Introduction to gauge transformations in electromagnetism
NEXT STEPS
  • Study the implications of gauge transformations in electromagnetism
  • Learn about the relationship between electric and magnetic fields in moving reference frames
  • Explore the role of potentials in electromagnetism, including electric and magnetic potentials
  • Investigate the concept of electromagnetic radiation and its relation to moving charges
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Physics students, educators, and researchers interested in advanced electromagnetism, particularly those exploring the dynamics of electric fields and the effects of charge motion.

Dr.Brain
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Growing up from my high school physics where I was taught a simple formula for calculating the electric field. On reading on many physics books , I realized that I could precisely write down the whole electrodynamics on one page.I learned that these electrical influences cannot possibly travel at a speed greater than 'c'.So that put me to think about whether Coulombs Law was true or not because it didnot have the 'correction-factor' as I term it.Later while reading Fynmann-I , i got what I was looking for , I read about a good formula which sums up the whole electric field into an equation comprising of three terms .

First one is the coulombs Law which is suppose is wrong, because it does not take into account for fact that electrical influences have a speed limit of 'c'.

Second one is the correction factor which I had been looking for

Thord term is a sort of description for radiation.

My first question:

1. So how does electric field behave? Does it:

i) propagate from the charge at 'c' and its strength decreases with distance while propagate.

ii) When a charge is created a field is always there with it, its their property, there is not initial propagate.

................

In the book , Feynman says that it is not possible to calculate the
electric field at a certain point because you don't know precisely where the charge is now and at what distance it is now. Also the electric field at a point is due to behavious of charge in the past, you cannot calculate what sthe ELECTRIC FIELD 'now'.So a correction-term is added to the formula of electric field at a point .

My Second question:

Is my explanation right?
"Coulomb's Law is perfect for finding electric field at a point only if the charge is at rest or the charge moves infinitesimally slowly, but when the charge is moving , we need to add some 'correction electric field' to make up for the 'r/c' time lost due to field propagate taking time"


BJ
 
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Dear Dr. Brain,

It sounds to me like you're delving into the details of "gauge transformations", or at least in the alternative gauges that the E&M fields can be defined with.

These are complicated issues. Please forgive me if I screw up an explanation.

Dr.Brain said:
My first question:
1. So how does electric field behave? Does it:

i) propagate from the charge at 'c' and its strength decreases with distance while propagate.

ii) When a charge is created a field is always there with it, its their property, there is not initial propagate.

Case (i) is correct. Case (ii) is impossible because charge is conserved. That is, when you create a + charge, you also have to simultaneously create a - charge. Thus the initially created situation is a dipole. When the pair is first produced, they are "infinitely" close together, and therefore the strength of the dipole is "infinitely" small. Note that this is all in the QM "position" representation. If you look at the problem in the momentum representation, you will be confused.


Dr.Brain said:
My Second question:
Is my explanation right?
"Coulomb's Law is perfect for finding electric field at a point only if the charge is at rest or the charge moves infinitesimally slowly, but when the charge is moving , we need to add some 'correction electric field' to make up for the 'r/c' time lost due to field propagate taking time"

This feels to me like a reasonable explanation, but it is not the one I would give.

Look at it from the point of view of relativity. If the charge is moving there is (at least instantaneously) a reference frame that moves with the charge. In that reference frame, the charge is not moving. So in that reference frame, the usual Coulomb's law will work.

So what you really want to know is how an electric field is transformed when you move from one reference frame to another. What happens is that the electric field becomes a mixture of an electric field and a magnetic field. So a moving charge produces an electric field (according to Coulomb's law) and a magnetic field (according to whoever's law that was).

If I recall correctly, the best way of looking at this is by using the potentials. That is, use the electric potential (voltage) whose gradient is the electric field, and the magnetic potential whose gradient is the magnetic field. The electric potential acts like the "time" component of a 4-vector whose "space" component is the potential of the magnetic field. This 4-vector transforms under the usual methods of transforming relativistic 4-vectors. This tells you how the potentials transform. (If I recall correctly, I admit it's been a quarter century since I was educated in the subject.) Since the potentials transform in this manner, it is fairly easy to find how the fields themselves transform.

Carl
 
Look at it from the point of view of relativity. If the charge is moving there is (at least instantaneously) a reference frame that moves with the charge. In that reference frame, the charge is not moving. So in that reference frame, the usual Coulomb's law will work.

In the reference frame you are talking about , no doubt charge will appear to be at rest , but the point on which we need to calculate the electric field will be moving , so Coulomb's Law is still not applicable unless the correction-factor is added.

Case (i) is correct. Case (ii) is impossible because charge is conserved. That is, when you create a + charge, you also have to simultaneously create a - charge. Thus the initially created situation is a dipole. When the pair is first produced, they are "infinitely" close together, and therefore the strength of the dipole is "infinitely" small. Note that this is all in the QM "position" representation. If you look at the problem in the momentum representation, you will be confused.

So should I say that " Electric field initially propogates from a charge at speed 'c' and its strength decreases with distance, and once it has propogated , there comes a static state where the field setup by above process remains as it is and then nothing else happens...(?)

Now let's say charge at rest has an electric field setup by above process, now when this charge moves from pt. A to pt.B , and it is still moving , when it is at B , does it carry the same electric field as A or does it again propagate a new electric field , becayse it is moving , does it have any effect on the field?

BJ
 

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