- #1
henry_arsenal
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Hello Everyone, I'm new to this forum, I've to say here is one of the greatest physics communities in the net and I'm really glad I have the opportuninty to use the huge amount of information presented here.
By the way, I have a question about Electromagnetics:
Consider two stationary charges, +Q1 and +Q2 and an observer beside them. This observer measures a columbian force on both of the charges and nothing else.
Now, imagine an observer getting close to the charges at a relative speed of V. He should measure an additional force (rather than the former columbian force) produced by the magnetic field which exists due to the fact the the moving observer finds out that the charges are getting close to him at a speed of -V, of course from his point of view.
Both of the observers obey the laws of physics and maxwell's equations correctly, but they measure different forces on the charges. What is the correct answer to this problem?
I guess it should be related to the special relativity transformations, but I don't know how. I'll be thankful if you help me understanding how to solve this problem.
Thanks
Arman.
By the way, I have a question about Electromagnetics:
Consider two stationary charges, +Q1 and +Q2 and an observer beside them. This observer measures a columbian force on both of the charges and nothing else.
Now, imagine an observer getting close to the charges at a relative speed of V. He should measure an additional force (rather than the former columbian force) produced by the magnetic field which exists due to the fact the the moving observer finds out that the charges are getting close to him at a speed of -V, of course from his point of view.
Both of the observers obey the laws of physics and maxwell's equations correctly, but they measure different forces on the charges. What is the correct answer to this problem?
I guess it should be related to the special relativity transformations, but I don't know how. I'll be thankful if you help me understanding how to solve this problem.
Thanks
Arman.