Charged Particle in Space: External Forces and Constant Velocity

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SUMMARY

A charged particle moving with constant velocity in a region devoid of external magnetic fields does not necessarily imply the absence of an external electric field. The discussion concludes that while the sum of external forces acting on the particle is zero, this can occur if gravitational and electric forces cancel each other out. The reasoning presented supports the conclusion that the external electric field may still exist, contradicting the textbook's assertion that the answer is simply "Yes." The participants emphasize the importance of considering all forces, including gravitational influences.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Basic knowledge of electromagnetic theory
  • Familiarity with the concepts of electric and magnetic fields
  • Awareness of gravitational forces and their interactions with charged particles
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  • Study the relationship between electric fields and charged particle motion
  • Explore the role of gravitational forces in electromagnetic systems
  • Investigate how moving charged particles generate magnetic fields
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Homework Statement


A charged particle, passing through a certain region of space, has a velocity whose magnitude and direction remain constant. If it is known that the external magnetic field is zero everywhere in this region, can you conclude that the external electric field is also zero?

Homework Equations


F=ma

The Attempt at a Solution


Constant velocity means no acceleration is occurring. No accelerations means that the sum of the external forces acting on the mass is zero. That can mean either no forces are acting on it, or that forces are acting on it, but they exactly cancel each other.

So in this case we have no magnetic field, but possibly an electric field. There might also be a gravitational field. So it is then possible that the gravitational force and the electric force exactly cancel each other out, so that the sum of the forces are zero. So then answer to the question then is "no".

Would you agree with my reasoning?
 
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I think the answer is yes.Cause the gravitational force is lower then electromagnetic force.##\vec{F}=Mmg/r^2##.Its a particle so I think the mass of particle will be very low.Then the force will be low.If we consider this situation the anwer will be yes.
 
RyanH42 said:
I think the answer is yes.Cause the gravitational force is lower then electromagnetic force.##\vec{F}=Mmg/r^2##.Its a particle so I think the mass of particle will be very low.Then the force will be low.If we consider this situation the anwer will be yes.
But you don't know an upper bound on the other mass.
 
haruspex said:
But you don't know an upper bound on the other mass.
I thought that way cause.If we consider sun and an electron the force will be very small like 10-11 ,##r=1##.If we think larger masses we need to thing something like black hole-electron maybe a neutron star electron.But I don't think that neutron star will pull just one electron.Maybe it does I am not sure.
 
I would say you are definitely correct to say no. Cathode ray tube televisions and ink jet printers balance gravity against EM all the time. What I am really curious about is how the moving charged particle is not creating its own magnetic field.
 
Wily Willy said:
I would say you are definitely correct to say no. Cathode ray tube televisions and ink jet printers balance gravity against EM all the time. What I am really curious about is how the moving charged particle is not creating its own magnetic field.
OP says the external magnetic field is zero.
 
This topic has 127 views. Any more thoughts? The reason that I'm asking is that the answer given in my textbook is simply a "Yes.", but no explanation whatsoever is given. Also, if the answer is yes, that would mean that my answer is incorrect, but my answer is based on my reasoning, and I don't see how my reasoning would be incorrect.
 
Fluxxx said:
This topic has 127 views. Any more thoughts? The reason that I'm asking is that the answer given in my textbook is simply a "Yes.", but no explanation whatsoever is given. Also, if the answer is yes, that would mean that my answer is incorrect, but my answer is based on my reasoning, and I don't see how my reasoning would be incorrect.
I don't think you were intended to invoke other forces, such as gravitation.
 
haruspex said:
I don't think you were intended to invoke other forces, such as gravitation.
This post and my posts makes the answer.
 

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