Charged Particle in Space: External Forces and Constant Velocity

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A charged particle with constant velocity in a region with zero external magnetic field does not necessarily imply that the external electric field is also zero. The particle's constant velocity indicates that the net external forces acting on it sum to zero, which could result from the cancellation of multiple forces, including gravitational and electric forces. Some participants argue that gravitational forces are typically much weaker than electromagnetic forces, suggesting that the electric field could still be present. The original poster expresses confusion over their textbook's answer of "yes," as they believe their reasoning supports a "no" conclusion. The discussion highlights the complexity of forces acting on charged particles and the need for clarity in understanding these interactions.
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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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