Accelerating a stationary proton

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Homework Help Overview

The discussion revolves around the question of how to accelerate a stationary proton, specifically whether to use an electric field or a magnetic field. Participants are exploring the fundamental principles of forces acting on charged particles in different fields.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the dependency of forces on the fields, questioning the conditions under which each field can accelerate a stationary proton. Some express uncertainty about the correctness of their reasoning.

Discussion Status

The conversation includes affirmations of the reasoning behind using an electric field over a magnetic field, with some participants providing justifications based on the nature of the forces involved. However, there remains a sense of uncertainty among some participants regarding their understanding.

Contextual Notes

Participants mention an upcoming test, indicating a time constraint and the pressure to understand the concepts thoroughly before the assessment.

jeffreyl
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Homework Statement



If you wanted to accelerate a stationary proton, would you use electric field or magnetic field? why?

Homework Equations





The Attempt at a Solution

 
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Oh, that's a nice question :D. Well, pretty simple actually. Just check and tell us, what does the force of each of the fields depend on? In this situation, one of them will be 0, and so this field will not accelerate a stationary proton.
 
actually that's the question. there's no other stuff. i think electric field would be used since magnetic field needs a existing velocity for a force to applied to the proton.
 
So what's the problem, since you already know the answer and can properly justify it?:P
 
i have no idea if it's right or not. and my test is in a few days T_T
 
Well of course it's right!
For electric field:
[tex]\vec{F}=q\vec{E}[/tex]
[tex]\vec{a}=\frac{q \vec{E}}{m_p} \neq 0[/tex] under assumption that there is an electric field ;)

For magnetic field:
[tex]\vec{F}=q\vec{v}\vec{B}[/tex]
[tex]\vec{a}=\frac{q\vec{v}\vec{B}}{m_p}=0[/tex], because v=0.

You could of course wirte it down more exactly, with all the angles considered and so on, but it doesn't change anything.
 
thanks :D
 

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