What is the final speed of a proton in an electric field after traveling 2.0 mm?

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Homework Help Overview

The discussion revolves around a problem in electrostatics, specifically involving the motion of a proton in a uniform electric field. The original poster seeks to determine the final speed of the proton after it travels a specified distance in the field.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between force, charge, and acceleration, with some questioning the need for the proton's mass and charge. Others suggest using energy concepts to relate potential energy and kinetic energy.

Discussion Status

There is an ongoing exploration of different approaches to the problem, including the use of classical mechanics and energy principles. Some participants express confusion regarding the appropriate energy terms to use, particularly in distinguishing between gravitational and electrostatic potential energy.

Contextual Notes

Participants note the importance of understanding the properties of the proton, such as its mass and charge, and the need to clarify the correct application of energy concepts in the context of electric fields.

spoonthrower
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A uniform electric field has a magnitude of 2.3*10^3 N/C. In a vacuum, a proton begins with a speed of 2.1*10^4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 2.0 mm please help. Thanks.
 
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What are your thoughts on it?
 
i know F=qE, so is there a value of q for a proton that i don't know about. once i know the force i guess i need to find the acceleration to calculate the final speed. so i would need the mass of a proton?
 
Yeah, you need the mass and charge of the proton (which you can look up anywhere). So now you know F (which is constant) and it's just a classical mechanics exercise, analogous to a particle in a uniform gravitational field.

It is more efficient to use an energy approach though. The potential energy lost in moving the 2mm is gained by the kinetic energy of the proton.
 
so I would use .5mvf^2-.5mvo^2=mgh

The mass cancels i know that. I use gravity?

I am still getting the wrong answer
 
spoonthrower said:
so I would use .5mvf^2-.5mvo^2=mgh

The mass cancels i know that. I use gravity?
No! The only thing analogous is that the force acting on the particle is constant. You mustn't use gravitational potential energy (which is proportional to the mass) but electrostatic potential energy (Which is proportional to the charge). I made the analogy to simplify the view on the problem, but I see it's only confusing. Forget I said the whole thing!

If you don't know electrostatic potential energy, just use the work-energy theorem. What is the work done by the field in moving the particle those 2mm?
 
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