Proton Motion in Electric Field

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Homework Help Overview

The discussion revolves around the motion of a proton in an electric field created by two large metal sheets separated by a potential difference of 40 volts. Participants are exploring the kinetic energy of the proton just before it strikes the negatively charged sheet and its velocity at that moment.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the kinetic energy formula and its relation to the proton's velocity. There are questions about the correctness of the calculated velocity and the kinetic energy value.

Discussion Status

Some participants have provided calculations for the kinetic energy and velocity, while others are seeking confirmation on these values. There appears to be a mix of agreement and uncertainty regarding the calculations presented.

Contextual Notes

Participants are working within the constraints of a homework problem, which may limit the information available for discussion. There is an emphasis on verifying calculations and understanding the underlying physics concepts.

XTEND
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Homework Statement


2 large metal sheets are separated by a potential of 40 volts by a vaccum. When the proton particle (mass 1.67x10e-27 kg)is released close to the (+) sheet it moves closer to the (-) sheet by way of electric field. What's the KE right before the proton smacks the other sheet? And what's the velocity of before striking the (-) sheet?


Homework Equations



KE=>q*v and EK=mv^2/2

The Attempt at a Solution



KE= 1.6x10e-19(40)=6.4x10e-18

v= ?
 
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What's the formula for the kinetic energy of an object (in terms of the object's velocity)?
 
Can anyone help me with this?

I think the velocity is 87548 m/s. Is this correct?
 
XTEND said:
Can anyone help me with this?

I think the velocity is 87548 m/s. Is this correct?

Your answer is correct.
 
what about the first part for KE?
 
XTEND said:
what about the first part for KE?

If that isn't correct, you wouldn't have gotten the right answer for v. So yes, KE=qV, and your calculation is right as well.
 

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