What Is the Speed of a Proton Between Two Plates with a 60V Difference?

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Homework Help Overview

The problem involves a proton accelerating between two plates with a potential difference of 60V. The original poster seeks to determine the speed of the proton upon reaching the second plate, having initially provided relevant equations and variables.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles to find the speed of the proton, but expresses uncertainty about their calculations. Some participants question the accuracy of the mass used for the proton and request clarification on units in the calculations.

Discussion Status

The discussion includes attempts to clarify the original poster's calculations and identify errors. While one participant pointed out a mistake regarding the mass of the proton, the original poster acknowledged the error and indicated that the issue was resolved.

Contextual Notes

Participants noted the importance of including units in calculations and the potential for confusion when mixing up particle properties, such as the mass of a proton versus an electron.

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1. The problem statement, all variables and 'given'/'known' data
A proton accelerates from rest from plate X to plate Y. If the potential difference between the 2 plates is 60.0V, what is the speed of the proton when it reaches plate Y?

2. Governing equations
V = E/q
Ekf - 1/2mv^2

The Attempt at a Solution


E = (60V)(1.6x10^-19C)
E = 9.6x10^-18 V/m

Ekf = 1/2mv^2
9.6E-18 J = 1/2(9.11x10^-31kg)v^2EDIT: added units

And when I solve for v my answer is not correct. So I'm making a mistake somewhere.

Thanks
 
Last edited:
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Good job posting the question. I'm having trouble checking your work because units are not shown throughout. Could you please include units on things like the 1.6x10^-19 term, etc.? Thanks.
 
added units
thanks
 
The mass of a proton is not 9.11e-31 Kg!
 
Oh dear, how could I mix that up.

I did it as an electron...

Problem solved, thanks
 

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