How Is the Kinetic Energy of a Proton Calculated in an Electric Field?

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Homework Help Overview

The discussion revolves around calculating the kinetic energy of a proton in a uniform electric field, specifically examining the values derived from the formula KE = q * E * s. The context includes the electric field strength and the distance over which the proton is accelerated.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the kinetic energy using the formula and expresses confusion over a significant discrepancy between their result and the book's answer. Some participants question the accuracy of the book's value, while others suggest that the calculations may be correct based on the provided parameters.

Discussion Status

Participants are exploring the calculations and discussing the potential for errors in the textbook. There is a recognition that the original poster's approach may be valid, and some guidance has been offered regarding the interpretation of the electric field and potential difference.

Contextual Notes

There is mention of a significant difference in the expected kinetic energy values, leading to discussions about potential errors in the textbook or assumptions made in the calculations. The original poster's calculations have been checked multiple times, indicating a thorough review of the problem setup.

Peter G.
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Hi,

A proton initially at rest finds itself in a region of uniform electric field of magnitude 5.0 x 106 Vm-1. The electric field accelerates the proton for a distance of 1 km.

Find the kinetic energy of the proton.

So, what I did was the following:

KE = q * E * s

I then converted the result from J to MeV. I, however, get 5000 MeV and the book gets 500 MeV. Is my line of thought incorrect?

Thanks once again!
 
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Your answer is right for the given numbers.
 
Thanks! I checked and rechecked my calculations and numbers several times before posting here and I was about to go crazy! I guess the book made a mistake (everyone has the right to!)
 
Given the numbers for the electric field and distance, you should be able to calculate the potential difference of 5000 MV between the two endpoints. Because the proton has charge e, to get the work done, you just stick an e in front of the V in the units. If you do that, you can see pretty easily 5000 MeV is correct. You don't need to worry that you made a mistake while converting units.
 

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