Speed of Proton (given electric potential difference)

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SUMMARY

The discussion centers on calculating the speed of a proton accelerated through a 500V electric potential difference. The correct approach involves using the relationship between electric potential energy and kinetic energy. The energy gained by the proton is equal to the charge of the proton multiplied by the potential difference, resulting in a kinetic energy of 8.0E-17 J. This energy translates to a speed of approximately 3.1E5 m/s, as confirmed by the professor. The initial attempt to calculate the speed using distance and time was incorrect due to a misunderstanding of the relevant equations.

PREREQUISITES
  • Understanding of electric potential difference and its relation to energy
  • Knowledge of kinetic energy equations
  • Familiarity with the charge of a proton (1.6E-19 C)
  • Basic principles of motion (distance, speed, time)
NEXT STEPS
  • Study the relationship between electric potential energy and kinetic energy
  • Learn about the conservation of energy in electric fields
  • Explore the concept of electric potential difference in more depth
  • Review the derivation of speed from kinetic energy equations
USEFUL FOR

Students in physics, particularly those studying electromagnetism and mechanics, as well as educators looking to clarify concepts related to electric potential and particle motion.

bec2008
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Homework Statement


A proton, initially at rest, is accelerated through an electric potential difference of 500V. What is the speed of the proton?


Homework Equations


V=k(q/r)
Solving for r: r=k(q/v)

v=d/t

known charge of a proton: 1.6E-19C
known value of k: 9E9Nm^2/C2


The Attempt at a Solution


I tried solving for r to get a distance.
r= (9E9Nm^2/C2)(1.6E-19C)/ 500V
r= 2.88E-12m

v=d/t
v= (2.88E-12m)/1 s
v= 2.88E-12 m/s

Our professor gave the answer as 3.1E5 m/s, but I cannot get to this answer. Any help would be greatly appreciated!
 
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Assume that all the work done by the source goes into increasing the speed of the proton (what sort of energy would it have when it moves?)
 
bec2008 said:

Homework Equations


V=k(q/r)
Solving for r: r=k(q/v)
That formula describes the potential from a point charge--not relevant here. (Further note that you are given potential difference, not potential.)

Hint: How does potential difference relate to energy?
 

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