Electric potential no clue and midterm in 2 and a half hours

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Homework Help Overview

The discussion revolves around a physics problem involving electric potential and the motion of a proton towards the nucleus of an iron atom. The original poster is attempting to determine the initial speed required for the proton to reach the nucleus, given specific parameters related to the nucleus and the proton's properties.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply conservation of energy principles and kinematic equations but expresses confusion regarding the spherical nature of the nucleus and the appropriate distance to use in calculations. Some participants suggest re-evaluating the arithmetic and converting units correctly.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach. There is acknowledgment of a potential arithmetic error, and a request for further clarification on a different topic related to wire diameter and current has been introduced, indicating a shift in focus.

Contextual Notes

The original poster appears to be under time pressure due to an upcoming midterm, which may influence the urgency of their inquiries and attempts to seek assistance.

cemar.
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Homework Statement


A proton is fired from far away toward the nucleus of an iron atom. Iron is the element number 26 and the diameter of the nucleus is 9.0fm. What initial speed does the proton need to just reach the surface of the nucleus? Assume the nucleus remains at rest.


Homework Equations


Conservation of energy which would get reduced to Ki=Uf
U=kq1q2/r
kinematic equation
K = 1/2 m v^2
Do some thing with the element number 26... has 26 protons so make a field?


The Attempt at a Solution


Oh gosh. K this is completely wrong but ill just throw it out there for kicks.
Ki=Uf
Since i don't knwo what r is since its sperical and that confuses me... ill guess you use the distance on the surface of the nucleus?
1/2mv^2 = kq1q2/r
1/2 (1.67E-27) v^2 = 9E9 (1.6E-19) (26*1.6E-19) / 4.5fm
v = 350192

the answer is 3.99E7

Thank you!
 
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cemar. said:
1/2mv^2 = kq1q2/r
1/2 (1.67E-27) v^2 = 9E9 (1.6E-19) (26*1.6E-19) / 4.5fm
This is perfectly correct, just redo your arithmetic. (Change fm to standard units.)
 
Seriously?!? thanks!
Im a mess! :S
You wouldn't happen to know and equation concerning the diameter of a wire with length and current and all that jazz would you??
 
cemar. said:
You wouldn't happen to know and equation concerning the diameter of a wire with length and current and all that jazz would you??
Not sure what you're asking. If you post a complete problem, I can take a look.
 

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