Repulsive force between 2 protons?

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Homework Help Overview

The discussion revolves around calculating the repulsive force between two protons positioned 1 femtometer apart, utilizing Coulomb's law. Participants explore the implications of this force in the context of atomic structure and the influence of the strong nuclear force.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants analyze the calculated force and question its magnitude relative to the strong nuclear force. Some express concern about the implications of their results, while others suggest focusing solely on the electrical force.

Discussion Status

There is ongoing exploration of the relationship between the calculated electric force and the strong nuclear force. Participants are questioning assumptions about the forces at play and discussing the significance of the results in a broader context.

Contextual Notes

Some participants note that the original poster's calculations may not align with typical scenarios in nuclear physics, and there is mention of the relevance of different distances in practical applications, such as in helium nuclei.

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Homework Statement


I have 2 protons 1 femtometer away from each other, what is the repulsive force between them?

q1 = 1.6x10-19C
q2 = 1.6x10-19C
r = 1x10-15m
K = 9x109Nm2/C2



Homework Equations


FE=Kqqqw/rw



The Attempt at a Solution


My answer comes out to be -230.4N but I feel that that is wrong because this is approximately the distance that 2 protons are from each other in an atom and that would mean the strong force is acting on them wouldn't it? So wouldn't the repulsive force be very high? Or is the strong force just much more powerful than it needs to be?
 
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I think you're over thinking the problem, unless your course is dealing with nuclear structure and the strong and weak nuclear forces. Just go with the electrical force.

The strong force is called "strong" for a reason :smile:
 
A quick check by order of magnitude:
q1=q2~10^-19
r^2 ~ 10-30
K ~ 10^10
So F ~ 10^(10-38+30)=100 ... 230~100 so you are fine.
(read "~" as "is of the order of")

The negative sign would usually indicate an attractive force.

? So wouldn't the repulsive force be very high?
... things can only be big or small in comparison with something else. So what are you comparing this force to, that you suspect it is not large?

consider: If the protons were held at 1fm apart, then released, what would be their initial acceleration?
If the protons came to rest 1fm from each other, how much kinetic energy did they start out with?
Is that sort of kinetic energy readily available to free protons in Nature? Say - from normal heating to room temperature? What sort of kinetic energy do two bound protons have? How does this compare with the coulomb potential energy at 1fm?
 
Nerdydude101 said:

Homework Statement


I have 2 protons 1 femtometer away from each other, what is the repulsive force between them?

q1 = 1.6x10-19C
q2 = 1.6x10-19C
r = 1x10-15m
K = 9x109Nm2/C2

Homework Equations


FE=Kqqqw/rw

The Attempt at a Solution


My answer comes out to be -230.4N but I feel that that is wrong because this is approximately the distance that 2 protons are from each other in an atom and that would mean the strong force is acting on them wouldn't it? So wouldn't the repulsive force be very high? Or is the strong force just much more powerful than it needs to be?
Actually, the equation for F you wrote in an equation for the magnitude of the force so it should read

F_E = K \, |q_1| \, |q_2| / r^2
so this always comes out positive.

You are right that at that distance, the strong nuclear force is important, on top of the electric force.
You seem to feel that the force you got is not as large as you expected, am I right?
It may not sound that big if you think about the force on a macroscopic object (like a baseball or a person) but such force on a tiny particle like a proton is huge, all proportions considered. You can see that it is huge if you calculate the acceleration of the proton (compared to the acceleration of, say, a person who experiences a force of the same magnitude). The point is that such a force concentrated on a single proton is quite impressive.
 
Nerdydude101 said:

Homework Equations


FE=Kqqqw/rw

What is w? Should it be 2 instead of w?
 
To have a more practical value i'd suggest we take distance 3.8 fm which is found in Helium nuclei. The magnitude of force then comes out to be ~15 N.
I agree with nrqed. To make more sense of the current magnitude we'll need to scale the values by translating the metrics to a more real life scenario.
 
@YashKumarSingh : Note that this thread is quite old, from September of 2014. The Original Poster, Nerdydude101, was last logged in on Jan 8, 2015. It is rarely of value to "wake up" old threads.
 

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