Repulsive force between 2 protons?

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SUMMARY

The repulsive force between two protons, each with a charge of 1.6 x 10-19C and separated by a distance of 1 femtometer (1 x 10-15m), is calculated using Coulomb's law, resulting in a force of approximately 230.4 N. This force is significant when considering the scale of subatomic particles, despite being perceived as small compared to macroscopic forces. The strong nuclear force also plays a critical role at this distance, indicating that while the electric repulsion is substantial, it is overshadowed by the strong force in atomic interactions.

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  • Coulomb's Law for electric force calculations
  • Understanding of subatomic particle interactions
  • Basic knowledge of nuclear forces, specifically the strong force
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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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