View Full Version : Finding a force from a simple potential
emoboya3
Nov20-08, 04:30 PM
1. The problem statement, all variables and given/known data
I need to find the acceleration of an atom, given the potential:
U(r) = ε[(σ/r)^12 - (σ/r)^6]
where r = |x1 - x2|, ε = 1e-20 J, and σ = 3e-10 m.
2. Relevant equations
I know F=-dU/dr and F=ma so
a=(-dU/dr)/m but I think my dU/dr is wrong.
3. The attempt at a solution
dU/dr = (6εσ^6/r^7) - (12εσ^12)/r^13
so F = (12εσ^12)/r^13 - (6εσ^6/r^7)
Hootenanny
Nov20-08, 04:37 PM
Welcome to Physics Forums.
The negative gradient of a scalar potential will give you the force acting on the atom. Note that this is not simply the same as taking the derivative of U with respect to r.
borgwal
Nov20-08, 04:38 PM
Yes, your dU/dr is wrong. Maybe if you write r^{-12} you'll see what your mistake is.
borgwal
Nov20-08, 04:40 PM
Welcome to Physics Forums.
The negative gradient of a scalar potential will give you the force acting on the atom. Note that this is not simply the same as taking the derivative of U with respect to r.
Since U only depends on r, that's NOT the mistake OP makes.
Hootenanny
Nov20-08, 04:46 PM
Since U only depends on r, that's NOT the mistake OP makes.
I assumed we were in spherical coordinates and r was the radial distance from the origin. Of course on later reflection after you pointed it out, I noticed that r was define as the segment between two points.
My mistake.
emoboya3
Nov20-08, 04:49 PM
sorry, the formula i gave for my solution should NOT BE dU/dr. it is the Force. dU/dr has the opposite sign. I will edit.
Hootenanny
Nov20-08, 05:08 PM
sorry, the formula i gave for my solution should NOT BE dU/dr. it is the Force. dU/dr has the opposite sign. I will edit.
dU/dr = (6εσ^6/r^7) - (12εσ^12)/r^13
so F = (12εσ^12)/r^13 - (6εσ^6/r^7)
Looks good to me now :approve:
Apologies for the mix-up before.
emoboya3
Nov20-08, 05:14 PM
Much appreciated to all. And sorry for my lack of etiquette. I'm still learning.
Hootenanny
Nov20-08, 05:17 PM
And sorry for my lack of etiquette.
There was nothing at all wrong with your posting etiquette, whatever gave you the impression that there was?
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