# Calculating force given a potential

1. Dec 11, 2012

### hahaha158

1. The problem statement, all variables and given/known data

Calculate the force on a particle at a position in space

r=[x,y,z]

V(r;A,B)=(A/r^14) - (B/r^8)
with A=1.5 and B=2/5

2. Relevant equations

3. The attempt at a solution
I know how to solve these when there is x,y,z involved, but i am not sure how to deal with something that has r in the equation, cna anyone explain?

Thanks

2. Dec 11, 2012

### Staff: Mentor

If the position has coordinates [x, y, z], what's the magnitude of the radius vector, r?

3. Dec 11, 2012

### hahaha158

(x^2,y^2,z^2)^.5?

4. Dec 11, 2012

### Staff: Mentor

Sorry, I don't understand the notation... are those commas meant to represent "+" operations?

5. Dec 11, 2012

### hahaha158

sorry yes i meant to use +.

However i think i may have done it, can you confirm whether or not this is correct?

F= -((-14Ar^-15)+(8Br^-9))*([x,y,z]/r)?

and is this an acceptable form to leave it in?

6. Dec 11, 2012

### Staff: Mentor

[x,y,z]/r doesn't match any mathematical notation I'm familiar with (that said, I'm not familiar with every variation of notation or formalism). But I would think that simply resolving r as $\sqrt{x^2 + y^2 + z^2}$ and substituting it for r in the function v(r;A,B) would get you where you want to be.

7. Dec 11, 2012

### hahaha158

i should probably have clarified a bit better

this is doing the nabla which i believe is just partial differentation, where you usually have 3 different functions (x,y,z) but since this is r i was a bit unsure. I think i understand it better now after looking some stuff online but thanks for the help regardless!

8. Dec 11, 2012

### Staff: Mentor

Okay. Truthfully, I didn't recognize any sign of the $\nabla$ operator being involved. Is the [x,y,z] or V(r;A,B) syntax diagnostic? I'd like to be able to spot the formalism for future reference.

9. Dec 11, 2012

### hahaha158

Well i'm not too sure of the syntax, i just tried to convey it as accurately as i can possible to the way i learned it.

starting from the top of the 2nd slide to where it says e)i)

10. Dec 12, 2012

### aralbrec

You can do E=-∇V in spherical coordinates or you can substitute r=√(x2+y2+z2) and do it in cartesian coordinates.

But there is a much quicker shortcut.

Recall that the electric field is perpendicular to a constant potential surface and the magnitude of the field is simply the rate of change of E in the direction of maximum rate of change.

Your potential function is constant for fixed r, which means the constant potential surfaces are spheres centered on the origin. This means the electric field is a vector pointing either away or toward the origin and its magnitude is just dV/dr -- the rate of change of the potential in the direction it is changing fastest (perpendicular to the sphere, along r).