# Force and Potential Energy

## Homework Statement

The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)= -C6/x^6, where C6 is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?

## The Attempt at a Solution

I'm thinking we have to take the derivative of the equation??

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this is what i have...
after taking the derivative of U(x), i got F(x)= -C6/6x^7. And this force is an attractive force.

Im not sure about the last part.

Just as comment on your derivative, I think it should actually be F(x)= -6C/x^7 because if you used the multiplication rule, the -6 is in the numerator

So you're absolutely right. It is the derivative of the given eqn. thus getting -6C_6/x^7

And considering atoms are neutral and give off no outward magnetism, they apply a really small attractive force on each other [will have to check but i think it's due to gravity, not sure. but it's definitely an attractive force]

THe derivative is the correct answer but can anyone explain why? This is potential energy which is supposed to be in figures equivalent to that of work vise-vis Force times distance or Newton meters (Joules). I would think that the answer would be to divide -C6/x^6 by the displacement x but that is the wrong answer. WHYY??

Force is equal to the negative slope of the U(x) (potential energy) function. Also, you get attraction when F > 0 and repulsion when F < 0.

I am sorry to tell you that the derivative is +6C/x^(7). I just looked it up. You can re-write the equation as -1C*x^(-6),so -6*-1= +6, and x^(-6) becomes x^(-7), so the derivative is +6C*x^(-7) or +6C/x^(7). Here's the website I found this at. It's really great if you have math questions:
http://www.wolframalpha.com/input/?i=derivative+of+-c/x^(6)

Hope this helps :)

They are approaching the problem using

$F(x) = -\frac{dU(x)}{dx}$

that is why they are giving the answer as negative. So, yes, the derivative is positive but the force does carry a negative.

When taking the derivative you are actually talking about infinitesimal displacements, so you can think of this as the force being exerted by the atoms at this distance. Exact.

Lets think of it in more familiar grounds. Standing on Earth always provides the same amount of acceleration, correct? Being this is the case, you always 'feel' the same force begin applied to you. This is because you are always the same distance away, radially. This is the same scenario except in electrostatics situation.

lightgrav
Homework Helper
F∙dx = dWork , and PE = -Work , do F = - dW/dx ... yes, there's a 6 on top, attractive since negative.
Physically: one atom's electron cloud will fluctuate (off-center) temporarily, becoming an electric dipole;
that will induce a dipole to form in the other atom; those two dipoles attract one another.
Gravity is not involved.