Let $F(x,y,z) = -c(r/||r||^3)$ be the force resulting from the inverse square law... $c$ is a constant and $r = (x,y,z)$. Show that $\displaystyle f(x,y,z) = \frac{c}{\sqrt{x^2+y^2+z^2}}$ is a potential function for $F$. What can be concluded from any path from point $A$ to point $B$ in $F$? What can be concluded about a simple closed path in $F$?
 A: A force field $F:\mathbb{R}^3\to\mathbb{R}^3$ is conservative if there exists a scalar function $f:\mathbb{R}^3\to\mathbb{R}$ such that $\nabla f=(f_x,f_y,f_z)=F$.  The scalar function $f$ is called a potential function for $F$.  So we only need to check that the given potential function satisfies the equation $\nabla f=F$
Notice that for the given potential function $f$, we have
$$f_x=-\frac{cx}{(x^2+y^2+z^2)^{3/2}}$$
See if you can show that this together with the other partial derivatives give the desired force field $F$.  Remember that $r=(x,y,z)$ so that $||r||=\sqrt{x^2+y^2+z^2}$.
Now, when force fields are conservative, the work done on a particle moving from point $A$ to point $B$ is independent of the path taken from $A$ to $B$.  This can be seen by the line integral analogue of the fundamental theorem of calculus:
$$\text{Work done by $F=\nabla f$ in moving a particle from point $A$ to point $B$ along curve $C$}=\\\int_CF\cdot dr=f(B)-f(A)$$
We see that only the endpoints of the curve ($A$ and $B$) determine the work done.
Finally, closed paths in $F$ have the same starting and ending point, that is, $A=B$.  Using the above formula, what does this say about the integral of a conservative force field $F$ on a closed path?
