# If $PE$ is defined the following way, then for arbitrary force field which tends to zero at infinity, will $\vec{F}=-\vec{\triangledown}(PE)$

Potential energy is defined as the work done by external force to move from infinity to that point.

$$PE=-\int^{r}_{\infty}\vec{F}.\vec{dr}$$

(for an arbitrary force field $\vec{F}$ which tends to zero at infinity)

$$=-\left[ \int\vec{F}.\vec{dr} \right]_{at\ r} +\left[ \int\vec{F}.\vec{dr} \right]_{at\ \infty}....(1)$$

Now whenever the force tends to zero at infinity, can we say the second term in $(1)$ is zero?? so that potential energy simplifies to:

$$PE=-\int\vec{F}.\vec{dr}$$ $$=-\int F_xdx-\int F_ydy-\int F_zdz$$

and hence:

$\vec{F}=-\vec{\triangledown}(PE)$

• Not necessarily. Assume a force $F \sim 1/r$ and symmetric on a sphere, this force goes to $0$ for $r \rightarrow \infty$ but not "fast" enough. There are still contibutions to the integral at infinity since your primitive integral is $ln(x)$. – Alpha001 May 11 '17 at 8:48