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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:

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

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  • $\begingroup$ 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 $ln(x)$. $\endgroup$ – Alpha001 May 11 '17 at 8:48
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First of all as @Alpha001 commented, it does not necessarily mean that the line integral is zero if the force tends to zero at infinity. There is another reason why this is not always true. The relation holds only for conservative forces, the class of forces for which the line integral is independent of the path chosen. This is a very crucial point to observe. A potential function "at a point" can be defined only if the function is single valued at that point. It does not make sense to define potential at a point differently for different paths chosen, in that case the potential is no longer a function.

So in summary, only conservative forces can be written as a gradient of a scalar potential.

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