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work = force x displacement. Here displacement is infinity? so then potential at a point?

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    $\begingroup$ Work is not always force times displacement. That's only for constant forces (the electrostatic force is not constant with distance!). The general definition of work is $W=\int_A^B \mathbf F.d \mathbf x $. Use that instead. And please try to write your question a little better next time. $\endgroup$ – Sahand Tabatabaei Nov 17 '17 at 1:20
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    $\begingroup$ I think the downvotes are a bit harsh - we were all beginners once. $\endgroup$ – John Rennie Nov 17 '17 at 7:30
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Yes, work is force times displacement, but the force can change with distance. I would guess you are talking about the gravitational or electrical potential and in both cases the force does decrease with distance.

If the force at some distance $r$ is $F(r)$ then we can define the work done as the force times the distance if we move a tiny distance $dr$ that is so small the force remains constant - technically this tiny distance $dr$ is an infinitesimal distance. In that case we get:

$$ dW = Fdr $$

So for example with the gravitational force we get:

$$ dW = \frac{GMm}{r^2}dr $$

To get the total work we have to add up all these infinitesimal $dW$s, and we do that by integrating:

$$ W = \int dW = \int \frac{GMm}{r^2}dr $$

In this case the work moving a particle from infinity to a distance $r$ is done by integrating between the limits $r$ and $\infty$ and we get:

$$ W = \int^r_\infty \frac{GMm}{r^2}dr = -\frac{GMm}{r} $$

which is the expression for the gravitational potential energy.

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