forces - Clarification of the definition of potential at a point

My textbook defines the potential at a point to be equal to the work done in bringing a unit positive charge from infinity to that point, and then explains that the point contains another unit positive charge and so it's the work done in opposing the repulsive like charge. However, shouldn't the work then be infinite? From $W = Fs\cosθ$, The $F$ = work done to oppose the unit positive charge by distance squared plus some force to move the charge (say $k N$), so $F=\frac{q}{r^2}+k$, and with the charge moving from infinity, so isn't $s=\infty$? Since the work is done in the direction opposite to the force $q$, $\theta = 180$ and that just makes the work negative, so $W = (frac{q}{r^2}+k) \times \infty \times -1.$

Is applying $W=Fs$ wrong here? Is $s$ not infinity, and if so, why? Are any other assumptions here wrong, and if so, why? And how does this equate to $W=qV$?

Answer

$W=FS$ is wrong because $F$ is not constant. You need to integrate. At infinity, the distance $S$ is tending to infinity but $F$ is tending to $\theta$. Then $F$ changes with $S$. You have to use integration.