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 $W=qV$?

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    $\begingroup$ Hi! Please, edit your question using MathJax (it's like LaTeX), to make your equations more readable! $\endgroup$ – JackI Feb 27 '17 at 7:09
  • $\begingroup$ "plus some force to move the charge (say k N)" - There is no extra force required other than force to counter electrostatic repulsion. This is because we don't impart any velocity so as to keep change in kinetic energy zero. So no acceleration. This means k= 0. This should solve your problem. $\endgroup$ – Ketan Chaware Feb 27 '17 at 7:29
  • $\begingroup$ @KetanChaware If no extra force is needed, then wouldn't both bodies remain in static equilibrium? Also, why is it necessary to keep change in kinetic energy zero? $\endgroup$ – Aryan poonacha Feb 27 '17 at 7:31
  • $\begingroup$ You are right in your argument that initially you need a little bit of kick so as to get the charge moving but once you get it moving you can reduce that force such that the particle is not having any acceleration. Now why we can't have any acceleration is because potential energy is defined to be the work done exactly against the field so if you are going from point A to B with 0 acceleration then all your mechanical energy is exactly converted into potential energy of the field, and potential is defined, now you can use that value to know about the energy of particle in fields... $\endgroup$ – Sarthak Sharma Feb 27 '17 at 11:33
  • $\begingroup$ And yeah if you give it some acceleration then when it goes from A to B , some of its mechanical energy is stored as PE of field, and other remains as kinetic energy, now you can define potential in this way too but see when you use those potential values to calculate various Energies, those will give wrong results because potential values don't take into consideration that extra energy that you supplied by giving the particle acceleration, $\endgroup$ – Sarthak Sharma Feb 27 '17 at 11:37

$W=FS$ 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.

  • $\begingroup$ I never looked into the integral definition, my bad. $\endgroup$ – Aryan poonacha Feb 27 '17 at 7:28
  • $\begingroup$ Please use mathjax to format mathematical expressions. To learn more about mathjax, please read MathJax basic tutorial and quick reference. $\endgroup$ – Yashas Feb 27 '17 at 7:49

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