# Work done by a force on a particle

I am doing a practice problem in my classical mechanics class and don't understand my results. The problem is to calculate the work done on a particle to move it from the origin to a distance $d$ away along the $x$ axis by a force $$\vec{F}=k \frac{x\hat i+y\hat j}{(x^2+y^2)^{3/2}}$$ where $k$ is a positive constant. I've fount that $\nabla \times \vec F=0$ so $\vec F$ is a conservative force. If I set $$V(x,y)= \frac{-k}{(x^2+y^2)^{1/2}}$$ then $\nabla V=\vec F$. So the work done is:

$$W=\int_C \vec F \cdot d\vec s=\int_C \nabla V \cdot d\vec s=V(d,0)-V(0,0)=\frac{-k}{d}+\infty$$ so the point $(0,0)$ corresponds to infinite energy? Am I doing something wrong or is there a problem with the question? I notice that the vector field $\vec F$ is undefined at $(0,0)$ so does it even make sense for a particle to travel through that point?

• using stokes theorem, $W= \oint_C \vec{F} \cdot \mathrm{d}\vec{s} = \int_S \left(\nabla \times \vec{F}\right) \cdot \mathrm{d}\vec{\Sigma}$ – Smilia Sep 18 '16 at 21:16
• @Smilia The path is not closed – Alex Sep 18 '16 at 21:18
• oups ... to ignore – Smilia Sep 18 '16 at 21:19
• This exercise is not well-formed. A particle starting off at the origin would have an infinite force on it. Are you sure that you're not supposed to start with the particle an infinite distance away? – user137731 Sep 18 '16 at 21:53
• It is as @Bye_World said. This problem also appears for example in electrodynamics, when you try to define the potential energy of e.g. a configuration of a finite number of charged point particles (if I remember well). – Daniel Robert-Nicoud Sep 18 '16 at 22:08

## 1 Answer

You are correct. Actually, this force field originates from a point charge located at the origin. Consequently, the vector field is only defined for points other than the origin. It doesn't make sense for a particle to travel through the origin, because that would involve going through the point charge.

• Perhaps you could address the mathematical "sense" (or lack thereof) for work on a particle traveling on a path through the origin. Simply agreeing with the OP's doubt "does it even make sense?" after a year's time seems not to add much information for future Readers. – hardmath Sep 29 '17 at 21:21
• Thanks for your comment. Mathematically, since the force field is singular at the origin, we can only talk about going arbitrarily close to the origin, and define "traveling through the origin" as the limit, which doesn't exist here. – Jiaqi Li Sep 29 '17 at 21:39