# 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