Question 6.8: Finding 3-D potential energies (a) Show that the uniform gra...

Since the potential energies are given, it is sufficient to evaluate − grad V in each case and show that this gives the appropriate F. Case (a) is immediate. In case (b),

$\frac{\partial H(r)}{\partial x}=\frac{d H}{d r} \frac{\partial r}{\partial x}=H^{\prime}(r) \frac{x}{r}=h(r) \frac{x}{r},$

since $r=\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2} \text { and } H^{\prime}(r)=h(r).$ Thus

$-\operatorname{grad}[-H(r)]=h(r)\left(\frac{x}{r} i +\frac{y}{r} j +\frac{z}{r} k \right)=h(r) \frac{ r }{r}=h(r) \widehat{ r },$

as required.

In particular then, the potential energy of the SHM field $F =-\alpha r  \hat{ r } \text { is } V= \frac{1}{2} \alpha r^{2}$, and the potential energy of the attractive inverse square field $F =-\left(K / r^{2}\right) \widehat{ r }$ is V = −K/r.