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if $\phi = \frac{1}{|\underline{r}|}$ and $E = -\operatorname{grad} \phi$ then show $\operatorname{div}E = 0$

the question does not say what $\underline{r}$ is explicitly, the solution shows the following steps:

$\underline{E} = - \nabla \phi = - \nabla \frac{1}{r} = \dfrac{\underline{r}}{r^3}$ then he computes the divergence. I am confused as to how he went from $- \nabla \frac{1}{r} = \dfrac{\underline{r}}{r^3}$, we defined $\nabla = \begin{pmatrix} \partial _x \\ \partial _y \\ \partial _z \end{pmatrix}$ but $\underline{r}$ is not defined explicitly and he does not compute it explicitly, similarly for finding $\nabla \cdot E$ he goes from $\nabla \cdot \left (\dfrac{\underline{r}}{r^3}\right ) = \dfrac{\nabla \cdot \underline{r}}{r^3} - 3 \dfrac{\underline{r} \cdot \underline{r}}{r^5}$ which I don't see how

Michael Hardy
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user144464
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1 Answers1

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$\underline r$ is perhaps not given explicitly but, from polar coordinates $\underline r = (x,y)$, ie : $$\frac{1}{|\underline r|} = \frac{1}{\sqrt{x^2+y^2}}$$

First I suggest you check what $\frac{\partial}{\partial x}\frac{1}{|\underline r|}$ is and you'll be able to deduce that $\nabla \phi = -\frac{\underline r}{r^3}$

Then to compute divergence, again use the fact that $\underline r = (x,y)$ with the previous result so you need to calculate : $$\nabla\cdot \Bigg(\frac{x}{(x^2+y^2)^{3/2}}, \frac{y}{(x^2+y^2)^{3/2}}\Bigg)$$

I'm sure all you needed was that $\underline r = (x,y)$. I was stuck in the same manner when I had the same question asked.

user88595
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