Calculate the divergence of the following vector function $$\vec{C}(\vec{r}) = \frac{r}{r^4}$$

Want to know if I am on the right track?

$$\hat{r}=\frac{\vec{r}}{|r|} \implies \frac{1}{r^4}\frac{\vec{r}}{|r|}=\frac{\vec{r}}{r^5} \therefore \nabla\cdot v=\nabla\cdot\frac{\vec{r}}{r^5}=\frac{\partial}{\partial x}(x(x^2+y^2+z^2)^{{-\frac{5}{2}}})+\frac{\partial}{\partial y}(y(x^2+y^2+z^2)^{{-\frac{5}{2}}}) \ +\frac{\partial}{\partial z}(z(x^2+y^2+z^2)^{{-\frac{5}{2}}}) \ ...$$

thus the answer is $$\frac{-4}{(x^2 +y^2 + z^2)^{\frac{5}{2}}}?$$

  • $\begingroup$ Yes, you are on the right track. $\endgroup$ – Feng Shao Sep 9 '19 at 4:11
  • $\begingroup$ @FengShao thanks, will the answer then be $$\frac{-2}{(x^2 +y^2 + z^2)^{\frac{5}{2}}}$$ $\endgroup$ – Robben Sep 9 '19 at 4:23
  • $\begingroup$ No. The numerator is$-4$ rather than $-2$. $\endgroup$ – Feng Shao Sep 9 '19 at 5:50

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