The example imI'm working on has this given identity: $\bigtriangledown \mathbf{\bar{r}}=3$ The question is; find the divergence of a vector field $\bar{\mathbf{E}}=\frac{\mathbf{r}}{r^{3}}$$\bigtriangledown \cdot \mathbf{\bar{r}}=3$.
The question is: find the divergence of a vector field $\bar{\mathbf{E}}=\frac{\mathbf{r}}{r^{3}}$.
So, using the product rule relationship, i arrange it as such:
$\bigtriangledown \cdot (\psi \bar{\mathbf{E}})=(\bigtriangledown \psi) \cdot E + \psi (\bigtriangledown \cdot \bar{\mathbf{E}})$
as then substitute in:
$\bigtriangledown \cdot (\psi \frac{\mathbf{\bar{r}}}{r^{3}})=(\bigtriangledown \psi) \cdot \frac{\mathbf{\bar{r}}}{r^{3}} + \psi (\bigtriangledown \cdot \frac{\mathbf{\bar{r}}}{r^{3}})$
I can then do $\bigtriangledown \cdot \frac{\mathbf{\bar{r}}}{r^{3}}$ to $\frac{-3}{r^{4}}$ by differentiation, (i think this is wrong)
And using the given relationship $\bigtriangledown \mathbf{\bar{r}}=3$
I can re-write as: $\bigtriangledown \cdot ( \frac{\mathbf{\bar{r}}}{r^{3}})=\frac{3}{r^{3}}-\frac{3}{r^{4}}$
This is where i'm a little lost, I have the list of solutions, and the solution is 0
I can see its there is something mathematically wrong I have done to get $\frac{1}{r^4}$ instead of what i presume should be $\frac{1}{r^3}$, but really cant find out what it is.
