Charge distribution related to electric field $E=A\frac{e^{-br}}{r}\hat{r}$ I'm going over some problems with a solution manual in order to brush up for a coming exam and one of the problems I came across was this:

What is the charge density related to the field $E=A\frac{e^{-br}}{r}\hat{r}$? 

Now obviously here I would use $\nabla\cdot E=\rho/\epsilon_0$, however taking the divergence of the field the solutions manual says I'm supposed to find a result that includes the dirac delta function $\delta(r)$. Now I fully realise that $\nabla\cdot\frac{\hat{r}}{r^2}=4\pi\delta(r)$, however I don't see how this is related as my field is proportional to $1/r$ and not $1/r^2$.
According to the solutions manual we use $\nabla\cdot u\mathbf{v}=\mathbf{v}\cdot\nabla u+u\nabla\cdot\textbf{v}$.
Applying this is supposed to give:
$\nabla\cdot E=A\bigg[\nabla(e^{-br})\cdot \frac{\hat{r}}{r^2}+e^{-br}\nabla\cdot(\frac{\hat{r}}{r^2})\bigg]$
However I'm not seeing where the extra factor of $1/r$ seems to be coming from.
I'd deeply appreciate some help here as I'm starting to suspect that this might just be a printing mistake.
 A: As you suspect, it's a printing mistake and the electric field should be:
$$\mathbf{E} = A \frac{e^{-br}}{r^2} \mathbf{\hat{r}}$$
and in that case the hint makes perfect sense.

Side note:
For misprinted electric field $\mathbf{E} = A \frac{e^{-br}}{r} \mathbf{\hat{r}}$ the solution you obtain by straightforward differentiation equals to the one computed in this answer, when you realize that $r \delta(r) = 0$.
Also note that is not field of Yukawa potential, but just one component of that field.
A: The formula is correct. Simple mathematics will get you the answer for this Yukawa potential-like field.
$$\mathbf{\rho = \epsilon \nabla \cdot E = A \epsilon \nabla \cdot \left(\frac{r\exp(-br)}{r^2} \hat r\right)}$$
We know that $$\mathbf{\nabla \cdot (\phi F) = \phi (\nabla \cdot  F) + (\nabla \phi) \cdot F}$$
Now, in spherical coordinates, divergence operator becomes $$\mathbf{\nabla \cdot F=\frac{1}{r^2 \sin\theta}\left[\frac{\partial}{\partial r}(r^2 \sin \theta \, F_r) + \frac{\partial}{\partial \theta}(r \sin \phi \, F_\theta) + 
\frac{\partial}{\partial \phi}(rF_\phi)\right]}$$
And gradient operator becomes $$\mathbf{\nabla = \frac{\partial}{\partial r} \hat r+ \frac{1}{r} \frac{\partial}{\partial \theta} \hat \theta+ \frac{1}{r \sin\theta} \frac{\partial}{\partial \phi} \hat \phi}$$
So, if you approach in this way
$$\mathbf{\rho = A \epsilon \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial r}\left(r^2 \sin \theta \cdot \frac{r\exp(-br)}{r^2} \right) = A \epsilon \frac{1}{r^2 }\frac{\partial}{\partial r} \left[ r\exp(-br) \right]}$$
You miss the singularity at $r=0$ and it is a wrong answer.
You need to go as
$$\mathbf{\rho = A \epsilon \left[ r\exp(-br) \left\{\nabla \cdot  \frac{\hat r}{r^2} \right\} + \nabla \left\{r\exp(-br)\right\} \cdot \frac{\hat r}{r^2} \right]}$$
$$\Rightarrow \mathbf{\rho = A \epsilon \left[ 4 \pi r\exp(-br) \delta(r) + \left(\frac{1-br}{r^2}\right) \exp(-br)  \right]}$$
$$\Rightarrow \boxed{\mathbf{\rho = A \epsilon \exp(-br) \left[ 4 \pi r \delta(r) + \left(\frac{1-br}{r^2}\right)   \right]}}$$
Hope this helps.
