How to calculate the divergence by the definition

Question

I am trying to find the divergence of this field: $E = \frac{q(1-\sqrt{\alpha r})}{4 \pi \epsilon r^2} \hat r$

I already found the surface integral, that is, $\int EdS = \frac{q(1-\sqrt{\alpha r_{1}})}{\epsilon}$

I am having trouble to evaluate how to calculate the divergence by the definition now. (It has to be by definition

Answer 1

If $\vec{E}=f(r)\hat{r}$ then integration over the radius-$r_1$ circle centred on the origin gives$$\int\vec{E}\cdot d\vec{S}=\int f(r)\hat{r}\cdot(dS)\hat{r}=\int f(r)dS=\int f(r_1)dS=f(r_1)\int dS=4\pi r_1^2f(r_1).$$

Answer 2

Please note that the vector field is undefined at the origin.

Here is how you would find divergence in spherical coordinates -

$\displaystyle \vec{E} = \frac{q(1-\sqrt{\alpha r})}{4 \pi \epsilon r^2} \hat r$

$\nabla\cdot\vec{E} = \displaystyle \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 E) + 0 \,$ as $\frac{\partial}{\partial \phi}, \frac{\partial}{\partial \theta} \,$ will be zero.

$\displaystyle = \frac{q}{4 \pi r^2 \epsilon}\frac{\partial}{\partial r}(1-\sqrt{\alpha r}) = -\frac{q \sqrt \alpha}{8 \, \pi \, \epsilon \, r^{5/2}}$

If you want to do it in Cartesian coordinates -

As $\displaystyle \hat{r} = \frac{\vec{r}}{r} = \frac{(x, y, z)}{r}, $

$\displaystyle E_x = \frac{q}{4 \pi \epsilon}\frac{\partial}{\partial x} \Bigg(\frac{x}{(x^2+y^2+z^2)^{3/2}} - \frac{\sqrt \alpha \, x}{(x^2+y^2+z^2)^{5/4}}\Bigg)$

$E_y, E_z$ will be similar to $E_x$. So once you find $E_x$, you can find,

$\nabla \cdot \vec{E} = E_x + E_y + E_z \, $ and you will get the same value as above in spherical coordinates.