Question

Obtain the divergence and curl of electric field due to a point charge at 

1- A far away point

2- It's own location

Answer 1

The electric field of the point charge $q$ at the distance $r$ from it has the following view (bold letters denote vectors):

$$\mathbf{E} = k\dfrac{q}{r^2}\hat{\boldsymbol r}$$

where $k = 9\times 10^9N\cdot m^2/C^2$ is the Coulomb's constant, and $\hat{\boldsymbol r}$ is the unit vector in radial direction. As one can see, the field has radial symmetry. Thus, in sherical coordinates it has only radial component:

$$\mathbf{E} = (E_r, E_\theta, E_\phi) =\left(k\dfrac{q}{r^2},0,0\right)$$

Then, the only non zero term in the expression for the divergence in spherical coordinates (https://en.wikipedia.org/wiki/Divergence#Spherical_coordinates) will be:

$$\nabla\cdot \mathbf{E} = \dfrac{1}{r^2}\dfrac{\partial}{\partial r} (r^2E_r) = \dfrac{1}{r^2}\dfrac{\partial}{\partial r} (kq) = 0$$

Thus, the divergence is zero at any point (except it's own location).

The curl in spherical coordinates is (see https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates):

$$\nabla\times \mathbf{E} = \frac{1}{r\sin\theta} \left(\frac{\partial}{\partial \theta} \left(E_\varphi\sin\theta \right) - \frac{\partial E_\theta}{\partial \varphi} \right) \hat{\boldsymbol r} \\ + \frac{1}{r} \left( \frac{1}{\sin\theta} \frac{\partial E_r}{\partial \varphi} - \frac{\partial}{\partial r} \left( r E_\varphi \right) \right) \hat{\boldsymbol \theta} \\ + \frac{1}{r} \left( \frac{\partial}{\partial r} \left( r E_{\theta} \right) - \frac{\partial E_r}{\partial \theta} \right) \hat{\boldsymbol\varphi}$$

As one can see, there are no non-zero terms in this expression. Thus

$$\nabla\times \mathbf{E} = 0$$

at any point (except it's own location).

At charge's own location neither divergens, nor curl is defined, since both expressions contain $\dfrac{1}{r}$ that becomes undefined at $r = 0$.

Answer. 1. Divergence and curle are both zero. 2. Undefined.