Electric field and Electric field intensity

* “ Electric field ”:Electric field due to a charge is the space around the charge in which any other charge experiences a force of attraction or repulsion.

* “ Electric field intensity ”:

The Electric field intensity at a point is defined as the force experienced by a unit positive charge placed at that point.

Now consider the figure given below where a charge $q$ is located at point $O$ in space.

If $\vec{F}$ is the force acting on test charge $q_0$placed at a point $P$ in an electric field then electric field at that point is$$\vec{E}=\frac{\vec{F}}{q_0}$$

Let us imagine a test charge 

$q_0$ to be placed at P. Now we find force on charge $q_0$ due to q through Coulomb's law.$$\vec{F}=\frac{1}{4\pi {ϵ}_0}\frac{q{q}_0}{r^2}\hat{r}$$Electric field at point P is$$\vec{E}=\frac{\vec{F}}{q_0}=\frac{1}{q_0}\left(\frac{1}{4\pi {ϵ}_0}\frac{q{q}_0}{r^2}\right)\hat{r}$$Or,$$\vec{E}=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}\hat{r}$$And$$\left|\vec{E}\right|=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}=9\times {10}^9\frac{q}{r^2}$$Above expression gives the magnitude of electric field intensity due to charge $q$at any point at a distance $r$ from it. 
The direction of the field is away from the charge q if it is positive. Electric field for either a positive or negative charge in terms of unit vector $\hat{r}$ directed along line from charge q to point P is$$\vec{E}=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}\hat{r}$$Here, $r$ is the distance from charge $q$ to point p .

Electric field is a vector quantity and since

$$ \vec{F}=q_0\vec{E}$$ \vec{E}$$ \vec{F}$$ (N.C^{-1})$$ \left[ML{T}^{-3}{A}^{-1}\right]$$