$$\vec{E} = \lim_{q_{0} \to 0} \frac{1}{q_{0}} \vec{F_{0}}$$
At each point P, the electric field set up by an isolated positive point charge q points directly away from the charge in the same direction as $\hat {r}$.
Note: Unit vector $\hat {r}$ points from source point S to field point P. $\hat {r}$ gives direction to $\vec {F}$ and $\vec {E}$. Note that $\hat {r}$ does not affect the magnitude since its length is 1 (unit vector).
Coulomb’s Law:
$$ \vec {F_{0}} = \frac{1}{4 \pi \epsilon _{0}} \frac{q q_{0}}{r^{2}} \hat{r}$$
Electric field:
$$ \vec {E} = \frac{1}{4 \pi \epsilon _{0}} \frac{q}{r^{2}} \hat{r}$$
