$$\begin{gathered} \mathrm{Electrostatic} \mathrm{force} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{F} =\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} = 9\times {10}^9\frac{(1.6\times {10}^{-19}{)}^2}{{\mathrm{r}}^2} \\ \mathrm{Gravitational} \mathrm{force} \mathrm{is} \mathrm{given} \mathrm{by} \mathrm{F}=\mathrm{G}\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{r}}^2}= 6.67\times {10}^{-11} \frac{(1.67\times {10}^{-27}{)}^2}{{\mathrm{r}}^2} \end{gathered}$$

Ratio of these forces are given by,

            $\frac{9\times {10}^9{\displaystyle \frac{(1.6\times {10}^{-19}{)}^2}{{\mathrm{r}}^2}}}{6.67\times {10}^{-11}{\displaystyle \frac{(1.67\times {10}^{-27}{)}^2}{{\mathrm{r}}^2}}}$

            =        $1.24\times {10}^{36}$.

Two essential conditions for this to happen is:
i) Both charges cannot be of the same sign.
ii)The point where electric field intensity has to be 0 is closer to the smaller charge as compared to the charge larger in magitude.

The electric field lines of the point charge are as given below:

Electric field line or the line of force is the path along which a unit positive test charge would move when kept in an electrostatic field.
In other words, a field line is an imaginary line drawn such that its direction at any point is the same as the direction of the field at that point. 
Line of field due to two equal positive charge is shown below: