Take into account that the magnitude of electric field at point P, produced by a charge q is given by:
[tex]E=k\frac{q}{r^2}[/tex]where k (9*10^9 N/m^2C^2) is the Coulomb's constant, r is the distance from the charge to the point P and q is the magnitude of the charge.
An illustration of the given situation is shown below:
As you can notice, at point P the electric fields generated by each charge point to the negative charge. The signs of the charge determine the direction of the electric field.
The magnitude of each electric field is the same, due to the distance to each charge is the same and the magnitude of the charge is equal. Then, The magnitude of the total electric field is:
[tex]\begin{gathered} E=E_1+E_2=2(9\cdot10^9\frac{N}{m^2C^2})(\frac{2.00\cdot10^{-6}C}{(0.05m)^2}) \\ E\approx1.44\cdot10^7\frac{N}{C} \end{gathered}[/tex]Hence, the electric field at a point midway the given charges is approximately 1.44*10^7N/C
