The magnitud of an electric field is given as:
[tex]E=\frac{1}{4\pi\epsilon_0}\frac{\lvert q\rvert}{r^2}[/tex]
For the charge 1 the magnitude is:
[tex]\begin{gathered} E_1=\frac{1}{4\pi\epsilon_0}\frac{\lvert-6.39\times10^{-9}\rvert}{(0.424)^2} \\ E_1=319.543 \end{gathered}[/tex]
Now, since charge 1 is negative this means that this field points towards the charge, in this case to the left, then the electric field for charge one is:
[tex]\vec{E_1}=-319.543[/tex]
For the charge 2 the magnitude is:
[tex]\begin{gathered} E_2=\frac{1}{4\pi\epsilon_0}\frac{\lvert3.22\times10^{-9}\rvert}{(0.636)^2} \\ E_2=71.565 \end{gathered}[/tex]
Now, since charge 2 is positive this means that this field points away from the charge, in this case to the left, then the electric field for charge two is:
[tex]\vec{E_2}=-71.565[/tex]
Now, the total field on point P is the sum of both electric fields, then the total electri field on this point is:
[tex]E=-391.108[/tex]
