The point midway between the two charges is at
[tex]d= \frac{20~cm}{2}=10~cm=0.1~m [/tex]
from both charges.
1) Let's calculate the electric field generated by the charge [tex]q_1=60~\mu C=60\cdot10^{-6}~C[/tex] at the point halfway between the two charges. It will be
[tex]E_1 = k_e \frac{q_1}{d^2} = 8.99\cdot 10^9 Nm^2C^{-2} \frac{60\cdot10^{-6}~C}{(0.1~m)^2} = 5.39\cdot10^7~NC^{-1}[/tex]
2) Instead, the electric field generated by the charge [tex]q_2=-12 \mu C =-12\cdot10^{-6}~C[/tex] will be
[tex]E_2 = k_e \frac{q_2}{d^2} = 8.99 \cdot 10^9 Nm^2C^{-2} \frac{-12\cdot 10^{-6}~C}{(0.1~m)^2} =-1.08\cdot10^7~NC^{-1}[/tex]
3) The total field is given by the resultant of the two fields. Now, if we assume that the first charge is on the left of the half-way point and the second charge is on its right, the resultant will be the algebric difference of the two fields, therefore the total field is
[tex]E_{tot}=E_1-E_2 = 5.39\cdot10^7~NC^{-1}-(-1.08\cdot 10^7~NC^{-1})=[/tex]
[tex]=6.47\cdot10^7~NC^{-1}[/tex]
and the verse is towards charge 2 (same verse of [tex]E_1[/tex]).
