you will have Ex(P)=(9E9)*(-3.9E-6)/(.067^2+.067^2).
this will give you the magnitude of the electric field produced by q1. all thats left is to multiply your answer by a sin(45) and voila, there's your answer!
This should be your final answer (-2764474.05 N/C)
for the second part you do the same, use the formula but this time q2 will affect it because you are looking for the y-component.
Ey(P)=(9E9)*(7.1E-6)/(.067^2) which equals out to 14234796.17 N/C.
to get the final answer just use the answer from part A and subtract it from the above answer and you are all set to go!
This should be the final answer for the second part. (11470322.12 N/C)
