Two charges, Q1= 3.60 μC, and Q2= 6.40 μC are located at points (0,-3.00 cm ) and (0,+3.00 cm), as shown in the figure.
What is the magnitude of the electric field at point P, located at (4.50 cm, 0), due to Q1 alone?
so I got 1.11×107 N/C which is right

now ,
What is the x-component of the total electric field at P?
here what I did
E2= KQ2/(L^2)
and then :
cos alpha = x2/l
alpha = 56.309
then Epx = (E1x)*sin alpha +(E2x)*sin alpha

my final answer was 9.2335*10^-6 N/C
but it did not work out with me :(

The x component of the electric field due to charge Q1 is indeed 9.22 x 10^6 N/C. But you need to add the component due to charge Q2. You should be using an equation like this:


E_x = k(\frac {Q_1} {{d_1}^2} cos\alpha_1+ \frac {Q_2} {{d_2}^2} cos\alpha_2)

OK , but, what is (alpha1, and what is alpha2)
Thanks :)

Aplha1 and alpha2 are the angles at point P between the horizontal and the charges Q1 and Q2 respectively. It's arccos(3.5/d), or about 33.7 degrees.