Electric field at a point near three charges in a line

[rinarez7]

1. Homework Statement
Consider 3 charges along a horizontal line as shown:
8.72e-6 C (+) ___4.16cm____2.45e-6 C (+)______3.18cm_____-1.46e-6C (-)
The acceleration due to gravity is 9.8 m/s^2
The Coulomb Constant is 8.98755e 9 Nm^2/ C^2
What is the electric field at a point 2.18 cm to the left of the middle charge? Answer in units fo N/C


2. Homework Equations
Q1= 8.72e-6 C , r= .0198m (.0416m-.0218m)
Q2= 2.45e-6 C, r= .0218m

E1= k Q1 / r^2
E2= k Q2/ r^2

3. The Attempt at a Solution
I converted all of my units first to C and meters.
Then I solved for E1 and E2 separately.
Once I had E1 and E2, I added the two together. But I can't seem to get the right answer.
Am I supposed to subtract them instead becasue Q1 and Q2 are both positive and therefore will repel each other?

 

[Delphi51]

You have THREE charges causing electric fields, so you need to find E1, E2 and E3.
Add them together, keeping direction in mind!

 

[rinarez7]

so I calculated E1 = 1.999e8, E2 as 4.633e5, and E3 as -3.657e6
Am I correct to say then that, E1 and E3 are negative (pointing to the left) and E2 is positive?
E1(+)<----X----->E2 (+)-----><------E3 (-)

 

[Delphi51]

E1 to the right (away from a positive charge), E2 to the left (away from a positive charge), E3 to the right (toward a negative charge).

 

[rinarez7]

Thank you!