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. Relevant 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?