Electric Field with three charges

1. The problem statement, all variables and given/known data
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?


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


Homework Helper
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).
Thank you!!!!

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