Calculating Net Electric Field

[BornSurvivor]

Homework Statement

Two charges are placed on the x axis. One charge (q1 = +8.36mC (microC)) is at x1 = +2.74cm and the other (q2 = -20.2mC (microC)) is at x1 = +9.05cm. Calculate the net electric field at x = 0cm.

Homework Equations

E= kq/r^2

The Attempt at a Solution

I tried calculating the electric field at that point for both charges and then tried adding them together. There was another topic with a very similar question, and that is the solution that was reached, but it doesn't work for this problem.

 

[BruceW]

It should work for this problem. One possible mistake: mC usually means milli-Coulombs, and \muC usually means micro-Coulombs

 

[BornSurvivor]

I think the mC meaning microC is just my weird homework interface.
When I calculate the individual electric fields, is q always positive? I know it is in Coulomb's Law calculations.

 

[BruceW]

No, in Coulomb's law calculations, you only get the magnitude of the electric field. You must decide on what direction it is in. (Or you can use the full vector form of the Coulombic force, which tells you the direction).

EDIT: Well, I should have said yes, q is always positive, because then this gives the magnitude of the electric field. But I meant no, the electric field isn't always positive.

 

[BornSurvivor]

Ok, so
E1= http://www.texify.com/img/%5CLARGE%5C%21%288.987E9N/C%2A8.36E-6C%29/.0274%5E2m.gif
E1= 1.0007E8 N/C
E2= http://www.texify.com/img/%5CLARGE%5C%21%288.987E9N/C%2A20.2E-6C%29/.0905%5E2m.gif
E2= -2.216E-7 N/C
And Enet = E1+E2
Enet= 7.79E7N/C
But that wasn't right.
Am I correct using the two charges given in the problem as q? Or does q actually refer to a test charge I have to make up?
Edit: Oh, and I realize that N/C is not the unit for k, but that doesn't make a difference if I know the units for electric field.

 

[BruceW]

E1 should be negative of what you've got. and E2 should be negative of what you've got. Remember, the electric field is defined as going in the direction that a positive test charge would be forced. And yes, q refers to the charges given in the problem.