Thanks so much!
Why should you disregard the negative for the 18 coulombs?
When you use formulae such as this one and Coulomb's law, you're interested in the magnitude of the electric field strength or the magnitude of the force. So, if you had a Coulomb's law question and you had one point charge of 2 µC separated from a second point charge of -2 µC by a distance of 2 m, then you would disregard the negative in your calculations for the magnitude of the force.
So you would have
F = ((9*109)(2*10-6)(2*10-6))/22
Rather than having a negative 2 for one of the Qs
F = ((9*109)(2*10-6)(-2*10-6))/22
This is because we are simply interested in the magnitude of the force (which will always be presented as a positive number). The signs of the charges will tell us whether the force between the charges will be attractive or repulsive. In this case, we know that the force between them will be attractive. This is because Coulomb's law tells us the opposite charges attract.
Similarly, when we are working with electric field strength we are interested in the magnitude. So we can leave the negative off of the -18µC charge. All it tells us is that the electric field from the negative point charge will oppose that of the positive 2µC point charge. Meaning they will cancel out. If you put the negative in the calculation, then when you do your simultaneous equation the will add together (instead of cancel out) and you will get the wrong answer for distance.
I hope this makes a little more sense