# (Two fixed charges add in a third), why does it matter which side I put the third charge if I change signs accordingly?

So the question is "Two charges, 1.0 micro-Coulombs and -3.0 micro-Coulombs, are 10 cm apart. Where can a third charge be located so that no net electrostatic force acts on it?". Now the instructor states that we are assuming the two charges with given values are fixed.

So when he chooses to solve it, he draws his diagram with the +1.0 micro-Coulombs on the left of (in line with) the -3.0 micro-Coulombs and places the third charge "q3" to the left of (in line with) the +1.0 micro-Coulombs charge which results in a net electrostatic force of 0 (assuming q3 is positive). For the purpose of my question he gets to this point: $$\frac{1.0\; x \;10^{-6}}{d^2} - \frac{3.0\; x \;10^{-6}}{(d+0.1)^2} = 0$$

$$And\; Thus \;\; d = 13.6 [cm]$$

I get that and I get how he solved for d, but what I don't get is that when I choose to assume that q3 is negative and put it to the right of the -3.0 micro-Coulombs charge I end up with: $$\frac{3.0\; x \;10^{-6}}{d^2} - \frac{1.0\; x \;10^{-6}}{(d+0.1)^2} = 0$$ And thus getting an incorrect answer and I can't figure out why.

Also a side note question: so he usually says to basically ignore the signs of the charges and then do some rational as to whether it is a repulsive or attractive force. So if I am choosing to assume that q3 is negative and I want to keep the signs, the force itself for the 2 negative charges will be repulsive (positive) and the force itself for q3 and the 1 micro-Coulomb charge would be attractive (negative). But then we can cancel out q3's and my mind tends to go a little dizzy trying to figure out what the resulting signs should be.

Thank you so much! I have done this problem over and over again and can't seem to understand why the charge placement matters if you're changing q3's sign accordingly.

Edit: Wait I think I see now. So it seems that his coordinate axis has the 1.0 micro-Coulomb charge as the "x = 0" and to the left is positive and to the right is negative. So the negative solution, -0.037 (which 0.037 < 0.1) means it would be in the middle which we know is not a valid solution (the net forces cannot = 0 in the middle no matter the sign of q3). So if I was to make my axis the same (with 1.0 micro-Coulomb being my 0) I get -0.237 or -0.063. And 0.062 < 0.1 so that would mean it's in the middle so not a valid solution.

That and the fact that it is further makes sense because 3.0 micro-Coulomb > 1.0 micro-Coulomb so if the charge is on the -3.0 micro-Coulomb side, then it has to be further from it (being pushed harder by stronger force) than a corresponding positive q3 near the smaller 1.0 micro-Coulomb charge. If anything I said is incorrect please let me know, otherwise I guess problem solved!