# (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!

## 2 Answers

1. The polarity of the test charge doesn't matter since you are looking for a net force of zero
2. It really helps if you clearly define your co-ordinate system. What exactly does d=0 mean in each of your cases and in which direction is d positive and negative.
3. How do you know your answer is actually wrong? This is not a symmetric problem and so the answer to the first part doesn't tell you anything about the answer to the second part.
• I definitely need to clearly define my coordinate axis, that's a problem. Well before I go any further, am I going to get different answers no matter how I define my coordinate axis because of the different values of the charges? Is there going to be a different magnitude of force to the left of the 1 micro-Coulomb charge than the right of the -3 micro-Coulomb charge? (ah I think I remember there is now. I am reviewing this material again and this is the first lecture). So I imagine there is not a scenario I can make the distance for my setup the same as his? – intwarrior Dec 12 '18 at 21:25
• So to the right of the negative charge, there is going to be a stronger repulsive force to adding in a -Q to the right of the -3 micro-Coulomb than a corresponding repulsive force if we put a +Q to the left of the 1 micro-Coulomb. Thus needing the added Q to be a different distance in order to have a net force on it of 0. Right? – intwarrior Dec 12 '18 at 21:28
• Thanks for your insight on the coordinate system, that definitely cleared some things up. I realized he had it backwards from your standard coordinate system and that threw me off a little! – intwarrior Dec 12 '18 at 23:26

In order to have a net force of zero, the two forces must have the same magnitude (and opposite directions, of course). You can have equal magnitude only if you test charge is closer to the small (in magnitude) charge and farther from the larger (in magnitude) charge. This way you can compensate the smaller charge by a smaller distance. This is true no matter what is the sign of the test charge. If you put the test charge on the other side, closer to the big charge, there is no way to have the two forces balancing. You have the force from a charge with a small magnitude and farther away, competing with the force from a large charge nearby. So your second equation should have no solution if you set it up correctly.

• Thank you for your clarifying about the fact that they're different magnitudes of forces and thus resulting in different distances needed to have a net force of 0! I'm apparently too much of a noob to give a +1 for helpful vote lol. – intwarrior Dec 12 '18 at 23:29