Basic Electrostatics - Why Did I Get This Multiple Choice Question Wrong

There is a line with region 1 to the left, region 2 in the middle and region 3 on the right. There is a charge Q1 = -3e, which is between regions 1 and 2. There is a charge Q2 = +5e, which lies between regions 2 and 3.

A negative charge of -e could be placed in various regions along the line with Q1 and Q2 fixed in their positions. Determine the direction of the net force by Q1 and Q2 on -e when the negative charge -e is located in region 1.

Answer: Depends on the position inside region 1

Why is this the answer? I thought the net force would be to the left since Q1 is closer, and distance has a significant effect in electrostatic force since force is inversely proportional to distance squared, and since Q1 and the new charge are both negative, Q1 would repel the new charge to the left. How would that not be the case? Q1 is even close in magnitude to Q2, and as mentioned, it's closer to the new charge.

Yes, but the effect from distance is not linear. Because Q2 has a greater charge, as you go farther from Q1, there will come a point where the Q2 has more of an effect than Q1.

Close to the charges there is a repulsion because the -e is so close to Q1. Farther from the charges the relative distances become (proportionally) more similar.

We can even solve this directly for the ratio of the -e to Q1 and -e to Q2 distances.
F(q1) = F(q2)
kQq/d^2 = kQq/d^2
q/d^2 = q/d^2
3/(x)^2 = 5/(y)^2
y^2/x^2 = 5/3
y/x = 1.29

So when -e is so close to Q1 that it is more than 1.29 times farther to Q2, then the Q1 charge dominates and there is a repulsion. When -e is so far from Q1 that it is less than 1.29 times farther to Q2, then the Q2 charge dominates and there is an attraction.

Lets say the distance from the -e charge at point A to Q1 is x and the distance between Q1 & Q2 is e.
Then the net force acting at A is 5/(x+e)² - 3/x²
When e = 0 the force will always be positive: 5/x²-3/x² = 2/x²