Position of a charge (out of 3) with a net force of 0

[doggydan42]

Homework Statement

Point charges $q_1=50 \mu C$ and $q_1=-25 \mu C$ are placed 1.0 m apart.
Where must a third charge $q_3=20 \mu C$ be placed so that the net force on it is zero?

Homework Equations

$$\vec F=\frac{1}{4\pi {\varepsilon}_0}\frac{|q_1q_2|}{r^2}\hat r$$

The Attempt at a Solution

Let $r_1$ be the distance from $q_1$ to $q_3$, and $r_2$ be the distance from $q_2$ to $q_3$
$$\vec F=0=\frac{1}{4\pi {\varepsilon}_0}(\frac{|q_1q_3|}{{r_1}^2}+\frac{|q_3q_2|}{{r_2}^2})
\\ -\frac{|q_2q_3|}{{r_2}^2}=\frac{|q_3q_1|}{{r_1}^2} \Rightarrow -\frac{|q_2|}{{r_2}^2}=\frac{|q_1|}{{r_1}^2}
\\ -{r_1}^2|q_2|={r_2}^2|q_1|$$
Plug in values:
$$-{r_1}^2|q_2|={r_2}^2|q_1| \Rightarrow -{r_1}^2(25 \mu C)={r_2}^2(50 \mu C)
\\ -{r_1}^2=2{r_2}^2$$

I am not sure if I missed something in my approach, but I do not know how to continue.

Thank you in advance

 

[mfb]

Did you draw a sketch?
Can you describe in words where the charge should be?

The approach is good so far.

 

[doggydan42]

Would the charge be between $q_1$ and $q_2$

If so, then I was able to get a quadratic equation, but I get a complex number.

$$r_1+r_2=1 m \Rightarrow r_1 = 1-r_2
\\ -{r_1}^2=2{r_2}^2 \Rightarrow -(1-r_2)^2=2{r_2}^2
\\ 1-2r_2+{r_2}^2=-2{r_2}^2 \Rightarrow 1-2{r_2}+3{r_2}^2=0$$
When using the quadratic formula, the discriminant becomes $\sqrt{4-4(1)(3)}=\sqrt{4-12}=\sqrt{-8}$

Did I make a mistake somewhere? How can the position be complex?

 

[mfb]

If the charge is in between, in which direction do the two forces point?
Does this explain why you didn't find a solution?

Strictly speaking you should have considered the direction for the initial equation with the forces already. You only calculated the magnitude.