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

## 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

## Answers and Replies

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

The approach is good so far.

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
Mentor
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.