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

  • Thread starter doggydan42
  • Start date
  • #1
166
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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

  • #2
34,956
11,143
Did you draw a sketch?
Can you describe in words where the charge should be?

The approach is good so far.
 
  • #3
166
18
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?
 
  • #4
34,956
11,143
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.
 

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