Net force on a charge $Q_3$

1. Jun 3, 2016

steroidjunkie

1. The problem statement, all variables and given/known data
Two point charges $Q_1 = 9 \mu$C and $Q_2 = -16 \mu$C are fixed in space on a distance r=7cm. At what distane $x_1$ from the first charge, and $x_2$ from the second charge, should we place the third charge $Q_3$ so that net force on $Q_3$ is zero? make a sketch with force diagram for third charge. Is $Q_3$ positive or negative? What distance from $Q_1'$ and $Q_2$ should we put $Q_3'$ at, if charge $Q_3'$ is $0.5Q_3$

2. Relevant equations
$Q_1 = 9 \mu C=9 \cdot 10^{-6}$C
$Q_2 = -16 \mu C=-16 \cdot 10^{-6}$C
r=7cm
$Q_3 =?$

3. The attempt at a solution
Charge $Q_3$ must be positive in order to be stationary in presence of charges $Q_1$ and $Q_3$. It's visible from the sketch in the attachment.

Since the net charge on $Q_3$ is zero it follows that $F_{13}=F_{23}$.
$F_{13}=k \cdot \frac{Q_1 Q_3}{x_{1}^2}$
$F_{23}=k \cdot \frac{Q_2 Q_3}{x_{2}^2}$
$x_2=r+x_1$ is visible from the picture attached.

Now I substitute $F_{13}$ and $F_{23}$ in the first equation with corresponding equations:
$k \cdot \frac{Q_1 Q_3}{x_{1}^2}=k \cdot \frac{Q_2 Q_3}{x_{2}^2}$

I divided it by $k \cdot Q_3$:
$\frac{Q_1}{x_{1}^2}=\frac{Q_2}{x_{2}^2}$

It is visible from this equation that no matter what is the value for $Q_3$ the distance from $Q_1$ and $Q_2$ won't be affected since there is no $Q_3$ in this formula. I can conclude that charge $Q_3'$ put in place of $Q_3$ will be at the same distance from charges $Q_1$ and $Q_2$.

I substitute $x_2$ with $r+x_1$:
$\frac{Q_1}{x_{1}^2}=\frac{Q_2}{(r+x_1)^2}$
$Q_1 \cdot (r+x_1)^2=Q_2 \cdot x_{1}^2$
$Q_1 \cdot (r+x_1)^2-Q_2 \cdot x_{1}^2=0$
$9 \cdot 10^{-6} \cdot (r^2+2r x_1+x_1^2)-(-16) \cdot 10^{-6} \cdot x_{1}^2=0$
$9 \cdot 10^{-6} \cdot (r^2+2r x_1+x_1^2)+16 \cdot 10^{-6} \cdot x_{1}^2=0~~~~/\cdot 10^{-6}$
$9 \cdot r^2+9 \cdot 2r x_1+9 \cdot x_1^2+16 \cdot x_{1}^2=0$
$25 \cdot x_{1}^2+18 \cdot r x_1+9 \cdot r^2=0$

If I try to solve this equation with r=7cm I get a result with real and complex part. If I insert r=0.07m I get something similar. How do I fix this?

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2. Jun 3, 2016

rpthomps

The charge on Q3 can be either negative or positive. Also, when you drew your FBD and setup your Fnet statement you already considered direction, so you don't have to include the signs on the charges in your calcs

3. Jun 3, 2016

haruspex

First, you mean there is no net force on Q3. Secondly, it helps to be consistent about which way is positive for forces. Think again about the claim that the forces are equal.

4. Jun 3, 2016

steroidjunkie

$F_{13}+F_{23}=0$ or
$F_{13}=-F_{23}$

If I follow through with this then:
$9 \cdot x_1^2+18 \cdot r x_1+9 \cdot r^2=-(-16) \cdot x_1$
$9 \cdot x_1^2+18 \cdot r x_1+9 \cdot r^2-16 \cdot x_1=0$
$-7 \cdot x_1^2+18 \cdot r x_1+9 \cdot r^2=0$
$-7 \cdot x_1^2+18 \cdot 7 x_1+9 \cdot 7^2=0$
$-7 \cdot x_1^2+18 \cdot 7 x_1+9 \cdot 7^2=0$
$-x_1^2+18 \cdot x_1+9 \cdot 7=0$
$-x_1^2+18 \cdot x_1+63=0$
$x_{1,2}=\frac{-18 \pm \sqrt{324-(-4)*63}}{-2}$
$x_{1,2}=\frac{-18 \pm \sqrt{324+252}}{-2}$
$x_{1,2}=\frac{-18 \pm \sqrt{576}}{-2}$
$x_{1,2}=\frac{-18 \pm 24}{-2}$
$x_1=21cm$
$x_2=-3cm$

This makes sense. Does $x_2$ mean I can also put the charge $Q_3$ between $Q_1$ and $Q_2$?

5. Jun 3, 2016

rpthomps

Think about your last statement logically by first putting a positive charge between the two other charges and then a negative charge. What will be the net force acting on the new charge?

6. Jun 3, 2016

Delta²

Yes that's what it means. In general negative value means towards the right of Q1, positive value towards the left of Q1. And this follows from the way you setup the equations and the diagram in the picture.

7. Jun 3, 2016

haruspex

that is correct for x1 but I do not understand how you get -3 for x2. You wrote originally, and correctly, x2=r+x2.
It would make no sense to put Q3 between a positive and negative charge. The fields from the two fixed charges would reinforce each other there, not cancel.

8. Jun 3, 2016

Delta²

Symbol ambiguity error (lol). When he says $x_2$ he means the second root of the equation.

But you are right in between the forces cant cancel out. So the second root of the equation does not correspond to solution in the problem. That's because $F_{13}$ changes direction if $x_1$ becomes negative but the way the equations are setup this cant be seen. The only way to account for the direction change is to write $F_{13}=k\frac{q_1q_3}{\mid{x_1}^3\mid}x_1$ and similar for $F_{23}$.

Last edited: Jun 3, 2016
9. Jun 3, 2016

SammyS

Staff Emeritus
You left off the exponent on $\ x_1\$ on the right hand side above, but corrected it after that.
That's not good notation to show two solutions for $\ x_1\ .\$ You have previously used $\ x_2\$ as the position of $\ Q_2\ .$
The two solutions only indicate the two values for $\ x_1\$ for which the magnitudes of the forces are equal. Use @rpthomps post to see if that's enough.

10. Jun 3, 2016

steroidjunkie

I see I've made some typing errors and my notation for distance of $Q_3$ from charge $Q_1$ isn't appropriate. Also, when $x_1=-3cm$ charge $Q_3$ must be negative because it is closer to the positive charge. If it was positive, it'd be repelled. Thank you all for your help and pointing that out. I hope I didn't make a mistake in this post :)

11. Jun 3, 2016

SammyS

Staff Emeritus
If you change the sign of Q3 that will change the direction of both forces, won't it?

12. Jun 3, 2016

steroidjunkie

I get it, but I switched the plus sign to minus in $k \cdot \frac{Q_1 \cdot (-Q_3)}{x_1^2}=-k \cdot \frac{Q_1 \cdot (-Q_3)}{x_1^2}$
Now I can divide equation by minus and I will get the same equation as was for positive $Q_3$. Right?

13. Jun 3, 2016

SammyS

Staff Emeritus
Right. The same position no matter the sign of the charge, Q3 . However, there is a difference in how stable that equilibrium will be.

In one case, if you move Q3 slightly from the equilibrium position, then the resulting force will be in a direction to push the charge back toward the equilibrium position. In the other case, move Q3 slightly from the equilibrium position and the net force will tend to push it farther from equilibrium.

14. Jun 3, 2016

steroidjunkie

I have one more question. Does it matter what charge I put at $x_1$? I just inserted $Q_3$ in the $F_{13}+F_{23}=0$ equation and it doesn't matter if $Q_3$ is positive or negative. I always get $0=0$. That would mean $Q_3$ can be positive or negative. Or not?

15. Jun 3, 2016

nrqed

To be more careful with notation, you meant $\vec{F}_{13} + \vec{F}_{23}=0$ (while the corresponding equation for the magnitudes is just $F_{13}=F_{23}$).

Since on can get rid of $Q_3$ completely (it disappears from the equation), it is correct that $Q_3$ can be anything at all (and yes, it can be either positive or negative).