# Where Should Charge Q3 Be Placed to Nullify Net Force Between Q1 and Q2?

• steroidjunkie
In summary, the problem involves three point charges: ##Q_1 = 9 \mu##C, ##Q_2 = -16 \mu##C, and ##Q_3##. The objective is to find the distances ##x_1## and ##x_2## from charges ##Q_1## and ##Q_2## respectively, where a third charge ##Q_3## can be placed so that the net force on it is zero. The net force is equal to the forces exerted by ##Q_1## and ##Q_2##, which can be represented by the equations ##F_{13}=k \cdot \frac{Q_1 Q_3}{x_{1}^
steroidjunkie

## Homework Statement

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##

## Homework 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 =?##

## 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?

#### Attachments

• IMG_20160603_152326.jpg
24.1 KB · Views: 478
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

steroidjunkie
steroidjunkie said:
Since the net charge on ##Q_3## is zero it follows that ##F_{13}=F_{23}##.
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.

##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##?

Delta2
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?

steroidjunkie
steroidjunkie said:
##F_{13}+F_{23}=0## or
##F_{13}=-F_{23}##This makes sense. Does ##x_2## mean I can also put the charge ##Q_3## between ##Q_1## and ##Q_2##?
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.

steroidjunkie said:
##x_1=21cm##
##x_2=-3cm##
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.

steroidjunkie
haruspex said:
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.
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 can't 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 can't 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:
steroidjunkie
steroidjunkie said:
##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##
You left off the exponent on ##\ x_1\ ## on the right hand side above, but corrected it after that.
...

##x_{1,2}=\frac{-18 \pm 24}{-2}##
##x_1=21cm##
##x_2=-3cm##
That's not good notation to show two solutions for ##\ x_1\ .\ ## You have previously used ##\ x_2\ ## as the position of ##\ Q_2\ .##
This makes sense. Does ##x_2## mean I can also put the charge ##Q_3## between ##Q_1## and ##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.

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 :)

steroidjunkie said:
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 :)
If you change the sign of Q3 that will change the direction of both forces, won't it?

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?

steroidjunkie said:
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?
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.

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?

steroidjunkie said:
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?
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).

steroidjunkie

## 1. What is net force on a charge?

The net force on a charge, also known as the resultant force, is the sum of all the individual forces acting on the charge.

## 2. How is the net force on a charge calculated?

The net force on a charge is calculated by adding all the individual forces acting on the charge, taking into account their magnitude and direction.

## 3. What factors affect the net force on a charge?

The net force on a charge is affected by the magnitude and direction of the individual forces acting on the charge, as well as the charge's mass and its velocity.

## 4. How does the net force on a charge relate to its acceleration?

In accordance with Newton's Second Law of Motion, the net force on a charge is directly proportional to its acceleration. This means that the greater the net force, the greater the acceleration of the charge.

## 5. Can the net force on a charge be zero?

Yes, the net force on a charge can be zero if the individual forces acting on the charge cancel each other out, resulting in a state of equilibrium where the charge remains stationary or moves at a constant velocity.

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