How Does the Charge Q Affect Forces Between Q1 and Q2?

[1729]

Homework Statement

A positive point charge $Q_1$ (4.0 µC) and a negative point charge $Q_2$ are placed near each other. When a negative charge $Q$ is placed in the middle between the two point charges, the magnitude of the resultant forces on $Q_1$ and $Q_2$ doubles. What is the charge of $Q$?
(answer: -3.0 microcoulombs)

Homework Equations

Coulomb's law ($|F|=k\frac{|Q_1||Q_2|}{r^2}$)

The Attempt at a Solution

I've written down the equations for the resultant forces for $Q_1$ and $Q_2$ when $Q$ is present and when it is not. It's quite a lot of stuff to type, so I'm going to summarise the gist of it:

$$2|F_{R,Q2}|=|F_{R',Q2}|\\
-2k\frac{|Q_1||Q_2|}{4r^2}=k\frac{|Q_2|}{r^2}(|Q|-\frac{|Q_1|}{4})\\
\Rightarrow |Q|=-\frac{|Q_1|}{4}=-1.0 \ \mathrm{\mu C}$$
where $2r$ is the distance between $Q_1$ and $Q_2$

I hope someone can help me out here.

Thanks in advance.

 

[TSny]

$$2|F_{R,Q2}|=|F_{R',Q2}|\\
-2k\frac{|Q_1||Q_2|}{4r^2}=k\frac{|Q_2|}{r^2}(|Q|-\frac{|Q_1|}{4})\\$$

Be careful with the signs. Note in the first equation above the left hand side is a positive quantity. In the second equation, the left hand side is negative. How did that come about?

 

[1729]

We are working with vectors that lie on the same line, but point into opposite directions. I've more or less defined the unit vector to point from Q_1 to Q_2, from left to right, that is.

 

[TSny]

We are working with vectors that lie on the same line, but point into opposite directions. I've more or less defined the unit vector to point from Q_1 to Q_2, from left to right, that is.

OK. So, your equation

$-2k\frac{|Q_1||Q_2|}{4r^2}=k\frac{|Q_2|}{r^2}(|Q|-\frac{|Q_1|}{4})$ is a vector equation where a negative force means that the force is to the left.

Note that the problem states that the magnitude of the force on Q₂ doubles. Does that necessarily mean that the force vector doubles?

 

[1729]

I believe so. We are multiplying a vector by a number, by doubling the magnitude the force vector would double along with it.

 

[TSny]

If a baseball initially has a momentum vector of 10 kg⋅m/s due east, and then later you're told that the baseball has a momentum vector with twice the magnitude (20 kg⋅ m/s), what can you conclude about the direction of the final momentum vector of the baseball?

It also might help to think about the following. Suppose charge Q (in the middle) has a very small amount of negative charge. Would placing Q between Q₁ and Q₂ cause the magnitude of the net force on Q₂ to increase or decrease?

 

[1729]

If a baseball initially has a momentum vector of 10 kg⋅m/s due east, and then later you're told that the baseball has a momentum vector with twice the magnitude (20 kg⋅ m/s), what can you conclude about the direction of the final momentum vector of the baseball?

It's eastbound? I think I don't understand your analogy.

It also might help to think about the following. Suppose charge Q (in the middle) has a very small amount of negative charge. Would placing Q between Q₁ and Q₂ cause the magnitude of the net force on Q₂ to increase or decrease?

The magnitude would very slightly decrease. Q_1 and Q_2 attract each other heavily, whilst Q and Q_2 would repel each other ever so slightly.

 

[TSny]

It's eastbound? I think I don't understand your analogy.

If all you know is the magnitude of a vector, then you don't know anything about its direction. So, even though the initial momentum of the baseball was eastward, there is no reason why the final momentum would also have to be eastward. Maybe the baseball got hit by a bat so that the final momentum is 20 kg⋅m/s westward. Or it got "popped up" so that it's final momentum is 20 kg⋅m/s upward. The only thing they tell you in the problem is that the magnitude of the final force on Q₂ is twice the magnitude of the initial force on Q₂.

The magnitude would very slightly decrease. Q_1 and Q_2 attract each other heavily, whilst Q and Q_2 would repel each other ever so slightly.

Yes, the net force on Q₂ would decrease. And if you made Q a little more negative, the magnitude of the net force on Q₂ would decrease even more. So, think about how it is possible for the magnitude of the net force on Q₂ to end up being twice as much as it was before bringing in Q.

 

[1729]

I understand now. Since magnitude is irrespective of direction, we shouldn't consider the vector's direction when talking about its magnitude. Hence:
$$|k\frac{|Q_1||Q_2|}{2r^2}|=|k\frac{|Q_2|}{r^2}|\cdot||Q|-\frac{|Q_1|}{4}|\\
\Leftrightarrow 0 \ \mathrm{N} = |k\frac{|Q_2|}{r^2}|\cdot||Q|-\frac{|Q_1|}{4}-\frac{|Q_1|}{2}|\\
\Rightarrow |Q|=\frac{|Q_1|}{4}+\frac{|Q_1|}{2}=|-3.0 \ \mathrm{\mu C}|$$
Thank you for taking the time to explain!

 

[TSny]

I understand now. Since magnitude is irrespective of direction, we shouldn't consider the vector's direction when talking about its magnitude. Hence:
$$|k\frac{|Q_1||Q_2|}{2r^2}|=|k\frac{|Q_2|}{r^2}|\cdot||Q|-\frac{|Q_1|}{4}|\\
\Leftrightarrow 0 \ \mathrm{N} = |k\frac{|Q_2|}{r^2}|\cdot||Q|-\frac{|Q_1|}{4}-\frac{|Q_1|}{2}|$$

The first equation looks good, but does the second equation necessarily follow from the first equation? Cancelling common factors, these two equations are
$$\frac{|Q_1|}{2} =\left | |Q|-\frac{|Q_1|}{4} \right |\\
0 = \left | |Q|-\frac{|Q_1|}{4}-\frac{|Q_1|}{2} \right |$$

From the first equation, you can get $$0 = \left | |Q|-\frac{|Q_1|}{4} \right | - \frac{|Q_1|}{2}$$ What is the justification for writing this as $$0 = \left | |Q|-\frac{|Q_1|}{4}-\frac{|Q_1|}{2} \right |$$ I'm not saying it's wrong in this case, but in general it is not true that $|x - y| - |z| = |x - y - z|$.