# Solving for Charge of Particles on X-axis

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1. Sep 14, 2015

### Matthew B.

• Homework template is missing because this was moved from a non-homework forum
Hello everyone. Hopefully someone would be able to help me with this problem.

The question states:
Two point charges qa = -12 micro coulombs and qb = 45.0 micro coulombs and a third particle with unknown charge qc are located on the x axis. The particle qa is at the origin placed so that each particle is in equilibrium under the particles. Find the required location and the magnitude and the sign of the charge of the third particle...

I used a sum of forces model to answer the first part..
starting with Fac + Fbc = 0... which got me to
kqaqc/r^2 = -kqbqc/(x+r)^2
some things canceled.. ended up with a quadratic...
came to the conclusion that r = 0.16 m or -.051 m.
I'm fairly confident in that answer but I am now not sure how to find the charge of qc. Modeling it as a sum of forces would just cancel the qc...

2. Sep 14, 2015

### BvU

Hi Mat, welcome tp PF !

Good thing you are confident your answers are right. Can you also convice me ? It seems to me that some position is still missing in the problem statement.

3. Sep 14, 2015

### Matthew B.

Oh you're right! Forgot a given distance... qb is located at x = 15.0 cm.

4. Sep 14, 2015

### BvU

Fair enough. And you solve $$k \; {q_a q_c \over r^2} = -k\; {q_b q_c \over (x+r)^2}$$If I draw a picture, I expect $q_c$ to be at a negative $r$. Do you agree ?

On the other hand, if you solve $$k \; {q_a q_c \over (x+r)^2} = -k\; {q_b q_c \over r^2}$$I expect $q_c$ to be at a positive $r$. Nonsense. as Mat says in post #5 below.

What do you do with $$k \; {q_a q_b \over x^2} \ \ ?$$

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Last edited: Sep 14, 2015
5. Sep 14, 2015

### Matthew B.

I used your first setup but I'm puzzled why you need the second setup.. Looking at how I drew it :

wouldn't you just need your first setup? Because regardless of what X is, the distance between charges qa and qc will always be simply r.. not x+r^2.. because qa is located at the origin? Looking at the attached photo, couldn't the charge c only be to the left of charge a or to the right of charge B? Or are there more possible positions for qc?

6. Sep 14, 2015

### Matthew B.

7. Sep 14, 2015

### ogg

If the THREE particles are "at equilibrium" then the net forces on each of them is zero. I am going to change the variables to Qm, Qr, Ql for the three charges (mid, right, and left) and the two distances are R and L. You can easily determine the m-r force, fmr, since you're given all values. The net forces (fmr+fml)= (fmr+frl)=(fml+frl) all are zero. Thinking about it another way, Ql can vary as long as it and the square of its distance to Qm (say) are in proportion to give the same NET force onto Qm. But there is also a force between it and Qr - and that is in proportion to 1/(L+R)². That is, you have two relationships and two variables to solve for: Ql and L. {R=15 of course}. You know by simple logic that the signs of Ql and Qr must be the same, (otherwise a particle in between couldn't be at equilibrium), so they are both positive. That's as far as I've gotten so far... But it seems the rest is simply algebra...?...and you've got enough information to solve this, I think.

Last edited: Sep 14, 2015
8. Sep 14, 2015

### BvU

Yes, you are right -- I got confused because x is the known one and r is the horizontal coordinate of $q_c\;$ .

But in your picture I then do not understand why you designate the position of $q_c\;$ with $x+r$.

And why you want to solve r from $$k \; {q_a q_c \over r^2} = -k\; {q_b q_c \over (x+r)^2}$$instead of $$k \; {q_a q_c \over r^2} = -k\; {q_b q_c \over (x-r)^2}$$

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9. Sep 14, 2015

### BvU

Ah ! There is something fundamentally wrong with the way we are looking at this. We calculate $|\vec E|$ and forget to consider the signs (at least I did until now ).

Taking your picture: $q_c\$ is at r, $q_a\$ is at 0, $q_b\$ is at x = 0.15.

(somewhat unfortunate choice ; $x_c$, 0 and $x_b$ would have been clearer)​

For a point with x-coordinate < 0 the field from $q_a\$ points to the right,
and if x-coordinate > 0 the field from $q_a\$ points to the left.​
For a point with x-coordinate < x the field from $q_b\$ points to the left,
and if x-coordinate > x the field from $q_a\$ points to the right.​

So the force on $q_c\$ can not be zero in the range 0 -- x .
Two solutions may be possible: a negative charge with r > x and a positive charge with r < 0

You write down the equations once more, but a little extra thinking might eliminate one of these two beforehand....
There is only one point where the fields from a and b cancel, no matter what the charge on c is -- and you probably already have the value of the coordinate - but I think not the sign

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