# Homework Help: Electric Charge homework

1. Jan 13, 2008

### Goldenwind

[SOLVED] Electric Charge

1. The problem statement, all variables and given/known data
Two point charges q and 4 q are distance L apart and free to move. A third charge is placed so that the entire three-charge system is in static equilibrium. What is the magnitude of the third charge?

2. Relevant equations
F = kq1q2 / r^2
k ~= 8.988x10^9

3. The attempt at a solution
Not a clue. All I've figured out is that for this to work in perfect equilibrium, the third charge must be directly in between the first two.

Not looking for an all-out solution from anyone (Although that'd be cool too), but more looking for just a push in the right direction.

In theory, it'd involve drawing free body diagrams for the three charges, and then working with the charge q3 to balance the equilibrium... but I can't grasp my head around that.

2. Jan 13, 2008

### mjsd

yes! just need to go ahread and write down some simultaneous equations and solve them... obviously the variables will be the charge for the unknown/3rd particle, and its distances from each one.

3. Jan 13, 2008

### Goldenwind

Particle 3:
Net Force = 0 = F1 + F2
-F1 = F2

-kq1q3/(r13)^2 = kq2q3/(r23)^2
The problem here is that I don't know the distance between 1 and 3, or between 2 and 3.
Also, the q3 cancels out on each side. /boggle

4. Jan 13, 2008

### mjsd

also need equations for particle 1 and 2, all particles must have net force add to zero, since particle 3 is in between the two, the the two distances (13, 23) should add to L which gives u another equation... I haven't solved this, but that's how you go about it.

5. Jan 13, 2008

### Goldenwind

Particle 1:
Net Force = 0 = F2 + F3
-F2 = F3

-kq1q2 / L^2 = kq1q3 / (r13)^2
-4q / L^2 = q3 / (r13)^2

Particle 2:
Net Force = 0 = F1 + F3
-F3 = F1 (Doing it the other way, since the force would be in the other direction)

-kq2q3 / (r23)^2 = kq1q2 / L^2
-q3 / (r23)^2 = q / L^2

L = r13 + r23
r13 = L - r23

-4q / L^2 = q3 / (L - r23)^2
Unsure where to go from here. Could simplify the above line a little, but then I'd still have two unknowns: q3 and r23.

Wait, I have an idea..
-Sqrt((q3*L^2) / -4q) + L = r23
Which means:
r13 = L - r23
r13 = Sqrt((q3*L^2) / -4q)

Then from the particle 3 formula... plugging in r13 and r23...
-q3/(r13)^2 = 4q3/(r23)^2
-q3/(Sqrt((q3*L^2) / -4q))^2 = 4q3/(-Sqrt((q3*L^2) / -4q) + L)^2
-1/(Sqrt((q3*L^2) / -4q))^2 = 4/(-Sqrt((q3*L^2) / -4q) + L)^2
-((q3*L^2) / -4q) = ((-Sqrt((q3*L^2) / -4q) + L)^2) / 4
q3*L^2/q = ((-Sqrt((q3*L^2) / -4q) + L)^2)

Just checking, am I on the right path? Computing the right side is going to be a HUGE mess, I don't want to do it if I'm already doing things wrong.

6. Jan 13, 2008

### mjsd

I haven't checked the last bit but once you have got those equations it is just a matter of solving them.... and method of substition seems to be the way to go.. so you are on right track

7. Jan 13, 2008

### Goldenwind

Sigh... here goes nothing.

q3*L^2/q = (-Sqrt((q3*L^2) / -4q) + L)^2
q3*L^2/q = 2L^2 - q3*L^2/4q3
q3*L^2/q = 2L^2 - L^2/4
q3/q = 2 - 1/4
q3 = 2q - 1/4q
q3 = 8/4q - 1/4q
q3 = 7/4q
q3 = 1.75q

Trying this result...
Incorrect.

8. Jan 13, 2008

### Goldenwind

I just wanna point out that I'm almost completely certain that the mistake is mine somewhere along the line. Just saying that the answer I came up with happened to be incorrect, not saying that your advice is off or that. I do appreciate you helping me at this ungodly hour.

9. Jan 13, 2008

### mda

I can't see where you've gone wrong, but it is.
It may be easier to first solve for L concentrating on the added particle.

10. Jan 14, 2008

### mjsd

ungodly hour? :rofl:

Couldn't you see my Avatar?