# Electric Charge equilibrium

1. Dec 18, 2008

### Air

1. The problem statement, all variables and given/known data
The figure below shows charged particle are fixed in place on an axis. In which situation is there a point to the left of the particles where an electron will be in equilibrium?

2. Relevant equations
$F=\frac{KQ_1Q_2}{r^2}$?

3. The attempt at a solution
The solution says 'In case of (a) and (b), the electron will be in equilibrium because both have equal magnitude of charges and have opposite polarity'. The problem is I don't see this. When I tried to do this mathematically, I didn't get $F=0$. Can someone explain how this works. Surely, the other charge $\pm 3q$ also plays a part and hence equilibrium will not be produced.

2. Dec 18, 2008

### buffordboy23

Your thinking right. The net force on the electron should equal zero if the electron is in equilibrium. Why not define your own lengths. Place the origin of your coordinate axes on the first particle to your left, and then let "a" be the distance between the two charges that are given and let x = -x' be the position of the electron at equilibrium. Then determine the individuals forces on the electron; set sum of the two forces equal to zero and solve for x'. When you found your answer, you can verify with Coulomb's law.

Last edited: Dec 18, 2008
3. Dec 18, 2008

### IsrTor

you don't need to get F=0. you need to make F=0. thats when the two forces will equal. Just from looking you get understand that in cases C and D you will not get equalibrium on the left side. In both cases A and B there will be. How do I know? well because you can assume the distance between q and 3q to be any value bigger then 0. Therefore, at some value of that distance bigger then 0, there must be an equalibrium somewhere to the left, where the two forces are equal. If you do the maths you will find a ratio between the distances to each charge(to charge q and to 3q). Then you will see it is possible. I hope that helps.