# Homework Help: Electrostatics - finding magnitude of a third charged particle

1. Feb 14, 2010

### Linus Pauling

1. Two point charges q and 4q are at x=0 and x=L, respectively, and free to move. A third charge is placed so that the entire three-charge system is in static equilibrium.
Two point charges q and 4q are at x=0 and x=L, respectively, and free to move. A third charge is placed so that the entire three-charge system is in static equilibrium.Two point

2. Coulomb's Law

3. I know the answer is 0.444q, and I know how to get it:

k*q*Q3/(0.333L)^2 = k*q*4q/(L^2)

But why is 1/3 *L used in the left side of the equality? If the magnitude of the charge on the far right is 4q and that on the left is q, why isn't the middle particle with charge Q3 1/4 as far from q1 as the 4q particle?

2. Feb 14, 2010

### kuruman

You may know the answer, but we don't know the question that the problem asks because you never posted it.

3. Feb 14, 2010

### Linus Pauling

Oops:

Two point charges q and 4q are at x=0 and x=L, respectively, 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?

4. Feb 14, 2010

### kuruman

What does this mean? I interpret it to mean that all three charges are free to move and that the net force on each charge due to the other charges is zero. This is impossible. To get the net force on q3 to be zero, you need to put it between the other two charges. Once you do that, the charges at the ends will experience net repulsive forces. Draw a force diagram and you will see what I mean.

5. Feb 14, 2010

### Linus Pauling

It simply means that you're dropping a third charge of unknown magnitude between two known charges, and the system goes to equilibrium, i.e. none of the particles is moving. To get the correct value for the magnitude of q3 I have to use 1/3*L, i.e. that q3 1/3 the distance from q1 that q1 is from q2. How do I obtain that 1/3*L value, though. My intuition would be that q3 is 1/4*L away from q1 since the difference in charge between q1 and q2 is 4q.

6. Feb 14, 2010

### kuruman

As I explained in the previous posting, this cannot happen if all charges are free to move.

7. Feb 14, 2010

### Linus Pauling

It's saying that the two charges q1 and q2 are free to move, then q3 is dropped in and equilibrium is reach, i.e. they're no longer moving. My question has to due with the distance from q1 and q2 that q3 is when the particles are in static equilibrium.

8. Feb 14, 2010

### kuruman

OK. Draw yourself a force diagram. If q and 4q are both positive, charge q3 must be negative. Place q3 at distance x from q. Can you write two equations saying that the net force on each of charges q and 4q is zero? You will then have two equations and two unknowns, x and q3.

9. Feb 14, 2010

### Linus Pauling

Ahhh, actually I got it already. The next question asked me to find the distance x q3 is located it, but it was much simpler to find x first ten the charge of q3. Thanks though.