Originally posted by Cyto
no.. can't say that i do...
Okay.
Let's start there.
The first thing you need to know about electric fields is that they just add right on top of one another.
Imagine that you have one particle all by itself -- a proton, say. It creates an electric field all around itself that tends to pull negative particles toward it. You know how to calculate the force that would be experienced by a particle in that field -- it's just Coulomb's law.
Now, imagine that you add a second particle (of any arbitrary charge). What happens? Well, the field due to the first particle stays exactly the same -- it is completely unaffected by the new charge.
The new charge, however, exerts a field of its own. It may be in opposition, or in concert, with the field of the first.
Imagine that the fields created by both particles are like the transparencies you'd use on an overhead projector. You can put down a tranparency with the field due to the first particle drawn on it. Then you can stack another transparency with the field due to the second particle right on top. The total field is just the sum of the two individual fields. This is called
the principle of superposition.
So we've established that the total force on test particle (the third, movable charge in this problem) is divided up into two parts: some due to the first particle, and some due to the second.
As you know from playing tug-o-war, if you pull on an object from both directions with equal force, the object won't move. This is what has to happen in this problem -- the first particle has to pull just as hard as the second, in the opposite direction.
Let's assume that the x-axis is horizontal, and goes through both particles. Let's assume that the first charge, \inline{q_1 = 3.8 \cdot 10^{-6} C} is at the origin of this coordinate system, while the second charge, \inline{q_2 = -2.0 \cdot 10^{-6} C} is at the position \inline{x_2 = 0.20 m}.
What's the total force experienced by the third, test particle?
It's the sum of two copies of Coulomb's law:
F_\textrm{total} = \frac{k q_1 q_3}{r_1^2} + \frac{k q_2 q_3}{r_2^2} = 0
where \inline{r_1} is the distance between the first and third charges, and \inline{r_2} is the distance between the second and third charges.
All you need to do is solve for either \inline{r_1} or \inline{r_2}. Does this make sense?
- Warren
