 Part B has the same setup as part A, so we'll use the same diagram, except now we know the distance between the particles. We want to find the charge of the positive particle, so we want to find Qp. Let's pretend that there is a pin keeping particle B in place, so we can look at the forces acting on A first. There is a force pulling particle A towards the positive particle, and there is a force pushing particle A away from particle B. Now, for particle A to remain stationary, it must have a net force of 0 acting upon it. Fb pushes A to the left, while Fp pulls A to the right. So a net force of 0 is certainly possible. Now we want Fb to equal Fp. That way A experiences no net force and thus remains stationary. The distance between particle A and the positive particle is smaller than the distance between particle A and particle B. In other words, Da is smaller than dt. So for the forces A experiences from these particles individually to be the same, then particle B must have a bigger charge than the positive particle to compensate for the differences in distance. Using Coulomb's law again, we get these two equations. Equating them and cancelling terms, putting in the numbers, and then solving for Qp. We get a negative number, which is telling us that the charge the positive particle must have is negative. So it is in fact a negative particle. That's weird. However, we can make sense of this by remembering that force is a vector, and so it has direction. Fb points in the opposite direction to Fp. Since we want them to point in opposite directions, but be equally strong, we actually wanted Fb equals negative Fp. If we carry this through the working out, we'll find that the charge the positive particle has is in fact positive 0.4 Coulomb's. Remember that due to the difference in distance, the positive particle must have a smaller charge than the negative particle. 0.4 Coulomb's is smaller than 3 Coulomb's. So this condition is satisfied, and our answer is plausible. Now, if we run through this whole process again with particle A fixed and try to balance the forces on particle B, we'll actually get the same answer for the charge of the positive particle. It isn't immediately obvious why this should be the case. But if we dig a bit deeper, we can find out why it must be so. So let's look at our system again. Remember how in part A we change the distance such that the positive particle experiences the same amount of force from both A and B. The same amount of force pulls particle A towards the positive particle, as discussed in previous videos. This is the same for the interaction between the positive particle and particle B. Particle B exerts a force on particle A, pushing it to the left. Let's call this force FC. Particle A also exerts the same amount of force on particle B. In part B of the question, we were trying to find the value QP must take such that FA equals FC. This gives us a net force of zero acting on particle A. But at the same time, it also gives a net force of zero acting on particle B. So for part B, the charge that the positive particle has, which would keep particle A stationary, would also keep particle B stationary.