 In this worked example, we're going to look at balancing forces between charged particles. So we have a particle A with the charge of negative one coulombs, and to its right, there is a particle B with the charge of negative three coulombs, eight metres away. Looking at part A, we want to find out where we can place a positive particle such that it will remain still. Newton's first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. So if we want a positive particle to be at rest and remain at rest, then we need a net force of zero to be acting upon it. We know that positive particles are attracted to negative ones, and so if we place the positive particle to the left of A, it will be pulled to the right by A and B. Thus, it will experience a net force to the right, not zero. Same if we place the positive particle to the right of B. The particle will experience a net force to the left. If we place the positive particle above A and B, then it will be pulled down. Same for placing it below A and B. It would be pulled up. But if the positive particle is placed between A and B, it would be pulled to the left by A, let's call this force FA, and to the right by B, FB. If we can make these two forces equal, then the positive particle will experience a net force of zero and thus be stationary. We know that B has a bigger negative charge than A, which means that it will pull the positive particle more strongly. However, we also know that distance affects attraction. So if we place the positive particle closer to A and further away from B, then this can compensate for the fact that B has a larger charge than A. So we expect the positive particle to be closer to particle A than particle B in our answer. If that is not the case, then we have done something wrong. So let's call the distance between the positive particle and particle A, DA, which means the distance between the positive particle and particle B will be 8 minus DA. Now we know that the force charged particles experienced due to each other is given by Coulomb's law. Let the positive charge have a charge of QP, P for positive. So FA equals K times QP times QA divided by DA squared. And FB equals K times QP times QB divided by 8 minus DA all squared. We want FA equals FB. So KQPQA divided by DA squared equals KQPQB divided by 8 minus DA squared. Now we can try to solve for DA and find where the positive charge must be located to balance the two forces. Well, the case and the QPs cancel, so we're left with this expression. Substituting in the right numbers. And now we just need to solve for DA. Using the quadratic formula, you get two answers for DA. 2.93 meters and negative 10.93 meters. Negative 10.93 meters is located to the left of particle A, which we have already previously eliminated and said was impossible. So the correct answer is 2.93 meters. Therefore the positive charge must be located 2.93 meters to the right of particle A for it to remain stationary. 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 zero acting upon it. FB pushes A to the left, while FP pulls A to the right. So a net force of zero 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 changed 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 0 acting on particle A. But at the same time, it also gives a net force of 0 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.