 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 meters 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-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-DA all squared. We want FA equals FB. So KQPQA divided by DA squared equals KQPQB divided by 8-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.