Static Electricity Legacy Problem #27 Guided Solution

This is a very difficult problem. Like many difficult problems, it's both conceptually and mathematically difficult. And as I talk through the problem, one of the things I'm going to do is exhibit the habits of an effective problem solver. I'm going to read the problem carefully and get a mental picture of what's going on, trying my best to visualize the situation. I'm going to identify known and unknown quantities. And most of all, I'm going to plot a strategy and do good thinking about concepts, principles, and formulas before I ever pick up my calculator. In fact, there's a lot of algebra that's going to go on in this problem. I'm going to pick up my calculator again and calculate the final answer, but I'm not going to rush into my calculator by any means. There's much too much thinking going on in this problem that I'll have to do. And that's where I'm going to begin, after reading and talking about the concepts. So, you've got two charges, Q1 and Q2. They're labeled for you in the diagram. You've got the amount of charge stated as negative 4.5 nanocoulombs on object 1 and 9.1 nanocoulombs on object 2. It's charge 1 that's negative, charge 2 is positive. What you wish to do is calculate the location along the axes here that connects these two charges, the location where the net electric field is going to be 0 newtons per coulomb. Now, both objects being charged create their own individual electric field. There's got to be some location where the strength of these two electric fields are equal in magnitude and opposite in direction. So, we're going to begin by just talking conceptually about where that location would be where E1 is equal to E2 and opposite in direction. So, let's just begin with suppose it's in between the objects. I'm just kind of starting somewhere. So, if it were in between the object, what you'd have to have is there'd have to be an electric field from Q1 pointed opposite of the electric field by Q2. Electric field is by definition the direction that a positive test charge would be pushed by the object creating the electric field. So, if you were to put a positive test charge in between objects 1 and 2, it would be pushed to the left by object 2 and pulled to the left by object 1. These are not opposite directions. So, the location can't be in between Q1 and Q2. It either must be to the left of Q1 or to the right of Q2. So, how do you figure this out? Well, you have to begin to look at the quantity of charge. It's less for object 1 than for object 2. So, in order for the electric field magnitude to be equal, what object 1 lacks in terms of its charge, it must make up for in terms of its proximity to this location. And so, we have to make it closer to object 1 than to object 2. And so, the only way to satisfy this need of the location not being in between the two objects, but being either to the left of Q1 or to the right of Q2, and the magnitude being equal, is to have our location to the left of Q1. Now, this location, I'm just going to put a little dot down to the left of Q1 on my axes, and I'm going to call the distance from this dot to Q1, I'm going to call that x, where x is a coordinate position in units of centimeters. That's what I'm trying to find. Now, the distance from this little dot to Q2 is going to be 40 plus x, and that is in units of centimeters. And now, I'm going to begin by stating E1 equals E2, and I'm going to kind of expand that E1 using my formula for electric field. I could rewrite it as kQ1 divided by x squared. And for E2, I can do something similar. I'm going to say k times Q2 divided by the quantity 40 plus x, the whole quantity squared. Now, notice in my equation, now I have kQ1 over x squared is equal to kQ2 over 40 plus x, the quantity squared, and I've got a k on both sides, so I can cancel out my value of k. Doing so means that it doesn't matter that my distance units being used here for x and 40 plus x is units of centimeters, because I'm going to have centimeters in the denominator of my lot squared, and on the right side as well, centimeters squared, and they'll cancel as centimeters. And in my numerator, I can just use nanocoulombs, so it doesn't, I don't have to convert to coulombs, and that simplifies our problem. So, instead of substituting numbers in right now, I'm going to keep doing some algebra. You might be tempted to substitute numbers in and expand this 40 plus x, and you'd end up getting a quadratic, and you'd have to, you know, do some fancy math to solve for the quadratic. I'm not going to do that. Instead, my trick is to take the square root of each side, and when I do take the square root of each side, I end up with square root of Q1 over x is equal to square root of Q2 over 40 plus x. Now, I don't have any need to use a quadratic. So, I'm going to cross-multiply, and I end up with square root of Q1 times quantity 40 plus x is equal to square root of Q2 times quantity x. I'm going to expand, I'm going to distribute the square root of Q1 to the 40 and to the x, and then I'm going to group my x terms on the same side of the equation. Now, I'm left with square root of Q1 multiplied by 40 is equal to square root of Q2 times x minus square root of Q1 times x, and I'm going to factor out an x, and I end up with the right side becoming the quantity square root of Q2 minus Q1 times x. I divide each side of the equation by this, square root of Q2 minus square root of Q1. I'll end up with this, square root of Q1 divided by the quantity square root of Q2 minus Q1 is equal to x. Now, finally, I get to use my calculator, and I substitute into the equation 4.5 for Q1. I ignore the negative, and I substitute in 9.1 for Q2, and I solve for my value of x, and it comes out to be 95 centimeters. I put a negative in front to indicate to the left of Q1, so my answer here is negative 95 centimeters.