Two small spheres spaced 20.0 cm apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33 * 10^-21 N?

Video transcript

Welcome back everybody. We are taking a look at two equally charged miniature small steel bearings. We are told that they are positioned away from each other by 16 cm of distance or .16 m. We are told that they have equal charges. So there is a repulsive force between them Of five times 10 to the negative 19th newtons. And we are tasked with finding the number of extra electrons on each ball. We have a formula for this that states that the magnitude of charge of one of the balls just equal to the number of extra electrons times the charge of an electron. I'm gonna go ahead and divide both sides by E. And then these terms cancel out. And we get that the end that we are looking for is just equal to the magnitude of charge over E. But what is the magnitude of charge for one of the balls? What we're going to have to do is we're going to have to use columns. Law states that the repulsive force is equal to a constant times the charge of the first ball, times the charge of the second ball all over the distance between them squared. This simplifies down to K. Q squared over R squared. So we need to isolate now this Q term. And here's how I'm going to do it. I'm first going to multiply both sides by R squared over K. R squared over K. And you're gonna see that these terms cancel out here on the right hand side. So we are left with U. Squared is equal to F. R squared over K. Now to isolate Q. I'm going to take the square root of both sides. And we get that Q is equal to the square root of F R squared over K. Now that we have our formula for Q. Let's go ahead and plug in our numbers and find the magnitude of charge for one of the balls we have that our force is five times 10 to the negative 19th. Newtons times are distance between them of 100.16 m squared, all divided by columns constant of nine times 10 to the ninth. Which when we plug into our calculator, we get 1.19 times 10 to the negative 15th columns. Now we are ready to go ahead and find the number of extra electrons. This is given by Q over E. So we have 1.19 times 10 to the negative 15th columns on top. The charge of a one electron particle is going to be 1.6 or the magnitude of charge of one electron particle, going to be 1.6 times 10 to the negative 19th columns. And this gives us that there are extra electrons on each steel ball corresponding to be. Thank you all so much for watching. Hope this video helped. We will see you all in the next one