 Hello, I'd like to talk to you about a Milikin-like experiment. Why is it Milikin-like? Well, because we're going to do a really simplified version of the calculation Milikin did. Where we ignore pesky things like air resistance or viscosity and things like that and just get to the force analysis of this oil drop. And then after we do that calculation, we're going to do another one because I think two is always better than one. Let's get of it a try. First of all, what does it look like? What does this experiment really do? Here's a good animation from the folks over at King's University College here in Edmonton that shows you the basics. Well, you've got yourself a oil drop in a container. It is allowed to go through this little hole in this top plate. There's a bottom plate and they're both connected to a power source. When the ball of oil is dropped, it does what you'd expect it to do. It speeds up because there's an unbalanced force acting on it. Not very interesting until you apply some voltage. So if you put a potential difference across this gap, you can see an electric field is created. This is a negatively charged oil drop. So the electric field going down will apply a force to it going up and now as that particle enters the field, it slows down. This one might even come to rest and then eventually turn around and go back up. Now what Milikin did was not maybe necessarily try to make them accelerate upwards. Although he certainly could have and watch it oscillate back and forth. What we're going to do is we're going to adjust that field until we get a force which is balanced by the force of gravity. Now how are we going to know the force is balanced while those two arrows are going to look the same? The oil drop is going to drop at a uniform velocity and we can see from our graph over here that the velocity time graph is going to be horizontal and you can play around with it until you get it horizontal. At that point the forces will be equal and you can record the voltage to find the charge in the oil drop. This is how Milikin went through and eventually calculated the charge on an electron using a few other techniques that we're not going to look at here today. So what we want to do is figure out how does he know the charge in the oil drop just based on the voltage? And the other pieces of info we need here are the mass the oil drop which he worked out using some crazy math who won't get into and how far apart the plates are. So how do we do that? Here's an example of a question while I'll show you. In an inversion of the experiment we got the mass the oil drop got the potential difference we got the separation uniform velocity, which means that these forces are balanced. So determine the charge in the drop first thing I want to do just make a little force free body diagram. The two forces are balanced. They're equal. So the net force acting on this oil drop is zero because there is uniform motion. It's balanced which means you can kind of just jump straight to the chase of saying the electric force is equal to the force of gravity. Or if you like you can do a little bit more work and make it a little bit more you know correct I guess and say the net force acting on the oil drop which is made up of the force of electricity and the force of gravity is zero which means you can rearrange this statement to give the negative force of gravity maybe equals the force of electricity. That's why they're equal when you look at it from a force point of view. It doesn't matter how you get to this really as long as you start off by setting those two forces equal to one another. Okay so now I think I will make myself some space and actually sub some numbers in and see what we get. Fine be that way I'll move everybody over there. There we go so we're going to go through now and do our calculation. Our force of gravity is going to be expressed using mass times acceleration due to gravity and our force of electricity charge times the electric field. There's a problem though we don't know the electric field but luckily we know that this is a parallel plate capacitor so the electric field which I'll do in green is equivalent to the separation between the plates or the voltage divided by the separation between the plates so we know that number 7.2 times 10 to the 4 volts is our potential difference and our separation 52 centimeters 0.52 meters. So our separation is get the calculator out here or rather our electric field is 138,461 let's write that down 138,461 Newtons per Coulomb. All right now we can do some substitution. Mass the well drop was given in the question 7.5 times 10 to the negative 15 kilograms. The acceleration due to gravity is 9.81 meters per second squared. Technically that's negative because the force of gravity is going down but remember also technically there's a negative sign here outside of the brackets so the two negatives become positive and if you don't want to put the negative in it's not a huge deal on this problem. The next one it is going to be so you know just bear that in mind 138,461. From here it's just a bit of math to plug and chug through on your calculator and so let us do that and we're getting for a charge about 5.3 times 10 to the negative 19 Coulombs. That seems pretty realistic it's about three and a bit electrons. The bit comes from the fact that you know we don't have a whole lot of precision in some of these measurements so there's some some error coming in. Now let's look at the second calculation similar but different. In this calculation the oil drop same mass same potential difference falls with an acceleration of 1.2 meters per second squared. You know I should have put a negative sign in there saying if it's falling just remind you that's going down. Well what does the free body diagram for this look like? Well there's still a force of gravity going down that hasn't changed any because the mass of the drop hasn't changed but the electric force pulling up on it isn't quite as big because now there's an overall force there is a net force acting on the particle and that net force is still equal to the sum of the force of electricity and gravity acting on the oil drop. I can express the net force any force really as the mass of the object times the acceleration of the object. We already know the formula for electric field we already know the formula for force of gravity. Now for the mass of this particle we're given that earlier I think it was 7.5 times 10 to the negative 15 it was and here comes the part that's going to throw some students off. Your acceleration has to be negative because the acceleration and the force are pointing downwards so that's where the negative sign came from. Lots of students will forget that and then we'll end up getting the wrong value. Our electric field is 138,461 newtons per coulomb and then our force of gravity. Do you notice how I put in a negative value on the 9181 meters per second squared as well? That's because the force of gravity is also a negative number. That free body diagram helps you remember that fact that those two numbers need to be negative and now it's just plugging and jugging from here so let's give it a shot see if we can type it into our calculators. All right so there's my left hand side now I'm going to add this term it's a negative term when I multiply those two numbers I'm going to add it to move it to the other side all right that is the electric force that I have now if I divide it by my electric field I'm going to get my charge. There we go it's about 4.7 times 10 to the negative 19 coulombs all right so just just a little under four electrons so there you have it two milikon problems if you've got problems with milikons problems check out my website www.ldindustries.ca there's more information about it there