11.3 Voltage

Ok, today we're gonna continue with voltage and look at a new formula and do some questions on how we can put these formulas together to solve a bunch of different types of questions. So first off, remember from yesterday we were talking about if we have a charged plate - one that's positive, one that's negative - we're going to get charges to move through that in a uniform electric field. So the idea then is to remember the key thing is to remember that whatever the charges - so if we had a positive charge inside that, right now it has high potential energy because it's going to be repelled from the positive and attracted to the negative. So it's going to move through those gaps from high potential down to low potential. Remember, the electric potential difference between those plates is what we call voltage. Remember, our voltage then was how much energy per Coulomb of charge is gained, whether it's kinetic. So in this case, the proton or positive charge will move downwards, so it's gonna be gaining kinetic energy as it goes. Or we could do the reverse and push that positive against the electric field, and that would be how much potential energy we're going to gain as we go the opposite direction. And the same thing would happen with a negative charge. If we put a negatively charged object in there, right, the negative is naturally going to go upwards, so that would be gaining voltage or gaining potential energy or kinetic energy, sorry. And if we had it the other way around, we push it against the electric field, we'd be gaining potential. So it doesn't really matter which type of energy we're talking about, the voltage tells you the same amount. It's gonna be how much energy per Coulomb, just depending on which direction you're going and whether it's positive, negative, and so on. So what we're gonna look at today is how can we get any formula? Because we're gonna be gaining energy. Remember, work is Force times distance, but we also know that work is change in energy. And if we're going, let's see, let's stick with potential. So if we were pushing this charge against the electric field, that change in energy then would equal the potential energy it gains, which would be equal to force times distance. I'm going to just add the change in energy on here, so our change in energy will be EP equals force times distance. And it doesn't matter what type of energy we use, potential or kinetic, it's the same answer anyway. But we also know that from before, we know that voltage equals change in energy over Q. So if we combine these two formulas, we can actually substitute the force times distance in for Q. Alright, for energy. So voltage equals force times distance over Q. But we also know that electric field equals force over Q as well. So you can see that within this new voltage formula, we have force over Q, we're dealing with electric force in this case. If there was two of the same things. So if we combine those two together, we actually get a new formula that is voltage equals electric field times distance. And your formula sheet changes it around a little bit, they write it as electric field as delta V over delta d. We don't care about positive or negative direction, we just worry about the charges, and we write it as change in voltage over change in distance. So our voltage, remember, is in joules per Coulomb or whatever we want to call it. But let's leave it in volts. And our distance would be the distance between the plates, between the locations that the charge is moving. So that'll be in meters. So you can see that we actually get our electric field can be in volts per meter or what we had before with this first one that we had Newton's per Coulomb. Okay, so that's the key thing to sort of remember is we've got two different formulas now for electric field. We can use force over charge, which will give us Newton's per Coulomb. Or now we can use voltage over distance, which gives us volts per meter. Okay, so let's do a couple examples. The first one is we got parallel plates sixteen millimeters apart and we want to get a 800 N/C (Newtons per Coulomb) electric field. So, the question is asking what is our voltage. So, if we just take this new formula, go electric field times distance and plug it in. So, we'd have 800 times 16 millimeters (16 x 10⁻³) and that'll give us the voltage of twelve point eight. Thank so it's pretty easy to work with the next part of the question now is how much work is done if we move an alpha particle from the positive plate to the negative plate. Okay, so we want to calculate the work or the energy. So, we're looking for the change in energy, know our voltage and we know how much charge enough a particle is so we want to basically use the old formula now. So, we know that voltage equals energy over charge. So, we can solve it for energy like going bolted x charge. So, we have twelve point eight and a alpha particle, remember, is two times our base charge. So, 2.2 x 10⁻¹⁹. Multiply those out, we would get 4.1 and 0 x 10⁻¹⁸ joules. Or if we wanted to have to leave it in electron volts, remember, we would just go twelve point eight times 2, right, 2 to the value of 2 electrons or two base units. The top point eight times two would give us twenty five point six electron volts. Those two answers for the same, just depending on which unit we want to use. The last part of this question is, is that energy gained is that 4.1 x 10⁻¹⁸, is that in kinetic or potential energy. So, we just got to kind of think about it if we have an alpha particle and it says it's moving from the positive plate to the negative plate, it's alpha particles positively charged. So, that's going to be the natural movement, so it should be kinetic. All right, we're decreasing potential increasing in kinetics, no, it would be kinetic energy that's getting gained. One more example, so this one had said they have enough particle moving between two parallel plates and they give you the potential, their friends to give us a voltage with the question is, what is our force. So same type of thing, we're gonna use our two formulas again. So we're going to use our electric field is voltage over distance first. So, you have 130 volts over 2.7 x 10⁻² and that gives us an electric field of 4815 V/m (volts per meter) or N/C (Newtons per Coulomb), however we want to rate it. And now the second part of the question is, what's our force. So we'd use our little formula electric field equals force over charge and that's electric force obviously. So, we would have electric field times two. Just take our answer we just figured out 4815 and times it by the charge of an alpha particle and that will get less than a force of 1.5 × 10^(-15). Okay, so pretty basic questions. We're just using the two formulas: the new voltage formula we did yesterday and then F = ma from before. So we can do lots of questions with these formulas. But remember, with the new formulas, we can also bring back old stuff that we learned in physics 20. Then we can do a variety of different types of questions, and that's really the only challenge. This stuff is sort of figuring out which way you're gonna go, which formulas you can use, and so on. With the new stuff, we can still use F = ma, but in this case, our force is dealing with electric force. Because we have acceleration, we can still use any of our old acceleration formulas. We can use conservation of energy stuff that we did in physics 20, all kinds of things. So all we have to basically do is think of this in terms of an object moving or something that has kinetic energy, potential energy, acceleration, velocities, all those kinds of things. Keep in mind that most of these questions are going to involve electrons, protons, alpha particles. So our charges are going to be pretty small, 1.6 x 10^(-19) or 3.2 x 10^(-19) if it's an alpha particle. And our mass of an electron is pretty small too. All of that data is on your formula sheet, so you don't need to memorize it. But as you use it over and over again, you're going to easily remember these numbers. Okay, so we're gonna do another example, and this time I'm going to show you two different ways or approaches you could solve it. The question is: we have an electron moving between parallel plates. So in this case, if we have an electron that's inside the parallel plates, and the way I drew it, obviously, it's gonna move upwards towards the positive plate. So we have a distance of 4 centimeters, and we know the voltage is 20 volts. We have everything that we need. The question is asking: what is its velocity when it hits the other side of the plate? It's starting from rest, so it's looking for the final velocity. First thing I jotted down here, one way you could solve this, is to first find the electric field. Electric field is voltage over distance, so that gives us 20 volts over 4 centimeters. That gives us an electric field of 500 V/m (volts per meter). Now, the next thing we could do is use our electric field formula to find the force. This electric field is force over charge (F = EQ), so our force would be our electric field times the charge of an electron. That gives us an answer of 8 × 10^(-17) Newtons. Okay, now we have the force. The next question is, if we have force, we must have acceleration. We could use F = ma, so acceleration would be our force answer that we just calculated divided by the mass of an electron. That gives us a fairly large acceleration, 8.78 × 10^(13) m/s^(2). So pretty fast acceleration, but because it's a particle or an electron moving over a relatively short distance, it should technically have a pretty good acceleration. Now, the last step is, now that we know the acceleration, you could be looking for our final velocity. Just like we would do in physics 20, our final velocity is what we're looking for. In this case, we know our initial velocity was zero. We just calculated our acceleration, and... We also know that the distance between the plates was 4 centimeters. So, now we can just pick one of our old acceleration formulas and solve it that way. The one I'm going to use is VF squared equals VI squared plus 2ad. Plug in all the numbers, and I won't bother playing it in, but we should get a VF of 2.7 times 10 to the 6. Thanks. You can see four different formulas here. We're combining electric formulas with our old acceleration and velocity stuff. And we do get a final velocity. The main difference between what we did in the physical NT now you can see the numbers are huge. We're talking billions of meters per second as that electron shoots across a gallon, which is always going to happen. Or in terms of electricity, the forces are stronger. That means our speeds are going to be a lot stronger as well. Netha - I just want to show you here maybe a quicker or easier way for you to understand this. And I want to just show you that you do get the exact same thing. So if we're method two, I'll do it over on the side here. The first thing we want to do is calculate our voltage, which we can do, right? We know our - sorry, we want to use the voltage to calculate energy. So we're going to use our voltage formula, which was energy over Q. So that means our energy would be voltage times Q. So our voltage was 2.20. Charge of an electron is 1.6 times 10 to the -19. And therefore, we get an energy of 3.2 times 10 to the -18 joules. Okay. And now that we know the energy, we know that that's going to be speeding up. So that energy is going to go from potential, that's our total potential, it's going to get converted to kinetic. So we can just simply use our energy formula from physics 20 for kinetic energy. You can just use energy equals one-half MV squared. So we can plug in our 3.2 times 10 to the -18 equals one-half times mass of an electron times V squared, and then solve and rearrange it and solve it for V, and you'll see you get exactly the same answer. Okay. So, either method, totally fine. The idea is just which one you prefer, which one do you think makes more sense to you, and it's just going to depend on your preference. So just like in physics 20, remember we'd always use the old acceleration formulas, but once we used conservation of energy, that worked as well. So the same sort of application still happens here. So Part B, the question now is, we just calculated the maximum velocity. So now the question is, what would happen if we only moved it one? So instead of going across the whole 4 centimeter plate, what happens if it's only gonna go one centimeter? The only thing we'd do differently is for the new method one where we used our VF squared formula. So the only thing that we do differently is we plug in a distance of 1 centimeter instead of 4 centimeters. VI was 0, our acceleration we calculated was 8.78, and our distance we put in as 1 centimeter instead of 4 centimeters. So we get a final velocity once you plug all that in and square root it, we get 1.3 times 10 to the 6. Or, we can kind of go back to what we did with the electric field stuff. So in this case, we'd have electric scales. So we want to find electric field, which is voltage over distance. If we solve it for voltage, we would get 500 times. One centimeter now, so that gives us 5 volts. So remember, in our formula, this is the change in voltage and change in distance. What that means is that to go only one centimeter, we've gone 5 volts out of the 20. Which makes sense, one centimeter is a quarter of 4, so our voltage should be 1/4 as well. Now that we've got our volts, we can figure out what the energy is by using the formula energy = voltage times charge (Q). So, 5 times our charge of an electron (8 times 10 to the power of -19) gives us 8 times 10 to the power of -18 joules. The last step is to take that energy and solve it for velocity. So, the kinetic energy will be equal to 1/2 MV squared. When we solve it for V, we get 1.3 times 10 to the power of 6 as well. You can see that it sort of just depends. The first one is a lot easier once we use the acceleration formulas. One step. This one we need about three point minutes to get. We do get the exact same answer either way. I also added a method 3 on here which I just called it ratio. The idea is just to keep in mind that if our original voltage is 20 volts and our original distance was 4 centimeters, then if our particle went halfway, it had a change in voltage of 10 volts. If it only went one centimeter, like we just did (and it doesn't matter where it is gonna be at the start; it could be at the ends), in this case, our change in voltage would be 5 volts. And then the final centimeter would also be 5 volts, and so on. So, you can always use ratios to help you figure out how much voltage it's gonna go through. For example, if our total was 100 volts and we had 5 centimeters, then it'd be 20 volts per centimeter, or however you want to break it up into proportions. You can always do your voltage like that. You can't do the velocity that way though because it's accelerating, so it's not gonna be equally proportioned. The energy would be right halfway, the energy we'd have, and so on. So, we could actually maybe do this Part B question a little bit easier. We went back to Part A and figured out what our energy was. So, we had an energy of 3.2 times 10 to the power of -18. That was for the full distance. We could just take that energy of 3.2 times 10 to the power of -18 and, you know, we're only going a quarter of the way, so we could divide it by four. That should give us the same answer that we had. Okay, so you could proportion the energy so we get the same answer. And then, once you've done that, you still have to do 1/2 MV squared to get velocity. We're going to do lots of practice questions in a book, so we'll spend a day or so working on these types of questions. You'll get the hang of when to use which formula and what all your different options are. And that's all!