Relationships Between Force, Field, Energy, Potential

Alright, guys. So we've seen a lot of stuff so far. We've seen electric forces seen electric fields, potential energies and potentials. And now we're basically gonna put it all together in an awesome formula sheet. That's gonna be really helpful for your tests. Let's go ahead and check it out. So so far, we've seen these four related things and often times it's really hard to keep track of all of them. Which ones? Which, Which one involves what variables. So this table is gonna be really awesome at putting all of these things in perspective together, Let's go ahead and take a look at it. So we've got our squares. We've got some equations involve R squared, some that involves single ours, some that involved multiple charges, and some of that involves single charges. So, basically, let's take it from the bottom. We know that this electric field here is a force field, and the best way that we can think about this is that it is a charge that sets up a field that another charge will feel and experience a force from now. I just want to point out really quickly that I know I'm using little cues here. What? I've used big queues, and I know that's potentially might be confusing. But the thing is, is that it won't matter. You just have to be sure that on tests and homework, you understand which charge is producing the field and which charge is feeling that force and feeling that electric force, because the reality is that some of your professors might be use. Q. One Q. Two. Some of your oppressors might use big queues and little cues. It's up to you to decide and figure out which one is the producing charge and which ones the feeling charge. In any case, we know that a charge that it produces a field that is e has a resulting force that on a feeling charge, we know that the relationship between these two formulas is f equals. Q Times E. So if one charge produces an electric field and another charge, which is this little Q right here feels that electric field it's gonna have a force now. Similarly, we know that a charge will also produce something called a potential, and this potential is basically just an energy field. Instead of telling other charges, how much force to feel. It's basically telling other charges. How much energy toe have We know that this relationship between these two formulas is given as U equals Q times V. Now what we haven't seen yet. So we haven't seen the relationships between these two formulas f and you. So basically the negative of the potential energy difference is gonna be f times Delta are. But most of the time you won't see this as Delta are. Ah, lot of times you'll see this as Delta X instead. But I wanted to. I just want to point out that I'm using this Delta are because it sort of helps reinforce the relationship between these ours right here. But ultimately these air distance variables, Sometimes you might see them as Delta, ours, Delta X's Delta DS or something like that. Thes air basically just distances okay, and we have a similar relationship between the electric field and the electric potential, and that's that. The negative of the potential difference right here it's just gonna be equal to we've got E times Delta are, or sometimes most of the time you'll see this as Delta X again, just reinforcing the fact that it's just a distance variable. Okay, now, with last thing we want to do is point out that these two quantities, these deltas actually have special names. Remember that this Delta V is defined as the voltage. It's the potential difference between these two points. So that's Delta V right here. That's the voltage and this delta you over here, the negative and the change of the potential energy difference is known as the work. So this is actually known as work. And this basically relates all of the things that we've seen so far work, potential energy, electric force, electric field, all of those things. All right, so basically, that's it for this video. Go ahead and print out this page. Take it to your exam and you'll be good to go. All right?

Alright, guys, we're gonna work this one out together, so we'll have the square of charges right here. They're not all equal charges, but we're supposed to do is find the potential at the center of this arrangement right here. So just right off the bat potential, which formula is that potential is the formula with one Q and one are so the words, the potential is K times the producing charge at some distance right here. And we're supposed to find out what the what is the total potential at this point right here. This is our point of interest. So basically, at the center of this arrangement right here, all we need to do is figure out what the distances is. The distances are to each one of these charges that are on the corners. So one of things we could do is sort of break this up into a little triangle, like a little mini triangle. And basically, we've cut this thing in half. So we know that this distance, which is half, is gonna be 2.5 millimeters. This is also gonna be 2.5 millimeters. So that means we can actually figure out by using the high pot news in the Pythagorean theorem that this is basically 3.54 millimeters. Right? And that's just something we get for the Pythagorean theorem. So we know that this our distance right here, which is 3.54 millimeters, is going to be 3.54 times 10 to the minus three. That's gonna be in meters. So basically, what the idea is is that all of these charges are gonna be producing potentials at the center. And these charges, these potentials that they produce are gonna be scaler, which means that we can just add them up together. So the total potential So that's one thing that you could possibly right is just v. Total is gonna be v the potential from the two nano Coolum charge, plus the potential from the 1.5 Nano cool in charge of the negative 1.5 and then so on and so forth, right? You could basically just out of all of those potentials together. So that means that the potentials here v total um, you would get by using cake you over our and the only thing that's changing is what? Just depending on the producing charge that you're putting inside of that equation. But if you add all these things up together, what ends up happening is that this K and this are end up being greatest common factors. So one of the shortcuts that you can use is say that this is gonna be Cavor are and then basically, just add up all of the charges that contribute a potential Q one Q two Q three and then so on and so forth. Right, Because basically, just you could just add those things up together. You don't have to keep sticking k over our into your calculator over and over. So what we can do is that the total potential is gonna be 99 times 10 to the ninth. Now, we've got the distance between all of these charges, and center is always going to be 3.54 times 10 to the minus three. And now we just add the charges up together individually. And it doesn't matter which order we add them in because Edition is just community. It doesn't matter which order you do it in. So that means I could do to nano columns minus three Nano columns. Let me see. Minus three narrow columns and then plus one nano Cool. Um and then we've got minus 1.5 narrow columns. Let me scoot down. So basically, what happens here is if you close off this parentheses, this to the negative three and the plus one are all gonna cancel out to zero on. All you have to do is just do one multiplication here 8.9 times 10 to the ninth, this K divided by r and then times this negative 1.5 nano columns, which is 10 of the negative nine, by the way. So don't forget to stick this in as 10 of minus nine, and you should get that the potential is equal to negative 3.81 times, 10 to the third. And that's gonna be in volts. So it's negative, by the way. And I just want to point out the fact that you could Onley do the shortcut here because all of these distances that we calculated to the charges are gonna be the same. So a lot of times, this is one way that symmetry can work out on here advantage. So if you know how to cancel these things out and take these shortcuts, then it's gonna be a great tool for you for you to use. Let me know if you guys have any questions in the comments, and I'll see you the next one.

Alright, guys, in this example, we're upping the ante. Now we're gonna figure out what the potential difference is not just a one charge, but due to two charges. So we've got this five Nano Coolum charge and a negative Nano three Coolum charge on the same line. So I'm just gonna draw a quick little sketch right here. I've got this five Nano Coolum charge Negative three Nano Coolum charge. And I know that they're on the same line and that distance is six millimeters now, in part A. What I'm asked for is a mask for the potential directly between them. So in other words, this is gonna be the potential at V A. So let me do a better job. It's sort of highlighting that. So that is gonna be the potential at V A. All right, So let's go ahead and do work that out. We've got the potential at V. A. Is basically going to be the potential due to both of these charges at this location. But one of the things we know is that if this is half of the distance between these two charges, then that means that this distance right here this three millimeters is actually going to be the same for both of them. So in other words, we've got the potential at point A is gonna be the potential due to the five Nano Coolum charge, plus the potential due to the negative three Nano column charge. We're just gonna add those things up together because their scale er's so we we have that the potential difference over here is just gonna be K Times Q one over R A plus K Times Q two over are a but one of things we notice is that we have the same exact case and raise for both of these things. So, in other words, since both of these are the same, we can actually use a little shortcut that we've used before in reducing these potentials down to, like even simpler forms we have. The potential is basically just equal to K. Over are a times, and then we're just gonna add the charges up together, and it's only because we have symmetry. It's only because thes things are the same distances that we can actually do. You use this rule so, in other words, that the potential appoint a is going to be 8.99 times 10 to the ninth, and then this is gonna be divided by the distance. We have to be careful because this is three millimeters, which means it's 0.3 and then we're just gonna add up the charges. We have five Nano columns minus the three Nano columns. So that means that the results here so both of these things together are just gonna add up to two Nano Coolum charges, right? So you're just gonna add in both of these charges and you get to nano columns, which, by the way, can be represented by two times 10 to the negative nine. That's gonna be in columns, right? So that means the potential at point A is just going to be, Let's see, I got six times 10 to the third. That's in volts. Alright. So we can use that shortcut for this case right here. So in part B, now in part B, we're supposed to figure out what is the potential at this point B, which is now halfway between the charges. But now it's, um, extra distance above that line. So, in other words, point B is somewhere over here we're supposed to be figuring What is the potential at this point? So basically we need to Dio is now that we have a different point, we actually have to calculate what the distance is to both of these charges. Now, fortunately, we've got here, we've got a three and we know that this line right here is four millimeters above so we can recognize this is a 345 triangle. If you wanted to figure out this our distance, if you didn't know it was 345 you could always just use the Pythagorean theorem. But in any case, we've got five millimeters right here for this distance. And this R B is actually gonna be the same for both of these things, right? Because it's symmetrically placed around around this, uh, this access so you can kind of use the same shortcut that we did here, but now we're just gonna sort of like, do it a little bit quicker. So this VB is just gonna be k over RB and then times again, the addition of the two charges which we already know, is to nano columns. So in other words, just gonna be five nano colognes minus three Nanako loans. And if you write it all out, if we write it all that we get that the potential at point B is 8.99 times 10 to the ninth. And now we've got the distance, which is 0.5 And now we've got two times 10 to the negative nine. And if you work that out, you should get the potential at this point is equal to 3.6 times 10 to the third, and that's in volts. So that is part A and part B. And now the last part over here, which I'm gonna do Let's say over here is I need to figure out what the work that's done on a one nano Coolum charge. And let's see, we've got from the first point to the second point. In other words, we're going from point A over two point B, and we know that a work that's done on a feeling charge through a potential difference is negative. Q. And such the feeling charge times the potential difference at this point, and this is the potential from a to be So, in other words, what happens is the work that's done is equal to negative Q times. The difference in final minus initial potentials. VB minus v a. If it had said the second point to the first that we would actually reverse that, so it's very important that you do the right step here. All right, so we've got the work that's done. It's gonna be negative. We've got one times 10 to the minus nine. That's the feeling charge that we have this negative one Nano Coolum charge is this feeling charge over here? And then we've got the potential difference. So VB was equal to 3.6 times 10 of the third V A was six times 10 to the third. And if you work all that out on your calculator, by the way, you could have actually just figure out what the potential difference is just, like actually gotten the subtraction. And then you would just plug that in here. But I'm just doing it sort of like the expanded way. And if you work this all out, you should get a work that's done, which is equal to 2.4 on That's times 10 to the negative six, and that's in jewels. So that's the work done in moving in charge from that first point over to the second. All right, let me know if you guys have any questions with this and I'll see you guys the next one.

Hey, guys, Hopefully we're gonna take a look at this one on your own, maybe try it out. But this is gonna be a little bit different than the problems that we've seen before. We've got a 5 g three Mike Racoon charge. It has some initial speed and it's moving away from some negative charge. And then basically, we're supposed to figure out how far can this thing travel before it stops? So I want to figure out just sort of like, what's going on here? So let's go ahead and draw a quick diagram. Now, I've got this negative five micro Coolum charge and then I got another three Coolum or three. Michael, um, charge. I know the mass of this charges 5 g, and I know the initial speed of this is equal to 20 m per second. So what ends up happening is that as this positive charge is moving away from this negative charge, the force on it wants to pull it back towards the negative charge because thes two things are opposite charges. Opposites attract. Right now, I know what the initial distance is between these two objects so that our initial is equal to let's see what that's what is that? Five centimeters. That's five centimeters. So what ends up happening is that as this thing is flying out with some velocity, the force is eventually going to bring it to a stop at some later distance over here. Now, I know at this point be while it stops, the final velocity is going to be zero. And basically what I need to figure out is what is the horizontal distance or just horizontal? I'm guessing. What is this distance here? That it can travel before it actually stops at this point. So if I can kind of think about this, this will be the final distance R f So that means that this X distance, if I could write it as an equation, is basically just going to be. I've got X is equal to our final minus our initial. And so this really is my target variable. How far can it travel before stopping? I know what the initial distance is Now. I just need to figure out what the final distance is. So in all this cases, I've got this final distance initial distance. I've got speeds, masses, and I'm also talking about energy. So what I need to use is I actually need to use energy conservation in order to solve this problem. So what I'm gonna do is look at all of my energies in the before and after case. So we know that our energy conservation formula is the initial kinetic energy, any potential and any initial potential energy which we know is gonna be electric potential energy. Now, there's no work that's done by non conservative forces. We know that already and then that's getting equal to the final kinetic energy, plus the final electrical potential energy. Okay, so we've got the work done by non conservative forces is zero. The electric force is conservative, and we know that when this thing finally stops, is there anything we can cancel out? Well, let's see. Um, we've got an initial initial kinetic energy because we know that object discharge is moving, and we also have an electrical potential energy because we have two charges and they're separated by some distance. So there's always potential energy. Now what happens is when this thing stops at this moment, that's right out here. We know that the kinetic energy is equal to zero, but there still is some final potential energy. Okay, so basically, we can go ahead and expand out all of these terms. I know that this initial kinetic energy is gonna be one half MV not squared. Now, what happens is I've got to do the, uh, the initial Connecticut or sorry, the initial electrical potential energy, which is K Q one q two divided by our initial. And then that's gonna equal the final electric potential Energy K Q. One Q two over our final. So, in other words, I'm actually looking for this our final because then I can take this our final and plug it back into this formula and figure out what this X distance is that it travels before stopping. So that's basically the game. I'm gonna try to try to figure out what this our final is. But what happens is I can't go around and like start manipulating this by flipping these fractions and flipping, flipping these formulas because this is like, uh, this is in addition and I got a formula. That's the fraction in here. It's gonna get really, really ugly. Fortunately, what I can do, though, is I know what the masses. I know what the initial velocity is. I know what all of these constants are and including the distance. So rather than trying to algebraic Lee manipulated, just start plugging in numbers for this stuff, just reduce it down to, like, a simple number and then worry about the algebra later. So, basically, I'm just gonna plug in this really long mess right here. We've got 5 g, so that's gonna be 50.5 Now I've got the initial velocity is 20 and they were to square that Now you've got plus and we've got 8.99 times 10 to the ninth. Now, the two charges that are involved, we've got a three micro cool OEMs. So that's gonna be three times 10 to the minus six and a Let's see, negative five times. 10 to the minus six. And now you've got the distance, which is 60.0. Uh, that's 0.5 That's in centimeters. Right? So that's the 0.5 m, and then we've got these formulas right here. We know what these charges are. Okay, so basically, if you plug all of this stuff in that should equal what k Q one Q two over our final is. And basically, if you plug all of this stuff in really carefully into your calculator, you're just gonna get one big number or doesn't have to necessarily be big. But this is actually gonna be negative 1.7 jewels and that's gonna equal K Q one q two divided by our final. So now what we can dio is now we can actually manipulate with this. Our final is and basically it's just gonna happen. Is this our final goes up to the top, This negative 1.7 comes down to the bottom, and then we can just go ahead and plug everything and again. So that's why it's gonna be easier to do that than actually work through all the algebra in that big step over there. Okay, so you've got the final distance is gonna be 8. times 10 to the ninth. Now I've got the two charges which are, Let's see of God's three times 10 to the minus six negative five times 10 to the minus six. By the way, there's some shortcuts that you could take and manipulating some of this algebra, but, um, yeah, so we've got this right here, and then we've got this negative 1.7, okay? And now, if you plug all of this stuff in, you should actually get 0.793 But remember, this is gonna be in meters, so this actually represents is this actually represents 7. centimeters. So this either one of these expressions, if you got this, if you got this, would be correct. But basically, just this is just a different way of expressing this. Um, But what we can do is we can stick this number. I'm just gonna go ahead and stick with centimeters because we're dealing with centimeters and then go back and plug that into this formula over here. So now what? My ex distance is what? My real target variable. How far can this thing traveled before it stops is gonna be 7. centimeters minus the initial distance of five centimeters. So that just means that this thing travels to 0.93 centimeters before stopping. So this right here is actually our final final final answer that 2.93 centimeters Okay, Let me know if you guys have any questions. This energy conservation stuff can actually become really useful in solving some problems. Sometimes you'll have to do it, so make sure you're comfortable with this. Go ahead, Watch the video again. If you didn't understand everything, drop me a link or a comment or a question, anything like that, and I'll see you guys later.