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Work due to Electric Force

Patrick Ford
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Hey guys. So in this video we're gonna talk about the work that's done due to the electric force. We've already talked about electric potential energies and potentials, we're gonna see how all of that stuff ties in to the work that's done on a charge. So remember that whenever a charge moves, it's basically changing its position. So that means that relative to another charge, its potential energy is changing. Remember that's that's that's energy as a result of an objects position. And so the reason that these charges are moving is that the electric force or the field which is basically a force or that's something that gives information to charges to field forces is the thing that's accelerating and moving these charges. So remember whenever there is a change in energy, whether it's potential or kinetic some work is done. And we know that the equation for that work is equal to uh in most cases W equals F. D. Times the co sine of theta. We've seen this sort of relationship before. Now. What I want to just refresh your memory on energy conservation. We're working with conservative forces here. The electric force is a conservative force. So that just means that any loss in potential energy is going to be a change or positive in kinetic energy and also vice versa. So it means if I lose one jewel of potential energy, I've gained one jewel in terms of kinetic. And we know that from the work energy theory, we know that the work that's done on an object is the change in its kinetic energy. So one of the things that we can do is because these two things are equal to each other. Another way that we can write the work is it's just the negative change in the potential energy, right? Just because these two things are again the same and that just comes from this formula right here. But we actually have what the negative, what the change in potential energy is due to a charge that's in a potential. We know that the change in energy as a charge is moving through a potential difference is just Q. V. So one of the things we can do is pull all of this stuff together and actually write that. The work that's done is equal to negative Q times delta v. Right? So if we have Q delta v and that's equal to negative to the delta U. We have this negative sign right here. We just stick that down there. Okay, so that means that let's take a look at what happens for point charges. So four point charges. One of the interesting things that happens here is we actually can't use so point charges, we can't use this. W equals FD cosine. Theta. The reason we can't do that is because the electric force is a result or is dependent on the distance between these two things. So, for instance, if I have this charge and it moves to another location, let's say like this is let's say this is Q two and it moves to a different the electric force is not constant because it depends on this our distance. So this W. D. This FD cosine Theta actually doesn't work in the cases of point charges. So let's look at what the potential looks like for these two things because we know that the work done is going to be the charge, the feeling charge, which by the way, I'm just gonna label this is key and this is Q. The feeling charge as it moves through a potential difference, it gains some energy. That's how we're going to tie this back into work. So for instance, if I have a potential here V one and a potential here V two, then that means the potential difference between those locations is going to be V two minus V one. Now we know that from a point charge. So a point charge over here that the V. Is equal to K. Times big Q over R. The producing charge over here. So that means that the difference in potential is just gonna be K times Q. And we can do one over R two minus one over R one. Now, what happens is this negative sign right here actually serves to reverse the direction of these two things here. Sorry, not to reverse the direction. It can basically flip the signs right here. So this negative sign can actually make it sort of that they're flipped one over. R. One instead of one over R. Two. So basically we can tie all of this stuff together and say that do two point charges, the work that's done, it's gonna be K. Q. Times this little Q. Over here and now we have one over R one minus one over R. Two. This is the amount of work that is done from a point charge when you move a charge from one location to another. Okay, so this is again this is gonna be actually won over our initial minus one of our final. And the reason that it's not final minus initial has to do the fact that we actually sort of absorbed this negative sign that was originally out here and all that did was just flip the fractions. Okay, so there's another actually situation, there's another situation where you have to calculate the work and where we actually can use FD cosine Theta is in cases where we have a constant electric field. So when we have, when whenever we have a constant electric field, we know that the charge. So let me go ahead and scoot down a little bit. We know that a charge that is placed inside a uniform electric field is going to experience a force in this direction. We know that this electric force here is equal to the feeling charge times the magnitude of the electric field. But what happens is unlike in this situation where the electric force is constantly changing with distance. We know that this electric force is going to be constant. So in this situation this FD cosine Theta actually does work and we can use it. And so all we need to do is that if this is changing some distance D here, in order to find out what the work is done, we just need to take the cosine of the angle between the force and the distance. Remember that this angle here is always between these two and that means we can simplify this as Q E times D cosine of Theta. It simplifies down to that easier expression. Well, all you have to do is just realize that this Q E here is actually the force that is done by the electric field. Okay, so this actually does look like FD Cosine theta. So basically these are the two equations that I want you to remember, this is for point charges and this is for a constant electric field. But in any case in either one of those situations, the work that is done by the electric force depends only on the change in the distance and not the actual path that you take. What I mean by that is that if I were to move this charge from this location to this location or if I were to have moved it down here and then up here or if I were to move the, like in some squiggly path and eventually end up here. In all of the situations, the work is the same because it depends only on the final minus initial, It doesn't depend on the path that you take. So this is something that you might see in your textbooks referred to as path independence. Just means it doesn't matter how you get there. Just the fact that you got there in your finest final minus initial. Alright, the last thing I want you to remember is that as charges get very far away or sometimes infinitely far away, the electric potential energy and the potential itself approaches zero. And that's something we can actually see just from these equations. So K Q Q over R or K Q over R as r gets really, really big the potential and the electric potential energy goes to zero. And that's basically all we need to know. So let's go ahead and take a look at some problems right here, we've got a two nano Coolum charge is initially five millimeters away. So I've got this sort of before. So I'm gonna label a before and after because the to Coolum Coolum charge is eventually moved closer to this. So there's gonna be before and after kind of thing here. So let's go ahead and draw. So let's go, I've got this 10 nano Coolum charge and then I've got this to nana Cool um Charge initially they're separated by a distance of five millimeters here and then what happens is that after I've got this 10 nano column charge and now this to nana Coolum charge and now the distance is equal to three millimeters. So that means that as it's traveled through this distance here, this small little distance, there's actually been some work done and we're basically just trying to find out what is the work done by the electric force. Oh sorry, this actually should be, yeah. Yeah. So what is the work done by the electric force? Okay. So we know that we're dealing with two point charges. So I've got two point charges here. So that means that I'm gonna use the work formula that is for point charges. I can't use the FD cosine theta because F is constantly changing. We did say is that it's kQ Q times one over our initial minus one over our final. So this is like sort of like a shortcut formula that you can use, you know, have to re derive everything. You don't have to find the potential differences or anything like that. All right, you can just start out from this formula. The work that's done is going to be 8.99 times 10 to the ninth. Now we've got the charges involved, let's say Q. Is gonna be it doesn't really matter in this case what big Q. And little Q. R. But just to sort of be consistent, this Q is gonna be the 10 times 10 to the minus nine because these are nano Coolum charges that we're working with and we got two times 10 to the minus nine. Now we've got the distances involved. So I've got one over our initial, it's moving from five millimeters away. So that's 50.5 and then the final distance is 1.1 over 0.3. So that's gonna be my final distance. So now the work that's done is just going to be negative 2.4 times 10 to the minus five jewels. Now, I want to point out that the negative sign is actually really important here. What does this negative sign mean In terms of work? That means that you've actually done work on the system instead of by the system. One way you can think about this is that these two charges are actually positive charges, which means that they actually want to fly away from each other. So the fact that you've moved this charge closer to the other one that wants to push it away means that you actually have to do work against it. So, it's kind of like you think about like how you have a magnet, another magnet and you want to push them closer together. The closer you get, the more you have to push you have to put work on the system to keep it closer. All right, and let's go ahead and take a look at this other one right here, we've got this one micro cooling charge that's now placed in this uniform electric field. Okay? So basically we're going to deal with the other kind of work. So we're trying to figure out what's the work that's done on this charge. So let's go ahead and draw a little diagram right here. So I've got let me go ahead and write it. Let's do it over here. So we've got this electric field that's over here and it's gonna be constant and we're gonna place a charge inside of this. Let's go ahead and make that black. So we've got a one micro column charge and now it's going to just have a displacement of two m at an angle of 30 degrees below the horizontal. So that's gonna be that distance right here. So what is the work that's done? So now we have constant E. Fields. So that's one thing you're gonna need to figure out in your problems. Are you working with a constant electric field or you're working with just point charges? Okay. So we've got point part a the work that's done is just gonna be Q. E. D. Cosine of theta. Remember this data here is between the distance and the force right? Which is gonna be in the direction. Um Got it. So this is the cosine of the angle that we're gonna use. That the angle that we're gonna use. So the work that's done is Q. Which is the one micro column, one times 10 to the minus six. Now we've got the electric field which is 1000. Now we've got let's see um And we've got the distance involved which is two m and now we've got the cosine of the angle so it's cosine of 30. If you plug all of this stuff in, you should get at work of 3.09 times 10 to the -4 and that's in jewels. And that's actually a positive number. Which means that this electric forces actually doing positive work on this charge. That makes sense because it's moving along the same direction. So this is work that's done by the electric field. And now in part b part B were asked if this three g charge initially starts from rest, then how fast is it going after the two m displacement. So remember that we relate the work to the speed by using the work energy theorem, work is equal to the change in kinetic energy. So in other words, work is equal to one half M. Then we have the final squared minus V initial squared but one of the things that were told is that this thing starts from rest. So it means this is equal to zero. So that means that the work that's done is just equal to one half and we've got the mass which is 10.3 and then actually wait, hold on a second. So we've got the mass right here um one half M. V final squared. Sorry, we're actually looking for the velocity right here. So let's go ahead and just rearrange this formula. We're going to bring the one half over to the other side. We're gonna divide by the M. And we're gonna get that two times the work from the one half. Right? That's the one half when it goes over, divided by the mass is equal to the final squared. So I'm just gonna scoot up for a second and I've got let me actually just go ahead and remove myself. So I've got that V final is going to equal this square roots of two times the work. Which is that work? 3.9 times 10 to the minus four divided by the mass, which is 40.3. So that means the final velocity is going to be 0.45 m per second. All right. So that's basically how you use the work from the electric force in order to figure out these kinds of problems. All right, let me know if you guys have any questions. Let's go ahead and check out some examples and I'll see you guys in the next one