Work From Electric Force - Video Tutorials & Practice Problems

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1

concept

Work due to Electric Force

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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

2

Problem

Problem

An electron moves from point A to point B. The potential difference between these two points is 100 V. What is

a. the point of higher potential? b. the work done on the electron? c. the final speed of the electron if its initial speed is zero?

A

(a) B (b) 1.6 × 10^{-17} J (c) 5.95 × 10^{6} m/s

B

(a) B (b) -1.6 × 10^{-17} J (c) 5.95 × 10^{6} m/s

C

(a) A (b) 1.6 × 10^{-17} J (c) 5.95 × 10^{6} m/s

D

(a) A (b) -1.6 × 10^{-17} J (c) 5.95 × 10^{6} m/s

3

example

Work to Bring Two Charges From Infinity

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4m

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Alright, guys. So let's work this one out together. This is actually a very common type of problem that you might see. It might pop up in homework or some tests or stuff like that. So we're basically asked to find how much work is done by the electric force in bringing a charge from infinitely far away to the origin of coordinate system. So let's go ahead and draw out with that coordinate system might look like. So it's gonna look like this. And basically, the idea is that you have a let's say, what is it? Five Cool in charge that's out here infinitely far away or very, very far away, and you're basically just gonna bring it towards the origin of this coordinate system. How much work does that actually take? What we're talking about? Basically like point charges. So let's go ahead and use our work formula for point charges. You know it's cake, one of her, our initial minus one of our final. The problem is, is that we actually don't have any other charge that's gonna interact with this five Coolum charge. It's basically just like bringing in something when it has nothing to interact with. So in other words, if there's no other charges, then there's gonna be no other forces that are acting on this charge. Which basically means that the work done is just gonna be zero. It doesn't take any work to move around a charge because there's nothing else to basically attract or repel it, which means there's nothing else around it to do work on it. Does that make sense? So basically, the if you just have one single charge and you're moving it around, it takes no amount of work. Now what does happen, though, is you have this second charge this not negative to Coolum charge that's floating away over here. And now, if you wanna bring this in towards a specific point now you have this charge that's at the center of the coordinate system, and you are gonna actually use this formula. So let's go ahead and see how this is gonna work out. So we've got basically, this is 3 m. This is 4 m. You're gonna bring it to a point right over here, in which the distance this our distance is basically gonna be the high pot news of this triangle right here. This is 3 m and this is 4 m. Then hopefully I should recognize this as a 345 triangle. That's gonna be 5 m over here. And by the way, this is actually gonna be the final distance once we take this charge from infinitely far away and then move it towards that specific location. So now the work done. So this is Let's let's say like this is gonna be with work done from the five Coolum charge. Now, the work that's done by the electric force on the negative to cool um, charge is gonna be K Q one. Start Que que que? So that's gonna be a big queue. Little Q one over our initial minus one over our final. Okay, so that's gonna be 8. 99 times 10 to the ninth. And now we've got the queue. The queue is just basically going to be that that one charge that that the center of the coordinate system, which is five cool homes Now we've actually have to incorporate the negative sign negative to cool homes. And now here's where we have to basically evaluate what are initial and final distances are Let's see, when you are initially infinitely far away So infinitely far away means that this initial distance right here is equal to infinity or just gonna be a really, really huge number. And what that means is that basically, this whole entire thing goes to zero because again, this our initial is just that infinity symbol. It just means it's a really, really, really big number. And then you have this minus one of our final in which that is equal to 5 m. So let's just go ahead and plug that in. So we're basically just gonna have zero for that first term and they're gonna have minus one over, and that's gonna be 5 m. So now we can go ahead and just plug this in, and we have the work that's done by the electric force because that's actually what we're trying to find here. How much work is done by the electric force. And that's just gonna equal. Let's see, I've got I've got positive 1.8 times, 10 to the 10th, and that's in jewels. Now, what I want to do is talk about basically the sign here because we actually got a positive sign, which means that this is work done by the electric force. And one way to think about this is being positive is the fact that if you were to just leave these things, you believe these two charges their natural state would be to try to attract one another because you have a positive and a negative charge you have, unlike charges. So basically, what happens is if you have this positive charge, it's gonna attract this negative charge from infinitely far away. And the fact the electric field, the fact that the electric force is gonna point in this direction and that the displacement is also gonna point in that direction means that you should have positive work that's done. If these were things were negative than you would have negative work because you actually have to force this charge in closer to the stationary charge, and that would actually be a negative work. That's one way to think about this. Let me know if you guys have any questions with this

4

Problem

Problem

What work is needed to assemble an equilateral triangle of side length 5 cm, with a 5µC charge at each vertex?

A

- 0.045 J

B

- 0.135 J

C

- 4.5 J

D

- 13.5 J

5

example

Speed of Electron in Electric Field

Video duration:

5m

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Hey, guys. So in this problem, if take a look at it, we have a new electron that is initially at rest and that it starts moving inside of a uniform electric fields. And then after some displacement, we're supposed to figure out what is the electrons Speed. Alright, so first things first and these kinds of problems where you're not really told the direction where the electric field points, just feel free to draw it yourself. So what I'm gonna do is I'm just gonna assume that the electric field points off to the right for this problem. So I mean, that is R E Field. And the other thing is that we know that this actually a constant e field. So that's really important, because that determines which work equation we're going to use. So let's figure out what's going on here. You have an electron, so we have sort of a charge like this, and we know that that charge little Q is equal to the negative elementary charge. So we know that the Q we're working with is negative 1.6 times 10 to the minus 19 cool homes, and we have a uniformed electric field and we know that the field strength there is 500 Newtons per Coolum. Now the displacement here or this travel distance is going to be D, and that's equal to 10 centimeters. So that's equal to 0.1 m. And we're supposed to figure out what is the electrons speed. So it means the variable that we're looking for is V. But how do we relate that Back to the work equation? Remember what happens is the work is equal to the change in the kinetic energy, and the change in the kinetic energy is just final kinetic energy minus initial. So that's gonna be one half mass of the electron V final squared, minus one half massive electron, the initial squared. Right, So it's just final minus initial. We're working with the kinetic energy equation. Now what happens is this electron is initially at rest inside of this electric field, so that means that the initial Velocity V not just equal to zero. So what that does is that basically just cancels out this term over here. So that means that the electrons speed after some displacement is actually this V final that we're looking for So that means that V Final is our target variable right here. And the way we solve that is relating it back to the work equation for a moving charge inside of a constant electric field. All right, so let's take a look at our equation. Right. So we know that the work equation we're gonna use is gonna be Q e d times the co sign of data. Remember, we have those two work equations. This is not point charges. This is actually constantly fieldwork. So we have the charge. That is the electron. We have the strength of the electric field. We have the displacement. Now we just need to figure out what the coastline of the angle is. And to do that, we need to look at our diagram and figure out which the displacement. What's that? Is that What direction is the displacement? Right. So we have this electron that is in a constant electric field. Now, whenever you have positive charges, positive charges want to move along the field lines. But negative charges always want to do the opposite of those field lines. So that means in this case, what happens is that the electron is gonna have a force that pulls it to the left. So that means that this distance over here is actually my displacement. Now, remember that the angle this CO sign of data over here is always the angle between the displacement and the electric field. Now, in this case, what happens is that my electric field points to the right and my displacement actually points to the left. So that means that the angle between these two things is exactly 180 degrees because they're totally opposite. So that means that this fatal here is 180 degrees. And what happens is that the co sign of this 180 degrees silly me, right that the co sign of 180 degrees is just equal to negative one. So what happens is that this work equation just picks up a negative sign. So it's negative. Q e d. Right. So we have that this, uh, this angle here is equal to 180 degrees. So now what happens is we're ready. Just plug everything and figure out the work. So this is just gonna be equal to negative. And now our charge is negative. 1.6 times 10 to the minus 19. Now we have the electric field of 500 the displacement is 0.1. So if you plug all of this stuff in, what happens is that the negative signs will cancel out in this equation and you should just get a work of eight times 10 to the minus 18 jewels. So now that we have the work here, we can relate that to the final velocity using this equation. So we're not quite done yet. So we have one extra step to Dio, and that is that the work that's done is equal to one half times the mass of the electron V final squared. So we're actually looking for this V final, so we have to move everything over to the other side. So the one half goes over and the mass of the electron goes underneath and gets divided. So we end up with is two times the work divided by the mass of the electron is equal to the final squared. And now the last thing we need to dio is just take the square root. So that means that the final is equal to the square roots of two times eight times 10 to the minus 18 jewels divided by the mass of the electron. Now, just in case you're not given this in a worksheet, most likely you will. The mass of the electron is 9.11 times 10 to the minus 31. So if you go ahead and work all that stuff out in your calculator, you get a final velocity of 4.19 times 10 to the sixth meters per second. And that is our final answer. That is the electron speed after it gets accelerated through this electric field. Our guys, that's it for this one. Let me know if you have any questions.

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