Hey, guys. So in this video, we're going to talk about the work that's done due to the electric force. We've already talked about electric potential energies and potentials. We're going to see how all of that stuff ties into 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 energy as a result of an object's position. And so the reason that these charges are moving is that the electric force or the field, which is basically a force or something that gives information to charge, guides us to feel 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, in most cases, W=Fdcos(θ). We've seen this sort of relationship before. Now what I wanted to do is 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 1 joule of potential energy, I've gained 1 joule in terms of kinetic. And we know that from a 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, 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 the negative change in potential energy 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 qV. So one of the things that we can do is pull all of this stuff together and actually write that the work that's done is equal to -qΔV. Right? So if we have qΔV, and that's equal to negative, to the ΔU, we have this negative sign right here. We'll just stick that down there. Okay?

So, that means, let's take a look at what happens for point charges. So for point charges, an interesting thing that happens is we can't use W=Fdcos(θ). The reason we can't do that is that the electric force is a result of 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 q2, and it moves to a different location, the electric force is not constant because it depends on this r distance. So this WdFdcos(θ) doesn't work in the cases of point charges. So let's look at what the potential looks like for these two things since we know that the work done is going to be the charge, the feeling charge, which by the way, I'm just going to label this as q 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 V1 and a potential here V2, then that means the potential difference between these locations is going to be V2-V1. Now we know that from a point charge, a point charge over here, that V is equal to k·q/r, the producing charge over here. So that means that the difference in potential is going to be k·q(r2-r1). Now, what happens is this negative sign right here actually serves to reverse these things here. Sorry, not to reverse the direction. It could basically flip the signs right here. So this negative sign can actually make it sort of that they're flipped to 1 over r1 instead of 1 over r2. So basically, we can tie all of this stuff together and say that due to point charges, the work that's done is going to be kq·q/r1-r2. 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 going to be actually one over r initial minus one over r final. And the reason that it's not final minus initial has to do with the fact that we actually sort of absorb this negative sign that was originally out here, and all that did was just flip the fractions. Okay?

There's another actually situation where there's another situation where you have to calculate the work and where we actually can use W=Fdcos(θ) in cases where we have a constant electric field. So when we have a constant electric field, 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. Well, 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 WdFdcos(θ) does work, and we can use it. And so all we need to do is that if this is changing some distance d here, to find out what the work is done, we 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 that we can simplify this as QE·dcos(θ). It simplifies down to that easier expression. All you have to do is realize that this QE here is actually the force that is done by the electric field. Okay? So this does look like Fdcos(θ). 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 it, like, in some squiggly path and eventually end up here, in all of those 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. It just means that it doesn't matter how you get there, just the fact you got there in your finest minus final. 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 낄ом#ес to 0. And that's something we can actually see just from these equations. So kqq/r, as r gets really big, the potential and the electric potential energy goes to 0. 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 2 nanocoulomb charge that is initially 5 millimeters away. So I've got this sort of before. So I'm going to label a before and an after, because the 2 nanocoulomb charge is eventually moved closer to this. So there's going to be a before and after kind of thing here. So let's go ahead and draw. So let's go. I've got this 10 nanocoulomb charge, and then I've got this 2 nanocoulomb charge. Initially, they're separated by a distance of 5 millimeters here, and then what happens is that after, I've got this 10 nanocoulomb charge, and now this 2 nanocoulomb charge. And now the distance is equal to 3 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 should be, Yeah. Yeah.uss#пашлалиΓέν#パタ иąязеь8 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 going to use the work formula that is for point charges. I can't use the Fdcos(θ) because f is constantly changing. What we did say is that it's kqq times 1 over r initial minus 1 over r final. So this is like sort of like a shortcut formula that you can use. You don't have to rederive everything. You don't have to find the potential differences or anything like that. Alright? You could just start out from this formula. The work that's done is going to be 8.99·109. Now you've got the charges involved. Let's see. Q is going to be it doesn't really matter in this case what big Q and little q are, but just to be consistent, this Q is going to be the 10·10-9, because these are nanocoulomb charges that we're working with. And we've got 2 times 10-9. Now we've got the distances involved. So I've got one over r initial. It's moving from 5 millimeters away, so that's 0.5. And then the final distance is 1 over 0.3. So that's going to be my final distance. So now the work that's done is just going to be negative 2.4·10-5J. Now I want to point out that the negative sign is actually 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 positive charges, which means that they 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 away means that you have to do work against it. So it's kind of like thinking about, like, how you have a magnet and 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. Alright?