Work From Electric Force - Online Tutor, Practice Problems & Exam Prep

Work done by the electric force relates to changes in electric potential energy as charges move. The work can be expressed as $w=-q∆v$ for point charges, while in a constant electric field, it is $w=qEdcos\theta$. The work is path-independent, depending only on the initial and final positions, and approaches zero as charges move infinitely far apart.

Work due to Electric Force

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?

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 × 10⁶ m/s
• B. (a) B (b) -1.6 × 10^-17 J (c) 5.95 × 10⁶ m/s
• C. (a) A (b) 1.6 × 10^-17 J (c) 5.95 × 10⁶ m/s
• D. (a) A (b) -1.6 × 10^-17 J (c) 5.95 × 10⁶ m/s

Work to Bring Two Charges From Infinity

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 the coordinate system. So let's go ahead and draw out what that coordinate system might look like. So it's going to look like this. And basically, the idea is that you have a 5 coulomb charge that's out here, infinitely far away or very, very far away, and you're basically just going to bring it towards the origin of this coordinate system. How much work does that actually take? Well, we're talking about point charges. So let's go ahead and use our work formula for point charges. We know it's kqq(1rinitial-1rfinal). The problem is, is that we actually don't have any other charge that's going to interact with this 5 coulomb 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 going to be no other forces that are acting on this charge, repel it, which means there's nothing else around it to do work on it. Does that make sense? So basically, 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 2 coulomb charge that's floating away over here. And now if you want to bring this in towards a specific point, now you have this charge that's at the center of the coordinate system, and you are going to actually use this formula. So let's go and see how this is going to work out. So we've got basically this is 3 meters. This is 4 meters. And you're going to bring it to a point right over here in which the distance, this r distance is basically going to be the hypotenuse of this triangle right here. This is 3 meters and this is 4 meters, then hopefully, you guys should recognize this as a 3, 4, 5 triangle. This is going to be 5 meters over here. And by the way, this is actually going to 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 say, like, this is going to be with work done from the 5 coulomb charge. Now, the work that's done by the electric force on the negative 2 coulomb charge is going to be kQq(1rinitial-1rfinal). So, that's going to be 8.99 times 109. And now we've got the Q, the Q is just basically going to be that one charge that's at the center of the coordinate system, which is 5 coulombs. Now, we've actually have to incorporate the negative sign, negative 2 coulombs. And now here's where we have to basically evaluate what our initial and final distances are. Well, 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 it's just going to be a really, really huge number. And what that means is that basically, this whole entire thing goes to 0 because again, this rinitial is just that infinity symbol. It just means it's a really, really, really big number. And then you have this minus 1 over r final in which that is equal to 5 meters. So let's just go ahead and plug that in. So we're basically just going to have 0 for that first term, and then we're going to have minus one over, and that's going to be 5 meters. 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 going to equal Let's see. I've got, I've got positive 1.8 times 1010, and that's in joules.

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. One way to think about this as being positive is the fact that if you were to just leave these things, you were to leave 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 going to attract this negative charge from infinitely far away. And the fact the electric field The fact that the electric force is going to point in this direction, and that the displacement is also going to point in that direction, means that you should have positive work that's done. If these were things were negative, then you would have negative work because you'd actually have to force this charge in closer to the stationary charge, and that would actually be a negative work. So that's one way to think about this. Let me know if you guys have any questions with this.

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

Speed of Electron in Electric Field

Hey, guys. So in this problem, if you take a look at it, we have an electron that is initially at rest and then it starts moving inside of a uniform electric field. And then after some displacement, we're supposed to figure out what is the electron's speed. Alright. So first things first, in 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 it means that is our E field. And the other thing is that we know that this is 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 -1.6 × 1019-19 coulombs. And we have a uniform electric field, and we know that the field strength there is 500 Newtons per Coulomb. 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 meters. And we're supposed to figure out what is the electron's speed. So it means the variable that we're looking for is v. But how do we relate that back to the work equation? Well, remember, what happens is the work is equal to the change in the kinetic energy, and the change in kinetic energy is just final kinetic energy minus initial. So that's gonna be 1/2 mvfinal2 minus 1/2 mvinitial2. 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 v₀ is just equal to 0. So what that does is that basically just cancels out this term over here. So that means that the electron 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 that we solve that is relating it back to the work equation for a moving charge inside of a constant electric field. Alright? So let's take a look at our equation. Right? So we know that the work equation that we're gonna use is gonna be qed times the cosine of theta. Remember, we have those 2 work equations. This is not point charges. This is actually a constant e field work. So we have the charge that is, you know, the electron. We have the strength of the electric field. We have the displacement. Now we just need to figure out what the cosine of the angle is. And to do that, we need to look at our diagram and figure out what's the displacement What's this is this 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 wanna do the opposite of those field lines. So that means that 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 cosine of theta 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 theta here is 180 degrees, and what happens is that the cosine of 180 degrees, so let me write that, the cosine 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 -QED, right? So we have that this, that this angle here is equal to 180 degrees. So now what happens is we're ready to just plug everything in to figure out the work. So this is just gonna be equal to negative, and now our charge is -1.6 × 1019-19. Now we have the electric field of 500 and 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 8 × 10^-18 joules. 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 do, and that is that the work that's done is equal to 1/2 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 2 times the work divided by the mass of the electron is equal to v final squared. And now the last thing we need to do is just take the square root. So that means that v final is equal to the square root of 2 times 8 times 10 to the minus 18 joules 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^-31. So if you go ahead and work all that stuff out in your calculator, should get a final velocity of 4.19 times 10⁶ meters per second, and that is our final answer. That is the electron speed after it gets accelerated through this electric field. Alright, guys. That's it for this one. Let me know if you have any questions.