Course: Physics library > Unit 11Lesson 3: Electric potential energy, electric potential, and voltage
Electric potential energy (part 2-- involves calculus)
Electric potential energy difference in a varying field. Created by Sal Khan.
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- why is it a 'minus K' while integrating ?(10 votes)
- because he considered the outward direction to be positive, so the force needed to move it is negative ("radially inward")(13 votes)
- At4:00instead of making the decision to use calculus could he not have used colombs law with r = 10 and then minused it from r=5 to get the energy of being 5m away from q1?(5 votes)
- No. That calculation would give you the difference between the force exerted on the test charge at a distance of 5 meters away and the force exerted on the test charge at a distance of 10 meters away. We want to know how much work it would take to move the test charge from a distance of 10 meters away to a distance of 5 meters away. Work = Force * Distance. We know the distance is 5 meters, because we want to move the test charge 5 meters toward the point charge. However, the force will constantly change as we do this, because the force is inversely proportional to the square of the distance. Therefore, we need to use calculus to account for the continuous change in force as we move the test charge toward the point charge.(14 votes)
- I don't understand what Negative voltage mean... Can anyone explain? This has confused me for a long time(6 votes)
- in simpler words, Voltage(potential) when calculated, is considered to be in the negative direction of Force applied. Thus it is taken negative(3 votes)
- in a medium, what is permittivity and permeability?(3 votes)
- Permittivity has to do with how a material reacts to an electric field and permeability has to do with how a material reacts to a magnetic field.(4 votes)
- A general question about electrical fields: I understand that the magnitude of the field at any given radius is the same, and as the radius increases the magnitude decreases, but does the magnitude ever reach 0?
I would assume obviously not and that it would just become so small that it's negligible, but that implies that EVERYTHING is exerting an electrical field on EVERYTHING in the entire universe... which is mind boggling.(3 votes)
- theoretically field spreads over infinity but as the distance reaches astronomical units it is practically zero(4 votes)
- Why is the work not negative?? At5:08, Sal mentions that the work done to push the positive test charge towards +Q1 should be negative...so after doing the calculus, why don't we get a negative in our answer?? Why does the total work become a positive value of kQq/10?(2 votes)
- The work done to push a positive toward positive is positive. Work is positive when the force is in the same direction as the displacement.
In the problem, he has defined outward as the negative direction. Therefore a force directed inward is a negative force, AND a displacement directed inward is negative. The negative force * the negative displacement will, after the multiplication is done, give you positive work(2 votes)
- Aren't we suppose to swap the bounds since the upper bound, 5 is smaller than the lower bound, 10?(1 vote)
- No,since we are moving a particle from 10 meters to 5 meters. Lower bound is initial,upper bound is final.(3 votes)
- We know that V= Cq/ r . If r=0 then the value of V reaches infinity. Does that mean we can never make the distance between two particles of positive charges zero?(2 votes)
- Yes. Then they would be in exactly the same place, and that's not possible, right?(1 vote)
- I think your integral should give you (r^(-1))/(-1), shouldn't it? I could be wrong but I'm almost positive.(1 vote)
- Yeah that correct. However that is the same as -1* r^(-1), which he got. He then proceeded to move the -1 to the outside of the brackets, and used it to cancel out the other negative. He was a little unclear on that in the video, good question.(2 votes)
- What does coulumb's law state?(1 vote)
- The force is proportional to the product of the charges and inversely proportional to the square of the distance between them. This is known as Coulomb's law.(2 votes)
In the last video, we figured out how much work or how much energy we have to put into a particle to move it within a constant electric field. Let's see if we can do the same thing now with a variable electric field, and actually, then we could figure out the electric potential of one position relative to another. So let's say we have a point charge. It doesn't have to be a point charge, but let's just say that there's some field being generated by this charge that is a positive q1 coulombs. And if we wanted to draw field lines-- and, of course, this is different than the example with the infinite uniformly charged plate because this will have a variable electric field, right? What is the electric field of this charge? The electric field is going to look like this. It's essentially the amount of force it will exert on any particle, and it's always going to be pointed outwards, because we assume that the test particle in question is always a positive charge, so a positive charge will be repelled from this positive charge. That field is Coulomb's constant times this charge q1 over the distance that we are from this charge. So if I were to draw this electric field close up, it's pretty strong. And then as we get further, it gets weaker a little bit. It's always radially outward from the charge. This a bit review for you. And this is a vector field that I'm drawing, where I'm just randomly picking points and showing the vector of the field at that point, that it's pointed outward, and at any given radius away from the circle, these vectors should be the same magnitude. I know mine aren't exactly, but hopefully, you get the point, right? And the magnitude of these vectors decrease with the square of the distance. So that's what the field is going to look like if I were to draw a bunch of vector field lines. And that makes sense, just as a review, right? Because we know the force-- if we had another test charge q2, we know that the force on that test charge is just q2 times this, and that's just Coulomb's Law, right? That the force on some other particle q2 is equal to the electric field times q2, which is equal to k q1 q2 over r squared, right? This is just Coulomb's Law, and actually this comes from Coulomb's Law. So since we know that, let's take some other positive particle out here, and let's call that q2. I'll write it up here since I already messed up down there. q2, and let's say this is a positive charge, so it is going to be repelled from this q1. And let's figure out how much work does it take to push in this particle a certain distance, right? Because the field is pushing it outward. It takes work to push it inward. So let's say we want to push it in. Let's say it's at 10 meters. Let's say that this distance right here-- let me draw a radial line-- let's say that this distance right here is 10 meters, and I want to push this particle in 5 meters, so it eventually gets right here. This is where I'm eventually going to get it so then it's going to be 5 meters away. So how much work does it take to move it 5 meters towards this charge? Well, the way you think about it is the field keeps changing, right? But we can assume over a very, very, very, very infinitely small distance, and let's call that infinitely small distance dr, change in radius, and as you can see, we're about to embark on some integral and differential calculus. If you don't understand what any of this is, you might want to review or learn the calculus in the calculus playlist, but how much work does it require to move this particle a very, very small distance? Well, let's just assume over this very, very, very small distance, that the electric field is roughly constant, and so we can say that the very, very small amount of work to move over that very, very small distance is equal to Coulomb's constant q1 q2 over r squared times dr. Now before we move on, let's think about something for a second. Coulomb's Law tells us that this is the outward force that this charge is exerting on this particle or that the field is exerting on this particle. The force that we have to apply to move the particle from here to here has to be an inward force. It has to be in the exact opposite direction, so it has to be a negative. And why is that? Because we have to completely offset the force of the field. Maybe if the particle was already moving a little bit, then our force will keep it from decelerating from the field, and if it wasn't already moving, we would have to nudge it just an infinitely small amount just to get it moving, and then our force would completely offset the force of the field, and the particle would neither accelerate nor decelerate. So this is the amount of work, and I just want to explain that we want to put that negative sign there because we going in the opposite direction of the field. So how do we figure out the total amount of work? We figured out the amount of work to get it from here to here, and I even drew it much bigger than it would be. These dr's, this is an infinitely small change in radius. If we want to figure out the total work, then we keep adding them up. We say, OK, what's the work to go from here to here, then the work to go from there to there, then the work to go from there to there, all the way until we get to 5 meters away from this charge. And what we do when we take the sum of these, we assume that it's an infinite sum of infinitely small increments. And as you learned, that is nothing but the integral, and so that is the total work is equal to the integral. That's going to be a definite integral because we're starting at this point. We're summing from-- our radius is equal to 10 meters-- that's our starting point-- to radius equals 5 meters. That might be a little unintuitive that we're starting at the higher value and ending at the lower value, but that's what we're doing. We're pushing it inwards. And then we're taking the integral of minus k q1 q2 over r squared dr. All of these are constant terms up here, right? So we could take them out. So this is the same thing-- I don't want to run out of space-- as minus k q1 q2 times the integral from 10 to 5 of 1 over r squared-- or to the negative 2-- dr. And that equals minus k-- I'm running out of space-- q1 q2. We take the antiderivative. We don't have to worry about plus here because it's a definite integral. r to the negative 2, what's the antiderivative? It's minus r to the negative 1. Well, that minus r, the minus on the minus r will just cancel with this. That becomes a plus r to the negative 1, And you evaluate it at 5 and then subtract it and evaluate it at 10. And then-- let me just go up here. Actually, let me erase some of this. Let me erase this up here. Valuable space to work on. So we said that the work is equal to-- I'll just rewrite that. We had the minus out here, but then we had a minus and we took the antiderivative and they canceled out, so we have k q1 q2 times the antiderivative evaluated at 5, so 1 over 5, right? r to the negative 1, so 1 over r minus the antiderivative evaluated at 10 minus 1 over 10 is equal to-- well, 1 over 5, that's the same thing as 2 over 10, right? So we have the work is equal to k q1 q2 times 2 over 10 minus 1 over 10 is 1 over 10, so that equals k q1 q2 over 10. That's the work to take the particle from here to here. And so similarly, we could say that the potential energy of the particle at this-- the potential difference of the particle at this point relative to this point, that the potential difference here is this much higher. It's going to be in joules because that's the unit of energy or work or potential, because potential is energy anyway. So the electric potential difference between this point and this point: at this point, it is this value higher. Let's do another example, and actually, you might just find this interesting, just from something to think about. The big difference between-- actually, let me just continue this into the next video because I realize I'm already at 10 minutes. I'll see you soon.