In the last video we figured out how much work or how much energy do we have to put into a particle to move it with a constant electric field. Let's see if we can do the same thing now with a variable electric field. And actually then we can figure out the electric potential of one position relative to another.
So let's say we have a point charge. alright, 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 I don’t know, let's say it’s a positive q1 coulombs. And if you wanted to draw a few lines, and of course this is different then, than the example with the infinite uniform charged plate because this will have a variable electric field. What is the electric field of this charge? Electric is going to look like this. Essentially, the amount of force that will exert to any amount of particle and it 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 and that field is coulombs constant times this charge q1 over the distance that we are from this charge.
If I were to draw this electric field close up, it’s pretty strong, close up is pretty strong and then as we get further, it gets weakened, weaker a little bit, it’s always radially outward from the charge. This is a bit of review for you. Alright, and these are just field, this is a vector field that I'm drawing where I'm just randomly picking points to join the vector field. At that point, it’s pointed outward and at any given radius away from the circle, this vector should be the same magnitude. I don’t mind our exactly but hope you get the same of the point. And the magnitude of these vectors decrease with a 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 a review. Because we know the force. If we have another test charge q2, we know that the force on that test charge is just q2 times this. And that’s just coulombs law. That the force on some other particle q2 is equal to the electric field times q2, no not q squared, q2 which is equal to kq1q2 over r squared. Alright, this is just coulombs law. And actually this comes from coulombs law.
Since we know that, let's take some other positive particle out here. Let’s call that q2, not the big q, I’ll do the small right up here since I already messed up down there. q2 and let's say this is a positive charge. So this is going to be repelled from q1. And let's figure out how much work does it take to push in this particle a certain distance. 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 a 10 meters distance right here. Let me draw a radial line. Let's say that this distance right here, let's say that that distance is I don’t know, 10 meters. And I want to push this particle in 5 meters. So eventually gets right here, alright. This is where I'm eventually going to get it. So it’s going to be 5 meters away. So 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, right, change in distance. 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. Let's just assume over this 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 coulombs constant q1 q2 over r squared times dr.
Now before we move on, let's think about something for a second. Coulombs law tells us that this is the outward force that this charge is exerting on this particle or that a field is exerting on this particle. The force that we have to apply to move the particle from here to here has to be inward force, has to be the exact opposite direction. So it has to be a negative. And why is that? Because we have to completely offset that the force of the field. And you know, maybe if the particles are already moving a little bit then our force will keep it from decelerating from the field and if it wasn’t already moving, we have to nudge it just a little infinitely small amount just to get it moving and then our force will completely offset that the force of the field and the particle would neither accelerate nor decelerate. So this is the amount of work. I just want to explain that we want to put that negative sign there because we’re going to the opposite direction of the field.
So how do we figure out the total amount of work? Alright, we figure out the amount of work to get it from here to here and I even drew it much bigger than it would be. This drs, 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, okay what's the work to go from here to here, the work to go from there to there then the work from there to there, all the way until we get to this 5 meters away from this charge. and what we do, we take the sum of these and we assume that these are, it’s an infinite sum of infinitely small increments. And as you learned that that is nothing but the integral. And so that is the total work is equal to the integral, it’s going to be a definite integral because we start at this point. We’re summing from our radius is equal to 10 meters, that’s the starting point, to our radius equal to 5 meters, starting at the higher value ending at the lower value. But that’s what we’re doing. Were pushing it inwards.
And then were taking the integral of minus kq1q2 over r squared, dr. all of these are constant terms, up here. Right. So we can take them all out. So this is the same thing, I don’t want to run out of space, as minus kq1q2 times the integral from 10 to 5. Well 1 over r squared or r to the negative 2, dr. And that equals minus k, I'm running out of space, q1q2. I'm going to take the anti derivative. We don’t have to worry about plus C because it’s a definite integral. r to the negative 2 is anti derivative, 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 evaluated it 5 and then subtract it evaluated it 10. And then let me just go up here. actually, let me erase some of this.
Erase this up here. valuable space to work on.
So let's see, we said that the work, I’ll just rewrite that. we had the minus out here. we have the minus, we took the anti derivative and they’ve cancelled out so we get kq1q2 times the anti derivative of evaluated 5. So 1/5, r to the negative 1. So 1/r minus the anti derivative evaluated 10 minus 1/10. They equal to, well 1/5, that’s the same thing as 2/10. Right. so we have the work is equal to kq1q2 times 2/10 minus 1/10 is 1/10. So it equals, you go kq1q2 over 10. That’s the work to take the particle from here to here.
And so similarly, we could say that the potential difference of the particle at this point, relative to this point. Right. That the potential difference here is this much higher. And it’s going to be joules, that’s right. 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, actually this is, you might just find this interesting just from something to think about. that the big difference between, actually, let me just continue into the next video. Because I realize I'm already at 10 minutes. I’ll see you soon.
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