If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:9:50

Video transcript

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 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 right it doesn't have to be a point charge but you know 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 Q one coulombs and if we wanted to draw a few lines and of course this is different then then the example with the infinite uniformly charged plate because this will have a variable electric field right what does the electric field of this charge the electric field is going to look like this essentially the amount of force it'll 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 positive charge will be repelled from this positive charge electric field is Coulomb's constant times this charge Q 1 over the distance that we are from this charge right so if I were to draw this electric field close up it's pretty strong close up it's pretty strong and then as we get further it gets weak and weaker a little bit it's always radially outward from the charge this is a bit review for you all right and these are just filled this is this is a would you call it 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 the same at the point right and the the magnitude these vectors decrease with the square of the distance right 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 Q 2 we know that the force on that test charge is just Q 2 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 not Q squared Q 2 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 let's let's since we know that let's let's take some other positive charge particle out here and let's call that q2 and not a big Q I'll do a small I'll write it up here since I already messed up down there the q2 and let's say this is a positive charge so it is going to be repelled repelled from this from from q1 and let's say how let's figure out how much work does it take to push in this particle to 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 and let's say it's at 10 meters let's say that this distance right here let me draw a radial 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 it eventually gets right here right 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 to move it five meters towards this charge well the way you think about it is the 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 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 D R now before we move on let's think about something for a second this is 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 has to be in the exact opposite direction right so it has to be a negative and why is that because we have to completely offset the force of the field and you know maybe that if the particle is already moving a little bit then our force will will keep it from decelerating from the field and if it wasn't already moving we would have to nudge it just the 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 or 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're going in an opposite direction of the field so how do we figure out the total amount of work right 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 you know these 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 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 this five meters away from from this charge and what we do and you know we take the sum of these and then we assume that these are 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 the total work is equal to the integral is 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 our rate is equal 5 meters you might not there might be a little unintuitive that we're starting at the higher value ending at the lower value but that's what we're we're pushing it inwards and then we're taking the integral of minus K q1 q2 over R squared D are 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 that's well 1 over R squared or R to the negative 2 d R and that equals minus K I'm running out of space q1 q2 we take the antiderivative don't have to worry about plus C because it's a definite integral R to the negative 2 of C antiderivative it's minus R to the negative 1 well that minus R I will just the minus on the - R will just cancel with this that becomes a plus R to the negative 1 and you've evaluated it 5 and then subtract it evaluated to 10 and then let me just go up here I know I'm actually let me erase some of this let me erase this up here valuable space to work on so what is so let's see we said that the work is equal to I'll just rewrite that we had the minus out here then we had a - we took the antiderivative and they cancelled 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 valued at 10 minus 1 over 10 equal to a 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 to over 10 minus 1 over 10 is 1 over 10 so it equals you go k q-1 q-2 over 10 that's the work to take this 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 right that the potential difference here is this much higher and the new it's going to be in joules right because that's the unit of energy or work or potential well 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 this is that you might just find this interesting just from something to think about that the big difference between I actually let me just continue this into the next video because I realize I'm already at 10 minutes I'll see you soon