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Current time:0:00Total duration:13:27

Video transcript

in this video we're going to study the electric field created by an infinite uniformly charged plate and why are we going to do that well one because we'll learn that the electric field is constant which is neat by itself and then that's that's kind of an important thing to realize later when we talk about parallel charged plates and capacitors because you know our physics books tell them that the that the field is constant but they never really prove it so we will prove it here and the basis of all of that is to figure out what the electric charge of an infinitely charged plate is so let's take a side view of the infinitely charged plate get some intuition so let's say that's the side view of the plate and let's figure out what it's and let's say that this this plate has a charge density of Sigma and what's charge density just says well that's Coulomb's per area all right charge density is equal to charge per area that's all charge charge Sigma is so we're saying this has a uniform charge density so before we we break into what what may be hard core mathematics and if you're watching this in the calculus playlist you might want to review some of the electrostatics from the physics playlist and that'll probably be relatively easy for you if you're watching this from the physics playlist and you haven't done the calculus playlist you should not watch this video because you will find it overwhelming but anyway let's proceed so let's say so once again this is my my infant so it goes off in every direction it even comes out of the video where you know this is a side view and let's say I have a point charge up here Q right point charge Q so let's think a little bit about if I have a point let's say I have an area here on my plate let's think a little bit about what it's what the net effect of of it is going to be on this point charge well first of all let's say that this point charge is at a height H above the field so let me draw that this is a height H and let's say this is a point directly below the point charge and let's say that this distance right here is R so first of all what is the distance between this part of our plate and our point charge what is this distance that I'll draw in magenta what is this distance what is that distance well Pythagorean theorem it's this is a right triangle so it's a square root of this side squared plus this side squared so this is going to be the square root of H squared plus R squared so that's the distance between this area and our test charge now let's get a little bit of intuition so if this is a positive test charge and if this plate is positively charged the the force from just this area on the charge is going to be radially outward from this area so it's going to be it's going to look let me do it in another color because I don't want to it's going to go in that direction right but since this is an infinite plate in every direction there's going to be another point on this plate that's essentially on the other side of this point right over here where its net force its net electrostatic force on the point charge is going to be like that and as you can see since you know we have a uniform charge density and the plate is symmetric in every direction the X or the horizontal components of the force the horizontal components of the force are going to cancel out and so that's true for Furley any point along the but along this this plate because if you pick any point along it we're looking at a side view but if we took a if we took a top view if that's the top view and of course the plate goes off in every direction forever and that's kind of you know where our point charge is if we if we said oh well you know there's this point on the plate and it's going to be you know it's going to have some Y component that's in on this top view coming out of the video but I'll have some X component this points X component affect will cancel it out right you can always find another point on the plate that's symmetrically opposite whose X component of electrostatic force will cancel out with the first one so given that that's just a long-winded way of saying that the net force on on this on this point charge will only be upward right I think it should make sense to you that all of the X components or the horizontal components of the electrostatic force all cancel out right because there are infinite points to either side of this test charge so with that out of the way what do we need to focus on well we just need to focus on the Y components of the electrostatic force so what's the y component so let's say that this is exerting let's say that this point right here and I'll keep switching colors let's say that this point and once again this is the side view is exerting a it's field at that point is let's say the electric field at that point is I don't know ee1 right and it's going to be going in that direction what is its Y component what is the component in that direction and of course it's pushing outwards if they're both positive so what is the Y component what is that well if we knew theta if we knew this angle if we knew this angle the Y component or the upwards component is going to be the electric field times cosine of theta right cosine is adjacent over hypotenuse so hypotenuse times cosine of theta is equal to the adjacent so if we wanted the vertical or the Y component of the electric field we would just multiply the magnitude of the electric field times cosine of theta so how do we figure out theta well that data is also the same as this data from our basic from our basic trigonometry and so what's cosine of theta cosine is adjacent over hypotenuse from sohcahtoa right cosine of theta is equal to adjacent over hypotenuse cosine of theta is equal to adjacent over hypotenuse so when we're looking at this angle which is the same as that one what's what's adjacent over hypotenuse this is adjacent that is the hypotenuse so what do we get we get that the Y component of the electric field due to just this little chunk of of our plate the electric field let in the Y component let's just call that you know sub one because this is just a little small part of the plate it is equal to the electric field generally the magnitude of the electric field from this point times cosine of theta which equals the electric field times the adjacent times height over the hypotenuse over the square root of H squared plus R squared fair enough so now let's see if we can figure out what the magnitude of the electric field is before and then and then we can put it back into this and we'll figure out the Y component from this point and actually well not we're not just going to figure out the electric field just from that point we're going to figure out the electric field from a ring that's surrounding this so let me give you a little bit of perspective or draw it with a little bit of perspective so this is my infinite plate again I'll draw it in yellow again since I originally drew it in yellow so this is my infinite plate it goes in every direction and then I have my charge floating above this plate someplace at a height of H and you know this point here this could have been right here maybe but what I'm going to do is I'm going to draw a ring that's of an equal radius around this point right here so this is our let's draw a ring because all of these points are going to be the same distance from our test charge right they all they all are exactly like this this one air point that I drew here you could almost view that this is a cross-section of this ring that I'm drawing so let's figure out what the O the Y component of the electric force from this ring is on our point charge so to do that we just have to figure out the area of this ring right we just figure out the area of this ring multiply it times our charge density and we'll have the net the total charge from that ring and then we can use Coulomb's law to figure out its force or the field at that point and then we could use this formula which you just figured out to figure out the Y component I know it's involved but it'll all be worth it because you'll know that we have constant electric field so let's do that so first of all Coulomb's law tells us well first let's figure out the charge from this ring so Q of the ring it equals what it equals the circumference of the ring times the width of the ring so let's say the circumference is 2 PI R 2 PI R and let's say this it's a really skinny ring it's really skinny it's d are infinitesimally skinny so its width is d R so that's the area that's the area of the ring and so what's its charge going to be its area times the charge density so times Sigma right that is the charge of the ring and then what is the electric field generated by the ring at at this point here where our test charges well Coulomb's law tells us that the force generated by the ring the force by the ring is going to be equal to Coulomb's constant times the charge of the ring times our test charge divided by the distance squared right well what's the distance between really any point on the ring and in our test charge well this could be one of the points on the ring and this could be another one right and this is like a cross section so the distance at any point this distance right here is once again by Pythagorean theorem right because this is also R this distance is the square root of H squared plus R squared right it's the same thing as that so if we want to do so it's the distance squared and that's just that's equal to K times the charge in the ring times our test charge divided by distance squared well distance is square root of H square root of a squared plus R squared so if we square that it just becomes H squared plus R squared and if we want to know the electric field created by that ring we just the electric field is just the force per test charge so if we divide both sides by Q we learned that the electric field of the Ring is equal to Coulomb's constant times charge in the ring / h squared plus R squared and now what is what is the y component of the charge in the ring well it's going to be this right what we just figured out is the magnitude of essentially this vector right but we want its Y component because all of the X components just cancel out so it's going to be x cosine of theta and we figured out that cosine of theta is essentially this so we multiply it times that so e the d the field from the ring in the y direction is going to be equal to its magnitude times cosine of theta which we figured out was H over the square root of H squared plus R squared and we could simplify this a little bit the denominator becomes what h squared plus R squared to the three-halves power and what's the numerator let's see we have K H K H and then the charge in the ring which we solved up here so that's 2 Pi Sigma R 2 Pi Sigma R make sure I don't lose anything 2 Pi Sigma R dr so we have just calculated the Y component the vertical component of the electric field at H units above the plate from and not not from the entire plate just the the electric field generated by a ring of radius R from the base of where we're taking this height and so I've already gone 12 minutes into this video just to give you a break and myself a break I will continue in the next but you can imagine what we're going to do now we just figured out the electric field the electric field created by just this ring right so now we can integrate across the entire plane we can essentially take the we could sum up all the rings of radius infinity all the way down to zero and that'll give us the sum of all of the electric fields and essentially the net electric field H units above the surface of the plate see you in the next video