Learn about Proof - Advanced: Field from infinite plate - part 1 Video

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Learn about Proof - Advanced: Field from infinite plate - part 1

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 learned that the electric field is constant which is neat by itself and then that’s kind of an important thing to realize later on when we talk about parallel charged plated and capacitors because our Physics books tell them 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 have an infinitely charge plate is. So let’s take a side view of the infinitely charged plate gets me intuition. So let’s say that’s the side view of the plate and let’s say that this plate has a charge density of sigma, and what’s charge density? It just says “Well, that’s Coulombs per area”, right? Charge density is equal to charge per area. That’s all charge sigma is. So we’re saying this has a uniform charge density. So before we break into what maybe hardcore Mathematics and if you’re watching this in the Calculus play list, you might want to review some of the electro static from the Physics play list, and that would probably be relatively easy for you. If you’re watching this from the Physics play list and you have not done the Calculus play list, you should not watch this video because you will find it overwhelming but anyway let’s proceed. So once again this is my infinite. It goes of in every direction and even comes out of the video where we lose the 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 the net effect of it is going to be on this point charge. At first I will say that this point charge is that a height h above the field. This is a height h and let’s say this is the point directly below the point charge. 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? Well Pythagorean Theorem, this is a right triangle so it is square root of these sides squared plus these sides squared. So this going to be the √h2+r2, so that’s the distance between this area and our test charge. And let’s get a little bit intuition. So if this is a positive test charge and at this plate is positively charge, the force from just this area on the charge is going to be radially outward from this area. 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 is essentially on the other side of this point over here where its nets force, its net electro static force on the point charge is going to be like that. As you can see since every uniform charge density, and the plate is symmetric in every direction, the x or the horizontal components of the force are going to cancel out. And so that’s true for any point along this plate because if you pick any point a longer we’re looking at the side view but if we took the top view, if that’s the top view and of course the plate goes off in very direction forever and that’s kind of where our point charge is. If we said, “Oh well, there’s this point on the plate and it’s going to have some y components that’s on this top view coming out of the video but has some x component.” These points x component effect will cancel out, right? You can always find another point on the plate that’s symmetrically opposite whose x component of electro static force will cancel out with the first one. So given that that is just a long winded way of saying that the net force on this point charge will only be upwards, right? I think it should make sense to you that all of the x components or the horizontal components of the electro static force all cancel out right because your infinite points to either side of this test charge. So with that out of the way, what we need to focus on? Well, we just need to focus on the y components of the electro static force. So what’s the y component? So let’s say that this point right is exerting its field at the point is E1, 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 is pushing out hours and they’re both positive, so what is the y component? What is that? Well if we knew theta, if we knew this angle, the y component or the upwards component is going to be the electric field times cos(Ø), right? Cosine is adjacent over hypotenuse so hypotenuse times cos(Ø) 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 cos(Ø). So how do we figure out Ø? Well that theta is also the same as this theta from our basic trigonometry and so what’s cos(Ø)? Cosine is adjacent over hypotenuse from SOHCAHTOA, right? Cos(Ø) is equal to adjacent over hypotenuse. So we’re looking at this angle which is the same as that one. What’s adjacent over hypotenuse? This is adjacent, that is a hypotenuse. So what do we get? We get that the y component of the electric field do to just this little chunk of our plate. The electric field in the y component, let’s just call that sub one because it’s 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 h2+r2, fair enough. So now let’s see if we can figure out what the magnitude of the electric field is and then we can put it back into this, and we’ll figure out the y component from this point. And actually 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 some place at a height of h and this point here is this could have been right here maybe. But what I'm going to do is I'm going to draw a ring that have an equal radius around this point right here. So this is r. Let’s draw a ring because all of these points are going to be the same distance from our test charge, right? They are all exactly like this one r point that I drew here. You can almost view that this is a cross section of this ring that I'm drawing. So let’s figure out what 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, multiply it times our charge density and we’ll have 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 is to figure out the y component. I know it’s involved but it will all be worth it because we know that we have a constant electric field. So let’s do that. So first for of all Coulomb's law tell us—well, first of all, let’s figure out the charge from this ring. So Q of the ring is equal to the circumference of the ring times the width of the ring. So let’s say the circumference is 2∏r, and let’s say its really skinny ring. It’s dr, infinitely skinny. So its width is dr. So that’s the area of the ring and so what is this charge ring going to be? It’s 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 this point here where our test charge is. Well, Coulomb's law tells us that the force generated by a 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? What’s the distance between really any point on the ring and our test charge? Well, this could be one of the points on the ring and this could be another, right? This is like a cross section. So the distance at any point is once again by Pythagorean Theorem because this is also r. This distance is the √h2+r2. It is same thing as that. So it’s d2 and that’s equal to k times a charge in the ring times our test charge divided by d2. Well d = √h2+r2 so if we squared that, it’s just becomes h2+r2. And if we want to know the electric field created by that ring, 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 the charge in the ring divided by h2+r2. Now what is the y component of the charge in the ring? Well, it’s going to be this, right? Well we just figured out is a magnitude of essentially this vector, right? But we want its y component because all of the x components just cancel out so this is going to be times cosine of theta and we figured out that cosine of that is essentially this, so we multiply times that. 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/√h2+r2. We could simplify this a little bit. The denominator becomes (h2+r2)3/2 and what’s the numerator? We have Kh and then the charge in the ring which we solve up here. So that’s 2∏r dr. So we have just calculated the y component, the vertical component of the electric field at h units above the plate, not from the entire plate just 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 of this video just to give you break and myself a break. I will continue in the next but you can imagine what we are going to do now. We just figured out the electric field created by just this ring, right? So now we can integrate across the entire plane. We could sum up all the rings of radius infinity all the way down to zero and that will 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.

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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 learned that the electric field is constant which is neat by itself and then that’s kind of an important thing to realize later on when we talk about parallel charged plated and capacitors because our Physics books tell them 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 have an infinitely charge plate is.

So let’s take a side view of the infinitely charged plate gets me intuition. So let’s say that’s the side view of the plate and let’s say that this plate has a charge density of sigma, and what’s charge density? It just says “Well, that’s Coulombs per area”, right? Charge density is equal to charge per area. That’s all charge sigma is.

So we’re saying this has a uniform charge density. So before we break into what maybe hardcore Mathematics and if you’re watching this in the Calculus play list, you might want to review some of the electro static from the Physics play list, and that would probably be relatively easy for you. If you’re watching this from the Physics play list and you have not done the Calculus play list, you should not watch this video because you will find it overwhelming but anyway let’s proceed.

So once again this is my infinite. It goes of in every direction and even comes out of the video where we lose the 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 the net effect of it is going to be on this point charge. At first I will say that this point charge is that a height h above the field. This is a height h and let’s say this is the point directly below the point charge. 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? Well Pythagorean Theorem, this is a right triangle so it is square root of these sides squared plus these sides squared. So this going to be the √h2+r2, so that’s the distance between this area and our test charge. And let’s get a little bit intuition.

So if this is a positive test charge and at this plate is positively charge, the force from just this area on the charge is going to be radially outward from this area. 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 is essentially on the other side of this point over here where its nets force, its net electro static force on the point charge is going to be like that.

As you can see since every uniform charge density, and the plate is symmetric in every direction, the x or the horizontal components of the force are going to cancel out. And so that’s true for any point along this plate because if you pick any point a longer we’re looking at the side view but if we took the top view, if that’s the top view and of course the plate goes off in very direction forever and that’s kind of where our point charge is.

If we said, “Oh well, there’s this point on the plate and it’s going to have some y components that’s on this top view coming out of the video but has some x component.” These points x component effect will cancel out, right? You can always find another point on the plate that’s symmetrically opposite whose x component of electro static force will cancel out with the first one.

So given that that is just a long winded way of saying that the net force on this point charge will only be upwards, right? I think it should make sense to you that all of the x components or the horizontal components of the electro static force all cancel out right because your infinite points to either side of this test charge.

So with that out of the way, what we need to focus on? Well, we just need to focus on the y components of the electro static force. So what’s the y component? So let’s say that this point right is exerting its field at the point is E1, 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 is pushing out hours and they’re both positive, so what is the y component? What is that?

Well if we knew theta, if we knew this angle, the y component or the upwards component is going to be the electric field times cos(Ø), right? Cosine is adjacent over hypotenuse so hypotenuse times cos(Ø) 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 cos(Ø). So how do we figure out Ø?

Well that theta is also the same as this theta from our basic trigonometry and so what’s cos(Ø)? Cosine is adjacent over hypotenuse from SOHCAHTOA, right? Cos(Ø) is equal to adjacent over hypotenuse. So we’re looking at this angle which is the same as that one. What’s adjacent over hypotenuse? This is adjacent, that is a hypotenuse.

So what do we get? We get that the y component of the electric field do to just this little chunk of our plate. The electric field in the y component, let’s just call that sub one because it’s 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 h2+r2, fair enough.

So now let’s see if we can figure out what the magnitude of the electric field is and then we can put it back into this, and we’ll figure out the y component from this point. And actually 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 some place at a height of h and this point here is this could have been right here maybe.

But what I'm going to do is I'm going to draw a ring that have an equal radius around this point right here. So this is r. Let’s draw a ring because all of these points are going to be the same distance from our test charge, right? They are all exactly like this one r point that I drew here. You can almost view that this is a cross section of this ring that I'm drawing.

So let’s figure out what 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, multiply it times our charge density and we’ll have 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 is to figure out the y component. I know it’s involved but it will all be worth it because we know that we have a constant electric field. So let’s do that.

So first for of all Coulomb's law tell us—well, first of all, let’s figure out the charge from this ring. So Q of the ring is equal to the circumference of the ring times the width of the ring. So let’s say the circumference is 2∏r, and let’s say its really skinny ring. It’s dr, infinitely skinny. So its width is dr. So that’s the area of the ring and so what is this charge ring going to be? It’s 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 this point here where our test charge is. Well, Coulomb's law tells us that the force generated by a 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?

What’s the distance between really any point on the ring and our test charge? Well, this could be one of the points on the ring and this could be another, right? This is like a cross section. So the distance at any point is once again by Pythagorean Theorem because this is also r. This distance is the √h2+r2. It is same thing as that. So it’s d2 and that’s equal to k times a charge in the ring times our test charge divided by d2. Well d = √h2+r2 so if we squared that, it’s just becomes h2+r2.

And if we want to know the electric field created by that ring, 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 the charge in the ring divided by h2+r2. Now what is the y component of the charge in the ring? Well, it’s going to be this, right? Well we just figured out is a magnitude of essentially this vector, right? But we want its y component because all of the x components just cancel out so this is going to be times cosine of theta and we figured out that cosine of that is essentially this, so we multiply times that.

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/√h2+r2. We could simplify this a little bit. The denominator becomes (h2+r2)3/2 and what’s the numerator? We have Kh and then the charge in the ring which we solve up here. So that’s 2∏r dr.

So we have just calculated the y component, the vertical component of the electric field at h units above the plate, not from the entire plate just 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 of this video just to give you break and myself a break. I will continue in the next but you can imagine what we are going to do now. We just figured out the electric field created by just this ring, right? So now we can integrate across the entire plane. We could sum up all the rings of radius infinity all the way down to zero and that will 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.

Transcription by: Scribe4you Transcription Services