Learn about Proof - Advanced: Field from infinite plate - part 2 Video

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Learn about Proof - Advanced: Field from infinite plate - part 2

So, where I left off, we had this infinite plate, this is just infinite plane and it’s a charge plate with a charge density sigma. And what we did is we said okay, what we’re taking this point up here, that’s H units above the surface of our charge plate. And we wanted to figure out the electric field at that point generated by ring of radius R essentially centered at the base of where that point is above. We want to figure out what is the electric field generated by this ring at that point. And we figured out that the electric field was this. And then because we made a symmetry argument on the last video, we only care about the y component because we figured out if the electric field generated from any point, the x components cancels out because if we have point here, it will have some x component, the field x component might be in that direction to the right then you have another point out here and its x component will just cancel it out. So, we only care about the y component. So at the end, we meticulously calculated what the y component of the electric field generated by the ring is at h units above the surface. So with that out of way, let’s see if we can sum up a bunch of rings going from radius infinity to radius zero and figure out the total y component or essentially the total electric field because we realize that all the x is canceled out anyway. The total electric field at that point h units above the surface of the plane. And this is pretty much all Calculus at this point. Watch the previous video if you forgot how it was derived. There you go. Okay. So, let me redraw a little bit so that we never forget what we’re doing here because that happens. So, that’s my plane goes off in every direction. I have my point above the plane where we’re trying to figure out the electric field and we’ve come to the conclusion that the field is going to point upwards so we only care about the y component. Its h units above the surface and we’re figuring out the electric field generated by a ring around this point of radius r. And what’s the y component of that electric field and we figured out it was this. So now, what we’re going to do is take the integrals. So, the total electric field is going to be the integral from a radius of zero through radius of infinity, so we’re going to take a sum of all of the rings starting with the radius of zero all the way to the ring that has a radius of infinity because it’s in infinite plane, we’re figuring out the impact of the entire plane. To take the sum of every rings, the charge or the field entered by every ring and this is the field during by each of the rings. Kh2 pi sigma r dr/(h2+r2)3/2. Now, let simplify this a little bit. Let’s take some constants out of it just so this looks like a slightly simpler equation. So let’s take the K, I'm going to leave the two there and you’ll see why in a second. Let me take all the other constants out that we’renot integrating across. So, it’s equal to Kh pi sigma times the integral from zero to infinity of what is this? What did I leave in there? I left a 2r so we could rewrite this as 2rdr/(h2+r2)3/2 or we could think of it as the negative 3/2, right? And so what’s the entire derivative of here? Well this is essentially the reverse chain rule, right? I could make a substitution here, if you’re more comfortable using the substitution rule but you might be able to eyeball this at this point. We could make the substitution that u is equal to h2+r2, h is just a constant, right, then du/dr=2r or we could say du=2rdr. And so if we’re trying to take the entire derivative of 2rdr/(h2+r2)3/2, this is the exact same thing. Let’s take the entire derivative with this substitution, 2rdr we just show right here, that’s the same thing as du, right? So, that’s du over and this is just u3/2 which is equal to the entire derivative of u-3/2 du. And now that’s easy, this is just kind of reverse exponent rules. So that equals -2u-1/2 and we can confirm if we take the derivatives minus ½ times minus 2 is 1 and then subtract one from here, we get -3/2 and then we could add plus c but since we’re eventually going to do a definitive roll, the c is all canceled out or we could say that this is equal to -2 over the square root of h2+r2, right? So all of this stuff I did in magenta was just to figure out the entire derivative of this and we figured it out to be this, -2 over the square root of h2+r2. So, that out of the way, let’s continue evaluating our definite integral. So, this expression simplifies two. This is a marathon problem satisfying. Kh pi sigma times -2 and all of that and we’re going to evaluate the definite integral with the two boundaries, one over the square root of h2+r2, evaluated, add infinity minus it evaluated at zero. Well, what is this expression equal? What is one over the square root of h2 plus infinity? What happens when we evaluate r infinity? Well, the square root of infinity is still infinity and one over infinity is zero. So, this expression right here becomes zero when you evaluate—infinity this becomes zero minus this expression evaluated zero. So what happens when its set zero? When r2 is zero, we get one over the square root of h2, right? This becomes -2Kh pi sigma (0-1/√h2), well this equals -2Kh pi sigma times -1/h, well this minus and that minus cancel out, fair enough. And then this h and this 1/h should cancel out, fair enough. And all we’re left with after doing all of that work and I'll do it in a bright color because we’ve done a lot of work to get here is 2K pi sigma. So, that’s the end of a lot of levels. First of all, what do we even do here? We might have gotten lost in the Math. This is the net electric field, the total electric field that appoint at height h above this infinite plate that has uniform charge and the charge density is sigma. But notice this is the electric field at that point, but there’s no h in here. So, that essentially is telling us that the strength of the field is a no way dependent on how high above the field we are which tells us it’s going to be a constant field. We can be anywhere above the plate and the field will be the same, , and if we have a test charge, the force would be the same. And the only thing that the strength of the field or the strength of the exerted electrostatic force is dependent on, is the charge density, Right? This is Coulomb's constant, pi is pi, 2pie. And I think it’s kind of cool that you involve pi but that’s something else to ponder but all that matters is a charge density. So hopefully, you found that reasonably satisfying and that the big thing that we learned here is that if I have an infinite uniformly charge plate, the field when I put some distance h above that plate, it does not matter what that h is. I could be here, I could be here, I could be here, and all of those points, the field has the exact same strength or the net electrostatic force on a test charge of those points has the exact same strength and that’s kind of neat thing. And now, if you do believe everything that occurred in the last two videos, you can now believe that there are such thing as uniform electric fields and they occur between parallel plates especially far away from the boundaries. See you soon.

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So, where I left off, we had this infinite plate, this is just infinite plane and it’s a charge plate with a charge density sigma. And what we did is we said okay, what we’re taking this point up here, that’s H units above the surface of our charge plate. And we wanted to figure out the electric field at that point generated by ring of radius R essentially centered at the base of where that point is above.

We want to figure out what is the electric field generated by this ring at that point. And we figured out that the electric field was this. And then because we made a symmetry argument on the last video, we only care about the y component because we figured out if the electric field generated from any point, the x components cancels out because if we have point here, it will have some x component, the field x component might be in that direction to the right then you have another point out here and its x component will just cancel it out.

So, we only care about the y component. So at the end, we meticulously calculated what the y component of the electric field generated by the ring is at h units above the surface.
So with that out of way, let’s see if we can sum up a bunch of rings going from radius infinity to radius zero and figure out the total y component or essentially the total electric field because we realize that all the x is canceled out anyway. The total electric field at that point h units above the surface of the plane.

And this is pretty much all Calculus at this point. Watch the previous video if you forgot how it was derived. There you go. Okay. So, let me redraw a little bit so that we never forget what we’re doing here because that happens. So, that’s my plane goes off in every direction. I have my point above the plane where we’re trying to figure out the electric field and we’ve come to the conclusion that the field is going to point upwards so we only care about the y component.

Its h units above the surface and we’re figuring out the electric field generated by a ring around this point of radius r. And what’s the y component of that electric field and we figured out it was this. So now, what we’re going to do is take the integrals. So, the total electric field is going to be the integral from a radius of zero through radius of infinity, so we’re going to take a sum of all of the rings starting with the radius of zero all the way to the ring that has a radius of infinity because it’s in infinite plane, we’re figuring out the impact of the entire plane. To take the sum of every rings, the charge or the field entered by every ring and this is the field during by each of the rings.

Kh2 pi sigma r dr/(h2+r2)3/2. Now, let simplify this a little bit. Let’s take some constants out of it just so this looks like a slightly simpler equation. So let’s take the K, I'm going to leave the two there and you’ll see why in a second. Let me take all the other constants out that we’renot integrating across. So, it’s equal to Kh pi sigma times the integral from zero to infinity of what is this? What did I leave in there? I left a 2r so we could rewrite this as 2rdr/(h2+r2)3/2 or we could think of it as the negative 3/2, right?

And so what’s the entire derivative of here? Well this is essentially the reverse chain rule, right? I could make a substitution here, if you’re more comfortable using the substitution rule but you might be able to eyeball this at this point. We could make the substitution that u is equal to h2+r2, h is just a constant, right, then du/dr=2r or we could say du=2rdr. And so if we’re trying to take the entire derivative of 2rdr/(h2+r2)3/2, this is the exact same thing. Let’s take the entire derivative with this substitution, 2rdr we just show right here, that’s the same thing as du, right? So, that’s du over and this is just u3/2 which is equal to the entire derivative of u-3/2 du.

And now that’s easy, this is just kind of reverse exponent rules. So that equals -2u-1/2 and we can confirm if we take the derivatives minus ½ times minus 2 is 1 and then subtract one from here, we get -3/2 and then we could add plus c but since we’re eventually going to do a definitive roll, the c is all canceled out or we could say that this is equal to -2 over the square root of h2+r2, right? So all of this stuff I did in magenta was just to figure out the entire derivative of this and we figured it out to be this, -2 over the square root of h2+r2. So, that out of the way, let’s continue evaluating our definite integral.

So, this expression simplifies two. This is a marathon problem satisfying. Kh pi sigma times -2 and all of that and we’re going to evaluate the definite integral with the two boundaries, one over the square root of h2+r2, evaluated, add infinity minus it evaluated at zero. Well, what is this expression equal? What is one over the square root of h2 plus infinity? What happens when we evaluate r infinity?

Well, the square root of infinity is still infinity and one over infinity is zero. So, this expression right here becomes zero when you evaluate—infinity this becomes zero minus this expression evaluated zero. So what happens when its set zero? When r2 is zero, we get one over the square root of h2, right? This becomes -2Kh pi sigma (0-1/√h2), well this equals -2Kh pi sigma times -1/h, well this minus and that minus cancel out, fair enough. And then this h and this 1/h should cancel out, fair enough. And all we’re left with after doing all of that work and I'll do it in a bright color because we’ve done a lot of work to get here is 2K pi sigma.

So, that’s the end of a lot of levels. First of all, what do we even do here? We might have gotten lost in the Math. This is the net electric field, the total electric field that appoint at height h above this infinite plate that has uniform charge and the charge density is sigma. But notice this is the electric field at that point, but there’s no h in here. So, that essentially is telling us that the strength of the field is a no way dependent on how high above the field we are which tells us it’s going to be a constant field. We can be anywhere above the plate and the field will be the same, , and if we have a test charge, the force would be the same.

And the only thing that the strength of the field or the strength of the exerted electrostatic force is dependent on, is the charge density, Right? This is Coulomb's constant, pi is pi, 2pie. And I think it’s kind of cool that you involve pi but that’s something else to ponder but all that matters is a charge density.

So hopefully, you found that reasonably satisfying and that the big thing that we learned here is that if I have an infinite uniformly charge plate, the field when I put some distance h above that plate, it does not matter what that h is. I could be here, I could be here, I could be here, and all of those points, the field has the exact same strength or the net electrostatic force on a test charge of those points has the exact same strength and that’s kind of neat thing.

And now, if you do believe everything that occurred in the last two videos, you can now believe that there are such thing as uniform electric fields and they occur between parallel plates especially far away from the boundaries. See you soon.

Transcription by: Scribe4you Transcription Services