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GL
University of California, Berkeley

In physics, a charged disc is a charged ring of radius.

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Elyse G.

Cornell University

Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Jared E.

University of Winnipeg

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Video Transcript

now the magnitude of the electric field generated by that little cut we've taken. E. It's simply K times dick over de Squared where d is this distance here. And, of course, this distance. Here's our so D squared. It's simply the square root of X squared plus r squared. We keep our K and we know are charged kick. It's sigma r D r d theta. And so we have RD the component of the electric field or the electric field generated by this little cut of the ring we've taken. We recall when we did the problem with the ring, we look at the top and bottom and we saw that the bottom point will generate an electric field in that direction, and the top point will generate an electric field in that direction. These white components cancel each other out so we only have an X component. And that's true all around the ring because whatever is not in the X direction generated by this point here will be canceled out by some point on the opposite side of the ring. So really, we only care about D E X, which is simply d E. Times CO sign of data which his case sigma R D r D theta and you'll know I made a mistake here. They should not be the square root because it's d squared where X squared plus r squared times cosine theta is X And here we have the square root of X squared plus are square or simplifying it. Okay, Sigma x r d r d theta over x squared plus r squared to the three halfs. So now we know the electric field which will be in the X direction more appropriately. I should write e of X because we're measuring from a distance X Okay, we'll just need to integrate this expression that we've come up with over here. First, I'll pull out the constants K, Sigma and X. And over here we have our over X squared plus r squared to the three halves D r D theta. Now we need to integrate twice. We need to integrate around each end of the each ring from 0 to 2 pi, and then we need to integrate the rings outward from zero tow R equals a well, we can see in this expression we have no data is just the D theta so that inter girls easy. We simply get to pie So we have two pi que sigma x Then we're gonna integrate from zero to a or D r of our over X squared plus r squared to the three halves. And when we saw this integral, we'll see the electric field generated by this disc given by this expression one over X minus one with square root of X squared plus a squared which we can simplify a little bit. Two pi que sigma one minus X over the square root of X squared plus a squared and in the extraction. And so we have found the electric field generated by a charge disk at a point along the axis passing through the center of that disk. We can also look at the units here. Notice in here. This is unit list is just one. And here we have a length over a length. So that's unit lists. Que is in terms as units of Newton's meter squared per Coolum squared and sigma is columns per meter squared, which gives us Newtons per Coolum exactly what we expect for the electric field. Now, as I mentioned in an earlier video, we had actually calculated the electric field generated by a single ring, and this expression in the box is what we got so we can utilize this as a ZAY said to do the disc problem. And all we have to do is take care of the queue, while the total charge is what What we got to remember now. The ring is a two dimensional object because we want to do a disk. So those two dimensions are circumference and its thickness D are all multiplied by the charge density in two dimensions sigma. And so our earring becomes two pi Okay, X sigma RDR over X squared plus R squared to the three halves in the X direction, and you'll notice this is the exact same expression we just got after doing the data integration for the full problem. So now we just have to integrate over our and we'll get this previous result we saw before. One last thing will add, So what happens when we move very far away from the disk? In other words, when X is much, much greater in the radius of the disk. So let's rewrite this well factor in X out of our square root so that we get one over one plus a over X squared. Remember, A over X is much, much less than one. And so we can tailor expand that term, Mr. And to First Order or to the lowest order in a Rx. We get one minus one half a squared over x squared. We see these winds cancel. And so we're just left with two pi ke sigma times a a squared over two x squared. Now the twos cancel and I'll write it like this. Okay, Sigma pi a squared over x squared. But notice this term here, sigma is our surface charge Density and pay squared is simply the area of the ring or of the disc. And so Sigma pi a squared. It's simply the total charge Q And so when X is much, much greater than a our field goes to que que over x squared. In other words, when we're really far away, this disk just looks like a point charge and we recover the equation for a point charge. One last thing to note is what happens when a goes to infinity. In other words, When our disk is very, very large, it's almost an infinite sheep. Well, if it becomes large, we'll see that this term here become zero because the denominator grows very rapidly. So if we get rid of that, we see our electric field becomes simply to pie que sigma or recalling the K is equal to 1/4 pipes on not our electric field sigma over two Epsilon. Not This is a very famous expression, and we will drive it again in a later video using a different method. But this is the electric field due to a knee infinite charge sheet of uniformed charge, and we'll know it doesn't matter how far away you get because you can't really get away from infinity.

GL
University of California, Berkeley
Elyse G.

Cornell University

Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Jared E.

University of Winnipeg