Electric Field due to Hollow Shell

by Patrick Ford
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Hey, guys, let's go ahead and check out this problem. This is actually a very common type of problem that you'll see in Gaza's law. And it's very important that you know that. So watch this video a couple times, just in case you didn't catch everything. But basically what we're gonna dio is there gonna be evaluating with the electric field is in these particular three regions surrounding this conducting spherical shell. One of them is gonna be inside of the shell, but outside the point charge, one of them part B, is gonna be within this conducting spherical shell, and then part C is going to be outside outside of this radius, this outer radius be here. So if we're trying to figure out what the electric field is, then one thing we can use we have, We have charges and we can use gas is law. God's law helps us figure what the electric field is. But first, what we have to do is we have to draw a Gaussian surface and that Gaussian surface for a point charge. Hopefully, you guys know by now is going to be a perfect sphere. We covered in the last in the previous video. But basically the reason for that is that the electric field lines for this point charge point radiantly outwards. And that's also where the normals of the surfaces also points. So basically, these electric field lines and the normals always point in the same direction, no matter where you are on the Gaussian surface. So that means that if we're trying to evaluate the flux, which is going to tell us what the electric field is due to the charge that's enclosed, that's gonna be the electric field times the area of this Gaussian surface times the cosine of theta. But what happens is because thes things always point in the same direction. That means the cosine of the angle is just one, because data at anywhere along the Gaussian surface between the electric field and the area vector, since they always point in the same direction, that data is just equal to zero. So coastline zero just one. Now, gasses Law tells us that the total amount of flux through this Gaussian surface that I've constructed is equal to the charge and closed, divided by epsilon not, and what we're supposed to find is what this electric field is. So the only thing we have to dio is figure out what the area of our Gaussian surfaces. So I'm just gonna go ahead and say that the Gaussian surface has a radius off our little art. It's just some arbitrary thing I picked. If I picked it out here and drawn it a little bit wider, then that would be our right. So it's whatever basically you pick and the area of that surface is just four pi r squared because that's the surface area of a sphere. Now all we have to do is just move that area over to the other side and we get that the electric field is equal to Well, the charge enclosed is just gonna be this little Q right here. So it's little Q divided by Epsilon Not, but we have to divide it by this one over four pi r squared. So one of four pi r squared, and we can actually rewrite this as one over four pi epsilon, not times Q. Over R squared or another way that you might see that is just cake you over our square. All of these are valid representations of the electric field, which is exactly what you would expect for just a point charge, because that's the in charge. That's the enclosed charge of this Gaussian surface. It's just a point charge. Okay, so this is the area does the answer to part A in part B. We're now going to be looking at inside of the spherical conducting shell. So we have to construct another Gaussian surface, and the Gaussian surface is gonna be a sphere just like it was before. So now we can actually use a shortcut, because if we're trying to figure out the electric field and we know that we're inside of a conductor, remember how we talked about the electric field inside of a conductor is always equal to zero. This is an important fact that you must remember if you ever trying to value the electric field and you're inside of a conductor, that's going to be equal to zero. So some of you might be now confused. So if this is inside a conductor so this is actually answer, by the way. Now, some of you guys might be really confused because you say the galaxy in surface that we have just constructed right? If you imagine this Gaussian surface, what is the charge that's enclosed? It's this plus. Q. So how could the electric field possibly be? Zero? Well, what? Turns out what happens is that the enclosed charge is equal to zero. Because even though you have this positive point charge that exists in the center of this galaxy in surface, what ends up happening is that the electric field gets set up inside of this conductor so that there is a basically a distribution of charges that equals negative que that gets set up inside of the inside walls of the conductor. So what happens is you have a plus Q from the point charge, right, so this plus Q is the point charge that's at the center. But what happens is that point charge basically sets up an electric field inside of the conductor that cancels it out. So this negative que here is on the inside wall of this conductor. So that means that if you take a look at this Gaussian surface right now, the total amount of charge that's enclosed within this Gaussian surface is the negative. Q. That's distributed evenly on the walls of the conductor and the positive. Q. That's a dissenter of this Gaussian surface. So what's the total amount of charge that's enclosed? It's just zero. So that's why the electric field is zero. This is a very, very important thing. Watch that explanation over again. So the positive Que sets up a field inside of the conductor that cancels out the electric fields to the electric field to zero. And because of that, there's a distribution of negative que that gets set up on the inside walls. Now what happens is that due to charge conservation, the fact that you can't just generate a negative que out of nowhere, that means that a positive Q has to get set up on the outside walls. So this is what happens. It basically just polarizes these charges so that they cancel out the electric field and it's equal to zero. Alright, so go ahead and watch those videos on the electric field inside of conductor. Just so you make sense of that. Okay, so basically, the answer is that inside of the electric inside of the conductor, electric field is equal to zero, as it always should be. All right, in the last part. Now, as we have to figure out what the E at part C is so er greater than be right. So we have to set up our galaxy and surface. The galaxy and surface that we're gonna use is gonna be all the way out here, right? And it doesn't matter how far we make it. But what we do know is that this distance out here is just gonna be our And so this is the Gaussian surface that we've that we've constructed. So that means that the fi, the total amount of electric flux is e a co sign of data. Right? We know that data is always just gonna be equal to one. And all we have to do is just that this e a here is equal to the Cuban closed over Absalon. Not so. Let's take a look. What is the amount of charge that's enclosed within our Gaussian surface? We have a plus Q that gets set up on the outer wall of the conductor theme. Inside of the wall gets a negative Q. And then we have this positive Q. That's at the center of that sphere. So that means that this Q eyes actually let's see this Q and closed right here is just equal to Q minus. Q plus Q. So let me go ahead and remove myself here. So I've got Q minus Q plus Q. So that means that these charges just go away and you end up with just a positive Q over here. So that means. But the electric flux is just gonna equal to, Let's see, the area of the spherical Gaussian surface that we've constructed is four pi r squared. Just it always was. And that means that the electric field, actually let's see, means you wanna move this over. That means theological field is equal to plus Q over Epsilon not and then divided by one one of the four pi r squared. So that means that that's just equal to cake You over r squared, just as it is for a point charge. So it turns out that this spherical conductor doesn't actually really do anything other than set up a charge or set up a distribution of charges so that it cancels out the electric field inside of this. But then anywhere else after that, it basically just starts to look like a point. Charge again. Right. So make sure you go un understand each one of these parts here. By the way, this is actually supposed to be part C over here. And let me know if any one of these processes or steps confused you just drop me a link in the comments and I'll answer your questions. All right, let me know if you have any questions. See you later.