Flux Through Spherical Shell due to Point Charge

by Patrick Ford
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Hey, guys. So for this example, we're gonna build off of something that we talked about in the last example. So were asked to find out what the electric flux is through a spherical shell of radius r do some point charge there that's in the center. So we're asked for the electric flux. Let's go ahead and start with our electric flux formula. We've got E a times the co sign of theta in which this data is between the normal vector and the electric field at that specific point. So the first thing is, first, where's the electric field? Points due to a point charge? Well, at any surface here, it always points away from that point charge. Remember, the electric field lines always points outwards, but at the same time, the normal vector of a spherical shell also always points outward directly at the surface. So this perpendicular lines here, the normals are always gonna point away from that spherical shell, which means that the cosine of the angle right here this data, wherever you look along the surface, thes these field lines and the normal always point in the same exact direction. Since and since the data is always equal to zero. That means this CO sign of data is always just going to be equal toe one, no matter where you was that you're looking at. So that means that the total amount of electric flux is gonna be the total amount of electric field times the total amount of area. So the electric field, let's see theoretic field due to a point charge is one or remember, it's k times what we're gonna use little Q right now divided by r squared. So at some are distance or actually, it's gonna be That's gonna be big r squared. So that's the electric field. Due to this point charge in the area of the spherical shell, the surface area of a sphere is just four pi times r squared. So if you go ahead and put those two things together, that means that the electric flux is gonna be K times Q divided by whoops. I got big r squared times four pi and I forgot to forget to make this little are a big are. So if you caught that, if you caught that I was supposed to write big are instead of little our That was good. So that's four pi big r squared. So anyways, I've got cake over r squared times 44 pi r squared. Now what happens is the r squared will cancel. And so we end up just getting that. The flux is equal to four pi times K Times Q. And this is the answer for pie. Que times Q. There's actually another way that we could write this because we know that this K has a relationship with that epsilon, not that permitted ity constant. Or remember that this K is equal to 1/4 pi times. Epsilon. Not the reason we wanna make this substitution is because now the four pies will cancel, so this actually will turn into this will just turn into five equals. Let's see, the K will turn into this. The four pies will cancel, and we'll just get Q divided by epsilon. Not this is actually really important is a really important result. We're gonna talk about it much later when we get to Galaxies law. That's just another way you could express this, by the way. So both of these things would actually be perfectly valid on this if you were given this on a test or anything like that. All right, So this is the answer, or this is the answer. Let me know if you guys have any questions.