Hey, guys. So for this example, we're going to build off of something that we talked about in the last example. We were asked to find out what the electric flux is through a spherical shell of radius r due to some point charge 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 times A times the cosine of θ, in which this θ 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 point outwards, but at the same time, the normal vector of a spherical shell also always points outward directly at the surface. So these perpendicular lines here, the normals, are always going to point away from that spherical shell, which means that the cosine of the angle right here, this θ, wherever you look along the surface, these field lines and the normal always point in the same exact direction. Since the θ is always equal to zero, that means this cosine of θ is always just going to be equal to one, no matter where you are that you're looking at. So that means that the total amount of electric flux is going to be the total amount of electric field times the total amount of area.

So the electric field, let's see, the electric field due to a point charge is kQ/R2. So at some distance R, that's going to be R2. So that's the electric field due to this point charge and the area of the spherical shell, the surface area of a sphere, is just 4πR2. So if you go ahead and put those two things together, that means that the electric flux is going to be kQ/R2×4πR2. Now what happens is the R2 will cancel. And so we end up just getting that the flux is equal to 4πkQ.

This is the answer for 4πkQ. There's actually another way that we could write this because we know that this k has a relationship with that ε₀, the permittivity constant. Remember that this k is equal to 14πε₀. The reason we want to make this substitution is that now the 4π will cancel, so this actually will turn into Q/ε₀. This is a really important result. We're going to talk about it much later when we get to Gauss's law. That's just another way you could express this, by the way. So, both of these things would actually be perfectly valid if you were given this on a test or anything like that.

All right, This is the answer, or this is the answer. Let me know if you guys have any questions.