In this video we're going to focus on electric flux. So let's talk about what it is and how to calculate it. Imagine if we have a surface with an electric field emanating out of the surface. And that surface has an area which we'll call A. The product of this perpendicular electric field, it's perpendicular to the surface. The product of the electric field times the area is equal to the electric flux. Phi is the angle between the normal line and the electric field. Because the normal line and electric field are parallel to each other, the electric flux the angle phi is equal to 0, and cosine 0 is 1. Now what if the electric field is not perpendicular to the surface? What happens in that case? So let's draw the normal line. And here's the line parallel to the surface. And let's say the electric field is at an angle. So this is the angle phi, this is the angle theta. So phi is between the normal line and the electric field vector. In this case, the electric flux is going to be equal to Ea cosine phi. The electric flux has its maximum value when this angle is zero. Now what happens if the electric field vector is parallel to the surface? or perpendicular to the normal line. So let's say the electric field vector is here. In this case, the angle between the normal line and the electric field is 90. The electric flux is equal to Ea cosine phi. Cosine 90 is equal to 0. So the electric flux is 0 in this case. Let's try this problem. What is the electric flux through a sphere of radius 4 meters that contains a 50 microclume charge and a negative 50 microclume charge at its center? Let's draw a picture. Here we have a sphere and there's a positive charge inside the sphere. The electric field will always be perpendicular to the surface. The flux is simply the electric field times the area. In the case of a sphere, it's always perpendicular to the surface. So it's going to be E times A, where A is the surface area of a sphere. So let's say the sphere has a radius R. The surface area is going to be 4 pi times R squared. Now, the electric field produced by a sphere, you can treat it as if it's a point charge, particularly if you want to calculate the electric field outside of a sphere. And so, that's equal to K times Q, which is the charge in the sphere, divided by R squared. So, notice that we can cancel R squared. And k is 1 over 4 pi times epsilon sub naught. So we can also cancel the 4 pi. So therefore, the electric flux in the sphere is equal to the charge enclosed by the sphere divided by epsilon sub naught. This equation is basically associated with Gauss's law, which states that the electric flux is proportional to the charge enclosed by the object. So, it's just going to be 50 times 10 to the minus 6 divided by 8.85 times 10 to the minus 12. So, you should get 5.65 times 10 to the 6th power. Newton's times square meters per coulomb. So that's the electric flux. Now here's a question for you. Is the answer positive or negative? The answer is positive because if you place a positive charge inside the sphere, the electric flux is an outward flux. For part B, the magnitude is the same. But if you put a negative charge inside the sphere, then the electric field enters the sphere. So we have a negative inward electric flux. So the answer for part B is the same value, but negative 5.65 instead of positive 5.65. Now, let's say if we have a horizontal disk. Let me draw it better. And it's the shape of a circle. And this disk has a radius of, let's say, 3 meters. Here's a normal line. And let's say the electric field. is in this direction. It passes through the disk all throughout the surface of the disk but in that direction. The angle between the electric field vector and the plane of the disk, let's say it's 30 degrees. Calculate the flux, the electric flux passing through the surface. Feel free to pause the video. And let's say the electric field, let's give it a value of 100 Newtons per Coulomb. So the electric flux is equal to the electric field times the area of the disk times cosine of phi. Now, the angle that we have is theta, not phi. Phi is this angle. It's the complementary angle. So phi is going to be 90 minus 30, so it's 60 degrees. The area of the disk is pi r squared. So this is going to be E times pi r squared times cosine phi. The electric field is 100. r squared is 9, 3 squared is 9, and then cosine 60. Cosine 60 is 1 half, and 100 times 9 is 900. Half of 900 is 450. So the electric flux is 450 pi. So this is the answer. Now let's try another problem. So let's say if we have a cube that looks like this, and we're going to place a positive charge at the center of the cube. What is the total electric flux that emanates outside of the cube? How can we calculate the total electric flux? Now, for a problem like this, you don't want to find the electric field that passes through each surface. That's going to be difficult because in some areas, the electric field will be at an angle. and so that's going to be very difficult the best way to answer this problem is to use Gauss's law where the electric flux is simply equal to the charge that is enclosed by the surface divided by epsilon sub naught this equation makes these problems a lot easier to solve So let's say the charge is 30 microcouplems. So all you got to do is take the 30 microcouplem charge, which is 30 times 10 to the minus 6 and divide it by epsilon to the naught. So that's going to give you the total electric flux that passes through the cube. And this is an outward electric flux, so the answer is positive. So you should get. 3.39 times 10 to the 6. So that's the electric flux. Now, what if we wanted to calculate the electric flux through one of the six faces in the cube? If you need to do that, simply multiply your answer by 1 over 6, or divide it by 6. So the electric flux that passes through each surface, or each face of the cube, is 5.65 times 10 to the 5 Newtons times square meters per coulomb. Now, let's draw another cube. So this time, there's not going to be any charge inside the cube. Now, the length of each side of the cube is 3 meters. Now let's say that there's an electric field. that passes through this sphere from the bottom face to the top face and let's say the electric field is also perpendicular to the plane of the top part of the cube in the bottom part of the cube which means that it's parallel to the normal line so what is the electric flux through this object how would you calculate it So the electric flux on the right side and on the left side is zero because it's not perpendicular to the plane of those sides. The electric flux in the front and in the back is also zero. Now what about the top and the bottom phase? What is the electric flux in those regions? So let's focus on the top part. The electric flux that passes through it is going to be the product of E times the area of that surface. So it's going to be positive Ea. The reason why it's positive is because the electric field is leaving the closed object. It's escaping from it. So it's an outward flux. So it's positive. Now what about the bottom surface? Notice that the electric field vector. enters the object so it's an inward flux so the electric flux is e times a but it's going to be negative EA so to find a total electric flux is the sum of these two values positive EA and negative EA they cancel and so the net electric flux is therefore equal to 0 because the flux that enters is equal to the flux that leaves And it makes sense, because based on Gauss's law, the total electric flux that passes through a closed object is equal to the charge enclosed divided by epsilon zero. Now there is no charge enclosed in this object, so Q is zero. And if Q is zero, the flux has to be zero.