 A cylindrical surface is coaxial with the y-axis and has its center at the origin. It's 4 meters in height, we can see it scores from 2 to minus 2, 4 meters in height with a base area of 10 meters square. So this area right here, this is 10 meter square and so is this area, this is also 10 meter square. A non uniform electric field which is 3yj cap is present in this region. What is a net charge enclosed by the cylinder? Alright, like always pause the video and give this one a try first. Alright, hopefully you have given this a try. Now we need to figure out the net charge q net and we know there is some electric field which is 3yj cap. So let's draw that electric field that will be that will be somewhat like this right 3y and it's non uniform so its magnitude will be changing, it will be different, it will be different in different areas right, it's non-uniform. So this is the electric field, it's non-uniform and we need to figure out the q net right. So if we think about what kind of principle would be applicable here, we know that there will be some net charge enclosed because there is an electric field inside and this electric field will pass through the top surface, it will be passing through the bottom surface. So there will be some flux, there will be a net flux. So if you think about the Gaussian equation, if you think about Gauss law that really that really is finite, the net flux which is equal to, this is equal to q enclosed, the enclosed charge or let's call it q net divided by epsilon 0. So we need to figure out the net flux through the cylindrical surface and when we do that, multiply that with epsilon 0 and you get your q 0, q net. So how do we figure out the net flux? Well let's look at the cylindrical surface. This electric field it's parallel to the y-axis right and there could be electric field passing through the top surface, the bottom surface, through its sides. So let's think about the sides first. Will there be any electric field passing through the sides? Considering the electric field is parallel to the y-axis think about it. Well not really right, the area vector of the surface on the sides, it will be outwards like this. This will be the area vector and you have electric field which is parallel to the surface or perpendicular to the area vector. So the angle between them, this used to be, this used to be theta right, this theta is really 90 degrees and when we figure out the flux, it's e dot a, there is a factor of cos theta, so cos 90, cos 90 is 0. So there is no flux through the sides. So phi net will really just be the addition of phi top, the flux to the top surface plus flux through the bottom surface. So let's figure out the flux, the area vector for the top surface will be pointing outwards and the area vector for the bottom surface it will be again pointing outwards but this is in the negative z direction, so we need to keep that in mind. So phi top flux to the top surface, this is electric field which is 3y and in this case y would be 2 because the surface lies at the coordinate 2. So 3 into 2, this is j cap into into the area, right into the area. Area is base area is 10, so area is 10 meter square j cap. So this comes out to be equal to 63 into 26 into 1060 Newton meter square per coulomb. This is a flux through the top surface. Now flux through the bottom surface that would be 3 into minus 2 because this surface, the bottom surface it lies at the coordinate minus 2. So 3 into minus 2 is minus 6, minus 6 j cap into 10 meter square, 10 meter square minus j cap, minus j cap because the area vector points in the negative, the negative y, the negative y direction. So this is again 60, so you add 60 Newton meter square per coulomb. The total flux comes out to be equal to 120 Newton meter square per coulomb. So this is a total flux and when we substitute the total flux in this equation we get q net. So q net comes out to be equal to 120 epsilon naught coulombs.