Skip to main content
Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Flux Through Angled Surface

Patrick Ford
Was this helpful?
Alright, guys, let's get some more practice with this electric flux stuff. So we have this electric field and we have the area of the surface and we're told to find with the magnitude of the electric flux is through the surface depicted. So let's go ahead and start with the equation. Three. Electric flux is just going to be e magnitude of the electric field times the area times the cosine of the angle feta. Now, if you take a look at this, we actually told what the electric field is. We have that and we're told the area of the surface is just 1 m squared and it's in the right units. So the only thing we have to find is we have to find it with the cosine of the angle is, And what I want you guys to do is remember that we were never We're dealing with co signs of angles, especially with this electric flux. This angle represents the angle between the area or the normal of the surface and the electric field. So if we sort of extend these electric field lines out like this than what we really need to Dio is we need to find what the normal of the surface is. The normal of the surface is always perpendicular to the surface itself. So it points out in that direction. And the reason we're told to find the magnitude of the electric flux is because we actually have no idea. We have no information of whether the normal is pointing this way or this way. So we're just gonna go ahead and assume that it's positive by calculating the magnitude of the electric flux anyways. So the angle that we really need is actually the angle between this vector and the normal of the surface, which is not the same as this 30 degrees. This 30 degrees is the angle between the electric field and the surface itself, not the perpendicular of that surface. So really, we're not using this 30 degrees. Instead, we're just going to use 30 plus 60 equals 90 so this angle is actually 60 degrees, which is this angle as well by the geometry, right? These things are all opposite angles. So what we really need to dio is that the electric flux is going to be 100 Newtons per Coolum. The area is just 1 m squared. And now we have to do just to co sign of 60 degrees. Now, cost of 60 is just one half. So that means the electric flux is just half of 100 which is 50. And that electric flux is Newton meters squared per cool. Um, just in case you needed the units. Alright, so that's the answer. Let me know if you guys have any questions.