Find the electric fluxes ΦA to ΦE through surfaces A to E in FIGU...

Question

Find the electric fluxes ΦA to ΦE through surfaces A to E in FIGURE P24.29.

Accepted Answer

Hello everyone. Let's go through this practice problem determine the electric flux passing through five different surfaces labeled P through T depicted in the diagram below. OK. So this problem isn't too long or too hard, but it does require you to really understand the rules of electric flux. Recall that electric flux usually symbolized with this P thigh is equal to and has a magnitude of the strength of the electric field multiplied by the surface area that it passes through, multiplied by the cosine of the angle theta where theta represents the angle between the, the direction of the E field itself and the vector normal to the surface of the area. What that's basically saying is that there is only ever electric, there's only ever electric flux if the E field is directly piercing the surface or if some component of it is passing right through the surface. But if the electric field lines are kind of passing by the surface, if there is a 90 degree angle between the normal to the surface and the electric field lines, then there is no flux whatsoever. And as long as you understand that, then right away, we can actually answer most of the problem pretty easily. Because if we look at the diagram, we can see that these electric field lines are pointing to the right. They're symbolized with these dashed purple arrows all pointing to the right. And what we can see is that three, three of these surfaces, QR and T are all positioned in such a way that the electric field line is not piercing them at all. Like this bottom face here are, we can see that it's normal has 100 90 degree angle with the electric field lines. And we get the same kind of thing going on with Q and T. With the way the diagram is set up, we can see that the electric field lines are only going to have any component at all that's piercing through sides P and S. So that tells us that right off the bat, we know that the electric flux through side R is zero, the electric flux through side Q is zero and the electric flux through side T is also zero. So right off the bat, we've answered three of the five parts of the problem. But the last two remaining sides P and S are going to be a little bit more involved because we can actually see that there will be some flux from that. So let's take a look at this and let's start with side S so the flex through side S so as we discussed earlier this is equal to the magnitude of the electric field or 350 new in per Coolum multiplied by the surface area that the lines are passing through. Now, we can see from the diagram, it's going to be this little square right here. And we're given the dimensions of the square, we can see that its width is going to be 5 m. And although we are not directly given the height, we know that's also gonna be 5 m because the angles given here tell us that these triangular faces are equilateral triangles. And since one side of them is 5 m, all sides must be 5 m. So the area of this square is just going to be 5 m squared. And lastly, we multiply this by the cosine of the angle between the electric field lines and the normal to the surface. So if we take a look at the diagram, if we, if we take a look at the way the diagram is laid out, we can see that these all these little angles are going to be 60 degrees due to the fact that it's an equilateral triangle and the normal from the square is going to be pointing kind of in this diagonal direction leaving this small angle here due to the fact that all the sides are tilted by 90 degrees. So we can find this angle theta by simply taking 90 degrees and then subtracting 60 degrees and seeing what's left as a result of the tilt and we're left with 30 degrees. So we're going to use a cosine of 30 degrees. And if we put this into a calculator, then we find an electric flux of about of approximately 7600 newton meters squared for coons. Or if we wanted to write this more succinctly 7. kg newton meters squared per coons. So that is the magnitude of the flux through side S but we need to take a second and talk about the sign on these answers because of this cosine theta whether or not the answer is positive or negative is going to depend on which way, which direction we choose to have the normal vector point by convention by convention, it's most common in for cases of shapes like this three dimensional shapes to have the normal be pointing away from the shape itself. So the normal for side S is gonna be pointing away from the shape and the normal for side P is going to be kind of pointing in the opposite direction once again away from the inside of the shape. This means that the flux for side S can stay positive. So positive 7.6 kg Newton meter squared per per Coolum is our is gonna be our final answer for Phi subs. But Phi sub P is going to be a little bit different now because of the nature of the shape, the angle that we're gonna use is the same. And since the magnitude of the electric field strength is also the same and the surface area for all sides of the shape are also going to be the same. That means that the magnitude for the flux here is going to be the exact same value 7.6 kg new meter squared per Coolum. However, because side P here has its normal vector pointing in the opposite direction kind of pointing more towards the left opposing the direction of the electric field lines. That means that this is going to have to be negative because it's opposing the electric field. So negative 7.6 kg new in meter squared per Coolum is our answer for flux of side P for Sahi sub P. So we have now answered all five parts of the problem. You found five sub R five sub Q by sub T five sub S and A now by sub P. So really that is it for this problem. I hope this video helped you out. And if it did, please consider checking out some of our other tutoring videos which will give you more experience with these types of problems. But that's all for now and I hope you all have a lovely day. Bye bye.

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Find the electric fluxes ΦA to ΦE through surfaces A to E in FIGURE P24.29.

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