A flat sheet of paper of area 0.250 m2 is oriented so that the no...

Question

A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. 
(a) Find the magnitude of the electric flux through the sheet. 
(b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?

Accepted Answer

Hi, everyone in this practice, we're asked to calculate the electric flux through the surface where we have a planer square surface with an area of 0.3 m squared immersed in a uniform electric field of manitou 20 newton per Coolum where the normal to the square surface forms an angle of 45 degrees with the electric field itself. After calculating the electric box were asked to determine why is the electric flux itself independent of the geometrical shape of the surface? So we're gonna just start with identifying all the things that is given in our problem statement. So first, we have E which is the electric field of 20 Newton per columns. And then we will also have the area which is 0.3 m squared. And lastly, we have the five value which is going to be the normal angle, which is going to actually be 45 degrees. And we want to recall that the way we want to calculate the electric flux E or five E is by multiplying the electric field with the area and the angle, the call sign off the angle between the normal to the square surface or the normal to the surface with the electric field. So with this formula recall that the phi is the angle between the electric field and normal to the surface, the ease the magnitude of the electric field and the A is the area of the planer surface. So we have everything that is given in this uh part. So we can actually solve part one. So the is going to be 20 newtons per cool. Um And then the area 0.3 m square and the coast sine is cosine of 45 degrees. And that all came out to be 4. newton meters squared per Coolum. That will actually be the answer for our first part. So we can kinda just eliminate option C and option D just like so and then next, we want to move on to the second part. So we know that the electric flux is given by this formula right here. And when the magnitude of the area is the same or where we still have a planer surface, a planer square surface with the same 0.3 m square area, then the variables that will change is only the electric field and the angle how. However, in this case, the equation or the formula for the electric flux does not show any dependence on shape. So the flux will actually be the same as long as we have the same area in which in this case, we do. And in that planer square surface, the flux will always be the same as long as the electric field and defy are constant. So I'm just gonna write down that the electric flux equation has no dependence on shape. As long as we are considering the same area which in this case, we are considering the same planer square surface. So I'm gonna write down as long as E and the five value are constant Just like. So, so five e as independent of sheep after sheet, as long as E N FI are constant at all points on the sheet. So essentially that will correspond to option be where it will be independent of the shape because the electric field and the angle phi are constant at all points on the surface. And that will be all for this particular video. If you guys have some sort of confusion on this particular topic, make sure to check out our other lesson videos on similar topics and that will be all for this particular video. Thank you.

A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?

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