Top view of a Gaussian surface in a uniform electric field

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In a uniform electric field, the electric flux through a Gaussian surface is calculated using the formula Φ_E = E · A, where the area vector A is normal to the surface. The professor's calculations for the electric flux through the faces of the surface yield a=EAcosθ for face 1, indicating that the angle θ is measured with respect to the normal (adjacent side). For face 2, the flux is negative, represented as b=-EAsinθ, reflecting the direction of the electric field relative to the area vector. The confusion arises from understanding which side of the triangle corresponds to the cosine or sine function, with cosine being used for the adjacent side and sine for the opposite. Overall, the key takeaway is that the angle θ's relationship to the surface's orientation determines the use of cosine or sine in the flux calculations.
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The field makes an angle θ with side 1 and the area of each face is A. In symbolic form, find the electric flux through (a) face 1, (b) face 2, (c) face 3, (d) face 4 and (e) top and bottom.

My professor got:
a=EAcosθ
b=-EAsinθ
c=-EAcosθ
d=EAsinθ
e= 0
I understant why e=0 but for the other sides, for some reason I am not understanding why she got for example a=EAcosθ and not sinθ. Which part of the triangle am I trying to find out? Would it be the adjacent for each or the opposite side for each?

I hope my question makes sense.

gaussiansurface_zps89f0cc05.jpg

 
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brojas7 said:
The field makes an angle θ with side 1 and the area of each face is A. In symbolic form, find the electric flux through (a) face 1, (b) face 2, (c) face 3, (d) face 4 and (e) top and bottom.

My professor got:
a=EAcosθ
b=-EAsinθ
c=-EAcosθ
d=EAsinθ
e= 0
I understant why e=0 but for the other sides, for some reason I am not understanding why she got for example a=EAcosθ and not sinθ. Which part of the triangle am I trying to find out? Would it be the adjacent for each or the opposite side for each?

I hope my question makes sense.

gaussiansurface_zps89f0cc05.jpg
For a uniform electric field, and a flat surface, the electric flux \Phi_E through a surface with area A is

\Phi_E = \vec E \cdot \vec A

Notice I'm using vector notation. The direction of \vec A is normal to the surface. And there's a loose convention that for a closed surface, the surface vector points out of the surface. (That's just a convention though, although that conversion is used here.)

Another thing to notice is that we're dealing with the "dot" product. Another way to represent the dot product is

\vec E \cdot \vec A = EA \cos \theta

The dot product measures (in part) how parallel two vectors are. If they are perfectly parallel, the dot product is simply EA. If they are perpendicular the dot product is 0. If they are in perfectly opposite directions, the dot product is -EA.

If it helps, imagine the situation where \theta is 0, and other situation where it is 90o. If \theta is 0, a maximum amount of flux would pass through side 1. How many flux lines would pass though side 1 when \theta is 90o? So would cosine or sine be used to represent that?
 
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