I Calculation of electric flux on trapezoidal surface

[annamal]

I am confused at this calculation of the electric flux through a trapezoidal surface. The flux in should equal the flux out.
The flux in equals -E*A1 where A1 is the area of the bottom of the trapezoid. The flux out equals E*A2 where A2 is the area of the top of the trapezoid. But the two fluxes aren't equal due to differing areas. We are only considering the bottoms of the trapezoid since the electric field only flows there as shown in the image.

 

[Ibix]

Why do you think the flux out is E×A2? You said the electric field only exists within the area of the bottom surface of the trapezoid.

 

[vanhees71]

Obviously you are looking at a homogeneous electrical field. Why do you think the flux calculated through only 2 boundary surfaces of your volume must cancel? You have to calculate the flux over the entire closed boundary surface of your volume. Of course due to Gauss's Law you know that it must be 0 for a homogeneous electric field, because $\vec{\nabla} \cdot \vec{E}=0$ everywhere.

 

[annamal]

Why do you think the flux out is E×A2? You said the electric field only exists within the area of the bottom surface of the trapezoid.

Because the electric field is E and area of the top surface where the flux is coming out of is A2.

 

[annamal]

Obviously you are looking at a homogeneous electrical field. Why do you think the flux calculated through only 2 boundary surfaces of your volume must cancel? You have to calculate the flux over the entire closed boundary surface of your volume. Of course due to Gauss's Law you know that it must be 0 for a homogeneous electric field, because $\vec{\nabla} \cdot \vec{E}=0$ everywhere.

Because the electric field only goes through the bottom of the trapezoid. There is no electric field to the left or right of it. That is, there isn't an electric field that goes through the whole surface only a certain part of it.

 

[Ibix]

Because the electric field is E and area of the top surface where the flux is coming out of is A2.

This must be wrong - your own calculations tell you so.

Either the field goes through the sides as well as the base, or the field lines diverge and go through all of the top at a lower value of $E$, or the field only goes through a part of the top. The situation you are describing, where you have a homogeneous electric field over a small area and then the same strength homogeneous field over a larger area is only possible if there are charges inside your trapezoid. That is what the non-zero integral is telling you.

 

[annamal]

This must be wrong - your own calculations tell you so.

Either the field goes through the sides as well as the base, or the field lines diverge and go through all of the top at a lower value of $E$, or the field only goes through a part of the top. The situation you are describing, where you have a homogeneous electric field over a small area and then the same strength homogeneous field over a larger area is only possible if there are charges inside your trapezoid. That is what the non-zero integral is telling you.

I am not sure how it is possible to go through all of the top at a lower value of E...

 

[DaveE]

How would you do it for this field? The e-field isn't always perpendicular to the surfaces it crosses, is it?

PS: Or this one, a simpler version

 

[Ibix]

I am not sure how it is possible to go through all of the top at a lower value of E...

The field lines spread out.

 

[annamal]

The field lines spread out.

Well then the areas of the other faces of the trapezoid have to be taken into account as well.

 

[annamal]

How would you do it for this field? The e-field isn't always perpendicular to the surfaces it crosses, is it?

View attachment 300460

PS: Or this one, a simpler version

View attachment 300463

Flux is a dot product so you would have to cos(angle in between)

 

[rude man]

Your E field is uniformly in the direction you show. Don't be confused by other E field orientations.

As post 11 says you need to take the dot product of the E field, and the local normal, at each point on each surface. On the top & bottom ones the E field and normal are aligned but on the sides they are not. Whence the cosine mentioned.