Calculation of electric flux on trapezoidal surface

In summary, the conversation discusses the calculation of electric flux through a trapezoidal surface and the confusion over the equality of the flux in and out due to differing surface areas. The concept of a homogeneous electric field and the use of Gauss's Law to calculate the total flux over the entire closed boundary surface is also mentioned. The importance of considering the dot product of the electric field and the local normal at each point on each surface is emphasized.
  • #1
annamal
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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.
Screen Shot 2022-04-22 at 10.16.09 PM.png
 
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  • #2
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.
 
  • #3
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.
 
  • #4
Ibix said:
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.
 
  • #5
vanhees71 said:
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.
 
  • #6
annamal said:
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.
 
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  • #7
Ibix said:
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...
 
  • #8
How would you do it for this field? The e-field isn't always perpendicular to the surfaces it crosses, is it?

20220423_122823.jpg


PS: Or this one, a simpler version

20220423_125126.jpg
 
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  • #9
annamal said:
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.
 
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  • #10
Ibix said:
The field lines spread out.
Well then the areas of the other faces of the trapezoid have to be taken into account as well.
 
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  • #11
DaveE said:
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)
 
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  • #12
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.
 
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Related to Calculation of electric flux on trapezoidal surface

1. What is electric flux and why is it important?

Electric flux is a measure of the amount of electric field passing through a given surface. It is important because it helps us understand the behavior of electric fields and how they interact with different surfaces and objects.

2. How is electric flux calculated on a trapezoidal surface?

To calculate electric flux on a trapezoidal surface, we use the formula Φ = E * A * cos(θ), where Φ is the electric flux, E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface.

3. What factors can affect the electric flux on a trapezoidal surface?

The electric flux on a trapezoidal surface can be affected by the strength and direction of the electric field, the size and shape of the surface, and the angle between the electric field and the surface. It can also be affected by the presence of other charges or objects nearby.

4. How is the direction of electric flux determined on a trapezoidal surface?

The direction of electric flux on a trapezoidal surface is determined by the direction of the electric field and the orientation of the surface. The flux will be positive if the electric field and surface are in the same direction, and negative if they are in opposite directions.

5. Can electric flux be negative on a trapezoidal surface?

Yes, electric flux can be negative on a trapezoidal surface if the electric field and surface are in opposite directions. This indicates that the electric field is passing through the surface in the opposite direction of the normal, resulting in a net outward flux.

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