Electric Flux confusion and help

In summary, Homework Statement In summary, the three problems ask for the flux through different parts of a box with an electric field on one side. The first problem asks for the flux through the entire surface in terms of the area vectors and the electric field E. The second problem asks for the flux through the left side of the box in terms of dAi and Ei. The third problem asks for the flux through the right side of the box in terms of dA and E.
  • #1
jlmccart03
175
9

Homework Statement


I have three questions spanning three different situations, but each deal with the single concept of electric flux.

Problem 1: Write an expression for the net electric flux through the entire surface in terms of the area vectors and the electric field E.

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Problem 2: Consider the left side of the box as consisting of N small pieces. Led dAi represent the area of the ith small surface element on the left side of the box, and let Ei represent the electric field on that surface element. Write an expression for the net electric flux through the left side of the box in terms of dAi and Ei.

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Problem 3: Suppose that the new charge located to the right of the loop had been negative instead of positive. How would your answer to part b change, if at all? Explain.

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Homework Equations


Φ = ∫E⋅dA
Φ = Qenclosed0

The Attempt at a Solution


So for the first problem I truly have no idea how to start it. I mean I have to use the vector definition for electric flux to start writing the expressions, but I feel like its just ∫E1⋅dA1 + ∫E2⋅dA2 + ∫E3⋅dA3 since there are three pieces to this one surface. However, I think this is wrong since the integral should be over the whole surface not individual pieces like I have. Thats what dA means correct?

For problem 2 its the same issue I am having with problem 1.I don't know if it is looking for what I am doing in problem one with all the subscript 1, 2, 3 scenario or just ∫En⋅dAn. Again confused.

For problem 3 I think I have the answer as positive since the electric field is coming toward the negative charge so the positive charge would have its field lines going through the surface as positive lines go away from the charge and then there would be negative field lines going toward the negative charge which come from the positive and elsewhere so the lines would point the same direction as my dA which is to the right as I drew in the first part of the problem. Hopefully that is correct?

I know its a lot, but the flux concept is having me confused and the equations are especially daunting.
 
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  • #2
Maybe your confusion is because you are mixing two diferent concepts. In your equations there are two diferent concepts and you need to understand them very well.

1) flux. You can think of it as being the number of field lines passing throw a surface. The surface integral is the general case but when E is constant and the surface is flat the flux is just perpendicular component of E times A and that is how every diferential of flux is build inside the integral. The surface can be open or a closed one it depends on the problem.

2) Gauss law. Flux of a closed surface is proportional to the enclosed charge.

So in the first problem you don't need any closed surface. Is ok to divide the flux in three parts and then sum.
 
  • #3
Diegor said:
Maybe your confusion is because you are mixing two diferent concepts. In your equations there are two diferent concepts and you need to understand them very well.

1) flux. You can think of it as being the number of field lines passing throw a surface. The surface integral is the general case but when E is constant and the surface is flat the flux is just perpendicular component of E times A and that is how every diferential of flux is build inside the integral. The surface can be open or a closed one it depends on the problem.

2) Gauss law. Flux of a closed surface is proportional to the enclosed charge.

So in the first problem you don't need any closed surface. Is ok to divide the flux in three parts and then sum.
So in problem 1 is it simply ∫E1⋅dA1 + E2A2 + ∫E3⋅dA3? And then in problem 2 it will use the Gauss Law concept since it is a cube? How does that equation differ though? I got confused when the problem started asking about N small pieces and the ith small surface element.
 
  • #4
jlmccart03 said:
So in problem 1 is it simply ∫E1⋅dA1 + E2A2 + ∫E3⋅dA3? And then in problem 2 it will use the Gauss Law concept since it is a cube? How does that equation differ though? I got confused when the problem started asking about N small pieces and the ith small surface element.
That is ok for the first one. And you will see that the other two integrals simplify also since the field and the surface vector are constant.

For the second there is no charge involved so no Gauss law is needed. The problem only mention one side of the cube so it is just the flux in that square. I think you just need to write one diferential dflux in terms Ei and dAi and then write the integral to get net flux

I see this two problems are intended to teach how flux is computed from simple cases so at the end you understand why a surface integral is the general form of flux. And your approach was to use the integral from the beginning. I recommend you to go back to a textbook and read again from the basics it would help a lot.
 
  • #5
Diegor said:
That is ok for the first one. And you will see that the other two integrals simplify also since the field and the surface vector are constant.

For the second there is no charge involved so no Gauss law is needed. The problem only mention one side of the cube so it is just the flux in that square. I think you just need to write one diferential dflux in terms Ei and dAi and then write the integral to get net flux

I see this two problems are intended to teach how flux is computed from simple cases so at the end you understand why a surface integral is the general form of flux. And your approach was to use the integral from the beginning. I recommend you to go back to a textbook and read again from the basics it would help a lot.
Ok, well we don't use a textbook so I will try and research online the best I can. Thank you for helping me!
 
  • #6
jlmccart03 said:
So in problem 1 is it simply ∫E1⋅dA1 + E2A2 + ∫E3⋅dA3? And then in problem 2 it will use the Gauss Law concept since it is a cube? How does that equation differ though? I got confused when the problem started asking about N small pieces and the ith small surface element.
The electric field is constant, so you do not need integrals. Φ=∫EdA = E⋅A You should know that E⋅A is the scalar product of the electric field E with the surface area vector which is the outward normal of the surface multiplied by the surface area. So the flux across the surfaces in problem (1) is Φ = (E A cos(θ) ) where θ is the angle between the normal of the surface and the electric field.
In case of the cube, you need the integral form as E is not said constant.
In the third problem, you have to decide if the direction of the flux from one charge is the same or opposite of the direction of the flux from the other charge.
 
  • #7
jlmccart03 said:
Ok, well we don't use a textbook so I will try and research online the best I can. Thank you for helping me!
Try this one
.
And there is a lot of other videos and examples on the web.
 
  • #8
ehild said:
The electric field is constant, so you do not need integrals. Φ=∫EdA = E⋅A You should know that E⋅A is the scalar product of the electric field E with the surface area vector which is the outward normal of the surface multiplied by the surface area. So the flux across the surfaces in problem (1) is Φ = (E A cos(θ) ) where θ is the angle between the normal of the surface and the electric field.
In case of the cube, you need the integral form as E is not said constant.
In the third problem, you have to decide if the direction of the flux from one charge is the same or opposite of the direction of the flux from the other charge.
For problem one I watched Diegor's video and would the answer also be just the magnitude of E ⋅ A cos(θ) of the three individual surfaces? I am starting to think we are not supposed to use the integral yet as that is Gauss's Law and that is for the next tutorial HW.
 
  • #9
jlmccart03 said:
For problem one I watched Diegor's video and would the answer also be just the magnitude of E ⋅ A cos(θ) of the three individual surfaces?
Yes, the answer is the sum of E ⋅ A cos(θ) of the three surfaces.
jlmccart03 said:
I am starting to think we are not supposed to use the integral yet as that is Gauss's Law and that is for the next tutorial HW.
The flux is defined as ##\int{\vec E \cdot \vec{dA}}##, and the second problem needs this definition. Gauss' Law is different, it is connected with the flux for a closed surface and the charge enclosed by it.
 

FAQ: Electric Flux confusion and help

1. What is electric flux and how is it calculated?

Electric flux is a measure of the electric field that passes through a given area. It is calculated by multiplying the strength of the electric field by the area it is passing through, and then taking the cosine of the angle between the field and the area. This can be written as Φ = E * A * cos(θ), where Φ is the electric flux, E is the electric field strength, A is the area, and θ is the angle between the electric field and the area.

2. How is electric flux different from electric field?

Electric flux and electric field are related concepts, but they are not the same. Electric field is a vector quantity that describes the strength and direction of the electric force at a given point. Electric flux, on the other hand, is a scalar quantity that describes the amount of electric field that passes through a given area. In other words, electric field tells us about the electric force at a specific point, while electric flux tells us about the overall effect of the electric field on a given area.

3. What does it mean to have a negative electric flux?

A negative electric flux indicates that the electric field is passing through the given area in the opposite direction of the normal vector of the area. This means that the electric field is pointing away from the area, rather than towards it, and the overall effect of the electric field on the area is reduced.

4. How can I visualize electric flux?

One way to visualize electric flux is by imagining a light shining through a window onto a surface. The amount of light that passes through the window and hits the surface represents the electric flux, while the window and the surface represent the area. Another way to visualize electric flux is by using electric field lines, which show the direction and strength of the electric field at different points in space. The number of field lines that cross a given area can represent the electric flux through that area.

5. What are some real-life applications of electric flux?

Electric flux has many practical applications, including in the design of electronic devices, electric motors, and generators. It is also important in understanding how electricity is distributed through power grids and how lightning strikes occur. Additionally, electric flux is used in medical imaging techniques such as electrocardiograms and electroencephalograms to measure the electric activity of the heart and brain, respectively.

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