# Electric Flux confusion and help

Tags:
1. Feb 4, 2017

### jlmccart03

1. The problem statement, all variables and given/known data
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.

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.

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.

2. Relevant equations
Φ = ∫E⋅dA
Φ = Qenclosed0

3. 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.

2. Feb 4, 2017

### Diegor

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. Feb 4, 2017

### jlmccart03

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. Feb 4, 2017

### Diegor

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 aproach was to use the integral from the beginning. I recomend you to go back to a text book and read again from the basics it would help a lot.

5. Feb 5, 2017

### jlmccart03

Ok, well we don't use a text book so I will try and research online the best I can. Thank you for helping me!

6. Feb 5, 2017

### ehild

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. Feb 5, 2017

### Diegor

Try this one
.
And there is a lot of other videos and examples on the web.

8. Feb 5, 2017

### jlmccart03

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. Feb 5, 2017

### ehild

Yes, the answer is the sum of E ⋅ A cos(θ) of the three surfaces.
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.