The Electric Flux on all sides of a cube

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Homework Help Overview

The discussion revolves around calculating the electric flux through the surfaces of a cube, utilizing the concept of electric fields and area. Participants are exploring the correct application of the electric flux equation and addressing potential errors in their calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the electric flux equation, questioning the use of area formulas and the significance of angles in relation to the electric field. There are attempts to clarify the meaning of flux and its dependence on the orientation of surfaces relative to the electric field.

Discussion Status

Some participants have provided hints and suggestions, such as sketching the electric field and reconsidering the definitions of area and flux. There is acknowledgment of the need to understand the underlying concepts rather than relying solely on equations. The discussion reflects a mix of interpretations and approaches without a clear consensus.

Contextual Notes

There are indications of confusion regarding the correct formula for area and the role of the distance component in the flux equation. Participants are encouraged to verify their calculations and consider the implications of their assumptions.

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Homework Statement


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Homework Equations


[itex]\phi[/itex] = [itex]\vec{E}[/itex] * A * d
phi is the electric flux
E is the Electric field from vector form
A is the area and A = [itex]\pi[/itex][itex]r^{2}[/itex]
d is the distance on the axis (x, y, or z)


The Attempt at a Solution


s1 = s3 = 0 *Since they are along the y-axis and there is no y-vector
s2 = 1.14e-1
s4 = - 1.14e-1
s5 = 2.69e-1
s6 = -2.69e-1

I don't know why I am getting these wrong but you need to arrange them from s1-s6 on the answer when you submit them and it tells me to check my sign on s5. Changing it to negative doesn't help.
I'll recheck my calculations again and try a different method. I probably forgot to put " [itex]\pi[/itex]" into the calculations but I will check on that.

Am I doing this right? Or is the equation to solve this problem totally different and I am off.
Pointing out where I went wrong helps, please do not post the answer as I just need a hint or tip.
Thank you!
 
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Area for a square is the length times width not [itex]\pi r^2[/itex] .
You need to consider that flux is dependent on angle and that not all the surfaces are perpendicular to the electric field.
Also what is the deal with the distance component in you're equation for flux?
 
I'd add as a hint: sketch the field.
 
Bhumble said:
Area for a square is the length times width not [itex]\pi r^2[/itex] .
You need to consider that flux is dependent on angle and that not all the surfaces are perpendicular to the electric field.
Also what is the deal with the distance component in you're equation for flux?

I really have no clue. This equation is from the book directly.
 
[itex]\varphi[/itex] = [itex]\vec{E}[/itex][itex]L^{2}[/itex]cos(θ)
or
[itex]\varphi[/itex] = -[itex]\vec{E}[/itex][itex]L^{2}[/itex]cos(θ)
Where E is the electric field from the [itex]\hat{n}[/itex] vector.
θ is either 90° or 180°

Is that the equation I am looking for?
 
Last edited:
Instead of trying to guess which equation to apply - why don't you start with what "flux" actually means? And seriously: sketch the field.
 
Were you able to figure it out?
That equation looks right. Like Simon said though you should make sure to understand what flux means and you'll know when to apply the appropriate equations.
 
Use $$\oint \vec{E} \cdot \vec{dA}$$ What is E and dA? What is the value of the electric field along S5?
 
Thanks. I solved it after looking at a few examples that people had on this site and eventually figured out how they got to the solution. I wouldn't guess it.
 
  • #10
Yeah - guessing seldom works without understanding.
 

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