Solving Flux of Electric Field through a Cube of Side L = 2m

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SUMMARY

The discussion focuses on calculating the electric flux through a cube with a side length of L = 2m, centered at the origin, in the presence of an electric field E = (15N/C)i + (27N/C)j + (39N/C)k. The total flux is derived using the equation phi total = (E * n) delta A, where n represents the unit normal vector for each face of the cube. The participants clarify that the flux through opposing faces cancels out, resulting in a total flux of zero, while emphasizing the importance of identifying the correct orientation of the normal vectors for accurate calculations.

PREREQUISITES
  • Understanding of electric flux and its calculation using the formula phi = E * A
  • Familiarity with vector operations, specifically the dot product
  • Knowledge of unit normal vectors for surfaces in three-dimensional space
  • Basic principles of Gauss's Law in electromagnetism
NEXT STEPS
  • Study the application of Gauss's Law for calculating electric flux through various geometries
  • Learn about vector calculus, focusing on dot products and their physical interpretations
  • Explore the concept of electric fields and their representation in three-dimensional space
  • Practice problems involving electric flux through different shapes and orientations
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Students studying electromagnetism, physics educators, and anyone interested in mastering electric flux calculations in three-dimensional geometries.

ibaraku
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Homework Statement


A cube of side L = 2m is centered at the origin, with the coordinate axes perpendicular to its faces. Find the flux of the electric field E = (15N/C)i + (27N/C)j + (39N/C)k through each face of the cube


Homework Equations



phi total = (E * n) delta A


The Attempt at a Solution



Ok, so I know that
phi total = phi 1 + phi 2 + ... + phi 6

what I am confused about is how do I know where the flux is parallel to the faces of cube, where is it zero?

Thanks
 
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ibaraku said:

Homework Statement


A cube of side L = 2m is centered at the origin, with the coordinate axes perpendicular to its faces. Find the flux of the electric field E = (15N/C)i + (27N/C)j + (39N/C)k through each face of the cube


Homework Equations



phi total = (E * n) delta A


The Attempt at a Solution



Ok, so I know that
phi total = phi 1 + phi 2 + ... + phi 6

what I am confused about is how do I know where the flux is parallel to the faces of cube, where is it zero?

Thanks

(E*n) is the dot product between \vec{E} and \hat{n}. You are given \vec{E}, so what is the definition of \hat{n} for any surface? What is \hat{n} for each surface of the cube? Once you find these, you just compute the dot product and then integrate over the area of each surface to get your phis.
 
gabbagabbahey said:
(E*n) is the dot product between \vec{E} and \hat{n}. You are given \vec{E}, so what is the definition of \hat{n} for any surface? What is \hat{n} for each surface of the cube? Once you find these, you just compute the dot product and then integrate over the area of each surface to get your phis.

Yeah I get that
[((15N/C)i + (27N/C)j + (39N/C)k] . i = (15N/C) a^2
"" "" "" "" . -i = -(15N/C) a^2
...
...

I am just wondering how do I know where one of this calculations will be zero, where will the flux be parallel to the sides of the cube, how do I recognize that?
 
ibaraku said:
Yeah I get that
[((15N/C)i + (27N/C)j + (39N/C)k] . i = (15N/C) a^2
"" "" "" "" . -i = -(15N/C) a^2
...
...

I am just wondering how do I know where one of this calculations will be zero, where will the flux be parallel to the sides of the cube, how do I recognize that?
Flux is a scalar quantity...it will be neither parallel nor perpendicular to any given side. Why do you have an a^2 there? And for which face is \hat{n}=+\hat{i}?Is it the same for all faces? Can you show your whole solution?
 
gabbagabbahey said:
What is \hat{i} \cdot \hat{i}, how about \hat{i} \cdot \hat{j} and \hat{i} \cdot \hat{k}?

1, 0, 0

I see, so for this problem

[(15N/C)a^2 + (-15N/C)a^2 + (27N/C)a^2 + (-27N/C)a^2 + (39N/C)a^2 + (-39N/C)a^2] = 0

Thanks
 
ibaraku said:
1, 0, 0

I see, so for this problem

[(15N/C)a^2 + (-15N/C)a^2 + (27N/C)a^2 + (-27N/C)a^2 + (39N/C)a^2 + (-39N/C)a^2] = 0

Thanks

If by a you mean L, then yes but you have calculated the TOTAL flux. The question asks for the flux through each side. You should draw a diagram, where you label each side and give the flux through each side.
 

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