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

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In summary: Use a for the length of a side, and then express the total flux in terms of a, as the question asks.In summary, the problem involves finding the flux of an electric field through each face of a cube with side length 2m, centered at the origin. The electric field is given as (15N/C)i + (27N/C)j + (39N/C)k and the equation used is phi total = (E * n) delta A. The dot product is used to compute the flux for each face, and the direction of the electric field and normal vector determine whether the flux is positive or negative. The total flux is found by summing the flux through each face.
  • #1
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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  • #2
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 [tex]\vec{E}[/tex] and [tex]\hat{n}[/tex]. You are given [tex]\vec{E}[/tex], so what is the defintion of [tex]\hat{n}[/tex] for any surface? What is [tex]\hat{n}[/tex] 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.
 
  • #3
gabbagabbahey said:
(E*n) is the dot product between [tex]\vec{E}[/tex] and [tex]\hat{n}[/tex]. You are given [tex]\vec{E}[/tex], so what is the defintion of [tex]\hat{n}[/tex] for any surface? What is [tex]\hat{n}[/tex] 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?
 
  • #4
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 [tex]\hat{n}=+\hat{i}[/tex]?Is it the same for all faces? Can you show your whole solution?
 
  • #5
gabbagabbahey said:
What is [tex]\hat{i} \cdot \hat{i}[/tex], how about [tex]\hat{i} \cdot \hat{j}[/tex] and [tex]\hat{i} \cdot \hat{k}[/tex]?

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
 
  • #6
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.
 

1. What is the formula for calculating the flux of electric field through a cube?

The formula for calculating the flux of electric field through a cube of side L is Φ = E * A * cosθ, where E is the electric field strength, A is the area of the cube's face, and θ is the angle between the electric field and the surface of the cube.

2. How do I determine the electric field strength for a given cube?

The electric field strength can be determined by dividing the electric flux by the area of the cube's face and the cosine of the angle between the electric field and the surface of the cube. This can be represented by the formula E = Φ / (A * cosθ).

3. Can the flux of electric field through a cube be negative?

Yes, the flux of electric field through a cube can be negative. This occurs when the angle between the electric field and the surface of the cube is greater than 90 degrees, resulting in a negative cosine value in the formula Φ = E * A * cosθ.

4. What is the unit of measurement for electric flux?

The unit of measurement for electric flux is Nm²/C, or Newton meters squared per Coulomb.

5. Can the flux of electric field through a cube be affected by the material of the cube?

Yes, the flux of electric field through a cube can be affected by the material of the cube. Different materials have different electric field properties, which can impact the strength and direction of the electric field passing through the cube.

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