Need help in Electric Flux within a cube with Gauss' Law

In summary: E0/L *(|i| |k|...|i+1|)In summary, Homework Statement: In order to find the electric flux of a surface, you need to first integrate the variables of dA and set one variable to 0 or L. Afterwards, you work out the flux through a small area on that surface.
  • #1
Blockade
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Homework Statement


I need help in figuring out if I have done this problem correctly. From what I understand ∫E * dA = E*A, where E is the electric field and A is the area of a side. My biggest concern is if I can plug in the length "L" for the "x" and "z" variables within "E = -5x * E0/L i + 3z * E0/L k" to get the electric flux of each side since I assume that "L" is the value of those two variable.

55e5506d86.jpg

Given: [ε0, L]

Unknown: [Electric flux of the top, bottom, front, back, left, right side, and the total]

Homework Equations


Electric Flux: φ = ∫E * dA
E = -5x * E0/L i + 3z * E0/L k

The Attempt at a Solution


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  • #2
My biggest concern is if I can plug in the length "L" for the "x" and "z" variables within "E = -5x * E0/L i + 3z * E0/L k" to get the electric flux of each side since I assume that "L" is the value of those two variable.
It would not be appropriate to set x=L and z=L ... that is not what that equation means.
The equation ##\vec E(x,y,z) = \frac{E_0}{L}\left[-5x\hat\imath +3z\hat k \right]## tells you the value of ##\vec E## at different x,y,z positions in space.
So the value of ##\vec E## at x=0 and z=0 is different from that at x=0 and z=L. That is why you had to do an integral.
However, the geometry in this problem is simple enough that you will always be setting one variable to 0 or L and the flux only depends on one of the other variables.

You proceed by picking a side, then defining a small area on that side and working out the flux through that area in terms of the position.
Then you add up the flux through all the small areas that make up that side of the cube.
ie for the side in the x-y plane: 0<x<L, 0<y<L, z=0
##d\vec A = -\hat k dxdy## is the area element between x and x+dx and y and y+dy.
The flux through that area is ##d\phi = -\vec E(x,y,0)\cdot \hat k\; dxdy## ... and the total flux is given by the integral.
For this surface, the only non-zero flux is parallel to the surface (there is no flux through the surface).
... it kinda looks like this is what you are doing but I figured I'd better check.

Understanding that it is the flux through a side that counts, you can shortcut the process considerably ... ie, there is no component in the y direction so the flux through surfaces parallel to the x-z plane is zero.
No further maths needed.
 
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  • #3
Simon Bridge said:
It would not be appropriate to set x=L and z=L ... that is not what that equation means.
The equation ##\vec E(x,y,z) = \frac{E_0}{L}\left[-5x\hat\imath +3z\hat k \right]## tells you the value of ##\vec E## at different x,y,z positions in space.
So the value of ##\vec E## at x=0 and z=0 is different from that at x=0 and z=L. That is why you had to do an integral.
However, the geometry in this problem is simple enough that you will always be setting one variable to 0 or L and the flux only depends on one of the other variables.

You proceed by picking a side, then defining a small area on that side and working out the flux through that area in terms of the position.
Then you add up the flux through all the small areas that make up that side of the cube.
ie for the side in the x-y plane: 0<x<L, 0<y<L, z=0
##d\vec A = -\hat k dxdy## is the area element between x and x+dx and y and y+dy.
The flux through that area is ##d\phi = -\vec E(x,y,0)\cdot \hat k\; dxdy## ... and the total flux is given by the integral.
For this surface, the only non-zero flux is parallel to the surface (there is no flux through the surface).
... it kinda looks like this is what you are doing but I figured I'd better check.

Understanding that it is the flux through a side that counts, you can shortcut the process considerably ... ie, there is no component in the y direction so the flux through surfaces parallel to the x-z plane is zero.
No further maths needed.

Thank you, I will keep in mind to integrate the variables of dA and to be mindful of the upper and lower bounds of each problem.
However to clear things up, let's say if I were to re-do the φTop, would it equal to:

Etop = -5x E0/L i + 3z E0/L k
= -5x E0/L *(|i| |k| cos(90°)) + 3z E0/L * (|k| |k| cos(0°))
= 3 E0/L * ∫ z dz
= 3 E0/L * z2/2 | z = L, z = 0
= 3 E0 * L/2
?
Therefore, in its whole form it looks like this:
Etop = -5 E0/L * (0) * ∫ x dx + 3z E0/L * (1) * ∫ z dz ?

Meaning my original answer of φTop = 3E0L2 is incorrect?
 
Last edited:

Related to Need help in Electric Flux within a cube with Gauss' Law

1. What is Electric Flux and how is it calculated?

Electric Flux is a measure of the amount of electric field passing through a given area. It is calculated by multiplying the electric field by the surface area and the cosine of the angle between the field and the surface.

2. How is Gauss' Law used to calculate Electric Flux within a cube?

Gauss' Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. To calculate Electric Flux within a cube, we can use Gauss' Law by choosing a Gaussian surface that encloses the cube and solving for the electric flux through that surface.

3. What are the units of Electric Flux?

The units of Electric Flux are Newtons per square meter, or Nm2/C.

4. Does the shape of the Gaussian surface affect the calculation of Electric Flux within a cube?

No, the shape of the Gaussian surface does not affect the calculation of Electric Flux within a cube as long as the surface encloses the cube completely and the electric field is perpendicular to the surface at all points.

5. How does the charge distribution within the cube affect the calculation of Electric Flux?

The charge distribution within the cube affects the calculation of Electric Flux as it determines the amount of charge enclosed by the Gaussian surface. The more charge there is within the cube, the higher the electric flux will be through the surface.

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