The Electric Flux through a Cube with a Uniform Electric Field

In summary, the conversation discusses the process of finding the electric flux through the right face of a cube with a uniform electric field. The answer for A) is 0, as all the inward and outward contributions cancel out. However, for B), the cancellation does not occur. The suggested method of finding the flux is by using Gauss' Law, using the known electric field. The surface integral of E*dA is used, where dA must be perpendicular to the surface. However, there seems to be an issue with the LaTeX graphics and the orientation of the cube is not specified.
  • #1
badtwistoffate
81
0
I got a cube withe edge length 1.4m and has a uniform electric field, i have to find the electric flux throught the right face for the following fields.

A) 2.00i

B)-3.00j

answer for a) is 0, i think because its uniform and all the inward and outward contribuitions cancel but then why doesn't b) also cancel? What should I do? I was thinking of a gaussian pill box but I am still pretty rough around the edges when applying it :devil:
 
Physics news on Phys.org
  • #2
If they're only asking you to consider the right face then you're really only looking at a single plane, not the whole box. Think about how flux is defined: it's the number of electric field lines that pass through the surface per unit area.
 
  • #3
so would that be |E|= sigma / 2Eo = (Q/Area of plane)/2Eo?
It didnt work ...
 
  • #4
You don't need to use the charge enclosed, you're already given the electric field. In this particular case all that you've been asked to find is the flux of the electric field across a surface, which is one way of using Gauss' Law. Another way of using Gauss' Law, which you attempted above, is to use the known quantity of charge enclosed to find the electric field.
 
  • #5
so do the surface integral of E*dA = |E|times area, area = s^2?
 
  • #6
If E is perpindicular to the area, yes. Technically it's:

[tex]\int_S \vec{\textbf{E}} \cdot d\vec{\textbf{A}}[/tex]

Where dA is defined as being perpindicular to the surface.
 
Last edited:
  • #7
the LaTeX Graphic is not loading i can't see what you said
 
  • #8
I got a cube withe edge length 1.4m and has a uniform electric field ? does that mean cube is generating its own field or what ? you did not give any information of the orientation of the cube ?
 
  • #9
Hm, I swear that TeX was working last night. Let's try again:

[tex]\int_S \vec{\textbf{E}} \cdot d\vec{\textbf{A}}[/tex]

Edit: It seems that even simple LaTeX isn't working? Is the tex tag broken?

[tex]a+b=c[/tex]
 
Last edited:

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the net charge enclosed by that surface.

2. How is Gauss' Law mathematically expressed?

Gauss' Law can be expressed as the integral of the electric field over a closed surface, equal to the total charge enclosed by that surface divided by the permittivity of free space.

3. What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given area. It is a vector quantity and is defined as the dot product of the electric field and the area vector.

4. What is the significance of Gauss' Law?

Gauss' Law is significant because it is a powerful tool for calculating the electric field in situations with high symmetry, such as when the charge distribution is symmetrical about a point or axis. It also has applications in determining the electric field inside and outside of charged conductors.

5. What are some real-world applications of Gauss' Law?

Gauss' Law has many practical applications, such as in determining the electric field and potential inside and outside of charged conductors, analyzing the electric field of a charged particle, and designing electronic devices such as capacitors and antennas.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
877
  • Introductory Physics Homework Help
Replies
1
Views
910
  • Introductory Physics Homework Help
Replies
26
Views
554
  • Introductory Physics Homework Help
Replies
10
Views
713
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
771
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
6K
Back
Top