Gauss' Law and electric flux

  • #1
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 doesnt b) also cancel? What should I do? I was thinking of a gaussian pill box but im still pretty rough around the edges when applying it :devil:
 

Answers and Replies

  • #2
307
1
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
307
1
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
307
1
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 cant 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
307
1
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:

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