The Electric Flux through a Cube with a Uniform Electric Field

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SUMMARY

The discussion focuses on calculating the electric flux through a cube's right face in a uniform electric field, specifically for fields 2.00i and -3.00j. The electric flux for the field 2.00i is determined to be 0 due to the uniform nature of the field, which causes inward and outward contributions to cancel. For the field -3.00j, participants clarify that the flux can be calculated using Gauss' Law, emphasizing the need to consider the surface area and the orientation of the electric field relative to the surface.

PREREQUISITES
  • Understanding of electric flux and its mathematical definition
  • Familiarity with Gauss' Law and its applications
  • Knowledge of vector calculus, particularly surface integrals
  • Basic concepts of electric fields and their properties
NEXT STEPS
  • Study the application of Gauss' Law in various geometries
  • Learn how to perform surface integrals in vector calculus
  • Explore the concept of electric field lines and their relation to flux
  • Investigate the implications of uniform electric fields on flux calculations
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Students and professionals in physics, particularly those studying electromagnetism, as well as educators looking to enhance their understanding of electric flux and Gauss' Law applications.

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I got a cube withe edge length 1.4m and has a uniform electric field, i have to find the electric flux through 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:
 
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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.
 
so would that be |E|= sigma / 2Eo = (Q/Area of plane)/2Eo?
It didnt work ...
 
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.
 
so do the surface integral of E*dA = |E|times area, area = s^2?
 
If E is perpindicular to the area, yes. Technically it's:

\int_S \vec{\textbf{E}} \cdot d\vec{\textbf{A}}

Where dA is defined as being perpindicular to the surface.
 
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the LaTeX Graphic is not loading i can't see what you said
 
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 ?
 
Hm, I swear that TeX was working last night. Let's try again:

\int_S \vec{\textbf{E}} \cdot d\vec{\textbf{A}}

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

a+b=c
 
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