The Electric Flux through a Cube with a Uniform Electric Field

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Homework Help Overview

The discussion revolves around calculating the electric flux through a cube's face in the presence of a uniform electric field. The specific cases involve different orientations of the electric field and their implications on the flux through the right face of the cube.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definition of electric flux and its dependence on the orientation of the electric field relative to the surface. Questions arise regarding the cancellation of contributions to flux and the application of Gauss' Law. Some participants consider using surface integrals to calculate flux.

Discussion Status

The discussion is active, with participants providing guidance on interpreting the problem and clarifying the application of concepts. There is an ongoing exploration of different approaches to calculating the electric flux, but no consensus has been reached on the correct method or interpretation of the problem.

Contextual Notes

Participants note the lack of information regarding the orientation of the cube and the electric field, which may affect the calculations. There are also mentions of issues with displaying mathematical notation, which could hinder communication of ideas.

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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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