The Electric Flux through a Cube with a Uniform Electric Field

AI Thread Summary
To find the electric flux through the right face of a cube in a uniform electric field, the relevant formula is the surface integral of the electric field across the area. For the field 2.00i, the flux is zero because the electric field is parallel to the face, resulting in no contribution. In contrast, for the field -3.00j, the flux does not cancel out since it is perpendicular to the face, allowing for a measurable contribution. The discussion emphasizes the importance of understanding how electric flux is defined and the application of Gauss' Law. Clarification on the cube's orientation and the electric field's direction is also noted as essential for accurate calculations.
badtwistoffate
Messages
81
Reaction score
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
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.
 
Last edited:
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
 
Last edited:

Similar threads

Back
Top