Net electric flux through a Gaussian cube

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SUMMARY

The discussion focuses on calculating the net electric flux through a Gaussian cube with an edge length of 0.160 m, subjected to a specific electric field described by the equation E = 9.6x - 6.4(y² + 4.8)y^. The calculations for the electric flux through the top, bottom, left, and back faces were provided, yielding values of -0.7906 Nm²/C, 0.7906 Nm²/C, -0.24576 Nm²/C, and 0 Nm²/C, respectively. The net electric flux through the cube remains unresolved, as the user struggles to integrate the electric field across the entire surface area of the cube.

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  • Familiarity with vector calculus and surface integrals
  • Knowledge of electric fields and their mathematical representations
  • Ability to manipulate mathematical equations and perform integrations
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Jrlinton
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Homework Statement


An electric field given by https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2560893entrance1_N10030.mml?size=14&ver=1486488694211 = 9.6https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2560893entrance1_N1004B.mml?size=14&ver=1486488694211 - 6.4(y2 + 4.8)https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2560893entrance1_N1006D.mml?size=14&ver=1486488694211 pierces the Gaussian cube of edge length 0.160 m and positioned as shown in the figure. (The magnitude E is in Newtons per coulomb and the position x is in meters.) What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back face? (e) What is the net electric flux through the cube?

Homework Equations

The Attempt at a Solution


I can compute the values for a-d simply enough but am having real trouble finding the net electric flux

A) finding the y component through the top face should be just substituting the edge length for y in the electric field equation and multiplying by the area of the face
-0.7906Nm^2/C

B) Should be the opposite of A because the electric field lines are entering through this side
0.7906 Nm^2/C

C) should be the integral ∫9.6dA with the result being negative as the electric field is entering through the left face
-.24576 Nm^2/C

D) There is no z component to the field so the flux upon this side is zero
0

E) This is were i become lost...
 
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I realized that the equation for electric field was lost in the post...
E= 9.6x^-6.4(y^2+4.8)y^
Also I misspoke on A and B as the electric field enters through the top so A is negative and B is positive
 
What about the figure mentioned in the statement of the question? Can you post that?
 
Jrlinton said:
I realized that the equation for electric field was lost in the post...
E= 9.6x^-6.4(y^2+4.8)y^
Also I misspoke on A and B as the electric field enters through the top so A is negative and B is positive
The images that you pasted are linked to an off-site location that is probably session-based and perhaps not public (requires site membership). It's better to upload images to the PF server using the UPLOAD feature (find the UPLOAD icon at the bottom right of the edit window), or use a cut & paste screen "snip" feature to capture the image off of your screen and paste it if the resolution will hold up.

If it's just math formulas you can take advantage of the x2 and x2 icons to produce subscripts and superscripts, and the ##\Sigma## icon provides a table of Greek letters and various math symbols you can insert by point-and-click. For heavy-duty math you can use LaTeX syntax to render professional quality math expressions (find the LaTeX / BBcode Guides links at the lower left of the reply edit window).
 

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