Net electric flux through a Gaussian cube

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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
 
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
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