# Net electric flux through a Gaussian cube

1. Feb 7, 2017

### Jrlinton

1. The problem statement, all variables and given/known data
An electric field given by https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2560893entrance1_N10030.mml?size=14&ver=1486488694211 [Broken] = 9.6https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2560893entrance1_N1004B.mml?size=14&ver=1486488694211 [Broken] - 6.4(y2 + 4.8)https://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2560893entrance1_N1006D.mml?size=14&ver=1486488694211 [Broken] 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?

2. Relevant equations

3. 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...

Last edited by a moderator: May 8, 2017
2. Feb 7, 2017

### Jrlinton

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

3. Feb 10, 2017

### kuruman

What about the figure mentioned in the statement of the question? Can you post that?

4. Feb 10, 2017

### Staff: Mentor

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