Net electric flux through a Gaussian cube

• Jrlinton
In summary, the net electric flux through a Gaussian cube is determined by the total amount of electric field lines passing through the surface of the cube. The electric flux is directly proportional to the charge enclosed within the cube and is also affected by the orientation of the surface relative to the electric field. The Gauss's Law is used to calculate the net electric flux and can be applied to various shapes and sizes of Gaussian surfaces. The net electric flux is a crucial concept in understanding the behavior of electric fields and is widely used in the field of electromagnetism.
Jrlinton

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?

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:
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).

1. What is net electric flux through a Gaussian cube?

The net electric flux through a Gaussian cube is the measure of the total electric field passing through the surface of the cube. It is a vector quantity that takes into account both the magnitude and direction of the electric field.

2. How is net electric flux through a Gaussian cube calculated?

The net electric flux through a Gaussian cube is calculated by taking the integral of the dot product of the electric field and the surface area vector over the surface of the cube. This is represented by the equation Φ = ∫E·dA.

3. What is the significance of a Gaussian cube in calculating electric flux?

A Gaussian cube is a hypothetical cube that allows for simplification of the calculation of electric flux. It is often used in situations where the electric field is not uniform, as it allows for the use of Gauss's law to calculate the flux.

4. How does the orientation of the Gaussian cube affect the net electric flux?

The orientation of the Gaussian cube does not affect the net electric flux, as long as it is completely enclosed by the electric field. This is because the electric flux through any closed surface is always equal to the net charge enclosed by that surface.

5. What are some real-world applications of calculating net electric flux through a Gaussian cube?

Calculating net electric flux through a Gaussian cube is useful in engineering applications, such as designing electrical circuits and systems. It is also important in understanding the behavior of electric fields in various materials and in studying the properties of charged particles and electromagnetic waves.

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