Gauss' Law problem: determine the electric flow through a square surface due to a nearby charge

In summary, the conversation discusses determining the electric flow through a square surface of side 2l due to a load + Q located at a perpendicular distance l from the center of the plane. The participants consider using Gauss's law and taking advantage of symmetry to simplify the calculation. It is suggested to apply Gauss's law to a cube with the assumption that the Y-axis passes through the center of the square. The conversation concludes that this approach is likely to be successful.
  • #1
Est120
51
3

Homework Statement


determine the electric flow through a square surface of side 2l due to a load + Q located at a perpendicular distance l from the center of the plane

I really don't know how to answer this question .i need help guys
Thanks

Homework Equations

The Attempt at a Solution


I ended up with a doble integral ,which i had to evaluate in order to calculate the flow of electric fiele in the whole square
 
Last edited:
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  • #2
Can you share the details of the the work you've done so far? Is there a figure that accompanies the problem that you can upload?
 
  • #3
upload_2018-10-6_8-0-3.png
upload_2018-10-6_8-5-50.png

this would be the image ,and i tried this:
im new to this kind of things also English is not my principal language
 

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  • #4
You may not need to do an integration for this. Take advantage of the symmetry
 
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Likes gneill
  • #5
Chandra Prayaga said:
You may not need to do an integration for this. Take advantage of the symmetry
Should i just apply Gauss law And multiply by 1/6 ?
 
  • #6
Est120 said:
Should i just apply Gauss law And multiply by 1/6 ?
If you can assume that the Y-axis passes through the center of the square so that symmetry is preserved, then yes.
 
  • #7
The question will be, what is the closed surface on which you are applying Gauss's law, before dividing by 1/6?
 
  • #8
Chandra Prayaga said:
The question will be, what is the closed surface on which you are applying Gauss's law, before dividing by 1/6?
A cube ,but i don't know if it's as easy as that
 
  • #9
Est120 said:
A cube ,but i don't know if it's as easy as that
Actually, it is as easy as that! (With the assumption described in post #6).
 
  • #10
I would say, just go ahead and try it.
 

1. What is Gauss' Law?

Gauss' Law is a fundamental law of electromagnetism that relates electric charges to the electric field they produce. It states that the electric flux through a closed surface is equal to the net electric charge enclosed by that surface.

2. How is Gauss' Law applied to determine the electric flow through a square surface?

To apply Gauss' Law to determine the electric flow through a square surface, we first draw a Gaussian surface which is a closed surface that encloses the square surface. Then, we calculate the net electric charge enclosed by the Gaussian surface and use the formula Q/ε0 to find the electric flux.

3. What is the formula for determining the electric flow through a square surface using Gauss' Law?

The formula for determining the electric flow through a square surface using Gauss' Law is E*A = Q/ε0, where E is the electric field, A is the area of the square surface, Q is the net electric charge enclosed by the Gaussian surface, and ε0 is the permittivity of free space.

4. How is the direction of the electric field determined in a Gauss' Law problem?

The direction of the electric field in a Gauss' Law problem is determined by the direction of the electric flux through the Gaussian surface. If the net electric charge enclosed by the Gaussian surface is positive, the electric field points outward from the surface. If the net electric charge is negative, the electric field points inward towards the surface.

5. Can Gauss' Law be applied to any shape of surface?

Yes, Gauss' Law can be applied to any closed surface, regardless of its shape. However, it is easier to apply to surfaces with high symmetry, such as spheres, cylinders, and cubes, as it simplifies the calculation of the electric field and flux.

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