Gauss's Law - field of infinite plane sheet

In summary: Where A, area, is the area of one cylinder end. i.e. ∏r^2Yes, so the 2 comes from the fact that the flux through each cylinder end is EAThanks very much ap123..!
  • #1
ZedCar
354
1

Homework Statement



Use Gauss's Law to calculate the field of an infinite plane sheet of surface charge density σ0.


Homework Equations





The Attempt at a Solution




The solution is, where A=area:

2EA=(σ/ ϵ0)A

E=σ/2ϵ0


Why is there a '2' in the solution? I know it's something to do with twice the electric field. Is it because there are two sides to the sheet surface?

Thank you.
 
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  • #2
The solution is, where A=area:

2EA=(σ/ ϵ0)A

E=σ/2ϵ0

What gaussian surface did you use to get this?
 
  • #3
ap123 said:
What gaussian surface did you use to get this?

Possibly a cylinder?

I'm not 100% sure, as I just have the solution as shown above, and no other 'working out'.
 
  • #4
ZedCar said:
Possibly a cylinder?

I'm not 100% sure, as I just have the solution as shown above, and no other 'working out'.

Sounds a good choice.
What will the flux be through the cylinder?
 
  • #5
ap123 said:
Sounds a good choice.
What will the flux be through the cylinder?

Not through it in one direction.

It's normal to the surface of the sheet.

As the sheet has two surfaces, the E-field is in both the two normal directions to the sheet.
 
  • #6
ZedCar said:
Not through it in one direction.

It's normal to the surface of the sheet.

As the sheet has two surfaces, the E-field is in both the two normal directions to the sheet.

Since the field is normal to the sheet surface, the flux through the sides of the cylinder is zero - so that only leaves the 2 cylinder ends. What's the total flux through these?
Hint : you've already got the answer :)
 
  • #7
ap123 said:
Since the field is normal to the sheet surface, the flux through the sides of the cylinder is zero - so that only leaves the 2 cylinder ends. What's the total flux through these?
Hint : you've already got the answer :)

2EA=(σ/ ϵ0)A

Where A, area, is the area of one cylinder end. i.e. ∏r^2
 
  • #8
Yes, so the 2 comes from the fact that the flux through each cylinder end is EA
 
  • #9
Thanks very much ap123..!
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the enclosed electric charge. It is based on the concept that electric fields emanate from positive charges and terminate at negative charges.

2. How does Gauss's Law relate to the field of an infinite plane sheet?

In the case of an infinite plane sheet, Gauss's Law states that the electric field is constant and perpendicular to the sheet. This means that the electric flux through any closed surface parallel to the sheet is equal to the electric charge enclosed by that surface divided by the permittivity of free space.

3. Can Gauss's Law be applied to finite plane sheets?

Yes, Gauss's Law can be applied to finite plane sheets, but only if the size of the sheet is much larger than the distance from the sheet. In this case, the electric field is still considered constant and perpendicular to the sheet, but the charge enclosed by the surface must also take into account the finite size of the sheet.

4. Does Gauss's Law apply to all closed surfaces?

Yes, Gauss's Law applies to all closed surfaces, as long as they enclose an electric charge. The shape and orientation of the surface do not matter, as long as it is a closed surface.

5. How is Gauss's Law used in practical applications?

Gauss's Law is used in various practical applications, such as calculating the electric field and flux through different geometries, designing capacitors and other electronic devices, and understanding the behavior of electric charges in different situations. It is also a fundamental concept in understanding the behavior of lightning and other atmospheric phenomena.

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