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Gauss's Law - field of infinite plane sheet

by ZedCar
Tags: field, gauss, infinite, plane, sheet
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ZedCar
#1
Mar28-13, 11:42 AM
P: 342
1. The problem statement, all variables and given/known data

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


2. Relevant equations



3. 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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ap123
#2
Mar28-13, 11:52 AM
P: 265
The solution is, where A=area:

2EA=(σ/ ϵ0)A

E=σ/2ϵ0
What gaussian surface did you use to get this?
ZedCar
#3
Mar28-13, 12:13 PM
P: 342
Quote Quote by ap123 View Post
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'.

ap123
#4
Mar28-13, 12:20 PM
P: 265
Gauss's Law - field of infinite plane sheet

Quote Quote by ZedCar View Post
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?
ZedCar
#5
Mar28-13, 12:24 PM
P: 342
Quote Quote by ap123 View Post
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.
ap123
#6
Mar28-13, 12:32 PM
P: 265
Quote Quote by ZedCar View Post
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 :)
ZedCar
#7
Mar28-13, 12:36 PM
P: 342
Quote Quote by ap123 View Post
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
ap123
#8
Mar28-13, 12:48 PM
P: 265
Yes, so the 2 comes from the fact that the flux through each cylinder end is EA
ZedCar
#9
Mar28-13, 12:51 PM
P: 342
Thanks very much ap123..!


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