# Gauss's Law - field of infinite plane sheet

1. Mar 28, 2013

### ZedCar

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.

2. Mar 28, 2013

### ap123

What gaussian surface did you use to get this?

3. Mar 28, 2013

### ZedCar

Possibly a cylinder?

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

4. Mar 28, 2013

### ap123

Sounds a good choice.
What will the flux be through the cylinder?

5. Mar 28, 2013

### ZedCar

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. Mar 28, 2013

### ap123

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?

7. Mar 28, 2013

### ZedCar

2EA=(σ/ ϵ0)A

Where A, area, is the area of one cylinder end. i.e. ∏r^2

8. Mar 28, 2013

### ap123

Yes, so the 2 comes from the fact that the flux through each cylinder end is EA

9. Mar 28, 2013

### ZedCar

Thanks very much ap123..!

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