# Gauss's Law - field of infinite plane sheet

by ZedCar
Tags: field, gauss, infinite, plane, sheet
 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.
P: 265
 The solution is, where A=area: 2EA=(σ/ ϵ0)A E=σ/2ϵ0
What gaussian surface did you use to get this?
P: 342
 Quote by ap123 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'.

P: 265
Gauss's Law - field of infinite plane sheet

 Quote by ZedCar 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?
P: 342
 Quote by ap123 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.
P: 265
 Quote by 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.
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?