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Gauss's Law  field of infinite plane sheet 
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#1
Mar2813, 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. 


#2
Mar2813, 11:52 AM

P: 265




#3
Mar2813, 12:13 PM

P: 342

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


#4
Mar2813, 12:20 PM

P: 265

Gauss's Law  field of infinite plane sheet
What will the flux be through the cylinder? 


#5
Mar2813, 12:24 PM

P: 342

It's normal to the surface of the sheet. As the sheet has two surfaces, the Efield is in both the two normal directions to the sheet. 


#6
Mar2813, 12:32 PM

P: 265

Hint : you've already got the answer :) 


#7
Mar2813, 12:36 PM

P: 342

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


#8
Mar2813, 12:48 PM

P: 265

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



#9
Mar2813, 12:51 PM

P: 342

Thanks very much ap123..!



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