Electric field intensity due to charged thin sheet

Electric field intensity due to charged thin sheet

consider a charged thin sheet has surface charge density +σ coulomb/metre. We are to find the electric field intensity due to this plane seat at either side at points P1 and P2. Here the line joining the point P1P2 is normal to the sheet, for this we can draw an imaginary cylinder of Axis P1P2 , length 2r  and area of cross section A.
           This behaves like a Gaussian surface it has three surface  S1, S2 and S3.
           It is clear from figure that the electric field is normal to the plane surface S1 and S2 . Thus  electric flux produce in them but the electric field is parallel  to the curved surface S3. Thus it has no flux.

By the definition of electric flux. Thus the electric flux passing through Gaussian surface:
   φε = φ1 + φ2 + φ3

  φε = ∫∫E ds1 cos0 + ∫∫E ds2 cos0                  +∫∫E ds3 cos90
 
  φε = E·S1 + E·S2 + 0

but  S1 = S2 = A

 φε = E·A + E·A
 φε = 2E·A-----------(1)

By gauss theorem
    φε = q/εo
This value of  φε is put in eqn(1)

  2E·A = q/εo
  E = q/2Aεo---------(2)

surfacecharge density=charge/Area
σ = q /A
∴ q = σ·A
Putting the value of q in eqn(2)
E = σ·A/2Aεo
[E = σ/2εo] N/C

NOTE = FOR CHARGED THICK SHEET
[E = σ/εo] N/C