Electric field intensity due to line charged or charged cylinderical conductor

“Electric field intensity due to line charged or charged cylinderical conductor”:

In a figure shows a line charge of linear charge density +λ coulomb/metre. We are to find the electric field intensity due to this line charge at a distance r from this line charge at point P. for this we can draw an imaginary cylinder with radius r and axis as line charge of length L. This cylinder has planes surface S1 and S2 and curved surface S3 . it is clear from figure the electric field intensity is parallel to the plane surfaces S1 and S2. Thus these surface has no electric flux but the electric field is passing normal to the curved surface S3 thus it has some electric flux .

By the definition of electric flux thus the electric flux passing through the Gaussian surface.
      φε = φ1 + φ2 + φ3
      φε = ∫∫ E ds1 cos90 + ∫∫ E ds2                     cos90 + ∫∫ E ds3 cos0

      φε = E·S3

∵[curved surface area (S3) = 2πrl ]

     φε = E×2πrl-----------(1)

By gauss theorem
     φε = q/εo

This value of  is put in eqn1
      E×2πrl = q/εo
      E = q/2πrl×εo--------(2)

Linear charge density                                                     = charge / length

       λ  = q/l
∴      q  = λ×l

putting the value of q in eqn 2

E = l×λ/2πrl×εo
[E = λ/2πr×εo ] N/C