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Consider, an uniformly charged infinite sheet having surface charge density σ . Let, A be the small area of the sheet containing charge q . As, charge lie on both the sides of the sheet, so electric field E is also on both the sides. using Gauss theorem, net electric flux through the cylinder enclosing the charge 'q' is -                       Φ    =  q/εο          ......(1) But,                Φ   =  Φ1 + Φ2 + Φ3                             = EACos0° + EACos0° + 0                        Φ  =  2EA      .......(2)           From equation (1) & (2)                 2EA = q/εο                  E    =   q/2Aεο  Also,          σ  = q/A  Therefore,                      [  E = σ/2εο  ] Still not clear click here to watch     by    www.advguruji.blogspot.com