Derivation of electric field intensity due to uniformly charged thin infinite plane sheet
Hi Akansha dear, Gauss concept could be utilised to derive the same
Let σ be the charge density. That is charge per unit area
Let us imagine a cylindrical portion being perpendicular to the plane sheet
Let A be the area of cross section.
Hence σA is the charge enclosed within that closed surface
By Gauss idea the flux coming out has to be 1/εo * (σA)
Now let us consider the two extreme flat faces of area A
Then by manipulation we have E A as the flux from each flat face
Hence 2 E A from both the faces.
Note there is no chance of getting a electrical lines of force crossing the curved surface area
Hence 2 E A = 1/εo * (σA)
==> E = σ/2εo
It does not have any distance of the point right from the charged sheet
Hence field has to be uniform on both sides of the sheet
Let σ be the charge density. That is charge per unit area
Let us imagine a cylindrical portion being perpendicular to the plane sheet
Let A be the area of cross section.
Hence σA is the charge enclosed within that closed surface
By Gauss idea the flux coming out has to be 1/εo * (σA)
Now let us consider the two extreme flat faces of area A
Then by manipulation we have E A as the flux from each flat face
Hence 2 E A from both the faces.
Note there is no chance of getting a electrical lines of force crossing the curved surface area
Hence 2 E A = 1/εo * (σA)
==> E = σ/2εo
It does not have any distance of the point right from the charged sheet
Hence field has to be uniform on both sides of the sheet