State Gauss's law in electrostatics. Using this law derive an expression for the electric field due to a uniformly charged infinite plane sheet.

Explanation

Electric field due to a uniformly charged infinite plane sheet : Suppose a thin non-conducting infinite sheet of uniform surface, charge density `sigma`.
Electric field intensity on either side of the sheet must be perpendicular to the plane of sheet having same magnitude at all points equidistant from sheet.
Let P be any point at a distance r from the sheet.
Let the small area element `vec(E) = ds hat(n)`
`vec(E) and hat(n)` are perpendicular, on the surface of imagined cylinder, so electric flux is zero.
`vec(E) and hat(n)` are parallel on the two cylindrical edges P and Q, which contributes electric flux.
`:.` Electric flux over the edges P and Q of the cylinder is
`2 phi = (q)/(in_(0)) rArr 2 oint vec(E).vec(ds) = (q)/(in_(0)) " " [ :' phi = oint vec(E).dec(ds)]`

`2 oint vec(E).vec(ds) = (q)/(in_(0))`
`2 oint E ds = (q)/(in_(0)) " " [ :' vec(E) bot vec(ds)]`
`2 E pi r^(2) = (q)/(in_(0)) rArr E = (q)/(2pi in_(0) r^(2))`
`:.` The charge density `sigma = (q)/(S) rArr q = pi r^(2) sigma` [where S-area of circle]
`E = (pi r^(2) sigma)/(2pi in_(0) r^(2)) E = (sigma)/(2 in_(0)) " " :.` vecotrically `vec(E) = (sigma)/(2in_(0)) hat(n)`
where `hat(n)` is a unit vector normal to the plane and going away from it.
When `sigma gt 0`, E is directed away from both sides. Hence electric field intensity is independent of r.
When `sigma gt 0`, E is directed away from both sides. Hence electric field intensity is independent of r.
Note : For conducting sheet, the surface charge density on both the surface of sheet will be same
`:. E = (sigma)/(in_(0))`