State Gauss' theorem in electrostatics. Derive an expression for the electric field intensity at any point of to an infinite plane sheet of charge.

Explanation

Gauss.s Theorem : It states that total electric flux through a closed surface enclosing a charge is equal to `(1)/(epsilon_(0))` times the magnitude of charge enclosed. Mathematical form of Gauss.s Las :
Electric flux =` phi = oint vec(E ).vec(dS) (q)/(epsilon_(0))`
Electric field intensity due to an infinite plane thin sheet of charge :
Consider a thin, infinite plane sheet of charge.
Let `sigma` = Surface charge density on both sides
Consider a point P at a distance r from the sheet.
Draw a cylindrical Gaussian surface of cross - section S through the point P with lenght r on each side of the cap.
Electric field E is perpendicular to its end circular Cpas I and II.
Charge enclosed by the Gaussian surface = `q = sigma S`
Using Gauss.s theorem
`oint vec(E). vec(dS) = (q)/(epsilon_(0)) = (sigma S)/(epsilon_(0))`
Gaussian surface is divided into three parts, I II and III

`therefore` Eqn. (i) becomes `underset(I)ointvec(E).vec(dS)+ underset_(II)ointvec(E).vec(dS)+underset(III)ointvec(E).vec(dS)=(sigmaS)/(epsilon_(0))`
Angle between `vec(E) and vec(dS)` is `90^(@)` for the curved III. In this case `underset(III)oint vec(E).vec(dS) = E.dS cos 90^(@) = 0`
Hence eqn. (ii) becomes `underset(I)oint vec(E).vec(dS) = undereset(II) oint vec(E).vec(dS) = (sigma S)/(epsilon_(0))`
Since angle between `vec(E) and vec(dS)` for surface I and II is zero, so
`underset(I)ointvec(E).dScos0^(@)+underset(II)ointE.dS cos 0^(@) = (sigma S)/(epsilon_(0)) or underset(I)ointE.dS+underset(II)oint E dS = (sigmaS)/(epsilon_(0))`
Since electric field intensity E is Constant at very point of the Gaussian surface so
`therefore Eunderset(I)ointdS+Eunderset(II)ointdS=(sigmaA)/(epsilon_(0))orES+ES=(sigmaS)/(epsilon_(0)) [therefore oint dS = S = "Area of end cap"]`
or `2ES = (sigma)/(epsilon_(0))`
or `E = (sigma)/(2 epsilon_(0))`
If the plane sheet has a finite thickness : then charges on both sides of the sheet are to be considered. Thus equation (i) is written as `int vec(E).vec(dS) = (2sigmaS)/(epsilon_(0))`
Then above eqn. becomes 2ES = `(2 simgma)/(epsilon_(0))` i.e., `E = (sigma)/(epsilon_(0)) " " [theregore int vec(E).vec(dS) = 2ES]`