Derive the expression of electric field intensity at a point 'P' which is situated at a distance x on the axis of uniformly charges disc of radius R and surface charge density `sigma`. Also, derive results for
(i) `x gt gt R` (ii) `x lt lt R`

Explanation

The disc can be considered to be a collection of large number of concentric rings. Consider an element of the shape of rings of radius r and of width dr. electric field due to this ring at P is
`dE=(K. sigma 2pir.dr.x)/((r^(2)+x^(2))^(3//2))`
Put, `r^(2)+x^(2)=y^(2)`
`2rdr=2ydy`
`:. dE=(K. sigma2pi y.dy.x)/y^(3) =2K sigma pi.x (ydy)/y^(3)`
Electric field at P due to all tings is along the axis :

`:. E=int dE implies E=2K sigma pi x underset(x)overset(sqrt(R^(2)+x^(2)))(int) 1/y^(2) dy=2K rho pi x. [-1/y]_(x)^(sqrt(R^(2)+x^(2)))`
`=2K sigma pi x[+1/x-1/sqrt(R^(2)+x^(2))]=2K sigma pi [1-x/sqrt(R^(2)+x^(2))]`
`= sigma/(2 epsi_(0)) [1- x/sqrt(R^(2)+x^(2))]` along the axis
Cases : (i) If `x gt gt R`
`E=sigma/(2epsi_(0)) [1-x/(x sqrt(R^(2)/x^(2)+1))]= sigma/(2 epsi_(0)) [1- (1+R^(2)/x^(2))^(-1//2)]`
`sigma/(2 epsi_(0))=[1-1+1/2 R^(2)/x^(2)+" higher order terms"]=sigma/(4 epsi_(0)) R^(2)/x^(2) =(sigma pi R^(2))/(4pi epsi_(0) x^(2))=Q/(4pi epsi_(0) x^(2))`
i.e., behaviour of the disc is like a point charge.
(ii) If `x lt lt R`
`E=sigma/(2 epsi_(0)) [1-0]=sigma/(2 epsi_(0))`
i.e., behaviour of the disc is like infinite sheet.