A uniformly charged disc of radius R with a total charge Q lies in the XY -plane. Find the electric field at a point at a point P, along the Z-a xis that passes through the centre of the discc perpendicular to its plane. Discuss the limit when `R gt gt z`.
Strategy `:` By treating the disc as a set of concentric uniformly charged rings, the problem could be solved by using the result obtained in example for ring. Consider a ring of radius of radius r' and thickness as shown in figure.

Explanation

By symmetry arguments the electric field at P points in the `+z` direction . Since the ring has charge `dq= sigma ( 2pir' dr')` , we see that the ring givesa contribution.

`dE_(z) = (1)/(4pi epsilon_(0))(zdq)/((r^('2)+z^(2))^(3//2))= (1)/(4piepsilon_(0))(z(2pisigmar'dr'))/((r^('2)+z^(2))^(3//2))`
integrating from `'=0` to `r'=R` the total electric fieldat P becomes.
`E_(z) = int dE_(z)= (sigmaz)/(2epsilon_(0))int_(0)^(R )(r'dr')/((r^('r)+z^(2))^(3//2))=(sigmaz)/(4epsilon_(0))int_(z^(2))^(R^(2)+z^(2))(du )/(u^(3//2))=(sigmaz)/(4epsilon_(0))(u^(-1//2))/((-1//2)))|_(z^(2))^(R^(+z^(2))`
`= - (sigmaz)/(2epsilon_(0))[(1)/(sqrt(R^(2)+z^(2)))-(1)/(z)]=(sigma)/(2epsilon_(0))[1-(z)/(sqrt(R^(2)+z^(2)))]`
`implies E_(z)= (sigma)/(2epsilon_(0))[1-(z)/(sqrt(R^(2))+z^(2))]`
To show that the "point-charge" limit the recovered for `z gt gt R `, we make use of the Binomial expansion
`1- (z)/( sqrt(z^(2)+R^(2)))=1-(1-(R^(2))/(z^(2)))^(-1//2)=1-(1-(1)/(2)(R^(2))/(z^(2))+.....) = (1)/(2)(R^(2))/(z^(2))`
This gives
`E_(z)=(sigma)/(2epsilon) (R^(2))/(z^(2))= (1)/(4piepsilon_(0))(sigmapi R^(2))/(z^(2))=(1)/(4piepsilon_(0))(Q)/(z^(2))`
which is indeed the expected "point-charge" result . On the other hand, we may also consider the limit where `R gt gt z`. Physically this means that the plane is very large , or the field point P is extremely close to the surface of the plane. The electric field in this limit becomes,in unit-vector rotation.
`vec(E)={((sigma)/(2epsilon_(0))hat(k)",",z gt0),(-(sigma)/(2epsilon_(0))hat(k)",",z gt 0):}`