Find the electric field on the axis of a charged disc.

Explanation

Consider a disc of uniform surface charge density `'sigma'`.
Let us calculate the electric field due to a charged ring of radius r. from the centre and having a width, dr. Here we are using the expression for electric field intensity for a charged ring of radius r at a point on the axis at a distance x from the centre.
`E=(kxq)/((x^(2)+r^(2))^(3//2))` directed along the axis outwards from the centre.
Thus electric field due to elementary ring.
`dE=(kxdq)/((x^(2)+r^(2))^(3//2))`, directed along the line OP.
Now, the area of the ring, `dS=2pir.dr`
Charge on the elementary ring
`implies dq=sigma2pirdr`,
Thus, electric field due to entire disc
`|E|_(p)=intdE=kxunderset(0)overset(R)int(sigma2pirdr)/((x^(2)+r^(2))^(3//2))=(1)/(4piin_(0))x.sigmapiunderset(0)overset(R)int(2r)/((x^(2)+r^(2))^(3//2))dr`
`E=(sigma)/(2in_(0))[1-(x)/((x^(2)+R^(2))^(1//2))]=(sigma)/(2epsilon_(0))(1-costheta)`, where `theta` = semi vertical angle subtend by the disc at P.
Also, as `Rrarroo,E=(sigma)/(2in_(@)`, which is the electric field in front of an infinite plane sheet of charge.