Calculate electric field at a point on axis, which at a distance x from centre of uniformly charged disc having surface charge density `sigma` and R which also contains a concentric hole of radius r.

Explanation

Consider a ring of radius `y (r lt y lt R)` and width dy concentric with disc and in the plane of the disc. Due to this ring , the electric field at the point P :

`dE=(K (dq)x)/([X^(2)+Y^(2)]^(3//2))`
`E_("net")=underset(r)overset(R)(int) (Kx. sigma (2pi y) dy)/([x^(2)+y^(2)]^(3//2))" "[ :'= sigma 2pi y dy]`
`E_("net")=(2pi sigma kx)/2 underset(x^(2)+r^(2)) overset(x^(2)+R^(2))(int) (d t)/t^(3//2)`, put `x^(2)+y^(2)=t, 2y. dy=d t`
`=(sigma x)/(2 epsi_(0)) [1/sqrt(x^(2)+r^(2))-1/sqrt(x^(2)+R^(2))]` away from centre
Alternate method
We can use superposition principle to solve this problem.
(i) Assume a disc without hole of radius R having surface charge density `+sigma`
(ii) Also assume a concentric disc of radius r in the same plane of first disc having charge density `-sigma`, Now using derived formula in last example the net electric field at the centre is :
`vec(E_("net"))=vec(E_(R))+vec(E_(r))=(sigmax)/(2 epsi_(0))[1/sqrt(r^(2)+x^(2))-1/sqrt(R^(2)+x^(2))]` away from centre.