A charge is distributed uniformly over a ring of radius 'a' Obtain an expression for the electric intensity E at a point on the axis of the ring . Hence show that for point a lerge distance from the ring it behaves like a point charge .

Explanation

Suppose that the ring is placed with its plane perpendicular to the x-axis as shown in Fig. Consider a small element dl of the ring .

As the total charge q is uniformly distributed the charge dq on the element dl is dq `=(q)/(2pia) .dl`
`:.` The magnitude of the field d `bar(E)` produced by the element dl at the field point P is
`bar(DE) =k .(dq)/(r^(2)) =(kq)/(2pia) .(dl)/(r^(2))`
The field `dbar(E)` has two components
(a) the axial component dE cos `theta` , and
(b) the perpendicular component dE sin `theta`
Since the perpendicular components of any two diametrically opposite elements are equal and opposite they all cancel out in pairs .Only the axial components will add up to produce the resultant field `bar(E)` at point P which is given by
`E = overset(2pia)underset(0)(int) dE " cos " theta" "[:' "only the axial components contribute towards" E]`
`=overset(2pia)underset(0)(int) (kq)/(2pia) ,(dl)/(r^(2)).(x)/(r ) =(kpx)/(2pia) .(1)/(r^(3)) overset(2pia)underset(0)(int) dl`
`=(kpx)/(2pia).(1)/(r^(3)) [l]_(0)^(2pia) =(kpx)/(2pia).(1)/((x^(2)+a^(2))^(3//2)).2pia " "[:' r^(2) =x^(2) +a^(2)]`
`"or "" "E =(kpx)/((x^(2) +a^(2))^(3//2))=(1)/(4piepsilon_(0)).(qx)/((x^(2) +a^(2))^(3//2))` ltbgt `" if " " "x gt gt a`
`x^(2) +a^(2) ~~x^(2)`
`E =(1)/(4pi epsilon_(0)) .(qx)/(x^(3)) rArr E=(1)/(4piepsilon_(0)).(q)/(x^(2))`
i.e., When the point is far away from the centre charged ring acts like a point charge .