A thin circular ring of radius r is charged uniformly so that its linear charge density becomes `lambda` . Derive an expression for the electric field at a point P at a distance x from it along the axis of the ring. Hence, prove that at large distances (r `gtgt` r), the ring behaves as a point charge.

Explanation

Let a thin circular ring of radius . is charged unifornmly having a linear charge density `lambda`.. Let Pbe a point on the axis of the ring at a distance .x. from its centre. Consider an element of length Al around a point A of the ring carrying a charge `Deltaq = lambdaDeltaI` Electric field at point P due to this charge element is
`|DeltavecE| =(1)/(4piin_(0)).(Deltaq)/((AP)^(2))=(1)/(4piin_(0)) .(lambda DeltaI)/((x^(2)+x^(2))`

The electric field is directed along AP and thus subtends an angle 0 with the axis of the ring and can be resolved into two components (i) `DeltaE` cos `theta` along the axis, and (ii) `DeltaE` sin `theta` normal to the axis of ring. It is clear from symmetry that normal components `DeltaE` sin `theta`. due to mutually opposite charge elements (say A and B) nullify each other, but their axial components `.DeltaE cos theta` are added up.
`:.` Net electric field at point P due to whole charged ring will be
`E =sumDeltaE cos theta= sum (1)/(4piin_(0)).(lambdaDeltal)/((r^(2)+x^(2))).(x)/(sqrt(r^(2)+x^(2)))`
`=(lambdax)/(4piin_(0)(r^(2)+x^(2))^(3//2))sum DeltaI= (lambda.x)/(4piin_(0)(r^(2)+x^(2))^(3//2)).2pir`
`= (lambdarx )/(2in_(0)(r^(2)+x^(2))^(3//2))`
The field `vecE` is directed along the axis OP of the charged ring If x `gtgt` r, then the above relation may be written as
`E = (lambdarx)/(2in_(0)(x^(2))^(3//2))=(lambdar)/(2in_(0)x^(2))`
As total charge on the ring q =` lambda.2pir ` so `lambdar= (q)/(2pi)` and hence E = `(q)/(4pi in_(0)x^(2))`
The above relation shows that electric field is same as that due to a point charge q situated at the centre of ring ie., the charged ring is now behaving as a point charge.