A ring of radius a contains a charge q distributed. uniformly ober its length. Find the electric field at a point. on the axis of the ring at a distance x from the centre.

Explanation

shows the situation. Let us consider a
. small element of the ring at the point A having a charge
. dQ. The field at P due to this element is
.`dE = dQ/4 pi epsilon_0 ((AP)) ^2`
By symmmetry, the field at P will be along the exis OP
. The comonent of dE along this direction is.
`dQ cos theta = dQ / 4 pi epslion_0 ( AP)^ 2 ((OP/AP)).`
`= x dQ / 4 pi epslion_0 ((a^2 + x^2)) ^3/2`
The net field at P is
. ` E = int dE cos theta = int x dQ/4 pi epslion_0((a^2 + x^2 ))^(3/2).`
`= 4 pi epslion_0 (( a^2 + x^2 )) ^(3/2) int dQ = xQ / 4 pi epslion_0 (a^2 + x^(3/2).`