Derive an expression for field intensity due to a uniformly charged ring at a point on the axis.

Answer

B, D

Explanation

A small section of rod is condered at x distance mass of the element
`=(M/Lxxdx=dm)`
`=dE_1=(G(dm)xx1)/((d^2+x^2))=dE_2`
Resultant `eE=2dE_1 sintheta`
`2xx(G(dm))/(d^2+x^2)xxd/(sqrt(d^2+x^2))`
`=(2xxGMxxd dx)/(L(d^2+x^2({(sqrt(d^2+x^2))})`
Total gravitatioN/Al field
`E=int_0^(L/2) (2Gmd dx)/(L(d^2+x^2)^(3/2))`
integrating the above the equation it can be found that
`E=(2Gm)/(dsqrt(L^2+4d^2)0`