Let us consider a small charge element of charge dq And, dq=\\dfrac{Q}{2\\pi R} .The field at point P due to this element is =E=\\dfrac{Kdq}{ r^2} E=\\dfrac{Kdq}{ (R^2+x^2)} Now, from figure we see that component of field normal to axis is cancelled by two diametrically opposite points.Hence, only component of field along axis is left which add up for all such elements. E_{net}=\\displaystyle\\int E\\cos\ heta where \ heta is same for all elements means \ heta=constant \\implies E_{net}=\\displaystyle\\int\\dfrac{K\\cos\ heta}{(R^2+x^2)}dq \\implies E_{net}=\\dfrac{KQ}{(R^2+x^2)}\\cos\ heta \\implies E_{net}=\\dfrac{KQ}{(R^2+x^2)}\\dfrac{x}{\\sqrt{R^2+x^2}} \\implies E_{net}=\\dfrac{KQx}{(R^2+x^2)^{3/2}} For x>>R ,we can say x^2+R^2\\approx x^2 E_{net}=\\dfrac{KQx}{(x^2)^{3/2}}=\\dfrac{KQx}{x^3} \\implies E_{net}=\\dfrac{KQ}{x^2} Answer-(B)