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

Consider an element of small angle `dphi` at an angle `phi` as shown.
Soordinates of element : `(R cos phi, R sin phi, 0)`
Coordinates of point : (0, 0, x)
Now electric field due to element :

`vec(dE)=(K(lambda Rd phi).[- R cos phi hat(i)-R sin phi hat(j)+x hat(k)])/((R^(2) cos^(2) phi+R^(2) sin^(2) phi+x^(2))^(3//2))implies E_(x)=Sigmad E_(x)=- underset(0)overset(pi)(int) (K lambdaR^(2) cos phi d phi)/((R^(2)+x^(2))^(3//2))=0`
`E_(y)=Sigmad E_(y)=-underset(0)overset(pi)(int) (KlambdaR^(2) sin phi d phi)/((R^(2)+x^(2))^(3//2))=(2KlambdaR^(2))/((R^(2)+x^(2))^(3//2))=(2KQR)/(pi (R^(2)+x^(2))^(3//2))`
`E_(z)=Sigmad E_(z)=underset(0)overset(pi)(int) (KlambdaRxd phi)/((R^(2)+x^(2))^(3//2))=(KQx)/((R^(2)+x^(2))^(3//2))`
`E_("net")=sqrt(E_(x)^(2)+E_(y)^(2)+E_(z)^(2))=(KQ)/((R^(2)+x^(2))^(3//2)) sqrt((4R^(2))/pi^(2)+x^(2))`
alternate solution :

Consider an element of charge dq at an angle `phi` on circumference of half ring. Due to this element electric field at the point on axis, which is at a distance x from the centre of half ring is dE. This electric field can be resolved into three component.

`E_(Z)= underset(-pi//2)overset(pi//2)(int) dE sin theta sin phi=0`
`E_(X)= underset(-pi//2)overset(pi//2)(int) dE cos theta=(KQx)/([R^(2)+x^(2)]^(3//2))` ...(1)
`E_(Y)= underset(-pi//2)overset(pi//2)(int) dE sin theta cos phi = int (Kdq)/(R^(2) +x^(2)) sin theta. cos phi =(2K lambdaR sin theta)/(R^(2)+x^(2))` ...(2)
`:. dq=lambdaRdphi, sin theta =R/sqrt(R^(2)+x^(2))" "implies" "E_("net")=sqrt(E_(X)^(2)+E_(Y)^(2))`