Semicircular ring of radius R=0.5 m is uniformly charged with q=1.4C.Find electric field at CENTRE OF CURVATURE??

consider a small element of length dl on the ring ..let @ is the angle be radious vector with base then angle substended

by this element at center is d@ ...

arc/radius = @

dl = Rd@     ..............1

let charge dencity = p then charge of this element is dq

dq = p(dl) = pRd@ ..............2

electric field due to this element at center dE (magnitude)= kdq/R²

dE = Kdq/R²cos@ (i) + kdq/R²sin@ (-j)          with direction

dE = kdq/R²[cos@i - sin@j]

dE = kp/R [ cos@d@ (i) - sin@d@ (j) ]            lim from 0 to pi

E = kp/R [ -2j ]

 now we have p (charge per unit length) = Q/piR so

E = 2kQ/piR² (-j)

putting all the values we get

E = 3.2x10¹² N/C

As E=q/4 pi epsilon knot R

R=0.5m  q=1.4C    1/4 pi epsilon knot =9*10 raise to power 9

 therefore, E = 9*10 raise to power 9*1.4/0.5

                    =25.2*10 raise to power 9 C per m