hey here E denotes episilon me 2 getting the ans as Q/1EL^2 acc 2 the ques pi r=L let l be the linear charge density => l=Q/L also since the charge is distributed over the semicircle thus lets consider a small element of length dx at an angle a frm 0 degree subtending angle da at the centernd let charge dq b distributed over it => dq= l* dx by symmetry we get the cos components of the field cancelled out and finally dfsin a=k.(dq/r^2) * sin a =>F= integration kldxsin a/r^2 =>F= klr/r^2 integration frm 0 to pi sin a da =>F=2kl/r replacing r by L/pi nd l by Q/L we get electric field = Q/2episilon L^2
i hope i cleared the doubt
there is no right way 2 do something wrong !!!!!!!!
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
44
