sharad
Last Activity: 7 Years ago
Just make a ring and mark it centre as o.Now from o take an theta of thickeness d(theta) so now the lenght taken on ring is rd(theta).As lameda=charge /lenght.And here lameda is not constant but changing with theta so we have to integerate .Let there be dq charge on the thickness rd(theta).So dq=lameda(not)×cos(theta/4)×rd(theta)And integerate left side 0 to q and right from 0 to pie.You will find answer 2(2)^1/2lameda(not)rBut net charge is 2 times of this as we integerate only a semicircle part.Therefore net charge is 4(2)^1/2lameda(not)r