# A rod of linear charge density A is bent in the form of an arc of radius R, what is the electric field subtended by it at it’s centre, if its central angle is ‘theta’?

Arun Kumar IIT Delhi
8 years ago
Hi

$\\charge density =A \\radius =R \\only horizontal component will exist \\dE={Kdqcos\theta \over R^2}={K*({ARd\theta })*cos\theta \over R^2} \\=>dE={K*({Ad\theta })*cos\theta \over R}={KAcos\theta d\theta \over R} \\=>E=\int_{-\theta/2}^{\theta/2}{KA \over R}cos\theta d\theta \\=>E=2*{KAsin{\theta \over 2} \over R}$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty