Question

The linear charge density on a dielectric ring of radius R varies with $\theta$ as $\lambda ={\lambda }_{0}cos\theta /2$, where ${\lambda }_0$is constant. Find the potential at the center O of the ring [in volt].

Answer 1

The charge on the infinitesimal element of length $Rd\theta$ is
$dQ=\lambda Rd\theta ={\lambda }_0\cos \frac{\theta }{2}Rd\theta$
Potential at the center of ring due to charge $dQ$ is
$dV=k\frac{dQ}{R}=k{\lambda }_0\cos \frac{\theta }{2}d\theta$
$V={\int }_0^{2\pi }k{\lambda }_0\cos \frac{\theta }{2}d\theta =2k{\lambda }_0[\sin \frac{\theta }{2}{]}_0^{2\pi }=0$