Question

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

Answer 1

Let dq be the small charge element on the ring.

The electric potential due to this element at the center of the ring will be
$dV=\frac{kdq}{r}$.............(i)
$dq=rd\theta \times \lambda$
$⟹dV=\frac{krd\theta \times \lambda }{r}$
$\lambda ={\lambda }_0\cos \frac{\theta }{2}$
$⟹dV=\frac{k{\lambda }_0\cos \frac{\theta }{2}rd\theta }{r}$
On integrating this equation we get
$V=k{\lambda }_0{\int }_0^{2\pi }\cos \frac{\theta }{2}d\theta$
On solving this integral we get
$V=0$