Question

The linear charge density on a dielectric ring of radius $R$ is varying with $\theta$ as $\lambda ={\lambda }_0\cos \left(\theta /2\right)$. The potential at the centre of the ring is

• A. $0$
• B. $\frac{{\lambda }_0}{2\pi {\in }_0}$
• C. $\frac{{\lambda }_0}{4\pi {\in }_0}$
• D. $\frac{{\lambda }_0}{\pi {\in }_0}$

Answer 1

The correct option is A $0$

Net charge on the ring
$d\theta =\frac{\lambda }{R}$
$\lambda =Rd\theta$
$dQ=\lambda Rd\theta$
According to the question,
$dQ={\lambda }_0\cos \frac{\theta }{2}Rd\lambda$

Potential due to elemental charge
$dV=\frac{1}{4\pi {ϵ}_0}\frac{dQ}{R}$
$=\frac{{\lambda }_0}{4\pi {ϵ}_{0}R}\underset{0}{\overset{2\pi }{\int }}R\cos \frac{\theta }{2}d\theta$
$=\frac{{\lambda }_0}{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}\cos \frac{\theta }{2}d\theta$
$=\frac{{\lambda }_0}{4\pi {ϵ}_0}{\left(\frac{\sin \frac{\theta }{2}}{\frac{1}{2}}\right)}_0^{2\pi }=0$