Question

A half ring of radius r has a linear charge density $\lambda$. The potential at the centre of the half ring is :

• A. $\frac{\lambda }{4{\epsilon }_0}$
• B. $\frac{\lambda }{4{\pi }^2{\epsilon }_{0}r}$
• C. $\frac{\lambda }{4\pi {\epsilon }_{0}r}$
• D. $\frac{\lambda }{4\pi {\epsilon }_0{r}^2}$

Answer 1

The correct option is A $\frac{\lambda }{4{\epsilon }_0}$

We know potential due to a charge q at a distance r is $\frac{kq}{r}$

Now potential due to $rd\theta$ element is

$\int dV={\int }_0^{\pi }\frac{k.rd\theta }{r}$

$v=k\pi \lambda$

$v=\frac{\pi \lambda }{4\pi {\epsilon }_0}$

$v=\frac{\lambda }{4{\epsilon }_0}$