Question

If $O$ is the center of a ring of radius $r$, then find the potential at point $O$ due to half ring that has a linear charge density $\lambda$.

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

Answer 1

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

Let us consider a small element on ring of length $dl$ as shown in figure.

Charge on the element,

$dq=\lambda dl=\lambda \left(rd\theta \right)$

Potential due to a charge element $dq$ at a distance $r$ is
$dV=\frac{1}{4\pi {ϵ}_0}\frac{dq}{r}$

Substituting the value of $dq$ we get,
$dV=\frac{1}{4\pi {ϵ}_0}\frac{\lambda rd\theta }{r}$

$\Rightarrow dV=\frac{\lambda d\theta }{4\pi {ϵ}_0}$

Now potential at the centre due to half ring will be,

$V=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{\lambda d\theta }{4\pi {ϵ}_0}$

$\Rightarrow V=\frac{\lambda }{4\pi {ϵ}_0}\theta {|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$\Rightarrow V=\frac{\lambda \pi }{4\pi {ϵ}_0}$

$\therefore V=\frac{\lambda }{4{ϵ}_0}$

Hence, option (a) is the correct answer.