A uniformly charged semicircular arc of radius ' $r$ ' has linear charge density $(\lambda)$. What is the…
A uniformly charged semicircular arc of radius ' $r$ ' has linear charge density $(\lambda)$. What is the electric field at its centre?
( $\epsilon_0=$ permittivity of free space)
$\frac{\lambda}{2 \pi \in_0 r}$
$\frac{2 \pi \epsilon_0}{\lambda}$
$\frac{\lambda}{4 \epsilon_0}$
$\frac{2 \epsilon_0}{\lambda}$
Solution
Electric field due to a circular arc at its center,
$\mathrm{E}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}} \sin \alpha$, where $\alpha$ is the angle which arc subtends at its center
For semicircle, $\alpha=90^{\circ}$
Hence, $E=\frac{\lambda}{2 \pi \epsilon_0 r}$