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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)
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( $\epsilon_0=$ permittivity of free space)
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The correct answer is:
$\frac{\lambda}{2 \pi \in_0 r}$
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, $E=\frac{\lambda}{2 \pi \epsilon_0 r}$
$\mathrm{E}=\frac{2 \mathrm{k} \lambda}{\mathrm{r}} \sin \alpha$, where $\alpha$ is the angle which arc subtends at its center
For semicircle, $\alpha=90^{\circ}$
Hence, $E=\frac{\lambda}{2 \pi \epsilon_0 r}$
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