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Question

(A) The electric field $E$ due to a charged arc of radius $r$ and linear charge density $\lambda$ at its center is given by the formula $E = \frac{2 k

Vedclass · Accepted Answer

$\frac{\lambda}{2 \pi \epsilon_0 r}$

Vedclass · Answer

(A) The electric field $E$ due to a charged arc of radius $r$ and linear charge density $\lambda$ at its center is given by the formula $E = \frac{2 k \lambda}{r} \sin \alpha$,where $2\alpha$ is the total angle subtended by the arc at the center.For a semicircular arc,the total angle subtended is $180^{\circ}$,so $2\alpha = 180^{\circ}$,which means $\alpha = 90^{\circ}$.Substituting $k = \frac{1}{4 \pi \epsilon_0}$ and $\alpha = 90^{\circ}$ into the formula:$E = \frac{2 (\frac{1}{4 \pi \epsilon_0}) \lambda}{r} \sin(90^{\circ})$Since $\sin(90^{\circ}) = 1$,we get:$E = \frac{2 \lambda}{4 \pi \epsilon_0 r} = \frac{\lambda}{2 \pi \epsilon_0 r}$.

$A$ uniformly charged semicircular arc of radius $r$ has linear charge density $\lambda$. What is the electric field at its centre? $(\epsilon_0 = \text{permittivity of free space})$

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