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(A) Consider a small element of length $dl = r d	heta$ at an angle $	heta$ with the vertical axis. The charge on this element is $dq = \lambda dl =

Vedclass · Accepted Answer

$\frac{\lambda}{2 \pi \varepsilon_0 r}$

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(A) Consider a small element of length $dl = r d	heta$ at an angle $	heta$ with the vertical axis. The charge on this element is $dq = \lambda dl = \lambda r d	heta$.The electric field $dE$ at the centre due to this element is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{r^2} = \frac{\lambda d	heta}{4 \pi \varepsilon_0 r}$.By symmetry,the horizontal components of the electric field cancel out. We only need to integrate the vertical component $dE \cos 	heta$ from $-\pi/2$ to $\pi/2$.$E = \int_{-\pi/2}^{\pi/2} \frac{\lambda \cos 	heta d	heta}{4 \pi \varepsilon_0 r} = \frac{\lambda}{4 \pi \varepsilon_0 r} [\sin 	heta]_{-\pi/2}^{\pi/2}$.$E = \frac{\lambda}{4 \pi \varepsilon_0 r} (1 - (-1)) = \frac{2 \lambda}{4 \pi \varepsilon_0 r} = \frac{\lambda}{2 \pi \varepsilon_0 r}$.

$A$ uniformly charged semicircular arc of radius $r$ has a linear charge density $\lambda$. What is the electric field at its centre? ($\varepsilon_0$ is the permittivity of free space)

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