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(A) Consider a small element of length $d\ell = R d	heta$ on the half ring.Charge on this element is $dq = \lambda d\ell = \lambda R d	heta$,where $

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$\frac{q}{2{\pi ^2}{\varepsilon _0}{R^2}}$

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(A) Consider a small element of length $d\ell = R d	heta$ on the half ring.Charge on this element is $dq = \lambda d\ell = \lambda R d	heta$,where $\lambda = \frac{q}{\pi R}$ is the linear charge density.The electric field at the centre due to this element is $dE = \frac{1}{4\pi \varepsilon_0} \frac{dq}{R^2} = \frac{1}{4\pi \varepsilon_0} \frac{\lambda R d	heta}{R^2} = \frac{\lambda}{4\pi \varepsilon_0 R} d	heta$.Due to symmetry,the horizontal components of the electric field cancel out,and only the vertical components add up.The vertical component is $dE_y = dE \cos 	heta$.Integrating from $-\pi/2$ to $\pi/2$ or $2 \int_{0}^{\pi/2} dE \cos 	heta$:$E = 2 \int_{0}^{\pi/2} \frac{\lambda}{4\pi \varepsilon_0 R} \cos 	heta d	heta = \frac{2\lambda}{4\pi \varepsilon_0 R} [\sin 	heta]_0^{\pi/2} = \frac{\lambda}{2\pi \varepsilon_0 R}$.Substituting $\lambda = \frac{q}{\pi R}$,we get $E = \frac{q/(\pi R)}{2\pi \varepsilon_0 R} = \frac{q}{2\pi^2 \varepsilon_0 R^2}$.

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

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