Charge q is uniformly distributed over a thin | vedclass

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(A) Consider a small element of length $dl = R d	heta$ on the half ring at an angle $	heta$ with the vertical axis.The linear charge density is $\la

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

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(A) Consider a small element of length $dl = R d	heta$ on the half ring at an angle $	heta$ with the vertical axis.The linear charge density is $\lambda = \frac{q}{\pi R}$.The charge on the element is $dq = \lambda dl = \lambda R d	heta$.The electric field at the centre due to this element is $dE = \frac{1}{4\pi \varepsilon_0} \frac{dq}{R^2} = \frac{1}{4\pi \varepsilon_0} \frac{\lambda R d	heta}{R^2} = \frac{\lambda d	heta}{4\pi \varepsilon_0 R}$.By symmetry,the horizontal components of the electric field cancel out,and only the vertical components $dE \cos 	heta$ add up.The total electric field at the centre is $E = \int_{-\pi/2}^{\pi/2} dE \cos 	heta = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4\pi \varepsilon_0 R} \cos 	heta d	heta$.$E = \frac{\lambda}{4\pi \varepsilon_0 R} [\sin 	heta]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4\pi \varepsilon_0 R} (1 - (-1)) = \frac{2\lambda}{4\pi \varepsilon_0 R} = \frac{\lambda}{2\pi \varepsilon_0 R}$.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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