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(C) Consider a small element of length $dl$ on the semi-circular ring at an angle $	heta$ with the vertical axis.The charge on this element is $dq =

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$\frac{2k\lambda}{R}$

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(C) Consider a small element of length $dl$ on the semi-circular ring at an angle $	heta$ with the vertical axis.The charge on this element is $dq = \lambda dl = \lambda R d	heta$.The electric field due to this element at the center is $dE = \frac{k dq}{R^2} = \frac{k \lambda R d	heta}{R^2} = \frac{k \lambda d	heta}{R}$.Due to symmetry,the components of the electric field perpendicular to the axis of symmetry cancel out.The net electric field is the integral of the components along the axis of symmetry: $E = \int_{-\pi/2}^{\pi/2} dE \cos 	heta = \int_{-\pi/2}^{\pi/2} \frac{k \lambda}{R} \cos 	heta d	heta$.$E = \frac{k \lambda}{R} [\sin 	heta]_{-\pi/2}^{\pi/2} = \frac{k \lambda}{R} [1 - (-1)] = \frac{2k \lambda}{R}$.

What is the electric field at the center of a semi-circular ring of radius $R$ having a linear charge density $\lambda$? $\left( k = \frac{1}{4\pi \varepsilon_0} \right)$

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