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(C) Let the radius of the semicircle be $R$. The length of the wire is $L = \pi R$,so $R = L / \pi$. The linear charge density is $\lambda = Q / L$.Co

Vedclass · Accepted Answer

$\frac{Q}{2 \varepsilon_0} \frac{\pi}{L^2}$

Vedclass · Answer

(C) Let the radius of the semicircle be $R$. The length of the wire is $L = \pi R$,so $R = L / \pi$. The linear charge density is $\lambda = Q / L$.Consider a small element of the wire 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 due to this element at the centre is $dE = \frac{1}{4 \pi \varepsilon_0} \frac{dQ}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda R d	heta}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda d	heta}{R}$.By symmetry,the horizontal components of the electric field cancel out. The net electric field is the integral of the vertical components $dE \cos 	heta$ from $-\pi / 2$ to $\pi / 2$:$E = \int_{-\pi / 2}^{\pi / 2} dE \cos 	heta = \frac{\lambda}{4 \pi \varepsilon_0 R} \int_{-\pi / 2}^{\pi / 2} \cos 	heta d	heta = \frac{\lambda}{4 \pi \varepsilon_0 R} [\sin 	heta]_{-\pi / 2}^{\pi / 2} = \frac{\lambda}{4 \pi \varepsilon_0 R} (1 - (-1)) = \frac{2 \lambda}{4 \pi \varepsilon_0 R} = \frac{\lambda}{2 \pi \varepsilon_0 R}$.Substituting $\lambda = Q / L$ and $R = L / \pi$:$E = \frac{Q / L}{2 \pi \varepsilon_0 (L / \pi)} = \frac{Q}{2 \pi \varepsilon_0 L} \cdot \frac{\pi}{L} = \frac{Q}{2 \varepsilon_0 L^2} \cdot \pi = \frac{\pi Q}{2 \varepsilon_0 L^2}$.Wait,checking the options provided,the standard result for a semicircle is $E = \frac{2k\lambda}{R} = \frac{2}{4\pi\varepsilon_0} \cdot \frac{Q/L}{L/\pi} = \frac{2\pi Q}{4\pi\varepsilon_0 L^2} = \frac{Q}{2\varepsilon_0 L^2}$.Thus,the correct option is $C$.

$A$ wire of length $L$ has a charge $Q$ distributed uniformly along its length. The wire is bent in the shape of a semicircle. The magnitude of the electric field at the centre of curvature of the semicircle is

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