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Two semicircular wires ABC and ADC, each of radius R, are lying on xy and xz planes, respectively.

If the linear charge density of the semicircular parts and straight parts is λ . The electric field intensity E at the origin is

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As we learned

Electric field due to line charge -

E_{x}=\frac{k\lambda }{r}\left ( \sin \alpha +\sin \beta \right )

E_{y }=\frac{k\lambda }{r}\left ( \cos \beta -\cos \alpha \right )

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Due to MA = \frac{\lambda }{4\pi \epsilon _{o}R}\left ( \hat{i} - \hat{k} \right )

Due to ADC = \frac{\lambda }{2\pi \epsilon _{o}R}\left ( - \hat{k} \right )

Due to NC = \frac{\lambda }{4\pi \epsilon _{o}R}\left ( -\hat{i} + \hat{k} \right )

Due to ABC = \frac{\lambda }{2\pi \epsilon _{o}R}\left ( - \hat{j} \right )

\begin{aligned} &\text { Net electric field is }\\ &\vec{E}_{0}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_{3}+\vec{E}_{4}=-\frac{\lambda}{2 \pi \varepsilon_{0} R}(\hat{j}+\hat{k}) \end{aligned}

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