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A thin glass rod is bent in a semicircle of radius \(\mathrm{R}\). A change is non-uniformly distributed along the rod with a linear charge density \(\lambda=\lambda_0 \sin \left(\lambda_0\right.\) is a positive constant). The electric field at the centre \(P\) of the semicircle is,
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Hint : 
\(\begin{aligned}
& \lambda=\lambda_0 \sin \theta \\
& \mathrm{dE}_{\mathrm{y}}=\frac{\mathrm{k} \lambda}{\mathrm{R}} \sin \theta \mathrm{d} \theta \\
& =\frac{\mathrm{k} \lambda_0}{\mathrm{R}} \int_{-\pi / 2}^{+\pi / 2} \sin ^2 \theta \mathrm{d} \theta=\frac{\mathrm{k} \lambda_0}{2 \mathrm{R}} \int_{-\pi / 2}^{+\pi / 2}(1-\cos 2 \theta) \mathrm{d} \theta \\
& \therefore \overrightarrow{\mathrm{E}}=-\frac{\lambda_0}{8 \varepsilon_0 \mathrm{R}} \hat{\mathrm{j}}
\end{aligned}\)
\(\begin{aligned}
& \lambda=\lambda_0 \sin \theta \\
& \mathrm{dE}_{\mathrm{y}}=\frac{\mathrm{k} \lambda}{\mathrm{R}} \sin \theta \mathrm{d} \theta \\
& =\frac{\mathrm{k} \lambda_0}{\mathrm{R}} \int_{-\pi / 2}^{+\pi / 2} \sin ^2 \theta \mathrm{d} \theta=\frac{\mathrm{k} \lambda_0}{2 \mathrm{R}} \int_{-\pi / 2}^{+\pi / 2}(1-\cos 2 \theta) \mathrm{d} \theta \\
& \therefore \overrightarrow{\mathrm{E}}=-\frac{\lambda_0}{8 \varepsilon_0 \mathrm{R}} \hat{\mathrm{j}}
\end{aligned}\)
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