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The charge distribution along the semi-circular arc is non-uniform . Charge per unit length $\lambda $ is given as $\lambda = {\lambda _0}\sin \theta $ , with $\theta $ measured as shown in figure. $\lambda_0$ is a positive constant. The radius of arc is $R$ . The electric field at the center $P$ of semi-circular arc is $E_1$ . The value of $\frac{{{\lambda _0}}}{{{ \in _0}{E_1}R}}$ is
$8$
$1$
$4$
$2$
Solution
$\mathrm{E}_{\mathrm{y}}=\int \mathrm{d} \mathrm{E} \cos \theta$
$\mathrm{E}_{\mathrm{y}}=0$
$E_{x}=\int d E \sin \theta$
$\mathrm{E}_{\mathrm{x}}=\int \frac{\mathrm{kdq}}{\mathrm{R}^{2}} \sin \theta$
$\mathrm{E}_{\mathrm{x}}=\frac{\mathrm{k} \lambda_{0}}{\mathrm{R}} \int_{-\pi / 2}^{\pi / 2} \sin ^{2} \theta \mathrm{d} \theta$
$\mathrm{E}_{\mathrm{x}}=\mathrm{E}_{1}=\frac{\mathrm{k} \lambda_{0}}{2 \mathrm{R}} \pi$
$\mathrm{E}_{1}=\frac{\lambda_{0}}{8 \epsilon_{0} \mathrm{R}}$
