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A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec{E}$ at the centre $O$ is
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Verified Answer
The correct answer is:
$-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}$
$-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}$
Linear charge density $\lambda=\left(\frac{q}{\pi r}\right)$
$E=\int d E \sin \theta(-\hat{\mathrm{j}})=\int \frac{\mathrm{K} \cdot d q}{\mathrm{r}^2} \sin \theta(-\hat{\mathrm{j}})$
$\mathrm{E}=\frac{\mathrm{K}}{\mathrm{r}^2} \int \frac{\mathrm{qr}}{\pi r} \mathrm{~d} \theta \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{K}}{\mathrm{r}^2} \frac{\mathrm{q}}{\pi} \int_0^\pi \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{q}}{2 \pi^2 \varepsilon_0 \mathrm{r}^2}(-\hat{\mathrm{j}})$
$E=\int d E \sin \theta(-\hat{\mathrm{j}})=\int \frac{\mathrm{K} \cdot d q}{\mathrm{r}^2} \sin \theta(-\hat{\mathrm{j}})$
$\mathrm{E}=\frac{\mathrm{K}}{\mathrm{r}^2} \int \frac{\mathrm{qr}}{\pi r} \mathrm{~d} \theta \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{K}}{\mathrm{r}^2} \frac{\mathrm{q}}{\pi} \int_0^\pi \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{q}}{2 \pi^2 \varepsilon_0 \mathrm{r}^2}(-\hat{\mathrm{j}})$
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