A thin glass rod is bent into a semicircle of radius $r$. A charge $+Q$ is uniformly distributed along the upper half and a charge $-\mathrm{Q}$ is uniformly distributed along the lower half, as shown in figure. Calculate electric field $E$ at $p$, the center of semicircle.

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Aicharya · Accepted Answer

Take PO as the $x$-axis and $P A$ as the $y$-axis. Consider two elements $E F$ and $E^{\prime} F^{\prime}$ of width $d \theta$ at angular distance $q$ above and below PO, respectively. The magnitude of the field at $P$ due to either element is $d E=\frac{1}{4 \pi \varepsilon_{0}} \frac{r d \theta \times \frac{Q}{\pi r / 2}}{r^{2}}=\frac{Q}{2 \pi^{2} \varepsilon_{0} r^{2}} d \theta$ Resolving the fields, we find that the components along $P O$ sum up to zero and hence, the resultant field is along $P B$. $\therefore$ Field at $P$ due to pair of elements $=2 d E \sin \theta$ $\begin{aligned} E &=\int_{0}^{\pi / 2} 2 d E \sin \theta \ &=2 \int_{0}^{\pi / 2} \frac{Q}{2 \pi^{2} \varepsilon_{0} r^{2}} \sin \theta d \theta=\frac{Q}{\pi^{2} \varepsilon_{0} r^{2}} \end{aligned}$

A thin glass rod is bent into a semicircle of radius \(r\). A charge \(+Q\) is uniformly distributed along…

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