question_answer

A thin glassrod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half, and a charge -Q is uniformly distributed along the lower half, as shown in fig. The electric field E at P, the center of the semicircle, is

A) \[\frac{Q}{{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}\]                     

B) \[\frac{2Q}{{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}\]

C) \[\frac{4Q}{{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}\]       

D)        \[\frac{Q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}\]

Correct Answer: A

Solution :

[a] Take PO as the x-axis and PA as the y-axis, Consider two elements EF and E?F? of width \[d\theta \]at angular distance \[\theta \] above and below PO, respectively. The magnitude of the fields at P due to either element is \[dE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{rd\theta \times Q/\left( \pi r/2 \right)}{{{r}^{2}}}=\frac{Q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}d\theta \] Resolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PB. Therefore, field at P due to pair of elements is \[2d\text{ }E\sin \theta \] \[E=\int_{0}^{\pi /2}{2d\,\,E\,\sin \theta }\] \[=2\int_{0}^{\pi /2}{\frac{Q}{2\pi {{\varepsilon }_{0}}{{r}^{2}}}\sin \theta d\theta =\frac{Q}{{{\pi }^{2}}{{\varepsilon }_{0}}{{r}^{2}}}}\]