Electrostatics Question 720
Question: 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
Options:
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}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Take PO as the x-axis and PA as the y-axis, Consider two elements EF and Ef 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/( \pi r/2 )}{r^{2}}=\frac{Q}{2{{\pi }^{2}}{\varepsilon _{0}}r^{2}}d\theta $ R
esolving 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 $ 2dE\sin \theta $
$ E=\int _{0}^{\pi /2}{2dE\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}}} $
