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 centre of the semicircle, is
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 centre 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.
$$\frac{Q}{{4{\pi ^2}{\varepsilon _0}{r^2}}}$$
Answer :
$$\frac{Q}{{{\pi ^2}{\varepsilon _0}{r^2}}}$$
Solution :
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
$$dE = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\frac{{rd\theta \times Q}}{{\left( {\frac{{\pi r}}{2}} \right)}}}}{{{r^2}}} = \frac{Q}{{2{\pi ^2}{\varepsilon _0}{r^2}}}d\theta $$
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 $$2d\,E\sin \theta $$
$$\eqalign{ & E = \int_0^{\frac{\pi }{2}} {2d\,E\sin \theta } \cr & = 2\int_0^{\frac{\pi }{2}} {\frac{Q}{{2\pi {\varepsilon _0}{r^2}}}\sin \theta d\theta } = \frac{Q}{{{\pi ^2}{\varepsilon _0}{r^2}}} \cr} $$
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
$$dE = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\frac{{rd\theta \times Q}}{{\left( {\frac{{\pi r}}{2}} \right)}}}}{{{r^2}}} = \frac{Q}{{2{\pi ^2}{\varepsilon _0}{r^2}}}d\theta $$
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 $$2d\,E\sin \theta $$
$$\eqalign{ & E = \int_0^{\frac{\pi }{2}} {2d\,E\sin \theta } \cr & = 2\int_0^{\frac{\pi }{2}} {\frac{Q}{{2\pi {\varepsilon _0}{r^2}}}\sin \theta d\theta } = \frac{Q}{{{\pi ^2}{\varepsilon _0}{r^2}}} \cr} $$
