Question

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 -Q is uniformly distributed along the lower half as shown in the figure. The electric field E at P, the centre of the semicircle, is

• A. $\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$
• B. $\frac{2Q}{{\pi }^2{\epsilon }_0{r}^2}$
• C. $\frac{4Q}{{\pi }^2{\epsilon }_0{r}^2}$
• D. $\frac{Q}{4{\pi }^2{\epsilon }_0{r}^2}$

Answer 1

The correct option is A $\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$

Take PO as the x-axis and PA as the y-axis. Consider two element EF and E'F' of width $d\theta$ at angular distance $\theta$ above and belove PO, respectively.


The magnitude of the field at P due to either element is
$dE=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{rd\theta \times Q/\left(\frac{\mathrm{\pi r}}{2}\right)}{r^2}=\frac{Q}{2{\pi }^2{\epsilon }_0{r}^2}d\theta \text{Resolving the fields,we find that the components along PO}\text{sums up to zero and hence the resultant field is along}\text{PB}\text{.}\mathrm{Therefore},\mathrm{field}\mathrm{at}P\mathrm{due}\mathrm{to}\mathrm{pair}\mathrm{of}\mathrm{elements}\mathrm{is}2dE\sin \theta .E={\int }_0^{\frac{\pi }{2}}2dE\sin \theta d\theta =2{\int }_0^{\frac{\pi }{2}}\frac{Q}{2{\pi }^2{\epsilon }_0{r}^2}\sin \theta d\theta =\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$