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 figure. The electric field E at P, the center 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

Answer: (A)

$\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$

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 field at P due to either element is

$dE=\frac{1}{4\pi {\epsilon }_0}\frac{rd\theta \times Q/\left(\pi r/2\right)}{r^2}=\frac{Q}{2{\pi }^2{\epsilon }_0{r}^2}d\theta$

Resolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PA.

Therefore, field at P due to pair of elements is 2 E sin $\theta$

$E={\int }_0^{\pi /2}2Esin\theta$

$=2{\int }_0^{\pi /2}\frac{Q}{2{\pi }^2{\epsilon }_0{r}^2}sin\theta d\theta =\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$