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 Calculate the electric field E at P, the centre of semicircle.

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

Answer 1

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

Take PO as the x-axis and PA as the y-axis
Consider two elements EF and E’F’ of width $dr$ at angular distance $\theta$ above and below PO, respectively.

The charge of these elements is $dq=rd\theta \times \frac{Q}{\pi r/2}$

The magnitude of the field at P due to either element is

$dE=\frac{1}{4\pi {ϵ}_0}\frac{dq}{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{rd\theta \times \frac{Q}{\pi r/2}}{r^2}=\frac{Q}{2{\pi }^2{ϵ}_0{r}^2}d\theta$
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 $=2dEsin\theta$
$E={\int }_0^{\pi /2}\frac{Q}{2{\pi }^2{ϵ}_0{r}^2}sin\theta d\theta =\frac{Q}{{\pi }^2{ϵ}_0{r}^2}$