Question

A thin Non-conducting 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. Find the electric field $E$ at $\text{P}$.

• A. $\frac{Q}{{\pi }^2{ϵ}_0{r}^2}$
• B. $\frac{2Q}{{\pi }^2{ϵ}_0{r}^2}$
• 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 $\text{PO}$ as the $\text{x-axis}$ and $\text{PA}$ as the $\text{y-axis}$.

Consider two small elements $\text{EF}$ and $\text{E'F'}$ of width $d\theta$ at angular distance $\theta$ above and below $\text{PO}$, as shown in figure.


The magnitude of the field at $\text{P}$ due to either elements is

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

$\Rightarrow dE=\frac{Q}{2{\pi }^2{ϵ}_0{r}^2}d\theta$

Resolving the fields, we find that the components along $\text{PO}$ sum up to zero, and hence the resultant field is along $\text{PB}$.

Therefore, field at $\text{P}$ due to pair of elements is $2dE\sin \theta$

Therefore, the total electric field at $\text{P}$ due the semicircular arc is,

$E=\underset{0}{\overset{\frac{\pi }{2}}{\int }}2dE\sin \theta$

Substituting the value of $dE$, we get

$E=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{Q}{{\pi }^2{ϵ}_0{r}^2}\sin \theta d\theta$

$\therefore E=\frac{Q}{{\pi }^2{ϵ}_0{r}^2}$

Hence, option (a) is the correct answer.

Alternate solution:

Electric field due to a quarter ring carrying charge $+Q$ is given by $${\overrightarrow{E}}_1=\frac{\sqrt{2}k\lambda }{r}\widehat{r_1}$$Electric field due to a quarter ring carrying charge $-Q$ is given by $${\overrightarrow{E}}_2=\frac{\sqrt{2}k\lambda }{r}\widehat{r_2}$$Since, $\widehat{r_1}$ is perpendicular to $\widehat{r_2}$



The net electric field at point $P$ is given by $$|E|=\sqrt{{\left(E_1\right)}^2+{\left(E_2\right)}^2}$$ $\Rightarrow |E|=\frac{2k\lambda }{r}$

Since, $\lambda =\frac{Q}{\left(\frac{\pi r}{2}\right)}$ , we can rewrite the above equation as,

$E=2\times 2\times \frac{Q}{4\pi {\epsilon }_0\times \pi r\times r}=\frac{Q}{{\pi }^2{ϵ}_0{r}^2}$