Question

A uniformly sharged insulating rod of length "has a total charge ${}^{\prime }{Q}^{\prime }$. It is bent into the shape of a semicircle as shown in the fig. Electric field through at $O$ can be expressed as

• A. $\overrightarrow{E}=\frac{Q}{2{\epsilon }_0{L}^2}\overrightarrow{j}$
• B. $\overrightarrow{E}=\frac{Q}{2{\epsilon }_{0}L}\overrightarrow{j}$
• C. $\overrightarrow{E}=\frac{Q}{2{\epsilon }_0{L}^2}(\overrightarrow{i}+\overrightarrow{j})$
• D. $\overrightarrow{E}=\frac{Q}{2{\epsilon }_0{L}^2}(\overrightarrow{i}-\overrightarrow{j})$

Answer 1

The correct option is A $\overrightarrow{E}=\frac{Q}{2{\epsilon }_0{L}^2}\overrightarrow{j}$

Length of wire $=L$

let radius $=R$

$\pi R=L$-----------------$\left(1\right)$

Now$,$

$\overrightarrow{E}=\frac{2KQ}{\varphi R^2}\sin \left(\frac{\varphi }{2}\right)$ where $\varphi =\pi$

$\Rightarrow \overrightarrow{E}\Rightarrow \frac{2KQ}{\pi R^2}\hat{j}$

$=\frac{2Q}{4\pi {\in }_0.\pi R^2}\hat{j}$

$=\frac{Q}{2{\pi }^2{R}^2{\in }_0}\hat{j}$

$\Rightarrow \overrightarrow{E}=\frac{Q}{2{\in }_{0}L}\hat{j}$

Hence,

option $\left(A\right)$ is correct answer.