Question

A wire, of length $L(=20cm)$, is bent into a semi-circular arc. If the two equal halves, of the arc, were each to be uniformly charged with charges $\pm Q,[|Q|=10^3{\epsilon }_0$ Coulomb where ${\epsilon }_0$ is the permittivity (in SI units) of free space] the net electric field at the centre $O$ of the semicircular arc would be :

• A. $\left(50\times 10^{3}N/C\right)\hat{j}$
• B. $\left(25\times 10^{3}N/C\right)\hat{i}$
• C. $\left(25\times 10^{3}N/C\right)\hat{j}$
• D. $\left(50\times 10^{3}N/C\right)\hat{i}$

Answer 1

Answer: (B)

$|\vec{E}|$ due to arc of a charged ring at the centre is given by $|\vec{E}|=\frac{Kq}{a^2}\frac{\sin (\varphi /2)}{(\varphi /2)}$

Here, $\left|{\vec{E}}_1\right|=\left|{\vec{E}}_2\right|=\frac{kQ}{a^2}\frac{\sin (\pi /4)}{(\pi /4)}$
as $\varphi =\pi /2$ for each quarter$=\frac{4KQ}{\sqrt{2}\pi a^2}$
Therefore, $|\vec{E}|=\sqrt{E_1^2+E_2^2}=\frac{4KQ}{\pi a^2}$
Also, $L=\pi a\Rightarrow a=\frac{L}{\pi }$
Therefore, $|\vec{E}|=\frac{4KQ}{\pi {\left(\frac{L}{\pi }\right)}^2}=\frac{4Q}{4/{\pi }^2{ϵ}_o\frac{L^2}{/{\pi }^{2^2}}}$
$=\frac{Q}{{ϵ}_o{L}^2}=\frac{10^3{ϵ}_o}{{ϵ}_o(0.2{)}^2}$
$=\frac{10^3}{4\times 10^{-2}}=25\times 10^{3}N/C$
$\vec{E}$ is directed along $x$ -axis as shown.