A wire of length L = 20   c m is bent into a semi-circular arc and the two equal halves of the arc are…

A wire of length $L=20 \mathrm{c}\mathrm{m}$ is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges $+Q$ and $-Q$ as shown in the figure. The magnitude of the charge on each half is $\left|Q\right|={10}^3{\epsilon }_0$, where ${\epsilon }_0$ is the permittivity of free the space. The net electric field at the centre $O$ is

1. $\left(25\times {10}^3\right)\hat{\mathrm{i}} \mathrm{N} {\mathrm{C}}^{-1}$
2. $\left(50\times {10}^3\right)\hat{\mathrm{i}} \mathrm{N} {\mathrm{C}}^{-1}$
3. $\left(25\times {10}^3\right)\hat{\mathrm{j}} \mathrm{N} {\mathrm{C}}^{-1}$
4. $\left(50\times {10}^3\right)\hat{\mathrm{j}} \mathrm{N} {\mathrm{C}}^{-1}$

Solution

$L=\pi R$

$\Rightarrow R=\frac{L}{\pi }=\frac{20}{100\pi }\text{m}=\frac{1}{5\pi } \text{m}$

$\Rightarrow$ due to a charge arc, electric field at centre is given by

$E=\frac{2K\lambda }{R}\sin \frac{\theta }{2}$



$E_1=E_2=\frac{2k⋋}{R}\sin \frac{90}{2} \{⋋=\frac{Q}{\pi R/2}\}$


undefined

Component along $\widehat{j}$ gets cancelled and 

$E_{net}=\sqrt{2}{E}_1$

$=\frac{4KQ}{\pi R^2}$

$=\frac{4\times 1\left({10}^3{ϵ}_0\right)}{4\pi {ϵ}_0 \pi R^2}=\frac{4\times {10}^3{ϵ}_0}{4{\pi }^2{ϵ}_0{R}^2}=\frac{{10}^2}{R^2}=\frac{100}{{\left(\frac{1}{5\pi }\right)}^2}=25\times {10}^3$

${\overrightarrow{E}}_{net}=25\times {10}^3 \mathrm{N} {\mathrm{C}}^{-1}\mathrm{ }\widehat{\mathrm{i}}$

Asked in: JEE Main 2015 (11 Apr Online)

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