Question 126 of 311

A wire of length $L=20cm$ 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

A$\left(25\times {10}^{3}\right)\hat{i}N{C}^{-1}$B$\left(50\times {10}^{3}\right)\hat{i}N{C}^{-1}$C$\left(25\times {10}^{3}\right)\hat{j}N{C}^{-1}$D$\left(50\times {10}^{3}\right)\hat{j}N{C}^{-1}$

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Answer: (A) $\left(25\times {10}^{3}\right)\hat{i}N{C}^{-1}$

$L=\pi R$

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

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

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



${E}_{1}={E}_{2}=\dfrac{2k⋋}{R}\sin ⁡\dfrac{90}{2}{⋋=\dfrac{Q}{\pi R/2}}$

undefined

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

${E}_{net}=\sqrt{2}{E}_{1}$

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

$=\dfrac{4\times 1\left({10}^{3}{ϵ}_{0}\right)}{4\pi {ϵ}_{0}\pi {R}^{2}}=\dfrac{4\times {10}^{3}{ϵ}_{0}}{4{\pi }^{2}{ϵ}_{0}{R}^{2}}=\dfrac{{10}^{2}}{{R}^{2}}=\dfrac{100}{{\left(\dfrac{1}{5\pi }\right)}^{2}}=25\times {10}^{3}$

${\vec{E}}_{net}=25\times {10}^{3}N{C}^{-1}\hat{i}$