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  4. A wire, of length L(= 20 cm), is bent into a semi-circular arc. If the two equal halves, of the arc, were each to be uniformly charged with charges ± Q, [ | Q | = 103 ε0 Coulomb where ε0 is the permittivity (in SI units) of free space] the net electric field at the centre O of the semicircular arc would be :

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Q. A wire, of length $L(= 20 \,cm)$, is bent into a semi-circular arc. If the two equal halves, of the arc, were each to be uniformly charged with charges $\pm Q, [ | Q | = 10^3 \varepsilon_{0}$ Coulomb where $\varepsilon_{0}$ is the permittivity (in SI units) of free space] the net electric field at the centre $O$ of the semicircular arc would be :Physics Question Image

JEE MainJEE Main 2015Electric Charges and Fields

Solution:

$|\vec{E}|$ due to arc of a charged ring at the centre is given by $|\vec{E}|=\frac{K q}{a^{2}} \frac{\sin (\phi / 2)}{(\phi / 2)}$
image
Here, $\left|\vec{E}_{1}\right|=\left|\vec{E}_{2}\right|=\frac{k Q}{a^{2}} \frac{\sin (\pi / 4)}{(\pi / 4)}$
as $\phi=\pi / 2$ for each quarter$=\frac{4 K Q}{\sqrt{2} \pi a^{2}}$
Therefore, $|\vec{E}|=\sqrt{E_{1}^{2}+E_{2}^{2}}=\frac{4 K Q}{\pi a^{2}}$
Also, $L=\pi a \Rightarrow a=\frac{L}{\pi}$
Therefore, $|\vec{E}|=\frac{4 K Q}{\pi\left(\frac{L}{\pi}\right)^{2}}=\frac{4 Q}{4 \not \pi^{2} \epsilon_{o} \frac{L^{2}}{\not \pi^{2^{2}}}}$
$=\frac{Q}{\epsilon_{o} L^{2}}=\frac{10^{3} \epsilon_{o}}{\epsilon_{o}(0.2)^{2}}$
$=\frac{10^{3}}{4 \times 10^{-2}}=25 \times 10^{3} N / C$
$\vec{E}$ is directed along $x$ -axis as shown.