question_answer

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 0 of the semi-circular arc would be :             [JEE MAIN 11-04-2015]

A) \[(50\times {{10}^{3}}N/C)\hat{J}\]

B) \[(25\times {{10}^{3}}N/C)\hat{i}\]

C) \[(25\times {{10}^{3}}N/C)\hat{j}\]

D) \[(50\times {{10}^{3}}N/C)\hat{i}\]

Correct Answer: C

Solution :

[c] Due to quarter ring electric field intensity is

\[E=\frac{2k\lambda }{R}\sin \frac{\theta }{2}\]when\[\theta ={}^{\pi }/{}_{2}\]

So, due to each quarter section, field intensity is  \[E=\frac{2k\lambda }{R}\times \sin \frac{\pi }{4}=\frac{\sqrt{2}k\lambda }{R}\]

So Net\[{{\vec{E}}_{Net}}=\sqrt{2}E\]  \[\therefore \]\[\lambda =\frac{\theta }{\pi R/2}\]

\[=\frac{\sqrt{2}\sqrt{2}k\lambda }{R}\]              \[\therefore \]\[\pi R=L\]

\[=\frac{2k.\lambda }{R}=\frac{2k\left( 2\theta  \right)}{\pi {{R}^{2}}}=\frac{4\theta }{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]\[\theta ={{10}^{3}}{{\varepsilon }_{0}}\]

So, \[{{E}_{Net}}=\frac{4\times {{10}^{3}}{{\varepsilon }_{0}}}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}=\frac{4\times {{10}^{3}}}{4{{\pi }^{2}}{{\left( \frac{L}{\pi } \right)}^{2}}}\]

\[=\frac{4\times {{10}^{3}}}{4.{{L}^{2}}}=\frac{4\times {{10}^{3}}}{4\times {{\left( 0.2 \right)}^{2}}}=\frac{4\times {{10}^{3}}}{4\times 0.04}=25\times {{10}^{3}}\]