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  • Need clarity, kindly explain! - Electrostatics - JEE Main-5

 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,\left [ \left | Q \right |=10^{3} \right \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 semi-circular arc would be :

  • Option 1)

    \left ( 50\times 10^{3}N/C \right )\: \hat{j}

  • Option 2)

    \left ( 25\times 10^{3}N/C \right )\: \hat{i}

  • Option 3)

    \left ( 25\times 10^{3}N/C \right )\: \hat{j}

  • Option 4)

    \left ( 50\times 10^{3}N/C \right )\: \hat{i}

 

Answers (1)

As we learnt in

Linear charge distribution -

\left ( \lambda \right ) - charge per unit length.

\dpi{100} \lambda=\frac{q}{L}=\frac{C}{m}=Cm^{-1}

- wherein

wire, circulating ring ex.

 

 

 

 

 E= \frac{2K\lambda }{R} \, \, sin\frac{\Theta }{2}

So due to each quarter section field intensity

E= \frac{2K\lambda }{R} \, \, \times sin\frac{\Theta }{4}= \frac{\sqrt{2}K\lambda }{R} \: \: \left ( \because \Theta = \frac{\pi}{2} \right )

So, \underset{E}{\rightarrow}_{net}= \sqrt{2}E\hat{i}= \sqrt{2}\times \frac{\sqrt{2}K\lambda}{R} \: \hat{i}= \frac{2K\lambda \hat{i}}{R}= \frac{2K\left ( 2Q \right ) \hat{i}}{R^{2}}= \frac{4Q\hat{i}}{4\pi^{2}E_{0}R^{2}}

Q=10^{3}E_{0}C, \: \: \: \pi R= L = 20cm

\underset{E}{\rightarrow}_{net}= \frac{4\times 10^{3}E_{0}}{4\times \left ( 0.2 \right )^{2}E_{0}}\, \hat{i}= \frac{4\times 10^{3}}{4\times \left ( 0.04 \right )}\hat{i}\,= 25\times 10^{3} NC^{-1}\hat{i}

 


Option 1)

\left ( 50\times 10^{3}N/C \right )\: \hat{j}

Incorrect Option

Option 2)

\left ( 25\times 10^{3}N/C \right )\: \hat{i}

Correct Option

Option 3)

\left ( 25\times 10^{3}N/C \right )\: \hat{j}

Incorrect Option

Option 4)

\left ( 50\times 10^{3}N/C \right )\: \hat{i}

Incorrect Option

Posted by

Vakul

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