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  • Need clarity, kindly explain! - Electrostatics - JEE Main-8

Two semicircular wires ABC and ADC each of radius r are lying in X-Y plane. If \lambda is linear charge density. Find \overset {\rightarrow } E at origin.

 

  • Option 1)

    \frac{\lambda }{2\pi \epsilon _{o}R} \hat{i}

  • Option 2)

    - \frac{\lambda }{2\pi \epsilon _{o}R} \left ( \hat{i} + \hat{j} \right )

  • Option 3)

    - \frac{\lambda }{2\pi \epsilon _{o}R} \left ( \hat{j} + \hat{k} \right )

  • Option 4)

    \frac{\lambda }{2\pi \epsilon _{o}R} \hat{k}

 

Answers (1)

As we learned

Electric field due to line charge -

E_{x}=\frac{k\lambda }{r}\left ( \sin \alpha +\sin \beta \right )

E_{y }=\frac{k\lambda }{r}\left ( \cos \beta -\cos \alpha \right )

-

 

 

 

 

Due to MA = \frac{\lambda }{4\pi \epsilon _{o}R}\left ( \hat{i} - \hat{k} \right )

Due to ADC = \frac{\lambda }{2\pi \epsilon _{o}R}\left ( - \hat{k} \right )

Due to NC = \frac{\lambda }{4\pi \epsilon _{o}R}\left ( -\hat{i} + \hat{k} \right )

Due to ABC = \frac{\lambda }{2\pi \epsilon _{o}R}\left ( - \hat{j} \right )

Solve it


Option 1)

\frac{\lambda }{2\pi \epsilon _{o}R} \hat{i}

Option 2)

- \frac{\lambda }{2\pi \epsilon _{o}R} \left ( \hat{i} + \hat{j} \right )

Option 3)

- \frac{\lambda }{2\pi \epsilon _{o}R} \left ( \hat{j} + \hat{k} \right )

Option 4)

\frac{\lambda }{2\pi \epsilon _{o}R} \hat{k}

Posted by

Vakul

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