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  • Need clarity, kindly explain! - Electrostatics - JEE Main-10

A long wire of \lambda charge density is bent into given structure . Find \overset {\rightarrow }E_{net} at point O .

  • Option 1)

    0

  • Option 2)

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

  • Option 3)

    \frac{\sqrt{2}\lambda }{2\pi \epsilon _{o}R}

  • Option 4)

    \frac{\lambda }{4\pi \epsilon _{o}R}

 

Answers (1)

As we learned

If Point P lies near one end of infinitely long wire. -

\alpha =0     \beta =\frac{\pi }{2}

E_{x}=E_{y}=\frac{k\lambda }{r}

E_{net}=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{2}\frac{k\lambda }{r}

- wherein

 

 

 

E = E_{1}+E_{2}+E_{3}

= \frac{\lambda }{4\pi \epsilon _{o}R} \left [ \hat{i} - \hat{j}\right ] - \frac{\lambda }{4\pi \epsilon _{o}R} \left [ \hat{i} + \hat{j}\right ] + \frac{\lambda }{2\pi \epsilon _{o}R}\hat{j} = 0


Option 1)

0

Option 2)

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

Option 3)

\frac{\sqrt{2}\lambda }{2\pi \epsilon _{o}R}

Option 4)

\frac{\lambda }{4\pi \epsilon _{o}R}

Posted by

Vakul

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