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A long wire with uniform charge density \lambda is bent in to given figure . Find \overset {\rightarrow }E at point O .

  • Option 1)

    0

  • Option 2)

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

  • Option 3)

    \frac{\sqrt {2}\lambda }{4 \pi \epsilon _{o}R }

  • Option 4)

    None

 

Answers (1)

best_answer

As we learned

If wire is infinitely long -

l\rightarrow \alpha ,    \alpha = \beta =90^{\circ}

E_{x}=\frac{2k\lambda }{r} ,  E_{y}=0 ,   E_{net}=\frac{\lambda }{2\pi \varepsilon _{0}r}

V=\frac{-\lambda }{2\pi \varepsilon _{0}}\log_{e}r+c

-

 

 

 

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

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

E_{net} = E = \frac{\sqrt{2}\lambda }{4\pi \epsilon _{o}R}

 


Option 1)

0

Option 2)

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

Option 3)

\frac{\sqrt {2}\lambda }{4 \pi \epsilon _{o}R }

Option 4)

None

Posted by

prateek

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