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  • How to solve this problem- - Electrostatics - JEE Main-3

The electric field near a conducting surface having a uniform surface charge density  \sigma is given by

  • Option 1)

    \frac{\sigma }{\varepsilon _{0}} and is parallel to the surface

     

     

     

  • Option 2)

    \frac{2\sigma }{\varepsilon _{0}}  and is parallel to the surface

  • Option 3)

    \frac{\sigma }{\varepsilon _{0}}  and is normal to the surface

  • Option 4)

    \frac{2\sigma }{\varepsilon _{0}}  and is normal to the surface

 

Answers (1)

best_answer

As we learned

 

infinite thin plate sheet of charge -

E=\frac{\sigma }{2\epsilon _{0}}\left ( E\propto r^{0} \right )

V=\frac{\sigma r }{2\epsilon _{0}}+c

- wherein

 

 

 Electric field near the conductor surface is given by \frac{\sigma }{\varepsilon _{0}} and it is perpendicular to surface


Option 1)

\frac{\sigma }{\varepsilon _{0}} and is parallel to the surface

 

 

 

Option 2)

\frac{2\sigma }{\varepsilon _{0}}  and is parallel to the surface

Option 3)

\frac{\sigma }{\varepsilon _{0}}  and is normal to the surface

Option 4)

\frac{2\sigma }{\varepsilon _{0}}  and is normal to the surface

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Avinash

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