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  • I have a doubt, kindly clarify. - Electrostatics - JEE Main-5

What  will be the electric field due to Uniformly charged disc  At a distance x from centre O on it’s axis if If X\rightarrow 0,

  • Option 1)

    E=\frac{\sigma }{4\varepsilon _{0}}

     

     

     

  • Option 2)

    E\simeq \frac{\sigma }{2\varepsilon _{0}}

  • Option 3)

    E= \frac{2K\lambda }{R}=\frac{Q}{2\pi ^{2}\varepsilon _{0}R^{2}}

  • Option 4)

    0

 

Answers (1)

best_answer

As we learned

 

Uniformly charged disc -

If x \rightarrow 0 

E\simeq \frac{\sigma }{2\epsilon _{0}}

- wherein

i.e Point situated near disc it behaves as an infinite sheet of charge.

 

 E=\frac{\alpha }{2\varepsilon _{0}}\left [ 1-\frac{x}{\sqrt{x^{2}+R^{2}}} \right ]Put x=0

 


Option 1)

E=\frac{\sigma }{4\varepsilon _{0}}

 

 

 

Option 2)

E\simeq \frac{\sigma }{2\varepsilon _{0}}

Option 3)

E= \frac{2K\lambda }{R}=\frac{Q}{2\pi ^{2}\varepsilon _{0}R^{2}}

Option 4)

0

Posted by

Avinash

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