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.An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

  • Option 1)

    10 times greater

  • Option 2)

    5 times greater

  • Option 3)

    smaller

  • Option 4)

    equal

 

Answers (1)

best_answer

As we have learned

Electric Field Intensity -

vec{E}=frac{vec{F}}{q_{0}}=frac{kQ}{r^{2}}

- wherein

 

FOR ELECTRON 

a_{e}=\frac{eE}{m_{e}}

t_{e}=\sqrt{(\frac{2h}{a})}= \sqrt{\frac{2hm_{e}}{eE}}

FOR PROTON :

a_{P}=\frac{eE}{m_{p}}\Rightarrowt_{p}= \sqrt{\frac{2hm_{p}}{eE}}

t_{e}/t_{p}= \sqrt{\frac{m_{e}}{m_{p}}}< 1

\therefore electron will take less time  

 

 

 

 


Option 1)

10 times greater

This is incorrect

Option 2)

5 times greater

This is incorrect

Option 3)

smaller

This is correct

Option 4)

equal

This is incorrect

Posted by

prateek

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