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An electron (mass = 9.1\times 10^{-31}kg and charge = 1.6\times 10^{-19}C) is sent in an electric field of intensity 1\times 10^6 V/m How long would it take for the electron, starting from rest, to attain one–tenth the velocity of light

  • Option 1)

    1.7\times 10^{-12} sec

  • Option 2)

    1.7\times 10^{-6} sec

  • Option 3)

    1.7\times 10^{-8} sec

  • Option 4)

    1.7\times 10^{-10} sec

 

Answers (1)

best_answer

As we have learnt,

 

when Charged Particle at rest in uniform field -

Velocity - 

\dpi{100} v=\frac{QEt}{m}=\sqrt{\frac{2Q\Delta V}{m}}

 

 

- wherein

\Delta V= Potential difference.

 

 By using 

v = \frac{QEt}{m}\Rightarrow \frac{1}{10}\times 3\times 10^8 = \frac{1.6\times 10^{-19}\times 10^6\times t}{9.1\times 10^{-31}} \Rightarrow t = 1.7\times 10^{-10} s

 


Option 1)

1.7\times 10^{-12} sec

Option 2)

1.7\times 10^{-6} sec

Option 3)

1.7\times 10^{-8} sec

Option 4)

1.7\times 10^{-10} sec

Posted by

Avinash

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