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  • An electron moving with the speed of  m/s is shot parallel to the electric field of

An electron moving with the speed of 5\times 10^6 m/s is shot parallel to the electric field of intensity 7\times 10^3 N/C. The field is responsible for the retardation of the motion of electrons. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of e = 9\times 10^{-31}kg charge =1.6\times 10^{-19}C)

Option: 1

1 cm


Option: 2

1 mm


Option: 3

1 m


Option: 4

1km


Answers (1)

best_answer

Electric force qE = ma \Rightarrow a = \frac{QE}{m} \therefore a = \frac{1.6\times 10^{-19}\times 7\times 10^3}{9\times 10^{-31}} = \frac{7\times 1.6}{9}\times 10^{15}

u = 5\times 10^6 and v = 0 

\therefore from v^2 = u^2 -2as\Rightarrow s = \frac{u^2}{2a}

\therefore Distance\;= s = \frac{(5\times 10^6)^2\times 9}{2\times 7\times 1.6\times 10^{15}} = 1 cm (approx)

 

Posted by

Gautam harsolia

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