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An electron moving with the speed 5\times 10^6 per sec is shot parallel to the electric field of intensity 1\times 10^3 N/C. Field is responsible for the retardation of motion of electron. 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)

    7m

  • Option 2)

    0.7 mm

  • Option 3)

    7cm

  • Option 4)

    0.7 cm

 

Answers (1)

As we have learnt,

 

when Charged Particle at rest in uniform field -

Force and acceleration

F= QE

a= \frac{QE}{m}

 

 

- wherein

m - mass

Q - charge

E - Electric field strength.

 

 Electric force qE = ma \Rightarrow a = \frac{QE}{m} \therefore a = \frac{1.6\times 10^{-19}\times 1\times 10^3}{9\times 10^{-31}} = \frac{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 1.6\times 10^{15}} = 7 cm (approx)

 


Option 1)

7m

Option 2)

0.7 mm

Option 3)

7cm

Option 4)

0.7 cm

Posted by

Vakul

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