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  • Confused! kindly explain, - Electrostatics - JEE Main-13

The acceleration of an electron in an electric field of magnitude 50 V/cm, if e/m value of the electron is 1.76\times10^{11} C/kg, is 

  • Option 1)

    8.8\times 10^4 m/s^2

  • Option 2)

    6.2 \times 10^{13}m/s^2

  • Option 3)

    5.4\times10^{12} m/s^2

  • Option 4)

    Zero

 

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.

 

 

a = \frac{eE}{m} \Rightarrow a = 1.76 \times 10^{11}\times 50\times 10^2 = 8.8 \times 10^{14}m/s^2

 


Option 1)

8.8\times 10^4 m/s^2

Option 2)

6.2 \times 10^{13}m/s^2

Option 3)

5.4\times10^{12} m/s^2

Option 4)

Zero

Posted by

Vakul

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