An electron moving with the speed 5×10^6 per sec is shooted parallel to the electric field of intensity 1×10^3N/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 electron =9×10^-31 ,charge =1.6×10^-19) ?

Question

Answer

Given:

Speed of electron (v) = 5 x 10^6 m/s

Electric field intensity (E) = 1 x 10^3 N/C

Mass of electron (m) = 9 x 10^-31 kg

Charge of electron (q) = 1.6 x 10^-19 C

Formula:

The force (F) acting on the electron due to the electric field is given by:

F = Eq

The acceleration (a) of the electron is given by:

a = F/m

The distance (d) travelled by the electron before coming to rest is given by:

d = (v^2)/(2a)

Solution:

We know that the force acting on the electron is given by:

F = Eq

= (1 x 10^3 N/C) x (1.6 x 10^-19 C)

= 1.6 x 10^-16 N

The acceleration of the electron is:

a = F/m

= (1.6 x 10^-16 N)/(9 x 10^-31 kg)

= 1.78 x 10^14 m/s^2

The distance travelled by the electron before coming to rest is:

d = (v^2)/(2a)

= [(5 x 10^6 m/s)^2]/[2 x (1.78 x 10^14 m/s^2)]

= 70.2 x 10^-6 m

= 7.02 x 10^-5 m

Therefore, the distance travelled by the electron before coming to rest is 7.02 x 10^-5 m.

Answer

q= charge of electron.

m= mass of electron.

let s be the dist. travelled

Retarding force on the electron= qE

therefore retarding acceleration on electron[a]= qE/m

initial speed of electron[u]= 5 x 10^6 m/s

final speed when it comes to rest[v]= 0 m/s

now just apply the equation of motion v^2 - u^2 = 2as and get the answer.

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