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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) ?
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An electron moving with the speed 5×10^6 per sec is shooted parallel t...
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.
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.
Community Answer
An electron moving with the speed 5×10^6 per sec is shooted parallel t...
q= charge of electron.
m= mass of electron.
let s be the dist. travelled
Retarding force on the electron= qEtherefore 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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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) ?
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