An electron moving with the speed `5xx10^(6)` per sec is shot parallel to the electric field of intensity `1xx10^(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= 9xx10^(-31)Kg` charge `= 1.6xx10^(-19)C)`

Options

A. 7m
B. 0.7mn
C. 7cm
D. 0.7 cm

Answer

C

Explanation

Electric force,`qE=ma`
`a=(qE)/m`
`a=(1.6xx10^(-19)xx1xx10^(3))/(9xx10^(-31))=(1.6xx10^(5))/9 ms^(-2)`
`u=5xx10^(6)` and v=0
From `v^(2)=u^(2)-2asimpliess=(u^(2))/(2a)`
Distance `s=((5xx10^(6))^(2)xx9)/(2xx1.6xx10^(15))=7cm`