Question

An electron is released with a velocity of `5xx10^(6)ms^(-1)` in an electric field of `10^(3)NC^(-1)` which has been applied so as to oppose its motion. What distance would the electron travel and how much time could it take before it is brought to rest?
A. `2.8xx10^(-8)s`
B. `1.8xx10^(-8)s`
C. `6.4xx10^(-8)s`
D. `4.2xx10^(-8)s`

Answer 1

Correct Answer - a
Since electric field is applied so as to oppose the motion of electron,
`a= -(eE)/(m)` (retardation)
`= -(1.6xx10^(-19)xx10^(3))/(9.1xx10^(-31))= 1.758xx10^(4)ms^(-2)`
Now, `u= 5xx10^(6) ms^(-1), v=0`
Using the relation: `v^(2)-u^(2)= 2`, as we get
`S= 7.11xx10^(-2)m`
Using the relation: `v=u+at`, we get,
`t= 2.844xx10^(-8)s`