Question

An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

Answer 1

(a). The speed of the electron is
`3.68*10^(4)\ m/s`

(b). The distance traveled by the electron is
`4.53*10^(-5)\ m`

Step-by-step explanation:

Given that,

Initial velocity = 50 km/s

Electric field = 50 N/C

Time = 1.5 ns

(a). We need to calculate the speed of the electron 1.5 n s after entering this region

Using newton's second law

`F = ma`.....(I)

Using formula of electric force

`F = qE`.....(II)

from equation (I) and (II)

`-qE= ma`

`a = (-qE)/(m)`

(a). We need to calculate the speed of the electron

Using equation of motion

`v = u+at`

Put the value of a in the equation of motion

`v = 50*10^(3)-(1.6*10^(-19)*50)/(9.1*10^(-31))*1.5*10^(-9)`

`v=36813.18\ m/s`

`v =3.68*10^(4)\ m/s`

(b). We need to calculate the distance traveled by the electron

Using formula of distance

`s = ut+(1)/(2)at^2`

Put the value in the equation

`s = 3.68*10^(4)*1.5*10^(-9)-(1)/(2)*(1.6*10^(-19)*50)/(9.1*10^(-31))*(1.5*10^(-9))^2`

`s=0.0000453\ m`

`s=4.53*10^(-5)\ m`

Hence, (a). The speed of the electron is
`3.68*10^(4)\ m/s`

(b). The distance traveled by the electron is
`4.53*10^(-5)\ m`