Question

An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 24 mm horizontally.

Answer 1

`y = 4.24 *10^(-4) m`

Step-by-step explanation:

the vertical accerlaration acting on the electron

`a = (f_y)/(m)`

`a = (4.9*10^(-16))/(9.11*10^(-31)) = 5.37 *10^(14) m/s^(2)`

the time taken by the electron to cover the horizontal distance is

`t =(24*10^(-3))/(1.9*10^(7))`

`t = 1.26 *10^(-9) s`

`v_i = o`

the vertical distance trvalled in time t is

`y = v_i*t +(1)/(2) a_y*t^(2)`

`y = 0 +(1)/(2)*(5.37 *10^(14))(1.26 *10^(-9))^(2)`

`y = 4.24 *10^(-4) m`

Answer 2

Step-by-step explanation:

It is given that,

Speed of the electron in horizontal region,
`v=1.9* 10^7\ m/s`

Vertical force,
`F_y=4.9* 10^(-16)\ N`

Vertical acceleration,
`a_y=(F_y)/(m)`

`a_y=(4.9* 10^(-16)\ N)/(9.11* 10^(-31)\ kg)`

`a_y=5.37* 10^(14)\ m/s^2`..........(1)

Let t is the time taken by the electron, such that,

`t=(x)/(v_x)`

`t=(0.024\ m)/(1.9* 10^7\ m/s)`

`t=1.26* 10^(-9)\ s`...........(2)

Let
`d_y` is the vertical distance deflected during this time. It can be calculated using second equation of motion:

`d_y=ut+(1)/(2)a_yt^2`

u = 0

`d_y=(1)/(2)* 5.37* 10^(14)\ m/s^2* (1.26* 10^(-9)\ s)^2`

`d_y=0.000426\ m`

`d_y=0.426\ mm`

So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.