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

Answer:

$y = 4.24 *10^{-4} m$

Explanation:

the vertical accerlaration acting on the electron

$a = \frac{f_y}{m}$

$a = \frac{4.9*10^{-16}}{9.11*10^{-31}} = 5.37 *10^{14} m/s^{2}$

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

$t =\frac{24*10^{-3}}{1.9*10^{7}}$

$t = 1.26 *10^{-9} s$

$v_i = o$

the vertical distance trvalled in time t is

$y = v_i*t +\frac{1}{2} a_y*t^{2}$

$y = 0 +\frac{1}{2}*(5.37 *10^{14})(1.26 *10^{-9})^{2}$

$y = 4.24 *10^{-4} m$

Answer 2

Explanation:

It is given that,

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

Vertical force, $F_y=4.9\times 10^{-16}\ N$

Vertical acceleration, $a_y=\dfrac{F_y}{m}$

$a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}$  

$a_y=5.37\times 10^{14}\ m/s^2$..........(1)

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

$t=\dfrac{x}{v_x}$

$t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}$

$t=1.26\times 10^{-9}\ s$...........(2)

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

$d_y=ut+\dfrac{1}{2}a_yt^2$

u = 0

$d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2$

$d_y=0.000426\ m$

$d_y=0.426\ mm$

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