Respuesta :
Explanation:
It is given that,
Speed of the electron in horizontal region, [tex]v=1.9\times 10^7\ m/s[/tex]
Vertical force, [tex]F_y=4.9\times 10^{-16}\ N[/tex]
Vertical acceleration, [tex]a_y=\dfrac{F_y}{m}[/tex]
[tex]a_y=\dfrac{4.9\times 10^{-16}\ N}{9.11\times 10^{-31}\ kg}[/tex]
[tex]a_y=5.37\times 10^{14}\ m/s^2[/tex]..........(1)
Let t is the time taken by the electron, such that,
[tex]t=\dfrac{x}{v_x}[/tex]
[tex]t=\dfrac{0.024\ m}{1.9\times 10^7\ m/s}[/tex]
[tex]t=1.26\times 10^{-9}\ s[/tex]...........(2)
Let [tex]d_y[/tex] is the vertical distance deflected during this time. It can be calculated using second equation of motion:
[tex]d_y=ut+\dfrac{1}{2}a_yt^2[/tex]
u = 0
[tex]d_y=\dfrac{1}{2}\times 5.37\times 10^{14}\ m/s^2\times (1.26\times 10^{-9}\ s)^2[/tex]
[tex]d_y=0.000426\ m[/tex]
[tex]d_y=0.426\ mm[/tex]
So, the vertical distance the electron is deflected during the time is 0.426 mm. Hence, this is the required solution.
Answer:
[tex]y = 4.24 *10^{-4} m[/tex]
Explanation:
the vertical accerlaration acting on the electron
[tex]a = \frac{f_y}{m}[/tex]
[tex]a = \frac{4.9*10^{-16}}{9.11*10^{-31}} = 5.37 *10^{14} m/s^{2}[/tex]
the time taken by the electron to cover the horizontal distance is
[tex]t =\frac{24*10^{-3}}{1.9*10^{7}}[/tex]
[tex]t = 1.26 *10^{-9} s[/tex]
[tex]v_i = o[/tex]
the vertical distance trvalled in time t is
[tex]y = v_i*t +\frac{1}{2} a_y*t^{2}[/tex]
[tex]y = 0 +\frac{1}{2}*(5.37 *10^{14})(1.26 *10^{-9})^{2}[/tex]
[tex]y = 4.24 *10^{-4} m[/tex]
