Respuesta :
Explanation:
It is given that,
Speed of the proton, [tex]v=3.6\times 10^5\ m/s[/tex]
Electric field, [tex]E=9.4\times 10^3\ N/C[/tex]
(a) Distance covered, d = 4 cm = 0.04 m
Let t is the time interval required for the proton to travel 4.00 cm horizontally. It can be calculated as :
[tex]t=\dfrac{d}{v}[/tex]
[tex]t=\dfrac{0.04}{3.6\times 10^5}[/tex]
[tex]t=1.11\times 10^{-7}\ s[/tex]
or
t = 111 ns
(b) Since, initial speed = 0 in vertical direction. So,
So, q E = ma
[tex]a_y=\dfrac{E_yq}{m}[/tex]
Displacement is given by :
[tex]y=ut+\dfrac{1}{2}a_yt^2[/tex]
[tex]y=\dfrac{1}{2}\dfrac{E_yq}{m}t^2[/tex]
[tex]y=\dfrac{1}{2}\times \dfrac{9.4\times 10^3\times 1.67\times 10^{-19}}{1.67\times 10^{-27}}(1.11\times 10^{-7})^2[/tex]
[tex]y=0.00579\ m[/tex]
(c) For vertical component of velocity, use equation of kinematics as :
[tex]v_y^2-u^2=2a_yd[/tex] (d = 4 cm)
[tex]v=\sqrt{2\dfrac{E_yq}{m}d_y}[/tex]
[tex]v=\sqrt{2\times \dfrac{9.4\times 10^3\times 1.67\times 10^{-19}}{1.67\times 10^{-27}}\times 0.04}[/tex]
[tex]v=2.74\times 10^5\ m/s[/tex]
For horizontal component of velocity,
[tex]v_x=\dfrac{d}{t}[/tex]
[tex]v_x=\dfrac{0.04}{1.11\times 10^{-7}}[/tex]
[tex]v_x=3.6\times 10^5\ m/s[/tex]
Hence, this is the required solution.
