Respuesta :
Answer:
The deviation in path is [tex]4.39 \times 10^{-3}[/tex]
Explanation:
Given:
Velocity [tex]v = 1 \times 10^{6}[/tex] [tex]\frac{m}{s}[/tex]
Electric field [tex]E = 500[/tex] [tex]\frac{V}{m}[/tex]
Distance [tex]x = 1 \times 10^{-2}[/tex] m
Mass of electron [tex]m = 9.1 \times 10^{-31}[/tex] kg
Charge of electron [tex]q = 1.6 \times 10^{-19}[/tex] C
Time taken to travel distance,
[tex]t = \frac{x}{v}[/tex]
[tex]t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }[/tex]
[tex]t = 10^{-8}[/tex] sec
Acceleration is given by,
[tex]F = qE[/tex]
[tex]ma = qE[/tex]
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }[/tex]
[tex]a = 8.77 \times 10^{13}[/tex] [tex]\frac{m}{s^{2} }[/tex]
For finding the distance, we use kinematics equations.
[tex]y = vt + \frac{1}{2} at^{2}[/tex]
Where [tex]v = 0[/tex] because here initial velocity zero
[tex]y = \frac{1}{2} at^{2}[/tex]
[tex]y = \frac{1}{2} \times 8.77 \times 10^{13 } \times (10^{-2} )^{2}[/tex]
[tex]y = 4.39 \times 10^{-3}[/tex] m
Therefore, the deviation in path is [tex]4.39 \times 10^{-3}[/tex]
