The Figure shows an electron passing between two charged metal plates that creat
ID: 1396841 • Letter: T
Question
The Figure shows an electron passing between two charged metal plates that create a 90 N/C vertical electric field perpendicular to the electron?s original horizontal velocity. The initial velocity of electron is 3.30x 10 m/s along the x direction, and the horizontal distance it travels in the uniform field is 4.30cm. a) What is in m/s^2 the vertical acceleration of the electron while between the parallel plates? SUBMIT b) For how long, in seconds, will the electron be between the plates? SUBMIT c) What is in m/s the vertical component of the velocity as the electron exits the metal plates? SUBMIT d) What is the vertical deflection (displacement) of the electrons due to the plates? SUBMIT e) At what angle 9 (in degrees) do the electrons exit the plates? SUBMIT Use 9.1 x 10^-31 kg for the mass of the electron. Neglect any edge effects. [OS]Explanation / Answer
a) force on electron due to electric field , F = e*E
also F = ma
so ma = e*E
a (vertical acceleration) = e*E / m
a = 1.6*10-19 * 90 / 9.1*10-31
a = 1.58*1013 m/s2
b) time, t = ? ( Velocity = distance/ time, t = d/v)
t = dx / v
t = 4.30*10-2 / 3.30*106
t = 1.30*10-8 sec
c) Vy = vertical component of velocity
Vy = uy + ay *t ( uy = 0 because initial velocity in y-direction is zero)
= 0 + 1.58*1013*1.30*10-8
= 205.4*103 m/s
d) y ( displacement in y-direction) = ?
from S = ut + (1/2) * a* t2
y = 0 + ( 1/2) * 1.58*1013 * ( 1.30*10-8 )2
y = 1.335*10-3 m
e ) tan @ = Vy / Vx
@ = tan-1 ( Vy / Vx )
= tan-1 ( 205.4*103 / 3.30*106 )
= 3.560
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