igure 1744 shows an electron passing between two charged met

igure 17.44 shows an electron passing between two charged metal plates that create an 100 N/C vertical electric field perpendicular to the electron\'s original horizontal velocity. (These can be used to change the electron\'s direction, such as in an oscilloscope.) The initial speed of the electron is 3.00 10⁶ m/s, and the horizontal distance it travels in the uniform field is 5.00 cm.

Solution

Force on electron F = qE = 1.6*10^(-19)*100 = 1.6*10^(-17) N

and direction is upward.so acceleration is

a = F/m = 1.6*10^(-17)/(9.1*10^(-31)) = 1.76*10^13 m/s^2 (upward)

using equation

D = ut + 1/2*at^2 for horizontal component

0.05 = 3*10^6*t + 0   [ax = 0]

t = 16.6*10^(-9) sec

now for vertical component

D = 0 + (1/2)*1.76*10^13*( 16.6*10^(-9)))^2 = 0.002424 m

a)

vertical deflection = 0.002424 cm

b)

using equation

v = u + at   for vertical component

v = 0 + 1.76*10^13*( 16.6*10^(-9)) = 292160 = 29.2*10^4 m/s

c)

theta = arctan (Vy/Vx) = arctan(29.2*10^4/(3*10^6)) = 5.559 degree