Answer to Question 9

Electrons are accelerated by a potential difference of 2000 V. They pass from left to right on the centreline between two horizontal plates that are separated by a distance of 3.0 cm The plates are 5.0 cm long and the top plate has a potential of 1000 V, while the bottom plate has a potential of 0 V. Calculate: (a) the speed of the electrons; (b) the force acting on each electron; (c) the acceleration of each electron; (d) the resultant velocity of the electrons in the beam; (e) the direction relative to the horizontal. (f) By considering the displacement of the beam, comment on whether the values you have worked out would actually be observed.
Mass of an electron = 9.11 × 10^{31} kg; Charge on an electron = 1.6 × 10^{19} C 
(a) Formula:
v^{2} = (2 × 1.6 × 10^{19} C × 2000 V) ÷ (9.11 × 10^{31} kg) = 7.025 × 10^{14} m^{2} s^{2}
v = 2.65 × 10^{7} m s^{1}
(b) Formula:
F = 2000 V × 1.6 × 10^{19} C = 1.067 × 10^{14} N (upwards) 0.03 m
(c) Use Newton II to work out the acceleration:
a = F/m = 1.067 × 10^{14} N ÷ 9.11 × 10^{31} kg = 1.171 × 10^{16} m s^{2} (upwards)
(d) Horizontal velocity remains the same, i.e. 2.65 × 10^{7} m s^{1} from left to right
We need to work out the time taken to travel between the plates:
t = 0.050 m ÷ 2.65 × 10^{7} m s^{1} = 1.887 × 10^{9} s
Work out the vertical velocity:
v_{y} = at = 1.171 × 10^{16} m s^{2} × 1.887 × 10^{9} s = 2.21 × 10^{6} m s^{1}
Resultant velocity^{2} = (2.65 × 10^{7} m s^{1})^{2} + (2.21 × 10^{6} m s^{1} )^{2} = 7.07 × 10^{14} m^{2} s^{}^{2}
Resultant velocity = 2.70 × 10^{7} m s^{1}
(e) Direction relative to the horizontal
q = tan^{1} (2.21 × 10^{6} m s^{1} ÷ 2.65 × 10^{7} m s^{1}) = tan^{1} 0.834 = 4.8^{o} above the horizontal
(f) Now use s = ut + 1/2at^{2} to work out the deviation:
s = 0 + 1/2 × 1.171 × 10^{16} m s^{2} × (1.887 × 10^{9} s)^{2} = 0.021 m = 2.1 cm.
The maximum deviation in this case is 1.5 cm, as the path of the electrons is equidistant between the plates. This means that the electrons will crash into the plates. 