Answer to Question 9

 

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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:

v2 = (2 1.6 10-19 C 2000 V) (9.11 10-31 kg) = 7.025 1014 m2 s-2

 

v = 2.65 107 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 1016 m s-2 (upwards)

 

 

(d)  Horizontal velocity remains the same, i.e. 2.65 107 m s-1 from left to right

 

We need to work out the time taken to travel between the plates:

 

t = 0.050 m 2.65 107 m s-1 = 1.887 10-9 s

 

Work out the vertical velocity:

 

vy = at =  1.171 1016 m s-2 1.887 10-9 s = 2.21 106 m s-1

 

Resultant velocity2 = (2.65 107 m s-1)2 + (2.21 106 m s-1 )2 = 7.07 1014 m2 s-2

 

Resultant velocity = 2.70 107 m s-1

 

 

(e) Direction relative to the horizontal

 

q = tan-1 (2.21 106 m s-1   2.65 107 m s-1) = tan-1 0.834 = 4.8o above the horizontal

 

 

(f) Now use s = ut + 1/2at2 to work out the deviation:

 

s = 0 + 1/2 1.171 1016 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.