answer part (B) After the electrons pass through the anode, they are moving at a
ID: 1645634 • Letter: A
Question
answer part (B)
After the electrons pass through the anode, they are moving at a speed of 9.290 times 10^6 m/s. They next pass between a pair of parallel plates [(A) in the figure]. The plates each have an area of 2.16 cm by 2.16 cm and they are separated by a distance of 0.532 cm. The uniform electric field between them is 1.00 times 10^3 N/C and the plates arc charged as shown. The electrons then pass between another set of parallel plates [(B) in the figure]. These plates also have an area of 2.16 cm by 2.16 cm and are separated by a distance of 0.532 cm. (a) In what direction must the field be oriented so that the electrons are deflected vertically upward? (b) If we neglect the gravitational force, how strong must the field be between these plates in order for the electrons to be deflected by 1.70 mm? N/CExplanation / Answer
a )
b )
x = 2.16 cm
x = 0.0216 m
Vi x = 9.29 X 106 m/s
Vi y = 0
dy = 1.7 mm = 0.0017 m
a = q E / m
a = 1.6 X 10-19 X E / 9.1 X 10-31
t = x / Vix
= 0 .0126 / 9.29 X 106
= 2.32 X 10-9 sec
dy = Vi y + 0.5 a t2
0.0017 = 0.5 X 1.6 X 10-19 X E X ( 2.32 X 10-9 )2 / 9.1 X 10-31
E = 0.0017 X 9.1 X 10-31 / 0.5 X 1.6 X 10-19 X ( 2.32 X 10-9 )2
E = 3592.72 N/C
b )
x = 2.16 cm
x = 0.0216 m
Vi x = 9.29 X 106 m/s
Vi y = 0
dy = 1.7 mm = 0.0017 m
a = q E / m
a = 1.6 X 10-19 X E / 9.1 X 10-31
t = x / Vix
= 0 .0126 / 9.29 X 106
= 2.32 X 10-9 sec
dy = Vi y + 0.5 a t2
0.0017 = 0.5 X 1.6 X 10-19 X E X ( 2.32 X 10-9 )2 / 9.1 X 10-31
E = 0.0017 X 9.1 X 10-31 / 0.5 X 1.6 X 10-19 X ( 2.32 X 10-9 )2
E = 3592.72 N/C
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