An electron at rest of mass 9.11x10^−31 kg is accelerated through a potential difference of 350 V. It then enters some deflecting plates of 50 V with dimensions as shown. Calculate the distance,x, or of the deflection of the electron. The charge on an electron is 1.6x10^−19 C). (20 marks)
2 answers
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€ = FE /q = 50/0.05 = 1000 N/C
Magnitude of force on electron: 1000 N/C * 1.6x10-19C = 1.6x10-16N
a = F/m = 1.6x10-16N/9.11x10-31 = 1.76x1014 m/s2
1/2mv2
= qv ½*9.11x10-31 *v2
= 350*1.6x10-19 v2
= 2*350*1.6x10-19 /9.11x10-31 v = 1.11x107
m/s time to travel through: t = d/v = 0.25 cm / 1.11x107
m/s = 2.25x10-8s
d = vi*t + ½*a*t2
= 0 + ½*1.76x1014 m/s2
*(2.25x10-8s)2
= 0.044m
The distance x is approximately 0.044m or 4.4 cm.
Magnitude of force on electron: 1000 N/C * 1.6x10-19C = 1.6x10-16N
a = F/m = 1.6x10-16N/9.11x10-31 = 1.76x1014 m/s2
1/2mv2
= qv ½*9.11x10-31 *v2
= 350*1.6x10-19 v2
= 2*350*1.6x10-19 /9.11x10-31 v = 1.11x107
m/s time to travel through: t = d/v = 0.25 cm / 1.11x107
m/s = 2.25x10-8s
d = vi*t + ½*a*t2
= 0 + ½*1.76x1014 m/s2
*(2.25x10-8s)2
= 0.044m
The distance x is approximately 0.044m or 4.4 cm.
2 answers