Answer to Question #94481 in Mechanics | Relativity for Annabelle Garduno
2019-09-13T17:18:58-04:00
An electron initially has a speed 19.4 km/s along the x-direction and enters an electric field of strength 24 mV/m that points in the y-direction and has a width of 18.5 cm. How far is the deflection?
1
2019-09-16T09:46:17-0400
d = v t d=vt d = v t
t = d v = 0.185 19400 = 9.536 ⋅ 1 0 − 6 s t=\frac{d}{v}=\frac{0.185}{19400}=9.536\cdot 10^{-6}\ s t = v d = 19400 0.185 = 9.536 ⋅ 1 0 − 6 s
The electrostatic force:
F = q E F=qE F = qE The acceleration:
a = F m = q E m a=\frac{F}{m}=\frac{qE}{m} a = m F = m qE
The deflection is
h = 0.5 a t 2 = q E 2 m t 2 h=0.5at^2=\frac{qE}{2m}t^2 h = 0.5 a t 2 = 2 m qE t 2
h = ( 1.6 ⋅ 1 0 − 19 ) ( 0.024 ) 2 ( 9.11 ⋅ 1 0 − 31 ) ( 9.536 ⋅ 1 0 − 6 ) 2 = 0.192 m h=\frac{(1.6\cdot 10^{-19})(0.024)}{2(9.11\cdot 10^{-31})}(9.536\cdot 10^{-6})^2=0.192\ m h = 2 ( 9.11 ⋅ 1 0 − 31 ) ( 1.6 ⋅ 1 0 − 19 ) ( 0.024 ) ( 9.536 ⋅ 1 0 − 6 ) 2 = 0.192 m
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