Answer to Question #94481 in Mechanics | Relativity for Annabelle Garduno

Question #94481
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
Expert's answer
2019-09-16T09:46:17-0400
d=vtd=vt

t=dv=0.18519400=9.536106 st=\frac{d}{v}=\frac{0.185}{19400}=9.536\cdot 10^{-6}\ s


The electrostatic force:


F=qEF=qE

The acceleration:


a=Fm=qEma=\frac{F}{m}=\frac{qE}{m}

The deflection is


h=0.5at2=qE2mt2h=0.5at^2=\frac{qE}{2m}t^2

h=(1.61019)(0.024)2(9.111031)(9.536106)2=0.192 mh=\frac{(1.6\cdot 10^{-19})(0.024)}{2(9.11\cdot 10^{-31})}(9.536\cdot 10^{-6})^2=0.192\ m


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