9:53 LTE Exercise 21.3 An electron is projected with an initial speed 1.30x106 m

Question

9:53 LTE Exercise 21.3 An electron is projected with an initial speed 1.30x106 m/s into the uniform field between the parallel plates in the figure K2.00 cm 1.00 cm Q Tap image to zoom Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. Part A If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the field? m/s Request Answer

Explanation / Answer

length for which electron travel horizontally =d=2 cm=0.02 m

horizontal speed through out the journey=v0=1.3*10^6 m/s

then time taken=t=distance/speed=d/v0=1.5385*10^(-8) seconds

vertical distance travelled=L=1 cm/2=0.01/2=0.005 m

if acceleration along vertical is a m/s^2,

then distance travelled=initial vertical speed*time+0.5*acceleration*time^2

as initial vertical speed=0

L=0.5*a*t^2

==>a=2*L/t^2

vertical speed of electron as it emerges from the field=initial vertical speed+acceleration*time

=0+a*t

=2*L/t

=6.5*10^5 m/s

so total speed=sqrt(horizontal speed^2+vertical speed^2)

=1.4534*10^6 m/s

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