Two 1 m x 1 m conducting plates are spaced 10 cm and oriented horizontally( each

Question

Two 1 m x 1 m conducting plates are spaced 10 cm and oriented horizontally( each plane is parallel to the floor). A 15 V battery is connected to the plates with the positive terminal connected to the lower plate. An electron enters the gap between the plates from the left midway between the plates (5 cm vertically from each plate & centered on the plate width), traveling horizontally at 500 m/s.

1. Which plate does the electron get closer to? Explain.

2. Does the electron leave the gap between the two plates before hitting one or the other plate? Explain.

Explanation / Answer

as lower plate is at higher potential, electric field is from lower plate to upper plate.

the electron , being of negative charge, will move in opposite direction to the electric field

so electron will move towards the lower plate.

Q2. electric field magnitude=potential difference/distance between plates=15/0.1=150 V/m

force on electron=charge magnitude*electric field=1.6*10^(-19)*150=2.4*10^(-17) N

acceleration of electron=force/mass=2.4*10^(-17)/(9.1*10^(-31))=2.6374*10^13 m/s^2

time taken to travel the length of the plates=1 m/(500 m/s)=0.002 seconds

vertical distance travelled in 0.002 seconds=initial vertical speed*time+0.5*acceleration*time^2

=0*0.002+0.5*2.6374*10^13*0.002^2=5.2748*10^7 m

as it is much higher than the distance from the gap to either of the plates,

so the electron will hit the lower plate before leaving the gap.

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