Question

The figure below shows an electron at the origin that is released with initial speed v0...

The figure below shows an electron at the origin that is released with initial speed v0 = 3.5 ✕ 106 m/s at an angle θ0 = 45° between the plates of a parallel plate capacitor of plate separation D = 2.0 mm. If the potential difference between the plates is ΔV = 185 V, calculate the closest proximity, d, of the electron to the bottom plate (in mm).

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Answer #1

Force on the electron

F=qE =q(V/D) =(1.6*10-19)(185/2*10-3)

F=1.48*10-14 N

acceleration

ay=F/m =(1.48*10-14)/(9.11*10-31) =1.6246*1016 m/s2

From

Vy2=Voy2+2ayy

0 =(-3.5*106Sin45)2+2*1.6246*1016*y

y=-1.885*10-4 m=-0.1885 mm

It will reach

d=(3/2) -0.1885

d=1.3115 mm

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