An electron is fired at a speed v0 = 5.3 ✕ 106 m/s and at an angle θ0 = −45° between two parallel conducting plates that are D = 2.5 mm apart, as in the figure below. The voltage difference between the plates is ΔV = 105 V. (a) Determine how close, d, the electron will get to the bottom plate. mm (b) Determine where the electron will strike the top plate. mm
Given that,
vo = 5.3*10^6 m/s
= 45 deg
d = 2.5 mm
V = 105 V
(a)
vertical component of velocity,
vy = vo*sin45
vy = 5.3*10^6*sin45 = 3.76*10^6 m/s
horizontal component, vx = vo*sin45
vx = 3.76*10^6 m/s
Electric field E = V / d
E = 105 / 2.5*10^(-3) = 42*10^3 N/C
acceleration of electron,
a = qE / m
a = 1.6*10^(-19)*42*10^3 / 9.1*10^(-31) = 7.38*10^(15) m/s^2
From kinematic equation,
v^2 - u^2 = 2*a*s
s = (3.76*10^6)^2 / 2*7.38*10^15
s = 0.962 mm
d = D/2 - s
d = (2.5 / 2) - 0.962
d = 0.287 mm
(b)
Let, Time taken by electron to reach the top plate = t
From kinematic equation,
-D/2 = vy*t - (1/2)at^2
-1.25*10^(-3) = 3.76*10^6*t - (1/2)*7.38*10^15*t^2
(3.69*10^15)t^2 - (3.76*10^6)t - 1.25*10^(-3) = 0
by solving the quadratic equation,
t = 1.124*10^(-9) s
Distance where electron will strike the top plate,
x = vx*t
x = 3.76*10^6*1.124*10^(-9)
x = 4.22 mm
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