An electron with kinetic energy K is traveling along the +x axis, which is along the axis of a cathode-ray tube as shown in the figure below. There is an electric field E = 9.00×10^4 N/C pointed in the +y direction between the deflection plates, which are d = 0.0500 m long and are separated by 0.0200 m.
1) Determine the minimum initial kinetic energy the electron can have and still avoid colliding with one of the plates. (Express your answer to three significant figures).
______________ x10^-15 J
Step1:
Using Force balance on electron in y-direction, to find acceleration in y-direction:
Force due to electric field = Fnet
Fe = Fnet
q*E = m*a
a = q*E/m
q = charge on electron = -e = -1.6*10^-19 C
a = -1.6*10^-19*9*10^4/(9.1*10^-31) = -1.58*10^16 m/sec^2
Now Using 2nd kinematic equation in y-direction:
y = Ui*t + 0.5*a*t^2
Ui = initial velocity in y-direction = 0 m/sec
y = -0.01 m, Since electron will travel downward with constant acceleration, So
-0.01 = 0*t + 0.5*(-1.58*10^16)*t^2
-0.01 = -7.9*10^15*t^2
t = sqrt (0.01/(7.9*10^15))
t = 1.125*10^-9 sec
Now Since there is no-acceleration in x-direction which means velocity in x-direction will remain constant (U0), So time taken by electron to cross the tube will be
t = d/U0
U0 = d/t = 0.05/(1.125*10^-9)
U0 = 4.444*10^7 m/sec
So for this minimum velocity kinetic energy of electron will be
KE = 0.5*m*U0^2
KE = 0.5*9.1*10^-31*(4.444*10^7)^2
KE = 9.00*10^-16 J
KE = 0.90*10^-15 J
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