a) 35063060.64 b) 45000 c) 7.2 x 10^-15 04. An AC power source heats the first n

Question

a) 35063060.64 b) 45000 c) 7.2 x 10^-15
04. An AC power source heats the first negative plate in the diagram Problem 04. to the right allowing electrons to be "boiled off" (easily removed)0 The beam of electrons accelerates to the positive plate and continues at a constant speed (through a hole in this positive plate) into the second electric field between the two horizontal plates. a.) What is the velocity of the beam of electrons as it passes -4500 V E2 0.00 cm through the hole in the first positive plate? b.) What is the magnitude of the electric field, E2. c.) What is the force on each electron in the second field? d.) What is the acceleration of each electron in the second field? e.) Sketch the path of the beam in E and calculate how long will it takes this beam to hit the positive plate. f) How far from the left edge of the positive plate will the beam of electrons exit the field (hit the positive plate)? g.) With what velocity will each electron strike the plate?

Explanation / Answer

mass of electron m = 9.1 e-31 kg

The energy gained by the lectron across pot. diff of 3500V = 3500ev = 3.5e+3 *1.61e-19 J

                                                                                              = 5.635e-16 J

d) electric field in the 2nd region E = 4500/0.1 = 45000 V/m

Force on the electron    F = Ee

acceleration of the lectron a = F/m = Ee/m

                                               = 45000*1.61e-19/9.1e-31

                                                = 7.96 e+15 m/s/s

e)acceleration in vertical direction 7.96e+15 m/s/s

distance to travel s= 10cm = 0.1 m

time t = sqrt(2s/a) = sqrt(2*0.1/7.96e+15) = 5.01e-9 s

f) horizontal vel . = 3.5e+7 m/s

    horizontal distance traveled in (t= 5.01e-9 s) = vt = 3.5e+7 *5.01e-9

                                                                              = 0.175 m

The electron hit he positive plate at a disatnce of 17.5 cm from the left edge

g) vertical vel. fo the electron at the time of hitting = at = 7.96e+15 *5.01e-9

                                                                                  = 3.99 e+7 m/s

    horizontal vel. = 3.5 e+7 m/s

resultant vel. of the electron as it strikes the plates

                                      = sqrt( (3.5e+7)2 + (3.99e+7)2 )

                                       = 5.3 e+7 m/s

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