The electron beam inside a television picture tube is 0.500mm in diameter and ca

Question

The electron beam inside a television picture tube is 0.500mm in diameter and carries a current of 46.0 uA u stands for mule. This electron beam impinges on the inside of the picture tube screen...
has 4 parts
part a) How many electrons strike the screen each second?the answer should be in electrons.

part b) What is the current density in the electron beam? answer should be in A/m^2

part c) he electrons move with a velocity of 4.00*10^7 m/s . What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.40 mm?answer should be in N/C

part d)Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the power delivered to the screen by the electron beam?answer should be in W.

please someone who is a physics expert answer this question, i cannot sleep at night trying to solve these questions..thanks so much in advance for your assistance.

Explanation / Answer

current i = 46*10-6 A charge of electron e = 1.6*10-19 C diameter of the television tube d = 0.5 mm radius of the tube r = d/2 = 0.25 mm                                         = 0.25*10-3 m  a) charge q = ne             it = ne    (since , current i = q/t   q = it) number of electrons strike the screen each second is         n = it/e            = (46*10-6 A)(1 s) / (1.6*10-19 C)            = 28.75*1013 electrons            = 287.5*1012 electrons            = 2.875*1014 electrons            = 287.5*1012 electrons            = 2.875*1014 electrons b) area of the tube A = r2 current density J = i/A                            = i / r2                            = (46*10-6 A) / (3.14)(0.25*10-3 m )2                            = 234.39 A/m2 c) mass of the electron m = 9.11*10-31 kg velocity of the electron v = 4*107 m/s initial velocity of the electron v0 = 0 m/s distance S = 5.4*10-3 m from kinematic equation's ,          v2 - v02 = 2aS          v2 - 02 = 2aS acceleration a = v2 / 2S                        = (4*107 m/s)2 / (2)(5.4*10-3 m)                        = 1.48*1017 m/s2 electrostatic force F = qE   ........... (1) from , Newton's second law of motion ,         F = ma  ............ (2) compare eq (1) and (2) , we get  ma = qE electric field E = ma/q                            =(9.11*10-31 kg)(1.48*1017 m/s2)/(1.6*10-19 C)                          = 8.43*105 N/C d) number of electrons = n
current i = nq/t         n/t = i/q     ................... (3)
total energy U = n(1/2)mv2     ................. (4)
power = total energy / time            = [n(1/2)mv2]/t            = (n/t)(1/2)mv2           = (i/q)((1/2)mv2)           = (46*10-6 A / 1.6*10-19 C)(1/2)(9.11*10-31 kg)(4*107 m/s)2           = 0.209 W           = 2.095*10-3 W           = 0.209 W           = 2.095*10-3 W                                   

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