The electron beam inside a television picture tube is 0.300 mm in diameter and c

Question

The electron beam inside a television picture tube is 0.300 mm in diameter and carries a current of 48.0 uA. This electron beam impinges on the inside of the picture tube screen.
1. What is the current density in the electron beam? (in A/m^2)
2. The electrons move with a velocity of 3.50 *10^7 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 4.70mm? (in N/C)
3. 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? (in W)
Please explain.
1. What is the current density in the electron beam? (in A/m^2)
2. The electrons move with a velocity of 3.50 *10^7 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 4.70mm? (in N/C)
3. 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? (in W)
Please explain.

Explanation / Answer

d = 0.300 mm, I = 48.0 uA. area A = pi*d^2/4 1. What is the current density in the electron beam? (in A/m^2) current density j = I/A = 4I/(pi*d^2) = 679 A/m^2 2. v = 3.50 *10^7 m/s, D = 4.70 mm acceleration a = v^2/(2D) force F = ma = qE E = ma/q = mv^2/(2qD) = 7.41*10^5 N/C 3. number of electrons = N I = Nq/t, so N/t = I/q total energy U = N*mv^2/2 power = U/t = (N/t)*mv^2/2 = (I/q)*mv^2/2 = 0.167 W

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