The electron beam inside an old television picture tube is 0.40 mm in diameter a

Question

The electron beam inside an old television picture tube is 0.40 mm in diameter and carries a current of 50 mu A. This electron beam impinges on the inside of the picture tube screen. The electrons move with a velocity of 3.2 times 10^7 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity In a distance of 5.0 mm? Express your answer with the appropriate units. 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? Express your answer with the appropriate units.

Explanation / Answer

C)

accelaration is a = F/m = q*E/m

Electric field strength is E = a*m/q =

initial speed is Vo = 0 m/sec

final speed is V = 3.2*10^7 m/sec

distance travelled is S = 5*10^-3 m

Using Kinematic equation

V^2 - Vo^2 = 2*a*S

(3.2*10^7)^2 - 0^2 = 2*a*5*10^-3

a = 1.024*10^17 m/s^2

then E = a*m/q = (1.024*10^17)*9.11*10^-31/(1.6*10^-19) = 5.83*10^5 V/m

D) P = V*I = E*l*I = 5.83*10^5*0.4*10^-3*50*10^-6 = 0.01166 W

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.