The electron beam inside an old television picture tube is 0.40 mm in diameter and carries a current of 50 μA. This electron beam impinges on the inside of the picture tube screen. Requried:
a. How many electrons strike the screen each second?
b. The electrons move with a velocity of 4.0m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 m?
c. 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?

a)     # _electron = 31 10¹³ electrons, b)    E = 9.1 10⁻¹² N / C , c)  P = 7.22 10⁻¹⁶ J

Explanation:

In this exercise we must examine the movement of electrons in an electric field.

a) they ask us for the number of electrons.

The electric current is

     i = q / t

     q = i t

let's calculate

      q = 50 10⁻⁶ 1

      q = 50 10⁻⁶ C

    For this we can use a rule of direct proportions, if the charge of an electron is 1.6 10⁻¹⁹ C, the number of electrons for the charge of 50 10⁻⁶ C

    #_electron = 50 10⁻⁶ C 1 / 1.6 10⁻¹⁹ C

   # _electron = 31.25 10¹³ electrons

b) For this part we use kinematics to find the acceleration

         v² = v₀² + 2 a x

electrons start from rest, so vo = 0

         a = v² / 2x

let's calculate

         a = 4²/2 5

         a = 1.6 m / s²

Having the acceleration we can use Newton's second law where the force is electric

          F = ma

          e E = m a

          E = m / e a

let's calculate

           e = 9.1 10⁻³¹ / 1.6 10⁻¹⁹   1.6

           E = 9.1 10⁻¹² N / C

c) what the power on the screen

            for this we must add the kinetic energy of all the electrons in the given time

           P = # _electron Km / y

           P = # _electron (½ m v²) / t

let's calculate

           P = 31 10 13 (½ 9.1 10-31 1.6²) / 1

           P = 7.22 10⁻¹⁶ J

f=ma newton

f=100x2.5=250n