Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.4cm apart with a 28kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.

1. What is the electric field strength between the plates?
2. With what speed does an electron exit the electron gun if its entry speed is close to zero?

Answer 1

a) F = 3.2 10⁻¹⁰ N , b) v = 9.9 10⁷ m / s

Step-by-step explanation:

a) The electric force is

F = q E

The electric field is related to the potential reference

V = E d

E = V / d

Let's replace

F = e V / d

Let's calculate

F = 1.6 10⁻¹⁹ 28 10³ / 1.4 10⁻²

F = 3.2 10⁻¹⁰ N

b) For this part we can use kinematics

v² = v₀ + 2 a d

v = √ 2 ad

Acceleration can be found with Newton's second law

e V / d = m a

a = e / m V / d

a = 1.6 10⁻¹⁹ / 9.1 10⁻³¹ 28 10³ / 1.4 10⁻²

a = 3,516 10⁻¹⁷ m / s²

Let's calculate the speed

v = √ (2 3,516 10¹⁷ 1.4 10⁻²)

v = √ (98,448 10¹⁴)

v = 9.9 10⁷ m / s