Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.0 cm apart with a 20 kV 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.

A. What is the electric field strength between the plates?
Express your answer to two significant figures and include the appropriate units.

B. With what speed does an electron exit the electron gun if its entry speed is close to zero?
Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

A) electric field strength between the plates;E = 2 x 10^(6) N/C

B) exit velocity;v = 8.39 x 10^(7) m/s

Explanation:

We are given;

Potential difference; V = 20 kV = 20000 V

Distance between the 2 parallel plates; d = 1cm = 0.01 m

A) The electric field strength will be gotten from;

E = V/d

E = 20000/0.01

E = 2000000

E = 2 x 10^(6) N/C

B) For exit speed, we'll use the formula for Kinetic energy; KE = (1/2)mv²

KE is also expressed as; V•q_e

Thus,

(1/2)mv² = V•q_e

Where;

V is potential difference = 20000 V

Q_e is charge of electron which has a constant value of; (1.6 x 10^(-19))C

m is mass of electron with a constant value of (9.1 x 10^(-31)) kg

v is the velocity

Thus, making v the subject, we have;

v = √((2V•q_e)/m)

v = √((2 x 20000•(1.6 x 10^(-19)))/(9.1 x 10^(-31)))

v = 83862786 m/s or

v = 8.39 x 10^(7) m/s