Question

The electron gun in an old CRT television accelerates electrons between two charged parallel plates (the cathode is negative; the anode is positive) 1.2 cm apart. The potential difference between them is 25 kV. Electrons enter through a small hole in the cathode, are accelerated, and then exit through a small hole in the anode. Assume the plates act as a capacitor.

a. What is the electric field strength and direction between the plates?
b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]
c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?
d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?

Answer 1

A. 2.083 MV/m from anode to cathode.

B. 93648278.15 m/s

C. 2.5x10^-5 C and there are about 1.56x10^14 electrons

D. 4x10^-15 Joules

Step-by-step explanation:

Voltage V across plate is 25 kV = 25x10^3 V

Distance apart x = 1.2 cm = 1.2x10^-2 m

A. Electric field strength is the potential difference per unit distance

E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m

= 2.083 MV/m

B. Energy of electron is electron charge times the voltage across

i.e eV

Charge on electron = 1.6x10^-19 C

Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules

Mass of electron m is 9.12x10^-31 kg

Kinetic energy of electron = 0.5mv^2

Where v is the speed

4x10^-15 = 0.5 x 9.12x10^-31 x v^2

v^2 = 8.77x10^15

v = 93648278.15 m/s

C. From Q = CV

Q = charge

C = capacitance = 1 nF 1x10^-9 F

V = voltage = 25x10^3 V

Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C

Total number of electrons = Q/e

= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons

D. To push electron from cathode to anode, I'll have to do a work of about

4x10^-15 Joules