Question

If an electron travels 0.210 m from an electron gun to a TV screen in 29.3 ns, what voltage was used to accelerate it? (Note that the voltage you obtain here is lower than actually used in TVs to avoid the necessity of relativistic corrections.)

Answer 1

If we assume that there are no electric fields over the 0.210 meters from the gun to the screen then the speed of the electron is constant and is simply the distance / time

velocity = 0.210/(29.3.0E-9) = 7.1 E+6 meters/sec

This is a tiny fracion of the speed of light in a vacuum so we can ignore relativistic effects.

The energy of the electron = 1/2 mv^2

= 1/2(9.1095E-31)*(7.1E+6)^2 = 2.296E-17 joules

Convert from joules to eV using the electron's charge (1.602E-19 coul)

voltage = 2.296E-17/1.602E-19 = 143 volts

(and that would be positive w/respect to the gun)

Seems like a low voltage for a TV screen, but there you have it