An old-style TV set places 30 kV across two plates of a parallel plate capacitor

Question

An old-style TV set places 30 kV across two plates of a parallel plate capacitor. An electron is released very close to one plate, and it accelerates rapidly towards the second plate which has a hole drilled through it. The electron exits the hole and hits a phosphor on the glass screen where the picture appears and the phosphor glows.

A.)If the plate where the electron is released is 0 V, what is the voltage of the plate with the hole in it?

B.)What is the speed of the electron when it exits the hole in the plate?

Explanation / Answer

here,

potential difference V = Vb - Va

V = +30kV - 0

V = + 30 kV (answer)

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Kinetic energy K = Potential energy

K = eV

0.5 mv^2 = e V

m of electron = 9.11 *10^-31 kgs

V is 30000 v

v^2 = 2eV/m

v^2 = (2 * 1.6*10^-19 * 30000)/(9.11*10^-31)

speed v = 1.026 *10^8 m/s (answer)

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