An electron is released from the negatively-charged plate of a parallel plate capacitor, initially at rest, and it is accelerated across the gap to hit the positively-charged plate. If the capacitor was powered by a 12 V battery, the capacitor was fully charged, what was the kinetic energy of the electron when it hit the positively-charged plate

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Physics

23 March, 16:04

An electron is released from the negatively-charged plate of a parallel plate capacitor, initially at rest, and it is accelerated across the gap to hit the positively-charged plate. If the capacitor was powered by a 12 V battery, the capacitor was fully charged, what was the kinetic energy of the electron when it hit the positively-charged plate

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Manuel Hutchinson · Answer 1

Answer: 1.9308 * 10^-18 J

Explanation: by using the work - energy theorem, the work done on the electron by the capacitor equals the kinetic energy of the electron.

Workdone on electron = qV

Kinetic eneegy = 1/2mv^2

q = magnitude of an electronic charge = 1.609*10^-19c

V = potential difference = 12v

Kinetic energy = qV

Kinetic energy = 1.609*10^-19 * 12

Kinetic energy = 19.308*10^-19

Kinetic energy = 1.9308 * 10^-18 J