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

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

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