Question

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? a) 144 eV b) 14 eV c) 12 ev d) 120 ev e) 360 eV

Answer 1

c) 12 eV

Step-by-step explanation:

The electron crosses a potential difference of 12 V in total, This means that its initial electric potential energy as it leaves the negative plate is equal to

`E=q\Delta V=e (12 V) = 12 eV`

due to the law of conservation of energy, when the electron moves towards the positive plate this electric energy is all converted into kinetic energy of the electron (in fact, the speed of the electron increases).

When the electron reaches the positive plate, all the electric potential energy has been converted into kinetic energy, which is therefore exactly 12 eV.