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Problem 2 The distance between the two plates of a parallel plate capacitor is 5 cm....

Problem 2

The distance between the two plates of a parallel plate capacitor is 5 cm. The electric potential difference between the two plates is V= 6.0 v.

  A proton is released from rest from the positive plate of a capacitor.

q = 1.60 × 10-19 C, m electron = 1.67 × 10-27 kg,

Show formulas, substitution, and calculation with units.

Hint: You need to write the relation between Electric potential energy ( q V)and kinetic energy. ( K= ½ mv2 ).

This Photo by Unknown Author is licensed under CC BY-SA
  1. Calculate how much the proton loses its electric potential energy once it reaches the negative plate.
  1. How much the kinetic energy of the proton will increase.

c) How fast will the electron be moving when it reaches to second plate.

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